Question on fractions:
Do fractions have to be less than 1? I missed a question on AIME 1991 #5 because I thought fractions could be greater than 1. Try it
Say the terms of the fraction, when put in lowest form, have the product of their numerator and denominator as $20!$. How many such fractions exist?

I think fractions can be greater than 1 but the only common factor of the numerator and denominator should be 1.

I'll take time to solve the problem

[hide="solution"]$20!$ has eight prime factors: 2,3,5,7,11,13,17,19.  Since the fraction is fully simplified, we cannot have a factor of 2 in both the numerator and the denominator.  We have to choose whether all of the factors of 2 go in the numerator or the denominator, and also for the other prime factors.  Therefore there are $2^{8}=256$ such fractions.[/hide]

Is this the right answer?

The actual problem stipulated that the fraction must be between 0 and 1; so you would have to divide by 2 since if the fraction would greater than 1, you could flip it over and get one less than 1; doing this for all fractions, you would count all the ones you want twice.

So to answer your question, 1234567890, fractions can be greater than 1, but this problem specifically asked for the fractions less than 1.  Thanks Elemennop!

Oh, I didn't read it, and got 256 also. :blush:

The real question was:

Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20!$ be the resulting product?

The answer was indeed 128.


Find the maximum value of $\frac{1}{x^{2}}+\frac{2}{y^{2}}+\frac{3}{z^{2}}$, where $x, y, z$ are positive reals satisfying $\frac{1}{\sqrt{2}}\leq z <\frac{ \min(x\sqrt{2}, y\sqrt{3})}{2}, x+z\sqrt{3}\geq\sqrt{6}, y\sqrt{3}+z\sqrt{10}\geq 2\sqrt{5}.$


geometry
3D geometry
sphere
analytic geometry
function
algebra
domain

has anybody ever thought about what a 4-d object would look like in our world? (in everything that follows I am only considering spatial dimensions. i'm ignoring time for now)

first imagine a 2-d world that we can fully see, like an infinitely expansive sheet of paper with 2-d beings living within it. note that the beings in this world only see a 1-d "projection" of their 2-d world. for example, suppose a being looks at a square in the 2-d world. he will only see a line even though the square is 2-d! there's no way to see depth, width, etc.

now imagine a 3-d object, like a sphere, that passes through the 2-d world. imagine what the 2-d beings would see. first they would see a very short line come out of nowhere (the tip of the sphere passing through the paper). then the line would increase (as the sphere approaches exactly halfway through). after reaching a maximum length, the line would begin to decrease again. when the sphere finally leaves the 2-d world, the 2-d beings no longer see anything. for a brief moment in time, the 3-d sphere exists as a 2-d object in their world! the beings in the 2-dim. world have effectively seen a 1-dimensional projection of a 3-dimensional object!

before moving on to the exact analogy realize that even though we live in a 3-d world, we see a 2-d projection of it. for example, think about what you see right now as a flat picture. it's 2-d projection!

now we can make the analogy. think about our 3-d world. suppose there is a 4-d object that begins to "pass" through our world. now, we cannot imagine what looking at our entire 3-d world would look like from a 4-d perspective, but we can draw the analogy as to what a projection of the passing 4-d object would look like in our 3-d world. with a 4-d "hypersphere" passing through our 3-d world, a 3-d sphere would temporarily appear. we would see this 3-d sphere as a 2-d projection, which would happen to be a circle (just like when you're looking at a sphere you are really seeing at a projection in the form of a 2-d circle) So we are visualizing a 4-d object via a 2-d projection! to summarize, a 3-d sphere would randomly appear in our world, begining with an infintismal radius and growing to a maximum radius. then the radius would begin to decrease again, and eventually the sphere would dissapear.

i know it sounds abstract and kind of out-there, but i believe there are some valid mathematical relationships that come into play here.
let me know what you think

I think that that idea (well, at least the first part) is presented in very similar form in [i]Flatland[/i].  The rest sounds familiar too.  So, I'd say that it's valid.

I think it's a very interesting topic. Using the analogy of 3d object on 2d world projection, we can deduce that the 4d object will look like a 3d object but moving(remember, that in science there are four dimensions: x,y,z and time :!: ). Now I am bad at anything above 3 dimensions, so could someone describe how a 4D "sphere"(a 4D object corresponging to a sphere in 3D or circle in 2D) would look in 3D world.

more like confusing, but interesting nonetheless

I think you could easily undrerstand the appearance of soem 4D objects in 3D world by just using coordinate geometry. :wink:

i think there was something abouut this in "A Wrinkle in Time".

yeah how would you use coordinate geometry

You can picture wrapping the surface of a hypersphere using a flat,infinites piece of 3-D space just like wrapping a 3-D sphere using a infinite big piece of 2-D foil.

sounds interesting. can anyone provide mathematical details, or is it too advanced?

how would you map a pt. in three dimensions to a pt. in two dimensions that would correspond to drawing the surface of a sphere in two dimensions? (basically making a map of the globe)

[url]http://casa.colorado.edu/~ajsh/sr/hypercube.html[/url]

Just google 'hypercube' and you'll find a million hits of various way to look at them.

I couldn't find it, but there is actually a four dimensional rubiks cube applet! That you can actually solve!

Read sphereland if you haven't, the sequel of flatland written by a different author.  I've actually thought about this many times.  You can think of projections, such as the projection of a hypercube into 3-d space, by analogy.  The example of a hypersphere: if a hypersphere were to move through our space, it would begin at a point, then expand until it would reach a maximal radius sphere, and then again contract down to a point.  You can consider many other things by analogy.

Rem, the concept of 4-dimensions we're talking about is 4 [b]spacial[/b] dimensions: not 3 space and 1 time...
I suppose you could consider time a dimension if you wanted to...if you go with Einstein, but it doesn't have to be so.  But if you do that, it's true that you get all of the cross-sections of the 4-d objects.

If a hypercube traveled through our space in a direction perpendicular to a face, we'd first see a cube for an instant, then the vertices of a cube for a while, and then a cube again.
If it weren't parallel, we'd have something much more complicated...first a point, then the vertices of an expanding tetrahedron (shape depending on the orientation of the hypercube), and slowly the polyhedron would get more and more complicated. (this is hard to imagine, but you can reason by analogy by thinking of a cube going through a plane.  You could also consider that a hypercube is just two cubes with theire 8 vertices linked in pairs by line segments perpendicular to all of the cubes' edges and equal in length.)

Lastly, I like the idea of a regular hypertetrahedron...We would have a tetrahedron, and then a fifth point equidistant from the other 4.  The projection into our space would look like a tetrahedron with a point in the center.  Then as the figure would rotate, the projection would get weird...

[quote="K81o7"]Rem, the concept of 4-dimensions we're talking about is 4 [b]spacial[/b] dimensions: not 3 space and 1 time...
I suppose you could consider time a dimension if you wanted to...if you go with Einstein, but it doesn't have to be so.  But if you do that, it's true that you get all of the cross-sections of the 4-d objects.

If a hypercube traveled through our space in a direction perpendicular to a face, we'd first see a cube for an instant, then the vertices of a cube for a while, and then a cube again.[/quote]

What a 4-D figure would look like with fourth dimension being time is essentially equivalent to what a 4-D figure would look like as it passed through our space perpendicular to an axis (since at each temporal instant, we would be looking at a single cross-section).  A $1m \times 1m \times 1m \times 60 sec$ hyper-box would simply be a $1m \times 1m \times 1m$ box which existed in our space for a minute and then vanished -- exactly what you would get if a $1m$ hypercube passed through our space at one meter per minute.

[quote="andy_tok"][quote="drunner2007"]
how would you map a pt. in three dimensions to a pt. in two dimensions that would correspond to drawing the surface of a sphere in two dimensions? (basically making a map of the globe)[/quote]

Use stereographic projection.
regard this projection as a function f with domain X  which is the sphere without the northpole and range  Y which is a plane parallel to the equator(
you can choose it to be the plane cross the equator).
f assigns to each point x of X to the point y of Y where the ray from the northpole through x meets the plane.
Then there is a 1-1mapping of a sphere without north pole into a plane[/quote]
wow, that is very interesting. thanks for the enlightment  :wink:
i made the idea you described more formal. it uses vectors mechanics, but its not too hard to follow. i'll post the pdf in here once i finish it.

I heard somewhere that because we can represent a 3D object on a 2D object (eg cube on paper), we can represent a 4D object on a 3D object... how would you do that?

my friend was telling me about a way. i'm not sure if this is right, but i'll try.

when you draw a 3D cube in 2D, you can think of the drawing as two squares in parallel planes, with corresponding vertices connected by line segments.
so when you jump to the case of drawing a 4D object in 3D, you can think of two cubes next to each other, and then connecting corresponding vertices with line segments.
not sure about this though. it was just an abstract way at looking at it. you can never truly visualize 4D objects; instead, you can just think of them mathematically (abstractly).

by the way. here is the stereographic projection paper i was working on. its not much (not completed), and i also realized that usually the plane i refer to in the paper is a tangent plane to the sphere.

like i said before, only requires vector calculus (but you can do it with simple geometry if you want)


[quote="lomos_lupin"]Well,I dont know why our instructors had chosen this problem for the second round but here it is :

Let $a,b \in \mathbb{N}$ and $P$ be a prime such that

$P=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$.

Find all values for the $P$.[/quote]

 :)

Are you sure its $a,b \in \mathbb{N}$? Cuase then there are none.

I seem to have proved there are NO integer solutions to this.


Explicitly defined sequences are like this: f(n)=n^2+2x+1
Recursive are like this: f(n+1)=f(n)+4; f(1)=3
Is there a method anyone can find to get the recursive from the explicit or vice-versa?
The way I've done it, to go from explicit to recursive, is computed f(n), f(n-1), f(n-2), etc... then combined them together.  For example:
f(x)=x^2+2x+1.  Then, finding f(x-1) and f(x-2) to be x^2 and x^2-2x+1, it comes out to be f(x)=2f(x-1)-f(x-2)+2.
However, if it were a complicated formula, just doing it by eye wouldn't work so well.
There are a couple of things I've noted: if it is a polynomial of degree one, the recursive formula only needs two functions in it.  If it's of degree two, the recursive needs three.
So if anyone knows a way to go from explicit to recursive or recursive to explicit please post it here.

This reads to me like an informational question rather than a challenge problem, so I'll let it stay here on Getting Started Forum, with the remark that if this were a challenge problem, I would move it to a higher-level forum INSTANTLY, as beginning problem solvers would hardly be expected to know this terminology or to have any experience with this level of math.

You're one of the MathLinks members who now posts on the merged forums, right? Welcome aboard.

I'm trying to keep things as easy as possible on the Getting Started Forum (because people who are looking for harder problems can always find them on Intermediate, Advanced, or College Playground), because there is no where else to go for the real beginners in problem-solving to find help online.

The recursive series $f(n+2)=f(n+1)+f(n)$ where $f(1) = f(2) = 1$ is the Fibonacci series. Explicitly, it is $f(n) = \displaystyle \frac{\phi^{n}-\left(1-\phi\right)^{n}}{2\phi-1}$, where $\phi = \displaystyle\frac{\sqrt{5}+1}{2}$. If we say $f(n) = (a^{n}-b^{n})c$ we can quickly find $a, b,$ and $c$ by plugging in our first 3 values of $n$ and $f(n)$. The key to finding the explicit is noticing what type of function it would be, or in the case of the Fibonacci, which type it wouldn't be, and then start playing around with the function. For example, the easier series: $f(n+1) = f(n) + 3$. We notice each consecutive term differs by 3, so it is clear the relationship is linear.

[quote="Wumbate"]Yes, it is an informational question.  I don't know the answer, but it seems to me like there is probably some method that people who have studied series would know.  I apolgize if it is harder than it seems.

No, I'm not a MathLinks member, I am a completely new member.  If there is an answer, but it's too complicated for this forum, move this to a harder one.[/quote]

The three methods are, in order of easiest to hardest, recognizing it as a common sequence, eg:

if f(n) = n!, then f(n+1) = (n+1)*f(n)

if f(n) = a + nd, then f(n+1) = f(n) + d

if f(n) = ar^n, then f(n+1) = rf(n)

....

If the sequence is linear and homogeneous, then we can use the characteristic polynomial (you can look that up at Mathworld)

....

If the sequence is not homogeneous, then we may want to use generating functions.

[quote="Wumbate"]What exactly is the definition of homogeneous?[/quote]

In this case, that there are no constants in the recursive equation.

Homogeneous means consisting of terms of the same degree or elements of the same dimension.


Hi,

I still do not understand so please check this one :

(n(n+1)(2n + 1)/6) + (2^n) -1) mod n = 1 if n [b]is prime >3[/b]

Example
59*(59+1)*(2*59 + 1)/6 + (2^59) - 1 mod 59 = 1

Thank you for any answer.

[u][b]I'm asking only for checking [/b][/u]because I do not have the tools to check for bigger numbers.

If $ n$ is a prime greater than $ 3$ then $ 6$ is invertible so $ n | n(n \plus{} 1)(2n \plus{} 1)$ and $ 2^n \minus{} 1 \equiv 1 \bmod n$ by [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url].  So this test isn't any better than Fermat's test (did you read my explanation of why that first term is useless in the thread where you brought it up)?  In particular, the $ 2$-pseudoprimes (which include the Carmichael numbers) not divisible by $ 3$ pass it.

Why do you like adding two terms that you don't understand to get something else you don't understand?  Wouldn't it be better to understand each of those terms individually instead?

Anyway, if you don't have the tools to run [b]this[/b] computation for large numbers it's because you're not computing $ 2^n \bmod n$ efficiently; use [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]binary exponentiation[/url] and reduce $ \bmod n$ at every step.  This is one of the computations that we [b]do[/b] know how to do quickly, so you're getting closer to something that looks like a real primality test.


function
Euler
trigonometry
complex numbers
complex analysis
complex analysis unsolved

Hi @ll!

Could someone help me with this exercise please?

Calculate Discrete Transformate Fourier from the sample:

$ x_0\equal{}1  ;  x_1\equal{}2  ;  x_2\equal{}3  ;  x_3\equal{}4  ;  x_4\equal{}5  ;  x_5\equal{}6$

I haven´t problems with N=4, but we have N=6.

Greetings!

[quote="Angus"]Hi @ll!

Could someone help me with this exercise please?

Calculate Discrete Transformate Fourier from the sample:

$ x_0 \equal{} 1 ; x_1 \equal{} 2 ; x_2 \equal{} 3 ; x_3 \equal{} 4 ; x_4 \equal{} 5 ; x_5 \equal{} 6$

I haven´t problems with N=4, but we have N=6.

Greetings![/quote]

The calculation is identical, you just need to apply the Discrete Fourier Transform formula for this particular case:
\[ X[k] \equal{} \sum_{n \equal{} 0}^5x[n]e^{ \minus{} \frac {j2\pi kn}{6}}.
\]
Another possibility is calculating first the Discrete-Time Fourier Transform $ X(\Omega) \equal{} \sum_{n \equal{} \minus{} \infty}^{ \plus{} \infty}x[n]e^{ \minus{} j\Omega n}$, where in your case $ x[n] \equal{} 0$ for $ n < 0$ and $ n\geq6$, and then evaluating at $ \Omega \equal{} \frac {2k\pi}{N}$.  This can be simplified by the use of the difference operator $ \Delta$, defining sequence $ y[n] \equal{} \Delta x[n] \equal{} x[n] \minus{} x[n \minus{} 1]$.  By the properties of the DTFT, we obtain that $ Y(\Omega) \equal{} X(\Omega)(1 \minus{} e^{ \minus{} j\Omega})$, $ y[0] \equal{} y[1] \equal{} y[2] \equal{} y[3] \equal{} y[4] \equal{} y[5] \equal{} 1$ and $ y[6] \equal{} \minus{} 6$.  Note then that, except for $ k \equal{} 0$, the contribution of the $ 6$ first terms will vanish.  Note also that the trick of using the difference operator allows you to find all coefficients except for $ k \equal{} 0$, since you would obtain a $ \frac {0}{0}$ indeterminancy.  Fortunately, $ X[0]$ is the easiest term to calculate, since it is just the sum of all values of $ x[n]$, $ X[0] \equal{} 21$ in your case.  For the rest of the cases, you would obtain $ X[k] \equal{} \minus{} \frac {6}{1 \minus{} e^{ \minus{} j\frac {2k\pi}{6}}}$, ($ k \equal{} 1,2,3,4,5$), done...

NOTE: In Fourier analysis, you may think that you need to spend a lot of time in calculations: Fourier Series coefficients and Fourier Transforms for continuous functions, and Discrete Fourier Series coefficients, Discrete-Time Fourier Transforms and Discrete Fourier Transforms for sequences.  My experience is that it is not so.  In most exercises and problems, using the properties of the different Series and Transforms (ie, what happens to the coeff or to the transform when we apply certain operations or transformations to the function or to the sequence), saves you precious time in calculations, appart from allowing you to doublecheck some of your results (for example, for even $ N$ and real valued sequences, your $ X\left[\frac {N}{2}\right]$ better be real, since for real sequences $ X[N \minus{} k]$ is the complex conjugate of $ X[k]$ for all $ k$).  I therefore advise you to learn well the properties of the series and transforms, this will improve your understanding of Fourier analysis, save you time in calculations, and help you avoid and correct many mistakes.

Hi daniel73,
Thank you very much for your quikly answer.

I am trying the first method but I have some problems with the next expression:

$ X[k]=1+6[e^{jk\frac{\pi}{3}}+e^{-j5k\frac{\pi}{3}}]+3e^{-j2k\frac{\pi}{3}}+5e^{-j4k\frac{\pi}{3}}$

Could I carry any simplification with Euler formula?: $ e^{-j\frac{\pi}{3}}=cos(-\frac{\pi}{3})-jsen(-\frac{\pi}{3})=\frac{1}{2}-j(-\frac{\sqrt{3}}{2})$

I consider this way very complicated.

Thanks in advance

Usually, the idea is precisely using Euler in order to simplify.  For example, in the problem at hand, the idea could be to group terms, such as for $ n \equal{} 0$ and $ n \equal{} 5$, as follows:
\[ 1 \plus{} 6e^{ \minus{} \frac {j5k\pi}{3}} \equal{} e^{ \minus{} \frac {j5k\pi}{6}}\left(7\frac {e^{\frac {j5k\pi}{6} }\plus{} e^{ \minus{} \frac {j5k\pi}{6}}}{2} \minus{} 5\frac {e^{\frac {j5k\pi}{6}} \minus{} e^{ \minus{} \frac {j5k\pi}{6}}}{2}\right) \equal{} 7e^{ \minus{} \frac {j5k\pi}{6}}\cos\frac {5k\pi}{6} \minus{} 5j\sin\frac {5k\pi}{6}.
\]
As you see, the idea is to "break up" terms so that you can get an exponential common factor, and then group terms with exponentials that add up to sines or cosines.  If you do the same with $ n \equal{} 1$ and $ n \equal{} 4$, and also with $ n \equal{} 2$ and $ n \equal{} 3$, you will be able to extract also the term $ e^{ \minus{} \frac {j5k\pi}{6}}$ as overall common factor...  It sounds complicated at the beginning, the only thing that I can recommend is practice, practice, and then some more...  Sorry, but that's what I went through - pretty much any of us that learnt Fourier analysis at one point or another.  The good thing is, practice works, and you'll learn it right once and for all...


superior algebra
superior algebra unsolved

the set S of the form a + b sqrt 2, such that a and b belong to quotient is a subfield of real numbers?

Verifying the field axioms is trivial.  The only one you might have difficulty with is multiplicative inverse; note that

$ \frac{1}{a \plus{} b \sqrt{2}} \equal{} \frac{a \minus{} b \sqrt{2}}{a^2 \minus{} 2b^2}$.

Therefore the set S of the form a + b sqrt 2 is not a subfield of the Reals?


A small wooden boat containing a man and an iron anchor floats in a swimming pool . The water level at the edge of the pool is marked . The man removes the iron anchor from the boat and drops it into the pool . As a result , the level of water

A . will rise
B . will fall
C . will stay the same
D . cannot be determined , unless the mass of the anchor is known
E . cannot be determined , unless the masses of the man,boat and anchor are known .

can you explain why ? because i think that it cannot be determined ?

Denote $m$ the mass of the anchor.

We have this equation:
$\rho_{water}. V_{water}=\rho_{anchor}. V_{anchor}=m$

Obviously, $\rho_{water}<\rho_{anchor}\Longrightarrow V_{water}>V_{anchor}$.

Therefore, the level of water in the pool will fall.  :P


The ants in a colony of African ants average one centimeter in body length. The ants in this colony, lined up end to end, stretch one kilometer. How many ants are there in the colony? Express your answer in scientific notation.

There are 100 centimeters in a meter, and 1000 meters in a kilometer, so there are 100000 centimeters in a kilometer. In scientific notation, this is $ 1 \cdot 10^5$.

I think that 1.0times10^5 should be acceptable for this.  Scientific notation usually has a decimal.


limit
calculus
integration
function
real analysis
logarithms
real analysis unsolved

Suppose $f: [a,b] \to \mathbb{R}$ is a bounded function. For any partition $\Pi: a = x_{0}< x_{1}< \cdots < x_{n}= b$ we let $M_{i}= \sup_{x\in [x_{i-1},x_{i}]}f(x)$ and form the upper Darboux sum $S_\Pi = \sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})$. We let the mesh of $\Pi$ be $|\Pi| = \max(x_{i}-x_{i-1})$. How do I show that $\lim_{|\Pi|\to0}S_{\Pi}$ always exists?

It is clear that the limit has to be $\inf S_\Pi$, but I can't show that this actually is the limit. This is a vital step to prove the well known equivalences between various definitions of the Riemann integral, but I can't figure it out. Probably it is simple and I am being stupid.

What is [quote="Kalle"] $\lim_{|\Pi|\to0}S_{\Pi}?$
[/quote]
i.e.?

I am being intentionally vague since I don't know exactly what can be shown. The form which is true for integrable functions is that there is an $S$ such that for any $\epsilon > 0$ there is a $\delta > 0$ such that for any partition $\Pi$ with $|\Pi| < \delta$ we have $|S_\Pi-S| < \epsilon$.

My lecturer mentioned that the limit exists in this sense, but he may be wrong. I know for sure that the limit exists in the sense that for any sequence of partitions with mesh converging to zero the corresponding upper sums converge to $S$. But I can't show this either.

Ah, the grubby details of Riemann integration.

Suppose that $|f(x)|\le M$ on $[a,b].$ As in your notation, let $S$ be the infinum of the upper sums. For any partition $P,$ use the notation $U(f;P)$ for the upper sum for that partition.

Given $\varepsilon>0,$ there exists a partition $P$ such that $U(f;P)<S+\frac{\varepsilon}2.$ This partition $P$ has $m$ points.

Now, fix a $\delta>0.$ (We will figure out later what $\delta$ needs to be.) Suppose $Q$ is any partition with $\|Q\|<\delta$ where I'm using $\|Q\|$ to mean the mesh of the partition. (It looks like you used $|Q|$ for this - same notion.)

Since $P\cup Q$ is a refinement of $P,$ we have $S\le U(f;P\cup Q)\le U(f;P)<S+\frac{\varepsilon}2.$

Now, we investigate the difference between $U(f;P\cup Q)$ and $U(f;Q).$ These two upper sum differ in at most $m$ terms, where points of $P$ lie within intervals of $Q.$ (Actually, $m-2,$ but who cares?) The difference between them for one of those terms can be no more than the range of possible values of $f,$ which is $2M,$ times the maximum width of the subinterval, which is $\|Q\|<\delta.$ So we have that $U(f;Q)<U(f: P\cup Q)+2mM\delta.$

Choose $\delta<\frac{\varepsilon}{4mM}$ and we get that

$S\le U(f;Q)<S+\varepsilon$ for $\|Q\|<\delta,$ which is what you wanted to prove.

Thanks Kent. I tried to do more or less the same thing, but I didn't realize I could use the number of points $m$ in my choice of $\delta$. I feel a little stupid  :blush:.

I have actually not seen a proper proof of the equivalence between definitions of riemann integration in any book. Everyone just refers to the well known fact. Why is this? It's not particularly hard (although maybe I shouldn't say this if I can't figure it out on my own, but still... ).

You say "grubby" details. I feel that in our presentation of measure theory there were similar messy difficulties in introducing concrete measures. For example in the Lebesgue measure when we had to show that the Lebesge outer measure is a metric outer measure or just to calculate the measure of a cube. Coincidentally, these two things were left as exercises in my book.

[quote]Coincidentally, these two things were left as exercises in my book.[/quote]
Coincidence? I think not. :cool:

Does the result remain true if $f$ is only assumed bounded from above?

mlok doesn't the answer follow from the definition of upper Stieltjes integral?

Can't a shorter solution be obtained by considering the least upperbound property of the set of upper sums for f over [a,b] - which is acceptable since the sequence of sums (S(n)) is decreasing according to the decrease in norm?

[quote]the sequence of sums (S(n)) is decreasing according to the decrease in norm[/quote]
Unfortunately, this isn't true. A partition with smaller norm may have a larger upper Darboux sum. For example, let $f\colon [0,1]\to\mathbb R$ be a function that is zero except for a narrow peak of height $1$ around $x=1/3$. The partition $[0,1/2]\cup [1/2,1]$ has upper Darboux sum $1/2$. The partition $[0,1/3]\cup [1/3,2/3]\cup [2/3,1]$ has upper Darboux sum $2/3$.

A guy in my class had a very interesting solution. As in Kent's solution we take a reference partion $\Pi: x_{0}< x_{1}< \cdots < x_{m}$ with sum close to the infimum. Now let $O$ be a finite open cover of $[a,b]$ consisting of the intervals $(x_{i-1},x_{i})$ plus some small intervals covering the points $x_{0}$,$x_{1}$ and so on. Let $\delta$ be a Lebesgue number for $O$.

Now take any other partition $\Theta: y_{0}< \cdots < y_{n}$ with $|\Theta| < \delta$. We can with each interval $I=(y_{i-1},y_{i})$ associate an interval $\tau(I)$ in $O$ such that $I \subset \tau(I)$. Now write $S_\Theta$ so that we sum in the following way: For every $J\in O$ we take all terms corresponding to intervals $I=(y_{i-1},y_{i})$ mapped onto $J$. Then it is very easy to see that the sum is not much bigger than $S_\Pi$.

You are right mlok. I was thinking of refinements when I said that.

So what about functions that are bounded from above only? For example, the upper Darboux sums for $f(x)=\log x$ on $(0,1]$ (and $f(0)=0$) have a limit, right?

Conjecture: the upper Darboux sums have a limit if and only if their infimum is finite.

[quote="Kent Merryfield"][quote]Then it is very easy to see that the sum is not [b]much[/b] bigger than ...[/quote]
Of course, to make for a complete and convincing proof, that word "much" absolutely must be quantitatively estimated.[/quote]
Do you think that I am not aware of this or that someone else is not aware of this? My intention was only to describe his proof, and for that purpose writing down epsilon and deltas is not worthwhile.


I'm taking the Introduction to Algebra class; Wednesdays at 4:30PM-6:00PM (Mountain Standard Time).

Two weeks ago, everything was fine; the class loaded quickly and I wasn't kicked out.
Last week, the class took about 10 minutes to load.
Today, the class took 10 minutes to load, and I've been kicked out approximately every 10-15 minutes. When I refresh the page, the classroom takes about 5 minutes to load.

I'd like to stick with Internet Explorer, but switch to Opera as a last resort. Please help!

My computer wouldn't let me download it.

What do you mean your computer won't let you download Opera?

When I clicked on the "download" button it said something like "unable to download"

[quote="hunter34"]Try Mozilla Firefox.[/quote]

Im on Firefox and I get kicked out a lot too.

yeah, I've had problems with firefox, for some reason, the room just freezes, but doesn't disconnect or anything, I downloaded opera for the classes, it works great, no problems

[quote="hunter34"]Try Mozilla Firefox.[/quote]
I have IE by the way...if you had problems w/ IE too let me know!

Generally, I don't like to use IE at all.  Not being in any classes, I couldn't tell you which browser to use for classes, but looking at the Online Classes page, it seems that Opera is your best bet.

[quote="rrusczyk"]Try again, and be more specific about what it says if it doesn't let you download.[/quote]
"error: initialaztion setup"


conics
ellipse
geometry
incenter
geometry proposed

Moreover, I think the intersection point is also the intersection point of the tangents in $C,D$ to the ellipse having $A,B$ as foci and passing through $C,D$, but my mom needs some time on the computer, so I'll have to come back later :).

Let $F=AD\cap BC$. Assume we have shown that the intersection point of the bisectors of $\angle DAC,\angle DBC$ (call this point $I$) is the intersection point of the tangents in $C,D$ to the ellipse mentioned in my previous post (call this fact $\mathcal P$). This means that it's the incenter of $FAC$, so the bisector of $\angle CFD$ also passes through it, meaning that three of the bisectors of $FDEC$ are concurrent, so $FDEC$ is circumscribed, meaning that the fourth bisector passes through $I$ as well.

I don't have time to prove $\mathcal P$ right now, but I think it's been discussed before on the forum.

I believe your solution is correct Grobber but my attempt to find
solution gives my a contradiction:
   Let bisectors of angles DBC, DAC, DEC intersect AC, BD, AB at points X, Y, Z respectively.Using Ceva theorem in trigonometric form BX, AY, EZ are concurrent iff
    sin EBX/sin ABX * sin BAY/sin EAY * sin AEZ/sin BEZ =1

We denote AB=a, BC=b, CD=c, AD=d, AC=e, BD=f. Using facts that

   sin EBX/sin ABX = sin CBX/sin ABX = a*CX/(b*AX)

   sin BAY/sin EAY = sin BAY/sin DAY = d*BY/(a*DY)

   sin AEZ/sin BEZ =1

we get that BX, AY, EZ are concurrent iff  a*CX/(b*AX) =a*DY/(d*BY)
or d*CX*BY = b*AX*DY.

But   AX =AE+EX, EX = [BE/(BE+b)]*EC, CX = [b/(BE+b)]*EC

        BY =BE+EY, EY = [AE/(AE+d)]*ED, DY = [d/(AE+d)]*ED hence

d([b/(BE+b)]*EC)(BE+[AE/(AE+d)]*ED)=b([d/(AE+d)]*ED)(AE+[BE/(BE+b)]*EC)

EC*( BE*(AE+d)+AE*ED) = ED*(AE*(BE+b)+BE*EC)     or

(e - AE)*(AE*f + BE*d) = (f - BE)*(BE*e + AE*b) but condition

d+f = b+e I think is not enough that lines will be concurrent.
Is there another method not Grobber's ?

I think I have a proof not using any conic geometry, but I'll be back later.

Is the problem really so difficult?

[quote="Zorro"]Let $ABCD$ be a convex tetragon. $AC$ and $BD$ intersect at $E$ and $AD+DB=AC+CB$. Prove that bisectors of angles $DAC,DEC,DBC$ intersect at one point[/quote]

Actually, the condition AD + DB = AC + CB can be rewritten as AD - BC = CA - DB, and this is equivalent to the assertion that the (self-intersecting) quadrilateral ADBC has an excircle, i. e. a circle touching its sidelines AD, DB, BC, CA. [The equivalence can be proven in the same way as one usually shows that a convex quadrilateral ABCD has an incircle if and only if AB + CD = BC + DA.] And now, considering this circle touching the sidelines AD, DB, BC, CA, it is clear that the center of this circle lies on the angle bisectors of the angles DAC, DEC, DBC. Problem solved!

I have not read Prowler's post but I believe the contradictions [b]may[/b] arise from the fact that he used a sketch where the points were arranged differently from those in the proposer's sketch. One always has to face such arrangement troubles unless one uses directed lines and directed circles.

  Darij

Let, as above, $F=AD\cap BC$. From the condition $DA+DB=CA+CB$ we get $FA+FC-AC=FB+FD-BD$, meaning that the incircles of $FAC,FBD$ touch $FA,FB$ in the same respective points, so the two triangles have one and the same incircle, and from here the conclusion follows.


geometry
geometric transformation
puzzles

[quote="bigboy82892"]All the kids who were in the classroom disappear[/quote]

What exactly does "disappear" mean?  Do you mean that nobody saw them ever again?  Did they die?

Wow, this is probably the weirdest brain teaser I've ever heard.

umm kid i didn't understand a word you said :maybe: speak english

[quote="moogra"][quote="bigboy82892"]A person walks into a classroom, and pulls out a gun, then tells them to do something. Then, he/she walks out and is arrested by the police. Nothing seems to be taken. Nobody is missing. No threats have been made. Everyone is perfectly healthy. $19$ years later, All the kids who were in the classroom disappear, but the person is still in jail. The person has not escaped and will never escape, nor will any plot be made to break him out. Why has this happened and who is this person and where did everyone go and what happened in the classroom :stretcher:  :huuh:  :ewpu:  :huh:  :what?:  :wacko:  :sleeping: ?[/quote]

hm... the person could just be walking in with a gun, and got arrested. After19 years, the people graduated, and have disappeared from the classroom. The person is just a normal bad guy, and the classroom was just aimed at. No idea.

@abcd.....pq, if you want to type faster, instead of saying shiznit you can just say shiz, stuffz could be stuff, peeps could be ppl, other than that, that made like no sense. Even worse than my spam.[/quote]
Good try but no.

I understand two languages: English, Korean. You are speaking none of them. Can someone please translate? :roll:

[quote]I understand two languages: English, Korean. You are speaking none of them. Can someone please translate?[/quote]

WHAT YA BE MEANIN IS, "I understand two languages: English AND Korean. You are speaking NEITHER of them. Can someone please translate?

I believe that I can come up with a translation, because I speak abcdefghijklmnopq also

[quote="abcdefghijklmnopq"]YO< CHECK IT> I AINT QUITE FIGURED OUT THAT NIFTY SHIZNIT WHERE U STICK DA ANSWER BEHIND DA PURTY BLUE LETTAHS AND THEN U BE CLICKIN DEM PRETTY STUFFZ AND THE ANSWER POPS OUT AND SCARES DA BEJEZUZ OUT OF YAZ AND MAKES YOU GO "WOAHH," BUT I DID STAY AT A HOLIDAY INN EXPRESS TEH LAST NIGHT AND ME BELIEVES DAT DAT FINE NIGHTS REST HAS GIVEN ME DE ANSWER, WHICH ME BRAIN TELLZ ME IS DAT THESE PEEPS IN DA CLASSROOM WERE ACTUALLY LITTLE MUNCHKINS FRESH OUTTA DA WOMB< AND THEY ALL BE TAKIN CLASS TOGETHER IN DA SAME ROOM TILL THEY 19. THEN THEY DECIDED WE DONE WID DIS - AND BE LIKE PEACE OUT. IF DEY HAD 19 YEARS OF SCHOOLINS TO FIGURE DAT OUT I THINK IT MUST BE GOOD SO I BE SAYINZ PEACE OUT TOO.[/quote]

Hello. I haven't learned how to hide my posts yet, but I would certainly enjoy learning, and I would be most grateful if you would teach me. I think that hiding my words is a lot of fun, and very cool. Last night, I stayed at a particularly nice hotel, (The Holiday Inn Express), and after my wonderful nights sleep, I have been refreshed enough to realize the answer. I believe that the people in the classroom were very young children who were going to school. Then, after 19 years, they didn't want to go to school anymore, so they decided to leave. I think that if it took them 19 years to decide that they didn't like school, school must be pretty good.

[quote="abcdefghijklmnopq"]AIGHT, CHECK IT> SO APPARENTLY TEH GUY DAT POSED DIS QUESTION DONT BE LIKIN MY PERFECT SOLUTION AND HE BE DISSIN ME AND SAYIN I IS DUMB AND GOTZ IT WRONGZ< HE IS IN FACTZ TEH DUMB ONE AND I CANT BRING MY MIND TO BE WORKIN AT SUCH A LOW LEVEL OF ITS FULL CAPICATORR BUT I WILL BE TRYIN MY HARDEST TO DO SO BY STICKIN A KNIFE TRHOUGH MY BRAIN AND CUTTIN OUT ALL BUT LIKEZ< LETZ SAYZ, ONE TENFTH> THATS 1/10 FOR U MATHEMAMTMICALLY CHALLENGED DAWGZ> AIGHT< SO ME DUMB SOLUTION IS DAT DESE LITTLE PEEPZ< WHO< AS I SAID< IS FRESH AT OF E WOMB< ALL BE GROWN UPZZER IN 19 YEARZ> SO< LOOKZIE< 19 YEARS LATER< DERE BE NO CHILD< CAUSE LIKE< DERE IS AN ADULT NOW AND DA CHILD IS HIDIN INSIDE OF DE ADULT< KNOWN AS TEH INNAH CHILD> YHO< PEACE OUT>[/quote]

I believe that while bigboy82892 has asked a very good question, and I have given a solution which I believe to be correct, bigboy82892 has said that my solution is, in fact, not correct. I take this as an insult, for I believe that if my answer was, in fact, wrong, then I would feel stupid. Because I seem to be slightly mentally challenged, I will retaliate to the perceived response in a like manner, and call bigboy82892 stupid. In fact, to help bring myself down to his level, I am seriously considering cutting out most of my brain, leaving behind only one tenth of it, approximately the amount of brains that I believe bigboy82892 to have. Anyways, my answer states that these very young children that are going to school at the moment will grow up into adults in 19 years, causing them to leave the school behind. There will, in fact, not be any children left, for they will have grown into adults. Goodbye.

I hope that this is a satisfactory translation for you. For any other translating needs, drop me a pm.

LOL!! that translation was awesome!
Hey...do the people in the classroom know the "person" or is he/she a complete stranger to them?

:rotfl: that person talks like a d***nig*** (no offense-term coined from a popular forum meaning a guy that spams the forums)

[hide="another try"]
Anyways, was the person arrested in a play, where you enter a classroom and leave, and everything is part iof the script, so even the threats wouldn't be threats, because it is part of the play.[/hide]

No answers?
[hide="Think STUPID!!!"]Think about all the places, real or not, this could happen.[/hide]

Did this happen in a movie, book, or dream?

[quote="mathnerd314"]YO DONT BE DISSIN NO http://en.wikipedia.org/wiki/Ebonics

Seriously, though, what is the answer to this puzzle?[/quote]

that place is awesome, reminds me of this koalas essay.
Yeah we need the answer now...

Drunken, he thought that the kids were customers, because all the desks looked like register counters to his inebreated eyes. Obviously, none of the kids spoke so he was confused. While he thought about why all the kids were silent, a policeman snuck up on him and arrested him.

OMGZ, THERE WERE NEVERZ ANY KIDZ IN DA DAMN CLAZZROOM

ZOMGWTFBBQ TEHCAPZ LOCKE ARE TEH SHIZZO UP IN THIS FIZZLE. LYKE US!NG DEM M@KE |\/|E C00LER TH@N J00.

OK, just stop. It's getting annoying.

Will the real answer please stand up.

[quote="bigboy82892"]Come on... The man was a renegade moonshiner breaking into an all-boys school. It was 1926, when the prohibition amendment was in effect. He was too drunk to think of holding the kids hostage. He was arrested, obviously. Then, the kids became all good friends and were all drafted into the war. Then they all were working on the U.S.S. Indianapolis in the section of the boat where it was torpedoed. nobody knows what happened to them, if they were killed by the torpedo or iof they were eaten by sharks. No bodies were ever found. If the wording seemed a little confusing or suggested an answer different than this, I'm sorry. :|[/quote]

yeah abcd...p he did. Glad to see you're back in typing in normal English, but that lingo you were using  was cool

[quote="abcdefghijklmnopq"]Wow, i just realized he posted the actual answer.  This question was retarded.[/quote]

Congratulations on finding the caps lock key. :coolspeak:

[quote="bigboy82892"]Come on... The man was a renegade moonshiner breaking into an all-boys school. It was 1926, when the prohibition amendment was in effect. He was too drunk to think of holding the kids hostage. He was arrested, obviously. Then, the kids became all good friends and were all drafted into the war. Then they all were working on the U.S.S. Indianapolis in the section of the boat where it was torpedoed. nobody knows what happened to them, if they were killed by the torpedo or iof they were eaten by sharks. No bodies were ever found. If the wording seemed a little confusing or suggested an answer different than this, I'm sorry. :|[/quote]

But where does the "pulls out a gun and tells them to do something" belong?


I know that the deadline to postmark solutions has passed, but we'd appreciate it if you'd wait until we've received the solutions to begin discussing the test.

Therefore, please do not post solutions or strategies for any of the questions on Admission Test B [b]until April 15.[/b]

I guess I don't mind if you say [i]how many[/i] you got, but please refrain from discussing individual problems.

Thanks,
Zac

will the number of remaining spots be announced?


calculus
integration
trigonometry
function
algebra
rational function
absolute value

Hi,
I think this may be a method of attack.
The first limit is simple since an antiderivative is $\arctan(nx)$ and thus the integral equals $\frac{\pi}{2}-\arctan(na)$, which converges to $\pi$, $\frac{\pi}{2}$ respectively $0$ according to whether $a<0$, $a=0$ respectively $a>0$.
For calculating the second limit first we make the following sequence of  transformations, $x=nt$, $1+t=y$ respectively $\frac{1}{y}=z$, and transform the integral to
$\int_{0}^{\infty}\frac{\sin \frac{n}{x}}{(1+\frac{x}{n})^{n}}dx=\int_{0}^{1}nz^{n-2}\sin(\frac{1}{z}-1)dz$.
We calculate this integral by parts and get that

$\int_{0}^{1}nz^{n-2}\sin(\frac{1}{z}-1)dz=\frac{n}{n-1}z^{n-1}\sin(\frac{1}{z}-1)|_{0}^{1}+$
$+\frac{n}{n-1}\int_{0}^{1}z^{n-3}\cos(\frac{1}{z}-1)dz=\frac{n}{n-1}\int_{0}^{1}z^{n-3}\cos(\frac{1}{z}-1)dz\rightarrow 0$.
The last limit being $0$ based on the Bounded Convergence Theorem since the sequence of functions $f_{n}(z)=z^{n-3}\cos(\frac{1}{z}-1)$ is bounded on [0,1] for all $n\geq 3$ and all $z\in [0,1]$ and the limit function is 0.
I hope that there are no computational mistakes.

Another attack on the second one. First, make the substitution $x=ny.$ The integral becomes

$\int_{0}^{\infty}n(1+y)^{-n}\sin y\,dy.$

Now integrate this by parts, integrating the rational function and differentiating the sine:

$=\left.\frac{n}{1-n}(1+y)^{1-n}\sin y\right|_{0}^{\infty}+\frac{n}{n-1}\int_{0}^{\infty}(1+y)^{1-n}\cos y\,dy$

$=0+\frac{n}{n-1}\int_{0}^{\infty}(1+y)^{1-n}\cos y\,dy$

Now estimate the absolute value:

$\left|\frac{n}{n-1}\int_{0}^{\infty}(1+y)^{1-n}\cos y\,dy\right| \le\frac{n}{n-1}\int_{0}^{\infty}(1+y)^{1-n}\,dy$

$=\frac{n}{(n-1)(n-2)},$ provided $n>2.$

Let $n\to\infty$ and the limit is zero.

Pretty sure these were designed to be done with the dominated convergence theorem.  The second requires a variant on the dominated convergence theorem; see if you can figure out what it looks like.

Sorry but I cant see the initial problem.
What was it?


MIT college Columbia Princeton Harvard Yale Stanford

MIT app goes out tomorrow. I don't like my chances, but hey, ya never know.

Godspeed, application. Godspeed.

I hope it all turns out well..

I'm getting butterflies just thinking about sending out applications next year.

Good luck...
i feel the same way..mine went out yesterday  :)

And mine will probably go out on Monday. Good luck all.

Good luck to all of you, MIT applicants.

By the way, how many colleges do you apply for? I thought about 10 when I was discussing this topic with my 10th grader (who will apply in two years).

Here's my list:

MIT* - U of Rochester - Cornell - UMass Amherst* - UVM* - Columbia U - WPI - RIT - Washington University - McGill

and *maybe* Princeton or Carnegie Mellon

* - Early Action

Yea I know, it's quite a mouthful  :icecream:

Hi, Curl, I take it you live in the Northeast? When the time comes (= when my oldest son is older, and when he is reasonably sure about what we wants to study), I expect my son's list will be a bit more regionally diverse, but it will have our state university on it as a "safety" (and not half-bad choice) almost for sure.

Good luck to the rest of you who are applying to MIT this year. An entering class full of AoPSers would be an exciting entering class.

When you apply for college, can you put more than one school for early action? (I noted in curl’s post, there were three marked for early action).

We made a tentative list of 10 schools, which include 2 state schools (honor’s program, not for “safety”, but for free-ride), 4 ivy leagues (Harvard, Yale, Princeton and Columbia, no particular order) and 4 non-ivy leagues (MIT, Stanford, Duke and Cal Tech, no particular order). I am sure the list will change over the next 2 years.

How many schools you can apply to early action (EA) depends on the most restrictive policy of the schools you might put on your list. MIT will accept early action applications from applicants applying early to other top-tier schools. (Notably, many applicants apply early to both Caltech and MIT at the same time.) But Harvard, Stanford, and Yale have what is called "single-choice early action," (SCEA) which means that the applicant agrees not to apply early to other domestic schools with a defined early action date. (Harvard allows early action applicants to apply for "rolling admission" to a state university as soon as they please, and that is a very good idea, and Harvard also allows early applicants to apply to foreign universities whenever it fits the foreign universities' schedules.) And of course an early decision (ED) school (e.g., Princeton) expects applicants not to apply early to any other school with an "early" admissions program, and expects admitted applicants who are admitted in the early round to withdraw all regular decision applications to other schools.

So it really depends on what colleges appear to be reasonable to put on your list. An instance of applying early, all at the same time, to MIT, Caltech, and Chicago might make sense to a would-be math major who wants to go to the most competitive schools, while an instance of applying ED to Princeton might make sense to an applicant who is really, really sure he wants to go to Princeton. You have to read the policies of each school carefully to avoid making inconsistent decisions about how to apply.

Yea, you can apply to as many schools as you want early [i]action[/i]; it's the early decision application that is binding and can only be thrown at one school. Early action just means I'm getting a decision sooner.

I always assume that a student apply for individual colleges on his/her list with no particular order (though you do have a preference in your heart), in other words, he/ she does not make an open/public ranks of the list, which is his/her first choice. He/she deals with each college individually. However, some of requirements regarding merit scholarships from certain colleges clearly state that you have to put them as your first choice on your list to qualify for specific type of scholarships. I wonder do you have to tell each school you applied that what else you also applied, and in what order?

I don't think a student always deals with each college individually.  It is a huge impropriety, in my opinion and sometimes also as a matter of law, for the universities to solicit or use information about applications to other schools.  In the case of the Ivy [League] Group financial aid meetings, the practice was stopped by a US government anti-trust prosecution.  Nevertheless, some university application forms used to inquire about the list of schools a student applied to, as did some college interviews.  If anybody knows the current situation I would be interested to hear about it.

About first-choice schools and their merit scholarships, the intention of the requirement is clear (school wants to offer the scholarship only to students who will accept it), but probably it is unenforcible.  Most applicants accept not just an offer of admission, but a combined package of admission and financial aid, so it is probably illegal for the school to solicit their agreement to the would-be offer before disclosing its financial terms.

well, during my MIT interview (last year) i was asked which other schools i'm applying to and how MIT ranks amongst them- not sure if that question is common though! i also dont think many people would have the courage to say their real likings even if MIT weren't their first choice...

Also, on the "yes i'm coming to your college" form, they ask where else you applied and how they compare in terms of financial aid and so on, but i guess that's different

From what I gathered from the MIT website/blogs, the decisions were sent out today... that means monday or tuesday could be the fated day. Good luck to all... I know I am going to need it.


Yes, can someone help? I got k / 2^k already. Just need 1 / k(k+1)(k+2). I know the final form is 1/4 - 1 / 2(n+1)(n+2) but don't know how to get to there. Hw is due this evening. Thanks in advance...

[quote="supastar33"]sum_{k=1}^{infinity} k / 2^k and
sum_{k=1}^{infinity} 1 / k(k+1)(k+2)[/quote]
Well,
\[
\sum_{k=1}^{\infty} \frac k{2^k}=x\left( \sum_{k=1}^{\infty} x^k\right)'|_{x=1/2}=x\left(\frac 1 {1-x}\right)'|_{x=1/2}=\frac {1/2}{(1-1/2)^2}=2,
\]
and
\[
\sum_{k=1}^\infty \frac 1 {k(k+1)(k+2)}=\sum_{k=1}^{\infty} \frac{1/2}k -\frac 1{k+1}+\frac{1/2}{k+2}%Error. "sybstackrel" is a bad command.
{*}=
1-\frac{1/2}{2}-\frac{1/2}{1}=\frac 1 4.
\]
(*) we reorganize every partial series to the coefficient ''$1/k$''.

RX the Crazy Math


Let $ x,y,z\geq0,t\in N$,prove that:

$ x^t(x \plus{} y) \plus{} y^t(y \plus{} z) \plus{} z^t(z \plus{} x) \geq \frac {6}{3^{t \plus{} 1}}(x \plus{} y \plus{} z)^{t \plus{} 1}.$

BQ

Let $ x_i\geq0,i\equal{}1,2,...,n,n\geq3,t\in N$,prove that:

$ \sum_{i \equal{} 1}^n{x_i^t(x_i \plus{} x_{i \plus{} 1})}\geq \frac {2n}{n^{t \plus{} 1}}(\sum_{i \equal{} 1}^n{x_i})^{t \plus{} 1}.$

BQ


geometry
trigonometry
rotation
geometric transformation

Let $1,\ \alpha ,\ \beta$ are roots of the cubic equation with real coefficients $x^{3}+ax^{2}+bx+c=0$ and are three distinct vertices of a regular hexagon with area of $6\sqrt{3}.$ Find the values of $a,\ b,\ c.$

[hide]If $\alpha$ and $\beta$ are real, then $1,\alpha,\beta$ will lie on the real axis, and cannot be three distinct vertices of a regular hexagon. Thus, they are complex, and moreover are complex conjugates of each other.
Since 1 is a vertex, $\alpha$ and $\beta$ are either $(\cos60+i\sin60,\cos60-i\sin60)$ or $(\cos 120+i\sin120, \cos120-i\sin120)$. From Viete, $a=-1-\alpha-\beta$, $c=-\alpha\beta$, $b=\alpha+\beta+\alpha\beta=-1-a-c$. Then $a=-1-2\cos60=-1-2\cos120=-2$, $c=-1$, $b=-1+2+1=2$. $\boxed{(-2,2,-1)}$[/hide]

[quote="Altheman"][quote="cincodemayo5590"][quote="pianoforte"][hide]Since 1 is a vertex, $\alpha$ and $\beta$ are either $(\cos60+i\sin60,\cos60-i\sin60)$ or $(\cos 120+i\sin120, \cos120-i\sin120)$[/hide][/quote]

That's not necessarily true. [hide="Because..."] either the radius might not be at $(0,0)$ (look at the area restriction), or the hexagon might be oriented differently, i.e. rotated about $(1,0)$.[/hide][/quote]

real coefficients[/quote]

the center would be like $(3,0)$ or something. (rotation of 180 degrees)


find a positive integer n so that 40n is a fifth power (of an integer) 500n is a sixth power, and 200n is a seventh power, or explain why it is impossible to do so (hint: let x = e2(n) and y = e5(n)  use the given assumptions about n to find systems of linear congruences that x and y must satisfy.  you may leave n in terms of its canonical prime factorization.  show as much work as possible

Hey.  i was just wondering if you are reading this could you possibly please answer my question or give me a hint.  i have no idea where to start.  thanks

This looks like a homework problem from an intro course in elementary number theory (as do your other posts)......
They're not very hard, either, so you'd probably be better off doing them yourself.

I think Intermediate Section is a proper place for this problem :)


by hand and complicated: about 15 minutes
by hand and simple: about 7 minutes
with a graphing calculator: about 2 minutes

Well, you'd need to take 5 4x4 determinants, each one is composed of 4 3x3 determinants, so that makes a total of 20 3x3 determinants.  I can do a 3x3 in about thirty seconds or a bit more, so I'd take at best 10 minutes to do that.  15 sounds like a reasonable guess.  If the system had lots of 0's, 1's and -1's, it would probably take me significantly less time.

This is why Cramer's rule is of more theoretical than practical interest.  I believe that the number of operations for calculating determinants of nxn matrices is on the order of n!, whereas Gaussian elimination and Gauss-Jordan elimination are on the order of 1/3n^3 and 1/2n^3 operations, respectively.


calculus
integration
function
real analysis
real analysis unsolved

Suppose $ f\in L^1(\mathbb{R}^3)$. Show that the integral

$ \phi(x)\equal{}\int_{\mathbb{R}^3}\frac{f(y)dy}{|x\minus{}y|}$

converges for a.e. $ x\in\mathbb{R}^3$.

Let $ \psi(x)\equal{}\int_{\mathbb{R}^3}\frac{|f(y)|}{|x\minus{}y|}\,dy.$

Let $ B_r$ be the ball centered at the origin of radius $ r.$

$ \int_{B_r}\psi(x)\,dx\le\int_{B_r}\int_{\mathbb{R}^3} \frac{|f(y)|}{|x\minus{}y|} \,dy\,dx$

$ \equal{}\int_{\mathbb{R}^3}|f(y)|\int_{B_r}\frac{1}{|x\minus{}y|}\,dx\,dy$

We can show that $ \int_{B_r}\frac{1}{|x\minus{}y|}\,dx\le\int_{B_r}\frac{1}{|x|}\,dx\equal{} Cr^2.$

(The constant $ C$ can be computed - it's actually $ 2\pi$ - but the exact value is immaterial.)

Hence, $ \int_{B_r}\psi(x)\,dx\le Cr^2\int_{\mathbb{R}^3}|f(y)|\,dy$

Since this is finite, we must have $ \psi(x)\le \infty$ almost everywhere on $ B_r.$ Since we can repeat this for any $ r$ (choose a countable sequence of such $ r$ tending to $ \infty$) we can say that $ \psi(x)$ is finite almost everywhere on $ \mathbb{R}^3.$ Hence the integral that defines $ \phi(x)$ converges absolutely for almost every $ x.$

Note the importance here of having the function $ \frac{1}{|x\minus{}y|}$ be locally integrable - in fact, uniformly locally integrable.

The integral swap works because the way that it is computed is absolutely integrable, so the two ways are equal by Fubini? There are so many ways of stating Fubini, it is often confusing what actually satisfies the hypothesis to me.

Call this Tonelli's theorem since I was switching the order of integration for nonnegative functions. (Tonelli is part of the same family of theorems as Fubini.)


Find all function such that
$ f: (0;\plus{}\infty) \rightarrow (0;\plus{}\infty)$
$ f(f(x))\equal{}\frac{1}{f(x)}\plus{}ax$
where $ a>0$ and  $ f(1)\equal{}2$

[quote="aktyw19"]Find all function such that
$ f: (0; \plus{} \infty) \rightarrow (0; \plus{} \infty)$
$ f(f(x)) \equal{} \frac {1}{f(x)} \plus{} ax$
where $ a > 0$ and  $ f(1) \equal{} 2$[/quote]

There are not solutions for any $ a$. For example $ a\equal{}\frac 32$ implies no solution.
There exist some values of $ a$ for which we have solutions: for example, $ a\equal{}\frac 12$ has solution $ f(x)\equal{}\frac 2x$

Before I continue to look for a general solution, could you kindly confirm us, aktyw19, that you are sure such a general solution exist for this problem ?

[quote="aktyw19"]I do not have a solution to this problem[/quote]

I did not ask you if you have a solution.
I asked you if you are sure a solution exists. In other words :

Is this problem from a serious source : a contest, a teacher, a book ? and it seems then this is the good forum

Or is this problem from a magic source : own-created, good friend, ... ? and better to post in "open"

I'm sorry for this question but you sometimes posted in "unsolved problems" some crazy invented problems. I just want to remember you that "unsolved" forum usage is "Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it."
So, posting here means that you claim that there is a solution.
Just wanted to be sure you understand this.
Sorry if this bothers you.


Elimination game of the top 10 (by alltime leaderboard) FTWers. Add or subtract one point per turn.

[color=green]Mewto55555-11[/color]
Superllamel-10
Kl2836-10
FantasyLover-10
Aime15-10
Isabella2296-10
#H3N41-10
Vallon22-10
Ernie-10
JOngao-10

[color=red]Mewto55555-10[/color]
Superllamel-10
Kl2836-10
FantasyLover-10
Aime15-10
Isabella2296-10
#H3N41-10
Vallon22-10
Ernie-10
JOngao-10

Mewto55555-10
Superllamel-10
Kl2836-10
FantasyLover-10
Aime15-10
Isabella2296-10
[color=red]#H3N41-9 <----Lies about his "nonmemming" every question![/color]
Vallon22-10
Ernie-10
JOngao-10

Mewto55555-10
Superllamel-10
Kl2836-10
FantasyLover-10
Aime15-10
Isabella2296-10
[color=green]#H3N41-10[/color]
Vallon22-10
Ernie-10
JOngao-10


function
parameterization
calculus
integration
advanced fields
advanced fields unsolved

Consider the boundary value problem

y" - 3y' - 4y = t^2
y(0) = 0
y(1) = 2

(a) Find ( if any) a Green's function associated with this equation.

(b) Use Green's function to find the solution ( if any) of the boundary value problem.

I don't know how to work this problem. Will someone please help me? Thanks.

hello, to solve the homogenous part of your equation set $ y(t)\equal{}e^{\lambda t}$.
Sonnhard.

Well, what do you know already?  One can derive the Green's function using variation of parameters:  The method gives you a general solution, and you can choose the integration constants to get zero boundary conditions.  Then you can rewrite the solution so that it is in the integral form you need to identify the Green's function.  By the way, this is not an easy exercise, unless you have a worked example to follow.


Let $x_0,y_0,z_0$ be three positive real numbers. Define the three sequences as follow
$x_{n+1}=y_n+\frac{1}{z_n}$
$y_{n+1}=z_n+\frac{1}{x_n}$
$z_{n+1}=x_n+\frac{1}{y_n}$
Prove that there exist $0<u<v$ such that : $u\sqrt{n}<x_n<v\sqrt{n}$  $\forall n\ge 1$

Isn't it rather : $z_{n+1}=x_n+\frac{1}{y_n}$   :?:

Oh yes I'm sorry  :(
I'll edit it right now


Let $ ABC$ be a triangle and construct squares $ ABED, BCGF, ACHI$ externally on the sides of $ ABC$. Show that the points $ D,E,F,G,H,I$ are concyclic if and only if $ ABC$ is equilateral or isosceles right.

Let $ A' \equiv HI \cap ED, B' \equiv ED \cap GF, C' \equiv GF \cap HI.$ Midpoints $ P, Q, R$ of $ AA', BB', CC'$ are on perpendicular bisectors $ p, q, r$ of $ ID, EF, GH$ as circumcenters of $ (ADA'I),$ etc. $ \triangle PQR \sim ABC$ with similarity center at their common symmedian point $ K.$ $ (p, PK)$ are isogonals WRT $ \angle RPQ,$ $ (q, QK)$ are isogonals WRT $ \angle PQR,$ $ (r, RK)$ are isogonals WRT $ \angle QRP$ $ \Longrightarrow$ $ p, q, r$ concur at the centroid $ M$ of $ \triangle PQR.$ Perpendicular bisectors $ u, v, w$ of $ ED, GF, HI$ concur at the circumcenter $ O$ of $ \triangle ABC.$ $ D, E, F, G, H, I$ are concyclic $ \Longleftrightarrow$ $ M \equiv O.$ As $ KM$ line of $ \triangle PQR$ coincides with $ KO$ line of centrally similar $ \triangle ABC,$ they need be isosceles, WLOG $ AB \equal{} AC \equal{} b,$ $ BC \equal{} a.$ Let $ h$ be A-altitude of isosceles $ \triangle ABC.$ $ KA \equal{} \frac {2b^2}{a^2 \plus{} 2b^2}\cdot h,$ $ AP \equal{} \frac {b^2}{a}$ $ \Longrightarrow$ similarity coefficient of $ \triangle PQR \sim \triangle ABC$ is $ k \equal{} \frac {KP}{KA} \equal{} 1 \plus{} \frac {a^2 \plus{} 2b^2}{2ah}.$ $ MP \equal{} OP \equal{} OA \plus{} AP$ $ \Longleftrightarrow$ $ \frac {2}{3}hk \equal{} R \plus{} \frac {b^2}{a} \equal{} \frac {a^2b}{2ah} \plus{} \frac {b^2}{a}$ $ \Longleftrightarrow$ $ 4ah^2 \plus{} 2ha^2 \plus{} 4hb^2 \equal{} 3a^2b \plus{} 6hb^2$ (substitute $ 4h^2 \equal{} 4b^2 \minus{} a^2$) $ \Longleftrightarrow$ $ (b^2 \minus{} a^2)(a \minus{} 2h) \equal{} 0$ $ \Longleftrightarrow$ $ b \equal{} a$ or $ a \equal{} 2h.$


MATHCOUNTS HMMT AMC AIME videos articles USA(J)MO

I gave up TV for the last week before AIME. (though i'm not too strict on myself)

A lot of things... especially these past few weeks.
Sleep (ah, so precious) to start with, a few tests in non-math classes, TV, computer games... what else?
I'm also gonna have to give up two school periods on Tuesday for the AIME.. not that I mind, though :)

Sleep, various ski meets, school classes, etc.
Last year one of my friends missed his graduation to go to ARML.

It is not my case, but given the dates when the OLCOMA is done, many of the participants skip their promt for it  :( .

Best regards,

[quote]It is not my case, but given the dates when the OLCOMA is done, many of the participants skip their promt for it Sad . [/quote]

what's OLCOMA????

and I never give up anything for MATH....

well,,

I would drop down school if those ugly school educators are willing to take away math .. :D

i have to miss a couple of midterms for a North Carolina Math Competition this coming wed.

has anyone gave up math for something?   :D

Ive had to give up math for math before(state mathcounts and HMMT were on the same day in 03  :( ), as well as giving up something for math alot, and theres been a few times when ive given up math for something(though normally not by my choice)

I'm giving up four class periods for the AIME. YA-err, darn.

I gave up a day of school for mathcounts this year :D .( It was a make-up school day, because we had snow earlier)

I give up piano competitions sometimes for math.  Usually, however, it is the other way around.

[quote="yif man12"][quote="plokoon51"][quote="yif man12"]Haha, this reminded me of the time I signed a contract saying I'd be at a state orchestra recital. It turned out to be on the same day as Mathcounts states. So, I broke the contract, me orchestra teacher was pretty mad at me for a while and I was banned from trying out for the recital the next year.[/quote]

i did the same thing with piano. trust me, you did the right thing :D[/quote]

Heh, how true.

The really freaky thing is that if I was thinking really hard about which to choose, because I hardly knew a thing about mathcounts and didn't think I had a chance to get top 4 at all. If I hadn't gone to mathcounts, my life would be so different right now...[/quote]

I've given up a good grade in Science to take care of math instead

[quote="yif man12"][quote="Dcadu9lvr"]Sleep and TV/Computer the week before state MATHCounts, and countless soccer games and dance rehearshals and few piano recitals to go to  various math competitions.

I gave up math (it was a district competition type thing...nothing big) for a soccer tournament in Las Vegas. It was worth it though, in the end, because our team ended up winning.  :)[/quote]

Did your coach let you guys gamble afterwards?[/quote]

If playing Blackjack with our coach as the dealer and betting Skittles in our hotel room counts, then yes.  ;)

my sanity... ahhh how true.  Cause I can prove that 2=1 if you don't notice something.  I'll post the thread address that shows how to do it, later :)

elvenchamp777

[quote="shinwoo"][quote=adidasty ]ive given up:
social life
girls
dates
prom
winter formal
sleep
money [/quote]

wow...

just wow..

I wouldn't give up all those stuff just for math...

I would give up most of them, but not the social life....[/quote]

You gave up dates, the prom and a winter formal for math?!?!?!

I understand the others, but wow.

I gave up computer games.

If i had adidasty's list, I'd keep the girls, dates, and money

i'm giving up state science fair (my project was in the category of mathematics) for a math competition this saturday, but assuming i do well, i'll be able to advance more and more... lol

i think the question must be what you're not giving up for math

My sanity, and, because of excessive AMC/AIME studying this year, good grades.

Here is a summary of what I don't do: (note, most are the result of computer gaming and foruming, not math)

I plan on acquiring my driver's license as late as possible, though my parents want me to acquire one as soon as possible
I never had a girlfriend
I never went to a dance nor did I go to a party since 7th grade
I never shop for anything. My parents buy everything for me and I'm heading to university next year
I'm vegetarian
I shower one day per week at the very most and if not urged by my parents, have gone for 3 weeks without showers
I never watch TV
I never have hanged out with classmates outside of school, with the exception of competitions
Out of all phone conversations I had, all were about something academic
I don't play sports or read, and only listen to computer game music
I determined to do something academic each and every one of my summers
I never join any non math/science clubs
My grades in school have dropped
I don't do chores nor do I have an allowance or any source of money. I'm just hoping that everything will go well before my first research experience

Here is what I do:

- collegeconfidential.com. My current incredible addiction
- artofproblemsolving.com more for looking at gifted student profiles than math
- accuweather.com, 9news.com, http://www.crh.noaa.gov/den  for Denver's weather in particular.
- cnn.com, seattletimes.com, msnbc.com for latest news on education.
- http://www.deathlist.net, http://www.deadoraliveinfo.com, obituaries on Wikipedia I overreacted to the deaths of Hans Bethe, Ernst Mayr, and George Kennan.
- Latest games played: Age of Kings(addicted for 3 years), Runescape, Wolfenstein 3D, Wizardry 8 demo (yes a demo is addicting), Pokemon through an emulator. Fortunately, my 4 year old computer precludes me from playing most recent games
- AIM when it's not uninstalled. I uninstalled it MANY times and reinstalled it MANY times
- Too much Wikipedia and random articles
- Gmail
- Education statistics when they come. Especially lists of national merit semifinalists and I love analyzing trends.

Yeah I've pretty much driven myself insane trying to impose self-control and to emulate Marie Curie. Only that I've repeatedly failed and have pretty much given myself up to playing more and more computer games. It may look gloomy right now but hopefully, university next year will rectify all of this. My parents won't buy my a laptop, I'll have rooms to study without distractions, and there aren't any very active graduate school forums to obsess over, although I'll probably become obsessed with physicsforums.com. I may also finally develop a social life: when people actually live with me and when there are people with ambitions similar to mine, I may look forward to social events.

Apparently, none of you remotely compare to what Archimedes, Sophie Germain, Marie Curie, Paul Erdos, and Ramanujan sacrificed.

Archimedes:

His life

Sophie Germain:

A husband and a life of being a woman otherwise. Oh yeah and lots of sleep all those hours of staying up!

Marie Curie:

Pretty much everythin except for her husband. Lots of sleep, ate very little food, considered everything a waste of time except time devoted to studies. I adopted her as a role model but only made myself insane in the process as I could not make the sacrifices she did because I'm still playing computer games and surfing the Internet incessantly.

Ramanujan

Pretty much his life as he did not attend to hygiene or any matters in taking care of himself.

Paul Erdos:

;)

I gave up 300 hours of AIM chatting to prepare for USAMO!  :lol:

I gave up a lot of stuff, but the most recent was dropping State Science Olympiad for the San Diego Math Olympiad.

Umm...lots of sleep.... for MATHCOUNTS 30 hours of time just working from chapter to state competition.... basketball games... tv...

<edit> forgot to include my sanity.  :)

[quote="elementsofmathkid"]
<edit> forgot to include my sanity.  :)[/quote]

Haha...yes...but math also gives me my sanity!

Yeah, it should be math [i]and[/i] the opposite gender, right? (Just a joke, before you harp on me about it - I agree with you, math is one of many focuses in my life, and I don't even put other school subjects above it.)


geometry
circumcircle
ratio
congruent triangles
perpendicular bisector
geometry solved

From an arbitrary point $M$ of the base $BC$ of an isosceles triangle $ABC$ we draw parallels to $BA$ and $CA$ which intersect $AC$ and $AB$ at $D$ and $E$ respectively. Prove that, when $M$ moves on $BC$, the perpedicular bisector of the segment $DE$ passes through a fixed point.
Also prove that the circumcircle of $ADE$ passes through the same point.

I have edited your post to correct a few language mistakes: A line passes [i]through[/i] a point, not [i]from[/i] a point (I am pointing this out since you have made this mistake a number of times).

Anyway, it is a nice problem. Let O be the circumcenter of triangle ABC. We will show that O is the required fixed point.

Since MD || BA, Thales yields $\frac{AD}{DC}=\frac{BM}{MC}$. Since ME || CA, Thales yields $\frac{BE}{EA}=\frac{BM}{MC}$. Thus, $\frac{BE}{EA}=\frac{AD}{DC}$.

As the triangle ABC is isosceles and O is its circumcenter, < BOA = < AOC. Also, BO = AO and AO = CO (as the point O is the circumcenter of triangle ABC). Hence, the two triangles BOA and AOC are congruent (note the correct order of the points!). The points E and D are corresponding points in these two triangles, since they lie on their corresponding sides BA and AC and divide these in the same ratio $\frac{BE}{EA}=\frac{AD}{DC}$. Since corresponding points in congruent triangles form equal lengths and equal angles, we thus conclude that OD = OE and < ODA = < OEB. From OD = OE, it follows that the perpendicular bisector of the segment DE passes through the point O. From < ODA = < OEB, we conclude that < ODA + < OEA = < OEB + < OEA = 180°, so the quadrilateral ADOE is cyclic, i. e. the circumcircle of triangle ADE also passes through the point O.

Since the point O is obviously fixed, this solves the problem.

  darij

[quote="darij grinberg"]Hence, the two triangles BOA and AOC are congruent (note the correct order of the points!). The points E and D are corresponding points in these two triangles, since they lie on their corresponding sides BA and AC and divide these in the same ratio $\frac{BE}{EA}=\frac{AD}{DC}$. Since corresponding points in congruent triangles form equal lengths and equal angles, we thus conclude that OD = OE  [/quote]
Can someone explain it to me please?Is there such a theorem?

[quote="silouan"][quote="darij grinberg"]Hence, the two triangles BOA and AOC are congruent (note the correct order of the points!). The points E and D are corresponding points in these two triangles, since they lie on their corresponding sides BA and AC and divide these in the same ratio $\frac{BE}{EA}=\frac{AD}{DC}$. Since corresponding points in congruent triangles form equal lengths and equal angles, we thus conclude that OD = OE  [/quote][/quote]
I understand it!Ok


calculus
integration
trigonometry
derivative
calculus computations

[quote="kunny"]Evaluate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}+\sqrt{\cos x}+3(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\ dx.$[/quote]

$\frac{\sqrt{\sin x}+\sqrt{\cos x}+3(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\\=\frac{(1+\cos 2x)\sqrt{\sin x}+(1-\cos 2x)\sqrt{\cos x}+2(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\\=\frac{(2\cos^{2}x)\sqrt{\sin x}+(2\sin^{2}x)\sqrt{\cos x}+2(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\\=\sqrt{\sin 2x}\left(\frac{(2\cos^{2}x)\sqrt{\sin x}+(2\sin^{2}x)\sqrt{\cos x}}{2\sin x\cos x}\right)+\frac{2(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\\=\sqrt{\sin 2x}\left(\frac{\cos x}{\sqrt{\sin x}}+\frac{\sin x}{\sqrt{\cos x}}\right)+\frac{2(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\\=2\sqrt{\sin 2x}\frac{d}{dx}\left(\sqrt{\sin x}-\sqrt{\cos x}\right)+(\sqrt{\sin x}-\sqrt{\cos x})\frac{d}{dx}\left(2\sqrt{\sin 2x}\right)\\=\frac{d}{dx}\left(2\sqrt{\sin 2x}\left(\sqrt{\sin x}-\sqrt{\cos x}\right)\right)$

 :)

Do you have other way ? Kunny ?

This is a terrible problem if there is no other way.

You may be right. I made this problem by the differentiation of $\sin x \sqrt{\cos x}+\cos x\sqrt{\sin x}$ as shyong showed.


geometry
geometric transformation
rotation
AMC
AIME
vector
complex numbers

Let the points a, b, and c lie on the unit circle of the complex plane and satisfy a+b+c=0. Prove that a, b, and c form the vertices of an equilateral triangle.

[quote="blahblahblah"]I seem to recall doing this problem before.

Is there a way to do it without assuming that, say, w1 = 1 (an argument that is justified by a properties of linear transformations)[/quote]

Yes.

[quote="Nukular"][quote]Is there a way to do it without assuming that, say, w1 = 1 (an argument that is justified by a properties of linear transformations)[/quote]

What's wrong with doing that? It makes the problem much more doable.

This rotation was used on an AIME problem concerning complex #'s, though I forget which year.[/quote]

It makes the problem almost trivial.  I was just wondering if there was a more aesthetic solution.

Set one of the three points equal to one.

Could someone explain to me why you're allowed to make that transformation?  It doesn't seem obvious to me, but maybe I'm just missing something.

Assume that you have a triangle with the three things summing to zero.  Rotation in the complex plane is just multiplication by a number with absolue value 1.  So when we have a sum of 0, rotating everything won't change the value of the sum.  Thus, just rotate until one of them is at 1.

It would probably be helpful if I knew stuff like that.  Thanks.

Well, have you ever done complex numbers in polar or exponential form?  It's really really obvious that this is true in exponential form, and it's fairly clear if you examine what happens if you multiply [tex](r_1 cis(\alpha)) \cdot cis(\theta)[/tex]

It's not necessary to make a rotation. It's straightforward to verify that [tex]|b-a|^2=|c-a|^2[/tex] as long as you know that [tex]|z|^2=z\bar{z}[/tex].

Try vectors.  Treat each point on the complex unit circle as a vector.

We are given that a+b+c=0.  The three vectors sum to 0 only if the figure made by joining the vectors tail-to-head is closed.

What do we call a closed figure with three equal-length straight sides?  See illustration.

[quote="fuzzylogic"]Nice interpretation! But they are not exactly the same triangle.[/quote]

Indeed, the equilateral triangle in the lower-right has a side-length of 1, while the equilateral triangle inscribed in the unit-circle in the lower-left has a side-length of [tex]\sqrt{3}[/tex].

I was trying to give a simple, visual interpretation of the problem.

Some geometry, along with the basic properties of vectors will show that if one of these triangles is equilateral the other triangle is also equilateral.


algebra
polynomial
function
Ring Theory
rational function
superior algebra
superior algebra unsolved

Suppose $ F$ is a field and $ F(t)$ is a field of rational functions.

1. Suppose $ p(t)/q(t)$ is an element of $ F(t)$ not in $ F$. Prove that $ p(t)/q(t)$ is not algebraic over $ F$.

2. For each such element $ p(t)/q(t)$, show that there is a unique endomorphism of $ F(t)$ that has $ F$ as a fixed field and takes $ t$ to $ p(t)/q(t)$.

3. Show that this endomorphism is an isomorphism iff $ p(t)/q(t) \equal{} (at \plus{} b)/(ct \plus{} d)$ for some $ a, b, c, d$ with $ ad \minus{} bc \equal{} 0$.

1. Assume there is some algebraic relationship concerning this element. Multiply with some "common denominator" to reduce it to a equation of polynomials in $ t$. Now proceed.

2. This is a combination of the universal properties of the polynomial ring and the quotient field (and can be called the universal property of the rational function field).

3. For it to be an isomorphism, there needs to be a inverse map. Let $ f$ be induced by $ t \mapsto \frac {p(t)}{q(t)}$ and let $ g$ be induced by $ t \mapsto \frac {r(t)}{s(t)}$. For the maps to be inverse, it's necessary and sufficient to have $ f(g(t)) \equal{} t \equal{} g(f(t))$. Now write it out.


ratio
quadratics
Vieta
search
algebra
polynomial
sum of roots

The eqn we write is $x - 1/x = 1$
Multiply by x and simplify, we get $x^2 - x -1$

The sum of the roots is $- - 1/1 =\boxed{1}$

the roots of the equation:

$x^2 - x - 1$

are

the 2 golden ratios

when you add them together u get root 5

no idea how u type the equations lol

Try using the quadratic formula, you will still get 1 as the answer.

I just tried it.

or you can use vieta's sums to get the sum right away, search if you don't know what that is  ;)

I used the sum of roots of a polynomial: -b/a, a and b are the coeff of the polynomial.

edit: just searched vietas and its the same thing :|

Vieta's is very important, but many people don't recognize that name right away.  See Sum of Roots (& Product of Roots) at the SC Math Team Website: http://www.geocities.com/scmathteam/polynomials.pdf

[hide="the correct answer"]
Agent is correct, the answer is $\sqrt{5}$.
I got one by subtracting the two equations and factoring...[/hide]

edit: now i notice that you cant get 1 with positive numbers.

So this is how i did it now
The discriminant is  $\sqrt{1 - 4(-1)(1)}$ which is $\sqrt5$ so thats the answer

[quote="xenon_nightmare152"]AMC 12 2002
#13

 Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b?[/quote]

[hide]the solution to the equation $x^2-x-1=0$ is indeed 1.

however

one of those roots is negatinve and one is positive

so we get the other equation $x-\frac{1}{x}=-1\Rightarrow x^2-x+1=0$ which has no root so there is only one positive root.

which  is from the first equation

did i do something wrong?[/hide]


MATHCOUNTS
quadratics
conics
parabola
analytic geometry
geometry
rectangle

Okay, I am going to host a little contest.  There are many ways to solve this problem and I am going to judge the nicest solution(s).

If the sum of two real numbers is 100, what is the maximum value of the product of the two numbers?

ooooooo, a contest.  8-)

In spoiler:

[hide]Let the two numbers be the roots of a quadratic x^2+bx+c. Since their sum is 100, by Viete's formula, b = -100, yielding the quadratic x^2-100x+c. Now for the roots to be real, the discriminant must be greater than or equal to 0. Thus, we have (-100)^2 - 4(1)c >= 0 <=> 2500 >= c, which is the product of the roots.[/hide]

I'm pretty sure you have to use limits and such to get the most credit, but I don't know any...

Chinaboy wrote:[hide]Ok... so it's

x(100-x) or -x^2+100x

So then... you use the formula to find the y-value of the vertex.

c - b^2/4a

So it's 0 - (10000/(-4)) = 10000/4 = 2500[/hide]

Now tell me why the formula works.

The formula works because-

[hide] a parabola is symmetric about its axis (the thing perpendicular to the directrix that passes through the vertex). So, its vertex is on the bisector of the parabola and is midway between the roots. quadratic formula = [-b- :sqrt:  (b :^2: -4ac)]/2a, and [-b+ :sqrt: (b :^2: -4ac)]/2a/ we just want to know the x-coordinate, which would also be the x coordinate of the midpoint. adding the two quad form things and div-ing by 2, we get -2b/4a, or -b/2a[/hide]

Can anyone here construct a proof using the sides of a rectangle?

How about using the arithmetic mean-geometric mean inequality?

AM-GM method (spoiler):

[hide]Let the two numbers be x and y. By AM-GM, we have (x+y)/2 = 100/2 = 50 >= Sqrt(xy) <=> xy <= 2500.[/hide]

[quote="Tare"]OK, you lost me on that last part. Can you explain what "arithmetic mean-geometric mean inequality" is?[/quote]

The AM-GM inequality says that if you have a great of nonnegative real numbers that the arithmetic mean of the group of numbers is always greater than or equal to the geometric mean of the group of numbers.  Equality only occurs when all the numbers are equal.

For instance, the arithmetic mean of 3 and 12 is 7.5.  The geometric mean of 3 and 12 is 6, so the arithmetic mean is at least the geometric mean.

The proof for the AM-GM inequality for two values goes like this:
( :sqrt: x -  :sqrt: y) :^2:  :ge: 0
This is because a square of a real number is never negative.
We can expand the left-hand side to get
x - 2 :sqrt: xy + y  :ge: 0
Thus
x + y  :ge:  2 :sqrt: xy
Dividing both sides by 2 we get
AM :ge: GM

This can be used to solve all kinds of cool problems.
If the perimeter of a rectangle is 40, what is the largest possible area of the rectangle?

The sum of the dimensions of the rectangle is 20.  The arithmetic mean of the dimension is 10.  The geometric mean of the dimensions is thus at most 10.  Since the area of the rectangle is the square of the geometric mean of the dimensions, the area is at most 10 :^2: = 100.  The largest area is produced by the square with side length 10.

There are a lot of other problems that can be solved with the AM-GM inequality.

See if you can solve this one: If x > 0, what is the smallest possible
value of  x + 1/x?

I will be going through a series of AM-GM inequality problems in my Algebraic Problem Solving course in a few weeks.  Some of them get quite difficult and are akin to problems that might be seen on the AMC exams, AIME, USAMTS, and perhaps even the USAMO.  But there are even a few MATHCOUNTS problems that can be solved more easily using the AM-GM inequality, even if they can always be solved by other methods.

The arithmetic mean-geometric mean inequality (also called AM-GM) states that if x_1, x_2, x_3, ..., x_n are positive, then 1/n*(x_1+x_2+x_3+...+x_n) >= (x_1*x_2*x_3*...*x_n)^(1/n). If n=2, then it reduces to 1/2(a+b) >= sqrt(ab).

[quote="Fierytycoon"]So Mr. Crawford, what solution to this problem do you think is most elegant?[/quote]

I'll give your first solution the highest grade thus far.

[quote="MCrawford"]See if you can solve this one: If x > 0, what is the smallest possible
value of  x + 1/x?[/quote]
Can you give me a hint as to what I should put as y? I tried x+1/x=y and used the AM-GM inequality, but it didn't get me anywhere (I got 1:ge:1)

Remember that the AM is never less than the GM. See if you can turn x+1/x into the AM of something. What would the GM be?

MCrawford wrote:See if you can solve this one: If x > 0, what is the smallest possible
value of  x + 1/x?

i solved it as follows.

Solution:[hide]so the arithmetic mean of this is (x+1/x)/2=(x^2+1)/x divided by 2. so then (x^2+1)/2x :ge: the geometry mean. the geometry mean is the square root of x times 1/x. so the square root of 1 (x times 1/x is 1) is 1

so then you know that the smallest possible value is 1. if you want to see for yourself, just see the quadratic x^2-2x+1. which is (x-1):^2:. i got there from multiplying (x^2+1)/2x :ge: 1 by 2x to get x^2+1 :ge: 2x so then subtracting 2x i got x^2+1-2x :ge: 0. so then it as 1[/hide]

dont get mad if i got it wrong cause this is the first problem ive tried the am-gm thingie on before

Quote:i solved it as follows.

Solution:[hide]so the arithmetic mean of this is (x+1/x)/2=(x^2+1)/x divided by 2. so then (x^2+1)/2x :ge: the geometry mean. the geometry mean is the square root of x times 1/x. so the square root of 1 (x times 1/x is 1) is 1

so then you know that the smallest possible value is 1. if you want to see for yourself, just see the quadratic x^2-2x+1. which is (x-1):^2:. i got there from multiplying (x^2+1)/2x :ge: 1 by 2x to get x^2+1 :ge: 2x so then subtracting 2x i got x^2+1-2x :ge: 0. so then it as 1[/hide]

dont get mad if i got it wrong cause this is the first problem ive tried the am-gm thingie on before

The only thing that you forgot to do was multiply back by two.  Remember, you divided the original function by 2 to get the AM.  Otherwise, way to go -- this is a great first step toward using the AM-GM Inequality as a monster tool for solving problems.

[quote="ComplexZeta"]You can solve it even more quickly as follows:

(x+1/x)/2 >= sqrt(x*1/x)=1.
x+1/x >= 2.[/quote]

That's what he did.  He just wanted to explain more about the solution I think.  But he didn't multiply the 2 back in.

indeed, but now i learned from me mistake ja. and i did do it like simon, except for the multiplying back part, i just wanted to explain to tare or whoever.

[hide=Why the bump]
It is clear that $\frac{d}{dx} \frac{x}{1+x^2} = \frac{1-x^2}{(x^2 + 1)^2}$. We solve this and get that it has roots $1, -1$. Plugging these in to the original equation we get that the maximum is $\frac{1}{2}$.
Intended solution is definitely not to take the reciprocal of both sides and minimize $\frac{x^2}{x} + \frac{1}{x}$
[/hide]

[quote=aops-g5-gethsemanea2][hide=Post #23 by SigmaPiE][url=aops.com/community/user/494837][b]SigmaPiE[/b][/url] · Mar 4, 2021, 10:34 AM [url=aops.com/community/p20733569](view)[/url][color=transparent]helo[/color]
[quote=MCrawford]Here's another one for you: What is the maximum value of the function  f(x) = x/(1+x^2)  where x is restricted to real numbers?

This one is a little tougher.  Try to manipulate the function until you have something you can apply the AM-GM inequality to.

I will be examining a lot of functional max/min problems like this in my Algebra class later this year.  :D[/quote]
[hide = ans]
$\frac{1}{2}$
I got this answer by essentially doing what you said, rearranging and modifying the equation until I got the AM GM inequality as something like this: $f(x) \le \frac{1}{2}$ (after applying it, of course) Therefore, the maximum is $\frac{1}{2}$ [s]bumping 18 year old threads are fun i guess[/s][/hide]

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[tip=@SigmaPiE][img]https://avatar.artofproblemsolving.com/avatar_494837.png?t=1606407020[/img] [center]AoPS User[/center][/tip] how did you find this thread[/quote]

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group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a group and $ N \triangleleft  G$, show that if $ \overline{M}$ is a subgroup of $ G/N$ and $ M\equal{}\left\{a \in G| Na\in\overline{M}\right\}$, then $ M$ is a subgroup of $ G$, and $ M\supset\ N$.

Not sure where to start.  I am new to abstract algebra.
Attempted solution:

Let $ \overline{M}\subset G/N$
Since $ N \triangleleft  G$,  $ Na \equal{} aN$, and $ N \equal{} a^{\minus{}1}Na$

$ \overline{M} \subset Na$
$ \overline{M} \subset a^{\minus{}1}Na$

also $ M\supset\overline{M}$ since $ M\equal{}\left\{a \in G| Na\in\overline{M}\right\}$
$ M\supset\ a^{\minus{}1}Na\supset Na$

therefore $ M\supset\ N$.

$ \overline{M}$ is not a subset of $ M$, and you have never really said what you mean by $ a$.  So I don't think you have really proved anything.

There are two things to show: that $ M$ is a group, and that $ N\subset M$.  You should clearly separate your proofs for both of these things.

For the former, you probably have some kind of theorem about subsets, and when they are subgroups.  Try to use that.

Here's how to prove the second statement.  You must first take an arbitrary element $ n\in N$, and show that it is an element of $ N$.  What are the elements of $ M$?  Look at what $ M$ is; it's defined very clearly in the problem.

So, what we need to show is that $ Nn \in \overline{M}$.  But you should know the basic fact that $ n\in N$ implies $ nN\equal{}N$, so rephrasing, we want to show that $ N\in \overline{M}$.

If you know the definition of $ G/N$, this should be very simple. (hint: what is the identity in this group?)

Thank you for the quick response and direct comments.  They helped clarify how to address this problem.
Would the following work...

Show $ M$ is a subgroup of $ G$ (closure, identity, inverse, non-empty).
Given $ a \in G$, let $ a \in M$ such that
$ ae \equal{} a \in M$ (where e = identity)
and $ a^{\minus{}1} a \equal{} e \rightarrow a^{\minus{}1} \in M$ (inverse)
also let $ a,b \in M$ then $ ab \in M$ (closure)

Show $ M \supset N$
Given $ G/N \equal{} \left\{Na | a \in G\right\}$ by definition.
Let $ n \in N$ and $ n \in G$ such that

for $ M \equal{} \left\{a \in G | Na \in \bar{M} \right\}$ and  $ nN \equal{} N$

$ M \equal{} Nn  \rightarrow  Nn \in  \bar{M}$ then $ Nn \equal{} N \rightarrow N \in \bar{M}$

$ N \subset \bar{M}$

$ G/N \equal{} \left\{Na|a \in G \right\}, G/N \equal{} Nn \equal{} N \rightarrow  N \subset G/N$

Also $ a \in M$ for all $ a \in G$
Also for all $ n \in N \rightarrow n \in M$ and $ N \subset M$

This shows N is contained in $ M, \bar{M}$ and $ G/N$ for all $ n \in N$

Be a little more careful with what you're writing.  $ N\subset \overline{M}$ is impossible, as this would imply $ N\subset G/N$, which doesn't make sense.  It might be that $ N\in \overline{M}$?

1. We wish to show $ M \leq G$.
(i) For all $ n\in N$, $ Nn \equal{} N$ is the identity element of $ G/N$ and since $ \overline{M} \leq G/N$, $ Nn \equal{}N$ must be the identity element of $ \overline{M}$. So we have $ Nn \in \overline{M} \implies n\in M \implies M \neq \emptyset$.

(ii) Choose $ x, y \in M$. Then $ Nx, Ny \in \overline{M}$.
Since $ \overline{M} \leq G/N$, $ Nxy^{\minus{}1} \equal{} (Nx)(Ny)^{\minus{}1} \in \overline{M} \implies xy^{\minus{}1} \in M$. And (i) and (ii) give $ M \leq G$.

2. We wish to show $ N \subset M$.
Choose $ n \in N$. Then $ n \in M \iff N \equal{} Nn \in \overline{M}$, which is true by (i). This proves $ N \subset M$.


Let $ a_{1},a_{2},...,a_{n}$ be reals. For each $ k \leq n$ let $ A_{k}$ be largest of the:
${ a_{k}, \frac{a_{k-1}+a_{k}}{2},...,\frac{a_{1}+a_{2}+...+a_{k}}{k}}$.
Show that, the smallest of: $ A_{1},A_{2},...,A_{n}$ is not larger than arithmetic mean of $ a_{1},a_{2},...,a_{n}$.

[quote="jasaczek"]Let $ a_{1},a_{2},...,a_{n}$ be reals. For each $ k \leq n$ let $ A_{k}$ be largest of the:
${ a_{k}, \frac {a_{k - 1} + a_{k}}{2},...,\frac {a_{1} + a_{2} + ... + a_{k}}{k}}$.
Show that, the smallest of: $ A_{1},A_{2},...,A_{n}$ is not larger than arithmetic mean of $ a_{1},a_{2},...,a_{n}$.[/quote]

 :rotfl: $ A_n = \frac{a_1+a_2+...+a_n}{n} \le \frac{a_1+a_2+...+a_n}{n}$

[quote="jasaczek"][quote="stephencheng"][quote="jasaczek"]Let $ a_{1},a_{2},...,a_{n}$ be reals. For each $ k \leq n$ let $ A_{k}$ be largest of the:
${ a_{k}, \frac {a_{k - 1} + a_{k}}{2},...,\frac {a_{1} + a_{2} + ... + a_{k}}{k}}$.
Show that, the smallest of: $ A_{1},A_{2},...,A_{n}$ is not larger than arithmetic mean of $ a_{1},a_{2},...,a_{n}$.[/quote]

 :rotfl: $ A_n = \frac {a_1 + a_2 + ... + a_n}{n} \le \frac {a_1 + a_2 + ... + a_n}{n}$[/quote]
 $ A_n = \frac {a_1 + a_2 + ... + a_n}{n}$?? how do you know that?[/quote]

Oh....Sorry. I misunderstood your question.

[b]Real Solution:[/b]

If $ A_n \le \frac {a_1 + a_2 + ... + a_n}{n} = A.M.$, then we are done.
If not, then there exists $ k_1$ so that $ \frac {a_{k_1 + 1} + a_{k_1 + 2} + ... + a_n}{n - k_1} > A.M.$

which leads to $ \frac {a_1 + a_2 + ... + a_{k_1}}{k_1} < A.M. .......(1)$

Now if $ A_{k_1} \le A.M.$, then we are done. If not , then we can find $ k_2 < k_1$ such that

$ \frac {a_{k_2 + 1} + a_{k_2 + 2} + ... + a_{k_1}}{k_1 - k_2} > A.M.$, which combines with $ (1)$ gives

$ \frac {a_1 + a_2 + ... + a_{k_2}}{k_2} < A.M. .......(2)$. The process can keep going infinitely many times if the statement is wrong.

But there is only finite number of positive integers less than $ n$ , contradiction. So the statement is true.


I am trying to install phpBB2 on my website, and I get the message:

The PHP configuration on your server doesn't support the database type that you chose

I installed PHP and MySQL 4. PHP Works, but I don't know how to fix this problem. Can someone help me?

If I read that correctly, it means that the server you are using doesnt support MySQL - some servers (webhosting) dont have PHP installed, so you can't use PHP on them. Some don't suport certain databases, in this case, MySQL. I take it that this is not on an Apache server, correct? If it is then I dont know what you could do about it, maybe reinstall Apache. If its on an outside server then I would either : 1. Switch servers, buy from someone else OR 2. Not use MySQL, but a different database that is supported. However, MySQL is one of the best databases, and most commonly used I believe. Also, if you still want that password gate or whatever you wanted, I'd need to use MySQL  :lol:

Of course, all of that is assuming that while installing phpBB2 that you chose at one point to use your configured MySQL, and then later on you provided the server on which you were going to install it, and it checked if all your configuration options work on it. I haven't installed phpBB2 yet, so I dont know, please confirm.


Prove that :

$ \underbrace{22...2}_{n}\plus{}\underbrace{(33...3)^2}_n\equal{}\underbrace{11...1}_{2n}\ ,\ (\forall)n\in \mathbb{N}$

[hide]$ 2\cdot{10^n \minus{} 1\over 9} \plus{} \left(3\cdot{10^n \minus{} 1\over 9}\right)^2 \equal{} {10^n \minus{} 1\over 9}\left(2 \plus{} 9\cdot{10^n \minus{} 1\over 9}\right) \equal{}{10^{2n} \minus{} 1\over 9}$[/hide]

May you please explain you solution? You only provided a bunch of numbers, without words explaining the process you used.

I'll describe it. 10^n-1 is 9999...9999 with n nines. dividing by 9 gives 1111...1111 and multiplying by 2 gives 222...222 with n 2s, or the number you are given. It's the same deal with the 333...333 and the 111111.......11111.

Squaring 3(10^n-1)/9 gives (10^2n-2*10^n+1)/9. Adding this to the previous 222..2222 expression gives
(10^2n-1)/9 or the answer.


To all the new AoPS Middle Schoolers!!!

when we answer questions, we must give our solutions!!!

and also, if you have any problems using the new forum, contact the administers or moderators

Hrm, if I was a moderator, I'd sticky this...

I think this applies for all of the math problems here :P .

[quote]Hrm, if I was a moderator, I'd sticky this...
[/quote]

Treething, you are a moderator, if you have noticed....

He's only a moderator of the MATHCOUNTS forum.  He cant sticky topics in any other forum besides that one.

[quote="shinwoo"]okay....

then who is the moderator for this Classroom Math forum?[/quote]For now there is only us, the Admins and rcv (for some forums). We'll name some other moderators in due time, have patience :)

I think there is an issue with reviving 10+ year-old threads over here

[quote="jack74"]I think there is a $\LaTeX$ issue over here[/quote]

Honestly, you need to leave. All you post are junk problems and spam you never contribute meaningful things to a thread or topic. Sorry but its the harsh truth.

Agree with the top two above.  Madeleine, you need to stop reviving.


superior algebra
superior algebra unsolved

Let $ 1\le i\le h$ be fixed integers and let $ F$ be the finite field with $ 2^h$ elements and let $ G\subset F$ be the set $ \{0,1,\alpha,\alpha \plus{} 1,\alpha^2,\ldots,\alpha^{i \minus{} 1} \plus{} \cdots \plus{} \alpha \plus{} 1\}$. Given $ F$-subsets $ (A_\lambda)_{\lambda \in G}$ and $ A_{\infty}$ we define sets $ B_\lambda \equal{} \{(x,y)\in F^2: x \plus{} \lambda y\in A_\lambda\}$ and $ B_\infty \equal{} \{(x,y)\in F^2: y\in A_\infty\}$.

Show that there exist such sets $ A_\lambda\subset F$ (for each $ \lambda\in G\cup\{\infty\}$), each of size $ 2^{h \minus{} i}$, such that any $ (x,y)\in F^2$ is contained in either $ 0$ or $ 2$ such $ B$-sets.

I have solved the two extremal cases: $ i \equal{} 1$ and $ i \equal{} h$: [list][*]For $ i \equal{} h$ we can take $ A_\lambda \equal{} \{\lambda^2\}$ and $ A_\infty \equal{} \{0\}$. [list][*]If $ y\neq 0$ then $ x \plus{} \lambda y \equal{} \lambda^2$ has either $ 0$ or two solutions $ \lambda_1,\lambda_2$, so $ (x,y)\in B_{\lambda_1}$ and $ (x,y)\in B_{\lambda_2}$ and in no other B-sets.[*]If $ y \equal{} 0$ then $ x \plus{} \lambda y \equal{} \lambda^2$ has $ 1$ solution $ \lambda$, then $ (x,y)\in B_\lambda$ and $ (x,y)\in B_\infty$ and in no other B-sets.[/list][*]For $ i \equal{} 1$ we can take $ A_0 \equal{} F\setminus G$, $ A_1 \equal{} G$, $ A_\infty \equal{} G$. Then $ (x,y)$ is... [list][*]in $ B_0$ and $ B_1$ if $ x\notin G,y\notin G$[*]in $ B_0$ and $ B_\infty$ if $ x\notin G,y\in G$[*]in $ B_1$ and $ B_\infty$ if $ x\in G,y\notin G$[*]in no B-set if $ x\in G,y\in G$[/list][/list]
I'm clueless about the rest though, so any help or suggestion would be very welcome!  :maybe:


first it is easy to see:
$(1-x_{n-1})(1-x_n) \leq (1-\sqrt{{x^2_n+x^2_{n-1}})}$
when $x_{n-1}+x_n \leq 1$
so if $n \geq 4$,the inequality can be reduces to $n-1$
so we should only need to prove $n=3$ which is
when $x^2_1+x^2_2+x^2_3=1$
$(1-x_1)(1-x_2)(1-x_3)\leq\frac{3}{2}-\sqrt{2}$
then somebody please solve it ;)

Thx for the ideea. I managed to solve it :)