calculus
integration
function
real analysis
probability
special factorizations
real analysis unsolved

Hello,

I'm trying to solve this integral:

$ \int^{\infty}_{ \minus{} \infty} e^{ \minus{} ax^2} e^{\left( \minus{} \frac {(x \minus{} \mu)^2}{2\sigma^2}\right)}dx$

Would it be fine to integrate by parts, with $ u(x) \equal{} e^{ \minus{} ax^2}$, $ du \equal{} \minus{} 2ax e^{ \minus{} ax^2}dx$,
$ dv \equal{} e^{\left( \minus{} \frac {(x \minus{} \mu)^2}{2\sigma^2}\right)}dx$, and $ v \equal{} \sqrt {2\pi\sigma^2}$?

I tried to do that and I got

$ \int^{\infty}_{ \minus{} \infty} e^{ \minus{} ax^2} e^{\left( \minus{} \frac {(x \minus{} \mu)^2}{2\sigma^2}\right)}dx \equal{} 2a^{ \minus{} 1}\sqrt {2\pi\sigma^2}$

Which appears to be incorrect.

[quote="kenn4000"]$ \int_{ - \infty}^\infty \frac {1}{\sqrt {2\pi \sigma^2}}e^{ - ax^2}e^{\frac { - (x - \mu)^2}{2\sigma^2}dx}$

$ = \int_{ - \infty}^\infty \frac {1}{\sqrt {2\pi \sigma^2}}e^{ - \frac {3a\sigma^2 + 1}{2\sigma^2}x^2 + \frac {2\mu}{2\sigma^2} x - \frac {\mu^2}{2\sigma^2}}dx$

$ y = \sqrt {\frac {3a\sigma^2 + 1}{2\sigma^2}}x, dy = \sqrt {\frac {3a\sigma^2 + 1}{2\sigma^2}}dx$

$ - y^2 + \frac {2\mu}{\sqrt {2\sigma^2}\sqrt {3a\sigma^2 + 1}}y - \frac {\mu^2}{2\sigma^2}$

$ = - y^2 + \frac {2\mu}{\sqrt {2\sigma^2}\sqrt {3a\sigma^2 + 1}}y - \frac {\mu^2}{2\sigma^2(3a\sigma^2 + 1)} + \frac {\mu^2}{2\sigma^2}(\frac {1}{3a\sigma^2 + 1} - 1)$

$ = - (y - \frac {2\mu}{\sqrt {2\sigma^2}\sqrt {3a\sigma^2 + 1}})^2 - \frac {3a\mu^2}{2(3a\sigma^2 + 1)}$

so we get $ \int_{ - \infty}^\infty \frac {1}{\sqrt {2\pi \sigma^2}}e^{ - (y - \frac {2\mu}{\sqrt {2\sigma^2}\sqrt {3a\sigma^2 + 1}})^2 - \frac {3a\mu^2}{2(3a\sigma^2 + 1)}}\frac {dy}{\sqrt {\frac {3a\sigma^2 + 1}{2\sigma^2}}}$

$ = e^{ - \frac {3a\mu^2}{2(3a\sigma^2 + 1)}}\frac {2\sigma^2}{3a\sigma^2 + 1}\frac {1}{\sqrt {2\pi \sigma^2}} \int_{ - \infty}^\infty e^{ - (y - \mu')^2}dy = \frac {2\sigma}{3a\sigma^2 + 1}e^{ - \frac {3a\mu^2}{2(3a\sigma^2 + 1)}}$

that seems a bit ugly[/quote]

Thank you for your help.

However, are you sure this is correct? I ran a simulation script in R to test your equation and the results do not appear to match:

[code]
sigma = 1
mu  = 0
a   = 1

X   = rnorm(1000000, mean=mu, sd=sigma)
Y   = exp(-a * X^2)

mean(Y)           #Returns 0.5771073
(2*sigma /(3 * a * sigma^2 + 1)) * exp((-3 * a * mu^2)/(6 * a * sigma^2 + 2)) #Returns .5
[/code]

EDIT: According to [url]http://www.mombu.com/science/mathematics/t-expectation-value-of-exponential-function-3094939.html[/url], completion of the square can be used to solve the integral.

EDIT2: I think I got it! By completing the square, I got
$ E[e^{ - a X^2}] = \frac {e^{\frac { - \mu^2}{2\sigma^2}}e^{\frac {\mu^2}{(2a\sigma^2 + 1) (2\sigma^2)}}}{\sqrt {2a\sigma^2 + 1}}$

And according to simulation script, this is correct,

[code]
sigma = 3
mu  = -2
a   = 1

X   = rnorm(1000000, mean=mu, sd=sigma)
Y   = exp(-a * X^2)

mean(Y)
exp(-mu^2 / (2 * sigma^2)) *  exp(mu^2 / ((2*a*sigma^2 + 1)*(2 * sigma^2))) / sqrt(2 * a * sigma^2 + 1)
[/code]

yeah i dont know why i had a three there, its obviously a two from the first time it appears. just sloppy.

That's a convolution. You can make the fact that it's a convolution clearer by writing it as

$ \int_{-\infty}^{\infty}e^{-ax^2}e^{-\frac{(\mu-x)^2}{2\sigma^2}}\,dx$

And with that being a convolution (we'll get a function of $ \mu$), there's a very strong incentive to use Fourier transform method.

Given the obvious probabilistic context, I'll call use the so-called "characteristic function" from probability theory: $ \chi(t)=E(e^{itX}).$

Multiply by the appropriate constants to make the problem

$ \frac1{\sqrt{2\pi}\sigma}\cdot\frac{\sqrt{a}}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-ax^2}e^{-\frac{(\mu-x)^2}{2\sigma^2}}\,dx$

Let $ X$ be the random variable whose density is $ \frac{\sqrt{a}}{\sqrt{\pi}}e^{-ax^2}$ and let $ Y$ be the random variable whose density is $ \frac1{\sqrt{2\pi}\sigma}e^{-\frac{y^2}{2\sigma^2}}.$

$ Y$ is a normal random variable with mean zero and standard deviation $ \sigma.$ It's fairly well known that its characteristic function is $ \chi(t)=e^{-\frac{\sigma^2t^2}2.}$

The fundamental fact there is that the Fourier transform of a Gaussian is a Guassian - we just have to work out the constant.

Then $ X$ is a normal random variable with mean zero and standard deviation $ \frac1{\sqrt{2a}}.$ That means that its characteristic function is $ \chi(t)=e^{-\frac{t^2}{4a}}.$

Now, suppose $ X$ and $ Y$ are independent. What is the distribution of $ X+Y?$ One way to answer that is to find the convolution of the densities of $ X$ and of $ Y$ - and that's precisely the integral we've been asked to find. The other think that you can say is that the characteristic function of $ X+Y$ is the pointwise product of the characteristic functions of $ X$ and $ Y.$ (Again, that's fundamental Fourier analysis: convolution on one side corresponds to pointwise multiplication on the other side.)

That gives the characteristic function of $ X$ and $ Y$ as $ \exp(-\frac{\sigma^2}2+\frac1{4a})t^2).$

That's the characteristic function of another normal distribution. In other words, the sum of independent normal distributions is a normal distribution. The variance of the sum is the sum of the variances; in this case $ \sigma^2+\frac1{2a}.$

From that I get the following:

$ \frac1{\sqrt{2\pi}\sigma}\cdot\frac{\sqrt{a}}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-ax^2}e^{-\frac{(\mu-x)^2}{2\sigma^2}}\,dx=\frac1{\sqrt{2\pi}\sqrt{\sigma^2+\frac1{2a}}}\exp(-\frac{\mu^2}{2(\sigma^2+\frac1{2a})}).$

Undoing the multiplication by constant, I have that

$ \int_{-\infty}^{\infty}e^{-ax^2}e^{-\frac{(\mu-x)^2}{2\sigma^2}}\,dx=\frac{\sqrt{2\pi}\sigma}{\sqrt{2a\sigma^2+1}}\exp(-\frac{\mu^2}{2(\sigma^2+\frac1{2a})}).$

[quote="Kent Merryfield"]That's a convolution. You can make the fact that it's a convolution clearer by writing it as

$ \int_{ - \infty}^{\infty}e^{ - ax^2}e^{ - \frac {(\mu - x)^2}{2\sigma^2}}\,dx$

And with that being a convolution (we'll get a function of $ \mu$), there's a very strong incentive to use Fourier transform method.

Given the obvious probabilistic context, I'll call use the so-called "characteristic function" from probability theory: $ \chi(t) = E(e^{itX}).$

Multiply by the appropriate constants to make the problem

$ \frac1{\sqrt {2\pi}\sigma}\cdot\frac {\sqrt {a}}{\sqrt {\pi}}\int_{ - \infty}^{\infty}e^{ - ax^2}e^{ - \frac {(\mu - x)^2}{2\sigma^2}}\,dx$

Let $ X$ be the random variable whose density is $ \frac {\sqrt {a}}{\sqrt {\pi}}e^{ - ax^2}$ and let $ Y$ be the random variable whose density is $ \frac1{\sqrt {2\pi}\sigma}e^{ - \frac {y^2}{2\sigma^2}}.$

$ Y$ is a normal random variable with mean zero and standard deviation $ \sigma.$ It's fairly well known that its characteristic function is $ \chi(t) = e^{ - \frac {\sigma^2t^2}2.}$

The fundamental fact there is that the Fourier transform of a Gaussian is a Guassian - we just have to work out the constant.

Then $ X$ is a normal random variable with mean zero and standard deviation $ \frac1{\sqrt {2a}}.$ That means that its characteristic function is $ \chi(t) = e^{ - \frac {t^2}{4a}}.$

Now, suppose $ X$ and $ Y$ are independent. What is the distribution of $ X + Y?$ One way to answer that is to find the convolution of the densities of $ X$ and of $ Y$ - and that's precisely the integral we've been asked to find. The other think that you can say is that the characteristic function of $ X + Y$ is the pointwise product of the characteristic functions of $ X$ and $ Y.$ (Again, that's fundamental Fourier analysis: convolution on one side corresponds to pointwise multiplication on the other side.)

That gives the characteristic function of $ X$ and $ Y$ as $ \exp( - \frac {\sigma^2}2 + \frac1{4a})t^2).$

That's the characteristic function of another normal distribution. In other words, the sum of independent normal distributions is a normal distribution. The variance of the sum is the sum of the variances; in this case $ \sigma^2 + \frac1{2a}.$

From that I get the following:

$ \frac1{\sqrt {2\pi}\sigma}\cdot\frac {\sqrt {a}}{\sqrt {\pi}}\int_{ - \infty}^{\infty}e^{ - ax^2}e^{ - \frac {(\mu - x)^2}{2\sigma^2}}\,dx = \frac1{\sqrt {2\pi}\sqrt {\sigma^2 + \frac1{2a}}}\exp( - \frac {\mu^2}{2(\sigma^2 + \frac1{2a})}).$

Undoing the multiplication by constant, I have that

$ \int_{ - \infty}^{\infty}e^{ - ax^2}e^{ - \frac {(\mu - x)^2}{2\sigma^2}}\,dx = \frac {\sqrt {2\pi}\sigma}{\sqrt {2a\sigma^2 + 1}}\exp( - \frac {\mu^2}{2(\sigma^2 + \frac1{2a})}).$[/quote]

Great way to solve it using Fourrier transform! Thank you!

P.S. the answer should be $ \frac {1}{\sqrt {2a\sigma^2 + 1}}\exp( - \frac {\mu^2}{2(\sigma^2 + \frac1{2a})}).$


vector
linear algebra
matrix
linear algebra unsolved

Let $T$ be an endomorfisma of vector space $V$, prove that:
$T$ is self adjoint iff all of the eigenvalues of $T$ is positive.

I thinks its false..
unless we add condition of $T$ must be positive definite..

Assume that $T$ is positive, since $T$ is self-adjoint the eigenvalues are all real numbers, and for any vectorspace $v_{i}$ we have $\langle{T(v_{i}),v_{i}}\rangle >0$  and so $\langle{\lambda_{i}v_{i},v_{i}}\rangle >0 \Longrightarrow \lambda_{i}\langle{v_{i},v_{i}}\rangle >0$ this conclude $\lambda_{i}> 0$.

For the converse, i thinks we can use spectral theorem, but maybe there are another elementary proof.

The matrix $\begin{pmatrix}-1 &0 \\ 0 &-1\end{pmatrix}$ is self-adjoint and has negative eigenvalues.

I thinks he forgot the condition that $T$ must be positive (see my first post in this thread)

The matrix $\begin{bmatrix}1&3\\0&2\end{bmatrix}$ has positive eigenvalues and is most assuredly not self-adjoint.

The problem as stated is very strange - so false as to be irrelevant. Being self-adjoint and having positive eigenvalues have very little relationship to each other.

"Positive eigenvalues" can be defined with no mention of either a basis or an inner product; to define "self-adjoint" requires either an inner product or an agreed-upon standard basis.

My suspicion is that bolpoin has misquoted his source in some way.

The problem say:

Let $T$ be a self adjoint endomorfisma, prove that:

$T$ is positive iff all of eigenvalues of $T$ is positive.

just like Beginner said.

OK. That's straightforward in finite dimensional spaces - just use the orthogonal (or unitary) diagonalization result. In an infinite-dimensional space, it's a little trickier, and may require us to talk about the spectrum rather than the eigenvalues.

So we really need spectral theorem here.

Let $y \in V$ then by spectral theorem since T is self- adjoint
$V$ is the direct sum of eigenspace of $T$, say $K(\lambda_{i})$, for different eigenbalues $\lambda_{i}$.

then $y=\sum v_{i}$ with $v_{i}\in K(\lambda_{i})$ so
\[\langle{ T(y),y\rangle}= \sum_{i}\sum_{j}\lambda_{i}\langle{ v_{i}, v_{j}\rangle}\]
since $\langle{v_{i},v_{j}\rangle}=0$ for $i \not= j$, and $\lambda_{i}\langle{v_{i},v_{i}\rangle}> 0$ the conclusion follow.


Let $ S$ be a set of $ n$ elements.  Prove that if $ n$ is even, the only clutter of size $ \binom{n}{\lfloor\frac{n}{2}\rfloor}$ is the clutter of all $ \frac{n}{2}$-combinations.

I don't know if everyone here knows what a "clutter" is - the more standard terminology as far as I know is "antichain", specifically an antichain of subsets of $ S$: this is a family of subsets none of which contains another.
[hide="Hint"]
This follows directly from one of the standard proofs of [url=http://en.wikipedia.org/wiki/Sperner%27s_theorem]Sperner's Theorem[/url], namely the one proving the stronger [url=http://en.wikipedia.org/wiki/Lubell-Yamamoto-Meshalkin_inequality]LYM inequality[/url].
[/hide]
It is also true that when $ n\equal{}2m\plus{}1$ is odd, the largest clutters are the one formed by all the $ m$-sets and the one formed by all the $ m\plus{}1$-sets. This requires a slightly more detailed analysis.


In the figure, triangle ABC has angle A=45 and angle B=30. A line DE, with D on AB and angle ADE=60, divides triangle ABC into two pieces of equal area. (Note: the figure may not be accurate; perhaps E is on CB instead of AC.) The ratio AD/AB is:

(A) $ \frac {1}{\sqrt{2}}$
(B) $ \frac {2}{2\plus{}\sqrt{2}}$
(C) $ \frac {1}{\sqrt{3}}$
(D) 1/cube root of 6
(E) 1/ fourth root of 12

[url]http://www.math.ksu.edu/main/events/hscomp/worksheets/amc12/worksheet11/Worksheet11.pdf[/url]

here is url, go to page 5 for diagram

can someone pm me on how to do a diagram WITHOUT geogebra on here?
I am having major problems with it.

For this problem, if someone can give some hints, that would be great.

Thanks in advance!

oh, i usually dont check that forum, thanks for the link

can people post some hints?
i really dont want to look at the solutions on that link, but im at my wit's end


[quote="GameBot"]Equilateral triangles are constructed on each side of a right triangle. The areas of the triangles constructed on the legs of the right triangle are 36 $ \text{cm}^{2}$ and 64 $ \text{cm}^{2}$. What is the number of square centimeters in the area of the triangle constructed on the hypotenuse?[/quote]

$ 36\plus{}64\equal{}100$.


[quote="psychicalchemist"]I was wondering, does anyone know if the criteria for being selected for the US physics Team varies depending on grade. I know for MOP, this is true since Red and Blue MOP are much easier to get into than Black MOP. Judging by the distribution of the past participants of te US Physics Team, I'm inclined to say that the team is chosen solely based on your score on the semifinal round, but I just wondered if anyone knew for sure.[/quote]
I have been wondering about the same thing. Where did you find the statistics/distribution of the past participants? Unlike AMC, there are hardly any info about the participants of the physics teams and training camp, although there are some info on the traveling team. Does anyone know how many students take part in the first round of the test each year on average? In other words, how many students are the 200 semifinalists selected from?

yeah, that was my question too!

also one more thing, what is the average score needed for being a semi-finalist?


function
limit
inequalities
logarithms
integration
real analysis
real analysis unsolved

Find all continuous functions $ f: \mathbb{R}\to [1,\infty)$ for wich there exists $ a\in\mathbb{R}$ and a positive integer $ k$ such that \[ f(x)f(2x)\cdot...\cdot f(nx)\leq an^k\]
for all real $ x$ and all positive integers $ n$.

[i]author :Radu Gologan[/i]

First of all easy to check that $ \lim_{n\to\infty }f(x)=1$
Suppose exist $ a$ such that $ f(a)>1$
Then exist an interval $ [a,b)$ such that $ f(x)>1,\forall x\in [a,b]$
$ M=inf f(x),x\in [a,b]$
For each $ x_0$  there are at least   for every $ k\in [\frac{a}{x+0}]-1,\frac{b}{x_0}]$ then $ kx_0\in [a,b]$
So $ f(x)....f(nx)\geq M^{n-k}$
But $ \lim_{n\to\+\infty}\frac{a^n}{x^m}=+\infty$ so the inequalites doesn't hold for all n ,
It mean that $ f(x)\equiv 1$

A slightly different take:

Take logarithms: $ \sum_{j \equal{} 1}^n\ln (f(jx))\le \ln a \plus{} k\ln n$. Dividing by $ n$, $ \frac1n\sum_{j \equal{} 1}^n\ln (f(jx))\le\frac {\ln a}{n} \plus{} k\frac {\ln n}{n}$. Set $ x \equal{} \frac {B}{n}$, and let $ n\to\infty$; we get $ \int_0^B \ln(f(t))\,dt\le \lim_{n\to\infty}\frac {\ln a}{n} \plus{} k\frac {\ln n}{n} \equal{} 0$.
Since $ \ln f(t)$ is nonnegative and continuous, it must be identically zero on $ [0,B]$. $ B$ was arbitrary, and $ f$ is identically $ 1$.


geometry
inradius
circumcircle
inequalities proposed
inequalities

Prove that

$ \frac{l_ah_a\minus{}1}{l_a^2}\plus{}\frac{l_bh_b\minus{}1}{l_b^2}\plus{}\frac{l_ch_c\minus{}1}{l_c^2}\geq\frac{12r^3\plus{}3r^2R\minus{}R}{3Rr^2}$

[quote="kuing"][quote="Ligouras"]Prove that

$ \frac {l_ah_a \minus{} 1}{l_a^2} \plus{} \frac {l_bh_b \minus{} 1}{l_b^2} \plus{} \frac {l_ch_c \minus{} 1}{l_c^2}\geq\frac {12r^3 \plus{} 3r^2R \minus{} R}{3Rr^2}$[/quote]

that's not a geometric inequality...

and, what are the $ l_a,l_b,l_c$ mean?[/quote]

My Friend

:   Let $ h_a, h_b, h_c$ be the heights, $ l_a, l_b, l_c$ the bisectors, $ r$ the inradius, and $ R$ the circumradius of a triangle $ ABC$.


geometry
geometric transformation
rotation
analytic geometry
linear algebra
matrix
trigonometry

I'm not always familiar with rotation of both points and line.

Let's say we have a point $ (1,2)$ in the line $ y \equal{} x\plus{}1$. Here are some things I want to do:

1) Rotate point $ (1,2)$ by 60 degree
2) Rotate point $ (i,2)$ using complex set? If so, how would this work for rotating by 60 degrees? $ n$ degrees?
3) Rotate line $ y \equal{} x\plus{}1$ by origin.
4) Rotate line $ y \equal{} x\plus{}1$ by x-axis/y-axis.
5) Rotate line $ y \equal{} x\plus{}1$ by $ n$ degrees.

Thanks a lot! I could never really get the concept of rotation (like how to do it..).

Two common ways are to use [url=http://mathworld.wolfram.com/RotationMatrix.html]rotation matrices[/url] or [url=http://mathworld.wolfram.com/PolarCoordinates.html]polar coordinates[/url].

[b]The Rotation Matrix:[/b] The rotation of a point $ \binom{x}{y}$ through an angle of $ \theta$ about the origin in $ \mathbb{R}^2$ is given by:

$ \left( \begin{array}{cc} \cos \theta & - \sin \theta & \sin \theta & \cos \theta \end{array} \right) \binom{x}{y}$

where the result is computed through standard matrix multiplication and the point $ (x,y)$ and it's image are expressed as column vectors.

[b]Basic Usage:[/b] Suppose we want to rotate $ (1,2)$ by 60 degrees. Just plug in:
$ \left( \begin{array}{cc} 1/2 & - \sqrt {3}/2 & \sqrt {3}/2 & 1/2 \end{array} \right) \binom{1}{2} = \binom{\frac {1}{2} - \sqrt {3}}{1 + \frac {\sqrt {3}}{2}}$

So, the resulting point is $ \left(\frac {1}{2} - \sqrt {3} , \; 1 + \frac {\sqrt {3}}{2}\right)$

[b]Rotating a Parametrizable Curve:[/b] Suppose we want to rotate the whole line $ y = x + 1$ by some angle -- let's do 30 degrees. The first step is to parametrize this in the minimum possible number of variables. For example, we can parametrize this as $ (x,y) = (t,t+1)$

Next, we apply the rotation matrix to the parametrized form:
$ \left( \begin{array}{cc} \sqrt {3}/2 & - 1/2 & 1/2 & \sqrt {3}/2 \end{array} \right) \binom{t}{t + 1} = \left( \begin{array}{c} \left( \frac {\sqrt {3} - 1}{2}\right) t - \frac {1}{2} & \left( \frac {\sqrt {3} + 1}{2}\right) t + \frac {\sqrt {3}}{2} \end{array} \right)$

A parametrization of the points on the rotated line is thus:
$ x = \left( \frac {\sqrt {3} - 1}{2}\right) t - \frac {1}{2}$
$ y = \left( \frac {\sqrt {3} + 1}{2}\right) t + \frac {\sqrt {3}}{2}$

Finally, if you want an explicit solution (as in this case), eliminate the parameter from the system through algebraic manipulation. In this case, we get $ y = (2 + \sqrt {3} ) x + (1 + \sqrt {3})$

[b]Alternate Solution:[/b] While the above is more representative of how to rotate a function in general, it is not the ideal method for rotating a line. We don't even need a rotation matrix.

[b]Background/Preparation:[/b] Recall that a non-vertical line in $ \mathbb{R}^2$ is uniquely defined by two parameters, which I will define in the most convenient way:
1. Slope: The appropriately-signed tangent of the angle the line makes with the positive x-axis
2. y-intercept: The unique value of $ r$ such that the polar point $ (r , 90^{\circ})$ is on the line.

[b]Procedure:[/b] The slope of $ y = x + 1$ is $ 1$, thus the tangent of the angle it makes with the x-axis is $ 1$. The slope of the new line will be the tangent of the angle 30 degrees more than the original angle it makes with the x-axis. Thus, by the tangent-addition formula, the slope of the new line is:

$ \frac {1 + \tan 30^{\circ}}{1 - 1 \cdot \tan 30^{\circ}} = \frac {1 + \frac {1}{\sqrt {3}}}{1 - \frac {1}{\sqrt {3}}} = 2 + \sqrt {3}$

Recall the definition of the y-intercept. The y-intercept of the new point takes the form $ (r, 90^{\circ})$, so its pre-image takes the form $ (r, 60^{\circ})$, where $ r$ is the same before and after the rotation. Since the pre-image has $ \theta = 60^{\circ}$, $ y = \sqrt {3} \cdot x$.

We solve $ \sqrt {3} x = x + 1$ to find the pre-image of the y-intercept point. We get $ x = \frac {1 + \sqrt {3}}{2} \implies y = \frac {3 + \sqrt {3}}{2}$. By the Pythagorean theorem, the $ r$ value of this point is $ 1 + \sqrt {3}$.

Thus, the new line is $ y = (2 + \sqrt {3} ) x + (1 + \sqrt {3})$

Thanks, TZF! That is exactly what I needed... It's been really long time since I saw matrice so thanks for refreshing it!  :)  :)


$f(x_{1}+x_{2})=f(x_{1})+f(x_{2})$ for evry real $x_{1}$ and $x_{2}$
f:R->R
f(x) is continuous. Prove that f(x)=ax.
Please excuse my bad english, and try to explain me this problem. Thank you..

From the initial equation derive

$f(\underbrace{x+x+\dots+x}_{m})=\underbrace{f(x)+f(x)+\dots+f(x)}_{m}\iff f(mx)=mf(x)\qquad(*)$

Then putting $x=1$ we get $f(m)=mf(1)=am$ for integer $m$, where $a=f(1)$

Now put $x={k\over m}, k,m\in\mathbb{Z}$ into $(*)$ to obtain

$f(k)=mf({k\over m})\iff f({k\over m})={f(k)\over m}=a{k\over m}$

since for integer $k$ we have $f(k)=ak$.

Therefore we've proven $f(x)=ax$ for all rational $x$.

Now use the continuity of the function and Dedekind-cuts representation of irrationals to expand that result to any real $x$, and the proof is complete.


[url]http://www.msnbc.msn.com/id/26884523?GT1=43001[/url]

Please discuss the house voting no on the largest government intervention since the depression.

Personally, I thought the package would easily pass...but I think part of the reason it didn't was the idea that people in congress were afraid of having something like this on their record, in case the plan fails tremendously. Again, I really think this is incredible - this was not supposed to happen at all.

I think this is another example of the struggles between Republicans and Democrats.  Republicans even said that it was Pelosi's speech about bipartisan work that "killed" the bill.
What is Congress thinking?  In this economy in the brink of another Depression, they refuse to pass this bill?
I think this is utter madness and that Congress is forcing the American economy to "commit suicide."
Last time I checked, the Dow was down 600 points.

[quote]Despite pressure from supporters, not enough members were willing to take the political risk just five weeks before an election.[/quote]
It's sad that the representatives care more about their own re-election than the American economy.

The vote here was less partisan than many; there were a lot of Democrats voting against it, and a lot of Republicans voting for it. The strongest indicator for voting against it was having a competitive reelection race, with nearly all such representatives voting no.

The public outrage is very real; the people aren't convinced that the plan is actually good for the economy. The stock and bond markets are not "the economy".

What went wrong? This plan started with an outrageously bad proposal from the Bush administration, which was then modified by the congressional Democrats. It would probably be a better idea to throw that out and start from scratch with a proposal from independent economists. That would include such radical ideas as reinstating the regulations that were dropped ten years ago, and allowed the mess.

that's not a very useful note. compare the percentage of democratic representatives who voted yes to that of the republicans. analyzing this on a strictly partisan level doesn't seem likely to help.

[quote="mewto55555"]Also, noting that fact that the House has a Democrat majority, one realizes that if the economy collapses, it would be blamed on Bush, severely hampering McCain.[/quote]

in addition to the obvious point that the bill was supported by the majority of democrats, allowing the economy to collapse to win an ground in an election is analogous to cutting off your nose to spite your face or however that stupid saying goes.

We still have to remember that the democrats didn't vote in 2:1 favor of it either...
[code]
                  Yes	  No
Democrats	      140	  95
Republicans	     65	 133
Total	          205	 228[/code]

So what do you think congress ought to do now - start from scratch, or revise this bill?

It seems the senate is trying to force the House to approve it, what do you guys think about this?  They just approved it 75-24 I believe, and sweetened the deal, hoping to attract those republican votes they didn't get last time

[quote="ProtestanT"]Just let it fail. Then start from scratch.

(This can be interpreted as refering to either the economy or the bill...I don't know which one I'm refering to more...I think I'm refering to both.)[/quote]

So you wouldn't mind living your entire life in an economic depression, not know whether the world would ever recover?  Notice how, in the words of Steven Colbert, we [b]are[/b] world leaders.  Look at how everyone else's markets are following ours!!!  And I'm sure those who [b]DIDN'T[/b] overextend themselves and [b]SAVED[/b] money would be glad that their lifetime of work amounted to nothing.

The only thing that saved us from the First Great Depression of the '30s (as it will probably be known as soon) was WWII.  FDR did tons to help people, yet still, he didn't remove the national debt.

What do you suppose people do in the transition state between the crash and rebuilding of the economy?  A barter system?

What about retired people?  "Oh, sorry sir, but you'll have to starve or beg from your family.  You don't have any family left?  I'm sorry sir, there's no welfare anymore, you'll just have to grow your own food.  What?  You're in a wheelchair?  You'll have to beg."


At $ 2: 15$ o'clock, the hour and minute hands of a clock form an angle of:

$ \textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$

[hide="Click for solution"]
In $ 15$ minutes the hour hand moves through an angle of $ \frac{1}{4} \cdot 30^\circ\equal{}\frac{15}{2}^\circ$.  Therefore, the angle between the hour hand and the minute hand is $ 22 \frac{1}{2}$, or $ \boxed{\textbf{(C)}}$.
[/hide]


Find all $ f: Z\minus{}>Z$ such that:
$ f(a^3\plus{}b^3\plus{}c^3)\equal{}f(a)^3\plus{}f(b)^3\plus{}f(c)^3$

[quote="Math pro"]Find all $ f: Z \minus{} > Z$ such that:
$ f(a^3 \plus{} b^3 \plus{} c^3) \equal{} f(a)^3 \plus{} f(b)^3 \plus{} f(c)^3$[/quote]
Here is a problem of Vietnam TST 2005.


group theory
abstract algebra
function
Euler
number theory
totient function
superior algebra

Suppose we have a bijection between two finite groups $ f: G \rightarrow H$ such that $ \forall g \in G \: |f(g)|\equal{}|g|$ is $ f$ an isomorphism?

($ \, |g|$ is the order of the subgroup which is generated by $ g$)

No. Consider a cyclic group of order five; any of the 24 permutations of the non-identity elements satisfies your condition, but only four of them are group homomorphisms.

Also, it's possible to construct non-isomorphic groups with the same numbers of elements of each order; I believe the smallest examples are 16-element groups.

oh ok, pretty obvious actually.

Thinking about partitions of groups, is the equivalence relation $ a\sim b$ iff $ <a>\equal{}<b>$ useful for much? I remember that it provides an identity with the eular-totient function $ n\equal{} \sum_{d|n} \varphi(d)$ where $ \varphi(d)$ is the Euler totient function. What else is this useful for? if anything..


factorial
floor function
number theory
prime factorization

For how many positive integers $ k$ does the ordinary decimal representation of the integer $ k!$ end in exactly $ 99$ zeros?

[hide]If it is possible for $ n!$ to end in 99 trailing zeros, then there are exactly 5 such numbers $ n$, which take the form $ 5k, \; 5k\plus{}1, \; 5k\plus{}2, \; 5k\plus{}3, \; 5k\plus{}4$ for some $ k$.

[hide="Justification"]The number of trailing zeros is the number of factors of $ 10$, or the lower of the number of factors of $ 2$ and the number of factors of $ 5$. The latter is necessarily lower, so we just want to count the number of factors of $ 5$.

$ (5k\minus{}1)!$ clearly has fewer factors of $ 5$ than $ (5k)!$, since $ 5k$ is divisible by $ 5$. Similarly, $ 5k\plus{}1,2,3,4$ all have no factors of $ 5$, so multiplying them in to $ (5k)!$ to create their respective factorials will not add any factors of $ 5$. However, $ (5k\plus{}5)$ is again divisible by $ 5$, therefore $ (5k\plus{}5)!$ has at least one more factor of $ 5$ than $ (5n)!$.[/hide]

We must now determine if there is a number $ n$ such that $ n!$ ends in 99 trailing zeros.

As detailed in the justification, the number of trailing zeros equals the exponent on $ 5$ in the prime factorization of $ n!$. Therefore, the number of trailing zeros is $ z(n)\equal{}\sum_{j\equal{}1}^{\infty} \left \lfloor \frac{n}{5^j} \right \rfloor$.

Observe that $ z(n) \leq \frac{n}{5^j} \equal{} \frac{n}{4}$, so a good place to start is $ n\equal{} 4 \times 99 \equal{} 396$. However, this has the same number of trailing zeros as $ 395!$, which is less than $ n/4$, so this must have fewer than $ 99$ trailing zeros (it actually has 97). So, we try the next multiple of $ 5$, which is $ 400$.

Indeed, $ 400!$ ends in $ 99$ zeros, hence there are $ \boxed{5}$ such numbers $ n$, and they are $ \boxed{400,401,402,403,404}$.[/hide]

$ \boxed{for\ a \ prime\ p: \ v_p(n) \equal{} a\iff p^{a}\parallel{}n\iff p^{a}|n\ and\ p^{a \plus{} 1}\not|n}$

we have $ v_2(k!) < v_5(k!)$ so $ k!$ end with exactly $ v_5(k!)$ zéro in dicimal representation,
here we have $ v_5(k!) \equal{} \sum_{i \equal{} 1}^{ \plus{} \infty}[k/5^i] \equal{} 99$
$ k < 5^{4}$ becsause $ v_5(5^{4}) > 99$ so we can write $ k \equal{} a \plus{} 5b \plus{} 25c \plus{} 125d$ where $ a,b,c,d\in\{0,1,2,3,4\}$
and we have $ v_5(k!) \equal{} [k/5] \plus{} [k/25] \plus{} [k/125] \equal{} b \plus{} 6c \plus{} 31d \equal{} 99$
and it's easy to find that the only solution is $ (b,c,d) \equal{} (0,1,3)$
then $ k \equal{} a \plus{} 25 \plus{} 125\times 3$ where $ a\in\{0,1,2,3,4\}$
there are $ 5$ solutions


Let A={1,2,3,...,30}.Find the number of subset of A that has 3-element such that sum of it's elements can be divisible with 5.

I suggest first taking $(mod5)$ of all of the integers in your set, and then considering which combinations of 3 can sum to 0(mod5):

0,0,0
0,1,4
0,2,3
etc.

Then use combinatorics to choose from the equivalence classes!

there are not so many combinations:
$(0,0,0),(0,1,4),(0,2,3),(1,2,2),(1,3,1)$
So then i find
$6\cdot 5 \cdot 4+6\cdot 6\cdot 6 + 6\cdot 6\cdot 6 +6\cdot 6\cdot 5 +6\cdot 6\cdot 5=912$ sets.

Could that be correct?


geometry
3D geometry
sphere
combinatorial geometry
geometry proposed

Finitely many points are given on the surface of a sphere, such that every four of them lie on the surface of open hemisphere. Prove that all points lie on the surface of an open hemisphere.

Let $S$ be the finite set of these points, situated on the surface of a sphere of centre $O$. Consider their convex hull $\operatorname{conv}(S)$. It is easy to see that $O\not \in \operatorname{conv}(S)$ if and only if $S$ is contained on the surface of an open hemisphere (by well-known results on separability of a point from a convex set).

Assume $O\in \operatorname{conv}(S)$. By the Caratheodory theorem, there exists a subset $T$ of $4$ points in $S$, such that $O\in \operatorname{conv}(T)$. By the given condition we however have that $T$ is contained on the surface of an open hemisphere, thus $O\not \in \operatorname{conv}(T)$, contradiction. Therefore $O\not \in \operatorname{conv}(S)$, and so by the previous remark $S$  is contained on the surface of an open hemisphere.

Notice the result stands no more for any infinite set of points, even a countable one. Say one point in $S$ lies on the equator, and each other point of the equator does not lie in $S$, but is an accumulation point for $S$. Then $S$ is not contained on the surface of an open hemisphere, but any $4$ points of it are.


function
calculus
integration
calculus computations

is the greatest integer function non differentiable at integral points

Yes the gretest integer function is non differentiable at integral points as they are not continous at those points . For an integer $ N$  $ lim_{x \to 0^\minus{}} {[x]} \equal{} N\minus{}1$ while $ lim_{x \to 0^\plus{} } { [x] } \equal{} N$ while $ f(N) \equal{} N$ , hence we see gthat it is not continous at integral points and hence not differentiable .


Prove that in any triangle $ABC$ the following inquality holds:

      \[ ab+bc+ca\geq 4S\sqrt{3} +\frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 .

  Darij


geometry
inequalities
triangle inequality

Heres a problem I cam across in one of my old geometry texts.

  Let $X$ and $Y$ be any two points contained in the disc of radius $r$ around $P$.  Using the traingle inequality, prove that

$d(X, Y) \le 2r$

  Two notes.

- $d(X, Y)$ means the distance from point $X$ to point $Y$.

-"disk" is defined as follows:  We define disk of center $P$ and radius $r$ to be the set of all points whose distance from $P$ is $\le r$.

[hide="solution"]From the triangle inequality (a property of all metric spaces) we have $d(X,Y)\le{d}(X,P)+d(P,Y).$ But from the definition of the disc, the distances from $P$ to $X$ and $Y$ are $\le{r}$. So $d(X,P)+d(P,Y)\le2r$.[/hide]

[hide="No triangle inequality..."]
The greatest possible distance between two points in a circle is the diameter, or $2r$. Therefore, the distance between any two points $X$ and $Y$ must be less than or equal to $2r$.
[/hide]

I don't think that's rigorous, is it? :(


The numbers from 1 to 9 are written in an arbitrary order. A move is to reverse the order of any block of consecutive increasing or decreasing numbers. For example, a move changes 916532748 to 913562748. Show that at most 12 moves are needed to arrange the numbers in increasing order.

Is it 12 needed for some permutation? or can always be done with 11 or even lower?


geometry
circumcircle
trigonometry
combinatorics unsolved
combinatorics

a) Suppose that three congruent disks with radius $R$ have their centers in the vertices of a given triangle which they cover. Find the smallest value of $R$ for a given triangle.

b) (this problem just asks itself to generalize :P) Suppose that $n$ congruent disks with radius $R$ have their centers in the vertices of a given convex $n$-gon which they cover. Find the smallest value of $R$ for a given convex $n$-gon.

for a), isn't the answer just the circumradius of the triangle itself? (at least for an acute triangle) because if R were to be smaller than the circumradius, then the circumcentre will never be covered by the disks....and i think for an obtuse triangle, min(R) is 1/2 of max{a,b,c}. :)

Your answer is correct for acute triangle but wrong for obtuse where it is $\frac{a}{2 \cos {\beta}}$ where $a$ is the second longest side and $\beta$ is the smallest angle in the given triangle.  :)

ahh i see, it's because if BC is the largest side, have to make sure that R extended from A covers the midpoint of BC. okay, thanx! :D


[quote="RoFlLoLcOpT"]It would be a great help if you knew the Ramanujan-Hardy number, $ 1729$ which is the first natural number that can be expressed as the sum of two distinct cubes in two different ways. That way, we can be sure none of the numbers we list will appear twice.

$ 1^3\plus{}1^3, 1^3\plus{}2^3, 1^3\plus{}3^3, 1^3\plus{}4^3, 1^3\plus{}5^3, 1^3\plus{}6^3, 1^3\plus{}7^3$
$ 2^3\plus{}2^3, 2^3\plus{}3^3, 2^3\plus{}4^3, 2^3\plus{}5^3, 2^3\plus{}6^3, 2^3\plus{}7^3$
$ 3^3\plus{}3^3, 3^3\plus{}4^3, 3^3\plus{}5^3, 3^3\plus{}6^3, 3^3\plus{}7^3$
$ 4^3\plus{}4^3, 4^3\plus{}5^3, 4^3\plus{}6^3, 4^3\plus{}7^3$
$ 5^3\plus{}5^3, 5^3\plus{}6^3, 5^3\plus{}7^3$
$ 6^3\plus{}6^3$

There are $ \boxed{26}$ numbers.[/quote]

Wait so if the number was >1729 it might be a repeat?

[quote="nmani2"][quote="RoFlLoLcOpT"]It would be a great help if you knew the Ramanujan-Hardy number, $ 1729$ which is the first natural number that can be expressed as the sum of two cubes in two different ways. That way, we can be sure none of the numbers we list will appear twice.

$ 1^3\plus{}1^3, 1^3\plus{}2^3, 1^3\plus{}3^3, 1^3\plus{}4^3, 1^3\plus{}5^3, 1^3\plus{}6^3, 1^3\plus{}7^3$
$ 2^3\plus{}2^3, 2^3\plus{}3^3, 2^3\plus{}4^3, 2^3\plus{}5^3, 2^3\plus{}6^3, 2^3\plus{}7^3$
$ 3^3\plus{}3^3, 3^3\plus{}4^3, 3^3\plus{}5^3, 3^3\plus{}6^3, 3^3\plus{}7^3$
$ 4^3\plus{}4^3, 4^3\plus{}5^3, 4^3\plus{}6^3, 4^3\plus{}7^3$
$ 5^3\plus{}5^3, 5^3\plus{}6^3, 5^3\plus{}7^3$
$ 6^3\plus{}6^3$

There are $ \boxed{26}$ numbers.[/quote]

Wait so if the number was >1729 it might be a repeat?[/quote]
@nmani2, yes, if the number was greater than $1729$, it would be a repeat because as Rofllolcopt said, $1729$ is the first number that can be expressed as the sum of two distinct cubes in two different ways and we can assume that there are more numbers greater than $1729$ that have the same or similar property in that it can be expressed as the sum of two cubes in two or more different ways.

[quote=GameBot]How many positive integers less than 500 can be written as the sum of two positive perfect cubes?[/quote]

https://artofproblemsolving.com/videos/counting/chapter2/186

[quote=ihatemath123]Yuh you aint need no chart the answer's just ${7 \choose 2} + 5$ :cool:[/quote]

Elaborating on this a bit, $\binom{7}{2}$ will choose two distinct numbers from $1-7$ ($8^3$ is already too large), but we have to take out $7^3+6^3$ since it's too large, and add back the $1^3 + 1^3$ up to $6^3 + 6^3$, thus $\binom{7}{2} - 1 + 6$.

Very nice solution, I just did the old fashioned chart


function algebra polynomial integration calculus geometry trapezoid

hi,
i will be glad if somebody helps to solve this problem :

[u]problem:[/u]

let f be a function that we can derive two times in [a;b]and f(a)=f(b)=0.
let x be a real number such that : a<x<b
prove that there exists a real number c such that :

$2f(x)=(x-a)(x-b)f' '(c)$
thanks in advance.

A prototype for all interpolating polynomial error estimates.

Fix $x\in(a,b).$ Define $M$ by $f(x)=M(x-a)(x-b).$

Let $g(t)=f(t)-M(t-a)(t-b)$ Note that $g(a)=g(x)=g(b)=0.$

By Rolle's Theorem, $\exists\,\alpha\in(a,x)$ such that $g'(\alpha)=0$ and $\exists\,\beta\in(x,b)$ such that $g'(\beta)=0.$ Then, using Rolle's Theorem again on $[\alpha,\beta],$ $\exists\,c\in(\alpha,\beta)$ such that $g''(c)=0.$

But $g''(t)=f''(t)-2M.$ Hence $2M=f''(c),$ which is what we desired.

Note that for general twice-differentiable $f,$ we can find the first-degree "interpolating polynomial" $P(x)=f(a)+\frac{(f(b)-f(a))(x-a)}{b-a}.$ From the result above it will follow that $f(x)=P(x)-\frac{f''(c)}2(x-a)(b-x)$ for some $c\in(a,b).$

Exercise: Suppose $f''$ is twice differentiable on $[a,b].$ The trapezoidal rule says that:

$\int_{a}^{b}f(x)\,dx=\frac{(b-a)}2\cdot(f(a)+f(b))+(\text{Error}).$

Use the last result in my post above to derive an estimate for the error in terms of bounds for $f''.$

thank you so much Kent Merryfield for your answer.For your

exercice ,i am sorry i can't answer because i haven't studied

integrals yet .

The exercise was intended to show one of the more important uses for error estimates for polynomial interpolation, which is to prove error estimates for numerical integration.

The particular side computation needed:

$\int_{a}^{b}(x-a)(b-x)\,dx=\frac{(b-a)^{3}}6.$

[Hint: yes, that's a polynomial and you can just multiply it out - but for efficiency, don't multiply it out and use integration by parts in the factored form.]

Now, we have that $f(x)=P(x)+R(x),$ where $P(x)$ is the equation of the line that connects $(a,f(a))$ to $(b,f(b)).$ By the result above we have that for each $x\in(a,b),$ there exists $\xi\in(a,b)$ such that $R(x)=-\frac{f''(\xi)}2(x-a)(b-x).$

$\int_{a}^{b}P(x)\,dx=\frac{(b-a)}2\cdot(f(a)+f(b)).$ In fact, that is simply the area of a trapezoid, hence the name of the rule.

Suppose $f''(x)\in[m,M]$ for all $x\in (a,b).$ Then

(Error) $= \int_{a}^{b}R(x)\,dx,$ where

$-\frac{M}2(x-a)(b-x)\le R(x) \le-\frac{m}2(x-a)(b-x).$

The side computation above then gives that

$-\frac{M(b-a)^{3}}{12}\le (\text{Error})\le-\frac{m(b-a)^{3}}{12}.$

Or, given the Darboux property of the derevative, there exists $\xi\in(a,b)$ such that

(Error) $=-\frac{f''(\xi)(b-a)^{3}}{12}.$

Consider this a prototype of proofs of the accuracy of many other numerical integration methods.


find all a,b,c reals such that
abc+2-a-b-c=0

when ab=1 we have a+b-2=0 so a=b=1 and we have the solution (1,1,c).
when ab<>1 solving out c wo have the solution (a,b,(a+b-2)/(ab-1)).

:(
i did this to but it's not what i want
anyway thx wpolly0419 :D.

So, what are you looking for Kuba????

Pierre.

i don't know what i wanted when i posted this message...:(
sorry

So, we may assume that the problem is solved, and move it....

Pierre.


Ktoś z was (polaków :)   ), oprócz mnie :D , będzie brał udział ??
Zadania wydają się dosyć trudne.     :huh:

Chyba wezme udzial, ... bo nie wiem czy moge dalej liczyc na OM :(
(zalosnie skwasiles zadanie kombinatoryczne i jestem ogolnie zdemoralizowany)
Poza tym zadania sa generalnie przepiekne na mathlinks contest :)

Zadanka rzeczywiscie bardzo fajniutkie acz trudne (co liczy sie raczej in plus).

Ja wystartuje (jak cos zrobie  :P )


linear algebra
matrix
combinatorial inequality
algebra
combinatorics
IMO
IMO 1971

Let $ A \equal{} (a_{ij})$, where $ i,j \equal{} 1,2,\ldots,n$, be a square matrix with all $ a_{ij}$ non-negative integers. For each $ i,j$ such that $ a_{ij} \equal{} 0$, the sum of the elements in the $ i$th row and the $ j$th column is at least $ n$. Prove that the sum of all the elements in the matrix is at least $ \frac {n^2}{2}$.

Take the the row or column (without loss of generality, a row) with the minimum possible sum $ S$ of its elements. Let $ z$ be the number of zeros in this row. We will assume $ S < \frac{n}{2}$ once the other case is obvious.  Clearly $ S \ge n \minus{} z \Rightarrow z \ge n \minus{} S$. The total sum $ T$ of all elements of the matrix is at least the number of zeros in this row multiplicated by $ n \minus{} S$ (because the sum of the elements of a row and a column meeting in a zero is $ \ge n$) plus the number of nonzero elements times $ S$ (that is the minimum possible sum), it is, \[ z(n \minus{} S) \plus{} (n \minus{} z)S.\] Note that, being $ z \ge n \minus{} S$ and $ n \minus{} S \ge S$, we can put $ z \minus{} 1$ instead of $ z$ and $ n \minus{} z \plus{} 1$ instead of $ n \minus{} z$ until we get $ z \equal{} n \minus{} S$, what makes the sum become smaller. So we have \[ T \ge (n \minus{} S)^2 \plus{} S^2 \ge 2(\frac{n \minus{} S \plus{} S}{2})^2 \equal{} \frac{n^2}{2}\] by AM - QM inequality.

If the main diagonal contains all zeroes, we can immediately deduce from the condition that the sum $S$ of the elements is at least $n^2/2$. This motivates us to consider the largest possible subset of elements, all zero, which are pairwise in different rows of columns - let the size of this subset be $k$. (If there is no such subset, $S\ge n^2$.)

Since the condition is invariant under the operation of switching rows or columns, we may rearrange the matrix such that these $k$ entries are on the main diagonal of the matrix, i.e. $a_{11}=a_{22}=\dots=a_{kk}=0$. Let $X=\sum_{1\le i,j\le k}a_{ij}$, $Y=\sum_{i\le k<j}a_{ij}+\sum_{j\le k<i}a_{ij}$ and $Z=\sum_{k<i,j\le n}a_{ij}$. Then applying the condition to $a_{11},\dots,a_{kk}$, we get $2X+Y\ge kn$.

If $i\le k<j$, then if $a_{ij}=a_{ji}=0$, then we may toss $a_{ii}$ out of our set and replace it with $a_{ij}$ and $a_{ji}$, to form a larger feasible set of elements, contradiction. Consequently, $a_{ij}+a_{ji}\ge 1$ for $i\le k<j$, which upon summation yields $Y\ge k(n-k)$. Finally, $Z\ge (n-k)^2$, trivially. Therefore,
$$2S=2(X+Y+Z)\ge 2X+Y+Y+Z\ge kn+k(n-k)+(n-k)^2=k^2+2k(n-k)+(n-k)^2=n^2.$$

 Denote $\ell_i,c_i$ the $i$th line and column and $s(\ell_i), s(c_i)$ the sum of the elements on that row. Let $M=\max_{\sigma \in S_n} | \{ 1\le i\le n| a_{i\sigma(i)}=0\} |$. For each permutation $\sigma\in S_n$, assign to it the number $$\displaystyle\sum\limits_{a_{i\sigma(i)=0}} \left ( s(\ell_i)+s(c_{\sigma(i)}) \right  )$$
Now pick a permutation $\sigma$ such that it generates $M$ and its assigned number is minimal. Suppose wlog that $\{ 1\le i\le n| a_{i\sigma(i)}=0\}=\{1,2,..,k\}$. If $k=n$, we are done as twice the sum of all elements is $$\displaystyle\sum\limits_{i=1}^{n} s(\ell_i) +\displaystyle\sum\limits_{i=1}^{n} s(c_i)=\displaystyle\sum\limits_{i=1}^{n} (s(\ell_i)+s(c_{\sigma(i)}))\ge n^2$$

So suppose $k<n$ and look at a $j$ such that $k+1\le j\le n$. By the maximality of $M$, $\ell_j$ can have $0$ only on the columns $\sigma(1),...,\sigma(k)$. Note that if $\ell_j$ has a $0$ on $c_{\sigma(r)}$ with $r\le k$, i.e. $a_{j\sigma(r)}=0$, by the minimality of the assigned number of $\sigma$, we have that $s(\ell_j)\ge s(\ell_r)$. However, if the column $c_{\sigma(j)}$ has a $0$ on $\ell_r$, i.e. $a_{r\sigma(j)}=0$, by the same argument we have $s(c_\sigma(j))\ge s(c_\sigma(r))$, hence $s(\ell_j)+s(c_{\sigma(j)})\ge s(r)+s(c_{\sigma(r)})\ge n$.

Suppose that $0$ appears exactly $t$ times on $\ell_j$. This means that $s(\ell_j)\ge n-t$. Suppose that the $0$s of $\ell_j$ appear on the columns $c_{\sigma(i_1)},...,c_{\sigma(i_t)}$. We have seen earlier that if $c_\sigma(j)$ has a $0$ on $i_1,..,i_t$ we have that $s(\ell_j)+s(c_{\sigma(j)})\ge n$. So suppose that it doesn't have $0$ on these lines. However, $c_{\sigma(j)}$ has $0$s only on the lines $\ell_1,..,\ell_k$ by the maximality of $M$, hence $c_{\sigma(j)}$ has at most $k-t$ zeroes, so $s(c_{\sigma(j)})\ge n-k+t$. We thus infer that $$s(\ell_j)+s(c_{\sigma(j)})\ge (n-t)+(n-k+t)=2n-k\ge n$$
so $s(\ell_j)+s(c_{\sigma(j)})\ge n,\ \forall j\ge k+1$. By the hypothesis this is also true for $j\le k$ so summing over all values of $j$ we get that the sum is at least $\dfrac{n^2}{2}$.


geometry
3D geometry
tetrahedron
vector
analytic geometry
trigonometry
pyramid

I believe it is $\frac{s^3\sqrt{2}}{12}$.

It's not easy to explain without a diagram, but...
[hide="proof"]
By Pythagorean theorem, the height of the tetrahedron is $\sqrt{l^2-\left(\frac23 \cdot \frac{\sqrt{3}}{2}l\right)^2} =\frac{\sqrt{6}}{3}l$.

So $V=\frac13\cdot \frac{\sqrt{3}}{4}l^2\cdot \frac{\sqrt{6}}{3}l =\frac{\sqrt{2}}{12}l^3$
[/hide]

[quote="drunner2007"]What is the formula for the volume of a regular tetrahedron of side length $l$ . I know of the vector method (using coordinates), but what are some other methods for deriving the result.[/quote]

To use vectors to find this we take one third of the box product, right???(I'm learning how to use vectors on geometry...)

frt, I see what your saying.
When the tetrahedron is regular, we drop a line from a given vertex to intersect the opposite face perpendicularly. This pt. of intersection is the centroid of the bottom face (not sure of how to prove that fact).
Now draw an altitude along this bottom face, which will clearly pass though the centroid and one vertex. Moreover, the centroid lies 2/3 of the length of the segment from the vertex that the altitude passes through.
The length of this altitude is $\frac{\sqrt{3}}{2}s$ , where s is the length of one side of the tetrahedron. Thus the length of the segement from the vertex to the centroid is $\frac{2}{3}\frac{\sqrt{3}}{2}s$ .
Apply the pythagorean thm. to the right triangle with one edge of the tetra. as a hypotenuse, and the original perpendicular line and the 2/3 segment as legs. This gives that the height $h$ of the tetrahedron is given by $\sqrt{s^2- (\frac{s}{\sqrt{3}})^2 }=s\frac{\sqrt{6}}{3}$ as frt said.
Then $V=\frac{1}{3}\frac{\sqrt{3}}{4}s^2\frac{\sqrt{6}}{3}s=\frac{\sqrt{2}}{12}s^3$

[quote="drunner2007"]pk, the volume is $\frac{1}{6}$ of the box product. Let $\vec{u}, \vec{v}$, and $\vec{w}$ be three vectors that represent three vectors lying along three adjacent sides of the tetrahedron with tails at a common corner. Then, taking the side determined by $\vec{u}$ and $\vec{v}$ to be the base, the tetrahedron's base has area

$\frac{1}{2}\left\|\vec{u}\times\vec{v}\right\|$

(1/2 the area of the corresponding parallegram, whose area is given by the cross product). Now notice that the height is given by the length of $\vec{w}$ times the sine of the angle $\theta$ between $\vec{u}\times\vec{v}$ and $\vec{w}$ . Thus the volume is given by

$\frac{1}{3}\frac{1}{2}\left\|\vec{u}\times\vec{v}\right\|sin\theta\left\|\vec{w}\right\|= \frac{1}{6}|\vec{u}\times\vec{v}\cdot\vec{w}|$[/quote]

I see, or else mine will give a pyramid with parallolegram as the base...


Twenty percent of the 20 boys in Ms. Schmidt's class live on a farm. Thirty percent of the 30 girls in her class also live on a farm. What percent of the students in the class live on a farm?

[quote="007math"]FTW says the answer is 22%.[/quote]

$26%%
$ is correct.

[hide]$\frac{20}{5} = 4$

$30 \cdot \frac{3}{10} = 9$

$\frac{13}{50} = \boxed{26 \%}$

If the answer is 22 please report the question[/hide]


A magazine advertisement states that a subscriber saves 75.6% off the cover price at news stands by purchasing a 17-issue subscription rather than purchasing the 17 issues separately at the cover price. If a 17-issue subscription costs 10.99 dollars, what is the average cover price of each magazine when bought at a news stand? Express your answer as a decimal to the nearest hundredth.

From the information given, we know that 10.99 is 75.6% of the cover price of 17 separate issues, so those separate issues cost $ 10.99 \div .756 \equal{} 14.54$ (approximately.) We then divide this number by 17, since we want the cover price of one magazine, to get $ 0.86$.


If $\frac{a!}{b!}$ is a multiple of 4 but not a multiple of 8, then what is the maximum value of $a-b$?

Can someone give me a good explanation to how to do this?
I read the answer in aops solutions manual but I still couldn't understand how to do it. Thanks

Is the answer 3? like when a=5 and b=2 or 7 and 4.
Don't you just know that you can't have 2^3. Isn't it becasue every 3rd multiple of 2 is divisible by 2... so then it cant be  even odd even.. because then one of the evens would be a multiple of 4 and that gives us 3 2s. so then it has to be odd even odd. where the even is not a multiple of 8. so i guess 19, 18, 17 or 21, 20, 19 work... I think that is right.. if the answer is 3...

[hide] We know that $\frac{a!}{b!}=(b+1)(b+2)(b+3)...(a)$.  Now, take every 4 positive, consecutive integers.  There is going to be exactly one that is divisible by 4, and one that is divisible by 2, but not 4.  Thus, when you multiply 4 consectutive integers, the product is going to be divisible by (4)(2)=8.

Now we know that it is divisible by 4, but not by 8.  Thus, there cannot be 4 consecutive integers in $\frac{a!}{b!}$.  However, there can be 3 consecutive integers.  So, a-b has the maximum possible value of $\boxed{3}$ [/hide]


linear algebra
matrix
LaTeX
function
abstract algebra
linear algebra unsolved

Suppose $A$ is a matrix from $M_n(C)$ and $B$ is a matrix from $M_p(C)$. Define the Kronecker product as $A*B_{ij}=a_{ij}B$. It is a $(n+p)x(n+p)$ matrix and I will let you see how it looks since I don't know how to write it in Latex (please, a moderator edit this).
   a) Compute $det(A*B)$ as a function of $detA, detB$.
   b) Prove that if $A,B$ are diagonalisable so is $A*B$.

As for the first one: it is $(detA)^m(detB)^p$. There is a more general fact: If we write the eigenvalues of $A$ as $\alpha_1,\alpha_2,...\alpha_m$ and for $B$: $\beta_1,...,\beta_p$, then the eigenvalues of $A*B$ are all the products $\alpha_i\beta_j$.
For the second one
more general: if we Jordanise our matrices with Jordan blocks $J_p(\alpha)$,$J_q(\beta)$ for matrices $A,B$ respectively, then their tenzor product in basis, which is product of two bases, which diagonalises our matrices, we get the tenzor product $J_p(\alpha) \otimes J_q(\beta)$, which is also diagonal(as it is not difficult to verify).
We can leave out our words about tensor product, assuming the matrix of tesor product of two linear maps to be Kronecker product $A*B$ and working just with Kronecker products.

Would you mind posting a proof, eugene? ;)

I've almost written it:for 1)
Let $J_p(\alpha)$ and $J_q(\beta)$ be Jordan blocks of $A$,$B$ corresponding to their eigenvalues $\alpha$,$\beta$. It is just a hand work(it is a bit difficult to write it fully in Latex) to verify that the Kronecker product $J_p(\alpha)*J_q(\beta)$ has uppertriangular form with diagonal $(\alpha\beta,...,\alpha\beta)$.

consider the endomorphism : $f: M_{n,p}(K) \to M_{n,p}(K)$, $M \to AMB$.
If you write the matrix of $f$ in the canonical basis of $M_{n,p}(K)$, what do you notice ?


[hide]
1. dunno :blush:
2. You put $3x+62=7x+30$ for the first two sides congruent, $7x+30=5x+50$ for second and third side congruent, $3x+62=5x+50$ for first and 3rd side congruent Then solve for x in each of the eqations, the biggest solution for X is the one you go with. So it would be $8,10,6$ Use 10. Plug it in and add up the sides to get $292$
3. $2^-1 +2^-2.... + 2^-10=\frac{2^9+2^8+2^7.....2^0}{2^{10}}$. The sum of the powers of two for $2^0+2^1+....2^k=2^{k+1}-1$. So $\frac{2^9+2^8+2^7.....2^0}{2^{10}}=\frac{1023}{1024}$
4. For any question like that you count up the number of leters in it. That word has 7. So the number of ways to arange it is $7!$ but there are repeated letters, A is used twice and E is used twice. So you take $7!$ and divide it by $2!*2!$ to get $1260$[/hide]

[quote="kidwithshirt"]1. Of the 400 eighth-graders at pascals middle school, 117 take algebra, 109 take advanced computer, 114 take industrial technology. Furthermore, 70 take both algebra and advanced computer, 34 take both algebra and industrial technology, and 29 take both advanced computer and industrial technology. Finally, 164 students take none of these courses. How many students take all three courses?

2.The length of the sides of isosceles TRIANGLE ABC are 3x+62, 7x+30 and 5x+50 feet. What is the least possible number of feet in the perimeter of TRIANGLE ABC?

Whew hard to type[/quote]
[quote="kidwithshirt"]3. 2^-1+2^-2+2^-3+.....+2^-10=?

4. What is the number of distinct ways of arranging the ltters in the word AVERAGE.[/quote]
[hide="1"]
$117+109+114-70-34-29+x+164=400, \boxed{x=29}$
[/hide]

[hide="2"]$3x+62$ is the longest length. So $3x+62<12x+80, x\ge-2$ If $x=-2$, we get lengths$16, 40, 56$which is perimeter $112$ but the problem didnt say anything about integer lengths so is it something like$112.0000001$? [/hide]
[hide="3"]$\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}=\boxed{\frac{1023}{1024}}$[/hide]
[hide="4"]$7$ letters, $2$ repeats of $2$ letters. $\frac{7!}{2!2!}=\boxed{1260}$[/hide]

[quote="kidwithshirt"]1. Of the 400 eighth-graders at pascals middle school, 117 take algebra, 109 take advanced computer, 114 take industrial technology. Furthermore, 70 take both algebra and advanced computer, 34 take both algebra and industrial technology, and 29 take both advanced computer and industrial technology. Finally, 164 students take none of these courses. How many students take all three courses?

2.The length of the sides of isosceles TRIANGLE ABC are 3x+62, 7x+30 and 5x+50 feet. What is the least possible number of feet in the perimeter of TRIANGLE ABC?

Whew hard to type[/quote]
[quote="kidwithshirt"]3. 2^-1+2^-2+2^-3+.....+2^-10=?

4. What is the number of distinct ways of arranging the ltters in the word AVERAGE.[/quote]

[hide="2"]510[/hide]
[hide="4"]7!=5040[/hide]
[hide="1"]im not sure i got 29[/hide]

[quote="Naonao"][hide]

4. For any question like that you count up the number of leters in it. That word has 7. So the number of ways to arange it is $7!$ but there are repeated letters, A is used twice and E is used twice. So you take $7!$ and divide it by $2!*2!$ to get $1260$[/hide][/quote]

This is definitely helpful.. I never figured out how to do permutations lol

[quote="Klebian"]Chinesechess, you're missing 2 dollar signs. One after "x=-2", and one after your equation in number 3.[/quote]

Oh. Didn't know dollar sign codes affected hides... Thanks

Any methodical (fast) way to do #2?

Number 2 shouldnt be very complicated... Its a spring round problem

[hide="Number 2"]$3x+62$, $7x+30$ and $5x+50$ are the side lengths, so the perimeter is $15x+142$.  Now, it's down to finding the smallest $x$:

If $3x+62=7x+30$, then $x=8$.
If $3x+62=5x+50$, then $x=6$.
If $5x+50=7x+30$, then $x=10$.

So, since $6$ is the smallest, the perimeter is $15(6)+142=232$.[/hide]

[hide="Number 3"]The formula is $\frac{a(r^n-1)}{r-1}$ where $r$ is the common ratio, $a$ is the first term, and $n$ is the number of terms.

$\frac{\frac12\left(\frac12\right)^{10}-1}{\frac12-1}=\frac{1023}{1024}$[/hide]

[quote="robinhe"]
The formula is $\frac{a(r^n-1)}{r-1}$ where $r$ is the common ratio, $a$ is the first term, and $n$ is the number of terms.

$\frac{\frac12\left(\frac12\right)^{10}-1}{\frac12-1}=\frac{1023}{1024}$[/quote]

That baby is definitely going on to my formula list =0  :lol:

[quote="ccy"]Just for the record, these problems are from the 2000 state sprint, not the 2001 one.[/quote]

ty i realized that

Sorry, I meant $\frac{\frac12\left[\left(\frac12\right)^{10}-1\right]}{\frac12-1}=\frac{1023}{1024}$

 :D  :oops:


Let $(x_n)$ be a sequence such that : $0 \leq x_i \leq 9$ , $x_2=x_1+1$ and $x_n \equiv x_{n-1}^2+x_{n-2}^2$ (mod 7) . Suppose $x_{2545}=1$ . Find $x_4$ ?

By the checking of all variants $x_1$ we can find, that $x_4\equiv1(7)$.


function
advanced fields
advanced fields theorems

My book contains the following sentence:

[quote]Since u(x,y) is a nonconstant harmonic function on $ B_R \equal{} \{(x,y) | x^2 \plus{} y^2 \leq R^2 \}$, it must be a nonconstant harmonic function in each subdisk.[/quote]

I don't see why it cannot be constant on some subdisk.

Harmonic functions, like analytic functions, satisfy the identity principle. If it's constant on some open region, it's constant on everything connected to that.

follows from
statement: http://planetmath.org/encyclopedia/RigidityTheoremForAnalyticFunctions.html
proof: http://planetmath.org/?op=getobj&from=objects&id=6282


limit
search
real analysis
real analysis unsolved

Let $ x_o \in (0,1]$ and $ x_{n\plus{}1}\equal{}x_n\minus{}arcsin(sin^3x_n) ,n \geq 0$
Compute limit : $ \lim_{n \to \plus{}\infty}{\sqrt{n}x_n}\equal{}?$

See here

http://www.mathlinks.ro/Forum/viewtopic.php?t=45424

and also here

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2109762428&t=319109


calculus
integration
trigonometry
logarithms
calculus computations

$ \int \frac{f'(x)}{f(x)}\ dx\equal{}\cdots$

hello, after the hint of kunny we get
$ \int_{0}^{\frac{\pi}{2}}\frac{\sin(2x)}{1\plus{}\sin^2(x)}\,dx\equal{}\ln(2)$.
Sonnhard.


geometry
Support
number theory
prime numbers

This guy claims he proved the Riemann Hyp.

Just out now.

http://arxiv.org/abs/0807.0090

terrence tao seems to have found an error in it.

http://terrytao.wordpress.com/2008/02/07/structure-and-randomness-in-the-prime-numbers/#comment-30714

Surely not [i]every[/i] paper that goes onto the internet with a claimed proof of the Riemann Hypothesis deserves to be posted here?  (Especially not with such a credulous title -- I've added a "?" at the end, I hope you don't mind.)

We should really wait until Clay Mathematics Institute verifies such a claim before we get excited.

[hide="Look I proved Riemann Hypothesis!"]

$ 1\plus{}1\equal{}3$

Yep, I did it. Now where's my money?
[/hide]

Even though my "proof" is quite ridiculous and is almost an insult to the Riemann Hypothesis (and for this, I apologize), you can see what happens when people "claim" they solved something.

man don't be an ass

apparently the guy that put forward this claim is actually quite knowledgeable in the subject area

and OP noted it was a claim so no need to beat up on him

I don't even know how to respond to the demon because seriously wtf

Lots and lots of very knowledgeable people have claimed to prove the Riemann Hypothesis.  A claim to have proved it is, frankly, uninteresting.  A claim to have proved it that has been subjected to some sort of peer review process (either formal, i.e. in the journal submission process, or informal) might be interesting, but a paper on the arxiv lacking the support of some experts in the area really just isn't.

So apparently Xian Li got his PhD under one of the most famous guys for a non-proof of the Riemann Hypothesis, but still a very famous guy for other great work in mathematics. He also published an apology paper for claiming to have proven it. Li was the student who found the error in his professor's paper, and now he claims to have proven the result.

Anyway, terry tao just said some comments about that paper

[quote="manuel"]So apparently Xian Li got his PhD under one of the most famous guys for a non-proof of the Riemann Hypothesis, but still a very famous guy for other great work in mathematics.[/quote]

just to elaborate, the aforementioned famous guy is [url=http://en.wikipedia.org/wiki/Louis_de_Branges_de_Bourcia]de Branges[/url], who's famous mostly for proving the Bieberbach conjecture (well, that and the numerous claimed proofs of RH)

These turn up practically once a month: http://eprintweb.org/S/article/math/0806.2361


given that a set whose elements are zero and even integers, positive &negative of the five operations applied to any pair of elements :(1)addition (2)subtraction (3)multiplication (4) division (5)finding the arthmetic mean (average) ,those operations that yields the elements of S are  :?:

$ S\equal{}${...$ \minus{}4$, $ \minus{}2$, $ 0$, $ 2$, $ 4$...}

Now apply each operation to two elements of $ S$.

1.) Addition: $ 0\plus{}4\equal{}4$, $ \minus{}2\plus{}2\equal{}0$. Since every time I add elements of $ S$ they equal an element of $ S$, addition is one of the answers.

2.) Subtraction: $ (\minus{}4)\minus{}(\minus{}2)\equal{}\minus{}2$, $ \minus{}4\minus{}0\equal{}\minus{}4$. Again, this operation always yeilds an element of $ S$ so it is another answer.

Now repeat this process for the remaining operations and your answers are the operations that yeild elements of $ S$. It appears to me that addition, subtraction, multiplication are the answers because in division odd numbers can result.(i.e. $ \displaystyle\frac{2}{\minus{}2}\equal{}\minus{}1$). And this is the same for averaging.


Solve the equation $ 3^x(4^x\plus{}6^x\plus{}9^x)\equal{}25^x\plus{}2.16^x$

Should be locked!

[url]http://www.mathlinks.ro/viewtopic.php?p=1382367#1382367[/url]

Japanese Communities Moderator

kunny


From IMO Team Selection Test 2000

* Let p be a prime number. For integers r,s such that rs(r^2-s^2) is not divisible by p, let f(r,s) denote the number of integers n in {1,2,3,...,p-1} such that {rn/p} and [sn/p] are either both less than 1/2 or both greater than1/2. Prove tht there exists N>0 such that for p >= N and all r,s,

The smallest integer greater or equal to (p-1)/3  <= f(r,s)  <= the greatest integer less than or equal to 2(p-1)/3.

Note: {x} : the farctional part of x.


Let $ G$ be a connected graph such that if any vertex of the graph is removed (and with it all the edges connected to it) the graph remains connected. Prove that for any two vertices, there exist two disjoint paths connecting the two vertices.

Suppose there is a pair of vertices X and Y such that any path between them contains an edge e=AB. If e is part of a cycle then there exists a path from A to B that does not include e and then we can find a new path from X to Y by traversing this new path around e - contradiction. If e is not part of a cycle then it is a cut-edge and since removing A or B removes e as well this is a contradiction. Note that if the component of A is empty after A is removed then we can instead remove B and if its component is empty as well then the graph is just a 2-vertex line and the statement doesn't hold in this case.

It's false; take $ G \equal{} K_2$. :)

This would suggest that the above argument needs to be refined a bit.

In what sense does the argument need to be refined?  dsadsa correctly noted that the statement fails for $ G \equal{} K_2$.   :maybe:

Hmm...

What if we define "disjoint paths" by "paths that share no common vertex"? Is the statement still true?

[quote="dsadsa"]If all paths from X to Y go through a vertex Z with Z =/= X and Z =/= Y then removing Z will result in no paths from X to Y implying that the graph is not connected.[/quote]

But for the statement to be false it's not necessary that all paths from $ X$ to $ Y$ go through a vertex $ Z$...

If two vertices are joined by three paths and each pair of paths intersect in a vertex (but the three paths don't have any vertex in common), I believe it's possible to patch bits and pieces of the paths together to form two disjoint paths joining the vertices.  (This is based on drawing a picture of what I hope is a generic situation.)

We induct on the distance between X and Y. If the distance is 1 then they are adjacent and X has at least one other vertex adjacent to it other than Y, call it Z. Then remove X and since G is still connected we can find a path from Z to Y and then add X to it to construct our second path. Suppose the statement holds for a distance of K.

Then for a distance of K+1 we consider the shortest path S between X and Y. Let A be the unique vertex adjacent to X on S. We can find two disjoint paths from A to Y, call them P1 and P2. Now remove A. There exists a path L from X to Y since G-A is connected. If L is disjoint from P1 and P2 then we are done. If not then let L intersect P1 first (WLOG) at Z as one travels down the path from X to Y. Now we can easily construct two disjoint paths from X to Y.

The first path goes from X, down part of L to Z and down P1 to Y.
The second path goes from X to A and then down P2 to Y.

By POMI the statement holds for all distances between X and Y.


inequalities
vector
real analysis
real analysis unsolved

Holder's inequality gives us the general norm inequality
\[{\| u \|}_{2}\le{\| u \|}_{1}^{1/3}{\| u \|}_{4}^{2/3}\, . \]
So in our situation we get the lower bound ${\| u \|}_{1}\ge \frac{1}{C^{2}}$.  I'm not sure whether that is the greatest lower bound.

Holder is sharp only for collinear vectors, that is, when $(u_{n})$ is a constant multiple of $(u_{n}^{4})$. This happens when all nonzero elements of $(u_{n})$ are the same. But you can't always arrange this under the stated assumptions on the norms.

Using that observation, I think we can show that my lower bound can be achieved for infinitely many values of $C$.  Namely, any $C$ of the form $\frac{1}{\sqrt[4]{N}}$, where $N$ is a positive integer.


Suppose that a,b,c,d are real numbers in descending order. Given
a^2 + d^2 = 1 ;
b^2 + c^2 =1 ;
ac+bd = 1/3 ;

Find the value of (ab-cd).

[hide="brute force"]
we have:
$ a^2\plus{}d^2\equal{}1$ and $ b^2\plus{}c^2\equal{}1$
Multiplying,we get:
$ a^2b^2\plus{}a^2c^2\plus{}b^2d^2\plus{}c^2d^2\equal{}1$   .............(i)

also,we have:$ {(ac\plus{}bd)}^2\equal{}a^2c^2\plus{}b^2d^2\plus{}2abcd\equal{}\frac19$  ........(ii)

Subtracting (ii) from (i),we get:
$ a^2b^2\plus{}c^2d^2\minus{}2abcd\equal{}\frac89$
$ {(ab\minus{}cd)}^2\equal{}\frac89$

 :D

[/hide]

OK this was the easiest method. No pun intended.

There are two more methods.The one I found really rips this apart in one line.

Well, this is very simple actually. Maybe this method is the one tht conjurer told rips the sum in one step.

Solution: We can write $ (a^2 \plus{} d^2) ( b^2 \plus{} c^2) \equal{} (ac \plus{} bd)^2 \plus{} (ab \minus{} cd)^2 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > (1)$.
Now , given that
$ a^2 \plus{} d^2 \equal{} 1 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > (i)$
$ b^2 \plus{} c^2 \equal{} 1 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > (ii)$
$ ac \plus{} bd \equal{} 1/3 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > (iii)$
Substituting the values  in the equation $ (1)$, we get that   $ (ab \minus{} cd)^2 \equal{} 1 \minus{} \frac {1}{9} \equal{} \frac {8}{9}$ yielding us that $ (ab \minus{} cd) \equal{} \pm \frac {2 \sqrt{2}}{3}$

Hey arvind, i thought u left Aops :rotfl:

neways, nice solution :D

Well you just generalized what skand already did. No pun intended here too.

My sol:

Define x = a$ {\hat i}$ + b$ {\hat j}$

y = c$ {\hat i}$ + d$ {\hat j}$

We are given x$ dot$y and we need to find x$ cross$y . Simple :)

Since there is a simple vector solution, there outta be a complex no solution as well. And a trignometric too. :p

Complex:  let $ x \equal{} a \plus{} di, y \equal{} b \plus{} ci$.  We are given that $ |x| \equal{} |y| \equal{} 1$ and $ xy$ has imaginary part $ \frac{1}{3}$...

Trig:  let $ a \equal{} \cos \alpha, d \equal{} \sin \alpha, b \equal{} \cos \beta, c \equal{} \sin \beta$.  We are given that $ \sin (\alpha \plus{} \beta) \equal{} \frac{1}{3}$...

(Of course, these two solutions are equivalent and pretty much equivalent to the "algebraic" solution as well.)


calculus
integration
vector
trigonometry
geometry
parameterization
analytic geometry

Hello all,

[img]http://img244.imageshack.us/img244/218/picture8ce5.png[/img]

I am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called).

any suggestions people?

i get that nabla x F or curl(F) = [b]i[/b] + [b]j[/b] + [b]k[/b]

cheers

-sarah :)

Well, a way to find N is to first find any normal to the surface and then normalize it. Some lead-in work: Put the surface S into vector form,

$S: \, \, \, \vec{r}(u,v)=\left< u\cos v,u\sin v, v\right> \, \, 0\leq u\leq 1,\, 0\leq v\leq\frac{\pi}{2}$

Also, recall that the vectors $\frac{d\vec{r}}{du}\mbox{ and }\frac{d\vec{r}}{dv}$ are tangent to S, and that the cross product of two vectors is normal to both vectors, so a normal vector to S is $\frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv},$ but we want a unit normal to S, so we normalize the above vector (divide it by its magnitude) to get

$\vec{N}(u,v)= \frac{\frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}}{\left| \frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}\right|}$

ok i do that and i get the unit normal to be $\frac{sin(v)i-cos(v)j+u k}{\sqrt{1+u^{2}}}$

how does that look?

and then how do i use that? that is what do i do next?

Write down the Stokes' theorem. You already have the curl and normal vector; your inifinitesimal area is $du dv$. Put the boundaries of double integral and you're essentially done.

how do i work out what dS is?, i think that is my main problem....

how does this look for the integral we need to evaluate?

[img]http://img151.imageshack.us/img151/5053/picture9fn3.png[/img]

that is, do we have to use the normal or the unit normal when doing this stuff?

$dS=\left| \frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right|\,du\,dv$, this is a kind of change-of-variables formula. This term cancels the denominator in the formula for unit normal vector.

mlok, is that forumula you gave for d[b]S[/b] or dS ?

That was $dS$. The vector form of the area element ($d\mathbf S$) is the vector of length $dS$ in the direction perpendicular to the surface. \[d\mathbf{S}=\left(\frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right)\,du\,dv\] $d\mathbf{S}$ can be used to compute surface integrals of vector fields directly, without reducing them to a surface integral of a function. Namely, \[\iint_{S}\mathbf{F}\cdot d\mathbf{S}=\iint_{D}\mathbf{F}(\mathbf{r}(u,v))\cdot \left(\frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right)\,du\,dv\] where $\mathbf{r}$ is the parametrization of $S$ and $D$ is the parameter domain. Look, no magnitude of vectors!

ok i see.  so how do you work out what dS or d[b]S[/b] is when you arnt given the parametric representation?

You have to find a parametric representation first. This may be the most difficult part of the problem. In the simplest case when your surface is a graph (say $z=g(x,y)$) you can take $x$ and $y$ to be your parameters, so that $\mathbf{r}(x,y)=\langle x,y, g(x,y)\rangle$. In other cases you can try a couple of cylindrical or spherical (ugh!) coordinates as parameters. Whether or not they work will depend on your ability to solve for the third coordinate.


Are any of you enrolled in Kumon, a math and reading program?
If so, what level are you in?  I'm in L on math and J for reading.

I used to do the math program and got up to J. Kumon is useful for building basic skills like factoring, but I think it emphasizes rote learning far too much. I can't speak for the english part of it, though.

I finished Reading (Level L) a couple of months ago and just finished Math level Q in July.

I would be lying if I said I enjoyed it. But upper levels were better. I heard in Japan math goes further to level R or S.

I think I remember my babysitter doing Kuman, and my mom wanted me to do it too, but my dad was really against it.  He says that it foocusses to much on knowing how, not why.  Being a math professor probably has something to do with it.

I did Kumon for maybe 5 years?  My school required it, so I did it from preschool (I think) to 4th grade (I left after that year).  I think I got up to level G-ish, H-ish; I can confidently say that I was faster in computations than I yam now.   :cool:.  It did help me boost my confidence to actually figure out that I was decent at math.

ah, man, the kids i was babysitting had to do kumon and it was HORRIBLE. i really didnt like making them do it, it seemed so busywork-esque, which i've ALWAYS ALWAYS hated. give me lots of homework, i dont care, as long as its not repetitive. its not a type of learning i necessarily agree with, but "different strokes for different folks."

I did Kumon once sometime in elementary school (I don't remember what level) and couldn't stand it.  They had me working on speed with basic arithmetic problems when I was well beyond that level and ready for more advanced material.  Needless to say, I didn't stick with it for more than a few months, if I remember right.

Just don't go. It's a waste of your money.

My parents made me do Kumon for about 2 to 3 years in elementary school. The program was good only for practice with basic arithmetic.  It might be good for a kid in elementary school who is in one of those "new math" curricula, but other than that, the program is essentially useless.  The problems are simple enough that you don't really learn anything new by doing them (especially if you have already seen the material).  I could rant and rave about Kumon for a long time, but instead I'll just say that if you are considering signing your kid up for Kumon don't.  Instead, sit down with him/her and a good math book which is ahead of what he/she is covering in school.  That way, by going through the book, your child will be forced to develop his/her understanding of the foundations of math in order to learn the new material.  At least this is my take on the whole issue.


Let $ \alpha$ be the positive root of equation: $ x^2\minus{}2008x\minus{}1\equal{}0$.
$ x_o\equal{}1, x_{n\plus{}1}\equal{}$[$ \alpha x_n$]
Find the remainder when we divide $ x_{2008}$ by 2008.
Everyone help me, please!! :|

First, we know that $ a_1\equal{}[\alpha]$, $ a_2\equal{}[\alpha^2]$ ...

So that $ a_2008\equal{}[\alpha^{2008}]$.

$ 2008<\alpha<2009$.

Therefore, $ 0<\alpha\minus{}2008<1$

So, $ [(\alpha\minus{}2008)^2008]\equal{}1$.

Therefore, the answer is $ 1$.

This may or may not be right...

[quote="BOGTRO"]First, we know that $ a_1 \equal{} [\alpha]$, $ a_2 \equal{} [\alpha^2]$ ...

So that $ a_2008 \equal{} [\alpha^{2008}]$.[/quote]  No.  Consider a sequence with $ x_0 \equal{} 1$, $ x_{n \plus{} 1} \equal{} \left\lfloor \frac {5}{2} x_{n}\right\rfloor$ -- we have $ x_1 \equal{} 2$, $ x_2 \equal{} 5$ but $ \left\lfloor \left(\frac {5}{2}\right)^2\right\rfloor \equal{} \left\lfloor\frac {25}{4}\right\rfloor \equal{} 6$.  The rest of your solution doesn't make much sense, either.

Anyone help me, please! :help:
I think it isn't a hard problem (not to me :oops: ).
So reply as soon as possible.
Thank you!

[quote="pythag011"][hide="Solution"]a = $ \frac {2008 \plus{} \sqrt {2008^2 \plus{} 4}}{2}$, which is > 2008 but < 2009. so [a] = a-2008. and $ a_2 \equal{} [a^2 \minus{} 2008a]$ But we know that $ a^2 \minus{} 2008a \equal{} 1$ So we get a_2 = 0. So $ a_2008 \equal{} 0.$ [/hide][/quote]
No, you are wrong.
[a] is the greatest integer number smaller than $ a$ and {a} = a - [a]

Nobody solve this problem so I'll post my teacher's solution.
[hide="Here"]$ x_{n+1}=[\alpha x_n]<\alpha x_n<[\alpha x_n]+1=x_{n+1}+1$
$ \Rightarrow \frac{x_{n+1}}{\alpha}<x_n<\frac{x_{n+1}}{\alpha}+\frac{1}{\alpha}$
$ \Rightarrow x_n-1<x_n-\frac{1}{\alpha}<\frac{x_{n+1}}{\alpha}<x_n$
$ \Rightarrow [\frac{x_{n+1}}{\alpha}]=x_n-1$
$ \Rightarrow [\frac{x_n}{\alpha}]=x_{n-1}-1$
We have $ \alpha^2-2008\alpha-1=0$.
$ \Rightarrow \alpha^2=2008\alpha+1$
$ \Rightarrow \alpha=2008+\frac{1}{\alpha}(1)$
In the other hand, we have:
$ x_{n+1]<\alpha x_n<x_{n+1}+1}$
$ \Rightarrow x_n+1<(2008+\frac{1}{\alpha})x_n<x_{n+1}+1$
$ \Rightarrow x_n+1<2008x_n+\frac{1}{\alpha}x_n<x_{n+1}+1$
$ \Rightarrow x_{n+1}=[2008x_n+\frac{x_n}{\alpha}]=2008x_n+[\frac{x_n}{\alpha}]$
$ =2008x_n+x_{n-1}-1$
$ \Rightarrow x_{n+1}\equiv (x_{n-1}-1)\ mod\ 2008$
$ x_{2008}\equiv (x_{2006}-1)\equiv (x_{2004}-2)\equiv ...\equiv (x_0-1004)\mod\ 2008$
Because $ x_0=1$ , the remainder is 1005[/hide]


How many triangles are in the figure to the right?

see the figure to the following pdf link  warm up -1 problem no 5

http://sch.ci.lexington.ma.us/~jfrost/MC/2004%20-%202005%20Warmups.pdf?FCItemID=S0133B27B

answer 8

Hi sri340.

Nothing happens when I click on your link.

For starters, you have the $ 1$ big triangle. You then have ecah of the $ 4$ interior triangles. The two most left triangles can form a triangle. The two most right triangles can form a triangle. Lastly, the three most right triangles can form a triangle. This is a total of $ 1\plus{}4\plus{}1\plus{}1\plus{}1\equal{}\boxed{8}$ triangles.


calculus
integration
superior algebra
superior algebra unsolved

So, I need to find an integral basis for O which is the ring of integer of K=Q(theta) where theta is a root of t^3+t+1.

That's my first question.

But I'm thinking about how a prime p in Z would factor(decomposes) in O. I know the decomposition has  a certain form and I wanted to find all the forms that this decomposition could take. For example, if I have a prime p then <p>O=pqr for three prime ideals is one example and <p>=(p^2)q is another example. I don't know these actually occur in this ring of integers, but I was hoping for a lot of help here.

I figured out the first part. Just need help with the decomposition, so you can ignore the first part of the question.


A train traveling from Aytown to Beetown meets with an accident after $ 1$ hr.  It is stopped for $ \frac{1}{2}$ hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown $ 2$ hr. late.  If the train had covered $ 80$ miles more before the accident, it would have been just $ 1$ hr. late.  The usual rate of the train is:

$ \textbf{(A)}\ \text{20 mph} \qquad
\textbf{(B)}\ \text{30 mph} \qquad
\textbf{(C)}\ \text{40 mph} \qquad
\textbf{(D)}\ \text{50 mph} \qquad
\textbf{(E)}\ \text{60 mph}$

[hide="solution"]
town $ A$ and $ B$ are seperated by distance $ d$ and the trian usually travels $ v$ miles/hr then the trip usually takes $ \frac{d}{v}$ hours. Now we can set up two equations;

$ \frac{d}{v} \plus{} 2 \equal{} 1 \plus{} \frac{1}{2} \plus{} \frac{d\minus{}v}{\frac{4v}{5}}$
$ \Rightarrow d\equal{}7v \qquad(1)$

$ 1 \plus{} \frac{d}{v} \equal{} \frac{80}{v} \plus{} \frac{1}{2} \plus{} \frac{5(d\minus{}80)}{4v}$
$ \Rightarrow d \equal{} 80 \plus{} 2v\qquad(2)$

Subbing (1) and (2)
$ 5v \equal{} 80 \Longrightarrow v \equal{} 40 \longrightarrow \boxed{C}$[/hide]

I think you are solving the question where the train travels 80 miles before the accident, as opposed to 80 miles MORE before the accident.

[hide]
The train gains an hour by traveling 80 miles at full speed instead of 4/5 speed. Let r=speed. Then
$ \frac{80}{r}\equal{}\frac{80}{\frac{4}{5}r}\minus{}1$
$ r\equal{}20\Rightarrow\boxed{A}$
[/hide]

[quote="SCHNOGGINVOGGINSCHWITZERW"]I think you are solving the question where the train travels 80 miles before the accident, as opposed to 80 miles MORE before the accident.

[hide]
The train gains an hour by traveling 80 miles at full speed instead of 4/5 speed. Let r=speed. Then
$ \frac {80}{r} \equal{} \frac {80}{\frac {4}{5}r} \minus{} 1$
$ r \equal{} 20\Rightarrow\boxed{A}$
[/hide][/quote]SCHNOGGINVOGGINSCHWITZERW has the correct answer to this problem.

ocha, you had everything right until your second equation, at which point you just threw every thing out the window.
The left side is right, but the right side is wrong...
you should get that the LHS=$ 1 \plus{} \frac {80}{v} \plus{} \frac {1}{2} \plus{} \frac {5(d \minus{} v \minus{} 80)}{4v}$

[quote=Bryanli]i got what ocha got hum...[/quote]

thank you, everyone was wondering if ocha was right or wrong. In fact, research mathematicians were pondering for 8 years whether "ocha" or "nittanylion" solved the big [i]train problem[/i] correctly!


Adam, God made out of dust
But thought it best to make me first,
So I was made before man
To answer God's most Holy plan.
A living being I became
And Adam gave to me my name.
I from his presence then withdrew
And more of Adam never knew.
I did my Maker's law obey
Nor ever went from it astray.
Thousands of miles I go in fear
But seldom on earth appear.
For purpose wise God did see,
He put a living soul in me.
A soul from me God did claim
And took from me the soul again.
So when from me the soul had fled
I was the same as when first made.
And without hands, or feet, or soul,
I travel on from pole to pole.
I labor hard by day, by night
To fallen man I give great light.
Thousands of people, young and old
Will by my death great light behold.
No right or wrong can I conceive
The scripture I cannot believe.
Although my name therein is found
They are to me an empty sound.
No feat of death doth trouble me
Real happiness I'll never see.
To Heaven I shall never go
Or to Hell below.
Now when these lines you slowly read,
Go search your Bible with all speed
For that my name is written there
I do honestly to you declare.

Who or what is the “I” referred to in the riddle?

[quote="K Sengupta"]

Who or what is the “I” referred to in the riddle?[/quote]

I'm pretty sure that it is a star...am I right?


alkene + bromine water

what would be the products of this reaction?

two products one would be the vicinal di bromide and other the bromohydrin (major)
see [url=http://www.chemguide.co.uk/organicprops/alkenes/halogenation.html]here[/url] .
also the mechanism has bromonium ion intermediate.

That reaction is a classic test used for the identification of double bonds C=C, although for that purpose it is more commonly used a CCl4 solution of bromine.

As Pardesi said, the products are the bromohidrin and the vicinal dibromide. The first step of the mechanism involves a complexation of one bromine molecule (acting as a Lewis acid) by one molecule of alkene (the Lewis base), which is converted to a bromide ion and a cyclic bromonium ion. This step is reversible. In the next step a nucleophile (a water molecule or a bromide anion) attacks one of the carbons of the "double bond" in a SN2 fashion to yield the products.

The active attacking species can also be hypobromous acid (HBrO), since it is know that the halogens disproportionate in water to halide and hypohalide ions.


algebra polynomial AMC AIME USA(J)MO USAMO inequalities

Is there a gener way of solving this kind of problem?

Given complxe numbers $a,b,c$ such that

$a+b+c=\omega$
$a^2+b^2+c^2=\beta$
$a^3+b^3+c^3=\phi$
find
$a^m+b^m+c^m$

where $\omega,\beta,\phi,m$ are all real.

i know it is almost impossible to solve for $a,b,c$ and i know you have to manipulate the equations in some way, but im am just not sure of a general way of doing so.
could someone please help me.

The quickest way to solve problems of the sort is to find the symmetric sum of $ab$, $a^2b$, and $abc$ by elementary manipulations of the given three identity's, then forming a polynomial with roots $a,b,c$ by relating the specific sums with Viete's formulas. From there, you just recursively apply Newton's Sums and get the desired sum.

could you please provide an example.
i do not understnad how you would use Viete's Sums.

I will provide an example which was on Mildorf's Mock AIME 3, which can be found here:

http://www.tjhsst.edu/~tmildorf/math/Adv.htm

Look at problem #4, which is a problem almost identitical to the one you gave. In the solutions, Mildorf solves it exactly as I stated earlier, in much greater detail than I'm willing to write out.

this tactic is also (briefly)  mentioned in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422380&highlight=#422380]www.artofproblemsolving.com/Forum/viewtopic.php?p=422380&highlight=#422380[/url]

Try this USAMO problem from 1973: Find all solutions, real or complex, to the system of equations $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=3$.

[hide="here is what i got"]
given $x+y+z=3$
we have $(x+y+z)^2=3^2=9=x^2+y^2+z^2+2(xy+yz+zx)$
so we find that $xy+yz+xz=3$
then
$(x+y+z)(x^2+y^2+z^2)=(3)(3)=9=$
$x^3+y^3+z^3+(x+y+z)(xy+yz+xz)-3xyz$ so $xyz=1$
so from this info. we can use Viete's Sums and find that
the solutions to the system are the roots to the equation
$x^3-3x^2+3x-1=(x-1)^3=0$
[/hide]

[quote="Altheman"]i am amazed how simple that problem is for an olympiad problem[/quote]

It is one of the first few years of the USAMO (2nd, I believe).

That's right, maokid. You could also use Newton's sums.

[quote="JohnConstantine"]Or simply note that x^2+y^2+z^2=xy+xz+yz which is true for rearrangement only for x=y=z[/quote]

Rearrangement doesn't hold for complex numbers... A similar mistake is applying the Power-Mean Inequality to the system.


function
real analysis
real analysis unsolved

If $\mu(E_n)<\infty$ for $n\in\mathbb{N}$ and $\chi_{E_n}\to f$ in $L^1$, then $f$ is a.e the characteristc function of a measurable set.

Firstly we have that $\chi_{E_k}(x)$ converges to $f$ in measure, hence by Riesz's Theorem, there is a subsequence of $\{\chi_{E_k}(x)\}$, say $\{\chi_{E_{k_i}}(x)\}$, converges to $f$ almost everywhere.
Let $P=\liminf_{i\to\infty} E_{k_i}$. Try to prove that $f=\chi_P$ a.e.

Thanks.  I couldn't figure out whether to work with $\limsup E_n$ or $\liminf E_n$.

I don't really see the necessity of $ \mu(E_n) < \infty$.
Where does it play role?

Otherwise $ \chi_{E_n}$ is not in $ L^1$...

Oh, I see now. That makes sense.
But it doesn't play any role whatsoever later one.
You just need to notice that $ \chi_{liminf_{n\to\infty} E_n}\equal{} liminf_{n\to\infty} \chi_{E_n} \equal{} lim_{n\to\infty} \chi_{E_n} \equal{} f$


geometry AMC AIME trigonometry USA(J)MO USAMO HMMT

My goal this year is to pass the AIME and make USAMO. To do this, it would help immensely to learn some geometry because that is definitely me weakest subject.

I can get almost any book, and I already have Challenging Problems in Geometry (though I find this really boring), Geometry Unbound (which is pretty difficult, but does this help very much with AIME?), Art and Craft of Problem Solving, including the geometry chapters, and AoPS Volume 2.

I think I might as well finish whatever sections there are on analytic geometry in Volume 2, but I am unsure what to move on to after that. Suggestions?

How far does Intro. to Geometry go? (as in how far through AMC 10, AMC 12, and AIME)

Actually, I have seen several difficult AIME problems (some of them like 8+) that can be solved using things from Introduction to Geometry.

For Geometry Revisited, which chapters would you recommend the most for AIME-prep? More specifically, would chapters 4, 5, and 6 (transformations, inversive geometry, and projective geometry respectively) help any?

Yeah in AoPS Vol. 2, make sure you master the Trig chapter, the Triangles chapter, and the Cyclic Quadrilaterals chapter.

Also, along the lines of trig, it's helpful to learn the Complex Numbers chapter (highlight is De Moivre's theorem, if you're focusing on trig).

[quote="AIME15USAMO"][quote="AwesomeToad"]rruscyk said most of the non-trig AIME geometry can be solved using stuff from Intro to Geometry. Although you may want to learn to solve things by coordinates.[/quote]
Doesn't Intro to Geometry have a chapter on analytic geometry?[/quote]

No, I didn't see one...

Wish it did though...

Yes, the second edition of Intro to Geometry does have a section on Analytic Geometry.

Chapter 17 of Introduction to Geometry is Analytic Geometry. These are the sections from it.

17.1 Lines
17.2 Circles
17.3 Basic Analytic Geometry Problems
17.4 Proofs with Analytic Geometry
17.5 Distance Between a Point and a Line
17.6 Advanced Analytic Geometry Problems
17.7 Summary

I guess, from what limac says, only the second edition has Analytic Geometry. I didn't know AoPS added or cut out chapters from the books, I thought the new editions only looked nicer and they fixed the minor typos.

actually the first edition of geo. is much different from the second edition
there is an addition of chapters on [b]analytic geometry and trigonometry[/b] real important.

I'm in a similar situation (trying to make USAMO, with geometry as a weak point).

I wanted to ask, do you think the geometry in Aops Vol. 2 is sufficient, accompanied with a heavy dose of past problems?  I seem to be improving well so far, but I want to be able to tackle the more difficult AIME geo questions (10-15 range).  And if not, would the Intro to geometry book be the best to get, assuming I could only get one more book?

Thanks

lol, I guess I should have mentioned...I cannot solve any AIME geometry problems. Though I can still manage to get almost every non-geo problem less than 10 right, and one of 11-15 right, this kind of sucks. The reason I am not looking for something like Intro to Geo is because I feel more mathematically mature, and able to attack harder stuff. Though, perhaps I need to humble myself...

But anyways, I will give Geometry Revisited a try. If it's too difficult, I'll figure something else out.

Well in that case, I can give you some advice since I was in that situation not too long ago.

Do past problems.

This is probably the single most helpful thing to do, especially for your weak points.  I learned some theorems I'd never heard of and some approaches that I'd never thought of trying.  Doing lots of past AIME geo questions has gotten me to where I can get most of the easy/mid questions right.  Personally, I'm wondering what to do from there.

Isn't that going a little far: I guess for the really late problems but is ACoPS needed for that?

[quote="worthawholebean"]The geo chapters in ACoPS were incredibly helpful to me. Go through those.

It really doesn't matter what you do, as long as you do a wide range of problems.[/quote]
Exactly what he said. Plow through ALL the HMMT geometry problems and the easier olympiads that have geometry (canada, ireland etc.). Very helpful.

[quote="SnowEverywhere"]Just curious. In what way do you find challenging problems in geometry boring? I thought it was a really great geometry book. Each question teaches new methods of construction. I personally liked it and found it to be quite engaging.[/quote]

Well...maybe it's just my own motivation and interests. I've never really been fond of drawing stuff. Perhaps the whole advent of Euclidean Geometry just doesn't appeal to me.

I like analytic geometry quite a bit though.

You're not fond of drawing stuff but you like analytic geometry? Isn't that kind of contradicting yourself?  :P

Well, you have to draw a lot less for analytic...generally, there are fewer auxiliary lines someone has to draw and they don't have to be nearly as accurate (at least not for me)

For me, I think that mastering the use of similar triangles, power of a point, cyclic quadrilaterals, and trig identities is VERY important in the AIME.  About 80% of AIME geo problems can be solved using those.  Personally, I don't think that mastering analytic geo is important, since I've never seen a AIME geo problem that can be most easily solved using analytic methods.  Of course, my strong point is Euclidean geo (and I've never been a fan of analytic geo), so I might be mistaken on that point.

Last year I was in a similar boat. AOPS 2 geometry and a lot of problems got me to the point where I can solve any geo problem up to 10 and about half of the ones above it. Geometry is about pattern recognition and using things that you've done in similar situations in the past, since it isn't an intuitive subject like algebra where you just do things and the answer should come out. These patterns include:
-Similarity
-Power of a Point
-Cyclic Quadrilaterals
-Angle Chasing
-Basic Trig (Law of sines, cosines, tangents, & sum-to-product)
-Triangle area formulas
-Inradius/Circumradius relations
-Basic transformations (see AOPS 1)
These are all in AOPS 2. The problems I did were all of the AIME geometry problems from 2000 up till 2009, HMMT problems from 2005 until 2009, and WOOT Problems. If you aren't in WOOT, I found ACOPS geometry really good, although you need to be patient, because they're much harder problems. However, ACOPS really lets you master and appreciate geometry at that level.

I'm not in WOOT, but could someone please post some WOOT AIME geometry problems please?

What about the trig questions on the AIME?  How should you learn trig?

I was doing a couple of the HMMT Geo problems and they were helpful.
Are the really hard HMMT problems around AIME level?

@mathwizarddude: I don't think we are allowed to post WOOT problems.. but there's the the olympiad resources section of the website that has a bunch of problems

@Truffles: The AOPS 2 Trig & Triangles Chapters are great- they cover basic definitions, identities, graphing, properties of graphs, sum-to-product, inverse functions, law of sines, law of cosines, law of tangents, and many area formulas that use trig. You can also use 103 problems in trigonometry- it's awesome at trig, but if you have limited time, go with the AOPS route.

@AlphaBetaTheta: The HMMT problems range from medium to hard AIME- they're the perfect level.

@simof:  Thanks! 103 Problems in Trigonometry is a great book!  I'm using it now :)


Look at other cubers pages to find the best algorithms.  I personally like Bob Burton's, but thats just my opinion.

I think that this is the best tutorial for anyone who wants to take up cubing. http://lar5.com/cube/index.html

[quote="TimReyn"]The only reason I disagree with starting by learning Petrus is that not many people do.  There's a ton of people using a cross-F2L-LL method, so if you ever need advice, there's plenty of people using the same method as you.  That said, Petrus has been proven fast.  Anthony Hsu (who got an 11.88 today at Captains Cove) uses Petrus F2L.  Others use it as well, though I cannot think of any off the top of my head (besides Lars Petrus).  There is no doubt in my mind that Petrus has potential to be good, but I'm not a huge fan.

math92:  It's hard to give tips when we don't know where you're at.  Can you solve the cube?  If so, how fast?  What method?  What algorithms do you know?  What are your splits?  What do you need advice on?[/quote]

I can solve the cube. Average of around 3 minutes. Most of the time is taken between algorithms.  I do know some finger tricks and triggers.  I need advice on how to get faster. I really want my time down into the 1 min - 1:30's.

[quote="nat mc"]Recognition comes with practice, and after that, you could start learning new algorithms.  Also, a good way to cut off a lot of time is to put the cross on the bottom.  This helps decrease recognition time and you can use triggers a lot.[/quote]

I'll try that out next time.  How long have you been cubing nat?

[quote="nat mc"]almost exactly one year[/quote]

I hope I can get as good as you in a year.

[quote="tjhance"][quote="nat mc"]Recognition comes with practice, and after that, you could start learning new algorithms.  Also, a good way to cut off a lot of time is to put the cross on the bottom.  This helps decrease recognition time and you can use triggers a lot.[/quote]

what do you mean by triggers?[/quote]

Instead of turning the U face with a whole hand, you can use one fast motion with your first finger.  Its the motion of firing a gun with a trigger.

The most common triggers are RUR', (RUR'U')(RUR'), and (RUR'U')(RUR'U')(RUR'), for some F2Ls and also the edge inserter: (RUR'U')(L'U'L) or (L'U'LU)(RUR').
There is also (RU2R'U')(RUR') and the nice OLL F(RU'R'U')(RUR')F' and many others

My left hand is so weak. I can barely do left hand triggers. GAH.

so u guys cheated [kind of]... u didnt really figure it out on you own and get this good with your timings?

[quote="kstan013"]I like wrongway.org, but that's cheating :D . Good if you get stuck though[/quote]

Haha I just used that. Thank you so much.  :)

I started off with Petrus because I didn't know better. Trust me, it's not good. The F2L is basically all intuitive so when you're starting out you know nothing. It's like when people start off learning the cross, they usually struggle for a bit before getting the hang of it. Starting off with Petrus is like 23094825 times that hard. My first solve was after like three days (maybe a week, I can't remember) on a virtual cube and I got 50 minutes something.

[quote="Cyrazeno"]ha.. i cheated! i got an account on rubiks.com and colored the rubiks cube the way mine looked and then followed their steps! soooo easy...[/quote]

Yeah, thats the same thing as wrongway.org.

What really is the point of the Rubic's cube if u cheat on it?
(Just wondering) :huh:

it was actually made as a complex puzzle that strengthens the mind

my best time is 97.04 seconds... i can do triggers but i need to learn finger tricks

LAWRENCE!
yeah, i joined AoPs...

my best time is 31 seconds, but that was lucky.
actual record is 38 seconds....

im using the LL of the beginner method of jasmin lee...very effective....

for F2L im using my own set of moves...

[quote="tjhance"][quote="moogra"][quote="tjhance"]Hey does anybody know an algorithm that keeps all pieces in place but just flips the orientation of all 12 edge pieces?[/quote]

R2L2U2B2(y)L2R2
makes a checkerboard pattern[/quote]

Eh I mean something that will end up with every piece in the same place, but I just want the [u]orientation[/u] to be flipped.[/quote]
just do a bunch of different commutators

If you actually wanted to learn it yourself, [url]http://www.learn2cube.com/[/url] is a good site.