logarithms
calculus
calculus computations

[quote="Nea"]$1.$ $d^{n}u ?$
$u = e^{ax+by}$

$2.$ $d^{5}u ?$
$u= \ln (x-y)$[/quote]

If $u=f(x,y)$ then $du=f_{x}(x,y)\, dx+f_{y}(x,y)\, dy$, where here the differential operator $d$ is given by $d=\frac{\partial}{\partial x}\, dx+\frac{\partial}{\partial y}\, dy$ so that $d^{n}=\left(\frac{\partial}{\partial x}\, dx+\frac{\partial}{\partial y}\, dy\right)^{n}$ and hence we have

$1.$ For $u = e^{ax+by}$ we have

\[\frac{\partial^{n}\left(e^{ax+by}\right)}{\partial x^{n-k}\partial y^{k}}=a^{n-k}b^{k}e^{ax+by}\]

and this gives

\[d^{n}u = \left(\frac{\partial}{\partial x}\, dx+\frac{\partial}{\partial y}\, dy\right)^{n}e^{ax+by}=\sum_{k=0}^{n}\binom{n}{k}\frac{\partial^{n-k}}{\partial x^{n-k}}\, dx^{n-k}\cdot\frac{\partial^{k}}{\partial y^{k}}\, dy^{k}\, \left(e^{ax+by}\right) \]
\[=\sum_{k=0}^{n}\binom{n}{k}\frac{\partial^{n}\left(e^{ax+by}\right)}{\partial x^{n-k}\partial y^{k}}\, dx^{n-k}dy^{k}= \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}e^{ax+by}\, dx^{n-k}dy^{k}\]
\[= e^{ax+by}\sum_{k=0}^{n}\binom{n}{k}a^{n-k}dx^{n-k}\cdot b^{k}dy^{k}= e^{ax+by}(a\, dx+b\, dy)^{n}\]

$2.$ Find $d^{5}u$ if $u= \ln (x-y)$ for which

\[\frac{\partial^{n}\ln (x-y)}{\partial x^{n-k}\partial y^{k}}=(-1)^{k+1}\frac{(n-1)!}{(x-y)^{n}}\]

and this gives

\[d^{n}u = \left(\frac{\partial}{\partial x}\, dx+\frac{\partial}{\partial y}\, dy\right)^{n}\ln (x-y) =\sum_{k=0}^{n}\binom{n}{k}\frac{\partial^{n-k}}{\partial x^{n-k}}\, dx^{n-k}\cdot\frac{\partial^{k}}{\partial y^{k}}\, dy^{k}\, \left(\ln (x-y)\right) \]
\[=\sum_{k=0}^{n}\binom{n}{k}\frac{\partial^{n}\ln (x-y)}{\partial x^{n-k}\partial y^{k}}\, dx^{n-k}dy^{k}= \sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}\frac{(n-1)!}{(x-y)^{n}}\, dx^{n-k}dy^{k}\]
\[=-\frac{(n-1)!}{(x-y)^{n}}\sum_{k=0}^{n}\binom{n}{k}\, dx^{n-k}\cdot (-1)^{k}dy^{k}=-\frac{(n-1)!}{(x-y)^{n}}\left( dx-dy\right)^{n}\]
\[=-(n-1)!\left(\frac{dx-dy}{x-y}\right)^{n}\]

set $n=5$ to get

\[d^{5}u =-4!\left(\frac{dx-dy}{x-y}\right)^{5}\]


The sides of a regular polygon with $n$ sides $(n>4)$ are extended to form a star. The number of degrees in the angle at each point of the star in terms of $n$ is?

For a regular n-gon, the general formula for each interior angle is $(n-2)180$. This can be easily proved, by sub-dividing each polygon into triangles. If we extend the sides of a regular n-gon, we form a triangle at each side, which is an isosceles triangle. This is because by using the geometry reasoning that the exterior angle equals the sum of the two interior angles, we see that the angle at each point of the star is $\frac{(n-2)180}{2n}$

Yeah... that's not what I get. Use one point of the star and the two closest vertices to form an isosceles triangle. The two exterior angles are both $\frac{360}{n}$ so the angle of the star is $180-\frac{720}{n}$. Notice that for $n = 4$ this is $0$ because they never meet, and it's easily verified for $n = 6$ (in case you have doubts).

eddie77, try plugging in $n = 5$ into your equation; it's clearly not true.


C’est moi! Je m’appelle Joseph. J’ai trente-trois ans, et je suis un plombier. J’aime manger la pizza et regarder la télévision. Le week-end, je faire du jogging au parc aussi. En général, je suis amusant, content, fort, et brun. Je ne suis ni grand ni petit. J’aime John McCain.

C’est ma femme, Chantal. Elle a trente-deux ans. Elle est gentille, intelligente, et brune. Chantal est grande et mince aussi. Elle est une joueur de golf. Elle aime aller au centre commércial surtout.

C’est mon fils, Jean. Il est brun, timide, et petit. Jean a dix ans. Il aime jouer à des jeux vidéo et faire du velo. Le week-end, il aime sortir avec ses amis.

C’est ma fille, Amélie. Elle est petite, blonde, et mignonne. Elle est curieuse aussi. Amélie a sept ans. Elle aime beaucoup jouer. Elle aime la MJC aussi.

C’est mon père , Charles. Il a soixante-huit ans. Il est brun, aimable, et diligent. Il est patient et sensuel aussi. Il est un architecte. Le week-end, il aime faire de l’ équitation.

C’est ma mère, Elizabeth. Elle a soixante-cinq ans. Elle est blonde, sérieuse, efficace, et pratique. Le week-end, elle aime voyager et parler au telephone.

Do you think I would write in French if it weren't my homework?

[quote="tinytim"]No hablo frances. :([/quote]

Shouldn't that be "No hablo francés"?

Then again, I haven't taken Spanish for over two years.

[quote]En général, je suis...brun.[/quote]
I think this means "I generally have brown hair."

It means "in general..." which can also be interpreted as when I'm talking about the ordinary features.

haha, i'm in french also.... i suck at it, i speak "franglais" too much in class and my teacher gets pissed at me. elle me deteste parce-que je suis noir. je ne suis pas noir, je suis brun (?indien?).

Is this French 1?

Also, why do you say aime so much in ur french?

Why do we use like so much in our English?

Ahh...I get it :D

So my usename is "LIKE15"?


What is the surface area of a sphere [color=green][mod edit: CUBE!][/color] with a volume of 2197? No calculators, as this is a sprint round question.

[hide]i got 6591/r, where r is the radius of the sphere. this is a pretty ugly problem for a sprint.[/hide]

[quote="IntrepidMath"][hide]
$\displaystyle (4/3)(\pi)(r^3)=2197\\r=\frac{3\sqrt[3]{2197-\pi}}{4}$

phew...
SA=$4\pi\frac{3\sqrt[3]{2197-\pi}}{4}=\pi3\sqrt[3]{2197-\pi}$

[/hide][/quote]

This is wrong.
First of all, it's $r=\sqrt[3]{\frac{3\cdot2197}{4\pi}}$

Then, SA should $4\pi{r}^{2} $

I was thinking this was ugly, and last night I suddenly realized I read the problem wrong  :rotfl:

I meant a cube. Easier?  :D

Ahahaha, that does make it much easier.  You can easily tell that the cube root of 2197 is 13, so surface area:

$S = 6 \cdot 13^2 = 1014$

I think that most of us were confused at the difficulty of this problem before there was cube in there...


Let $ f(x^2 \plus{} 1) \equal{} x^4 \plus{} 5x^2 \plus{} 3$  What is $ f(x^2 \minus{} 1)$?

[quote="Truffles"]Let $ f(x^2 + 1) = x^4 + 5x^2 + 3$  What is $ f(x^2 - 1)$?[/quote]

Given the information you have, we try to derive $ f(x)$.  We can see that $ x$ is a second degree expression.  We can experiment a bit by assuming that the first term of $ f(x)$ is of the form $ ax^2$.  We square $ (x^2+1)$ and get $ x^4+2x^2+2$.  Compensating for the $ 2x^2$, we need to add $ 3x^2$.  Therefore, we have to add $ 3(x^2+1)$.

So far we have $ (x^2+1)^2+3(x^2+1)$.  To compensate for the $ 3$, we simply subtract $ -1$.

We now have: $ f(x^2+1)=(x^2+1)^2+3(x^2+1)-1 \implies f(x)=x^2+3x-1$.

Now we substitute $ (x^2-1)$ into $ f(x)$ and get $ \boxed{x^4+x^2-3}$.


A crate with mass 10 kg is connected by a massless rope to a cylinder. The massless rope is wrapped around the cylinder several times, en the cilinder has a radius of 40 cm and a mass of 100 kg.
The coëfficient of friction between the crate and the inclined plane is 0,38.

Find the angular acceleration of the cylinder, the acceleration of the crate, and the tension in the rope.

[hide="Answer"]Let $\theta$ be the inclination of the plane (37º), M the mass of the cylinder, m the mass of the crate, $\mu$ the coefficient of friction, g = 9.8 m/s2 the acceleration of gravity, $\alpha$ the angular acceleration, a the liner acceleration, and T the tension. Then:

$a = \frac{g(\sin \theta-\mu \cos \theta)}{1+M/2m}= 0.487 m/s^{2}$
$\alpha = \frac{a}{r}= 1.22 rad/s^{2}$
$T = \frac{Ma}{2}= 24.4 N$
[/hide]


Hmmm... I"m not too sure about using abstract mathematics to gain ground on the war on terror is plausible, but reading the article, I now believe even more that the Patriot Act is unjust.

what is this Patriot Act that everybody is talking about?

Interesting that I didn't see [i]any[/i] mention of the NSA in the article -- seems like it deserves a mention in this article.

As for the Patriot act (if you don't want to read the several hundred pages that white_horse linked to), I can give you the short (and a-political) version:

The U.S.A. PATRIOT Act (the name is actually an acronym for something or other) was a massive security bill passed by the United States Congress and signed into law by President Bush shortly after September 11.  It was passed by an overwhelming majority in both the House and Senate (95 or more of the 100 senators voted in favor of it).  Since then, it's come under increasing attack by various civil liberty organizations, especially the ACLU, for allowing various actions which allegedly are unconstitutional.  Actually, some of the first groups to vocally oppose it were library and librarian's associations, interestingly enough.  (One portion of the bill obligated libraries to turn over customer records to Federal agents without a warrant and without being allowed to inform anyone, especially the individual whose records were seized, of what had happened.)


Provide a synthesis of m-nitrotoluene starting with benzene.

Again, I am just starting so lots of help needed:

Maybe:

1. Nitrating Mixture to create nitrobenzene (because $ \minus{}NO_2$ is a $ 3,5$ directing group)
2. Friedel-Craft Alkylation using $ AlCl_3$ and $ CH_3Cl$ to obtain the product.

Some comments:

1. -NO2 is called a meta directing group, not 3,5 directing. However, in your initial reaction that doesn't matter because the aromatic ring is unsubstituted.

2. The Fridel-Crafts reactions (alkylation and acylation) only work with activated aromatic rings, that is, they don't work in aromatic rings bearing a deactivating group (like the nitro group, which is a very powerfull deactivating group). So your second reaction certainly wouldn't work.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I like it:
Let $G$ be infinite group,and we know that for any subgroup $H$ of $G$ ,$|H|\neq 1$ ,we have $[G:H]<\infty $prove that $ G$ is cyclic.

It follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43092]this one[/url].

Take any element $a\ne 1$ of $G$ and consider the cyclic group $H=\langle a\rangle$ generated by $a$. It has finite index, so let $x_1=1,x_2,\ldots,x_n$ be representatives of all the left cosets of $H$. Since for each $k\in\overline{2,n}$ there is an $m_k$ s.t. $a^{m_k}\in\langle x_k\rangle$, it means that $a'=a^{m_2\ldots m_n}$ commutes with $x_1,\ldots,x_n$, and it's clear now that $a'$ is a non-trivial element of $G$ belonging to the center of $G$.

We have thus shown that $Z(G)\ne \{1\}$. This means that it has finite index in $G$, so, according to the problem cited above, $G'$ is finite. Given the imposed conditions on $G$, this is impossible unless $G'$ is trivial, which is equivalent to $G$ being abelian.

We can now use the structure theorem for finitely generated abelian groups ($G$ is clearly finitely generated) to conclude that $G$ is $\mathbb Z^n$ for some $n$, and it's easy to see that $n$ must be $1$.

All that's left now is for us to solve that other problem :).


One side of a square A is a diagonal of square B. What is the ratio of the area of square B to the area of square A? Express your answer as a common fraction.

[asy]draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
draw((0,0)--(1,1)--(2,0)--(1,-1)--cycle);[/asy]

We can clearly see that $ \frac{1}{2}$ of square $ B$ is the same as $ \frac{1}{4}$ of square $ A$.  Thus, the desired ratio is $ \frac{\frac{1}{4}}{\frac{1}{2}}\equal{}\boxed{\frac{1}{2}}$.

Additionally, let the side length of square $ B$ be $ x$.  Then the side length of square $ A$ is $ \sqrt{x^2\plus{}x^2}\equal{}x\sqrt{2}$.  The area of square $ B$ is $ x^2$, and the area of square $ A$ is $ (x\sqrt{2})^2\equal{}2x^2$.  The desired ratio is thus $ \frac{x^2}{2x^2}\equal{}\boxed{\frac{1}{2}}$.


geometry
parallelogram
trigonometry
rectangle
geometry solved

Let ABCD be a parallellogram. Let Q be the midpoint of [DA], and let F be the foot of the perpendicular on QC trough B. Show that |AF| = |AB|.

You know, this problem is just easy using cosine rules and other trigonometric stuff ... But I was wondering, is there an elegant - purely geometric - solution for this problem ? The only solution I found is the trig solution.

I think I have a ?pure? soln  :)  :

Let the foot of the perpendicular from A to CQ be E, the midpoint of BC be P and the pt of intersection between BF and AP be M. AE parallel to BF (they are both perpendicular to CQ), AP parallel to CE (obvious), so you get AMFE is a rectangle, so AM perpendicular to BF and AE=MF. Since 2AQ=BC and AQ || BC (|| means parallel) we also have 2AE=BF. Because AE=MF we get 2MF=BF, and since AM perpendicular to BF the median AM is also an altitude in triangle ABF, so it is obviously an isosceles triangle.

Let E be the midpoint of BC and G the intersection of AE with BF.Since AE || CQ (obviously) we have that AG is perpendicular to BF. But also , since AE || CQ , then EG || CF so in the triangle BCF , EG is a middle line . Thus BG=GF . From here it follows that ABF is isosceles since BG is a height and a median.

AB intersects QC  in T.  A is the mid-point of BT.
 AF is the median from right angle BFT in triangle BFT.

  Maj.  Pestich

   Polish olympiad 2001,  2nd round.

Translate $ \triangle DQC$, so CD is on AB, Q goes to E. Then QEBF is a rectangle, so since DQ=AQ=AE, AF=BF.


inequalities
IMO Shortlist
number theory unsolved
number theory

Let $p$ be an odd prime. Determine the positive integers $x$ and $y$ with $x\leq y$ for which the number $\sqrt{2p}-\sqrt{x}-\sqrt{y}$ is non-negative and as small as possible.

This is actually IMO Shortlist 1995 problem N8 ( http://www.kalva.demon.co.uk/short/soln/sh95n8.html ).

Can anyone check my solution below? It looks very different from the proposed solution, it has less ugly computations than the proposed solution, and it uses the primality of p only at the very end (it wouldn't use it at all if we would replace "non-negative" by "positive" in the condition of the problem), so I am wondering whether it can be correct...

[color=blue][b]Problem.[/b] Let p be an odd prime. Determine the positive integers x and y with $x\leq y$ for which the number $\sqrt{2p}-\sqrt{x}-\sqrt{y}$ is non-negative and as small as possible.[/color]

[i]Solution.[/i] We claim that the required integers x and y are $x=\frac{p-1}{2}$ and $y=\frac{p+1}{2}$.

In fact, first note that

$\left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}=\frac{p-1}{2}+\frac{p+1}{2}+2\cdot\sqrt{\frac{p-1}{2}}\cdot\sqrt{\frac{p+1}{2}}$
$=p+\sqrt{\left(p-1\right)\left(p+1\right)}=p+\sqrt{p^{2}-1}$.

Now, $\frac{p-1}{2}$ and $\frac{p+1}{2}$ are integers (since p is odd, and thus p - 1 and p + 1 are even), and $\sqrt{2p}-\sqrt{\frac{p-1}{2}}-\sqrt{\frac{p+1}{2}}$ is indeed non-negative (this is equivalent to

$\sqrt{2p}\geq\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ 2p\geq\left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}$
$\Longleftrightarrow\ \ \ \ \ 2p\geq p+\sqrt{p^{2}-1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ p\geq\sqrt{p^{2}-1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ p^{2}\geq p^{2}-1$,

what is true).

So it remains to prove that $\sqrt{2p}-\sqrt{\frac{p-1}{2}}-\sqrt{\frac{p+1}{2}}$ is indeed the smallest possible nonnegative sum of the form $\sqrt{2p}-\sqrt{x}-\sqrt{y}$ for positive integers x and y with $x\leq y$ and achieved only for $x=\frac{p-1}{2}$ and $y=\frac{p+1}{2}$.

In order to prove this, we assume that x and y are positive integers with $x\leq y$ and

$0\leq\sqrt{2p}-\sqrt{x}-\sqrt{y}\leq\sqrt{2p}-\sqrt{\frac{p-1}{2}}-\sqrt{\frac{p+1}{2}}$.

If we succeed to show that $x=\frac{p-1}{2}$ and $y=\frac{p+1}{2}$, then we will be done.

The inequality chain $0\leq\sqrt{2p}-\sqrt{x}-\sqrt{y}\leq\sqrt{2p}-\sqrt{\frac{p-1}{2}}-\sqrt{\frac{p+1}{2}}$ rewrites as

$\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\leq\sqrt{x}+\sqrt{y}\leq\sqrt{2p}$.

We square this inequality (it's not the time to worry about positive and negative yet ;) ):

$\left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}\leq\left(\sqrt{x}+\sqrt{y}\right)^{2}\leq 2p$
$\Longleftrightarrow\ \ \ \ \ p+\sqrt{p^{2}-1}\leq x+y+2\cdot\sqrt{x}\cdot\sqrt{y}\leq 2p$.

Now, we have $x+y<x+y+2\cdot\sqrt{x}\cdot\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)^{2}\leq\left(\sqrt{2p}\right)^{2}=2p$, but the QM-AM inequality for the positive reals $\sqrt{x}$ and $\sqrt{y}$ yields

$x+y=\left(\sqrt{x}\right)^{2}+\left(\sqrt{y}\right)^{2}\geq 2\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{2}=\frac12\left(\sqrt{x}+\sqrt{y}\right)^{2}$
$\geq\frac12\left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}=\frac12\left(p+\sqrt{p^{2}-1}\right)$.

Since $\sqrt{p^{2}-1}>\sqrt{p^{2}-2p+1}=p-1$, this yields $x+y>\frac12\left(p+\left(p-1\right)\right)=p-\frac12$, and thus, since x and y are integers, this entails $x+y\geq p$. So we have $p\leq x+y<2p$. Thus, x + y = p + s for some integer s satisfying $0\leq s\leq p-1$. Denoting x - y = k, we note that $k\leq 0$ (since $x\leq y$) and $x=\frac{x+y}{2}+\frac{x-y}{2}=\frac{p+s}{2}+\frac{k}{2}$ and $y=\frac{x+y}{2}-\frac{x-y}{2}=\frac{p+s}{2}-\frac{k}{2}$. Hence, $p+\sqrt{p^{2}-1}\leq x+y+2\cdot\sqrt{x}\cdot\sqrt{y}\leq 2p$ becomes

$p+\sqrt{p^{2}-1}\leq p+s+2\cdot\sqrt{\frac{p+s}{2}+\frac{k}{2}}\cdot\sqrt{\frac{p+s}{2}-\frac{k}{2}}\leq 2p$
$\Longleftrightarrow\ \ \ \ \ p+\sqrt{p^{2}-1}\leq p+s+\sqrt{\left(p+s+k\right)\left(p+s-k\right)}\leq 2p$
$\Longleftrightarrow\ \ \ \ \ \sqrt{p^{2}-1}-s\leq\sqrt{\left(p+s+k\right)\left(p+s-k\right)}\leq p-s$       (we subtracted p + s from the last inequality chain)
$\Longleftrightarrow\ \ \ \ \ \sqrt{p^{2}-1}-s\leq\sqrt{\left(p+s\right)^{2}-k^{2}}\leq p-s$.

Since $\sqrt{p^{2}-1}>\sqrt{p^{2}-2p+1}=p-1\geq s$, the left hand side of this inequality, $\sqrt{p^{2}-1}-s$, is positive; thus, we can square this inequality:

$\left(\sqrt{p^{2}-1}-s\right)^{2}\leq\left(p+s\right)^{2}-k^{2}\leq\left(p-s\right)^{2}$
$\Longleftrightarrow\ \ \ \ \ \left(p^{2}-1\right)+s^{2}-2\sqrt{p^{2}-1}\cdot s\leq p^{2}+s^{2}+2ps-k^{2}\leq p^{2}+s^{2}-2ps$
$\Longleftrightarrow\ \ \ \ \;-1-2\sqrt{p^{2}-1}\cdot s\leq 2ps-k^{2}\leq-2ps$
$\Longleftrightarrow\ \ \ \ \ 1+2\sqrt{p^{2}-1}\cdot s\geq k^{2}-2ps\geq 2ps$
$\Longleftrightarrow\ \ \ \ \ 1+2\sqrt{p^{2}-1}\cdot s+2ps\geq k^{2}\geq 4ps$.

Since $\sqrt{p^{2}-1}<\sqrt{p^{2}}=p$, we even have

$1+2p\cdot s+2ps>k^{2}\geq 4ps\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ 1+4ps>k^{2}\geq 4ps$.

Since $k^{2}$ is an integer, we thus have $k^{2}=4ps$. Thus, $p\mid k^{2}$, what, since p is prime, yields $p\mid k$, so that $p^{2}\mid k^{2}$. Thus, $p^{2}\mid 4ps$, so that $p\mid 4s$. Since p is odd, this yields $p\mid s$. Since $0\leq s\leq p-1$, this leads to s = 0, and thus $x=\frac{p+s}{2}+\frac{k}{2}=\frac{p}{2}+\frac{k}{2}$ and $y=\frac{p+s}{2}-\frac{k}{2}=\frac{p}{2}-\frac{k}{2}$. Now,

$\left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}\leq\left(\sqrt{x}+\sqrt{y}\right)^{2}$
$\Longleftrightarrow\ \ \ \ \ \left(\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}\right)^{2}\leq\left(\sqrt{\frac{p}{2}+\frac{k}{2}}+\sqrt{\frac{p}{2}-\frac{k}{2}}\right)^{2}$
$\Longleftrightarrow\ \ \ \ \ \left(\sqrt{p-1}+\sqrt{p+1}\right)^{2}\leq\left(\sqrt{p+k}+\sqrt{p-k}\right)^{2}$
$\Longleftrightarrow\ \ \ \ \ \left(p-1\right)+\left(p+1\right)+2\cdot\sqrt{p-1}\cdot\sqrt{p+1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq\left(p+k\right)+\left(p-k\right)+2\cdot\sqrt{p+k}\cdot\sqrt{p-k}$
$\Longleftrightarrow\ \ \ \ \ 2p+2\sqrt{p^{2}-1}\leq 2p+2\sqrt{p^{2}-k^{2}}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ 1\geq k^{2}$.

Since k is an integer and $k\leq 0$, this yields either k = 0 or k = -1. For k = 0, we have $x=\frac{p}{2}+\frac{k}{2}=\frac{p}{2}$, what is impossible, since p is odd and x must be an integer. Thus, k = -1, so that $x=\frac{p}{2}+\frac{k}{2}=\frac{p-1}{2}$ and $y=\frac{p}{2}-\frac{k}{2}=\frac{p+1}{2}$, and the problem is solved.

Oh dear, you are still reading this here? I apologize for being even more boring than usual, but I tried to write every single step to find a mistake. I don't see any. Do you?

  Darij

Denote $s=\sqrt{y}+\sqrt{x}, t=\sqrt{y}-\sqrt{x}$

Case #1: $x=y$, we will get the answer easily: $\sqrt{2p}-\sqrt{2(p-1)}$

Case #2: $x<y$, then consider this:

$(\sqrt{2p}-\sqrt{x}-\sqrt{y})(\sqrt{2p}+\sqrt{x}+\sqrt{y})(\sqrt{2p}+\sqrt{x}-\sqrt{y})(\sqrt{2p}-\sqrt{x}+\sqrt{y})=(2p)^{2}+(x-y)^{2}-4p(x+y)\geq 1$

Also

$(\sqrt{2p}-\sqrt{x}-\sqrt{y})(\sqrt{2p}+\sqrt{x}+\sqrt{y})(\sqrt{2p}+\sqrt{x}-\sqrt{y})(\sqrt{2p}-\sqrt{x}+\sqrt{y})=(2p-s^{2})(2p-t^{2})$

so, $(2p-s^{2})(2p-t^{2})\geq 1$ and $(st)^{2}=(x-y)^{2}\geq 1$

thus $(2p-s^{2})(2p-\frac{1}{s^{2}})\geq 1$

From this, we will get $s^{2}\leq p+\sqrt{p^{2}-1}\to s\leq \sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}$

then the answer is $\sqrt{2p}-\sqrt{\frac{p-1}{2}}-\sqrt{\frac{p+1}{2}}$

It seems that $p$ needn't to be a prime. Is there anything wrong with my solution?


The following inequality is written in Muirhead notation.
$ (a,b,c)\plus{}(d,e,f)\ge 2(g,h,i)$
Suppose $ a\plus{}b\plus{}c\equal{}d\plus{}e\plus{}f\equal{}g\plus{}h\plus{}i$, please find the maximum of $ g$ such that the above inequality is always true for non-negative real numbers $ x_k$, $ k\equal{}1,2,3$.

see the long discussion about a similar question [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=597177138&t=56369]here[/url].
For the case $ [a,b,c] \plus{} [d,e,f]\ge 2[g,h,i]$ (please use rather square brackets for symmetrical sums) it holds of course if $ a \plus{} d \equal{} 2g,b \plus{} e \equal{} 2h,c \plus{} f \equal{} 2i$ by AM-GM.

Schur yields a sharper ineq of this type, viz. $ [n \plus{} 2,0,0] \plus{} [n,1,1]\ge2[n \plus{} 1,1,0]$.

Supposing $ a\ge b\ge c$ etc. and $ a > d$, it seems that $ g$ can come much closer to $ a$, though. E.g., $ [6,1,0] \plus{} [3,3,1]\ge 2[5,1,1]$ seems to hold.
A more precise, interesting question would thus be if there is a minimum for $ \frac {a \minus{} g}{a \minus{} d}$.  :maybe:


function
vector
integration
calculus
geometry
3D geometry
sphere

Stokes theorem is usually stated as follows.
Let $S$ be a smooth surface in $\mathbb{R}^{3}$ parameterized by a $C^{2}$,$1-1$ function $\phi : T\rightarrow \mathbb{R}^{3}$, where $T$ is a plane region bounded by a piecewise $C^{1}$ Jordan curve$\Gamma$. Denote by $C$ the image of $\Gamma$ under$\phi$. Let $F$ be a $C^{1}$ vector field on an open set containing $S$ and $C$. Then we have$\int \int_{S}(\nabla\times F).ndS=\oint_{C}F.d\alpha$,
where the line integral is taken in the direction inherited via$\phi$ from the positive direction of$\Gamma$.
[b]a[/b])What is the injectivity of $\phi$ good for?
[b]b[/b])How do we conclude, rigorously, from (the extended version) of this theorem that for every closed orientable object, say a sphere, both integrals are equal to $0$?
PS For the former I guess it has something to do with orientablity.
And for the latter I have read that the boundary of closed objectes is empty, so the result follows!! But this does not content me.
Thank you in advanced for your reply.[A typo fixed in edit.]

If we're integrating over the image, we want that image to be traced only once. If $\phi$ were not injective, the image might not even be a surface.

For the second question, we need a topological theorem:
Every compact smooth orientable surface can be dissected into finitely many consistently oriented pieces, each parametrized as in the statement.

Then we integrate over the whole surface by adding up the pieces. The boundary integrals will directly cancel (each part of the boundary is traced twice in opposite directions), and the whole integral becomes zero.

In the general manifold treatment of Stokes' theorem, there are no tricks needed for the no-boundary special case. We need tricks here because we are restricted in the original statement to surfaces diffeomorphic to a compact subset of $\mathbb{R}^{2}$.

[quote="jmerry"]If we're integrating over the image, we want that image to be traced only once. If $\phi$ were not injective, the image might not even be a surface.[/quote]
Dear, jmerry. We have a definition for a [i]smooth curve [/i]as follows; Let $W$ be a region in the $uv$ plane, we call the function $\phi : W\rightarrow \mathbb{R}^{3}$ a smooth curve when these two hold;
[b]a[/b])$\phi$ is $C^{1}$ [b]b[/b]) $\forall (u, v) \in W, \phi_{u}\times \phi_{v}$ is non zero.
On the other hand we have defined integration over a smooth curve(ie density integral) without mentioning the condition of surjectivity. So the first question is still open for me. :)

Please check if this satisfies Stokes' conditions (and/ or its results)?
$\phi (u, v)=\rho (\sin u\cos v,\sin u\sin v, \cos u)$, where $0\leq u <\pi$, $0\leq v<2\pi$ with $F=d\alpha$.
And this one is about the second question.

:help: Hey,It's also my problem.Has anyone a solution?


What is the last 3 digits (hundreds, tens and units digits) of 3^1983.

Thanks.

Usually they won't ask you for the hundreds digit, just the tens and the ones (they're easy too).

I see the pattern way works, but can't you use mods which would make it faster and easier.

I suppose a combinatorics/number theory approach might work for this... but it gets awfully complicated... (read AoPS vol 2)

$3^{1983}=3*9^{991}$

I'm guessing that if we find the last three digits of $9^{991}$ , and multiply that by 3, then we will have the last three digits for $3^{1983}$ .

So...

$9^{991}={(10-1)}^{991}={991\choose991}{10^{991}}-{991\choose990}{10^{990}}+{991\choose989}{10^{989}}\ldots-{991\choose2}{10^2}+{991\choose1}{10}-{991\choose0}$

To find the last three digits, we take ${(mod 1000)}$ since we only want the last three terms, everything else will be modded out.

The last three terms are:

$-{991\choose2}*{10^{2}}+{991\choose1}*{10}-{991\choose0}$

Now...

$-{991\choose2}*{10^{2}}=-\frac{(991)(990)(100)}{2}\equiv{500 (mod 1000)}$

${991\choose1}*{10}=-(991)(10)={9910}\equiv{910(mod1000)}$

$-{991\choose0}\equiv{-1(mod1000)}$

we add it all up:
${{(500)}+{(910)}+{(-1)}=1409}$
${1409}\equiv{409(mod1000)}$

Multiply 409 by 3, because we took out the 3 at the beginning to make it a power of 9,
we get ${1227}\equiv{227(mod1000)}$

Therefore the last three digits are 227.

Here is another solution from Andrew Westerdale (Stoneman Douglas HS, Florida):

so close too far: i did it using euler's theorem, but it happens to be really messy/ugly
so close too far: just becasue of the numbers
so close too far: m^phi(n) = 1 mod n
so close too far: m^phi(n) = 1 mod n is really effective
so close too far: for n = 1000
so close too far: phi(n) = 1000 (1-1/2)(1-1/5) = 400
so close too far: so 3^400 = 1 mod 1000
so close too far: so 3^1600 = 1 mod 1000
so close too far: 3^1983 = 3^1600 * 3^383 mod 1000
so close too far: =3^383 mod 1000
so close too far: the way you get phi(n) is by evaluating the product of n and (1-1/f) for all factors f
so close too far: which is why phi(1000) = 1000(1-1/2)(1-1/5)
so close too far: anyways
so close too far: so we want to find 3^383mod1000
so close too far: break 383 into "binary":
so close too far: 2^8+2^6+2^5+2^4+2^3+2^2+2^1+2^0
so close too far: so 3^383 = 3^(2^8)*3^(2^6)*...*3^(2^0)
so close too far: the good thing about the sequence we created is that each term is the square of the one before it, with some terms, 3^(2^7), left out
so close too far: but this makes the it easier to do the arithmatic
so close too far: since we just keep squaring the term, from 3 to 9 to 81, etc. only keeping the last 3 digits
so close too far: we get 3^383 = 3 * 9 * 81 * 561 * 721 * 841 * 281 * 521 mod 1000
so close too far: where you break out the crazy arithmatic skills, combine pairs of them and keep the last 3 digits as you go
so close too far: and you end up with 227 mod 1000
so close too far: it usually works out nicer because the problem will be like 3^1220 so you can take out the 3^400's and are left with something more manageable


calculus
integration
calculus computations

i need a way to solve this integral:
integral[(x^2-2)/(x^5-x^3+x)]dx
thanks.

Hint
$ \frac {x^2 \minus{} 2}{x^5 \minus{} x^3 \plus{} x} \equal{} \frac {\frac {1}{2}(4x^3 \minus{} 2x)}{x^4 \minus{} x^2 \plus{} 1} \minus{} \frac {2}{x}$

and then how do i integrate the left one?

$ \int\frac{f'(x)}{f(x)}\,dx \equal{} \ln|f(x)| \plus{} C$.


Sa se determine toate morfismele de la (Q , +) la (Sn , o), unde Sn este grupul permutarilor de grad n

de la definitivat 2001

$\mathbb Q$ e grup divizibil (adica pentru orice $x\in\mathbb Q$ si orice numar natural nenul $n$ exista $y\in\mathbb Q$ astfel incat $ny=x$), si un morfism de la un grup divizibil la unul finit trebuie sa fie trivial.


AMC
USA(J)MO
USAMO
SFFT
special factorizations

EDIT: Blah, this solution was both ugly and had a subtle flaw.

To answer the below question about this being too easy for MOP, team contests have plenty of very easy problems on them. This because of the way team contests are scored, which is a little unorthodox.

Unless I'm doing it wrong, isn't this way too easy for MOP?
[hide]
We can rearrange the first equation to get $ xy\plus{}x\plus{}y\equal{}z$, and then use Simon's Favorite Factoring Trick to get

$ (x\plus{}1)(y\plus{}1)\equal{}z\plus{}1$.

Similarly, using the other equations,

$ (y\plus{}1)(z\plus{}1)\equal{}x\plus{}1$

and

$ (z\plus{}1)(x\plus{}1)\equal{}y\plus{}1$.

Substitute $ a\equal{}x\plus{}1$, $ b\equal{}y\plus{}1$, and $ c\equal{}z\plus{}1$ for convenience.  Multiplying the three equations together, we get

$ a^2b^2c^2\equal{}abc$.

First, we consider what happens when one of the variables, WLOG $ a$, is $ 0$.  Since $ bc\equal{}a$, either $ b$ or $ c$, WLOG $ b$, is $ 0$.  Since $ ab\equal{}c$, we then have $ c\equal{}0$.  So $ (a,b,c)\equal{}(0,0,0)$, which corresponds to $ (x,y,z)\equal{}(\minus{}1,\minus{}1,\minus{}1)$, works.

Now, assume that all variables are nonzero.  We can divide through by $ abc$ to get $ abc\equal{}1$.  Since $ ab\equal{}c$, we get $ c^2\equal{}1$.  Similarly, $ a^2\equal{}1$ and $ b^2\equal{}1$.  Since $ abc\equal{}1$, we can either have $ 0$ or $ 2$ of $ a,b,c$ be $ \minus{}1$.  So we find that $ (a,b,c)\equal{}(1,1,1)$, which corresponds to $ (x,y,z)\equal{}(0,0,0)$, is a solution, along with all permutations of $ (a,b,c)\equal{}(1,\minus{}1,\minus{}1)$, which gives all permutations of $ (x,y,z)\equal{}(0,\minus{}2,\minus{}2)$ as solutions.
[/hide]

[quote="MellowMelon"]To answer the below question about this being too easy for MOP, team contests have plenty of very easy problems on them. This because of the way team contests are scored, which is a little unorthodox.[/quote]

OK, but isn't this a little too easy to serve any purpose at MOP?  I mean, this is a 5 minute problem at most for anyone of MOP caliber.

[quote="towersfreak2006"]I think it's from Reid Barton's compilation:
http://web.mit.edu/rwbarton/Public/mop/ [/quote]I know that, but how'd he find that?

It caught my eye because I made up an identical system for a math team lecture 2 years ago while teaching factoring.

I found it really funny that something like that could be found around MOP, let alone, one that is coincidentally identical to an exercise used at a high school with no USAMO qualifiers.

To answer the above question, I found it clicking random links on Wikipedia. I somehow found myself at [i]"Frank and Brennie Morgan Prize for Outstanding Research in Mathematics by an Undergraduate Student"[/i], from where I clicked a link to go to [i]"Reid W Barton"[/i], which had a link to his collection of files.

[quote="The Zuton Force"]Find all ordered triples of real numbers $ (x,y,z)$ that satisfy $  \{\begin{array}{c} xy = z - x - y \\
yz = x - y - z \\
zx = y - z - x \end{array}  \;$

[u][b]Source:[/b][/u] [url=http://web.mit.edu/rwbarton/Public/mop/tc4.pdf]Shamelessly borrowed from here[/url][/quote]
Here's my solution, which is slightly different than E^(pi-1)'s solution above:
Plus all of three equations we have: xy+yz+xz=-(x+y+z). Therefore:xy+x+y=-(yz+xz+z)
Replace it into 1st equation we have: z=-(yz+xz+z)
If z=0 we easily solve the rest of systems
Now z#0 then divide both sides by z, we get: y+x=-2
Doing similarly for two more equations we rewrite the original system:
xy=z+2(1)
yz=x+2
zx=y+2
Now, just substitute y=-2-x and z=-2-x into(1) we get: (-2-x)(x)=-x
The rest is simple :)


Prove or disprove:

$a,b>0$

$\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge2$

[quote="minsoens"][hide]
$\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge2$
Counterexample: $a=b=2$ $\Longrightarrow$ $\frac{2^2}{2^3}+\frac{2^2}{2^3}\le2$
[/hide][/quote]

You are right minsoens, maybe I should write $\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge2$ for $a,b\le1$. Do you think it is true?

Or the other one $\frac{a^2}{b^3}+\frac{b^2}{a^3}\le2$ for $a,b\ge1$...?

$\frac{a^2}{b^3}+\frac{b^2}{a^3}\le2$ for $a,b\ge 1$
isn't true because if you make one variable really big and the other really low, like:
$a=100$ and $b=1$..

$\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge2$ for $a,b\le 1$
Okay, we use the AM-GM inequality:
$\frac{\frac{a^2}{b^3}+\frac{b^2}{a^3}}{2}\ge \sqrt{\frac{a^2}{b^3}\cdot\frac{b^2}{a^3}}$
$\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge \frac{2}{\sqrt{ab}}$
Since $a,b\le 1$, $\frac{1}{\sqrt{ab}}\ge 1$ $\Longrightarrow$ $\frac{2}{\sqrt{ab}}\ge 2$
$\frac{a^2}{b^3}+\frac{b^2}{a^3}\ge \frac{2}{\sqrt{ab}} \ge 2$

I found that if $b$ is kept constant, then the minimum value of $\frac{a^2}{b^3} + \frac{b^2}{a^3}$ is $\approx 1.96 \cdot \frac{1}{b}$.


I am uneasy about approaching a professor for a recommendation for an REU? Have you ever heard of a professor refusing to recommend a student when asked?

Thanks

I say "yes" the majority of the time but I have said "no." I write different letters for different people - if I didn't, my letters would be pretty much worthless. Why would I refuse? Perhaps because I don't think I know the student very well. (I may very well suggest, "Why don't you ask Professor A whom you had for course B?") But perhaps because I think the letter I would write wouldn't do the student much good.

All you can do is ask. But if you encounter reluctance, don't push or try to persuade - you might be pushing for a lukewarm letter.

When you ask for a recommendation letter one of the following is likely to occur:

1.  The professor writes an enthusiastic letter full of details that are helpful in evaluating the application.   A significant effort should be made by all serious students to cultivate relationships with a few professors who will write such letters on their behalf.

2.  The professor agrees to write a letter but it is either poorly written, "lukewarm," or lacking in specific details that tend to be helpful in evaluating the application.

3.  The professor forgets to write the letter.  Students often need to follow up or provide gentle reminders.

4.  The professor says "no."  This is far better than #2 or #3 above.

When you ask the professor for a recommendation use language that makes it more likely that your letter will fall into category #1.  "Professor X, I'm applying for (whatever), it's quite competitive, and I need a very strong recommendation by (date).  Would you be able to write such a recommendation?"

If the professor says "yes" then the next step is to make sure the right details get into the letter.  Don't assume that the professor will remember details of projects or work that you've done in classes or extracurriculars.  Have copies of your resume and old projects/work that you've done for the professor on hand to jog his/her memory.


Compute the Taylor series of $ \frac{1}{1\plus{}x\plus{}x^2}$ about the point $ \minus{}\frac{1}{2}$ and find the interval of convergence.

Can someone please help. Thank you very much. :P

$ \dfrac{1}{1\plus{}x\plus{}x^2}\;\equal{}\;\dfrac{1}{\dfrac{3}{4}\left(1\;\plus{}\;\dfrac{4}{3}\left(x\plus{}\dfrac{1}{2}\right)^2\right)}\;\equal{}\;\dfrac{4}{3}\left(1\;\plus{}\;\dfrac{4}{3}\left(x\plus{}\dfrac{1}{2}\right)^2\right)^{\minus{}1}\;\equal{}\;\;\sum_{n\equal{}0}^\infty\;(\minus{}1)^n\;\left(\dfrac{4}{3}\right)^{n\plus{}1}\;\left(x\plus{}\dfrac{1}{2}\right)^{2n}$

now,
$ \dfrac{|(n\plus{}1)^{th}\;\;term|}{|n^{th}\;\;term|}\;\equal{}\;\dfrac{\bigg|(\minus{}1)^{n\plus{}1}\;\left(\dfrac{4}{3}\right)^{n\plus{}2}\;\left(x\plus{}\dfrac{1}{2}\right)^{2(n\plus{}1)}\bigg|}{\bigg|(\minus{}1)^n\;\left(\dfrac{4}{3}\right)^{n\plus{}1}\;\left(x\plus{}\dfrac{1}{2}\right)^{2n}\bigg|}\;\equal{}\;\bigg|\minus{}\dfrac{4}{3}\left(x\plus{}\dfrac{1}{2}\right)^{2}\bigg|$

the series will converge absolutely (and hence converge) [b]iff[/b] $ \bigg|\minus{}\dfrac{4}{3}\left(x\plus{}\dfrac{1}{2}\right)^{2}\bigg|\;<\;1\;\;\;\quad\Rightarrow\;\;\;\bigg|x\plus{}\dfrac{1}{2}\bigg| \;<\; \dfrac{\sqrt 3}{2}$

hence, interval of convergence is $ \left(\dfrac{\minus{}\sqrt 3\;\minus{}\;1}{2}, \;\;\dfrac{\sqrt 3\;\minus{}\;1}{2}\right)$

The radius of convergence of a series for a "usual" function (meaning analytic function) should always be the distance between the center point for the series and the nearest singular point for the function.

The singular points for $ \frac{1}{z^2\plus{}z\plus{}1}$ are the poles at $ \minus{}\frac12\pm\frac{i\sqrt{3}}{2}.$ These points lie at a distance of $ \frac{\sqrt{3}}2$ from $ \minus{}\frac12,$ hence the radius of convergence should be $ \frac{\sqrt{3}}2.$

Of course, this is exactly what misan found.

Had we expanded this in a power series centered at zero, we would have had a radius of convergence of $ 1.$


Let $a,b,c>0$ such that $a+b+c+abc=4$.Prove that:

    $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq\frac{\sqrt{2}}{2}\cdot (a+b+c) $ ;)

IMHO, naming each inequality cool and new and nice and magnificient is a LITTLE proof of narcisism, but after all everyone with his taste. Anyway, again IMHO the purpose of an inequality is to be strong and quite relevant. This one does not meet the last two points and about new, absolutely no (I will prove that immediately). So it remains the cool quality. :D
    Now, let's solve this cool inequality. The method? Obviously, Cauchy (by the way, now it's really simple to solve your inequalities, since they are all with this method). In fact the problem is un unhappy combination of two ugly inequalities and one that I am fond of (appeared in Belarussian olympiad or India). The first two are Cauchy. Actually, after applying twice Cauchy we reduce it to the very nice inequality $a+b+c\geq ab+bc+ca$. For that one you can search the forum. Although it follows immediately from Schur.

Sorry!!!
I am not good in English. So Harazi can you write clearly your solution!!!!! :blush:

Ok, by Cauchy we have $\sum{\frac{a}{\sqrt{b+c}}}\geq\frac{(a+b+c)^2}{\sum{a\sqrt{b+c}}}$. Next, $\sum{a\sqrt{b+c}}\leq\sqrt{2(a+b+c)(ab+bc+ca)}$ again by Cauchy. Thus it remains to prove that $ab+bc+ca\leq a+b+c$. Suppose that $a+b+c<ab+bc+ca$. Write Schur inequality in the form $\frac{9abc}{a+b+c}\geq 4(ab+bc+ca)-(a+b+c)^2$. It follows that $\frac{9abc}{a+b+c}>(a+b+c)(4-(a+b+c))$ and thus $ a+b+c<3$. But then $abc<1$ and this is impossible since $a+b+c+abc=4$.

It follows that \frac{9abc}{a+b+c}>(a+b+c)(4-(a+b+c)) and thus a+b+c<3. But then abc<1 and this is impossible since a+b+c+abc=4.

Harazi  I really don't understand that I think I get some troubles from this.
More clearly, can you?
(Sorry!!  :blush: )

I don't understand you, man! Schur inequality is $\frac{9abc}{a+b+c}\geq 4(ab+bc+ca)-(a+b+c)^2$. Assume that $ab+bc+ca>a+b+c$ then $ 4(ab+bc+ca)-(a+b+c)^2>4(a+b+c)-(a+b+c)^2=(a+b+c)(4-(a+b+c))=abc(a+b+c)$ since $ a+b+c+abc=4$. Thus,  it follows that $a+b+c<3$ and by AM-GM we have $abc<1$ impossible. Is it finally clear?

Oh yes ti is really great , I misunderstand your solution!!!!!!! It's right. ( Because of some English words that I misunderstand).
How about you "ceza lupu" do you have any solution for this problem. I think this solution is not a natural one.

Harazi can you help me to give a nice solution for this one:

(a^2 + b^2 + c^2)^2>3(a^3b + b^3c + c^3a)
I have seen a solution in one post of you with Sasha( his picture is a cat ) but it is not really nice. Please help me!!!!
I wait you....
Loading...............

Do you mean that his solution is not nice enough? :?

What is the ugly problem you are reffering to, Arne? Because I think Vasc's inequality is one of the most elegant and hard and nice inequalities I have ever seen (and I like it so much since Muierhead is useless there!!!!). Unfortunately, the problem is doomed to have a bloody computational solution and I'm sorry to say this, but I really don't think a nice solution will ever be given for this problem.


I was trying to figure out which teams at PUMA-C had NYC representation on them, but they all have funny names and so it wasn't clear -- anyone care to enlighten me?

Team Colbert, Team Pig, and Team Goat were all from NYC. Team Pig was named as such because everybody on the team except for me went to Hampshire this past summer. Team Colbert was named as such in an attempt to get them on the show. I think Mr C. had a reason other than Mr. Novikoff's obsession with Goats for that name, but I'm not sure.

In addition, Florida had registered a full B-team when they only had 2 people for it, 6 NYC kids were on Florida B.

Here are the team rosters. Florida B was formed after Mr. C sent these out, so I'm not sure of everyone that was on it.

TEAM COLBERT

KAM  CHAN—BROOKLYN TECH

TAO RAN CHEN—BAYSIDE

ABE CHIEN—STUYVESANT

BEN LERNER—STUYVESANT

KRISHANU SANKAR—HORACE MANN

KEN SUZUKI—STUYVEANT

ALLAN ZHAO—BROOKLYN TECH

DANNY ZHU—STUYVESANT

TEAM GOAT

LOUIS EVANS—HUNTER

ARTUR DMOWSKI—STUYVESANT

ELIOT KIM—STUYVESANT

YOUN KIM—STUYVESANT

KEVIN KWOK—HUNTER

JENNY KWUN—STUYVESANT

CONNIE PARK—HUNTER

GABE ZUCKER—HUNTER

TEAM PIG

BENJAMIN HIRSCH—STUYVESANT

RACHEL KOGAN—BREARLEY

HARRIET LI—HUNTER

BENNO MIRABELLI—POLY PREP

DAVID MOON—HORACE MANN

ADAM SEALFON—STUYVESANT

DANNY YUAN—STUYVESANT

JIN WAN—BROOKLYN TECH

Thanks!  And many congratulations are due, since you all managed to do very well (rocking the power test in particular)!  Unrelatedly, I'm pleased to say that it only took me a minute or two to figure out who you were from the team roster ... otherwise, I might have gone on for ever believing you were from Nebraska.


One incorrectly remembered Fermat's little theorem as follows :
"For any prime n, an+1 - a is divisible by n". (*)
Describe the set of integers (not just primes) for which (*) is true.

A (not really convinced  :() try. This problem proved rather challaging. I'm surprised there are only finite number of such numbers  :?

First let's notice that n=1 (and n=2) works !

Let's suppose n != 1.

. Let n = prod pi^ai with pi prime.
If ai > 1, applying (*) with pi gives a contradiction so n = prod pi.

. "True" Fermat theorem : a^(pi-1) = 1 [pi] if (a,pi) = 1.
a^n = 1 [n] implies pi-1 | n for all i.

. So we have to solve pi-1 | prod pj for all i.
Wolog p1 < p2 .. < pj
. p1-1 < p1 so p1-1=1 or p1=2
. if j > 1, p1 <= p2-1 < p2  so p2-1=p1 or p2 = 3
. if j > 2 or 3, continuing gives p3 = 7, p4 = 43.
. if j > 4, p5 should divide 1806 but there are no divisor 'a' of 1806, greater than 43, such that a+1 is prime so there are no p5.

So the numbers satisfying the incorrect Fermat theorem are :
1, 2, 2*3=6, 2*3*7=42, 2*3*7*43=1806.

That is entirely correct !

I really enjoyed this problem.


AMC
AIME
blogs
geometry
probability
trigonometry
floor function

[b]2006 Mock AIME 1 Results[/b]

Total # of Test Takers:

16

Top Scorers:

13 - gotztahbeazn, Pakman2012
12 - scorpius119
11 - Elemennop
10 - krustyteklown

Hardest Questions:

1 correct answer - 14, 15
2 correct answers - 13

Easiest Questions:

14 correct answers - 3
13 correct answers - 4

Median Score:

8

Mean Score:

8

Mode Score:

8

There you go. Special congratulations to gotztahbeazn and scorpius119 to be the only people with correct answers to #15 and #14, respectively.

The solution to #15 can be found on my [url=http://wangsblog.com/jeffrey/]blog[/url].

[b]Answers and Topic Distribution[/b]

1. 668 (G)
2. 050 (NT/PC)
3. 470 (G)
4. 107 (NT)
5. 391 (A/NT)
6. 990 (NT)
7. 112 (NT)
8. 368 (SS)
9. 516 (A/SS)
10. 900 (G)
11. 702 (G)
12. 267 (G)
13. 193 (PC)
14. 322 (A)
15. 096 (G/T)

Topic Distribution:
Algebra - 3
Geometry - 6
Number Theory - 5
Probability & Counting - 2
Sequences and Series - 2
Trigonometry - 1

[quote]Score: 6

Distribution:

1 1 1 1 1 0 0 1 0 0 1 0 0 0 0[/quote]

 :(
Exactly as I feared
Got 6 and 7 wrong due to arithmetic errors, and apparently 9 and 12, too. :(
I feel dumb.

[quote="eryaman"]Hmmm looks like I got an 8.

On number 1:
What I did was the floor(2006/6)=334 to find the number within on radius on one side.  But  this include the orginal point, so there are actually 333.  So to find the number within the certain radius on the other side, multiply by 2 gives 666.

It looks like in the answer it didn't subtract 1.  Why is this not necessary?[/quote]

It actually doesn't include the original point; for example, try it with 7 points. The answer would be 2, or $2 \cdot \left\lfloor \frac{7}{6} \right\rfloor$.

I missed #1. I loved my logic on that one; It needs to be [b]six[/b]ty degrees, so the angle is pi over [b]six[/b].

And for #7, Is the answer not 448? f(n) is the number of divisors of n, which includes both negative and positive divisors. That would thus double the answer twice, and make it four times your given answer. I believed that you probably meant positive divisors only, but the problem didn't state it so I solved it as such.

Great test over all; I generally liked the questions, though some like the Collatz one (#8) is just plain computation. Yay for brute-forcing #13.

wow, i am really could have done better, i got 8, and for problems 2,5,6,8, i made some tiny computation error that messed me up, i have to work on my careless errors, if i had a cheap handheld calculator, i would have been fine... :wallbash_red:, but at least i did this before anything official, also i am thinking about making a mock aime...but it will be a while before i actually do anything with it, it takes a while to think of problems

Anyone have an elegant way to do #10? I had nasty computation, which I ultimately made an arithmetic error on.

How abouut a solutions thread?

I really want to see a solution for #8. :dry:

hmm, i got a 3. i got 2, 3, and 9. missed 1 and 4.  :(

[quote="dakyru"][quote="chess64"]I don't understand why my method is incorrect. :? I think I'm misreading the problem...

Let the radius of the circle be 1, so that the circumference is $2\pi$. The distance between each consecutive point is $\frac{2\pi}{2006}=\frac{\pi}{1003}$. The number of points $k$ that have distance $\le 1$ from one point are $\frac{\pi}{1003} \cdot k \le 1$, so $k = 319$...[/quote]

It's not distance along the arc, it's distance from one point to another.[/quote]Haha chess64, I did that exact same thing at first...along with the correct method (with the -1 part wrong), then I started wondering why the answers were so different.  I finally figured out what was going wrong, though, luckily.  :D

Looks like everyone made dumb mistakes.

I answered 1-12, missed 1 from that dumb mistake, 4 and 8 from arithmetic errors.

Anyone have a solution for number 11?

[quote="eryaman"]Anyone have a solution for number 11?[/quote]

Just use the Pythagorean Theorem (and know the centroid divides the medians in a $2: 1$ ratio) to get the height of the tetrahedron. Then use Pythagoras again to find the circumradius. Height - circumradius = inradius, which we get circumradius/inradius = $3$. So the ratio of volume of the spheres is $27$.

Easier with a cube. Circumradius is half the space diagonal = $\frac{s\sqrt{3}}{2}$. Then the inradius is half of a side = $\frac{s}{2}$. So circumradius/inradius = $\sqrt{3}$. Ratio of volumes is $3\sqrt{3}$.

$27^2 - (3\sqrt{3})^2 = 729-27 = 702$.

[quote="dakyru"][quote="chess64"]I don't understand why my method is incorrect. :? I think I'm misreading the problem...

Let the radius of the circle be 1, so that the circumference is $2\pi$. The distance between each consecutive point is $\frac{2\pi}{2006}=\frac{\pi}{1003}$. The number of points $k$ that have distance $\le 1$ from one point are $\frac{\pi}{1003} \cdot k \le 1$, so $k = 319$...[/quote]

It's not distance along the arc, it's distance from one point to another.[/quote]

OMG I should have realized that :wallbash:

[quote="eryaman"]Anyone have a solution for number 11?[/quote]

The altitudes to the faces of the tetrahedron intersect at $\frac{1}{4}$ their length from each face, so the radius of the inscribed circle is $\frac{1}{3}$ if the radius of the circumscribed circle is $1$.


Find all$P(x)\in R[x]$such that
                  $P(x^2-1)=P(x)^2-1$      $\forall x\in R$
I have a nice solution  :blush:
[u][b]another question[/b][/u]:
Let $Q(x)\in R[x]$;$degQ\ge 1$.Find all $P(x)\in R[x]$such that
                $P(Q(x))=Q(P(x))\forall x\in R$
I can solve it with :$Q(x)=x^n-1$

Oh,nobody. :( Why?

[color=red]Edit by Megus: because you posted it in wrong section  ;) - moved[/color]

I think you should post your nice solutions for everyone.It's very nice :D

[quote="Omid Hatami"]Discussed before for some special cases.

http://www.mathlinks.ro/viewtopic.php?p=376258#p376258[/quote]
The answer of my problem with$Q(x)=x^n-1$and $degQ\ge 1$:
$P(x)\in (P_k(x))_{k=0}^{\infty}: P_0(x)=x;P_{k+1}(x)=P_k^n(x)-1$ :)

[quote="Bluesea"]can you post your solution nthd[/quote]
Please wait some days! :)

[b][u]Lemma 1[/u][/b]:If$P(x)\in R[x],degP>1$ satisfy this equality$\rightarrow\exists P_1(x)$ such that:$P(x)=P_1(x^2-1)$
[b][u]Lemma 2[/u][/b]: there is only one polynomial $P(x)$ with $degP$ is odd satisfied is $P(x)=x$
 :)  ;)   good luck.


logarithms
symmetry
calculus
calculus computations

If $ u\plus{}\ln{u}\equal{}xy$, find $ \frac{\partial^2 u}{\partial y\partial x}$.

$ 0\equal{}\frac{\partial }{\partial x} \left[u\plus{}\ln u \minus{}xy\right]\equal{}u_x\plus{}\frac{u_x}{u}\minus{}y$

$ 0\equal{}\frac{\partial }{\partial y}\left[ u_x \plus{}\frac{u_x}{u}\minus{}y\right]\equal{}u_{xy}\plus{}\frac{u_{xy}}{u}\minus{}\frac{u_xu_y}{u^2}\minus{}1$

$ u_{xy}(1\plus{}u^{\minus{}1})\equal{}u_xu_yu^{\minus{}2}\plus{}1 \to u_{xy}\equal{}\frac{u_xu_yu^{\minus{}2}\plus{}1}{1\plus{}u^{\minus{}1}}$

$ u_x(1\plus{}u^{\minus{}1})\equal{}y \to u_x\equal{}\frac{yu}{u\plus{}1}$
and by symmetry of the original, we know we will get $ u_y\equal{}\frac{xu}{u\plus{}1}$
so
$ u_{xy}\equal{}\frac{\frac{xy}{(u\plus{}1)^2}\plus{}1}{1\plus{}u^{\minus{}1}}\equal{}\frac{u[xy\plus{}(u\plus{}1)^2]}{(u\plus{}1)^3}\equal{}\frac{u[u\plus{}\ln u\plus{}(u\plus{}1)^2]}{(u\plus{}1)^3}$


ratio
trigonometry
geometry unsolved
geometry

In a triangle $ABC$ with $AB\neq AC$, the internal and external bisectors of angle $A$ meet the line $BC$ at $D$ and $E$ respectively. If the feet of the perpendiculars from a point $F$ on the circle with diameter $DE$ to $BC,CA,AB$ are $K,L,M$, respectively, show that $KL=KM$.

The circle with diameter DE is the A-Apollonius circle of $\triangle ABC,$ the locus of points F with constant ratio of distances $\frac{FB}{FC}= \frac{AB}{AC}$ from the vertices B, C. The pedal triangle $\triangle KLM,\ K \in BC, \L \in CA,\ M \in AB$ of any point F on the A-Apollonius circle is isosceles with KL = KM. For example, quadrilaterals FKCL, FKMB are both cyclic and FC, FB are diameters of their circumcircles. Therefore, $KL = FC \sin C,\ KM = FB \sin B.$ Then

$\frac{KL}{KM}= \frac{FC}{FB}\cdot \frac{\sin C}{\sin B}= \frac{AC}{AB}\cdot \frac{\sin C}{\sin B}= 1$

by the sine theorem for $\triangle ABC.$

[b]An interesting remark.[/b] $m(\widehat{LKM})=\left|A-m(\widehat{BFC})\right|$ and the points $F,K,L,M$ never are concyclically.

In fact, it is the solution of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3459&sid=a2e3248bc6b2689da51f66d7b8f98b16#p3459]IMO 1996/2[/url].

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=644&t=356195&start=20]IMO 2010/4[/url] is also similar to the problem above. All uses Apollonius circle for angle bisectors.


This site is really cool, it has tons of ebooks (pdf's) for free, check it out.

[url]http://www.freecomputerbooks.com/[/url]

Hmm... It doesn't have what I am taking. T.T

just be thankful that you have a lot of free books.


function
integration
inequalities
calculus
derivative
calculus computations

does this hold for convex functions $ f$ in $ [a,b]$
$ (\int _{a}^{b}f^{n}(x) dx)^{\frac{1}{n}}$ is increasing in $ n$, where $ n \geq 1$

If $ f$ is constant and the interval is of size larger than 1, it fails.  Maybe if you fix the interval as $ [0, 1]$ or something?

Excuse my arrogance, but shouldn't a function satisfy $ f''(x)<0$ on an interval to be convex on in? A constant doesn't satisfy this condition. Or am I wrong?

How could such a question be arrogant? :)  I don't know whether the "usual" definitions of concave and convex involve strict or weak inequalities on the second derivative.  Even if it is strict, we could add a very very very tiny perturbation to a constant function (e.g. $ f(x) \equal{} 2 \minus{} \frac {x^2}{1000}$ over $ [ \minus{} 1, 1]$) to get a function with negative second derivative that violates the condition for small enough $ n$ -- I don't know whether the resultant sequence will be decreasing for all $ n$.  (For the particular example I've given, it decreases until $ n \equal{} 17000$ or so, but then begins to increase, at least according to Mathematica.)

I was going to give you $ f(x)\equal{}x^2$ on $ [0,10].$ The sequence decreases for $ n<14$ and then increases.

If you add in the stipulation that $ [a,b]$ have length $ 1,$ then the sequence is nondecreasing (the proof uses Hölder's inequality) and you don't need the hypothesis of convexity.

Question: does the sequence have to be [i]eventually[/i] increasing? And would convexity matter in that case?

Well, the $ L^p$-norm of $ f$ is a convex function of $ 1/p$ and it tends to the $ L^\infty$ norm as $ p\to\plus{}\infty$, so the only chance for it not to be eventually increasing is to be decreasing all the way. But this would mean that we can have a function bounded by $ 1$, say, such that $ \int_a^b{f^n}\ge 1$ for all $ n$, which is possible  if and only if the measure of the set where $ f\equal{}1$ is at least $ 1$. This, of course, is never the case for strictly convex functions.

sorry fedja...i am not competent enuf to understand what u wrote...
yes Dr.Kent will it be evnetually increasing


function
limit
integration
real analysis
real analysis unsolved

This is a fun little problem. Prove or come up with a counterexample.

If $ f\in L^2(\mathbb{R})$, then $ \lim_{n\to\infty}\sqrt{n}\int_n^{n\plus{}1}|f|dx\equal{}0$.

Let $ f(x) \equal{} \frac1k$ on the interval $ [k^2,k^2 \plus{} 1]$ and zero otherwise. Then $ \|f\|_2^2 \equal{} \sum_{k \equal{} 1}^{\infty}\frac1{k^2} < \infty.$

But if we let $ a_n \equal{} \sqrt {n}\int_n^{n \plus{} 1}|f|,$ then $ \limsup_{n\to\infty}a_n \equal{} 1.$

Further challenge: If we require $ |f|$ to be monotone decreasing on $ [1,\infty),$ does that change the answer?

(One comment: this all seems to really be about $ \ell^2$ rather than $ L^2$ - at least I haven't yet seen any features of the problem that I couldn't answer with a sequence.)


let $a.b.c$ real numbers and $x,y,z>1$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$    .what's the maximal value of $xyz$? :?:

no!
if $z\to\infty$ so $(\frac{1}{x}+\frac{1}{y})\to2$ and we have $x>1,y>1$ ;)

Yes!  :P
to be very explicit, put $x=y=1+\frac1{2N-1},z=N$ :D

can you explain? i think that you are mistaked :oops:

I think he isn't...
You should review your problem's statement... What/where are $a, b, c$?

first , what's the maximal value of $x+y+z$?

I guess I should not interfer in such an intelligent and meaningful conversation, but if someone is interested in proofs or something like that:

Okay, I assume the problem is to find the maximum of xyz, where x, y, z are real numbers which are > 1 and satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. In fact, such a maximum does not exist, because we can always take $x=y=\frac{1}{1-s}$ and $z=\frac{1}{2s}$ for a small enough $s<\frac12$ (yes, these x, y, z are > 1 and do satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$) and get $xyz=\frac{1}{\left(1-s\right)^2\cdot 2s}$ which can become as large as you want. The same counterexample shows that x + y + z has no maximum.

The question about the minimum of xyz makes more sense, but follows trivially from AM-GM:

$\frac{1}{x}\cdot\frac{1}{y}\cdot\frac{1}{z}\leq\left(\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}\right)^3=\left(\frac23\right)^3$
$\Longrightarrow\ \ \ \ \ xyz=\frac{1}{\frac{1}{x}\cdot\frac{1}{y}\cdot\frac{1}{z}}\geq\frac{1}{\left(\frac23\right)^3}=\frac{27}{8}$,

with equality for $x=y=z=\frac32$.

The minimum of x + y + z is an AM-HM application:

$\frac{x+y+z}{3}\geq\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac32\ \ \ \ \ \Longrightarrow\ \ \ \ \ x+y+z\geq 3\cdot\frac32=\frac92$,

with equality for $x=y=z=\frac32$.

  Darij

thanks  :lol:
now i want the maximal value of $x+y+z$            no exist???


trigonometry
function
geometry
3D geometry
complex numbers

Rewrite the complex numbers in the form $ a\plus{}bi$

1. $ 4\sin^2\frac{\pi}{18}$

2. $ \ 4\sin^2\frac{5\pi}{18}$

3. $ \ 4\sin^2\frac{7\pi}{18}$

show steps.

I don't follow.  All of those numbers are real.

Yes, they are (and I guess we only have to find a since b is 0), but in the post "Neat Cubic", Dr Sonnhard said that $ 4\sin^2(\pi/18) \equal{} \left(\frac{1}{2}(\minus{}1\minus{}i\sqrt{3})(\frac{1}{2}(\minus{}1\plus{}i\sqrt{3}))^{\frac{1}{3}}\plus{}\frac{1}{2}(\frac{1}{2}(\minus{}1\minus{}i\sqrt{3}))^{\frac{1}{3}}(\minus{}1\plus{}i\sqrt{3})\right)^2$, but do you know how they were derived, and probably simplifying the expression into a single real number with no $ i$?  How about the other two?  I asked that in the post but no one has replied yet, and so I decide to start a new topic.  :)

hello, to derive $ \sin\left(\frac{\pi}{18}\right)$ you will need this formula
$ \sin(3x)\equal{}3\sin(x)\minus{}4\sin(x)^3$
Sonnhard.

I guess you meant to substitute $ \pi/18$ for $ x$ and then solve for $ \sin(\pi/18)$, but is there a clean way to solve the resulting cubic?

That depends on what you mean by "clean."  Dr. Sonnhard is referring to the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url], which here is a tautology:  think about what the cube root of $ \frac{\minus{}1 \plus{} i \sqrt{3}}{2}$ is in terms of complex exponentials.  The identity Dr Sonnhard stated is essentially a trigonometric identity.

hello, yes, we have $ \sin(3\frac{\pi}{18})\equal{}\frac{1}{2}\equal{}3\sin(\frac{\pi}{18})\minus{}4\sin(\frac{\pi}{18})^3$, substituting $ t\equal{}\sin(\frac{\pi}{18})$ then we have
to solve $ \frac{1}{2}\equal{}3t\minus{}t^3$.
Sonnhard.

Nice link.  [url=http://mathforum.org/dr.math/faq/faq.cubic.equations.html]Here[/url] is another one.
One quick question: There should be three complex roots to the cubic, but how do we determine the desired real root?

There are two cube roots in the expression.  As long as you take the "same" cube root in both of them, the end result will be real three times (in this case) and will give you all three real roots.


function
trigonometry
calculus
derivative
integration
analytic geometry
real analysis

I got a question of a friend about defining the sin en cos as the solution of 2nd order diff. equation
y''(t) = z(y), the function y is continuous with z:R-->R
So i worked it out a bit, but I don't know how to prove the identities
sin'(x)=cos x, sin(x)^2+cos(x)^2=1 and cos(a+b)=cos(a)(cosb)+sin(a)sin(b)

The hint was to use Picard iteration and banah Theorem

Can someone help plz

What are you asking? In particular, why do you have both $ y$ and $ z$ in what you wrote? As written, it doesn't make sense.

I defined the cos and sin as solutions of the differential eq.
y''(t)=z(y(t), with f continuous and f:R->R
I mean an equation like y''(t)=y(t), but then in a general form. It is easy to see that sin(t) is a solution

Now I have to prove the identities 1. sin'(t)=cos(t) 2. sin(t)^2+cos(t)^2=1 and 3. cos(a+b)=cos(a)cos(b)+sin(a)+sin(b)
But the difficult thing is, because I defined the sin as a solution of the diff eq. I have to use the definition to show this simple identities. I can use the banah theorem to show that the solution is unique

I don't know how to use microsoft object or something on this site, or else I can can give more explanation if needed

You have to use the solution to that differential equation as your definition of the sine function? Most texts I have seen prove Taylor's Theorem first, define sine and cosine by their Taylor series, and prove that the derivative of sine is cosine once some properties of uniform convergence of sequences of functions are established. As for proving that $ sin^{2}\theta\plus{}cos^{2}\theta \equal{} 1$, this is fairly easy once you can show that $ cos\theta \equal{}\frac{e^{i\theta}\plus{}e^{\minus{}i\theta}}{2}$ and $ sin\theta \equal{}\frac{e^{i\theta}\minus{}e^{\minus{}i\theta}}{2i}$.

Yeah but the problem is to define it in this way

So i have the diff equation y''(t)=z(y(t), I take the integral on both sides and get y(t)=y0+y1(t-t0)+double integral z(y(t)). so it is easy so far.

I define sin as the solution of this equation with solution y(t,0,0,1)
so this means the solution y(t, t0, y0,z0) with this points as boundary values. so it means that the sin starts in the origin, in (0,0) . and the cos gives y(t,0,1,0), so it has coordinates (0,1) with t zero and y 1.
Now I have to prove
1. sin'(t)=cos(t)
2. cos(t)^2+sin(t)^2=1
3. cos(s+t)=cos(s)cos(t)+sin(s)sin(t)

I know that the solution is unique, so cos(s+t) and cos(s)cos(t)+sin(s)sin(t) are the same solutions for the equation y''(t)=z(y(t), . But how do I prove that they have the same solutions, so that they are equal to each other.

Plz help,

Cause I am a bit stuck here  :(


Show that if $ a|b$ and $ a>0$, $ b>0$, then $ a<\equal{}b$

$ a|b\Rightarrow b \equal{} ka$

$ k \ge 1$
$ ka\ge a$
$ b \equal{} ka \ge a$

[quote="SimonM"]$ k \ge 1$[/quote]

Next, try to prove that $ k > 0 \implies k \ge 1$.  Use the fact that any subset of the natural numbers has a least element.

If $ 0<x<1$, then $ 0<x^2<x<1$ ?!

If $ x$ happens to be our least element (which exists due to WOP (?) ) then we have a contradiction. Hence [as our set of natural numbers is non-empty], $ x\ge 1$...but $ 1$ is in the set, so we are done.

[quote="Santamaria"]Well,

we can assume that sinA/cosA = tanA

Then, we know that e^pi = i^i

0 < 1 < x for all z values in the z/y polar cylinder, when finding the antiderivative of a polar-quadratic function with height sinBpi.

If you work out the problem using pi, and euler's theory, we can assume that the answer will be a factor of 9, iff and only iff y > z.[/quote]

Now that is an example of how to kill a fly with a super laser bazooka! :P


Find $(x,y,z,r)$ satisfying

${\{ \begin{array}{lll} {\displaystyle{ {r+y}\choose r}} = \log_yx \\ \\{{ \log_yz} = 4 + \log_xz} \\ \\ {\displaystyle {{r+y}\choose y}} = \log_xz + \log_zz \end{array}.$

[hide]$log_yz=4+log_xz \implies log_yz=4+\frac{log_yz}{log_yx}$ but since: $log_yx=C_{r+y}^{r}$ we get:$log_yz=\frac{4C_{r+y}^{r}}{C_{r+y}^{r}-1}$ and by the last system's term:

$C_{r+y}^{y}=C_{r+y}^{r}=log_xz+log_zz \implies C_{r+y}^{r}=\frac{log_yz}{log_yx}+1=\frac{4}{C_{r+y}^{r}-1}+1 \implies {(C_{r+y}^{r}-1)}^2=4 \implies C_{r+y}^{r}=3$
we can check the Pascal's triangle and get:$C_{3}^{1}=C_{3}^{2}=3$ in consequence:

$y+r=3 \implies \{y=1$ and $r=2\}$ or $\{y=2$ and $r=1\}$ note that the first set is not possible because $y$ is a log base (ie. y>1), now we  just need to substitute the values in the system:

$C_{r+y}^{r}=log_28=log_2x \implies x=8$

$\frac{4C_{r+y}^{r}}{C_{r+y}^{r}-1}=log_264=log_2z \implies z=64$

Answer:r=1, y=2,  x=8,   z=64[/hide]


how do you factor this completely??

x^8-y^8

pls and thx!!

[hide] \begin{eqnarray*}x^{8}-y^{8}&=&\left(x^{4}-y^{4}\right)\left(x^{4}+y^{4}\right) \\ &=&\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right) \\ &=&\left(x-y\right)\left(x+y\right)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)\end{eqnarray*} Also, if you wanted to factor this further (with coefficients in $\mathbb{R}$ instead of $\mathbb{Z}$), use the fact that \[x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2x^{2}y^{2}=\left(x^{2}+y^{2}-\sqrt{2}xy\right)\left(x^{2}+y^{2}+\sqrt{2}xy\right).\] [/hide]


probability
Putnam
advanced fields
advanced fields unsolved

Suppose that n+1 points are chosen uniformly and independently from inside a square. Show that the probability that the points are in convex position is [C_n/(n!)]^2

(C_n is the nth Catalan number)

This sounds a bit like Putnam 2006 A6, but it's probably nothing like it. Help?

I'm pretty sure that the square is easier to work with than the circle, but it's still very hard. I have no idea how to approach many points.

But the case seems a lot more general, so it may turn out to be harder overall?

So I emailed Daniel Kane about this problem and he gave me a pretty nice solution, so I thought it'd be useful to share it:

[quote]

In order to produce a set of points in convex position, look at the
vectors along the edges of the convex hull.  These are some n vectors
that sum to 0 and are sorted by slope.  You can produce set a set of
points by picking n vectors that sum to 0, shorting by slope and using
them as differences of the points.  One question is whether they fit
in the square and if so, how much "area" do such configurations cover.
 It is not hard to show that they fit iff the sum of the positive x
coordinates (say X) and the sum of the positive y coordinates (say Y)
are both at most 1, and in this case, the area of the configuration is
(1-X)(1-Y).  But now we are looking at n random vectors summing to 0
and looking at some function of the x-coordinates times some function
of the y-coordinates.  Hence we can compute this expectation by
handling each coordinate separately.  In other words we look at n
numbers that add to 1 and see how often the sum of the positive ones
are at most 1 and look at the average value of 1-X in this case.  This
integral is not too hard.  It involves splitting into cases based on
how many of the numbers are positive and gives a combinatorial sum
yielding the desired answer.

[/quote]

refer to this journal:
[b]P. Valtr, Discrete Comput. Geom. 13 (1995), 637-643.[/b]

and
[b]Combinatorica 16 (1996), 567-573[/b]
for the case of triangle...

Or please read this:
[url]http://edocs.fu-berlin.de/docs/servlets/MCRFileNodeServlet/FUDOCS_derivate_000000000261/1994_01.pdf;jsessionid=6F9D8493BB7F228747429A4CB486F0FD?hosts=[/url]


inequalities
algebra
polynomial
Cauchy Inequality
inequalities unsolved

$ a,b,c,d$ are positive. prove that :

$ \frac{a\plus{}b\plus{}c\plus{}d}{4}\geq\sqrt[3]{\frac{abc\plus{}abd\plus{}acd\plus{}bcd}{4}}$

if u have Problem-solving strategies then see its 3rd problem in inequalities,it's the same

You are right letmast but it can be addressed by simple Cauchy inequality

Here are steps

$ \frac {abc \plus{} bcd \plus{} acd \plus{} bcd}{4} \equal{} \frac {ab(c \plus{} d) \plus{} cd(a \plus{} b)}{4} \le \frac {(a \plus{} b)(c \plus{} d)(a \plus{} b \plus{} c \plus{} d)}{16} \le \left(\frac {a \plus{} b \plus{} c \plus{} d}{4}\right)^3$

It is just AM-GM
[hide="solution"]
It is equivalent with $ (a\plus{}b\plus{}c\plus{}d)^3 \geq 16(abc \plus{}abd \plus{}acd \plus{}bcd)$

$ 16(abc \plus{}abd \plus{}acd \plus{}bcd ) \equal{}16ac(b\plus{}d) \plus{}16bd(a\plus{}c) \leq 4(b\plus{}d)(a\plus{}c)^2  \plus{}4(a\plus{}c)(b\plus{}d)^2 \equal{}4(a\plus{}c)(b\plus{}d)(a\plus{}b\plus{}c\plus{}d) \leq (a\plus{}b\plus{}c\plus{}d)^3$ [/hide]

[quote="mahanmath"]$ a,b,c,d$ are positive. prove that :

$ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [3]{\frac {abc \plus{} abd \plus{} acd \plus{} bcd}{4}}$[/quote]

Define polynomial $ P(x) \equal{} (x \plus{} a)(x \plus{} b)(x \plus{} c)(x \plus{} d)$ , so $ P(x)$ deegre 4

Define $ s_i$ be the coefficient of $ x^i$ , so $ s_1 \equal{} abc \plus{} bcd \plus{} cda \plus{} dab$ , and $ s_3 \equal{} a \plus{} b \plus{} c \plus{} d$, so by maclaurin inequality we have

$ \frac {abc \plus{} bcd \plus{} cda \plus{} dab}{4}\ge\sqrt [3]{\frac {a \plus{} b \plus{} c \plus{} d}{4}}$

I think there are some mistakes here, can somebody tell me please ?


geometry
geometric transformation
rotation

By catalytic hydrogenation of (+)-3-buten-2-ol $ \left([\alpha]^{27}_{D}=+33.2^{o}\right)$, (+)-butan-2-ol $ \left([\alpha]^{27}_{D}=+13.5^{o}\right)$ is obtained. Classify each of the following statements as true or false:

(a) As both compounds have the same sign of rotation (they rotate the plane of polarized light in the same direction), then they have necessarily the same absolute configuration

(b) Both compounds have the same sign of rotation because they have the same absolute configuration.

(c) Both compounds have configuration R.

(d) Both compounds have configuration S.

1.false
2.false
3,4 since nothing about one configuration is given can by just lookinga t alpha can one tell?
but one thing is sure since the bond at chiral centre is not broken both have same absolute configuration

That's all correct, because...

[quote="Pardesi"]since nothing about one configuration is given can by just lookinga t alpha can one tell?[/quote]

...there is no relation between sign of rotation and absolute configuration.


Prove the following. I know for certain (a) is true, and hopefully (b) is true also.  The inequality is for positive reals $a,b$.
(a)
$a^7 b^3 + a^5 b^8 + b + a^5 b^2 + a^2 b^3 + a^2 + 8 a^3 b^3 \geq 2 a^5 b^5 + 2 a^3 b^4 + 2 a^4 b^3 + 4 a^6 b^4 + 4 a^4 b^6$
(b)
$c + c^4 b + b^4 + b^3 + b^2 c^3 + c^2 + 6 ( b^2 c + b c^2 + b^2 c^2 ) \geq 8 (bc + b^3 c + b c^3 )$

To Davron: I've changed the question statement. Hope you can solve it.

Sorry what are the numbers : {real, postitive real, integer... ?}.

davron


[quote="jjefwfef"]Hi I was wondering if anyone was familiar with the nav4all navigation system for cell phones. I have been told it is pretty good, any comments on that?[/quote]

[url=http://ar.search.yahoo.com/search?p=nav4all+navigation+system&sm=Buscar&fr=FP-tab-web-t&toggle=1&ei=ISO-8859-1&meta=all%3D1]This has been modified by the moderator because it is extremely annoying to paste long urls directly into your message as they cause the screen to stretch horizontally.  Please don't do it.[/url]


function
algebra
polynomial
Euler
limit
number theory
totient function

Let $ \Phi_{n}(x)$ be the nth cyclotomic polynomial in x, $ \phi(n)$ be the Euler totient function, and $ \mu(n)$ be the Möbius function. Call $ P(n)=\prod_{prime}\frac{p^{\phi(n)}}{\Phi_{n}(p)}$. Prove that:

$ \mu(n)=1$ iff $ P(n)$ diverges.
$ \mu(n)=0$ iff $ P(n)$ converges to a positive finite real.
$ \mu(n)=-1$ iff $ P(n)$ converges to 0.

Say $ \mathbb{P}= \{p \in \mathbb{N}: p \mbox{ is prime}\}$, $ \mathbb{P}(x) = \{p \in \mathbb{P}: p \le x\}$ and $ \mathbb{N}(x) = \{q \in \mathbb{N}: \mbox{gpf}(q) \le x\}$, for any $ x \in \mathbb{R}$. Here $ \mathbb{N}= \{1, 2, \ldots\}$ and $ \mbox{gpf}(q) = \max\{p \in \mathbb{P}: p \mid q\}$, for any $ q \in \mathbb{N}$.

Trivially, $ P(1) = \prod_{p \in \mathbb{P}}\frac{p}{1-p}= \infty$, and indeed $ \mu(1) = 1$. So in the following let $ n \ge 2$ and set $ r = \sigma_{0}(n)$ and $ d_{1}, d_{2}, \ldots, d_{r}\in \mathbb{N}$ to be the (pairwise distinct) divisors of $ n$, with $ 1 = d_{1}< \ldots < d_{r}= n$. Then $ \Phi_{n}(x) = (1-x)^{\mu(n)}\cdot \prod_{k=2}^{r}(1-x^{d_{k}})^{\mu(n/d_{k})}$ and $ n = \sum_{k=1}^{r}\phi(d_{k})$, yielding $ \phi(n) = \sum_{k=1}^{r}d_{k}\cdot \mu(n/d_{k})$ through the Möbius inversion formula. Now fix $ x \in \mathbb{R}$ and let $ P_{n}(x) = \prod_{p \le x}\frac{p^{\phi(n)}}{\Phi_{n}(p)}$, where the product is over any $ p \in \mathbb{P}$ such that $ p \le x$. Then obviously $ P_{n}(x) = (-1)^{M(x)}\,Q_{n}(x) \cdot \prod_{p \le x}\!\left(\frac{p}{p-1}\right)^{\!\mu(n)}$, where $ M(x) = \pi(x) \cdot \sum_{k=1}^{r}\mu(n/d_{k}) = 0$ (prove it!) and $ Q_{n}(x) = \prod_{p \le x}\prod_{k=2}^{r}\!\left(\frac{p^{d_{k}}}{p^{d_{k}}-1}\right)^{\!\mu(n/d_{k})}= \prod_{k=2}^{r}\!\left(\prod_{p \le x}\sum_{i=0}^\infty \frac{1}{p^{id_{k}}}\right)^{\!\mu(n/d_{k})}= \prod_{k=2}^{r}\!\left(\sum_{m \in \mathbb{N}(x)}\frac{1}{m^{d_{k}}}\right)^{\!\mu(n/d_{k})}$. Since $ d_{k}> 1$, for any $ k = 2, \ldots, r$, it follows the limit $ \lim_{x \to+\infty}Q_{n}(x)$ exists, and particularly $ Q(n) = \prod_{k=2}^{r}\zeta(d_{k})^{\mu(n/d_{k})}$.

Hence the behaviour of $ P_{n}(x)$ as $ x \to+\infty$ is [i]dominated[/i] by the factor $ \prod_{p \le x}\left(\frac{p}{p-1}\right)^{\mu(n)}$. Actually, here are the cases:
[list]1) $ n$ is squareful. Then $ \mu(n) = 0$ and $ P(n) = \lim_{x \to+\infty}P_{n}(x) = Q(n)$;
2) $ n$ is squarefree and $ \omega(n)$ is odd. Then $ \mu(n) =-1$ and $ P(n) = \lim_{x \to+\infty}P_{n}(x) = Q(n) \cdot \prod_{p \in \mathbb{P}}\frac{p-1}{p}= 0$;
3) $ n$ is squarefree and $ \omega(n)$ is even. Then $ \mu(n) = 1$ and $ P(n) = \lim_{x \to+\infty}P_{n}(x) = Q(n) \cdot \prod_{p \in \mathbb{P}}\frac{p}{p-1}= 0$,
[/list]
where $ \omega(n)$ denotes the number of distinct primes $ p \in \mathbb{P}$ appearing in the Euclidean decomposition of $ n$, for any $ n \in \mathbb{N}$. Then the proof is complete.


[color=blue] [size=150]$16{th}$  Iberoamerican Olympiad[/size]
Minas, URUGUAY. [2001]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 2.[/b] El incirculo $ (I)$ de $ \triangle ABC$ es tangente a $ BC,CA,AB$ en $ X,Y,Z.$ Las rectas $ BI,CI$ cortan a la recta $ YZ$ en $ P,Q.$ Mostrar que $ \triangle ABC$ es isósceles si y solo si $ XP \equal{} XQ.$[/quote]
Es bien sabido que la paralela media a $ AB,$ la recta $ YZ$ y la bisectriz $ BI$ de $ \angle ABC$ concurren en $ P.$ Así mismo la paralela media a $ AC,$ la recta $ YZ$ y la bisectriz $ CI$ de $ \angle ACB$ concurren en $ Q.$ Una sencilla demostración puede apreciarse en el tópico [i]Another unlikely concurrency[/i] en el sitio [url]http://pagesperso-orange.fr/jl.ayme/[/url].

Consecuentemente los ángulos $ \angle BPC$ y $ \angle CQB$ son rectos, es decir que los cuadriláteros $ PQBC,PIXC$ y $ QIXB$ son cíclicos. Así se tienen las siguentes relaciones angulares:

$ \angle QPX \equal{} \angle QPI \plus{} \angle IPX \equal{} \angle ICX \plus{} \angle ICX \equal{} \angle ACB$

$ \angle PQX \equal{} \angle PQI \plus{} \angle IQX \equal{} \angle IBX \plus{} \angle IBX \equal{} \angle ABC$

Por lo tanto $ \triangle ABC \sim \triangle XQP$ y evidentemente $ \triangle ABC$ es isósceles si y solo si  $ \triangle XQP$ es isósceles.

[quote][b]Problema 6.[/b] Probar que es imposible cubrir un cuadrado de lado 1 con 5 cuadrados iguales de lado < 1/2.[/quote]
Considérese los siguientes 9 puntos en el plano cartesiano $(0,0),$ $ (\frac{_1}{^2}, 0),$ $ (1,0),$ $ (0,\frac{_1}{^2}),$ $ (\frac{_1}{^2},\frac{_1}{^2}),$ $ (1,\frac{_1}{^2}),$ $ (0,1),$ $ (\frac{_1}{^2},1),$ $ (1,1).$ Cualquier cuadrado de lado menor que $\frac{_1}{^2}$ solo podrá cubrir uno de de estos puntos. Asi, ello es cierto para 8 cuadrados.


What is the value of $ ((\frac{1}{2} \div \frac{1}{2})\div \frac{1}{2}) \div \frac{1}{2}$?

hello, we have $ \left(\left(\frac{1}{2}: \frac{1}{2}\right): \frac{1}{2}\right): \frac{1}{2}\equal{}\left(1: \frac{1}{2}\right): \frac{1}{2}\equal{}2: \frac{1}{2}\equal{}4$.
Sonnhard.


calculus
derivative
logarithms
calculus computations

One that I got by guessing by wasn't able to figure out exactly:

Given that ln(x*e^y) = e^(x*ln(y)), find dy/dx. Assume x and y are both positive.

${\log \left(x e^{y}\right)=e^{x \log (y)}}$
${\frac{d\log \left(x e^{y}\right)}{dx}=\frac{de^{x \log (y)}}{dx}}$
${\frac{e^{-y}\left(e^{y}x \frac{dy}{dx}+e^{y}\right)}{x}=y^{x}\left(\frac{x \frac{dy}{dx}}{y}+\log (y)\right)}$
${\frac{dy}{dx}+\frac{1}{x}=y^{x}\left(\frac{x \frac{dy}{dx}}{y}+\log (y)\right)}$
${\frac{dy}{dx}=-\frac{y \left(x y^{x}\log (y)-1\right)}{x \left(x y^{x}-y\right)}}$

Log = Ln


So, early decision comes out very soon (my Duke ED comes out monday).  Anyone excited/anxious?

I don't know any numbers for this year but last year, I heard that Duke received a really small ED pool so they got nervous and accepted more than usual.  Then, during RD, they got a record number so their RD acceptance rate was really low.  Not that I care about Duke anyway (go Blue Jays!!!).

ya, i got really happy when i checked out their class of 2011 profile, they accepted 40% ED and 20% RD.  Then my happiness got ruined when i checked out their 2010 profile, which was 20% ED and 20% RD.

O well, lets just hope Duke freaks out this year again!


maximize Z=15x+10y

subject to

4x+2y=100

Can you determine a solution to this model that will achieve objective?  :maybe:

Without any additional constraints on the variables, $ Z$ can be made arbitrarily large. Just put $ y\equal{}50\minus{}2x$ to make $ Z\equal{}500\minus{}5x$ which tends to infinity as $ x$ tends to minus infinity. If the variables are supposed to be non-negative, then $ x\equal{}0, y\equal{}50$, $ Z\equal{}500$ is obviously the maximum.

In either case this doesn't seem to belong in this forum.

[b]randomgraph[/b]

Thank you very much for the reply   :)

I thought the solution couldn't be found without any other information

yes, variables are non-negative...  and (0 50) is a maximum  yes?and  there is no another point that would maximize the equation. Z canit be more than 500

 sorry if I've opened the topic in the wrong place
I am a new girl here and have loved getting to know this forum  :roll:


Dear mathlinker. i would like to know which is the [b]Cristea's transversal theorem[/b] ?.I've been searching on several websites but i cannot find it.I know that this theorem is similar to Van Aubel's theorem.

Where I can find material about of Cristea´s theorem?
Thanks.

Dear Mathlinkers,
searching among the archive,
see
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2065798600&t=15764
Sincerely
Jean-Louis


inequalities
trigonometry
inequalities proposed

Prove the inequality $ \begin{equation*} \tan \left( \frac{\pi \sin x}{4\sin \alpha }\right) +\tan \left(\frac{\pi \cos x}{4\cos \alpha }\right) >1\end{equation*}$ for all values of $x$ and $\alpha $ satisfying the conditions $\begin{equation*}
0\leq x\leq \frac{\pi }{2},\;\;\;\;\;\;\;\;\;\;\frac{\pi }{6}<\alpha <\frac{%
\pi }{3}.\end{equation*} $

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


John has 12 marbles of different colors. In how many ways can he choose 4 marbles, if he chooses exactly one of the red, green, and blue marbles?

After taking those three, there are 9 left, and one to choose so its 9c1=9.

... no there are 252 ways, 9C3 for 3 marbles aside from the red green and blue ones, and then *3 for the color choice

There is 9, because order doesn't matter. He has a red, green, blue, and (9 other choices).

... read the question ... and i got this right  ... so ...

Please make your solution clearer then, because your method doesn't make sense to me. Thanks!

[hide="Correct Solution"]We will divide the $12$ distinct marbles into $2$ groups:  the first group will contain $1$ red, $1$ green, and $1$ blue marble.  The second group will contain the other $9$ marbles.  Among the $4$ selected marbles, we need to have exactly $1$ of them to be from the first group.  There are $\dbinom{3}{1}=3$ ways.  We need to select exactly $3$ marbles from the second group.  There are $\dbinom{9}{3}=84$ ways here.  Since the $2$ groups do not interfere with each other/affect the order that the marbles are chosen, the answer is $3\times84=\boxed{252}$.[/hide]

I am so dumb. I added instead of multiplyed.

[quote=pinkpig]:(
Lost my 1625 problem streak along with 0.022 C&P rating[/quote]

rip
wait how is this lvl 22
[hide = sol]There are 9 C 3 ways to choose the 3 other marbles, and 3 ways to choose one of the marbles that need to be either red, green, or blue. 3 * 84 = 252[/hide]

but how do you know that there aren't more red marbles? Nowhere does it say there is only one red marble. "INCLUDING"

What if ten of them were red, and the others were green and blue? This is possible, given that it just says there is at least one red, one blue, and one green


$ ABC$ is right-angled at $ A$ the incircle $ (I)$ touches the hypotenuse $ BC$ at $ X$ ,the perpendicular to $ BC$ at $ X$ cut the circumcircle of $ ABC$ in $ P$, $ P$ can be in the same o different side of $ A$. Prove that the square that has side $ PX$ has the same area that $ ABC$

we have $ BX\equal{}(BC\plus{}AB\minus{}AC)/2$  and $ CX\equal{}(BC\plus{}AC\minus{}AB)/2$
we know also that $ CX.BX\equal{}PX^{2}$
so it's done


summer program
MathPath
modular arithmetic
number theory
puzzles

[quote="mfs50"]For what number can you move the 1's digit to the front and create the same effect as doubling the number?

For example: 234 becomes 423, but we need one that gets doubled.

All I can think of is zero, but there must be a better answer.[/quote]

[hide="hint"]This was one of the MathPath qualifying problems. What can the ones digit be? Now that you know that, what do you know about the tens digit? And so on.[/hide]

The ones digit should be twice the first digit.  The tens digit should be twice the ones digit.  Or could the ones digit be one more than twice the first digit, as in 15......3 changing to 315......?
I'm still stuck.

[hide="Hint"]It cannot be 0,1 only. Try any others.[/hide]

The ones digit.
The answer is quite long.

1.....842

On the right track?  I don't know what comes next.

Yeah, I did it from 2->4->8->6->...
Write it like this
[code]......6842
         2
       684[/code]

No, you need to do until you get another 2.

I understand that.  But the digits are not the same anymore.  I can 't start with .....136842 and end up with .....273684.  It won't be the right digits.

I appreciate you helping me, but I'm super busy and I've already spent hours on this.  Do you think you could give me the answer if you have it?

Cool.  Did you do it like this?
2
42
842
1684
336842
6736842
134736842
2694736842
53894736842

How can you do this with a standard calculator or even a TI-82?  Is this just dog work to be done by hand?  Is there a faster way to get it?

I did it by hands.
[hide="Another way of doing it"]This may be harder, but it gives some information.
Do it by writing that number as $10x+y$ where $x$ is a natural number and $y$ is $0,1,...,9$.
So $2(10x+y)=10^{k}y+x$ where $k$ is the number of digits of  $x$. Therefore, $19x=(10^{k}-2)y$.
But $19$ does not divide $y$, so it divides $10^{k}-2$.
By using calculator OR trying to divide $99999....9998$ by $19$ by hands OR using the information from the answer we got above, $k$ can be $17$.
So $x=5263157894736842y$ which works for $y=2,3,...,9$ because we need $x$ with $17$ digits.[/hide]

[quote="OHO"]
But $19$ does not divide $y$, so it divides $10^{k}-2$.
By using calculator OR trying to divide $99999....9998$ by $19$ by hands OR using the information from the answer we got above, $k$ can be $17$.
[/quote]
What we want to know  the value of $k$ such that $10^{k}\equiv 2 \bmod 19$
which is same as $10^{k+1}\equiv 20 \equiv 1 \bmod 19$
Since 19 is prime, Fermat's little theorem tells us $10^{18}\equiv 1 \bmod 19$
(k=17 is one solution, there are other numbers with k=35  or 53 etc..)

Yeah, and there are no solution else other than $17, 35, 53,...$

[quote="Gyan"]Yeah,and  each of them give multiple solutions, like:

$105263157894736842$
$157894736842105263$
$210526315789473684$

Also please notice $1/19 = .052631578947368421 \ 052631578947368421\dots$[/quote]Nice  :)

[size=150][color=blue]Summary[/color][/size]

$\text{number}=10x+y$, where $0\leq y\leq 9$

$\text{obtained number}=10^{k}y+x$, where $k$ is the number of digits of  $x$.

$19x=(10^{k}-2)y \Rightarrow k\equiv-1 \pmod{18}$

Now, take the case $k=17$.

$10^{18}\equiv 1 \pmod{19}\Rightarrow$

$10^{18}-1 = 19m\Rightarrow \frac{1}{19}= \frac{m}{10^{18}-1}= \frac{m}{999999999999999999}$

It's well known that the last number is a periodic decimal number at the form $0.mmm...$, where $m$ fills 18 decimal places in each time it appears.

According to the previous posts, $m=052631578947368421$

Now we have $10^{17}-2 = \frac{10^{18}-20}{10}= \frac{10^{18}-1-19}{10}= \frac{19m-19}{10}= 19\cdot \frac{m-1}{10}$

$19x=(10^{k}-2)y \Rightarrow 19x = 19\cdot \frac{m-1}{10}y \Rightarrow x = \frac{m-1}{10}y \Rightarrow$

$10x = (m-1)y\Rightarrow$

$10x+y = my$

So, all the solutions for $k=17$ are $my$, where $m=\frac{10^{18}-1}{19}$ and $y$ is some digit of $m$. But we can see that $y$ takes all the values in $\{0,1,...,9\}$

Notice that $10x+y = \frac{10^{18}-1}{19}y = \left(10^{18}-1\right)\frac{y}{19}$

[size=150][color=blue]Conclusion[/color][/size]
[color=brown]So there is a quick way to find the solutions:
$\blacktriangleright$ Take all the numbers $\frac{y}{19}$, where $y\in\{0,1,...,9\}$ (we can include $0$ of course)
$\blacktriangleright$  Find the repeating part (the 18 first decimal digits)[/color]


integration
function
calculus
calculus computations

The problem is $ \Gamma{(1)} = \int{t^{z-1}e^{-t}dt}$ on the interval 0 to $ \infty$.

Also:
A.  Calculate $ \Gamma{(1)}$.
B.  Determin $ \Gamma{(z+1)}$ in terms of $ \Gamma{(z)}$.
C.  Based on your answer, explain how $ \Gamma{(z)}$ relates to another function you already know about.

I did:
A.  $ \Gamma{(1)} = \int{t^{z-1}e^{-t}dt}$ = 1

B.  (z+1) = n
     $ \Gamma{(z)}$ = (n-1)!
     n(n-1) = n!
     $ \Gamma{(z+1)}$ = (z+1)($ \Gamma{(z)}$)

C.  f(z+1) = f(z)(z+1)

Parts A and C I believe are right, but I think somewhere in B I made a mistake? Thanks for any help.

You did the problem backwards.  You're not supposed to already know that the Gamma function relates to factorials.; that comes out of your work in B.

[quote="JRav"]$ \Gamma(z \plus{} 1) \equal{} \int_{0}^{\infty}e^{ \minus{} t}t^{z}\ dt \equal{} \minus{} e^{ \minus{} t}t^{t}\bigg|_{0}^{\infty} \plus{} \int_{0}^{\infty}ze^{ \minus{} t}t^{z \minus{} 1}\ dt$
$ \equal{} z\int_{0}^{\infty}e^{ \minus{} t}t^{z \minus{} 1}\ dt \equal{} z\Gamma(z)$[/quote]

So does the $ \minus{} e^{ \minus{} t}t^{t}\bigg|_{0}^{\infty}$ drop out since it is undefined ($ 0^0$ is undefined?) and that is why you are left with $ z\int_{0}^{\infty}e^{ \minus{} t}t^{z \minus{} 1}\ dt$?

[quote="jmerry"]In fact, your answers to B and C would make more sense switched, and you should really proofread what you wrote for A. You should devote some space to the actual integral calculations here, not just repeating stuff you know from somewhere.[/quote]

For part A

$ \Gamma(1) \equal{} \int_{0}^{\infty}e^{ \minus{} t}t^{z}\ dt \equal{} \int_{0}^{\infty}e^{ \minus{} t}t^{1}\ dt$
or $ \minus{} (t \plus{} 1)e^{ \minus{} t}\bigg|_{0}^{\infty}$
so $ \Gamma(1)$ = 1

I'm not just repeating, I just don't completely understand this problem.

$ \Gamma(1)$ is actually $ \int_0^{\infty}t^0e^{\minus{}t}\,dt\equal{}\int_0^{\infty}e^{\minus{}t}\,dt.$

The calculation you did, of $ \int_0^{\infty}te^{\minus{}t}\,dt,$ is $ \Gamma(2).$

Of course, $ \Gamma(1)\equal{}\Gamma(2)\equal{}1.$

[quote="Kent Merryfield"]$ \Gamma(1)$ is actually $ \int_0^{\infty}t^0e^{ \minus{} t}\,dt \equal{} \int_0^{\infty}e^{ \minus{} t}\,dt.$

The calculation you did, of $ \int_0^{\infty}te^{ \minus{} t}\,dt,$ is $ \Gamma(2).$

Of course, $ \Gamma(1) \equal{} \Gamma(2) \equal{} 1.$[/quote]

 :blush: oops...yeah, that's what it should be. Guess I had a problem on part A as well. Thanks for pointing that out.

[quote="butterz"]So does the $ \minus{} e^{ \minus{} t}t^{t}\bigg|_{0}^{\infty}$ drop out since it is undefined ( is undefined?) and that is why you are left with $ z\int_{0}^{\infty}e^{ \minus{} t}t^{z \minus{} 1}\ dt$?
[/quote]
It's a typo. That should be $ t^z$, and the value is zero for $ z>0$.

OK,  that makes sense. And thanks for explaining my mistakes.

Now, can you prove that for $ z\in\mathbb{N}$,
$ \Gamma(z)\equal{}(z\minus{}1)!$?

How about $ \Gamma\left(\frac{1}{2}\right)$?
[hide="You May Need This"]
$ \int_{0}^{\infty}e^{\minus{}t^{2}}\ dt\equal{}\frac{\sqrt{\pi}}{2}$[/hide]

$ \Gamma(1)\neq\frac{1}{2}$.  It's already been pointed out that $ \Gamma(1)\equal{}1$.  And your method for $ \Gamma\left(\frac{1}{2}\right)$ does not seem correct.  You didn't even use my hint.

The starting point is wrong.

$ \Gamma(z) \equal{} \int_0^\infty e^{\minus{}t} t^{z\minus{}1}dt$.
Note that -1 in the exponent of t.
We get that:

$ \Gamma(\frac{1}{2}) \equal{} \int_0^\infty e^{\minus{}t} t^{\minus{}\frac{1}{2}} dt \neq \int_0^\infty e^{\minus{}t} t^{\frac{1}{2}} dt.$

[hide="Here is a Hint"]
You do indeed need to make a substitution.  The substitution will be of the form $ u\equal{}t^{x}$, so $ du\equal{}xt^{x\minus{}1}dt$.  What is that $ x$?  I'll let you try to figure that out.[/hide]


calculus
integration
trigonometry
function
limit
derivative
real analysis

Guys, can you help me with this improper integral?

$ \int^{\pi}_{0} \frac{dt}{\sqrt{t} \plus{} \sin t}$ thanks very much.

i tried integration by parts, substitution...but i seem to not find a way through this.

I tried Mathematica for $ \int \frac {dt}{\sqrt {t} \plus{} \sin t}$ and indeed it seems to have no simple antiderivative. I wonder how can you solve the integral when limits of integration are set up? :maybe:

By the way, I think this post should be moved to the Computations and Tutorials folder :)

well, since the function is undefined at $ x\equal{}0$, we can change it into: $ \lim_{x\minus{}>0^{\plus{}}} \int^{\pi}_{x} \frac{dt}{\sqrt{t}\plus{}\sin{t}}$ but i dont know what's after this...

hello, i have got $ \int_{0}^\pi \frac{1}{\sqrt{t}\plus{}\sin(t)}dt \approx 2.40899073296947082180$
Sonnhard.

[quote="fedja"]Well, what exactly do you want of it? :? It certainly converges by the comparison test (compare with $ 1/\sqrt t$) and almost certainly its value cannot be expressed in terms of $ e$, $ \pi$, $ \sqrt 2$, and such.[/quote]

I know that it converges...but to what value? Is it possible to do so? thanks!

[quote="Dr Sonnhard Graubner"]hello, i have got $ \int_{0}^\pi \frac {1}{\sqrt {t} \plus{} \sin(t)}dt \approx 2.40899073296947082180$
Sonnhard.[/quote]

Did you use an approximation? Can you please show me how you dealt with $ x\equal{}0$ and the limits involved? thanks!!!:D

[quote]Did you use an approximation? Can you please show me how you dealt with  and the limits involved?[/quote]

Since the function has a singularity at zero, some of the usual techniques of approximate integration (the ones that require a value at an endpoint) are problematic to define, and the worst case error estimates for all methods, which depend on norm estimates of certain derivatives, tell you nothing at all. There are several ways to deal with this. One possibility is to relieve the singularity through a change of variables.

Let $ x\equal{}\sqrt{t}$ so that $ t\equal{}x^2$ in the integral:

$ \int_0^{\pi}\frac{dt}{\sqrt{t}\plus{}\sin t}\equal{}\int_0^{\sqrt{\pi}}\frac{2x\,dx}{x\plus{}\sin(x^2)} \equal{}2\int_0^{\sqrt{\pi}}\frac{1}{1\plus{}\frac{\sin(x^2)}{x}}\,dx.$

Note now that $ f(x)\equal{}\frac{1}{1\plus{}\frac{\sin(x^2)}{x}}$ has a removable singularity at zero so we may safely set $ f(0)\equal{}1$ and use that value in any system that uses it (composite Simpson's rule, for instance, or the method I'd probably use myself if I were sitting at my computer with only a spreadsheet open, namely Romberg integration.) In fact, $ f(x)$ is a $ C^{\infty}$ function, so most methods should be well-behaved.

[Sonnhard Graubner isn't in the habit of explaining his methods here. It's likely he put this into some large general package, like Mathematica or Maple.]

[color=green][Unnecessary quote removed by moderator.][/color]

thanks for the solution mr kent merryfield. I was also about to use approximation because wolfram online integrator couldn't find a function for this....anyway, do you think there's no antiderivative for this? thanks very much!!!:D

As a default assumption, I just assume that a function does not have an elementary antiderivative unless I have some good reason to assume it does. Here, I have no good reason to assume it does.

That's not a proof - and there's a whole subject out there that others can comment on - but my main point is that you should regard [i]not[/i] having an elementary antiderivative as being the normal state of affairs, not some sort of surprising betrayal of the fabric of the universe.


$ \binom{n}{r\minus{}1}\equal{} \frac{n!}{(r\minus{}1)!(n\minus{}r\plus{}1)!} \equal{} r(r\plus{}1)$

$ \binom{n}{r}\equal{} \frac{n!}{r!(n\minus{}r)!}\equal{} (r\plus{}1)(n\minus{}r\plus{}1)$

$ \binom{n}{r\plus{}1}\equal{}  \frac{n!}{(r\plus{}1)!(n\minus{}r\minus{}1)!}\equal{}(n\minus{}r)(n\minus{}r\plus{}1)$

I don't get the last steps of these equations.

Clearly something's missing.  I don't think any of those work for $ n\equal{}2$, $ r\equal{}1$.

None of the last steps are correct in general.  It is true that the second column of each equation is proportional to the third.  Are you trying to solve a problem?


Suppose that four consecutive terms of a geometric sequence with common ratio r are the sides of a quadrilateral.What is the range of all posible values for  r?

i think $\frac{1}{x'} \le r \le x'$ where $x'$ is the real root of the equation $1+x+x^2=x^3$. If one of the equalities hold, the quadrilateral will degenerates to a line segment.


[quote="Smartguy"]MAML is going to be Oct. 28, 2008. Tues.

How are people going to prepare for it?[/quote]

Am I going to do it? Yes.
Am I going to prepare for it? Probably not.

It's about AMC 10-12 difficulty so if you prepare for other math comps you should be fine...

[quote="worthawholebean"]They should totally let Connecticut people do it.[/quote]
I don't think it's outside the realm of possibility for the MAML board to let Connecticut students take the first level MMO.  You could probably have your scoresheets machine-scored with the Massachusetts schools'.  But you wouldn't be ranked with Massachusetts students and you wouldn't be eligible for the second level MMO.

Mr. Yanco,

Hi, do you have any other MAMLs in your possession?
Can you post some online?

What is MAML?
im from Rhody- could I do it?

[quote="Smartguy"]Hi, do you have any other MAMLs in your possession?[/quote] I put ten online last year, which I think is enough for you to get a feel for what the contest might be like.  It varies from year to year, because a different person selects the questions each year, and each contest director has a different idea of what a contest should be like.
There's also a book of MAML Olympiads that you can buy from MAML; I think they charge 10 dollars or so.  It has the oldest ~10 years of contests.  If you're in the state your coach should be able to order it for you.

do you have the answers to years 1972-1975 for MAML?

has anyone done the 2005-2006 MMO round one? I need the answers for it...

haha...i need to redeem myself after my phail at the first mml meet

How high of a score do you need to go to the next round?

You guys should wait until Mr. Yanco says we can until discussion. I'm pretty sure you're allowed to take it any time today, and some boarding schools (Deerfield, Andover, probably some others) may do it in the evening.

Thank you Catalyst!!  I'd rather not have to go through deleting messages all the time, but you can't talk about a test until AFTER the testing window is over!!  Thanks.

Hello,
Sorry, I realized that i wasn't supposed to post after i had already posted it.  :oops:

Once again, sorry.

Hmm...why do people need to do this all the time?

The "official" testing window is over.  If the contest was not begun between 7:30 and 9:30 AM today, the results are not considered official.  However, at least one school was planning to take the contest unofficially after school.  Their results will be machine scored by MAML and ranked separately; so posting information about the contest could affect that ranking.

Thanks.

[quote="perfect628"]Yeah, we didn't finish up until around 7:15...[/quote]
Make that at least [i]two[/i] schools that participated unofficially.

[quote="Phelpedo"]Wait...Phillips Academy's results aren't going to be considered official?[/quote]
Only if PA started after 9:30 AM.  MAML's rule that all Olympiad administrations across the state need to begin within a fairly tight window in the morning has been around for quite some time; but this is the first year that a consequence for late administration was stated in advance.


[quote="mgao"]Consider an ordered set of six consecutive integers in increasing order.  To create a new set of six integers, the first, third and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two.  The median of this new set is four more than the median of the first set.  What is the sum of the median of the first set and the median of the second set?[/quote]

[hide]Okay, so the first thing I did to answer this question was I started with

$1,2,3,4,5,6$

Now we do the rule that corresponds to the list.

$3,4,5,8,7,12$
$3,4,5,7,8,12$

The median of the first one is 3.5 and the median of the second is 6, so we have to make the numbers bigger.  Of course this means that our answer will be in the form $2n+4$ because it is the same median but with 4 more added on to one of them.

Next, let's try it starting with 2 so we can maybe find a pattern.

$2,3,4,5,6,7$

$4,6,6,10,8,14$
$4,6,6,8,10,14$

The median ofthe first is 4.5 and the median of the second is 7, so the sum didn't change.

$3,4,5,6,7,8$

$5,8,7,12,9,16$
$5,7,8,9,12,16$

The median of the first one is 5.5, and the median of the second is 8.5.  So for each increasing odd number, the median of the second set is .5 greater than the median of the first set.  So to get from a difference of 3 to a difference of 4, we need to skip to 7,8,9,10,11,12 as the set of numbers.  The median of this set of numbers is 9.5, so in the form $2n+4$, we see that the answer is 23.[/hide]

I think I'm right, if not, I hope this helps you get started.

i did this problem before and i remember the answer isn't 23 but i forgot how to do it.

Somebody posted this problem recently-using inequalities is the most effective strategy, I believe.