Find all positive integer z such that the equation
$ x(x\plus{}z)\equal{}y^{2}$
has no positive integer solution $ (x,y)$

[quote="Yuriy Solovyov"]$ x \equal{}\frac{(z^{2}\plus{}4y^{2})^{\frac{1}{2}}\minus{}z}{2}$
So, $ z^{2}\plus{}4y^{2}\equal{} m^{2}$
And $ z \equal{} p^{2}\minus{}q^{2}, y \equal{} pq, m \equal{} p^{2}\plus{}q^{2}$, then $ x \equal{} q^{2}$
So.... $ z$ shouldn`t be $ p^{2}\minus{}q^{2}$ for all $ p\geq q; p,q\in Z^{\plus{}}$[/quote]
When the natural number can written in the form
$ p^{2}\minus{}q^{2}$

A natural number can't written in the form $ n \equal{} p^{2}\minus{}q^{2}$ when $ n\equiv 2(mod 4)$ because
$ p^{2}\equiv 0,1(mod 4),q^{2}\equiv 0,1(mod 4)\Rightarrow p^{2}\minus{}q^{2}\equiv 0,1,3(mod 4)$
-) If $ n \equal{} 2k\plus{}1$ then $ n \equal{} (k\plus{}1)^{2}\minus{}k^{2}$
-) If $ n \equal{} 4k$ then $ n \equal{} (k\plus{}1)^{2}\minus{}(k\minus{}1)^{2}$
Is it OK?

[quote="Yuriy Solovyov"]$ x \equal{}\frac{(z^{2}\plus{}4y^{2})^{\frac{1}{2}}\minus{}z}{2}$
So, $ z^{2}\plus{}4y^{2}\equal{} m^{2}$
And $ z \equal{} p^{2}\minus{}q^{2}, y \equal{} pq, m \equal{} p^{2}\plus{}q^{2}$, then $ x \equal{} q^{2}$
So.... $ z$ shouldn`t be $ p^{2}\minus{}q^{2}$ for all $ p\geq q; p,q\in Z^{\plus{}}$[/quote]
yes this is opinion

[quote="TTsphn"]Find all positive integer z such that the equation
$ x(x\plus{}z) \equal{} y^{2}$
has no positive integer solution $ (x,y)$[/quote]
let $ D(n)\equal{}\{d\in\mathbb{N}|\ d|n\}$
if i tell you: find $ S\equal{}\{(x,y,z)\in\mathbb{N}|\ xy\equal{}z\}$
certainely you find that $ S\equal{}\{(d,\frac{z}{d},z)|z\in\mathbb{N},d\in D(z)\}$

here it's the same
we have $ \{(x,y,z)\in\mathbb{N}|\ x(x\plus{}z)\equal{}y^{2}\}\equal{}\{(x,y,\frac{y^{2}}{x}\minus{}x)|\ y\in\mathbb{N},x\in D(y^{2}),x\le y\}$
if we take for exemple $ x\equal{}y,z\equal{}\frac{y^{2}}{x}\minus{}x\equal{}0$ or $ x\equal{}2,y\equal{}2k,z\equal{}\frac{4k^{2}}{2}\minus{}2\equal{}2(k^{2}\minus{}1)$

[quote="Yuriy Solovyov"]$ x \equal{}\frac{(z^{2}\plus{}4y^{2})^{\frac{1}{2}}\minus{}z}{2}$
So, $ z^{2}\plus{}4y^{2}\equal{} m^{2}$
And $ z \equal{} p^{2}\minus{}q^{2}, y \equal{} pq, m \equal{} p^{2}\plus{}q^{2}$, then $ x \equal{} q^{2}$
So.... $ z$ shouldn`t be $ p^{2}\minus{}q^{2}$ for all $ p\geq q; p,q\in Z^{\plus{}}$[/quote]
You are wrong this equation has this  solution if $ gcd(x,y,z)\equal{}1$

[quote="TTsphn"]Find all positive integer z such that the equation
$ x(x\plus{}z) \equal{} y^{2}$
has no positive integer solution $ (x,y)$[/quote]

The equation is equivalent to $ z \equal{}\frac{y^{2}}{x}\minus{}x$. Let $ \gcd(x,y) \equal{} g, x \equal{} gm,y \equal{} gn,(n,m) \equal{} 1$. Let us find those positive integers $ z$ which can be expressed in the form $ \frac{y^{2}}{x}\minus{}x$. Then $ \frac{y^{2}}{x}\equal{}\frac{gn^{2}}{m}\in\mathbf N$. This implies $ m\mid g$, as $ (n,m) \equal{} 1$. Let $ g \equal{} km$. So $ x \equal{} km^{2}, y \equal{} kmn$ and $ z \equal{}\frac{k^{2}m^{2}n^{2}}{km^{2}}\minus{}km^{2}\equal{} kn^{2}\minus{}km^{2}\equal{} k(n^{2}\minus{}m^{2})$.

Now we return to our problem. If $ z$ does not satisfy the given equation, then $ z$ cannot be written in the form $ k(n^{2}\minus{}m^{2})$. It is well known that all positive integers except those of the form $ 4t\minus{}2$ can be written as a difference of two squares. So we must have $ z \equal{} k(4t\minus{}2) \equal{} 2k(2t\minus{}1)$. Thus we conclude that if $ z$ can be written as the product of an even integer and an odd integer, then it will not satisfy $ x(x\plus{}z) \equal{} y^{2}$.

Something is probably wrong in this argument, but I can't find it. :blush:


calculus
derivative
geometry
3D geometry
calculus computations

If $ x$ is the sidelength, then the volume is $ V(x)\equal{}x^3$. We don't know the precise value of $ x$. One way (but not the only one) to estimate $ V(x)$ is to write $ V(x)\approx V(5)\plus{}V'(5)(x\minus{}5)$, where $ \approx$ indicates that we are not going to get $ V(x)$ exactly. Here $ V(5)$ and $ V'(5)$ are not hard to find. As for $ x\minus{}5$, the only thing we know about it is that it's at most $ 1\%$ of $ 5$ in absolute value. Now put this together.

$ V=s^3.$

$ dV=3s^2\,ds$

$ \frac{dV}{V}=\frac{3s^2\ds}{s^3}=\frac{3\,ds}{s}.$

The relative error of $ V$ is three times the relative error of $ s.$ If the relative error of $ s$ is $ 1\%$ then what is the relative error of $ V?$

[quote="MIA6"]The edge of a cube has length 5 with a possible error of 1%. What's the possible error in the volume of the cube?  I really have no idea about this problem. It's impossible to do like this (1%)^3. It must relate to calculus, so hope you can give me hint. thanks.[/quote]

Thanks. but why we need to find V'? like why V(x)=V(5)+V'(5)(x-5)

I did not write that with "$ \equal{}$". The tangent line to the graph of $ V(x)$ at $ x\equal{}5$ has equation $ y\equal{}V(5)\plus{}V'(5)(x\minus{}5)$, and we can hope that the graph stays close to its tangent line near $ x\equal{}5$.

[quote="mlok"]I did not write that with "$ \equal{}$". The tangent line to the graph of $ V(x)$ at $ x \equal{} 5$ has equation $ y \equal{} V(5) \plus{} V'(5)(x \minus{} 5)$, and we can hope that the graph stays close to its tangent line near $ x \equal{} 5$.[/quote]

But the derivative of x^3, should be 3x^2?


[quote="pohoatza"]Let $a,b,c$ three positive numbers with $abc=1$. Prove the following inequality:

$a^{2}+b^{2}+c^{2}+9(ab+bc+ca) \geq 10(a+b+c)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197


Rationalize: $ \frac{1}{\sqrt{1\plus{}\sqrt{2}}}$
And $ \frac{1}{2\minus{}\sqrt[4]{2}}$

For the first one, I would multiply both sides by the denominator and then multiply both sides by $ 1\minus{}\sqrt{2}$


For each real value of $ t$, determine whether or not there exist solutions to the simultaneous equations
\[ \begin{array}[r]{c}x x + y + z = t, \
tx + 2z = 3, \
3x + ty + 5z = 7,
\end{array}\]exhibiting the most general form of such solutions when they exist.

If $ A \equal{} \begin{bmatrix}1 & 1 & 1\\ t & 0 & 2\\ 3 & t & 5\end{bmatrix}$ then det(A) = 0 iff t = 1 or t = 6. With this we know that the system has a single non-zero solution for every real t except 1 and 6. Performing Gauss elimination, we can conclude that for t =1 the system has infinitely many solutions, and for t = 6 the system is impossible.

Oh I did Gaussian elimination first without finding $ \det A$.. wrong move. Thanks!

hello, by Gauss elemination we get the solutions as
$ x\equal{}\minus{}\frac{2t\minus{}1}{t\minus{}6}$
$ y\equal{}\minus{}\frac{5t\minus{}8}{t\minus{}6}$
$ z\equal{}\frac{t^2\plus{}t\minus{}9}{t\minus{}6}$
for all $ t$ with $ t\ne6$
Sonnhard.


say there are 3 cards.  behind one of them there is a car.  Behind the other 2, there are goats.  The car is in the middle.  A person who doesn't know this is asked to pick one.  He picks the middle but he doesn't know it is a car.  Then I turn over the one on the right revealing a goat.  If I let him change his pick, should he?

source: numb3rs

[quote="jli"]say there are 3 cards.  behind one of them there is a car.  Behind the other 2, there are goats.  The car is in the middle.  A person who doesn't know this is asked to pick one.  He picks the middle but he doesn't know it is a car.  Then I turn over the one on the right revealing a goat.  [color=red]If I let him change his pick[/color], should he?

source: numb3rs[/quote]
should he what?

[hide]yes, he should.  i read this in a book before.  It was [u]The Curious Incident of the dog in the Nightime[/u][/hide]

that was an amazing book.

and aren't his chance 1/2 if he picks again, not 2/3?

I don't think that's right. The chances of picking a car originally was $\frac{1}{3}$ and if you eliminate one then you would have to eliminate from the total number of choices, also. I believe the probability will then be \[\frac{1}{2}\]. So it won't matter if he chooses again or not..

[quote="seamusoboyle"]Yes. The chances of him origianally picking the car are $1/3$ Just because one of the doors is opened dosnt change this. That means the chances are the the car is behind the other door is $2/3$[/quote]
seamusoboyle is correct.
this problem was posted before, about a person finding a treasure, somewhere.

It's a variation of the famous Monty Hall problem....

http://www.comedia.com/hot/monty.html

The biggest difference lies in the fact that you choose a card [i]before[/i] one that does not contain the car is shown to you...If you only had two doors, then you were told to choose...the probability would be 50%.

Great problem to ask people to get them thinking about probabilities. I, too, first herad of it when I read the book math92 referred to.


Given that $\log _3 2=0.631$, find the smallest positive integer a such that $3^a>2^{102}$.

Taking the log of both sides of the inequality, we obtain $a\log3>102\log2$

Then, we have $a>\frac{102\log2}{\log3}=102\log_3 2$

Since $\log_3 2=0.631$, $102\log_3 2=102(0.631)=64.362$

Hence $a=65$

So that no one gets the wrong idea, I would like point out that $\log_3{2}\approx .631$ but isn't exactly equal.  In fact, $\log_3{2}$ is irrational.

Just thought I would clarify  ;)


The surface area of a cube is 48 square units. How many units, expressed in simplest radical form, are in the length of an edge of the cube?

Let $ s$ be the side length of the cube. So, $ 6s^2\equal{}48$, and $ s^2\equal{}8$. Thus the length of an edge of the cube, in simplest radical form, is $ \boxed{2\sqrt{2}}$.


[quote="Identity"]$ \frac {x^4 \minus{} 7x^2 \minus{} 9}{x \minus{} \frac {3}{x} \plus{} 1} \equal{} 0$

$ x^4 \minus{} 7x^2 \minus{} 9 \equal{} 0$

Let $ u \equal{} x^2$

$ u^2 \minus{} 7u \minus{} 9 \equal{} 0$

$ u^2 \minus{} 7u \plus{} \frac {49}{4} \minus{} \frac {49}{4} \minus{} 9 \equal{} 0$

$ (u \minus{} \frac {7}{2})^2 \equal{} \frac {85}{4}$

$ u \equal{} \frac {7}{2}\pm \frac {\sqrt {85}}{2}$[/quote]

oops there was a problem in the problem.  lol

[quote="sonny"]Find the zeroes of $ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1}$[/quote]
[hide]$ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1}\equal{}0$

$ x^4 \minus{} 7x^2 \plus{} 9\equal{}0$

$ x^2\equal{}\frac{7\pm\sqrt{49\minus{}36}}{2}$

$ x\equal{}\boxed{\pm\sqrt{\frac{7\pm\sqrt{13}}{2}}}$[/hide]

[quote="i_like_pie"][quote="sonny"]Find the zeroes of $ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1}$[/quote]
[hide]$ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1} \equal{} 0$

$ x^4 \minus{} 7x^2 \plus{} 9 \equal{} 0$

$ x^2 \equal{} \frac {7\pm\sqrt {49 \minus{} 36}}{2}$

$ x \equal{} \boxed{\pm\sqrt {\frac {7\pm\sqrt {13}}{2}}}$[/hide][/quote]
We still have to check for extraneous solutions, which doesn't look like too much fun.

[quote="i_like_pie"][quote="sonny"]Find the zeroes of $ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1}$[/quote]
[hide]$ \frac {x^4 \minus{} 7x^2 \plus{} 9}{x \minus{} \frac {3}{x} \plus{} 1} \equal{} 0$

$ x^4 \minus{} 7x^2 \plus{} 9 \equal{} 0$

$ x^2 \equal{} \frac {7\pm\sqrt {49 \minus{} 36}}{2}$

$ x \equal{} \boxed{\pm\sqrt {\frac {7\pm\sqrt {13}}{2}}}$[/hide][/quote]
[hide="Rationalized Denominator and Simplified Radical"]
$ x \equal{} \pm\sqrt {\frac {7\pm\sqrt {13}}{2}}$
$ x \equal{} \pm\sqrt {\frac {14\pm 2\sqrt {13}}{4}}$
$ x \equal{} \pm\frac {\sqrt {14\pm 2\sqrt {13}}}{2}$
Let $ a \plus{} b\sqrt {c} \equal{} \sqrt {14 \pm 2 \sqrt {13}}$
$ a^2 \plus{} 2ab\sqrt {c} \plus{} b^2c \equal{} 14 \pm 2 \sqrt {13}$
$ c \equal{} 13$
I'll just solve for the following, and take the opposite of the $ b$ value for the other case, since $ ab \equal{} 1$
$ a^2 \plus{} 13b^2 \plus{} 2ab\sqrt {13} \equal{} 14 \plus{} 2 \sqrt {13}$
$ a^2 \plus{} 13b^2 \equal{} 14$
$ 2ab \equal{} 2$
$ b \equal{} \frac {1}{a}$
$ a^2 \plus{} \frac {13}{a^2} \equal{} 14$
$ a^4 \plus{} 13 \equal{} 14a^2$
$ a^4 \minus{} 14a^2 \equal{} 13$
$ (a^2 \minus{} 13)(a^2 \minus{} 1) \equal{} 0$
$ a \equal{} \pm \sqrt {13}, \pm 1$
$ a$ is supposed to be an integer, so $ a$ could be $ 1$ or $ \minus{} 1$
$ b$ would also be $ 1$ or $ \minus{} 1$
so $ a \plus{} b\sqrt {c} \equal{} \pm1\pm1\sqrt {13}$, or $ \pm1\pm\sqrt {13}$
simplified, $ x \equal{} \pm\frac {1 \pm \sqrt {13}}{2}$

also, roots that won't work are roots that satisfy $ x \minus{} \frac {3}{x} \plus{} 1 \equal{} 0$
$ x \minus{} \frac {3}{x} \plus{} 1 \ne 0$
$ x^2 \minus{} 3 \plus{} x \ne 0$
$ x^2 \plus{} x \minus{} 3 \ne 0$
$ x \ne \frac { \minus{} 1\pm\sqrt {13}}{2}$
in other words, $ x$ can be neither $ \frac { \minus{} 1 \plus{} \sqrt {13}}{2}$ nor $ \frac { \minus{} 1 \minus{} \sqrt {13}}{2}$, so
that eliminates half the roots. that means that $ x \equal{} \boxed{\frac {1\pm\sqrt {13}}{2}}$

(It also means that $ x \minus{} \frac {3}{x} \plus{} 1$ was a factor of $ x^4 \minus{} 7x^2 \plus{} 9$ *cough* we coulda saved alot of work by multiplying the fraction by $ 1$ in the form of $ \frac{x}{x}$ and doing long division on $ x^4\minus{}7x^2\plus{}9$)
[/hide]


geometry
perimeter
combinatorics unsolved
combinatorics

A square paper of side n is divided into n x n unit square cells. A maze is drawn on the paper with unit walls between some cells in such a way that one can reach every cell from every other cell not crossing any wall. Find, in terms of n, the largest possible total length of the walls.

First erect all possible$ ($length:$ 2n^2\plus{}2n)$ walls.Now start from a corner sqare and draw a path through all the squares.We have total $ n^2$ Sqares.So we have to demolish  :P  $ n^2\minus{}1$ walls. Thus we must have at most $ (2n^2\plus{}2n)\minus{}(n^2\minus{}1)\equal{}(n\plus{}1)^2$ walls. Thus max lenth is $ (n\plus{}1)^2$.
P.S. I have assumed that walls can be obtained in the perimeter too.

This is not a proof, but it could be developed into one. Good job, I guess.

[quote="JBL"]All one needs to formalize this is to note that any connected graph on $ n^2$ vertices has at least $ n^2 \minus{} 1$ edges.  (Vertices are cells, edges join two adjacent cells not blocked by a wall.)[/quote]
Thank you JBL.


calculus
integration
function
real analysis
real analysis unsolved

Find:
$ \int (\sqrt [3]{x^{3} \minus{} \sqrt {x^{3} \plus{} x^{6}}} \plus{} \sqrt [3]{x^{3} \plus{} \sqrt {x^{3} \plus{} x^{6}}})\,dx$

I think it is not for normal people and I will be happy to see some short and/or nice solution or even solution. (I have one. It is again from The Book.)

--
Bradistilov's Fan

I think that these days, many problems posted in this forum should be moved to the "Calculus Computations and Tutorials" forum? Isn't it ??

And vice versa  :roll: Unfortunately, the beginning of the semester is a fairly busy time for most of the mods  :P, so we'll sort it out somehow a bit later. After all, the people who attend Computations and Tutorials and Unsolved and Proposed Problems are almost the same ones, so there is not too much harm made. As to this particular problem, I don't know: I have to solve it first and right now I hardly have any time for it :).

In acordance with Cardano's formula the function under integral sign is a root to the equation: $ y^{3} \plus{} 3xy \plus{} 2x^{3}$. We may substitute: $ z \equal{} \frac {y}{x}$.

P.S. I don't like this integral...


Give $ a,b,c \ge 0$.Prove that:
$ \sqrt{\frac{a}{a\plus{}2b}}\plus{} \sqrt{\frac{b}{b\plus{}2c}}\plus{} \sqrt{\frac{c}{c\plus{}2a}} \le \frac{a\plus{}b\plus{}c}{\sqrt{ab\plus{}bc\plus{}ca}}$

from CS inequality
$ \sum_{cyc} \frac{\sqrt{a}}{\sqrt{a\plus{}2b}}\equal{}\frac{\sum_{cyc} \sqrt{(ab\plus{}2ca)(c\plus{}2a)}}{\sqrt{(a\plus{}2b)(b\plus{}2c)(c\plus{}2a)}}\leq \frac{\sqrt{9(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)}}{\sqrt{(a\plus{}2b)(b\plus{}2c)(c\plus{}2a)}}$
it remains to prove $ 9(ab \plus{} bc \plus{} ca)^2\leq (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)(a \plus{} b \plus{} c)$ or
$ 2\sum_{sym} a^3b \plus{} 2\sum_{cyc} ab^3\geq 3\sum_{cyc} a^2b^2 \plus{} 3\sum_{cyc} a^2bc$ which is obviously true


Now we know that

$ z_1,z_2,z_3,z_4$ are known complex numbers.

For each complex number $ z_p$,

We define $ z_{12}$ (Look at the figure in the attachment)

As well,we can define $ z_{13},z_{14},z_{23},z_{24},z_{34}$.

Let

$ p \equal{} z_{12} \plus{} z_{34}$,
$ q \equal{} z_{13} \plus{} z_{24}$,
$ r \equal{} z_{14} \plus{} z_{23}$,

Prove that $ \frac {p \minus{} q}{p \minus{} r}$ is a invariant for variant $ z_p$.(Although $ z_p$ can be any complex number).

$ \frac{p-q}{p-r} $ is a invariant for variant $ z_{p} $
	\[\frac{p-q}{p-r}=\frac{(\overline{z_{1}}-\overline{z_{4}}) \cdot (\overline{z_{2}}-\overline{z_{3}})}{(\overline{z_{1}}-\overline{z_{3}}) \cdot (\overline{z_{2}}-\overline{z_{4}})}\]
Method: name $z_{1}=a,z_{2}=b,z_{3}=c,z_{4}=d$ and $\overline{z_{1}}=e,\overline{z_{2}}=f,\overline{z_{3}}=g,\overline{z_{4}}=h$
Name also $z_{p}=v,\overline{z_{p}}=w$
Calculate every $z_{ij}, p , q, r$.
Then the numerators of $p-q$ and $p-r$ are the same; the denominators lead to our result.


Find the mistake in the following reasoning:

$ \minus{}1 \equal{}\sqrt[3]{\minus{}1}\equal{} (\minus{}1)^{\frac{1}{3}}\equal{} (\minus{}1)^{\frac{2}{6}}\equal{} ((\minus{}1)^{2})^{\frac{1}{6}}\equal{} (1)^{\frac{1}{6}}\equal{} 1$

[quote="gauss202"]Find the mistake in the following reasoning:

$ \minus{}1 \equal{}\sqrt [3]{\minus{}1}\equal{} (\minus{}1)^{\frac{1}{3}}\equal{} (\minus{}1)^{\frac{2}{6}}\equal{}\boxed{ ((\minus{}1)^{2})^{\frac{1}{6}}}\equal{} (1)^{\frac{1}{6}}\equal{} 1$[/quote]

Generally, $ a^{bc}\equal{} (a^{b})^{c}$ does not hold over $ \mathbb{R}$.  It is precisely because of problems like this, and generally because the function $ a^{b}$ is not single-valued over $ \mathbb{R}$.

In some sense, however, the mistake is in the very last deduction; it is, in fact, true that $ \minus{}1$ is [b]one of[/b] the sixth roots of $ 1$, but the last deduction implies the most straightforward definition of exponentiation, which is not necessarily the one we want, as per the previous objection.

@t0rajir0u: I think you mean "does not hold over $ \mathbb{C}$".
----
I would explain why it doesn't hold (what t0rajir0u says), but I don't think it will make sense to you. Have you ever studied the [url=http://en.wikipedia.org/wiki/Complex_logarithm]complex logarithm function[/url]? Because that's how exponents are defined in general.

[quote="sylow_theory"]@t0rajir0u: I think you mean "does not hold over $ \mathbb{C}$".[/quote]

Well, it doesn't hold over $ \mathbb{R}$ either  :P

But it DOES hold over $ \mathbb{R}^{\plus{}}$ :D

what is $ \mathbb{R}$
i know it is the real numbers but what is real numbers ?
what is $ \mathbb{R}^{\plus{}}$
what is positive real numbers?

It's as torajirou says, another (much simpler) case where you can do screwy things is

$ 2 \equal{}\sqrt{4}\equal{}\sqrt{\minus{}2\cdot\minus{}2}\equal{}\sqrt{\minus{}2}\cdot\sqrt{\minus{}2}\equal{}\minus{}2$

[quote="binomial_4eva"]what is $ \mathbb{R}$
i know it is the real numbers but what is real numbers ?
[/quote]
It basically means all the numbers that aren't imaginary. So it includes both rational and irrational numbers. For example, $ \sqrt{2},0,,1,\pi$ are all real numbers. However, $ i,\pi i,$ arent real numbers.

And $ \mathbb{C}$ means the complex numbers. Complex number are both real and imaginary numbers together.

[quote]
what is $ \mathbb{R}^{\plus{}}$
what is positive real numbers?[/quote]
Same thing just positive.

Are you asking rather "How do you mathematically define $ \mathbb{R}$"? And "how do you mathematically define $ > 0$". That's a much more difficult question to answer. Once you take a course in Real Analysis you might talk a little about this. Google and read about [u]Cauchy Sequences[/u] and [u]Complete Ordered Fields[/u]. But I got to warn you, if you are new to analysis you won't understand the approaches used to define these concepts.

[quote="t0rajir0u"]Well it doesn't hold over $ \mathbb{R}$ either.[/quote]
I see what you mean. But you are still secretly working in $ \mathbb{C}$ when you take roots of negative numbers, if you know what I mean.

[quote="binomial_4eva"]$ \mathbb{R}\equal{}{\mathbb{Q}}\plus{}{\mathbb{Z}}\plus{}{\mathbb{N}}$ rite?

so is it any set of numbers except the set of imaginaries?

thankx loads i think i get it now.[/quote]

You forgot $ \mathbb{J}$ - the irrationals.

actually, $ \mathbb{R}\equal{}\mathbb{Q}\plus{}\mathbb{J}$.
The rationals include the integers, and the integers include the natural numbers, so you're counting the integers twice and the natural numbers three times in your equation.

[quote="binomial_4eva"]so is it any set of [i]numbers[/i] [/quote]

This is bordering on tautology because you haven't defined what, exactly, you mean by "number."  Usually when we say "number" we mean "real number," so you're not actually saying anything here.

[quote="Prime factorization"]Is there any place where we can find the definition of each symbol? For instance, I'm unfamiliar with what $ \mathbb{Q}$ stands for.[/quote]

$ \mathbb{Q}$ is the set of rational numbers. Try [url=http://en.wikipedia.org/wiki/Blackboard_bold]Wikipedia[/url] for what the black board bold symbols stand for.

[quote]I'm unfamiliar with what $ \mathbb{Q}$  stands for.[/quote]

$ \mathbb{Q}$ stands for quotient, and quotient means fraction.


calculus
derivative
function
vector
algebra
domain
linear algebra

sorry guys but this forum was the best place i could find

neways

i have the following doubt

when we differentiate we generally take it with respect it one variable only
is it true for even advanced calculus
(ignore the total differential?)

if by advanced calculus you mean multivariable caluculus, then you're essentially "differentiating" with respect to several variables. However, the method of differentiating when dealing with several variables is simply to differentiate with respect to one variable while holding the other variables constant.

To add on, in multivariable calculus, since we are dealing with functions of 2, 3, sometimes more variables, we have to differentiate with respect to several variables, although usually not at the same time.  This is called the partial derivative, and essentially all it is is treating every variable except the one we want as a constant, and differentiating normally.  In fact, the ordinary derivative is a special case of the partial derivative, and the partial derivative is a special case of the directional derivative.

thanks for the information
but one more question

wat is an directional derivative
is it the derivative multiplied by a unit vector?

Directional derivative is the approximation (derivative) of the function along a fixed direction, say vector $ v$. You can look it up [url=http://en.wikipedia.org/wiki/Directional_derivative]here[/url]

[quote="badwaik.jayesh"](ignore the total differential?)[/quote]No, [i]don't[/i] ignore it. The derivative in a multivariable context is a linear operator, mapping the tangent space of the domain to the tangent space of the codomain. If those are $ \mathbb{R}^m$ and $ \mathbb{R}^n$, the tangent spaces are also identified with $ \mathbb{R}^m$ and $ \mathbb{R}^n$.

Now, if you want to do any computations, you need a matrix for this linear operator. The entries of that matrix are partial derivatives, and taking partial derivatives is the backbone computational tool. Directional derivatives are partial derivatives after a change of basis in the domain; there's nothing sacred about the standard basis.

the information it has certainly helped me

thanks a lot for hearing this guy here


Evaluate:

cos squared (1 degree) + cos squared (2 degrees) + cos squared (3 degrees) + cos squared (4 degrees) ... + cos squared (90 degrees)

^^ Correct.

Try This :

cos1*cos3*cos5*cos7*cos9.......cos87*cos89 = 2^n
Find n.

[quote="1=2"][hide="hint"]cos(x)=sin(90-x)[/hide][/quote]
Not to be rude , but i know this one.
Was simple , just basic identities PS: after u have performed that step it gets tricky
there's also one in terms in sine.


calculus
vector
number theory
group theory
algebra unsolved
algebra

Okay, i have this problem. first i'm really good in math. i'm 19, i have learned calculus I,II,III all by myself and know it well. i have learned an entire course on elementary number theory and know it well. i also learned group theory (sylow theorem are among my favorite) and understood it perfectly all by myself. i'm familar a little with ring and field theory and vector calculus. again i have learned all these things by myself without any help.

but :( when i look at IMO problems few of them i can do, but most i cant, maybe i didnt spend enough time on them. does that mean im not talented in math, because i really think i am, im not trying to show off i just wrote that above to explain why i think i am talented at math. but these IMO problems really make me depressed. i tell myself i need special training, like the real olympics. and maybe if i brought myself a problem solving book (art and craft of problem solving) i am get really good because i already have a talent. am i right? or do i have to face the sad truth, i just cant those problems :(.

Olympiads are just a certain kind of math.  They emphasize taking basic tools and doing unintuitive things with them.  Olympiad math is rarely a matter of just knowing the right topics - at its heart, it tests problem-solving abilities (the subject of the site!).

But don't be discouraged because the IMO is a very high-level competition.  It's not supposed to be easy to do be able to do most of the problems...

I'm sure with the right training, though, that you could learn to do them  :)

And this is in algebra because...

In college at age 19, can you still do the IMO? Or are you just doing it for yourself/.?

[quote="t0rajir0u"]Olympiads are just a certain kind of math.  They emphasize taking basic tools and doing unintuitive things with them. [/quote]
doing unintuituve things i am used to, in fact, i feel more comfortable working formally than informally. the math i'm learning now is completely unintuitive.
[quote]
I'm sure with the right training, though, that you could learn to do them  :)[/quote]
and where do i get training. if i buy a book on problem solving will it help? i'm not doing this to compete i'm doing it because i love math and because i think being able to solve those problems will help me solve problems from advanced math.

[quote]In college at age 19, can you still do the IMO? Or are you just doing it for yourself/.?[/quote]
i'm doing it for myself.

guys i got one more question. this site is for problem solving, but do you guys do: group theory, field theory, algebraic number field? i'm asking because i want to discuss these  topics.

You might benefit from [url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Problem-Solving Strategies[/url] by Arthur Engel.  He occasionally dips into some advanced stuff, but it's largely Olympiad math.

As to your second question, the Superior Algebra forums are [url=http://www.artofproblemsolving.com/Forum/index.php?f=8]here[/url].

[quote="t0rajir0u"]You might benefit from [url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Problem-Solving Strategies[/url] by Arthur Engel.  He occasionally dips into some advanced stuff, but it's largely Olympiad math.
[/quote]
What about this,
http://www.amazon.com/o/ASIN/0471789011/ref=s9_asin_image_2/105-1150425-6720460

That book also comes highly recommended, from what I hear.

It is indeed an excellent book. Where Engle is brief, Zeitz shows what kind of mindframe you should be in and how to explore a problem. Engle usually just gives the answer with little explanation, for you to fill in the gaps. Fun!


calculus
integration
function
logarithms
calculus computations

Can someone explain to me what improper integration is? I know what it is used for, but have no clue as to how to do it. Any good websites that explain it really well?

Thanks a lot in advance.

I'm assuming you are referring to improper integrals.  No problem:

Let's say you have a definite integral.  Now, under normal circumstances, we should be able to compute the integral (of course assuming the function is integrable).  However, suppose the function is not continuous over the region of integration.  Then what?  Well, clearly, we can't actually evaluate the integral, but we can evaluate the limit.  That is, instead of considering the integral at the point, we consider the integral as we approach the point.  This is especially useful if we are integrating with $ \pm\infty$ as a limit.

Hope that makes sense.  Check wikipedia for more.

Calculus textbooks do cover this and do have problems that illustrate what's going on.

One thing you should do: ask the question "for how many different reasons is this integral improper?" Then split it into pieces where each piece has only one reason for the integral to be improper.

For instance, $ \int_0^{\infty}e^{\minus{}x}\ln x\,dx$ is improper for two reasons: $ \infty$ and $ 0.$ The logarithm tends to $ \infty$ (well, actually $ \minus{}\infty$ - same thing) at $ 0$ and any integral to $ \infty$ is always improper. Therefore we should find a convenient number in between (say, $ 1,$) to split at, and separately consider

$ \int_0^1e^{\minus{}x}\ln x\,dx\plus{}\int_1^{\infty}e^{\minus{}x}\ln x\,dx.$

In this case, we can make different useful comparisons on each piece to show that both converge (even though we can't write down an antiderivative). The standards for convergence are that each piece must converge separately for us to accept this as convergent.

Another example: $ \int_{\minus{}3}^5\frac1x\,dx.$ We say this is improper for two reasons: the left side of zero and the right side of zero. We break this up as

$ \int_{\minus{}3}^0\frac1x\,dx\plus{}\int_0^5\frac1x\,dx.$

This time both integrals diverge; the first to $ \minus{}\infty$ and the second to $ \infty.$ We call this a divergent improper integral.

What about $ \int_0^1x\ln x\,dx?$ I would not technically call this an improper integral, because the integrand can be extended to be continuous on the closed interval $ [0,1].$ But that's not an important distinction; there is no harm in behaving as if this were improper, and that's what we'll do to compute it in any case.

Well thank you [b]Kent Merryfield [/b] for a really nice explanation for improper integrals. :) well there is a really nice disscussion in improper integrals about their Convergece or divergence thats really cool and i think one of the most beautiful parts of calculus...(well its funny cuz calculus has many beautiful things to learn and discuss!!! :lol: )  i think i can give a nice example here for covering some of it... i dont know i've posted this before or not but lets just try...! :D
Find the values of $ \alpha$ and $ \beta$ such that the following improper integral (for exercise and reviewing what [b]Kent[/b] said above you can say why is it an improper integral!) converges:

$ I \equal{} \int^{1}_{0}\frac {(\ln x)^{2}}{x^{\alpha}(1 \minus{} x)^{\beta}}\text{d}x$

I can say it is improper because the function is not defined at $ x\equal{}0$ or $ x\equal{}1$.


What is the "thou" conjugation of "to have"?

Thou has
or
Thou hast
or
Thou have

I always thought it was "hast" until I read Carroll's Jabberwocky:
... Has thou slain the Jabberwock? ....

Hmm...

According to the Oxford English Dictionary, the preferred forms are "hast" in present tense and "hadst" in past tense, but there is quite a lot of variation:

Indicative present:
[quote] b. 2nd pers. sing. hast (hæst, h{schwa}st). Forms: {alpha}1 hafast, hæfst, 1-3 hafest, 2-6 (hauest), havest, 3 hafuest, hæfuest, hæuest, hafust, (afest, auest), hafst, 3- hast, (4 hest, 5-6 haste, 7- 'st). {beta}north. 3-5 haues, 3- has, hes, 3-4 hauis, (3-5 as), 4 habbes, -ez, 5 hauys, hais, 5-6 hase, (6 hess).
[/quote]

Indicative past:
[quote] b. 2nd pers. sing. hadst. Forms: {alpha}1 hæfdes, -est, 2-3 hefdest, 3-4 heuedest, haddist, 4-6 haddest, 4 hadest, 6- hadst. {beta}north. 3-5 hade, 3- had.[/quote]

Something to keep in mind is that this list provides most (if not all) of the attested forms starting with Beowulf.

I can post the quotations here, if you would like, but that's probably overkill.

Modern English:

(pres. indic., pres. subj., past indic., past subj.)
thou hast/have/hadst/had
(imperative) have thou

Old English:

thu haefst/haebbe/haefdest/haefde
(imperative) hafa thu


[quote="Slizzel"]Show that $\log_{3}{(\log_{3}{5})}+\log_{5}{(\log_{5}{7}})}+\log_{7}{(\log_{7}{3})>0$[/quote]

  I think I have solved this inequality in the general case  for $a=3,b=5,c=7$ and $a,b,c>1.$

This problem was given at Constantza Olympiad in 2000.

Maybe. I didn't participate at that phase because i'm not from Constantza and in 6th grade I didn't knew logs :D. Your solution is?


Do you think you could email some of your old semifinal exams to me?

I would really appreciate that for preparing for USABO. Thank you.

chancellorphoenix2@hotmail.com

What is USABO? is it like the USAMO for but bio?

Yes, I believe so.

I'd like any old exams as well. My e-mail is vince1123581321@gmail.com. Thanks.

If u don't mind can i plz have some as well

Thanks

yintex@msn.com

I could desperately use some practice too. If you could send some to evanryanlin@gmail.com, I'd really appreciate it.

Okay, here's the deal. A few months ago my computer caught a super-bad virus and pretty much died. I lost all of the files on my computer. I don't have the exams anymore but I am trying to get them from my old biology teacher again. I do, however, have last year's semifinal exam and I'll post some questions from it soon (I don't have the exam on my computer.) Good luck to everyone!

P.S. If anybody was wondering, I did [i]not[/i] qualify for the USABO last year. I got extremely, extremely close though.


The farmer asked each of his sons to fence in two sections of land. To the first son he said, "Go and fence in the land over there that is half a square mile." To the second son he said, "Go and build a fence around the land over there that is a half mile square." Each son did exactly as he had been told, but the farmer couldn't believe that the sons had different results.

Which son fenced in the larger area?

this belongs in the Puzzles and Brainteasers forum.

[quote="colts18"]The farmer asked each of his sons to fence in two sections of land. To the first son he said, "Go and fence in the land over there that is [b]half a square mile[/b]." To the second son he said, "Go and build a fence around the land over there that is [b]a half mile square[/b]." Each son did exactly as he had been told, but the farmer couldn't believe that the sons had different results.

Which son fenced in the larger area?[/quote]
That gives you the answer

[hide]
first one is $\frac{1}{2}$ square miles and second one is $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ square miles, i presume.[/hide]

[hide]the first son fenced in $\frac{1}{2}$ square miles and the second son fenced in $\frac {1}{4}$ square miles.[/hide]


I heard that there was going to be asymptote in the classroom. How do you use it?

[quote="rrusczyk"]Just like on the forum, but you start your post with a semicolon, just as with LaTeX in the classroom.  So,

;[asy]stuff[/asy][/quote]

Was it supposed to be

[code]
;[asy]
stuff
[/asy]
[/code]?


probability
ARML
search
expected value
conditional probability

As always, if this "topic" has been posted elsewhere, please notify me.

There are two envelopes. One of them contains twice the amount of money in the other envelope. Both of them contain greater than zero dollars. You open one envelope randomly, and you find 100 dollars.

Should you switch or not to the other envelope if you want to maximize your profit, and explain why you do or don't do so.

The expected value would be (200 + 50)/2. SO, you should always switch. We were discussing about the original Monty problem in another thread and we noticed that the show host could be evil and purposely reveal to you the one with less money knowing that probability wise, you would switch.

[quote]What if one envelope was inside the other? Then technically, if you had Envelope 1 which is the one with 100 dollars and Envelope 2 which has 100 dollars and Envelope 1 inside of it then that might get you somewhere. Iono.[/quote]
i don't think it makes any difference

Here's a question: what if we do the same thing, but you don't open the envelope to see how much is inside it?  Should you keep teh one you have, or switch?

Regarding the original problem, I would say that it is underspecified, and hence we can't answer.  The problem does not specify what probability distribution was used to stuff the envelopes.  We can easily create a probability distribution where the answer goes one way, and another distribution where the answer goes the other way.

The posts that say we should switch assume that, given that we see 100 dollars, the probability is 50-50 whether the other envelope holds 50 or 200 dollars.  But that's not necessarily so; that conditional probability depends on the underlying probability distribution.

Joel's question helps show why switching as a universal strategy makes little sense.

I'm glad you're enjoying the problem.

I am not quite understanding the probability distribution that you request. The envelopes are randomly selected and randomly "stuffed" with the cash - if that is what you mean.

At ARML, last year we went to this thing on mathematical paradoxes and this was one of the featured paradoxes.  There were many other paradoxal phenomenons.  Did anyone else to ARML at the Iowa City region?

[quote="banblink182"]I am not quite understanding the probability distribution that you request. The envelopes are randomly selected and randomly "stuffed" with the cash - if that is what you mean.[/quote]
I agree that we can assume the envelopes were randomly shuffled -- but what amounts were put there in the first place?  If we know, for example, that the amounts were 50 dollars and 100 dollars, then we can solve the problem.  If we know that the amounts were x dollars and 2x dollars, where x is a random integer from say 1 to 100, then again we can solve the problem.  But if we are not told how the original numbers were chosen, then we don't have enough information to solve the problem.

As written, this problem is not much different from the following: "My friend put a certain amount of money in an envelope.  What's the probability that he put in 100 dollars?".  I hope you agree there's not enough information to solve that problem.  And the original problem is in essence the same.

Basically, to solve any probability problem, we need to be given the underlying probability distribution -- informally speaking, what random process was performed.  Otherwise we can't calculate anything.

I don't think you understand it.  There are [size=150][b]two[/b][/size] envelopes, with amounts $x$ and $2x$.  One has $\$$100$.  There is a $\displaystyle \frac{1}{2}$ probability it is the $x$ envelope and a $\displaystyle \frac{1}{2}$ probability it is the $2x$ envelope.

[quote]There is a 1/2 probability it is the x envelope and a 1/2 probability it is the 2x envelope.[/quote]
We most certainly do understand the question, but what you say here isn't necessarily true.  The number x can come from all the positive real numbers, right?  You must explain the probability that any given x was chosen in order for the question to be answerable.  (That is, you have to give the probability distribution from which you were choosing x.)  The problem is that "at random" needs to be more precisely defined in the case where the options come from an infinite set.

If you don't believe me, consider the question I asked above.  By the reasoning that people have provided, no matter which envelope you get, you should switch (to generate 25% profit).  But then suppose that I gave you the opportunity to switch again.  If your reasoning is really correct, then you'll make another (expected) 25% profit if you switch a second time.  In fact, if we just keep switching back and forth forever, that analysis leads to the conclusion that you should have an unbounded expected profit from our experiment.

An interesting, somewhat related question:  let's play a game.  I'm going to pick 2 numbers from the real line, according to a process that you aren't allowed to know, and tell you one of them.  Your goal is to try and figure out which is the larger of the two numbers.  You may use any strategy you choose, but you have to (honestly) tell me what it is, and I am allowed to adopt my own strategy to counteract yours.  Give a strategy that will let you win on average more than 50% of the time.

A similar, more classical discrete problem, with apologies for political incorrectness: a young man is to choose a wife from among 1,000 possible brides.  Each bride has a distinct dowry, measured in some unit with which the man is unfamiliar.  The potential brides will walk into his room, one at a time, and tell him the value (in this unit with which he is unfamiliar) of their dowries.  Upon hearing each dowry, he may either choose to marry the woman currently in the room or to send her away and bring in the next.  If his goal is to marry the woman with the largest possible dowry, give a simple strategy that he can use that will allow him to pick correctly with at least 25% probability.  Can you do even better?

The thing is that you know 100 dollars is either the x or the 2x so their is a probability distribution.  There is only 2 possible outcomes.  This is a paradox because your solution tells you to switch every time, but if you had gotten the smaller of the two, switched twice, then you would have lost the profit you would have gotten if you had switched once.

In your problems, we are talking about $\infty$ so you could not possibbly have a way of saying if it is higher or lower.  You could say that, if the number said is greater than 0, then it is the bigger number.  The probability cannot be defined that it is because there are $\infty$ numbers greater than the number and $\infty$ numbers less than the number.

I see where you are coming from, but when we talk about these paradoxes and $\infty$, there really can be no solution that we could possibly prove since $\infty$ is essentially forever.

I understood the problem.  I was giving another problem where it was even more clear that there is not enough info.

The key question, as Joel said, is how x is chosen.  If x is fixed at 50, for example, then we shouldn't switch.  If x is fixed at 100, then we should switch.  If x is chosen to be a random integer from 1 to 75, then we shouldn't switch.  If x is chosen to be a random integer from 1 to 100, then we should switch.  If x is chosen to be a positive integer n with probability $\frac{1}{2^n}$, then we shouldn't switch.  And there are infinitely more scenarios we could devise for how x is chosen.  Because we get different answers depending on the scenario, the correct answer is that there is not enough information given.

I agree that before we saw the 100 dollars, we had a 50-50 chance of holding the smaller amount.  But once we saw there was 100 dollars, we have a conditional probability, and the probability of holding the smaller amount is no longer necessarily 1/2.  What the conditional probability is depends on how x is chosen.  It's true that at that moment we know that there are only two possible outcomes (x = 50 or x = 100), but that doesn't mean the two outcomes each have probability 1/2.

Tomorrow, I might win the Nobel prize, or I might not, but that doesn't mean my probability of winning is 1/2.  :)

For your first paragraph:  yes, there are two outcomes, and depending on the probability distribution with which the values in the envelopes were chosen, the chance that you hold the x will be greater, equal to, or less than 50%.

For your second paragraph, you are wrong.  There do exist strategies that will work.  In fact, you came rather close to one of them with your comment about 0.  If I were using a distribution that gave me equal likelyhood of choosing x and -x, then the strategy "take the number if it's positive, take the other number if it's negative" gives you better than 50% likelyhood of getting the larger number.  (75%, in fact -- not bad!)  However, since I get to know your strategy, I could always just pick two positive numbers and give you the smaller one, giving you a chance of 0%.  But you have the right kind of idea.

For your last comment, I would suggest not worrying too much about the meaning of infinity :)  You could probably drag up several old conversations about it with a search of this forum, if you were so inclined.

Explain how if it is positive choose it, if it is negative choose the other gives you a probability of 75%

Say that, whatever my distribution is, I have an equal probability of each number being positive or negative and that I will show you one of the two numbers at random.
This gives me a 25% chance that both numbers are negative.  If you get the larger, you'll be wrong, and if you get the smaller, you'll be right.  This gives you 12.5% total contributed from this case.
There is a 25% chance that both are positive.  If you get the smaller, you'll be wrong, but if you get the larger, you'll be right.  That contributes another 12.5%.
Now, there is a 50% chance that one is larger than 0 while one is smaller.  In this case, you will always be right (as you'll take the positive one no matter which I give you).  So your total probability of being right is 12.5% + 12.5% + 50% = 75%.

[quote="Ravi B"]If we know, for example, that the amounts were 50 dollars and 100 dollars, then we can solve the problem.  If we know that the amounts were x dollars and 2x dollars, where x is a random integer from say 1 to 100, then again we can solve the problem.[/quote]

Your second example, "If we know that the amounts were x dollars and 2x dollars, where x is a random integer from say 1 to 100," is how you should interpret the question. There is no restriction that x must be between 1 and 100 - in fact there is no restriction.

So now it should be solvable, as you yourself said.

[quote]Tomorrow, I might win the Nobel prize, or I might not, but that doesn't mean my probability of winning is 1/2. [/quote]

Excellent!

By the way, a lot of these types of "paradoxes" are described in Mosteller's [url=http://www.amazon.com/exec/obidos/tg/detail/-/0201048671/qid=1114465847/sr=1-21/ref=sr_1_21/103-5240224-0658208?v=glance&s=books]book.[/url]

[quote="Ravi B"]There is a big difference between the phrases "a random integer from 1 to 100" and "a random integer".  The first phrase has meaning (a uniform distribution on 1..100, where each such integer has probability 1/100).  The second phrase has no meaning (unless you specify a probability distribution).  How do you (or the person who stuffs the envelope) define or create a random integer?  There is no such thing as a uniform distribution on the set of all integers.

If you don't know what a probability distribution is, then you can't really understand this problem.[/quote]

Ok, then I will word it this way. A random integer between 1 and 200 for x.

Okay, now you have a well-defined question.  The answer is indeed, yes, you should switch.  The previous analysis of alan and others now applies.

More generally, the strategy of the envelope opener should be: If he sees at most 200 dollars, then he should switch; otherwise, he shouldn't switch.

(When you said "between", I assume you mean in the inclusive sense.)

The answer to Joel's original variant is that it doesn't matter whether we switch or not.


Given $2006$ points in the plane with no three colinear. One wishes to color some $1003$ points by [color=red]red[/color], the others by [color=blue]blue [/color] such that there exists a line satisfying:
1, There is no point given on this line.
2, All [color=red]red[/color] points are on the same side from this line, all [color=blue]blue[/color] points the other side.

How many ways one can color?

I'm pretty sure I have a solution but it's kinda lame.  Whats a nice solution?

One solution of this problems nearly bases on discret continuation. Given a point, you can uniquely choose another point such that the line connect two separates all points givens into 2 part equal. And...
The result I found is $1003\times 2$ (include the exchanging red and blue)


factorial geometry rectangle MATHCOUNTS parallelogram

I don't remember the answer for this, but I was wondering if anyone had a quick solution.

Alice, Bob, Cindy, Dave, Edith, Fred, Gretchen, Harry, Irene, and Jacob all sit down in a classroom, with 2 rows of 5 desks. Alice and Bob are a couple, Cindy and Dave are a couple, Edith and Fred are a couple, Gretchen and Harry are a couple, and Irene and Jacob are a couple. The couples all want to sit together, that is, next to each other or front and back. How many ways are there for them to sit down?

I got

[hide]$2^5$ * 5! * 8 = 30720[/hide]

I totally messed this question up at Nats, I somehow ended up with 5 million something, which is more than 10!.

I'm pretty sure this was #8, I think #6 was the change question, anyway, your answer is correct.
[hide]8 ways for couples to sit down, 5! to pick where the couples sit, 5C0 + 5C1 + 5C2... for them to be arranged, so 8*120*32 or 30720. [/hide]

I got it wrong because i did 8*5*2...he-heh, forgot factorial and fifth power... I felt really stupid
and yes, this was number 8.

[quote="mathgeniuse^ln(x)"][quote="Ignite168"][quote="mathgeniuse^ln(x)"]Don't remind me about this target.  I could've gotten an 8/8, or a 7/8.

The questions I got wrong were #3 and #8 on Target.  I blanked out on #3 because I didn't see something important in the problem.

I got #8 wrong because I forgot to multiply by 5!. :mad:[/quote]
I got #8 wrong because I ran out of time as I didn't know that xC0 + xC1 + ... xCx = 2^x[/quote]

Just wondering, but why would you need that in the problem? ;)[/quote]

Well there are 5 couples to choose whether to swap seats or not, you can choose 0, 1, 2, 3, 4, or all 5. So I was busy calculating them for all 6 when time ran out.

I forgot the $2^5$. :wallbash:  :oops_sign:

[hide="my complete solution"]

First, we want to find the arrangements of "dominoes" so to speak, or equivalently, lines.

By looking at a 5 by 2 rectangle, we see that all can be vertical: ||||| and that if one is horizontal, another pair has to be horizontal too: =

So, we have the following arrangements for the lines:

||||| (1)
|||= (2)
||=| (3)
|=|| (4)
=||| (5)
|== (6)
=|= (7)
==| (8)

So we have 8 ways to arrange the lines. In order to get the total seating arrangements, notice that any of the 5 pairs can be seated at each of the lines, so we have to multiply by $5!$, and that any of the 5 pairs can be switched, so we have to multiply by $2^5$ (each pair is either "switched" or "not switched").

This gives our final answer of $5! \cdot 2^5 \cdot 8 = 30720$
[/hide]

I really hate myself for target.  #1. Forgets how to count (starts using advanced techniques, then after time is called, realizes that you could just count them...)

#2.  x=8, y=6 (it's written on my paper).  3*8 + 4*6 + 2(6+8) = 24 + 24 + 2(10) = 68.

#3.  right

#4. RIght

#5. right

#6. Rong.

#7. Very rong

#8. I go through I whole counting book so I can get these problems, the reward, RONG.

Are you implying something about "Rong"?

Anyways, how did you get #7 "wrong"?

It was just two concentric circles the area between them is shaded in, find the radius of the circle with that exact same area.

[quote="SplashD"][quote="jclarkemathL314159"]First, we determine, by fibonacci, that there are 8 distinct arrangements of the couples (so that they sit together), regardless of who is actually sitting in the desks. This is determined by drawing a 2-by-5 rectangle, then putting in 1-by-2 rectangles. We then find that there are 5! ways for the 5 couples to occupy the smaller rectangles, and there are two arrangements within each couple.

Thus, there are $2^5 * 5! * 8 = 30720$ ways for everyone to sit.[/quote]

by fibonacci? :?[/quote]Yes, it's known that the number of ways to tile a $2 \times n$ rectangle with dominos is a Fibonacci number.  That's because on the left end, you can start with either a vertical domino or two horizontal dominos.  In the first case, you are left with a $2 \times n-1$ rectangle.  In the second case, you are left with a $2 \times n - 2$ rectangle.

By the way, Target #1, the question about open drawers, can also be killed with Fibonacci numbers.  Do you see how?

I was surprised to see two Target questions that can both be solved with the same Fibonacci technique.

Oh, really, I can't see it yet, but I will find out.

I just did $\binom{5}{1}+\binom{4}{2}+\binom{3}{3}=12$, and that was the correct answer. :lol:

hunh I actually thought this target round was easy, I didn't have to think too much on any of the problems, except maybe the problem with the change and the seating one...

On the other hand I thought the Sprint was incredibly tough, so it all evens out .-.

I actually did it a different way after getting 8 - I did 10 * 1 * 8 * 1 * 6 * 1 * 4 * 1 * 2 * 1, then times 8. It's the same thing as $5!(2^5)(8)$, but it's slightly different...

[quote="Treething"]I did it by counting, but in problems such as this one you can find the n/2-th fibonacci number, where n is the amount of people in the group. This number will be the amount of ways they can be seated like so.[/quote]

The 5th fibonacci number is 5 though.

I guess the sequence goes like

1,2,3,5,8,13,...

I remember going over to michigan and hearing allen (and neil) ranting about the dumb mistakes he made...

good times. :)

I hate you.
It's wasn't like we did that bad.
But, if I just got that one target question about the tests, then I would've probably made it to countdown.

Nice number of posts carl, you right now have 173.

Approximately $100\sqrt3$

If I just got a decent score (not 3), I would have been in the top 57.  If I had gotten "An excellent" score, I could have been in countdown.

[quote="XxHarryxX"]Yeah, well, if I had learned how to count to 50, I would've been in the top 57. (came up with 321 on the multiples of 3 target instead of 318...)

Oh well, at least I went to nationals twice!  :P[/quote]

Yay, I did too.

[quote="mathgeniuse^ln(x)"][quote="jclarkemathL314159"]I'm not saying you did bad...it was just funny because you made it sound like you did bad.

Neil, if you got 8/8 on target you would have been 2nd or 3rd. On the other hand, if I got 8/8 on target, I  would probably be around 40th.[/quote]

Actually my score would've been (23/16), but it wasn't, so I would've been either 5th, 6th, or 7th.

But, I mean I knew how to do at least 3 of the Sprint Problems that I got wrong and both of the Target I got wrong.

I am going to take that test and see how well I did.[/quote]

I knew how to do at least 8 of the problems on the Sprint I got wrong such as the 3 I got wrong in the first 10, a lot of the problems I skipped as I thought they would take too much time, etc. (I got 16 on sprint) and all the target problems I got wrong.
That would've given me a bye in the first round of the countdown.

Yeah, the questions I got right.

Sprint
1,2,3,4,5,6,7,8,9,10,11,12,13,_,15,_,17,18,19,_,21,_,23,_,_,26_,,28,29,30.

Target
1,2,_4,5,6,7_.

Team
1,2,3,_,_,6,7,8,9_.

Countdown.
I am not even going to answer.
If you count the number of questions that I got unofficial, I got about 95% of the questions right. grr.

Well good luck to all of you guys next year, particular Michigan (cough bias cough).

[quote="Ignite168"]Mine:
1, 2, 3, _, 5, 6, _, 8, _, 10, 11, 12, _, 14, _, _, 17, 18, 19, 20, _, _, _, _, _, 26, 27, _, _, _

I spent 15-20 minutes working on 2 problems, (the one with the boards and the one with tan pink brown tiles, I broke it down into 6 cases and got 5 denominators as the same and 1 denominator as different and started working on changing it into 1 fraction and got a very ugly number, I should've just assumed they all had the same denominator. I ended up getting that one wrong and the boards one right. The boards one I have no idea why it took so long, when I checked, I just kept getting different answers so I had to do it like 7 times but I ended up getting it right)

1, 2, 3, 4, 5, 6, 7, _
Ran out of time adding 5C0, 5C1, 5C2.. 5C5, as I didn't know xC0 +.. xCx = 2^x

1, 2, 3, _, 5, 6, 7, 8, _, _
Did I mention that we had the answer and first number of #9 written down but started arguing with around 10 seconds left and then time was called? I'm still pissed over that.

Countdown
Didn't get 1 problem wrong, was undefeated[/quote]

Even the problem that Neal Wu said 252, but the answer was 251.  I mean I got that right, but yeah.

If I had just concentrated on number 3, I would've made myself look good at nationals, but whatever.

[quote="Treething"][quote="mathgeniuse^ln(x)"]
Even the problem that Neal Wu said 252, but the answer was 251.  I mean I got that right, but yeah.

If I had just concentrated on number 3, I would've made myself look good at nationals, but whatever.[/quote]

lmao, I said like "69? idk"[/quote]

Actually you said 126, not noticing that order does matter in the problem, but it is not like that matters since you won the entire thing.

It's funny, if the referees in soccer were the referees of Mathcounts they probably would give a red card (not a yellow card) to Treething for timewasting, as many people already have
in the World Cup, but they received Yellow Cards.

Combinatorics killed me on sprint.....

I think I got something like

1, 2, 3, 4, 5, 6, 7, _, 9, 10, _, _, 13, 14, 15, _. _. _. 19. _. _. _. 23, _, _, _, 26, _, _. _, _.

In the last 15 questions, abouit 8 were combo, and I got them all wrong.

Bad day.
I swear, I don't remember any questions (except for 1 target) and I'll probably get at least 32/33 once I take it again.

Actually I am not sure if those are the exact question numbers I got wrong, but I think those are.

I got one with surface area, one with finding the number of points that make it an isosceles triangle, one with a fraction relating to boys and girls, one with splitting MATHCOUNTS, one with finding sum of mode and median something like that, one with finding number of ways to brick a house, and one with two boards overlapping eachother find the area of the intersection.

I should've gotten some of these right.

[quote="Treething"]brickahouse one was easy, I made a list, it turned out to be around 80-something so it wasn't that bad

board thingy was just a parallelogram (parallelagram? paralelogram?) i thihnk[/quote]

I know I got the answers I put 48 instead of 84 for the first one.  Yes, I was dyslexic. :mad: I had 84 written down too but I just read it wrong.

The second one was $16\sqrt{3}$, I also got that but I multiplied the answer by 3 when I shouldn't have.

If I only I could reverse time, not really, but yeah.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

A cute result: in a noetherian ring, any radical proper ideal is a finite intersection of prime ideals.

i'm probably terribly wrong

bus isn't the radical the intersection of all prime ideals containing it, and you can eliminate these to a finite number cause it's a noetherian ring

I agree with the first part, however I don't think it's so obvious you can eliminate them.

it's indeed not that obvious; again I will probably say something incorrect

We have that $\sqrt J=\bigcap_{P \supset J}P$. Since $R$ is Noetherian every ideal is finitely generated. So let $\sqrt J = (a_{1},...,a_{n})$. Now let $\{P_{i}\}_{i \in I}$ ($I$ some indexed set) be the prime ideals containing $J$. By assumption $\sqrt J=\bigcap_{i \in I}P_{i}$. We want to show that we can find a finite subset of $I$ such that $\sqrt J= P_{1}\cap .... \cap P_{n}$. Certainly $\{a_{1},...a_{n}\}\subset P_{i}$ for all $i \in I$. All these $P_{i}$ are finitely generated too. So assume that $P_{i}=(a_{1},...,a_{n},b_{i1},...b_{ik_{i}})$. If we really need all $P_{i}$ to form $\sqrt J$ (if they would indeed overlap) then the $b_{ij}$, $j=1,...k_{1}$ and $i$ fixed, can't belong to all of the $P_{i}$. So this gives us a chain $(a_{1},...,a_{n}) \subset (a_{1},...,a_{n},b_{11},...b_{1k_{1}}) \subset (a_{1},...,a_{n},b_{11},...b_{1k_{1}},b_{21},...b_{2k_{2}}) \subset$ ... which contradicts the A.C.C. So there indeed must be a finite subset of $I$ such that $\sqrt J= P_{1}\cap .... \cap P_{n}$.

p.s. I'm going to be honest. I'm new to Noetherian rings and comm algebra and some things were based on intuition :P

In a Noetherian ring every ideal $\frak a$ has a [url=http://planetmath.org/encyclopedia/PrimaryDecomposition.html]primary decomposition[/url], and hence can be written as a finite intersection of primary ideals $\frak q_{i},\ i=\overline{1,n}$. If you take radicals in the identity $\frak a=\bigcap\frak q_{i}$, you find that the radical of $\frak a$, which coincides with $\frak a$ when $\frak a$ is radical, is the intersection of the radicals of $\frak q_{i}$, all of which are prime.

What I did was to consider a maximal ideal for which it is not true. It's clearly not prime and then use some $x,y$ for which $xy\in I$ and $x,y$ are not in $I$ to obtain contradiction, by working with the larger ideals $\sqrt{I+(x)}$ and $\sqrt{I+(y)}$, which are still proper and radical.

there is generalization due to Anderson which says:Let $R$ be a communtative ring and $I$ an ideal of $R$. We show that if all the prime ideals minimal over $I$ are finitely generated, then there are only finitely many prime ideals minimal over $I$.


Minimize \[ h^T R h\] subject to \[ C h \equal{} d\] and $ h\in \mathbb{R}^N$.
Here $ R$ is a real $ N\times N$ matrix, $ C$ is a $ M\times N$ complex matrix. So I want to find $ h$ which solves the above equality constrained least squares problem and I want $ h$ always to be real. Any idea? Please help me!

bhupala!

We minimize $ f(x)\equal{}x^TRx$ subject to $ g_1(x)\equal{}L_1x\minus{}k_1\equal{}0,g_2(x)\equal{}L_2x\minus{}k_2\equal{}0$ (Considering real part and imaginary part of $ C,d$). Then the condition is $ g(x)\equal{}Lx\minus{}k\equal{}0$ where $ L$ is a $ 2M\times{N}$ matrix.
Thus there exists $ u\in\mathbb{R}^{2M}$ s.t. $ df\plus{}u^T{dg}\equal{}0$ or for all $ h,h^TRx\plus{}x^TRh\plus{}u^T{Lh}\equal{}0$. We obtain the system $ x^TR^T\plus{}x^TR\plus{}u^T{L}\equal{}0,Lx\equal{}k$ which is a linear system of $ N\plus{}2M$ equations with $ N\plus{}2M$ unknowns $ x,u$. The rest is your problem. :P


quadratics
inequalities
algebra
polynomial
Vieta
USA TST
number theory

A solution, including all the process that I got there with:

Part 1: Experiments.
Since I didn't see anything right away, I just looked for any $ m$ that worked with small values of $ n$. The results are here:
$ \begin{array}{ccl}n&n^2\minus{}n\plus{}1&\text{Pairs}\\1&1&(2,1)\\2&3&(1,2),(2,2),(5,2)\\3&7& \\4&13& \\5&21&(2,5),(10,5)\\6&31& \\7&43& \\8&57& \\9&73& \\10&91&(5,10),(17,10)\end{array}$

Those pairs $ (5,2),(2,5)$ and $ (10,5),(5,10)$ looked suspicious. I tested $ (10,17)$; $ 17\cdot 10\minus{}1\equal{}13^2$ divides $ (289\minus{}17\plus{}1)^2\equal{}(13\cdot 21)^2$

Part 2: OK, we now have a plan. Conjecture: if $ (m,n)$ works, $ (n,m)$ does as well. Formally, we want to show that if $ mn\minus{}1$ divides $ (n^2\minus{}n\plus{}1)^2$, it also divides $ (m^2\minus{}m\plus{}1)^2$.
We do some algebra mod $ mn\minus{}1$:
$ 0\equiv (n^2\minus{}n\plus{}1)^2\equiv(n^2\minus{}n\plus{}1\plus{}mn\minus{}1)^2\equiv(n^2\plus{}mn\minus{}n)^2\equiv n^2(m\plus{}n\minus{}1)^2$
Since $ n$ is invertible mod $ mn\minus{}1$ (its inverse is $ m$), $ (m\plus{}n\minus{}1)^2\equiv 0\mod mn\minus{}1$.
Similarly, $ (m^2\minus{}m\plus{}1)^2\equiv (m^2\plus{}mn\minus{}m)^2\equiv m^2(m\plus{}n\minus{}1)^2$; by the preceding, this is zero, and the conjecture is proved. Call it Lemma 1.
I originally started by working with $ (m^2\minus{}m\plus{}1)^2\minus{}(n^2\minus{}n\plus{}1)^2$; that was dropped for clarity, after it led me to the above manipulation.

Part 3: Now, we use Lemma 1 to build a ladder. In building the table, we notice that the solutions come in pairs at each $ n\ge 2$, corresponding to complementary divisors. Formalize this as Lemma 2:
Lemma 2: If $ (m,n)$ is a solution, $ (m',n)$ is also a solution, where $ (n^2\minus{}n\plus{}1)^2\equal{}(mn\minus{}1)(m'n\minus{}1)$. Moreover, if $ n\ge 2$ and $ m<n$, $ m'\ge n$, and if $ n\ge 2$ and $ m>n$, $ m'\le n$.
Proof of Lemma 2: By hypothesis, $ \frac{(n^2\minus{}n\plus{}1)^2}{mn\minus{}1}$ is a positive integer. Mod $ n$, this positive integer is $ \frac{1}{\minus{}1}\equiv \minus{}1$, and can be expressed as $ m'n\minus{}1$ for some $ m'>0$. If $ m<n$ and $ n\ge 2$, $ mn\minus{}1\le (n\minus{}1)n\minus{}1<n^2\minus{}n\plus{}1$ and thus $ m'n\minus{}1> n^2\minus{}n\plus{}1$, making $ m'\ge n$. If $ m> n$ and $ n\ge 2$, $ mn\minus{}1\ge n^2\plus{}n\minus{}1>n^2\minus{}n\plus{}1$ and thus $ m'n\minus{}1< n^2\minus{}n\plus{}1\le n^2\minus{}1$, making $ m'\le n$.

By Lemma 2, if $ m>n$ and $ n\ge 2$, we can make a smaller solution with the same $ n$. By Lemma 1, if $ m<n$, we can make a solution with smaller $ n$. Any solution thus descends uniquely to a solution with $ n\equal{}1$ (which must be $ (2,1)$ by inspection), unless it hits a solution with $ m\equal{}n$ somewhere. We close that loophole with the next lemma.
Lemma 3: The only solution of the form $ (n,n)$ is $ (2,2)$.
Proof: Mod $ n^2\minus{}1$, $ (n^2\minus{}n\plus{}1)^2\equiv (\minus{}n\plus{}2)^2\equiv n^2\minus{}4n\plus{}4\equiv \minus{}4n\plus{}5$. In order to have $ n^2\minus{}1$ divide $ 4n\minus{}5$, we must have $ n^2\minus{}1\le 4n\minus{}5$, or equivalently $ n^2\minus{}4n\plus{}4\equal{}(n\minus{}2)^2\le 0$. That makes $ n\equal{}2$ the only possibility. It is easily confirmed to work. This argument fails where $ 4n\minus{}5<0$, but in that case $ n\equal{}1$ and $ 0$ does not divide $ 1$.

Part 4: Enumerating the solutions.
The solution $ (2,2)$ stands alone.
We also have an infinite family of solutions of the form $ (a_{k\minus{}1},a_k)$ or $ (a_{k\plus{}1},a_k)$, where $ a_k$ is the infinite increasing sequence $ (1,2,5,10,17,\cdots)$. We would like to give a rule for this sequence. A little guesswork suggests that these are the squares plus 1; that should be easily testable.
$ \left((k^2\plus{}1)^2\minus{}(k^2\plus{}1)\plus{}1\right)^2\equal{}(k^4\plus{}k^2\plus{}1)^2$
$ \equal{}(k^2\plus{}k\plus{}1)^2(k^2\minus{}k\plus{}1)^2$

$ (k^2\plus{}1)((k\minus{}1)^2\plus{}1)\minus{}1\equal{}(k^2\plus{}1)(k^2\minus{}2k\plus{}2)\minus{}1$
$ \equal{}k^4\minus{}2k^3\plus{}3k^2\minus{}2k\plus{}1\equal{}(k^2\minus{}k\plus{}1)^2$

$ (k^2\plus{}1)((k\plus{}1)^2\plus{}1)\minus{}1\equal{}(k^2\plus{}1)(k^2\plus{}2k\plus{}2)\minus{}1$
$ \equal{}k^4\plus{}2k^3\plus{}3k^2\plus{}2k\plus{}1\equal{}(k^2\plus{}k\plus{}1)^2$
They do work, so $ a_k\equal{}k^2\plus{}1$. Done.

[quote="MellowMelon"]Find all pairs of positive integers $ (m,n)$ such that $ mn \minus{} 1$ divides $ (n^2 \minus{} n \plus{} 1)^2$.

[i]Source unknown.[/i][/quote]
I got an other solution but I also use Viete jump method . From the condition we have $ mn \minus{} 1|(n^2 \minus{} n \plus{} 1)^2$ and $ mn \minus{} 1|(m^2 \minus{} m \plus{} 1)^2$ , therefore $ (mn \minus{} 1)^2|[(n^2 \minus{} n \plus{} 1).(m^2 \minus{} m \plus{} 1)]^2$ or $ mn \minus{} 1|(m^2 \minus{} m \plus{} 1)(n^2 \minus{} n \plus{} 1)$ (*).
Notice that $ mn\equiv 1 (\mod mn \minus{} 1)$ , from (*) we obtain the fact that :$ mn \minus{} 1|(m \plus{} n \minus{} 1)^2$ .
[b]Lemma[/b] : Let $ k \equal{} \frac {(m \plus{} n \minus{} 1)^2}{mn \minus{} 1}$ then $ k\in \{3,4\}$ . We will use Viete jump method here . Let $ (m_0,n_0)$ be the smallest positive integer root of this equation , we can assume that $ m_0\geq n_0$ . The equation can be rewritten in the form :
\[ m^2 \minus{} m(kn \plus{} 2 \minus{} 2n) \plus{} (n \minus{} 1)^2 \plus{} k \equal{} 0\]
This is quadratic equation of $ m$ so it has an other solution $ m_1$ . By the Viete law , we have $ m_o \plus{} m_1 \equal{} kn \plus{} 2 \minus{} 2n$ , notice that $ m_0$ is an integer so $ m_1$ is . We also have :
$ m_1 \equal{} \frac {(n_0 \minus{} 1)^2 \plus{} k}{m_0}$
By the definition of $ m_0$ ,we obtain the fact that $ m_0\leq m_1 \equal{} \frac {(n_0 \minus{} 1)^2 \plus{} k}{m_o}$ . Expand this inequality ,we have :
\[ k\geq (m_0 \minus{} n_0 \plus{} 1)(m_0 \plus{} n_0 \minus{} 1) (**)\]
We have two case here :
[i]Case 1[/i] : $ m_0 > n_0$ ,then we have $ k\geq 2(m_0 \plus{} n_0 \minus{} 1)$ . Replace it to (*) ,we have :
\[ \frac {(m_0 \plus{} n_0 \minus{} 1)^2}{m_0\dot n_0 \minus{} 1}\geq 2(m_0 \plus{} n_0 \minus{} 1)\]
This inequality is equivalent to the following inequality :
\[ m_0 \plus{} n_0 \minus{} 1\geq m_0n_0 \minus{} 1 \leftrightarrow (m_0 \minus{} 1)(n_0 \minus{} 1)\leq 1\]
Obvious this inequality holds iff $ (m_0,n_0) \equal{} (m_0,1)$ (pair $ (m_0,n_0) \equal{} (2,2)$ also satisfies condition ,but notice that we are considering $ m_0 > n_0$ ) . When $ n_0 \equal{} 1$ ,we have $ m_0 \equal{} 2$ and $ k \equal{} 4$ . That ends case 1 .
[i]Case 2 [/i] $ m_0 \equal{} n_0$ . From the condition we have $ m_0^2 \minus{} 1|(2m_0 \minus{} 1)^2$ , it is not very hard to show that only $ m_0 \equal{} 2$ satisfies this condition . When $ m_0 \equal{} n_0 \equal{} 2$ we have $ k \equal{} 3$
Thus ,lemma has been proven . We will solve equation for $ k \equal{} 3$ and $ k \equal{} 4$.
For $ k \equal{} 4$ ,the equation can be rewritten in the form :
\[ (m \minus{} n)^2 \minus{} 2(m \plus{} n) \plus{} 5 \equal{} 0\]
Let $ m \minus{} n \equal{} 2t \plus{} 1$ then we have $ m \plus{} n \equal{} 2t^2 \plus{} 2t \plus{} 3$ .Solve this equation we have $ m \equal{} t^2 \plus{} 2t \plus{} 2$ and $ n \equal{} t^2 \plus{} 1$ . It is not hard to check that pair $ (m,n) \equal{} (t^2\plus{}2t\plus{}3 1,t^2\plus{}1) ,t\geq 1$ satisfies condition .
For $ k \equal{} 3$ , the equation only has $ (m_0,n_0) \equal{} (2,2)$ as solution ,this equation can be solved by some inequality ...

You've got the answer wrong; both $ m$ and $ n$ should be quadratic in $ t$.

The difference here: my method jumps from $ (a_k,a_{k\minus{}1})$ to $ (a_{k\minus{}1},a_k)$ and then to $ (a_{k\plus{}1},a_k)$, while yours goes from $ (a_k,a_{k\minus{}1})$ to $ (a_k,a_{k\plus{}1})$ and then to $ (a_{k\plus{}2},a_{k\plus{}1})$. You only get half of the solutions directly that way, so you need to do it twice. The bit "we can assume that $ m_0\ge n_0$" is false for your argument.
It would probably have helped to actually write out the argument that you have everything.

jmerry,

  TTsphn's solution is correct. His $ m$ and $ n$ are indeed quadratic(Take a careful look at his $ m$ and $ n$). The only 'silly' mistake he made was assuming $ m\minus{}n \equal{} 2t\plus{}1$. Instead, he should have assumed $ m\minus{}n \equal{} 2t\minus{}1$. Then $ m \plus{} n \equal{} 2t^2 \minus{}2t \plus{}3$ which gives $ m\equal{} t^2 \plus{}1$ and $ n\equal{} (t\minus{}1)^2 \plus{} 1$.  Then his solution also produces the same set of solutions.

TTsphn edited his post after I wrote that. It's right now, aside from the silly typo of "3 1" rather than "3-1".
It does still only list half the solutions, and misses the base solution $ (2,1)$ as well.

[i]A good solution , Tung , congratulation :lol:

  But i think , that problem is a easy problem ,The trick used to solve has become very familiar[/i]

de Vieta jump method will work on following problems :[b]
Problem 1[/b]: Let $ p;q$ be positive integers such that $ p\ge q$.
Find all natural numbers $ a;b$ which satisfy $ \frac {(a \minus{} b)^2}{pab \minus{} q} \equal{} k (k\in \mathbb N)$
[b]Problem 2[/b]: If there exists $ a;b \in \mathbb Z^{ \plus{} }$ satisfying $ \frac {(a \minus{} b)^2 \plus{} m}{pab \minus{} q} \equal{} k (m \in \mathbb N^*)$ with $ p;q$ are defined above then $ k$ (its value) belongs to a finite set.

[quote="MellowMelon"]Find all pairs of positive integers $ (m,n)$ such that $ mn \minus{} 1$ divides $ (n^2 \minus{} n \plus{} 1)^2$.

[i]Aaron Pixton.[/i][/quote]
An old problem  :D
I had wirte condition $ mn \minus{} 1$ divides $ (n^2 \minus{} n \plus{} 1)^2$ equivalent $ mn\minus{}1$ divides $ (m\minus{}1)^2\plus{}(n\minus{}1)^2\plus{}1$
Put:$ a\equal{}x\minus{}1$,$ b\equal{}y\minus{}1$,we have :$ (ab\plus{}a\plus{}b)|(a^2\plus{}b^2\plus{}1$.
Put:$ a^2\plus{}b^2\plus{}1\equal{}k(ab\plus{}a\plus{}b)$ (1) . Using  Vieta jump method we have $ k\equal{}2$
When:(1) equivalent $ (a\plus{}b\minus{}1)^2\equal{}4ab$.(2)
We have :$ gcd(a;b)\equal{}1$ so $ a\equal{}k^2;b\equal{}(k\plus{}1)^2$
If equation (2) be equation $ (a\plus{}b\minus{}5)^2\equal{}9ab$ then it is problem 2,USATST 2001,Day 3  :)

[color=#c00][b]A brief tutorial for my solution:[/b][/color]

Fully work and ignore the trivial solutions (namely $(2,2)$). Check the symmetry and ignore duplicated solutions.

As you say $(m,n)$ is the solution with minimal $2< n < m$ you will try to find $k < n$ such that $(n,k)$ is also a solution. You need $(n^2-n+1)^2/N \equiv -1/1 \pmod n$ with $x\geq n^2-1$. Suppose that such $x$ doesn't exist. It yelds the chain of inequalities
$$n^2-1 \leq \frac{(n^2-n+1)^2}{mn-1} \leq \frac{(n^2-n+1)^2}{n^2-1} \implies $$
$$\frac{(n^2-n+1)^2}{n^2-1} \geq n^2-1 \iff n^2-n+1 \geq n^2-1 \iff n \leq 2$$
and you obtained that all solutions are given by recurrences particular to a limited amount of solutions. More precisely

[color=#38761D][b]Lemma 1.[/b][/color] Fixing particulars $a_0$ and $b_0$ with $b_0 \leq 2$, all the solutions are given by,
$$(a_i,b_i)_{i=0}^{\infty}\ |\ a_{i} = b_{i+1},\ b_{i} = \frac{1}{b_{i+1}}\left[ \frac{(b_{i+1}^2 - b_{i+1} + 1 )^2}{a_{i+1}b_{i+1}-1} +1 \right] \ \forall \ i \in \mathbb{N}^*.$$

[b]Remark:[/b] You can go over the solutions you find with $a_0$ and $b_0$ satisfying $n \leq 2$.

Working on Lemma 1 you find that the solutions are $(2,2)$ and $(m,n) = (b_{\ell-1},b_{\ell}), \ (b_{\ell},b_{\ell+1})\ \forall \ \ell \in \mathbb{N}^*$. Cleaning up Lemma 1 and adding the minimal solutions (notice that $(2,2)$ does not increase solutions in recurrence)
$$b_{i+2} = \frac{b_{i+1}^3-2b_{i+1}^2+3b_{i+1}+b_{i}-2}{b_{i}b_{i+1}-1}\ \forall\ i \in \mathbb{N}^*,\ b_0 = 2,\ b_1 =5$$
which yields the sequence $(b_i)_{i=0}^{\infty} = 2,5,10,17,\dots$ and it is not hard (or maybe not) to guess
$$b_i = (i+1)^2+1$$
and you are done as you proved that this contains all the solutions. $\blacksquare$

Hi, I'm reading TTsphn's Vieta Jump method, and I do not understand why $m-n=2n$ or why jmerry would say that it should be $2n-1$. Maybe there's a step in the solution that I overlooked, but to me, this implies that $m-n$ is odd, and where did that come from?

We can reduce this to $$mn -1 \mid (m+n-1)^{2}$$
Now for this condition, we will vieta jump to get the minimal pair $(m,n)$ for some fixed value of $k$, where $$k = \frac{(m+n-1)^{2}}{mn -1}$$
Now consider the pair with minimal $m+n$ which satisfies this, assume $m>n$, now vieta jump to get the other root $m'$, which will be $m' = (k-2)n + 2 - m = \frac{(n-1)^{2}+k}{m}$, note that $m' \geq m$, so we get that $$m^{2} - (n-1)^{2} \leq k = \frac{(m+n-1)^{2}}{mn -1}$$, now simplify to get that $$(m-n+1) \leq \frac{m+n -1}{mn -1} \leq \frac{2}{n}$$, simplify further to get that $$n(m-n+1) \leq 2$$, through this, we get that the minimal pairs that for some fixed value of $k$ are $(m,n) = (2,2),(2,1)$, so the values of $k$ are $3,4$, for $k=4$, the solutions are of the form $((t+1)^{2}+1,t^{2}+1)$ and for $k = 3$ the only solution that works is $(2,2)$, checking both is just a matter of algebra and vieta jumping, so we're done.

[quote=anurag27826][u]We can reduce this to $$mn -1 \mid (m+n-1)^[/u]{2}$$
Now for this condition, we will vieta jump to get the minimal pair $(m,n)$ for some fixed value of $k$, where $$k = \frac{(m+n-1)^{2}}{mn -1}$$
Now consider the pair with minimal $m+n$ which satisfies this, assume $m>n$, now vieta jump to get the other root $m'$, which will be $m' = (k-2)n + 2 - m = \frac{(n-1)^{2}+k}{m}$, note that $m' \geq m$, so we get that $$m^{2} - (n-1)^{2} \leq k = \frac{(m+n-1)^{2}}{mn -1}$$, now simplify to get that $$(m-n+1) \leq \frac{m+n -1}{mn -1} \leq \frac{2}{n}$$, simplify further to get that $$n(m-n+1) \leq 2$$, through this, we get that the minimal pairs that for some fixed value of $k$ are $(m,n) = (2,2),(2,1)$, so the values of $k$ are $3,4$, for $k=4$, the solutions are of the form $((t+1)^{2}+1,t^{2}+1)$ and for $k = 3$ the only solution that works is $(2,2)$, checking both is just a matter of algebra and vieta jumping, so we're done.[/quote]  how can we reduce this to mn-1|(m+n-1)^2 ?


Two similar rectangular solids have volumes of 96 cm :^3:  and 324 cm :^3: . The length of each side of each solid is a whole number. Find the dimensions of the two solids. Explain your work.

Before anyone answers this, kaiku, are you looking for [i]one[/i] answer? Because there is more than one answer to this question.

I know there’s more then one answer. You can find one or another. This is a problem I made up and I want to see if anyone can solve it, before testing it on a few students.

Also thanks for noticing there was more then one answer.

Okie dokie :D I just didn't want any confusion.

No problem..umm anyone want to give my question a shot?

[hide]
The ratio of the volumes is 27/8 and therefore the ratio of the dimensions is 3/2 because the cuberoot of 27/8 = 3/2. Now we only have to find factors of 96 that come out even when multiplied by 3/2. This would mean all of the even factors. 96 factored out is 2^5 * 3. From this, we can find that there are 10 even factors because anything * 2 will be even, so we need to take a 2 out of the factorization and use the property where 1 + the exponents multiplied will equal the number of factors. We only need to pick out 2 numbers, and the remaining will multiply out. This means there is more than one answer. One such answer would be 8 * 2 * 6 and 12 * 3 *9. We can find the rest by knowing that there are only 5 factors of 2 in 96. The maximum number factors of 2 that can be in any one of the dimensions is 3, because there needs to be at least 2 to be distributed among the rest of the dimensions and to make them even. The only combination that has an 8 in it would be 8, 2, and 6. With 4, there are 3 factors of 2 and 1 factor of 3 remaining. Knowing this, we can see that 4, 4, 6, and 4, 2, and 12 are the dimensions that begin with 4. Starting with 2 there are 4 factors of 2 and 1 factor of 3 remaining. We can see that there the combinations that begin with 2 are 2, 8, 6    2, 4, 12, and 2, 2, 24. We do not want to count overlaps, so there are 4 answers:<br>
8, 2, 6 and 12, 3, 9
4, 4, 6 and 6, 6, 9
4, 2, 12 and 6, 3, 18
2, 2, 24 and 3, 3, 36[/hide]

A good edit to the problem would be: How many possible combinations are there for the dimensions of the rectangular solid?

You mean how many ways can we make 2 solids with the necessary vlumes, or the congruent solids?

Basically adding to the problem: Instead of listing the dimensions, they list how many combinations there are, using the given info already provided.


hi I dont know how hard this is or in which forum it should be placed but im gonna post it here, hope its ok.

how do you derivate $(1-x)ln(1-x)$?
I initially thought it would be a product deivate but I dont seem to get it right...

thanks

why not? it is a product of (1-x) and ln(1-x). The only problem is that ln(1-x) is a composition of functions.
So you have

f(x)=(1-x)ln(1-x)

f'(x)=-ln(1-x)-1 , becouse the derivate of ln(1-x) is -1/(1-x), and you have to moltiplicate this for (1-x), becouse of the rule to derivate a product

say: $y=(1-x)ln(1-x)$

and that this means $y=f(x)g(x)$

then $y'= f(x)g'(x)+f'(x)g(x)$

furthermore $y'=(1-x)(\displaystyle\frac{1-x}{x})+(-1)(ln(1-x))$

ia this right?
that $(ln(1-x))' = (\displaystyle\frac{1-x}{x})$

or is it g(u) where u=1-x which would mean one more operation?

thanks

[quote="salah"]$(ln(1-x))' = (\displaystyle\frac{1-x}{x})$
[/quote]

no, this is wrong, becouse if $f(x)=ln(g(x))$, then $f'(x)=\frac{1}{g(x)}*g'(x)$. So, in this case we have that $g(x)=(1-x)\rightarrow g'(x)=-1$

and $(ln(1-x))'=-\frac{1}{1-x}$


$ ABCD$ is convex $ AD\parallel BC$, $ AC\perp BD$. $ M$ is interior point of $ ABCD$ which is not a intersection of diagonals $ AC$ and $ BD$ such that $ \angle AMB \equal{}\angle CMD \equal{}\frac{\pi}{2}$ .$ P$   is intersection of angel bisectors of $ \angle A$  and $ \angle C$. $ Q$ is intersection of angel bisectors of $ \angle B$ and $ \angle D$. Prove that $ \angle PMB \equal{}\angle QMC$.

Denote intersection point of diogonals,intersectıon poınt of angle bısectors $ \angle B$ and $ \angle A$ and intersection point of angle bısectors $ \angle C$ and $ \angle D$ by $ L$, $ Y$ and $ X$,respectıvely.
$ AD\parallel BC\Rightarrow\angle CXD=\angle AYB=\frac{\pi}{2}$.
So five points $ B,L,Y,M$ and $ A$ are cyclıc. As the same $ C,L,X,M$ and $ D$ are also cyclıc.Then
$ \angle XMY=\angle XML+\angle YML=\angle XDL+\angle YAL=\angle XCL+\angle YAL=\angle XPY$
$ \Rightarrow X,Y,Q,M,P$ is cyclıc.
It ıs easy to show that $ XY\parallel AD$.So
$ \angle YPQ=\angle YXQ=\angle QDA\Rightarrow PQDA$ is cyclic.
From all of these
$ \angle PMB=\angle PMY-\angle BMY$   $ (1)$
$ \angle QMC=\angle QMX-\angle CMX$   $ (2)$
$ \Rightarrow\angle PMB-\angle QMC=\angle PMY-\angle QMX+\angle CMX-\angle BMY=\angle PMX-\angle QMY+\angle XDC-\angle YAB=\angle PQX-\angle YPQ+\angle XDA-\angle YAD=0$
$ \Rightarrow\angle PMB=\angle QMC$


triangle ABC has circumradius 1
any point M on the circumcircle, MA*MB*MC<=2
show it is equilateral

The answer to this one (although nobody seems interested) can be found on Mathlinks. I'm not posting a link because I'm too lazy to look it up. I'm not even sure whether there is a proper answer there or not, but I know for a fact that this problem was discussed there. I wouldn't call it 'easy' though... :D

all ideas i've seen so far was just bashes with algebra... (thats why I call it 'easy' ;) anybody can bash with algebra


:o the convex 2000 gon M  satisfies following prp. the max. dist. bet. 2 vertices equal to 1.its known among all convex 2000 gons with same prp'
M  has max. area. show some 2 diagonals of M  are perpendicular.
prp mans property

Suppose first that some vertex of our polygon $\mathcal P$ is not adjacent to any diameters (diagonals or sides of length $1$). Then pick any diagonal having this vertex as endpoint, and move the vertex slightly, so as to elongate the diagonal. Clearly, if the elongation is small enough, all  diagonals with our vertex as endpoint stay smaller than $1$, but the area increases strictly, and this contradicts the maximality property of $\mathcal P$. We have thus shown that every vertex of $\mathcal P$.

There are now two cases: either (a) every vertex is an endpoint of at least two diameters, or (b) there is a vertex which is an endpoint of exactly one diameter.

(a)

Notice that every two diameters must intersect (in an endpoint or otherwise). We’ll cal this assertion $(*)$.

Suppose there are vertices $x,y,z$ such that $xy,xz$ are diameters, but $y,z$ are not consecutive vertices of $\mathcal P$. Then, if $t$ is a vertex between $y$ and $z$ and such that $yz$ separates $x$ and $t$, then one of the (at least two) diameters through $t$ would be disjoint from one of the two diameters $xy,xz$, and this contradicts $(*)$. This proves that in the situation we’re considering now, every vertex is an endpoint of [i]exactly[/i] two diameters, and, moreover, that whenever $xy,xz$ are diameters, $y,z$ are consecutive vertices of $\mathcal P$. It easily follows from here (just play around with a drawing) that $\mathcal P$ has an odd number of vertices, and this contradicts the fact that it actually has $2000$ vertices.

Hence, (a) cannot occur.

(b)

Let $xy$ be the only diameter having $y$ as endpoint, and let $u,v$ be vertices immediately to the left and right of $y$. If $xy$ is not orthogonal to $uv$, we can rotate $y$ around $x$ slightly such that (i) $xy=1$ throughout the process, (ii) all other distances between vertices of $\mathcal P$ and $y$ stay $<1$ (or $\le 1$; it doesn’t matter), and (iii) the angle ($<\frac\pi 2$) between $xy$ and $uv$ increases, which is easily seen to make the area of $\mathcal P$ increase strictly. This contradicts the maximality of the area of $\mathcal P$, so we have what we wanted: the diagonals (well, $xy$ could be a side and not a diagonal in the strictest sense, but I assume this is allowed, i.e. by “diagonals” you mean “sides or diagonals”? :?) $xy$ and $uv$ must be orthogonal.


calculus
integration
floor function
trigonometry
logarithms
calculus computations

Let $ n\geq 1$ be a positive integer. Find

$ I\equal{}\int\limits_{0}^{1}\frac{\left\{nx\right\}}{1\minus{}x\plus{}x^{2}}dx$,

where $ \left\{*\right\}$ denotes the fractional part of $ *$.

OK let's solve it  :wink: ;
we have $ \int_{0}^{1}\frac{nx\minus{}[nx]}{x^{2}\minus{}x\plus{}1}dx\equal{}\int_{0}^{1}\frac{nx}{x^{2}\minus{}x\plus{}1}dx\minus{}\int_{0}^{1}\frac{[nx]}{x^{2}\minus{}x\plus{}1}dx$
I am sure you can find first integral I mean this
$ int_{0}^{1}\frac{nx}{x^{2}\minus{}x\plus{}1}dx$
OK and whats about seccond ? this is main moment here
let's consider
$ \int_{0}^{1}\frac{[nx]}{x^{2}\minus{}x\plus{}1}dx$
lets do substitution:
$ nx\equal{}y$ then we have
$ \int_{0}^{n}\frac{[y]}{y^{2}/n\minus{}y\plus{}n}dy$
More general ( I don't need interval $ (0,n)$ I can't write this long expression with interval $ (0,n)$ therefore I  find without this interval ) or let's find integral
$ \int\frac{[y]}{y^{2}/n\minus{}y\plus{}n}dy$

at first let's find
$ \int\frac{1}{y^{2}/n\minus{}y\plus{}n}dy\equal{}\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}$
OK
let say $ m\leq y <m\plus{}1$
then
$ \frac{[y]}{y^{2}/n\minus{}y\plus{}n}\equal{}\frac{m}{y^{2}/n\minus{}y\plus{}n}$ where $ m \in Z$
therefore :
$ \int\frac{[y]}{y^{2}/n\minus{}y\plus{}n}dy\equal{}m\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}\plus{}C_{m}$ where $ m\leq y <m\plus{}1$
because our antyderivetive is continuoe we have equality in point $ y\equal{}m\plus{}1$ or this means that :

$ |m\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}\plus{}C_{m}|_{y\equal{}m\plus{}1}\equal{}|(m\plus{}1)\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}\plus{}C_{m\plus{}1}|_{y\equal{}m\plus{}1}$
by this you can find $ C_{m}\equal{}\alpha (m)\plus{}C$ (use fact that $ y\equal{}nx$ it will help to find $ \alpha (m)$
where $ C\equal{}C_{0}$
and in finally
$ \int\frac{[y]}{y^{2}/n\minus{}y\plus{}n}dy\equal{}m\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}\plus{}\alpha (m)\plus{}C$ where $ m\leq y <m\plus{}1$.
and because  $ y$ can be change in above interval therefore always  $ m\equal{}[y]$
thus
$ \int\frac{[y]}{y^{2}/n\minus{}y\plus{}n}dy\equal{}[y]\frac{2arctan (\frac{\minus{}n\plus{}2y}{y\sqrt{3}})}{\sqrt{3}}\plus{}\alpha ([y])\plus{}C$

Extremal,

Thank you for attaching the problem, please check your calculations,

the integral does not depend on $ n$, that is the nice part about this Riemann integral.

Yes I think so it's don't depends on $ n$. I am sure  if you calculate $ \alpha (m)$ and put instead $ y \equal{} nx$  you will take something wich will not depend on $ n$ also don't forget about interval $ |_{0}^{n}$

[quote="Virgil Nicula"]$ \int_0^1f(x)\{nx\}dx \equal{}$ $ \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_{\frac kn}^{\frac {k \plus{} 1}{n}}f(x)\{nx\}dx \equal{}$ $ \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_0^{\frac 1n}f\left(x \plus{} \frac kn\right)\{nx\}dx \equal{} 0\ .$[/quote]
I don't understand this step
$ \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_0^{\frac 1n}f\left(x \plus{} \frac kn\right)\{nx\}dx \equal{} 0\ .$
Why ? :huh:
I don't agree with this
Let $ n\equal{}1$ then you will have :
$ \int_{0}^{1}f(x)\{x\}dx\equal{} 0$
but in other hends
$ 0\equal{}\int_{0}^{1}f(x)\{x\}dx\equal{} \int_{0}^{1}f(x)xdx$ because $ x\in (0,1)$
or you mean
$ \int_{0}^{1}f(x)xdx\equal{}0$ for all $ f(x)$ ?? it's false

Let $ S(n) \equal{} \int_{x \equal{} 0}^1 \frac {\{nx\}}{1 \minus{} x \plus{} x^2} \, dx$.  Then the correct answer is

$ S(n) \equal{} \frac {\pi}{3 \sqrt {3}}$.

An elementary solution follows.

[hide]
The first observation is to partition the interval of integration in the obvious way, and note that $ \{nx\} \equal{} nx \minus{} \lfloor nx \rfloor$:

$ S(n) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} \int_{x \equal{} k/n}^{(k \plus{} 1)/n} \frac {nx \minus{} \lfloor nx \rfloor}{1 \minus{} x \plus{} x^2} \, dx$.

Then we employ the substitution $ t \equal{} nx$, $ dt \equal{} n \, dx$, to obtain

$ S(n) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} \int_{t \equal{} k}^{k \plus{} 1} \frac {t \minus{} \lfloor t \rfloor}{1 \minus{} (t/n) \plus{} (t/n)^2} \cdot \frac {1}{n} \, dt$.

Next, since we have now partitioned the interval of integration in such a way that each summand is an integral over an interval that does not contain a neighborhood of an integer, we can simply note that $ \lfloor t \rfloor \equal{} k$, giving

$ S(n) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} n I(k)$,

where $ I(k) \equal{} \int_{t \equal{} k}^{k \plus{} 1} \frac {t \minus{} k}{t^2 \minus{} nt \plus{} n^2} \, dt$.

We then find

$ I(k) \equal{} f(k)\left(\tan^{ \minus{} 1}f(k) \minus{} \tan^{ \minus{} 1}f(k \plus{} 1)\right) \plus{} \frac {1}{2}\left(\log g(k \plus{} 1) \minus{} \log g(k)\right)$,

where $ f(k) \equal{} \frac {2k \minus{} n}{n \sqrt {3}}$, and $ g(k) \equal{} k^2 \minus{} kn \plus{} n^2$.  (We leave the computation of the above as an exercise for the reader.)  Then it is easy to see that the g's form a telescoping series:

$ S(n) \equal{} \frac {n}{2}\left(\log g(n) \minus{} \log g(0)\right) \plus{} n \sum_{k \equal{} 0}^{n \minus{} 1} f(k)\left(\tan^{ \minus{} 1} f(k) \minus{} \tan^{ \minus{} 1} f(k \plus{} 1)\right)$.

But $ g(n) \equal{} g(0) \equal{} n^2$, so this part of S vanishes completely.  We now collect like terms in the inverse tangent, obtaining

$ S(n) \equal{} n\left(f(0)\tan^{ \minus{} 1}f(0) \minus{} f(n \minus{} 1)\tan^{ \minus{} 1}f(n) \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} \left(f(k) \minus{} f(k \minus{} 1)\right) \tan^{ \minus{} 1}f(k)\right)$.

Then substituting in $ f$ and simplifying, we find

$ S(n) \equal{} \frac {\pi}{3\sqrt {3}} \plus{} \frac {2}{\sqrt {3}}\sum_{k \equal{} 1}^{n \minus{} 1} \tan^{ \minus{} 1}\left(\frac {2k \minus{} n}{n\sqrt {3}}\right)$.

The final step is to show this last sum is zero.  To this end, we observe that

$ f(n \minus{} k) \equal{} \frac {2(n \minus{} k) \minus{} n}{n\sqrt {3}} \equal{} \frac {n \minus{} 2k}{n\sqrt {3}} \equal{} \minus{} f(k)$.

Furthermore, since $ \tan^{ \minus{} 1} ( \minus{} x) \equal{} \minus{} \tan^{ \minus{} 1} x$,

$ \sum_{k \equal{} 1}^{n \minus{} 1} \tan^{ \minus{} 1} f(k) \equal{} \sum_{k \equal{} n \minus{} 1}^{1} \tan^{ \minus{} 1} f(n \minus{} k) \equal{} \minus{} \sum_{k \equal{} 1}^{n \minus{} 1} \tan^{ \minus{} 1} f(k)$,

and so this sum must be zero as claimed.  Therefore, $ S(n) \equal{} \frac {\pi}{3\sqrt {3}}$, which is also half of the definite integral $ \int_0^1 \frac {dx}{1 \minus{} x \plus{} x^2}$.
[/hide]

That is the correct answer. Here is my solution:

First observe that $ \left\{n(1\minus{}y)\right\}\equal{}1\minus{}\left\{ny\right\}$ for all $ y\in[0,1]$ except for $ 0$, $ \frac{1}{n}$, $ \cdots$, $ \frac{n\minus{}1}{n}$ and $ 1$, and since the value  of the Riemann integral  does not change over finite sets we get, using the substitution $ x\equal{}1\minus{}y$, that

$ I\equal{}\int\limits_{0}^{1}\frac{\left\{nx\right\}}{1\minus{}x\plus{}x^{2}}dx\equal{}\int_{0}^{1}\frac{dy}{1\minus{}y\plus{}y^{2}}dy\minus{}I$, from which the result follows.

Or, divide the interval $ [0,1]$ in intervals of the form $ [k/n, (k\plus{}1)/n]$ and on each such interval the preceding fractional formula holds, and the result folows.

Now why didn't I think of that....  *shrug*   Makes my solution look clumsy.   :roll:

Thank you [b]Extremal[/b], I don't know how appeared "$ 0$" there ! Maybe I was tired at that time. I think that I confound for a moment $ \{x\}$ with $ [x]$ ! I corrected.

[quote="Virgil Nicula"]$ \int_0^1f(x)\{nx\}dx \equal{}$ $ \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_{\frac kn}^{\frac {k \plus{} 1}{n}}f(x)\{nx\}dx \equal{}$ $ \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_0^{\frac 1n}f\left(x \plus{} \frac kn\right)\{nx\}dx\ .$[/quote]
it's good idea !!!
now we can say
$ \int_0^1f(x)\{nx\}dx \equal{} \sum_{k \equal{} 0}^{k \equal{} n \minus{} 1}\int_0^{\frac 1n}f\left(x \plus{} \frac kn\right)nxdx$ because$ x \in (0,\frac {1}{n})$ :)
or
[quote="didilica"]

Or, divide the interval $ [0,1]$ in intervals of the form $ [k/n, (k \plus{} 1)/n]$ and on each such interval the preceding fractional formula holds, and the result folows.[/quote]
already have post
atomicwedgie Thanks.
didilica your solution is nice perfect idea   :what?:


What is the side length of an equilateral triangle ABC for which there is a point P

inside it such that AP = 6, BP = 8, and CP = 10?

[color=darkblue]Let $ D'$ outside the triangle, such that $ \triangle{BDA}\equiv \triangle{BD'C}$.
Then $ \angle{ABD} \equal{} \angle{CBD'} \Leftrightarrow \angle{ABC} \equal{} \angle{DBD'} \equal{} 60^\circ$.

$ \left\|\begin{array}{cc} BD \equal{} BD' \\
\angle{DBD'} \equal{} 60^\circ \end{array} \right\| \Rightarrow \triangle{DBD'} \minus{}$ equilateral $ \Rightarrow DD' \equal{} 8$.

In $ \triangle{DD'C}$ we have $ DD' \equal{} 8, CD' \equal{} 6, CD \equal{} 10$, so $ DD'^2 \plus{} CD'^2 \equal{} 36 \plus{} 64 \equal{} 100 \equal{} CD^2 \Rightarrow \angle{DD'C} \equal{} 90^\circ$.

Then $ \angle{BD'C} \equal{} \angle{BD'D} \plus{} \angle{DD'C} \equal{} 150^\circ$.

By cosine's theorem in $ \triangle{BD'C}$ we have $ BC^2 \equal{} BD'^2 \plus{} CD'^2 \minus{} 2BD'\cdot CD'\cos{150^\circ} \equal{} 100 \plus{} 48\sqrt {3}$.

So the side of the triangle equals $ 2\sqrt {25 \plus{} 16\sqrt {3}}$[/color].

[hide="Image"][img]http://ipicture.ru/uploads/080611/6omEsW8fUP.png[/img][/hide]


[quote="jackie-chan01"]$ \ Prove \ \frac {1}{2^n}\sum_{2}^{\infty}{\tan \frac {\pi}{2^n}} \equal{} \minus{} \frac {1}{\pi}$[/quote]

Please try to be careful when you ask problems :(

1) what's the meaning of the $ \frac 1{2^n}$  [u][b]before [/b][/u]the sum sign ("$ n$" is not defined) ??

2) $ \tan(\frac {\pi}{2^n}) > 0$ $ \forall n\geq 2$ and so each term in the LHS sum is positive, so the sum would certainly not be negative ...

[quote="jackie-chan01"]$ \ Prove \ \frac {1}{2^n}\sum_{2}^{\infty}{\tan \frac {\pi}{2^n}} \equal{} \minus{} \frac {1}{\pi}$[/quote]

What can easily be proved is that $ \sum_{n \equal{} 0}^{ \plus{} \infty}\frac {x}{2^n}\tan(\frac {x}{2^n}) \equal{} 1 \minus{} \frac {2x}{\tan(2x)}$

And setting $ x\to\frac {\pi}4$ in this equality, we get $ \sum_{n \equal{} 2}^{ \plus{} \infty}\frac {\pi}{2^n}\tan(\frac {\pi}{2^n}) \equal{} 1$ and so $ \sum_{n \equal{} 2}^{ \plus{} \infty}\frac {1}{2^n}\tan(\frac {\pi}{2^n}) \equal{} \frac 1{\pi}$


geometry
3D geometry
tetrahedron
pyramid
probability
expected value
geometry unsolved

On the Scholastic Aptitude Test (an American college entrance exam), students were once asked to determine the number of faces of the polyhedron obtained by gluing a regular tetrahedron to a square pyramid along one of the triangular faces. The answer expected by the test authors was 7, since the two polyhedra have 9 faces, 2 of which are
removed by gluing, but a student taking the exam pointed out that this is incorrect. What is the correct answer?

$5$? :) If $VABCD$ is the pyramid with $ABCD$ being the square, place another square pyramid $UABXY$ with $ABXY$ being the square and s.t. it's coplanar with $ABCD$. It's pretty clear now that $Y,A,D,U,V$ are coplanar, and $ABUV$ is a regular tetrahedron. From these last remarks we find that $U,V,A,D$ are coplanar. The same holds for $U,V,B,C$, so two more faces vanish from our expected number of $7$ faces.


ux+y+2z=0
x+uy+z=0
2x+2y+uz=0

n E N
n=?
unique solution different by 0

Please help me..:)

[hide]
add them all up and fator we get:

$(u+3)(x+y+z) = 0$

So either $u =-3$, or $x+y+z = 0$
[/hide]


Hi, plz help me solve this:

if a and b are odd, then 2^n | a^n + b^n for at most finitely many n.
Tnx

I did think about it for a while, until I realized that it was sort of easy (I hope there are no stupid mistakes :D).

For $n\geq 2,\ 4|2^n|a^n+b^n\Rightarrow n$ is odd because otherwise we would have $a^n\equiv b^n\equiv 1\ (mod\ 4)\Rightarrow a^n+b^n\equiv 2\ (mod\ 4)$.

On the other hand, $a{}^{2{}^{n-1}}\equiv b{}^{2{}^{n-1}}\equiv 1\ (mod\ 2^n)\Rightarrow 2^n|a{}^{2{}^{n-1}}-b{}^{2{}^{n-1}}$. From here we find that for $n\geq 2$, ${2^n|(a^{2n}-b^{2n},\ a{}^{2{}^{n-1}}-b{}^2{}^{n-1}})=a^{(2n,\ 2^{n-1})}-b^{(2n,\ 2^{n-1})}=a^2-b^2$, and this can obviously happen for only finitely many $n$.

if n is even, then a^n+b^n==2(mod 4)

if n is odd, then a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+...+b^{n-1})
a^{n-1}-a^{n-2}b+...+b^{n-1} is odd
therefore 2^n|(a^n+b^n) ==> 2^n|(a+b), the number of such n is finite


Let $ a_1\equal{}a_2\equal{}1$  If $ a_n\equal{}\frac{1}{a_{n\minus{}1}}\plus{}\frac{1}{a_{n\minus{}2}}$ then find the limit of $ a_n$ and the covgerge rate .

Let $ a_n\equal{}\sqrt 2 (1\plus{}x_n)$, then $ x_n\equal{}\frac{\minus{}1}{2}(\frac{x_{n\minus{}1}}{1\plus{}x_{n\minus{}1}}\plus{}\frac{x_{n\minus{}2}}{1\plus{}x_{n\minus{}2}})$.
Let $ y_n\equal{}g(x_n), g(x)\equal{}\frac{\minus{}x}{1\plus{}x}$. Note, that $ g(g(x))\equal{}x$. Let $ f(x)\equal{}\frac{2ix}{(1\minus{}i)x\plus{}2}$, then $ g(x)\equal{}f(f(x))$.
Let $ z_n\equal{}f(x_n)$, then $ z_n\equal{}f^{\minus{}1}(x_n)\equal{}f^{\minus{}1}(\frac{f(z_{n\minus{}1})\plus{}f(z_{n\minus{}2})}{2}).$
For big n $ x_n\equal{}O(2^{\minus{}n/2}).$


geometry
rectangle
calculus
calculus computations

A farmer has 480 meters of fencing.  He wishes to enclose a rectangular plot of land and to divide the plot into three equal rectangles with two parallel lengths of fence down the middle.  What dimensions will maximize the enclosed area?

Let x & y be the length and breadth of the rectangle.
Given 2x+4y = 480, Find the maximum value of xy.

Answer
[hide]Max. Area = 7200 sq.m[/hide]


Let be given p>0, q<0, p+q<1 and a sequence of nonegative real {a_n}, satisfying:  a_(n+2) <= p.a_(n+1) + qa_n for n = 0, 1, 2, ...
Show that sequence {a_n} converge and find it limit.

<i>Source: VUMC 2001 </i>

Namdung

We can suppose that a(1) \leq a(0) by induction with the given inequality
a(n) is decreasing positive so lim a(n)=L \geq 0 exist
Suppose that L>0 with the given inequality we have
L \leq (p+q)L < L wich is a contradiction.
L=0