Prove .9bar = 1

Solution (possibly incorrect) in spoiler below:

[hide]x = .9 bar

10x= 9.9bar

9x=9

x=1[/hide]

I know it's true, but my seventh grade head can't find a simple, concise, and beautiful proof...

Let $0.99999999\dotsc$ = $\displaystyle \sum^{\infty}_{i=1} \frac {9}{10^{i}}$.  This is a geometric series with sum $\frac {\frac{9}{10}}{1-\frac{1}{10}}$.  The latter fraction clearly equals $1$.

Let $n=.\overline{9}$. Multiplying by ten, we get $10n=9.\overline{9}$. Subtract $n$ from $10n$, and we find that $9n=9$, and $n=1$.

Edit: sorry, didn't see that this is what you had in spoiler.

Or just try subtracting $.\overline{9}$ from $1$

Although I doubt that counts as a proof...

Sam3.14's solution was what I was taught first.

I really don't like dealing with things like $\ 0.\overline {a_1a_2\dotsc a_k}$, because they aren't really 'well-defined'.  Geometric series, on the other hand, are.  While the proofs you guys have presented are quite ingenious, rigor is also important.

For example, it's tempting to consider the complex number $i$ negative, because $i^2=-1$, but $i$ is neither positive nor negative.

Like ALL  rational fraction, this is a case of repeating decimals..  and you have  two forms for 1/1

either 1.0000000000000..............................

or   0.99999999999999999..................................

Take your pick, either repeat  0's or 9's. (some thimes,  lazy people do make rules like, you don't have to actually write those zeros to  "repeat" but they do repeat., if you really want to ....)

Nothing to get exitied about.  :)

I don't know whether there was a post mentioning this site but if you're still looking for these issuek, check out this page:

http://mathforum.org/dr.math/faq/faq.0.9999.html

Great page to look on. Hope this helps!

[quote="LynnelleYe"]The solution I was taught was...
$\frac{1}{3}=0.\bar{3}$
Multiply both sides by 3.
$3\times\frac{1}{3}=3\times 0.\bar{3}$
$1=0.\bar{9}$[/quote]

You can't just assume that $\frac{1}{3}=0.\bar{3}$.

Well that's EPGY's fault then. Ask them.

Has anyone created a nonstandard, consistent number system where 0.999... is (infinitesimally) less than 1?  In that case, I guess 0.333... would be infinitesimally less than 1/3 too, and 0.111... would be just less than 1/9.

[quote="gameworld7"]well technically no its not equal to 1 BUT it could be 1 IF it was 0.9 with a bar.  its so close to 1. so close, yet so far.[/quote]
lim
k --> 1=.9999...?
thats what jueyan told me at the harker contest
but i dont like calculusy limits
and
1-.999999...
=0.000000...1
since the zeroes are infinite,
1-.9999...=0
so .99999999...=1

[quote="3X.lich"][quote="towbomb"]On the subject of infintely small differences, I've always said that the smallest possible number is 0.0(bar)1.  I don't know how you can prove that it is equal to zero, but it's still kind of fun.[/quote]

0.0(bar)1 is not well-defined. You can see the infinity? After an 'infinite' string of zeroes the number ENDS with a digit 1? Definitely so wrong and ill-defined. People, stop making things up, it's making things more trouble and confusion than help.

[quote="gameworld7"]well technically no its not equal to 1 BUT it could be 1 IF it was 0.9 with a bar. its so close to 1. so close, yet so far.[/quote]

Simply wrong. yet it is not far. The only thing that sets 0.9bar and 1 apart is the existence of some hypothetical infinitesimal quantity. Besides, no one can actually see the infinity, it is an imaginative notion.

I'm not bashing anyone, just wanted to break midschoolers' misconceptions about numbers/math.[/quote]

I think gameworld was referring to the subject of the topic - since the second sentence there is that 'but it could be 1 if it was 0.9 with a bar' ;)

But yeah, I agree - I think people don't realise exactly what the bar means at times.

The reason why

.9999... can be "proved" to equal 1

is that our manner of dealing with infinity is ever so slightly inaccurate.  This inaccuracy is so infinitesimally small that it can be disregarded for most practical intents and purposes.  But when manipulated in a certain manner, this inaccuracy can be magnified to become more clear--specifically, in the form of an infinitely small difference between two commonly recognizable numbers.

[quote="TripleM"]0.0(bar)1 is not well-defined. You can see the infinity? After an 'infinite' string of zeroes the number ENDS with a digit 1?[/quote]
Yeah, I guess I hadn't exactly thought that over too well.  It's still kind of fun to think about, though.

[quote="3X.lich"]I'm not bashing anyone, just wanted to break midschoolers' misconceptions about numbers/math.[/quote]
I am a senior in high school, thank you very much.

[quote="3X.lich"]People, stop making things up, it's making things more trouble and confusion than help.[/quote]
I don't think making things up is so bad.  You never know when something "made up" is the start of a groundbreaking theory, although I'll admit what I "made up" was probably a long shot for doing so.

There are other problems - no closure under multiplication, etc.

Mind you, if we let 0. (n zeros) 1 generate a set of numbers using only addition, it is necessarily isomorphic to the naturals.

oh come on guys. the only post here which had a real sense was that one of blahblahblah(sorry for saying so). the problem is that most of you don't know what the number $0.a_1a_2a_3...$ is meaning; you ever saw a definition for it? it is just defined as the limit $\sum_{n=0}^{\infty} \frac{a_n}{10^n}$. and now, how do you want to prove anything about the number 0.999... without looking the definition? with the definition, it is obviously clear that 0.999...=1 as blahblahblah proved.

however, for those who don't like them to be equal: besides the standard analysis, there is the so-called "non-standard analysis"; in that theory there exist infinitesimally small numbers - we actually have $0.999...\neq 1$ there. the definition for the numbers used there is something similar to the cauchy sequences modulo series which are almost everywhere 0(well not really; the condition is that they are 0 in a set which is in an "ultrafilter" ( http://mathworld.wolfram.com/Ultrafilter.html ) of the natural numbers); not cauchy sequences modulo series which converge to 0, the last one being the normal real numbers.
however i don't know much about that theory and it has, as far as i know, no applications in other areas of mathematics.

Non-standard analysis is indeed a way of dealing with this sort of thing. But even in NSA, 0.99999... is arguably still equal to 1, though [0.9,0.99,0.999,0.9999,...] is certainly only infintesimally close to 1 (they differ by the infintesimal [0.1,0.01,0.001,0.0001,...]).

Oh, and to correct a minor technical detail of Peter Scholze's post: the definition of a non-standard real is a (not necessarily Cauchy) sequence of reals modulo an ultrafilter. This lets you also make various "unlimited" numbers, which are needed so that infintesimals have multiplicative inverses and you have a field.

But anyway, it's important to remember that non-standard real numbers ARE NOT REAL NUMBERS. The real number 0.99999... is undeniably equal to the real number 1, as has been repeatedly demonstrated in this thread.

[quote="towbomb"][quote="3X.lich"]I'm not bashing anyone, just wanted to break midschoolers' misconceptions about numbers/math.[/quote]
I am a senior in high school, thank you very much.[/quote]

I don't think he was referring to you specifically, more so to the general thoughts that people in middle school thought at that time, or were taught.

Hey everyone check out this explanation to the original question. I hope it helps.

http://mathforum.org/dr.math/faq/faq.0.9999.html

This is known as [b]Zenos Paradox of the Tortoise and Achilles[/b]. Some references:

- http://www.mathacademy.com/pr/prime/articles/zeno_tort/index.asp
- http://mathworld.wolfram.com/ZenosParadoxes.html

the real numbers is defined by the completeness axiom, and the reals being dense implies that there is no smallest positive real number.

Some things should be taken in face value (such as .999... =1)
Anyone's insistance on that being false will only confuse yourself.

I remember having an argument with someone who says sqrt(9) is always 3 and -3...

bugzpodder's avatar contains the answer to this dilemma  :D  :D
one of the coolest avatar's ever seen.

orl: didn't you have a German flag? now you are UK?  :?  :lol:

[quote="Wumbate"][quote="towbomb"]On the subject of infintely small differences, I've always said that the smallest possible number is 0.0(bar)1.  I don't know how you can prove that it is equal to zero, but it's still kind of fun.[/quote]

I agree.  I've thought about it before but it isn't the [i]smallest possible[/i] number.  Consider 0.0(bar)0(bar)1, which I guess you could argue is 0.00(bar)1 or 0.0(bar)1.  Or consider 0.0(bar)01.  Or consider whatever (0.0(bar)1)/2 is.

Interesting, to say the least.[/quote]

You realize that that's (a priori) meaningless, right? The "bar" notation denotes an infinite geometric sum. What index set are you summing over if not the integers?

This is a common topic... [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25371&postorder=asc&start=1]www.artofproblemsolving.com/Forum/viewtopic.php?t=25371&postorder=asc&start=1[/url]


I don’t remember if I posted this problem before; if I did, please tell me or lock the topic…

Let be $a$ and $b$ two differents real numbers such that $2a^2+2b^2=5ab$. Find all possible values of $\frac{a+b}{a-b}$.

Posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=431092#p431092[/url] (post #62) and solved in the next post (#63)


Hey guys,

even if that topic has not so much to do with the IMO 2008, i decided to ask here.

Does anyone in this forum participate at the International Physics Olympiad 2008 in Ha Noi? And will maybe someone go to IMO and IPhO? Because i qualified for the German Team for the IPhO a few days ago...

Naphthalin

I wish to do so... But i'm going to be sure after the TST next week :P

do you know now?

as you can see in the German IMO Team 2008 thread, there are two german members who leave on 19th of july to ha noi...

Naphthalin

Hey, i'm a member of the Moldovan team for the IPhO 2008 , if you want , my yahoo id is: victorbuza
You can add me and we can talk, also, other people who are going to IPhO this year, i'd love to chat with you too, so don't hesitate to add me .

--
Regards,
Victor

What about the results for this year? Does anyone has them?

Yay... They're out on the site:

[url]http://ipho2008.hnue.edu.vn/Competition/Results/tabid/82/Default.aspx[/url]

Congratulations to all medalists and competitors.

Please, Andrada Ianus from Romania got the silver medal or the gold one? Thank you.


In a plane are written all naturals between $1$ and $2006$.
Operator "DW" - We pick any two numbers in the plane, delete them and write instead of them their sum modulo $13$.
After some actions with operator "DW" , now in the plane there are only 2 numbers.
One of them is $100$.
Find the other one.

Wrong subforum (solved problems)!
Very similar question was in the german BWM some time ago...
Just note that the sum of all numbers $\mod 13$ remains the same all the time and that the last number is $<13$.


Prove that exist real number $ \alpha$ such, that $ \frac{1}{3}\leq \left\{\alpha ^ n\right\}\leq \frac{2}{3}$ for all natural $ n.$

P.S. $ 0\leq\left\{x\right\}\equal{}x\minus{}\left[x\right]<1$

The proof is evident. It's enough to construct a sequence of segments $ [a_1;b_1]\supset[a_2;b_2]\supset[a_3;b_3]\supset\ldots$ where $ a_n\equal{}(c_n\plus{}1/3)^{1/n}$, $ b_n\equal{}(c_n\plus{}2/3)^{1/n}$ with some $ c_n\in\mathbb N$. Then one can take $ \alpha\in\bigcap_{n\equal{}1}^\infty[a_n;b_n]$. The sequence is constructed inductively. Take $ c_1\equal{}4$. Suppose that we've already constructed $ c_1,\ldots,c_n$. Since
\[ (b_n^{n\plus{}1}\minus{}2/3)\minus{}(a_n^{n\plus{}1}\minus{}1/3)>a_n(b_n^n\minus{}a_n^n)\minus{}1/3\equal{}\frac{a_n\minus{}1}3\geqslant\frac{a_1\minus{}1}3>1\]
there's an integer $ c_{n\plus{}1}$ such that $ (c_{n\plus{}1}\plus{}1/3)^{\frac1{n\plus{}1}}\geqslant a_{n}$ and $ (c_{n\plus{}1}\plus{}2/3)^{\frac1{n\plus{}1}}\leqslant b_{n}$.

Is it possible to narrow the boundaries to :
$ \frac{1}{2}\minus{}\epsilon$; $ \frac{1}{2}\plus{}\epsilon$ ?

Example was for $ \epsilon\equal{}\frac{1}{6}$

[quote="mszew"]Is it possible to narrow the boundaries to :
$ \frac {1}{2} \minus{} \epsilon$; $ \frac {1}{2} \plus{} \epsilon$ ?
[/quote]
Of course. The proof is essentially the same. It's even possible to construct in this way $ \alpha\in\mathbb R$ such that for $ n\in\mathbb N$ one has $ \beta_n\leqslant\{\alpha^n\}\leqslant\gamma_n$, for any sequence of subsegments $ [\beta_n;\gamma_n]\subset[0;1)$ satisfying $ \inf\limits_{n\in\mathbb N}(\gamma_n\minus{}\beta_n)>0$. In fact a little modification of the proof admits to construct continuum of such $ \alpha$.


From point $P$ outside of circle O, tangent segments $PA$ and $PB$ are drawn. If $\frac{1}{(OA)^2}+\frac{1}{(PA)^2}=\frac{1}{16}$, compute the length of chord $AB$.

[hide]by pythagorean theorem, $PA^2+AO^2=PO^2$ (1)
ptolemy: $2(AO)(PA)=PO*AB\implies 2*\frac{AO*PA}{PO}=AB$  (2)
(right angles add up to 180 and tangent segments are congruent)
multiply the given by $AO^2*PA^2$ and use (1)
$PA^2+AO^2=\left(\frac{AO*PA}{4}\right)^2=PO^2 \implies \frac{AO*PA}{PO}=4$ (3)
then (3) into (1) implies $\boxed{AB=8}$[/hide]

Mine's wierd.
[hide]
Let $AB=z$,$AP=x$, and $AO=y$.
By Ptolemy's theorem: $\2xy=z\sqrt{x^2+y^2}\Rightarrow \frac{1}{z^2}=\frac{x^2+y^2}{4x^2y^2}$.
We have: $\frac{1}{z^2}=\frac{1}{4}(\frac{1}{x^2}+\frac{1}{y^2})\Rightarrow \frac{1}{z^2}=\frac{1}{64}$. Therefore $z=AB=\boxed{8}$
[/hide]


vector
linear algebra
linear algebra unsolved

This has stuttered me. Find the angle between the plane x + y = 3 and the line of intersection of the planes x -z = -5 and x - y + z = 1

Any help would be appreciated

Find the acute angle $\phi$ between the line and a normal vector to the plane. Then the angle between the line and the plane will be $\frac{\pi}{2} - \phi$


geometry
3D geometry
sphere
tetrahedron
analytic geometry
Pythagorean Theorem
angle bisector

A sphere is inscribed in a tetrahedron with side length eight. What is the surface area of the sphere?

=[hide]
With simple geometry (30-60-90 triangle rules), we can figure out that the height of one of the sides of the tetrahedron is 4 :sqrt: 3.  That means the apothem is  :sqrt: 3, and the segment connecting the center and a vertex is 3 :sqrt: 3. Using the Pythagorean Theorem: (3 :sqrt: 3 ) 2 + h 2 = 8 2 . h= :sqrt: 37 . Again, the apothem is ( :sqrt: 37)/4, making the segment connecting the center and the vertex (also the radius of the sphere) (3 :sqrt: 37)/4. So now you have the radius of the sphere.

Surface area of a sphere is 4(pi)r 2 .
That yields: (333/4)pi, or 83.25pi
[/hide]

I [i]think[/i] this is correct.

I had a solution there at first but then I realized I was wrong, so I put "nevermind" in there while I was fixing it. :P

[hide]The sphere will be tangent to the tetrahedron at the point where each of the angle bisectors of each of the faces of the tetrahedron meet (four points total). Sorry, I don't remember whether these points are orthocenters, centroids, or something else. :P

This next part is probably easiest to do on a coordinate plain. Put one vertex of the tetrahedron at $(0,0,0).$ To find where the other vertices go, do the following:

$8$ is the long side of a 30-60-90 triange in x-y-z space, and $a$ is the side that is in the x-y place, so $8=2a,$ and $a=4$
$a$ is the long side of a 30-60-90 triangle on the x-y plane.
$x=\frac{a}{2}=2$
$y=\frac{a\sqrt{3}}{2}=2\sqrt{3}$

So, the vertice that is not on the x-y plane is at $(2,2\sqrt{3},4\sqrt{3})$

The other two vertices are simply those of an equilateral triangle with one point at the origin and a second on the x-axis; namely, $(8,0,0)$ and $(4,4\sqrt{3},0).$

Now, since the sphere touches the tetrahedron at the center point of each of these triangles, we need to find the coordinates of the center points. The easiest will be the triangle that is entirely in the x-y plane. By simply drawing the angle bisectors and then the perpendicular bisectors, we divide the triangle into 6 30-60-90 triangles, and we can see that the center is at $(4,\frac{4}{\sqrt{3}},0).$

Next, we find the coordinates of the center point of one other triangle. For convenience, we'll use the one with two points on the x-axis, and the final point being the one in x-y-z space. The x-coordinate will be the same as the previous center point, so we need to simply divide the y-coordinate into y and z components.
$y=\frac{4}{\sqrt{3}}cos60=\frac{2}{\sqrt{3}}$
$z=\frac{4}{\sqrt{3}}sin60=2$

So, the coordinates of the second center point are $(4,\frac{2}{\sqrt{3}},2).$

We can discard the x-coordinates, since they are the same. Now, we draw a line between these two given points, and the length will be $(y_1-y_2)^2+(z_2-z_1)^2=l^2$
$\displaystyle\left(\frac{4}{\sqrt{3}}-\frac{2}{\sqrt{3}}\right)^2+(2-0)^2=l^2$
$\frac{4}{3}+4=\frac{16}{3}=l^2$
$l=\frac{4}{\sqrt{3}}$

The two centers also cover an arc on the circle of 120 degrees, since the angle on the tetrahedron is 60 degrees. So, we see that there $l$ is a chord of the sphere, and we make a 120-30-30 isosceles triangle, with $l$ as the long side. Make a perpendicular angle bisector to divide it into two 30-60-90 triangles, each with a medium side of $\frac{2}{\sqrt{3}}.$ The radius of the circle will be the long side of the triangles, which is $\frac{4}{3}.$

Now, to find the surface area, plug the radius into the formula for the surface area of a sphere.
$4\pi\displaystyle\left(\frac{4}{3}\right)^2=\frac{64\pi}{3}$[/hide]

I hope I did that right... that was pretty complicated.


geometry
3D geometry
tetrahedron
vector
geometry proposed

Let $OX,$ $OY$ and $OZ$ be three rays in the space, and $G $ a point ''between these rays'' (i. e. in the interior of the part of the space bordered by the angles $YOZ,$ $ZOX$ and $XOY$). Consider a plane passing through $G$ and meeting the rays $OX,$ $OY$ and $OZ$ in the points $ A,$ $B,$ $C,$ respectively. There are infinitely many such planes; construct the one which minimizes the volume of the tetrahedron $OABC.$

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

I think this happens when we choose the plane s.t. $G$ is the cantroid of $ABC$:

Let $\vec i,\vec j,\vec k$ be the unit vectors directed along $(OX,(OY,(OZ$ . There exist unique positive $x_i,x_j,x_k$ s.t. $\vec v=\overrightarrow{OG}=x_i\cdot\vec i+x_j\cdot\vec j+x_k\cdot\vec k$. The vectors $\overrightarrow{OA},\overrightarrow{OB},\overrightarrow {OC}$ are $\alpha\cdot\vec i,\beta\cdot\vec j,\gamma\cdot \vec k$ for some positive $\alpha,\beta,\gamma$ satisfying the condition $\frac{x_i}{\alpha}+\frac{x_j}{\beta}+\frac{x_k}{\gamma}=1\ (*)$. The volume we want minimized is $\alpha\cdot\beta\cdot\gamma\cdot|\vec i\cdot(\vec j\times\vec k)|$, which is minimized when $\alpha\cdot\beta\cdot\gamma$ is minimal. From $(*)$ we see that this happens only when $\frac{x_i}{\alpha}=\frac{x_j}{\beta}=\frac{x_k}{\gamma}=\frac 13$ or, in other words, $\vec v=\frac{\alpha\cdot \vec i+\beta\cdot \vec j+\gamma\cdot\vec k}3$, which is equivalent to what we want to prove.


If $x$ is a positive rational number, show that $x$ can be uniquely expressed in the form \[x=a_{1}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where $a_{1}a_{2},\cdots$ are integers, $0 \le a_{n}\le n-1$ for $n>1$, and the series terminates. Show also that $x$ can be expressed as the sum of reciprocals of different integers, each of which is greater than $10^{6}$.

Here is part 1: http://www.mathlinks.ro/viewtopic.php?search_id=1412560634&t=21118

For part 2 we can use an alike arguement as here: http://www.math.leidenuniv.nl/~naw/home/ps/pdf/uitwerking-2005-4.pdf

Part 2 is also closely related to the Egyptian fractions topic, with an extensive array of methods and results.


Ring Theory
superior algebra
superior algebra unsolved

Can we prove?
Let $ R$ be a commutative ring with unit.
Prove that $ R$ is a Noether Ring if and only if every prime ideal of $ R$ is finite generated.

And find a counterexample if we weaken the condition as every maximal ideal of $ R$ is finite generated.

True - since maximal elements in the set of non-finitely generated ideals are prime ideals.


Suppose there is a $ 6$-digit integer $ \overline{ABCDEF}$ . If the left-most digit moves to the right-most place ,i.e. $ \overline{BCDEFA}$,then the new one is triple of the original number.
What is $ \overline{ABCDEF}$?

Let $ x\equal{}\overline{BCDEF}$. Then we have $ 3\cdot10^5\cdot A\plus{}3x\equal{}10x\plus{}A\implies(3\cdot10^5\minus{}1)A\equal{}299999A\equal{}7x\implies x\equal{}42857A$. Since $ 1\le A\le9$ and $ 10000\le x\le99999$, $ A\equal{}\{1,2\}\implies x\equal{}\{42857,85714\}\implies\overline{ABCDEF}\equal{}\boxed{\{142857,285714\}}$.

Ahh, I remember this problem. Though I forget the correct way to solve it.(Which already showed by others), but I remember:
1/7=0.142857repeat
2-7=0.285714repeat
3/7=0.428571repeat
4/7=0.571428repeat
5/7=0.714285repeat
6/7=0.857142repeat
So, if there is a problem like this, then it most likely has something to do with this.

A similar problem is to try it with 5-digit integers.

[hide]
In this case, we have exactly the same setup: $ 3(10000A \plus{} x) \equal{} 10x \plus{} A\implies 30000A \plus{} 3x \equal{} 10x \plus{} A \implies 7x \equal{} 29999A$.

It becomes apparent that $ 29999$ is not divisible by 7, and $ A$ clearly can't be 7 because then 3 times $ ABCDE$ would be too large, so therefore the only possible value of $ x$ that makes $ A$ an integer is zero, but then $ A$ would also be zero, which causes the number to be $ 0$, not a 5-digit integer.
[/hide]

[quote="dgreenb801"]Here is an awesome way from a similar USAMTS solution!
Let 0.abcdefabcdefabcdef......=K be an infinite repeating decimal.
Then 10K-a=0.bcdefabcdefabcdef.....=3K
So $ K \equal{} \frac {a}{7}$, we find only k=1,2 works. :D
This explains what tomwolf wrote.[/quote]

Good idea!! :P


function
limit
calculus
calculus computations

I've found in a book this statement:
$ f: \mathbb{R} \to \mathbb{R}$ is a bijective function and for every convergent sequence $ x_n \Rightarrow f(x_n)$ is also convergent.

They say that from here is very easy to prove that if $ \lim_{n\to \infty}x_n\equal{} l \Rightarrow \lim_{n \to \infty} f(x_n)\equal{}f(l)$, so the function is continuous.

I can't understand why they should have the same limit. Thanks

Just consider the sequence $ x_1,\ell,x_2,\ell,x_3,\ell,\dots$ that also converges to $ \ell$ and apply $ f$ to it. The only chance to get a convergent sequence again is to have it convergent to $ f(\ell)$.


function
algebra
domain
calculus
calculus computations

Does anyone know any examples of functions obeying
$ f(T \minus{} x) \equal{} 1 \minus{} f(x)$ for some $ T\in \mathbf{R}$.

Please give as many examples as possible!

Assume $ T \neq 0$.  For any such function, define a function $ g(x)$ by $ g(x) \plus{} \frac {x}{T} \equal{} f(x)$.  Then we have $ g(T \minus{} x) \plus{} \frac {T \minus{} x}{T} \equal{} 1 \minus{} g(x) \minus{} \frac {x}{T}$, so $ g(T \minus{} x) \equal{} \minus{} g(x)$.   Given [i]any[/i] function $ h$ with domain $ \left[\frac {T}{2}, \infty\right)$ such that $ h(T/2) \equal{} 0$, we can extend $ h$ to a function $ g$ of the desired sort, and thus get a function $ f$ out as well.  (Actually, that was unnecessary: it's clear we can define $ f$ arbitrarily on $ \left(\frac {T}{2}, \infty\right)$, and then extend it uniquely to the whole real line.)


[color=black]Να λυθει στους φυσικους

$ a^3\minus{}b^3\plus{}ab\equal{}a^2\plus{}b^2\plus{}4$[/color]

Βρηκα μια λυση αλλα θελει αρκετο γραψημο, ισος να υπαρχει πιο συντομη λυση με καποιο θεωρημα που δεν γνωριζω. Δειτε την μια και πειτε μου αν εχω πουθενα κανενα λαθος

[hide]
αν $ b \equal{} 0$ τοτε η εξισωση γινεται: $ a^3 \minus{} a^2 \equal{} 4 \leftrightarrow a^2(a \minus{} 1) \equal{} 4$, και επηδη ο αριθμος $ a^2$ ειναι τελειο τετραγωνο ακεραιου, για να ειναι διαιρετης του 4 πρεπει να ειναι ισος με 1 η 4, για $ a^2 \equal{} 1$ καταληγουμε σε ατοπο, ενω για $ a^2 \equal{} 4$ η σχεση επαληθευεται και λαμβανουμε a=2, επομενος μια λυση της εξισωσης ειναι η $ (a, b) \equal{} (2, 0)$

αν $ b \equal{} 1$ τοτε η εξισωση γραφεται: $ a^3 \minus{} 1 \plus{} a \equal{} a^2 \plus{} 5 \leftrightarrow a(a^2 \minus{} a \plus{} 1) \equal{} 6$, οι φυσικοι διαιρετες του 6 ειναι οι: 1, 2, 3, 6, δοκιμαζοντας για $ a \equal{} 1,2,3,6$ βρησκουμε οτι μονο για $ a \equal{} 2$ επαληθευεται η εξισωση, αρα αλλη μια λυση της εξισωσης ειναι η $ (a, b) \equal{} (2, 1)$

Θα δειξουμε τωρα οτι για $ b \geq 2$ η εξισωση ειναι αδυνατη. Η εξισωση γραφεται $ a^3 \minus{} b^3 \plus{} 3ab \equal{} a^2 \plus{} b^2 \plus{} 2ab \plus{} 4 \leftrightarrow a^3 \minus{} b^3 \plus{} 3ab \plus{} 1 \equal{} a^2 \plus{} b^2 \plus{} 2ab \plus{} 5 \leftrightarrow a^3 \minus{} b^3 \minus{} 3a( \minus{} b)1 \plus{} 1 \equal{} a^2 \plus{} b^2 \plus{} 2ab \plus{} 5 \leftrightarrow \frac {1}{2}(a \minus{} b \plus{} 1)[(a \minus{} 1)^2 \plus{} (b \plus{} 1)^2 \plus{} (a \plus{} b)^2] \equal{} (a \plus{} b)^2 \plus{} 5 \leftrightarrow (a \minus{} b \plus{} 1)[(a \minus{} 1)^2 \plus{} (b \plus{} 1)^2 \plus{} (a \plus{} b)^2] \equal{} 2(a \plus{} b)^2 \plus{} 10 \leftrightarrow (a \minus{} b \plus{} 1)(a \minus{} 1)^2 \plus{} (a \minus{} b \plus{} 1)(b \plus{} 1)^2 \plus{} (a \minus{} b \plus{} 1)(a \plus{} b)^2 \minus{} 2(a \plus{} b)^2 \equal{} 10 \leftrightarrow (a \minus{} b \plus{} 1)(a \minus{} 1)^2 \plus{} (a \minus{} b \plus{} 1)(b \plus{} 1)^2 \plus{} (a \minus{} b \minus{} 1)(a \plus{} b)^2 \equal{} 10$
Θετουμε $ a \minus{} b \plus{} 1 \equal{} u$, $ u$: ακεραιος, και η εξισωση γραφεται:
$ u(a \minus{} 1)^2 \plus{} u(b \plus{} 1)^2 \plus{} (u \minus{} 2)(a \plus{} b)^2 \equal{} 10$
Αν $ u > 1$, τοτε επηδη δουλευουμε για $ b \geq 2$, το πρωτο μελος γινεται μεγαλυτερο απο 18 (ατοπο)
Αν $ u < 1$ τοτε προφανος το πρωτο μελος γινεται αρνητικος αριθμος (ατοπο)
Αρα πρεπει $ u \equal{} 1 \leftrightarrow a \minus{} b \plus{} 1 \equal{} 1 \leftrightarrow a \minus{} b \equal{} 0 \leftrightarrow a \equal{} b$, και τοτε η αρχικη εξισωση γινεται $ a^2 \equal{} 2a^2 \plus{} 4 \leftrightarrow \minus{} a^2 \equal{} 4$ η οποια ειναι αδυνατη

Αρα οι μονες λυσεις ειναι: $ (a, b) \equal{} (2, 1)$, και $ (a, b) \equal{} (2, 0)$.[/hide]

Ωραια λυση,πολυ διαφορετικη απο τη δικη μου.
Μπορεις να αποδειξεις απο την αρχη οτι $ a>b$
Πως? :)

εχεις
[hide]$ a^3\minus{}b^3\plus{}ab\equal{}a^2\plus{}b^2\plus{}4>0<\equal{}>
a^3\minus{}b^3 \minus{}ab\equal{}a^2\plus{}b^2\plus{}4\minus{}2ab\equal{}(a\minus{}b)^2\plus{}4>0
<\equal{}>
(a\minus{}b)(a^2\plus{}b^2\plus{}ab)\minus{}ab>0$
ή
$ (a\minus{}b)(a^2\plus{}b^2\plus{}ab)>ab\geq0$=>$ a>b$[/hide]

Ωραια. :)
Αλλιως
$ 2(a^3\minus{}b^3)\equal{}a^2\plus{}b^2\plus{}(a\minus{}b)^2\plus{}8>0$
δηλαδη $ a^3>b^3$
δηλαδη $ a>b$

ΓΙΑ ΣΑΣ! :wink: Set $ a \plus{} b \equal{} x$, $ a \minus{} b \equal{} y$ then u have:$ a^3 \minus{} b^3 \plus{} ab \equal{} a^2 \plus{} b^2 \plus{} 4 \equal{} >$ .......$ y(3x^2 \plus{} y^2) \equal{} x^2 \plus{} 3y^2 \plus{} 16 \equal{} > x^2(3y \minus{} 1) \plus{} y^2(y \minus{} 3) \equal{} 16$ If $ y > 3$ the equation have no solutions the same if $ y < 1$ so search for $ y \equal{} 1,y \equal{} 2$ and $ y \equal{} 3$....

Πολυ ωραια!
Στην λυση μου θετω
$ a\minus{}b\equal{}x>0$
$ ab\equal{}y$
και μετα απο πραξεις προκυπτει
$ x^3\minus{}x^2\minus{}4\equal{}y(1\minus{}3x)$
αρα ο $ 1\minus{}3x$ διαιρει τον $ x^3\minus{}x^2\minus{}4$ και ετσι βρισκουμε τον x αρα και τον y
οποτε και τους a,b :)


Anyone have experience with Worcester Polytechnic Institute? What about things you've heard from others?

They have an amazing actuarial and pure mathematics program.  I'm applying this year and it's my first choice.  I visited a few months ago and the professors all seem really good.  I also like the project oriented program.


[quote="Rodolphe2005"]Using convexity of $x\rightarrow x^{\frac{n-1}{n}}$ we have:
\[a_1^{n-1}+...+a_n^{n-1}\geq n \Bigg( \frac{a_1^n+...+a_n^n}{n} \Bigg)^{\frac{n-1}{n}}\]
We need to prove
\[\frac{a_1^n}{a_2}+...+\frac{a_n^n}{a_1}\geq a_1^{n-1}+...+a_n^{n-1} \Leftrightarrow a_1^{n-1}\Big(\frac{a_1}{a_2}-1\Big)+...+a_n^{n-1}\Big(\frac{a_n}{a_1}-1\Big)\geq 0\]
I don't succeed to continue ...[/quote].

rearrengement with ${a_i^n}$ and ${\frac{1}{a_i}}$.. it's clear that the worst ordering is $a_1^n\frac{1}{a_1}+\dots +a_n^n\frac{1}{a_n} = \sum {a_i^{n-1}}$ and the claim follows...

[quote="Rodolphe2005"]Using convexity of $x\rightarrow x^{\frac{n-1}{n}}$ we have:
\[a_1^{n-1}+...+a_n^{n-1}\geq n \Bigg( \frac{a_1^n+...+a_n^n}{n} \Bigg)^{\frac{n-1}{n}}\]
We need to prove
\[\frac{a_1^n}{a_2}+...+\frac{a_n^n}{a_1}\geq a_1^{n-1}+...+a_n^{n-1} \Leftrightarrow a_1^{n-1}\Big(\frac{a_1}{a_2}-1\Big)+...+a_n^{n-1}\Big(\frac{a_n}{a_1}-1\Big)\geq 0\]
I don't succeed to continue ...[/quote]

hi, i wonder why your claim is true.
By the famous power mean inequality you have the reverse sign of your inequality.

Thanks alot ,Manlio and Vasc!
I'm realize my solution is wrong in general case. :blush:

I have proof in case n=3
And for $n \ge 4$
$\sum\frac{a_{i}^n}{a_{i+1}} \ge n\sqrt[n]{\frac{4}{n}}(\frac{a_1^n+a_2^n+...+a_n^n}{n})^{\frac{n-1}{n}}$

And
$\sum\frac{a_{i}^k}{a_{i+1}} \ge n\sqrt[k]{\frac{4}{n}}(\frac{a_1^k+a_2^k+...+a_n^k}{n})^{\frac{k-1}{k}}$
still satisfy for all k integer.

Great problem this last one! Taking $ a_1^k+...+a_n^k=1$ we need to prove that $\sum {\frac{a_i^k}{a_{i+1}}\geq 4^{\frac{1}{k}}}$. Using Cauchy it suffices to prove that $\sum{x_i\cdot\sqrt[k]{4x_{i+1}}\leq 1}$ if $\sum {x_i}= 1$. This is true since $\sqrt[k]{4x_{i+1}}\leq\frac{4x_{i+1}+k-1}{k}$ and $ 4\sum {x_i\cdot\ x_{i+1}}\leq 1$.

Very clearly Solution,Harazi.
 :)  :)  :)  :)  :)


Recently, there has been a lot of spam in this forum.  Several topics have been locked and any further spam will be deleted.  The goal of this forum is not as a place to spam, but a place to talk about math in North Carolina.

[quote="A Good Thing To Read"]Netiquette (neologism, a portmanteau formed from "Internet etiquette") is a catch-all term for the conventions of politeness recognised on Usenet, in mailing lists, and on other electronic forums such as internet message boards. These conventions address group phenomena (such as flaming) with changes in personal behaviour, such as not posting in all uppercase, not (cross-)posting to inappropriate groups, refraining from commercial advertising outside the biz groups and not top-posting. RFC 1855 is a fairly lengthy and comprehensive set of such conventions.

The rules of netiquette are slightly different for newsgroups, web forums and IRC (Internet Relay Chat). For example, on Usenet it is conventional to write in standard English and not use abbreviations such as "u" for "you" or "ne1" for "anyone". These abbreviations are only slightly more likely to be tolerated on web forums, but are almost universal on IRC where, since discussion is real-time, they serve the practical purpose of speeding the flow of conversation. Many IRC users look down on this form of conversation, though. Issues such as the level of tolerance for off-topic discussion or spoilers may also vary from one newsgroup, forum, or channel to another. The rule of thumb in any of these discussion mediums is to "lurk before you leap"—get a feel for the local conventions before diving into conversation and inadvertently embarrassing oneself. Also, read the FAQ if there is one.

The following is a list of some of the more common rules associated with netiquette on Usenet.

* The most important rule of netiquette is, "Think before you post". If what you intend to post will not make a positive contribution to the newsgroup and be of interest to several readers, do not post it! Personal messages to one or two individuals should not be posted to newsgroups—use private e-mail instead.
o This is sometimes stated in other forms, such as "Remember the human."

* When following up on an article, quote the minimum text necessary to give some context to your reply and be careful to attribute the quote to the correct person. If the article to which you are responding was cross-posted to several groups, edit the distribution ("Newsgroups:") header to contain only those groups which are appropriate to your reply, especially if the original message was posted to one or more inappropriate groups in the first place.

* Re-read and edit your posting carefully before you post. Check the spelling, grammar, and capitalization. Typing in all capital letters usually denotes screaming or yelling. Keep your lines to fewer than 70 characters. Do not post test messages except in their appropriate groups; wait until you have something to say. When posting humorous or sarcastic comments, it is conventional to append an emoticon, but do not overuse them.

* Before asking a question, read the messages already in the group and read the group's FAQ if it has one. When you do post a question, follow it with, "Please reply by mail and I will post a summary if requested," and make sure you do post a summary if requested; or, if only a few people were interested, send them a summary by mail. This avoids umpteen people posting the same answer to the group and umpteen others posting "me too"s.
o However, many newsgroups follow the rule "Post here, read here" (that is, they explicitly discourage posters from requesting offline answers to topical questions). This is because there may be lurkers who would benefit from seeing the discussion unfold online, and because questioners often do forget to post summaries of their findings.

* Be proud of your postings, but do not post just to see your name in pixels. Remember: your future employer may be reading.

There is also a netiquette rule for how to deal with someone who has violated one of the rules of netiquette:

* If you believe someone has violated netiquette, send him or her a message by private e-mail; do not post a follow-up to the offending post. Be polite. The author may not have realized his or her mistake, may be a beginner, or may not even have been responsible for the "crime" — his or her account may have been used by someone else, or the address may have been forged. Furthermore, a person who breaks netiquette over and over may be doing it intentionally to disrupt the group (see Internet troll), in which case public flaming over the violation would amount to what is termed feeding the troll. [/quote]

If you have any questions, please ask.

Agreed. Arnav and I spam too much  :(

Maybe I should stop -- I have too much time to kill in math class.

[quote="Go Around the Tree"]if poems are considered spamming , then i did a few times lol  :P[/quote]

It's not so much the poems as the topics as a whole.  They didn't add much value to the forum.

dude what is this a duke conspiracy that is what I suspect

If you would like to discuss sports, use the appropriate NCAA thread.

huh what about other sports what about a specific sport-related event what about dude you wouldn't want us to get off-topic in the appropriate NCAA threads by bringing in other stuff not strictly relevant to the poll question sheesh c'mon now guys let's not be canadian here by just sitting around going eh? all the time I mean I do not actually have a problem with sitting around going eh? but it doesn't help you go get a fields medal now does it see now this brings to mind a funny story because just today before math class some guy was talking about getting the fields medal which comes around every four years and I was like huh I'm going to leave now but then I came back before class started and it was all good see doesn't it suck when you get off topic I sincerely hope that you can see how sucky it is by virtue of the example I have just presented man this is pretty good stream-of-consciousness writing well not really if it were really stream-of-consciousness there'd be more cursing and stuff I guess this is just how I normally type yeah

I find NC threads very amusing.  :rotfl:

Nothing useful has been posted in a LONG time, and we have all these outsiders (and Arnav) trashing the forum. I'm surprised I actually checked it to find this thread. At the rate things are going the whole forum deserves to be deleted. Oh yeah and Arnav (from the first post):

[quote]Remember: your future employer may be reading.[/quote]

[quote="not_trig"][quote="chess64"]If I was the "future employer" I would hire Arnav right away[/quote]

um... you are not an employer, so don't talk  :wink:

If any respectable employer saw all this spam concerning Canada and Duke>UNC, they would be like  :what?:[/quote]

um... you are not an employer, so don't talk  :wink:

If any respectable employer saw all this spam concerning buffaloes (this, btw, IS the real spelling), they would be like  :what?:

hey i have an idea

let's boycott this forum :lol:

[quote="not_trig"][quote="MysticTerminator"]fine, fine, I'll leave
sheesh you guys
now it'll just revert to being all stupid and get deleted b/c nobody will ever post here[/quote]
no you don't have to leave, but you need to stop spamming
and Marylanders and Canadians can boycott this forum -- sure! That would be great![/quote]
I didn't spam though, atleast I think I didn't. You and Mystic on the other hand did.

[quote="not_trig"]hmm... this is however, the NORTH CAROLINA forum.
So there. Canadians can go to the Canadian forum...
No, I'm not trying to throw you out, I'm just kidding. :D[/quote]
oh, ok. :)
You can come to the Canadian forum too. :lol:

i am sorry i am a hack who spams in their free time and will try to stay out of your forum until i visit NC again. :(

[quote="not_trig"][quote="MysticTerminator"]fine, fine, I'll leave

sheesh you guys

now it'll just revert to being all stupid and get deleted b/c nobody will ever post here[/quote]

[b]no you don't have to leave, but you need to stop spamming[/b]

and Marylanders and Canadians can boycott this forum -- sure! That would be great![/quote]

arnav + forum -> spam   (1)

you propose not spam, [assume this is possible/true for sake of contradiction]

lets consider the contrapositive of statement (1)

not spam-> not (arnav + forum)

hence: arnav is not at the forum; arnav left the forum

CONTRADICTION

SNAP

this is not spam

this is a valid mathematical proof...

lol

[quote="not_trig"]Guys... note the title of this thread. Thanks. (sadly our moderator is at AMP... hopefully he will come back and clean out all this spam)[/quote]

I'm back, and a bunch of spam has indeed been deleted (I have been making use of the option to delete many posts at once  :diablo:

it must be noted that "spam = not good" is to date [i]only a theory[/i] with significant holes in any attempted exposition

In case you had not noticed the title of the thread in which you have been posting, Spam = Not Good.

I have deleted the current spam, so please refrain from posting more.

Okay I'm back now, and I am done locking things. Spam will be deleted.

um just a thought, but when you lock a topic, and post in it to say "locked" that should be considered spam because everyone can tell its locked, and it doesn't contribute to the topic.

This time, I'm posting a not so friendly reminder that spam is not good, due to the sudden increase in spam in the forum, most of which has been deleted.

[quote="rt_08"]hey may i also join
 ur community![/quote]

Anyone is free to post in the North Carolina Forum.


Help would be greatly appreciated because I need to know the answers to the following questions in order to complete a big local mathematics tournament. I'm having trouble with the structure of my document.

First of all, I need a lot of space between my title and my introduction (which gives # of questions, etc). I know that "\\" skips a line, but how do I skip more lines?

Also, how do I indicate the start of a new page. For example, after the introduction, is there a command that sends all subsequent writing onto the next page or do I have to space it out myself somehow?

How can I apply something like "\doublespace" to a particular portion of the document? Is this possible? Actually, I couldn't get doublespace to work at all. Do I put curly brackets around the portion of the document that I want to double space?

Final question. Does something in the preamble (the stuff at the top) have to affect the entire document? Could I have "documentclass[20pt]" later in a document that has "documentclass[12pt]" in its preamble? I'm wondering about this because I want the test to have 12 pt. font, but the "\Huge" command doesn't make the title of the test big enough.

Lots of questions, but thankyou to anyone who can help.

\vspace{length}

\pagebreak

I don't know.

Yes, unfortunately, but probably you can't get 20pt in the first place. Try \usepackage{scalefnt} ...\scalefont{decimalnumber}{bigtext} (I think; the brackets may need to include the \scalefont command too.)

ok thanks. i'll try it tomorrow. too late right now. :roll:


A student enters an integer, $ n$  > $ 20$ into her calculator and multiplies it by $ 123$. She then takes the square root of it twice $ (\sqrt \sqrt n)$. She takes this new number and repeats this process for about $ 8$ to $ 15$ times. Eventually, that number stabilizes. To the nearest whole integer, what is that number?

[hide="Solution"]Consider $ n \equal{} 123\implies \log_{123}n \equal{} 1$. The process is equivalent to adding one to the log and then dividing by 4.

Listing out a few logarithm values after a few iterations:

0: $ \frac {1}{2}$

1: $ \left(\frac {1}{2} \plus{} 1\right)/4$

2: $ \frac {\left(\frac {1}{2} \plus{} 1\right)/4 \plus{} 1}{4} \equal{} \frac {\frac {1}{2} \plus{} 1 \plus{} 4}{4^2}$

And so on. After $ k \plus{} 1$ iterations,

$ \log_{123}n \equal{} \frac {\frac {1}{2} \plus{} \sum_{i \equal{} 0}^{k}4^i}{4^{k \plus{} 1}}$

$ \equal{} \frac {\frac {1}{2} \plus{} \frac {4^{k \plus{} 1} \minus{} 1}{3}}{4^{k \plus{} 1}}$.

In the limit as $ k\to\infty$, this clearly goes to $ \frac {1}{3}$.

The answer, then, is $ 123^\frac {1}{3}$. Note that $ 5^3 \equal{} 125$, which is very close, so our answer should be $ \boxed{5}$.[/hide]

This looks kind of long and inefficient, but it went pretty quick when I came up with it...

[hide]
We have $ (123n)^{\frac{1}{4}}\equal{}n \implies 123n\equal{}n^4 \implies 123\equal{}n^3$. So n is about 5.[/hide]


Define $ a_n$ as $ a_0\equal{}1, a_{n\plus{}1}\equal{}a_n\plus{}[\sqrt{a_n}]$. Find all perfect square in the sequece $ a_n$.

$ a_0 \equal{} 1, a_3 \equal{} 2^2,..$ If $ a_n \equal{} k^2$, then next perfect square is $ a_{n \plus{} 2k \plus{} 1} \equal{} (2k)^2.$ Therefore all perfect squares give by
\[ a_{2^{k \plus{} 1} \plus{} k \minus{} 2} \equal{} (2^k)^2.
\]


The length of a common internal tangent to two circles is 7, and a common external tangent has length 11. Compute the product of the radii of the two circles.

[hide="Full Sorution"]
Label the radii $r_{1}$ and $r_{2}$. Labal the distance from the center of each circle $d$. We can set up the following systems of equations:
\[(r_{2}-r_{1})^{2}+11^{2}= d^{2}\]
\[(r_{2}+r_{1})^{2}+7^{2}= d^{2}\]
So we find $r_{2}\cdot r_{1}$ to be
\[4r_{2}r_{1}= 11^{2}-7^{2}\Rightarrow r_{2}\cdot r_{1}=\boxed{18}\]
[/hide]


Call $ x\equal{}\sqrt[3]{a\plus{}\sqrt{b}}$,  $ y\equal{}\sqrt[3]{a\minus{}\sqrt{b}}$.  Then, $ x\plus{}y\equal{}1$, and $ 1\equal{}\left(x\plus{}y\right)^3\equal{}x^3\plus{}y^3\plus{}3xy\left(x\plus{}y\right)\equal{}2a\plus{}3\sqrt[3]{a^2\minus{}b}$, or $ b\equal{}a^2\plus{}\left(\frac{2a\minus{}1}{3}\right)^3$.  Obviously, since $ a,b$ are naturals, $ 2a\minus{}1$ must be a multiple of 3, or we may write $ a\equal{}3m\plus{}2$ for any natural $ m$, and consequently $ b\equal{}8m^3\plus{}21m^2\plus{}18m\plus{}5$.  These are all the possible natural solutions.

[hide="1"]
(Any solution to this equation must be at least 5 to define all square roots.) Rewrite the first equation as
\[ \sqrt{5x^2+14x+9}-7\sqrt{x+1}=\sqrt{x^2-x-20}-2\sqrt{x+1}\]
\[ \iff \frac{5x^2-35x-40}{\sqrt{5x^2+14x+9}+7\sqrt{x+1}}=\frac{x^2-5x-24}{\sqrt{x^2-x-20}+2\sqrt{x+1}}\]
Now $ x-8$ factors out of each of the quadratics. We find $ x=8$ is a solution.

Any solution other than $ x=8$ must satisfy (after dividing the factor and taking reciprocals)
\[ \frac{\sqrt{5x^2+14x+9}+7\sqrt{x+1}}{5(x+1)}=\frac{\sqrt{x^2-x-20}+2\sqrt{x+1}}{x+3}\]
Now a factor of $ \sqrt{x+1}$ cancels in the left hand side. Clearing denominators, it's equivalent to
\[ (x+3)(7+\sqrt{5x+9})=10(x+1)+5\sqrt{(x+1)(x^2-x-20)}\]
\[ \iff (x+3)\sqrt{5x+9}=5\sqrt{(x+1)(x^2-x-20)}+3x-11\]
But also, multiplying the original equation by $ \sqrt{x+1}$ gives
\[ (x+1)\sqrt{5x+9}=\sqrt{(x+1)(x^2-x-20)}+5x+5\]
Multiplying the second equation by 5 and subtracting off the first (2 factors out)
\[ (2x+1)\sqrt{5x+9}=11x+18=2(5x+9)+x
\Rightarrow 0=( \sqrt{5x+9}-x) ( \sqrt{5x+9}-2)\]
The second factor can't even have a positive solution, the first has solutions $ \frac{5\pm \sqrt{61}}{2}$, of which only $ \frac{5+\sqrt{61}}{2}$ is at least 5. This number in fact solves the original equation.

So our solutions are $ \boxed{8,\frac{5+\sqrt{61}}{2}}$.
[/hide]

Call $ a=\sqrt{4x^2+5x-1}$, $ b=\sqrt{4x^2-12x+12}$.  It is easy to verify that for $ x\geq\frac{1}{2}$ and for $ x\leq-\frac{3}{2}$, $ 4x^2+5x-1>0$, while $ x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0$ is true for all real $ x$.

The equation is written as $ 7\left(a-b\right)=a^2-b^2$, yielding solutions $ a-b=0$ and $ a+b=7$.  The first solution is equivalent to $ a^2=b^2$ since $ a,b$ are non negative, or $ x=\frac{13}{17}$, which is larger than $ \frac{1}{2}$, and thus a valid solution.

The second solution $ a+b=7$ requires that $ 4x^2+5x-1=4x^2-12x+12+49-28\sqrt{x^2-3x+3\right}$, or $ 62-17x=28\sqrt{x^2-3x+3}$.  In turn, this requires that $ 3844-2108x+289x^2=2352-2352x+784x^2$, or $ 1492+244x-495x^2=0$, with solutions $ x=\frac{244\pm\sqrt{244^2+1980\cdot1492}}{990}=\frac{244\pm1736}{990}$, ie $ x=2>\frac{1}{2}$ and $ x=-\frac{746}{495}<-\frac{3}{2}$.  So the three values $ -\frac{746}{495}$, $ \frac{13}{17}$ and 2 are the possible roots of the given equation.  In the first case, substitution yields $ 7\frac{367}{495}-14\frac{1549}{495}=-17\frac{746}{495}-13$, which may be proved to be true, in the second case substitution yields $ 7\frac{\sqrt{1492}}{17}-14\frac{\sqrt{373}}{17}=0$, which is also true since $ 1492=4\cdot373$, and in the third case $ 35-14=34-13$, also true.


calculus integration algebra function domain polynomial vector

Why every field F with p elements (p is a prime number) is isomorphic to the field of       all residues modulo p (Z/pZ), and why such an isomorphism is only one?

I know that it is simple, but I cannot build an EXACT proof...  :blush:

---------------

Thank you

[quote="MeKnowsNothing"]Let $ F$ is the subfield of $ \mathbb{Z}_p$. By using Lagrange's theorem to the additive group one gets $ |F|\mid p$. As $ |F| > 1$, the only possibility is $ |F| \equal{} p$.[/quote]
We already know that number of elements of F is p. The question is why F is ISOMORPHIC to Z/pZ.

We have a theorem:

Every field $ K$ contains an integral domain $ D \equal{} \{n1|n\in\mathbb{Z}\}\simeq \mathbb{Z}_p$ if $ \operatorname{char}(K) \equal{} p$.

Proof. Consider the mapping
\[ f: \mathbb{Z}\to K,\quad f(n) \equal{} n1.
\]
This is a ring homomorphism so the homomorphism theorem yields $ \mathbb{Z}/\ker (f)\simeq \operatorname{Im}(f)\subset K$. Here $ \operatorname{Im}(f) \equal{} \{n1|n\in\mathbb{Z}\}$ and
\[ \ker (f) \equal{} \{n\in\mathbb{Z}|n1 \equal{} 0\} \equal{} p\mathbb{Z}.
\]
As $ \mathbb{Z}/p\mathbb{Z}$ is an integral domain, the result follows.

Note that if $ \operatorname{char}(K) \equal{} 0$ then $ K$ contains an integral domain $ \mathbb{Z}$.

In general, suppose that a finite field  $ F$ contains $ q \equal{} p^{n}$ elements where $ p$ is a prime and $ n \in Z^{\plus{}}$.  Consider the polynomial $ f \equal{} x^{q} \minus{} x$. If $ \alpha \in F$ then $ f(\alpha) \equal{} 0$ since $ \alpha^{q} \equal{} \alpha$. Thus all elements of $ F$ are roots of $ f$.  Furthermore, $ f$ contains exactly $ q$ roots. Thus it follows that $ F$ is a splitting field over $ F_p$, a finite field with $ p$ elements (it's a splitting field since if we remove any element from $ F$ then $ f$ no longer splits since all elements of $ F$ are roots).

Since $ Z/qZ$ and $ F$ will both be splitting fields, and splitting fields are isomorphic, we are done.

[quote="t0rajir0u"]I'm not very familiar with field theory, so an elementary question:  why can the order of a finite field only be a power of a prime?[/quote]
Look at the characteristic of the field, which must be a prime (else it contains the rational numbers), say $ p$. If the order is not a power of $ p$, say it contains another prime factor of $ q$, then by Cauchy's theorem there exists a nonzero element x so that $ qx \equal{} 0$. But then $ px \equal{} 0$ as well which implies that $ x \equal{} 0$.

Or more standard: every finite field is a (finite dimensional) vector space over $ \mathbb F_p$ for some prime $ p$:


What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?

$\left(A\right) 0$
$\left(B\right) 1$
$\left(C\right) 2$
$\left(D\right) 2003$
$\left(E\right) 4006$

[hide]
You have:
$2+4+6+8+\cdots$
$1+3+5+7+\cdots$
When you subtract these two series, you see that the difference between each term is 1. There are 2003 terms, so the answer is $\boxed{\text{D}}$
[/hide]

[hide]since the sum of the first N odd numbers is N^2 and the sum of the first N even numbers is N^2+N, the difference is N.
Hence, D[/hide]


Give $ a,b,c\geq0$, prove that:
$ \frac {a^2 \minus{} ab \plus{} b^2}{c^2 \plus{} ab} \plus{} \frac {b^2 \minus{} bc \plus{} c^2}{a^2 \plus{} bc} \plus{} \frac {a^2 \minus{} ac \plus{} c^2}{b^2 \plus{} ac}\geq\frac3{2}$

If $ a,b,c$ are the lengths of a triangle $ ABC$, we have
$ \frac {a^2 \minus{} ab \plus{} b^2}{c^2 \plus{} ab} \plus{} \frac {b^2 \minus{} bc \plus{} c^2}{a^2 \plus{} bc} \plus{} \frac {a^2 \minus{} ac \plus{} c^2}{b^2 \plus{} ac}<3$

All we need to do is to prove the following inequality
$ (a^2\plus{}b^2\plus{}c^2)\left(\sum_{cyc}\frac{1}{a^2\plus{}bc}\right) \ge \frac{9}{2}$
But we also obtain the inequality
$ 2(a^2\plus{}b^2\plus{}c^2) \ge \sum_{cyc}{a^2\plus{}bc}$
We are done!


limit function search logarithms calculus integration Euler

Let $ a>0$ and let $ (x_{n})_{n\in \mathbb{N}}$ be the sequence defined by $ x_{1}>0$ and

$ x_{n\plus{}1}\equal{}x_{n}\plus{}\frac{a}{x_{n}}$,  for all $ n\geq 1$.

Find the limit $ \lim\limits_{n\rightarrow \infty}\left(x_{n}\minus{}\sqrt{2an}\right)$.

e can say that :
$ x_{n}\equal{}\sqrt{2an}\plus{}C\plus{}\frac{c_{1}}{\sqrt{n}}\plus{}\frac{c_{2}}{n}\plus{}o(\frac{1}{n})$
we can find $ c_{1}$ and $ c_{2}$ (by standart way.
but whats about $ C$?? (orit's same as    your limit)))
easy:
let's wrote your equation :
$ x_{n\plus{}1}\equal{}x_{n}\plus{}\frac{a}{x_{n}}$
or:
$ \sqrt{2a(n\plus{}1)}\plus{}C\plus{}\frac{c_{1}}{\sqrt{n\plus{}1}}\plus{}\frac{c_{2}}{n\plus{}1}\plus{}o(\frac{1}{n})\equal{}\sqrt{2an}\plus{}C\plus{}\frac{c_{1}}{\sqrt{n}}\plus{}\frac{c_{2}}{n}\plus{}o(\frac{1}{n}) \plus{}\frac{a}{\sqrt{2an}\plus{}C\plus{}\frac{c_{1}}{\sqrt{n}}\plus{}\frac{c_{2}}{n}\plus{}o(\frac{1}{n})}$
1) $ \sqrt{2a(n\plus{}1)}\equal{}\sqrt{2an} \plus{}\frac{\sqrt{a}}{\sqrt{2n}}\plus{}o(\frac{1}{n})$
2) $ \frac{c_{1}}{\sqrt{n\plus{}1}}\equal{}\frac{c_{1}}{\sqrt{n}}\plus{}o(\frac{1}{n})$
3)$ \frac{c_{2}}{n\plus{}1}\equal{}\frac{c_{2}}{n}\plus{}o(\frac{1}{n})$
4)$ \frac{a}{\sqrt{2an}\plus{}C\plus{}\frac{c_{1}}{\sqrt{n}}\plus{}\frac{c_{2}}{n}\plus{}o(\frac{1}{n})}\equal{}\frac{a}{\sqrt{2an}}\minus{}\frac{C}{4n}\plus{}o(\frac{1}{n})$
hence by substituting in our equation we will get:
$ \frac{C}{4n}\plus{}o(\frac{1}{n})\equal{}0$ hence $ C\equal{}0$
or limit is equal to $ 0$

[quote="Extremal"]e can say that :
hence by substituting in our equation we will get:
$ \frac {C}{4n} \plus{} o(\frac {1}{n}) \equal{} 0$ hence $ C \equal{} 0$
or limit is equal to $ 0$[/quote]

That doesn't follow to me, I multiply both sides by 4*n and get:

$ \ C \plus{} o(1) \equal{} 0$

$ \ C \equal{} o(1)$

Which just confirms that C is a constant

def:
$ o(1)$ is function such as
when $ n\to \infty$ then $ o(1) \to 0$
in our case let's $ n\to \infty$ then $ C \to 0$

[quote="DavidShor"]
That doesn't follow to me, I multiply both sides by 4*n and get:

$ \ C \plus{} o(1) \equal{} 0$

$ \ C \equal{} o(1)$

Which just confirms that C is a constant[/quote]
I believe that Extremal is right , since $ o(1)$ is an expression which tends to $ 0$ as $ n \rightarrow \infty$ .Maybe you thought of $ O(1)$.

[quote="Extremal"]e can say that :
$ x_{n} \equal{} \sqrt {2an} \plus{} C \plus{} \frac {c_{1}}{\sqrt {n}} \plus{} \frac {c_{2}}{n} \plus{} o(\frac {1}{n})$
[/quote]

Can someone please explain why is that ? From where do we know that the general term of the sequence is of that form ? It's clear why it should be with $ \sqrt{2an}$ (because $ \frac{x_n}{\sqrt{2an}} \rightarrow 1$) ,but what about the rest ?Thank you

in thie case it's not important:
if $ x_{n} = \sqrt {2an} + C + \frac {B}{n^{1/4}}$
or
$ x_{n} = \sqrt {2an} + C + \frac {B}{n^{\alpha}}$
then $ x_{n + 1} - x_{n}\approx \frac {A}{\sqrt {n}} + \frac {D}{n^{1 + \alpha}}$
for me is important coeficient of $ \frac {1}{n}$
because it's will be equal to $ CF$
wherer we need to find $ C$ but all coeficient of $ 1/n$ is equal to $ 0$, Hence $ C=0$
really also for:
${ \frac {a}{\sqrt {2an} + C + 1/n^{a}}\approx \frac {a}{\sqrt {2an}} - \frac {C}{4n} + \frac {N}{n^{1 + a}}}$

A generalization

Let $ k\geq 1$ be a natural number and let $ (x_{n})_{n\in \mathbb{N}}$ be the sequence defined by $ x_{1}=a>0$ and

$ x_{n+1}=x_{n}+\frac{a}{x_{n}^{k}}$ , $ n\geq 1$.

Find the limit $ \lim\limits_{n\rightarrow \infty}\left({x_{n}-\sqrt[k+1]{a(k+1)n}\right)}$.

same idea  also works here :
it works because $ \frac{k}{k\plus{}1}\plus{}\frac{1}{k\plus{}1}\equal{}1$  :)
or I mean :
if we suppose :
$ x_{n} \approx (a(k\plus{}1)n)^{ \frac{1}{k\plus{}1} }\plus{}C\plus{}\frac{ a_{1} }{ n^{\alpha_{1}} }\plus{}\frac{a_{2}}{n^{\alpha_{2}}}\plus{}...$
then
coeficient of $ \frac{1}{n}$ in difference :
$ x_{n\plus{}1}\minus{}x_{n}$ will be equal to $ 0$
on other hends
$ \frac{a}{( (a(k\plus{}1)n)^{\frac{1}{k\plus{}1}}\plus{}C\plus{}\frac{a_{1}}{n^{\alpha_{1}}}\plus{}\frac{a_{2}}{n^{\alpha_{2}}}\plus{}...  )^{k}} \approx \frac{a}{(a(k\plus{}1)n)^{\frac{k}{k\plus{}1}}}\minus{}\frac{Ck}{(k\plus{}1)n}\plus{}...$or coeficient of $ \frac{1}{n}$ is equal to $ \minus{}\frac{Ck}{(k\plus{}1)}$ where
$ k \neq \minus{}1,0$ and supposed that $ C\neq 0$
on other hands we know that
$ ...\plus{}\frac{1}{n}0\plus{}...\equal{}x_{n\plus{}1}\minus{}x_{n}\equal{} \frac{a}{( (a(k\plus{}1)n)^{\frac{1}{k\plus{}1}}\plus{}C\plus{}\frac{a_{1}}{n^{\alpha_{1}}}\plus{}\frac{a_{a}}{n^{\alpha_{2}}}\plus{}...  )^{k}} \approx \frac{a}{(a(k\plus{}1)n)^{\frac{k}{k\plus{}1}}}\minus{}\frac{Ck}{(k\plus{}1)n}\plus{}...$
hence $ C\equal{}0$

Perhaps someone can translate what Extremal wrote into material understandable to English speakers  :?:

[quote="perfect_radio"]Here's a big generalization  :P

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=361632543&t=58513[/quote]
 :)  Roughly speaking same idea works also here.
if we suppose
$ a_{n}\approx F^{ \minus{} 1}(n) \plus{} C \plus{} o$
then
$ f(a_{n})\approx f(F^{ \minus{} 1}(n)) \plus{} B \plus{} o$ it's enough to show that $ B \equal{} 0$
1) $ \frac{1}{f(a_{n})}\approx \frac {1}{f(F^{ \minus{} 1}(n))} \minus{} \frac {B}{f(F^{ \minus{} 1}(n))^{2}} \plus{} o$
and $ F^{ \minus{} 1}(n \plus{} 1) \minus{} F^{ \minus{} 1}(n)\approx \frac {1}{f(F^{ \minus{} 1}(p))} \plus{} o$
hence $ a_{n} \minus{} b_{n}\approx \frac {1}{f(F^{ \minus{} 1}(n))} \minus{} \frac {B}{f(F^{ \minus{} 1}(n))^{2}} \plus{} o$
on other hands :
$ a_{n} \minus{} b_{n}\approx \frac {1}{f(F^{ \minus{} 1}(p))} \plus{} o$
hence we have:
$ \frac {1}{f(F^{ \minus{} 1}(p))} \plus{} o\approx \frac {1}{f(F^{ \minus{} 1}(n))} \minus{} \frac {B}{f(F^{ \minus{} 1}(n))^{2}} \plus{} o$
it's enough to show that :
as $ n \to \infty$
$ \frac {1}{f(F^{ \minus{} 1}(p))} \minus{} \frac {1}{f(F^{ \minus{} 1}(n))} \equal{} o(\frac {1}{ f(F^{ \minus{} 1}(n))^{2} })$ where $ p\in [n,n \plus{} 1]$ which is easly to seen.

The extremal's answer is more a draft than a proof; moreover it's a false draft.
1) $ x_n$ is increasing and $ l\equal{}l\plus{}a/l$ hasn't any solution; then $ x_n\rightarrow\infty$.
2) Let $ \alpha$ a real. $ x_{n\plus{}1}^\alpha\minus{}x_n^\alpha\sim\alpha{ax_n^{\alpha\minus{}2}}$. We choose $ \alpha\equal{}2$; then $ x_{n\plus{}1}^2\minus{}x_n^2\sim{2}{a}$. Thus $ x_n^2\sim{2an}$ and $ x_n\sim\sqrt{2an}$.
3)$ x_{n}^2\minus{}x_{n\minus{}1}^2\equal{}2a\plus{}a^2/x_{n\minus{}1}^2\plus{}O(1/x_{n\minus{}1}^4)\equal{}2a\plus{}a^2/x_{n\minus{}1}^2\plus{}O(1/n^2)$.Then $ x_n^2\equal{}2an\plus{}a^2\sum{1/x_{n\minus{}1}^2}\plus{}o(\sum{1/x_{n\minus{}1}^2})\plus{}O(1)\equal{}2an\plus{}(a/2)ln(n)\plus{}o(ln(n))$.
We deduce easily that: $ x_n\equal{}\sqrt{2an}\plus{}\sqrt{a/32}ln(n)/\sqrt{n}\plus{}o(ln(n)/\sqrt{n})$.
In particular the form of the extremal's development is false!

[quote="loup blanc"]The extremal's answer is more a draft than a proof; moreover it's a false draft.
1) $ x_n$ is increasing and $ l = l + a/l$ hasn't any solution; then $ x_n\rightarrow\infty$.
2) Let $ \alpha$ a real. $ x_{n + 1}^\alpha - x_n^\alpha\sim\alpha{ax_n^{\alpha - 2}}$. We choose $ \alpha = 2$; then $ x_{n + 1}^2 - x_n^2\sim{2}{a}$. Thus $ x_n^2\sim{2an}$ and $ x_n\sim\sqrt {2an}$.
3)$ x_{n}^2 - x_{n - 1}^2 = 2a + a^2/x_{n - 1}^2 + O(1/x_{n - 1}^4) = 2a + a^2/x_{n - 1}^2 + O(1/n^2)$.Then $ x_n^2 = 2an + a^2\sum{1/x_{n - 1}^2} + o(\sum{1/x_{n - 1}^2}) + O(1) = 2an + (a/2)ln(n) + o(ln(n))$.
We deduce easily that: $ x_n = \sqrt {2an} + \sqrt {a/32}ln(n)/\sqrt {n} + o(ln(n)/\sqrt {n})$.
In particular the form of the extremal's development is false![/quote]

mm..  as I note it's not important, we don't need terms after $ C$ or it's enough to say that:
$ x_{n} = \sqrt {2an} + C + o$ where $ o \to 0$ as $ n \to \infty$

(read my last post about more generalization !!!)
Now I only repeat :
if $ x_{n} = \sqrt {2an} + C + o$ then :
$ \sqrt {2a(n + 1)} = \sqrt(2an)(1 + \frac {1}{2n} - \frac {1}{8n^{2}} + o(\frac {1}{n^{2}}))$
hence
$ x_{n + 1} - x_{n} = \frac {\sqrt {a}}{\sqrt {2n}} + O(\frac {1}{n^{1.5}}) + (o_{n + 1} - o_{n})$
hence $ \frac {\sqrt {a}}{\sqrt {2n}} + O(\frac {1}{n^{1.5}}) + (o_{n + 1} - o_{n}) = x_{n + 1} - x_{n} = \frac {a}{\sqrt {2an} + C + o} = \frac {\sqrt {a}}{\sqrt {2n}} - \frac {C}{2n} + O(\frac {1}{n^{1.5}})$
hence :
$ \frac {\sqrt {a}}{\sqrt {2n}} + O(\frac {1}{n^{1.5}}) + (o_{n + 1} - o_{n}) = \frac {\sqrt {a}}{\sqrt {2n}} - \frac {C}{2n} + O(\frac {1}{n^{1.5}})$ or
$ (o_{n + 1} - o_{n}) + \frac {C}{2n} = O(\frac {1}{n^{1.5}})$
note that $ (o_{n + 1} - o_{n})$ it's can't be $ \approx \frac {B}{n}$ if it's true then $ o\approx \ln n$ contrary about $ o\to 0$
and also note  that $ (o_{n + 1} - o_{n})\approx o(\frac {1}{n} )$becaose in other case we have contrary about equation :
$ (o_{n + 1} - o_{n}) + \frac {C}{2n} = O(\frac {1}{n^{1.5}})$
hence $ (o_{n + 1} - o_{n})\approx o(\frac {1}{n}$
and $ C = o(1)$ hence $ C = 0$

[quote="loup blanc"]
3)$ x_{n}^2 \minus{} x_{n \minus{} 1}^2 \equal{} 2a \plus{} a^2/x_{n \minus{} 1}^2 \plus{} O(1/x_{n \minus{} 1}^4) \equal{} 2a \plus{} a^2/x_{n \minus{} 1}^2 \plus{} O(1/n^2)$.Then $ x_n^2 \equal{} 2an \plus{} a^2\sum{1/x_{n \minus{} 1}^2} \plus{} o(\sum{1/x_{n \minus{} 1}^2}) \plus{} O(1) \equal{} 2an \plus{} (a/2)ln(n) \plus{} o(ln(n))$.
We deduce easily that: $ x_n \equal{} \sqrt {2an} \plus{} \sqrt {a/32}ln(n)/\sqrt {n} \plus{} o(ln(n)/\sqrt {n})$.
In particular the form of the extremal's development is false!
[/quote]

we know :
1.)
     $ x_{n} \equal{} x_{n \minus{} 1} \plus{} \frac {a}{x_{n \minus{} 1}}$
     then :
     $ x^{2}_{n} \equal{} (x_{n \minus{} 1} \plus{} \frac {a}{x_{n \minus{} 1}})^{2} \equal{} x_{n \minus{} 1}^{2} \plus{} 2a \plus{} \frac {a^{2}}{x^{2}_{n \minus{} 1}}$ or
$ x^{2}_{n} \minus{} x_{n \minus{} 1}^{2} \equal{} 2a \plus{} \frac {a^{2}}{x^{2}_{n \minus{} 1}}$
and you have :
[quote="loup blanc"] ...$ x_{n}^2 \minus{} x_{n \minus{} 1}^2 \equal{} 2a \plus{} a^2/x_{n \minus{} 1}^2 \plus{} O(1/x_{n \minus{} 1}^4)$ ...[/quote]
false.
2.) after summating from $ n \equal{} 2$ to $ n \equal{} N$ we have
$ \sum_{n \equal{} 2}^{N}(x_{n}^{2} \minus{} x_{n \minus{} 1}^{2}) \equal{} 2a(N \minus{} 1) \plus{} a^{2}\sum_{n \equal{} 1}^{N}\frac {1}{x^{2}_{n \minus{} 1}}$
about:
$ a^{2}\sum_{n \equal{} 1}^{N}\frac {1}{x^{2}_{n \minus{} 1}} \equal{} \frac {a}{2}\ln N \plus{} C' \plus{} o(1)$
hence
$ x^{2}_{N} \minus{} x^{2}_{1} \equal{} 2aN \minus{} 2a \plus{} \frac {a}{2}\ln N \plus{} C' \plus{} o(1)$
let $ B \equal{} x^{2}_{1} \plus{} C' \minus{} 2a$
then
$ x^{2}_{N} \equal{} 2aN \plus{} \frac {a}{2}\ln N \plus{} B \plus{} o(1)$

[quote="loup blanc"]Hi extremal,
you don't want to understand that the $ x_n$-development doesn't contain any constant $ C$; it's a structural property; then when you show that this ectoplasm is 0, it's a total nonsense.[/quote]
What do you mean? :|
I don't agree with your this commetn :
[quote="loup blanc"]..The extremal's answer is more a draft than a proof; moreover it's a false draft.
..[/quote]
I explain why I don't agree with this your comment.
I already  said : In my proof we don't need this terms :
$ x_{n} \equal{} \sqrt {2an} \plus{} C \plus{} \underbrace{?}_{this} \plus{} \underbrace{??}_{this} \plus{} \underbrace{???}_{this} \plus{} ...$
in my proof is enough to say that
$ \underbrace{?}_{this} \plus{} \underbrace{??}_{this} \plus{} \underbrace{???}_{this} \equal{} o(1)$
aim of this problem is to prove that $ C \equal{} 0$
I don't understand you.
I know that approximation like this :
$ x_{n} \approx (a(k\plus{}1)n)^{ \frac{1}{k\plus{}1} }\plus{}C\plus{}\frac{ a_{1} }{ n^{\alpha_{1}} }\plus{}\frac{a_{2}}{n^{\alpha_{2}}}\plus{}...$ is false
or false is $ ..\frac{ a_{1} }{ n^{\alpha_{1}} }\plus{}\frac{a_{2}}{n^{\alpha_{2}}}\plus{}.....$ when I wrote it I didn't do any calculation because I don't need it, or I don't need exact asymptotic expansion for $ x_{n}$  I only use fact that after constant every terms tend to $ 0$ . it's enough.

Your position: you assume that $ x_n\equal{}\sqrt{2an}\plus{}C\plus{}o(1)$; you prove that $ C\equal{}0$; yet it's possible that $ x_n\equal{}\sqrt{2an}\plus{}\mu{ln(n)}\plus{}o(ln(n))$.
My position: I prove that $ x_n\equal{}\sqrt{2an}\plus{}\lambda{ln(n)/\sqrt{n}}\plus{}o(ln(n)/\sqrt{n})$ and I haven't any problem with a constant $ C$ because I haven't seen it.

[quote="loup blanc"]Your position: you assume that $ x_n \equal{} \sqrt {2an} \plus{} C \plus{} o(1)$; you prove that $ C \equal{} 0$; yet it's possible that $ x_n \equal{} \sqrt {2an} \plus{} \mu{ln(n)} \plus{} o(ln(n))$.
[/quote]
Thanks, now I understand Why you don't agree with me.
Yes I assume that :
$ x_n \equal{} \sqrt {2an} \plus{} C \plus{} o(1)$ and don't explain why I assume it  :lol: . it's elementary standart problem :because
$ x_{n}' \approx x_{n \plus{} 1} \minus{} x_{n} \equal{} \frac {a}{x_{n}}$ hence  after "integration"
$ x_n \equal{} \sqrt {2an} \plus{} C \plus{} o(1)$
or it's can be cheked easly (there exist man y way ):

if $ x_{n} \equal{} \sqrt {2an} \plus{} \epsilon_{n}$ and suppose $ \epsilon_{n}$ is increasingly, hence after substituting in our equation we will get
$ \epsilon_{n \plus{} 1} \minus{} \epsilon_{n}\approx \minus{} \frac {\epsilon_{n}}{2n} \plus{} \frac {\sqrt {2a}}{8n^{1.5}}$ and if $ \epsilon_{n}$ increas then
$ \epsilon'_{n}\approx \epsilon_{n \plus{} 1} \minus{} \epsilon_{n}\approx \minus{} \frac {\epsilon_{n}}{2n}$
hence after "integration" we see that $ \epsilon_{n}\approx .... \minus{} > C \plus{} o(1)$
I can post another way .. :wink:
Hard is to calculate $ C$ therefore I don't explain why $ \epsilon_{n}\nrightarrow \plus{}\infty$
for example if
$ x_{n}\equal{}1\plus{}\frac{1}{2}\plus{}..\plus{}\frac{1}{n}$ then we can say (by standart method):
$ x_{n}\approx \ln n \plus{} C\plus{}\frac{a_{1}}{n}\plus{}\frac{a_{2}}{n^{2}}\plus{}...$
(if we consider : $ x'_{n}\approx x_{n\plus{}1}\minus{}x_{n}\equal{}\frac{1}{n}$ then after "integration" we will get $ x_{n}\approx \ln n \plus{}C\plus{}o(1)$ similary for $ a_{1},...,a_{m}$ we can calculate all $ a_{j}$ but we can't say nothing about $ C$ Euler constant.

Hi Extremal, your assertion:
"$ x_{n}' \approx x_{n \plus{} 1} \minus{} x_{n} \equal{} \frac {a}{x_{n}}$ hence after "integration"
$ x_n \equal{} \sqrt {2an} \plus{} C \plus{} o(1)$"
 is false; if you know that $ x_n\sim\sqrt{2an}$ then the summation of $ x_{n \plus{} 1} \minus{} x_{n} \sim{ \frac {a}{x_{n}}}$ gives: $ x_n\sim\sqrt{2an}\plus{}o(\sqrt{n})$ which is the same result.
Same remark for: $ x_{n \plus{} 1} \minus{} x_{n} \sim{1/n}$; the summation gives only: $ x_n\equal{}ln(n)\plus{}o(ln(n))$.
Example: $ x_{n\plus{}1}\minus{}x_n\equal{}1/n\plus{}1/(nln(n))$ [equality]; then $ x_n\equal{}ln(n)\plus{}ln(ln(n))\plus{}o(ln(ln(n)))$.

[quote="loup blanc"]Hi Extremal, your assertion:
"$ x_{n}' \approx x_{n \plus{} 1} \minus{} x_{n} \equal{} \frac {a}{x_{n}}$ hence after "integration"
$ x_n \equal{} \sqrt {2an} \plus{} C \plus{} o(1)$"
 is false; if you know that $ x_n\sim\sqrt {2an}$ then the summation of $ x_{n \plus{} 1} \minus{} x_{n} \sim{\frac {a}{x_{n}}}$ gives: $ x_n\sim\sqrt {2an} \plus{} o(\sqrt {n})$ which is the same result.
Same remark for: $ x_{n \plus{} 1} \minus{} x_{n} \sim{1/n}$; the summation gives only: $ x_n \equal{} ln(n) \plus{} o(ln(n))$.
Example: $ x_{n \plus{} 1} \minus{} x_n \equal{} 1/n \plus{} 1/(nln(n))$ [equality]; then $ x_n \equal{} ln(n) \plus{} ln(ln(n)) \plus{} o(ln(ln(n)))$.[/quote]
 :blush:  I know all of them :
but in this case don't forget about error when I wrote $ x'_{n}\approx x_{n \minus{} 1} \minus{} x_{n}$  it's very small and can be calculated exact.
or also  I have cheked my estimate

[quote="Extremal"][

if $ x_{n} \equal{} \sqrt {2an} \plus{} \epsilon_{n}$ and suppose $ \epsilon_{n}$ is increasingly, hence after substituting in our equation we will get
$ \epsilon_{n \plus{} 1} \minus{} \epsilon_{n}\approx \minus{} \frac {\epsilon_{n}}{2n} \plus{} \frac {\sqrt {2a}}{8n^{1.5}}$ and if $ \epsilon_{n}$ increas then
$ \epsilon'_{n}\approx \epsilon_{n \plus{} 1} \minus{} \epsilon_{n}\approx \minus{} \frac {\epsilon_{n}}{2n}$
hence after "integration" we see that $ \epsilon_{n}\approx .... \minus{} > C \plus{} o(1)$
[/quote]
anyway it's obviusly that :
$ x_{n}\approx \sqrt {2an} \plus{} C \plus{} o(1)$


[b]A set of $ 17$ points is chosen on the circumference of a circle. Find the maximum number of segments whcih can be drawn connecting pairs of points such that each segment intersects all others inside, not on, the circle. [/b]

does anybody have any thoughts on this problem?

I'm not sure about the intersecting part, but...

[hide="Partial solution, maybe the whole thing."]Since every distinct point on the circumference of a circle cannot have a line passing through it and hit any other point the circumference we may disregard that statement.

Then, each point has 16 other points that it may connect to, and once one has connected the next point need not reconnect to the original point.

This follows a pattern which yields $ \sum_{x\equal{}1}^{16}(x)$ or $ 136$ possible lines.

[hide="I may have something..."]If I [i]do[/i] understand the intersecting lines thing, then all the lines that go through the circle [i][b]must[/i][/b] intersect all other lines passing through the circle. This [i]should[/i] mean that all lines must pass through the center of the circle. The only points that can yield such a feat are opposite points through which the radius of the circle exists. All points must therefor be in pairs, and each pair gives 1 line. We have 8 total airs and one leftover point which is useless, so the answer is $ 8$.

[/hide]

[/hide]


prove that  there is an infinite numbers (x;y;z) such that:
$ 1/x\plus{}1/y\plus{}1/z\equal{}1/1990$  where  x, y and z  $ \in N$*

Bad title http://www.mathlinks.ro/viewtopic.php?t=135914 . Locked!


It is well know that it's just a matter of time (anywhere between 15-80 years basically - pick your side: optimist or pesimist) until the Oil exctraction will become more expensive than it's exploatation (energy-wise at least). So, as almost all transport industry in our days relies [b]heavily[/b] on Oil, as well as all modern agriculture, are we facing a certain doom to the fragile thing we call "western civilization"? ...

[quote="shobber"]When the oil is going to be end, people will have worked out new energies or they will start wars.[/quote]Ok, and what if, Oil would end in say ... umm ... 10-15 years?!

I went to University of Michigan because I did really good on the SAT and ACT, for Midwest Academic Talent Search, and a science professor from U of M lectured us on science stuff.

One thing he said was that the oil of the world will run out in 20 years. That's not good.  :?

Let's just hope we'll find a new way.

```````````````````````````````````

I also read an internet article that said that the US has already found a substitute for fossil fuels, but they don't want the public to know.. (funny how this info was on the internet.) It also said that they'll introduce it once the oil runs out. Hmm.... :roll:

Electric cars are now being made and they shall eventually replace gas cars. Now the fact that one natural resource ends, does not bring an end to the world economy, it is partially an economics question. There will be new resources available when oil is done, that is most important, maybe they will cost more, I dont know, that is a matter of the work of markets (and maybe some monopolies working in those markets like the huge gas companies), I think we will prevail.

15 years or less.
im kinda pessimistic :P

[quote="hello"]Electric cars are now being made and they shall eventually replace gas cars. [/quote]

that will increase our time before oil is gone. But, can electric cars replace in time?

the entire us economy and many others are entirely based on gas, without gas, we would all die..hopefully they will finally make their nuclear fusion reactors work in 20 years or so.

Oil and Gas will be used up one day. It's only a matter of time. But there are infinity $\text{H}_2$!  :roll: Hope they will get the hydrigon ready before people start wars.

[quote]In plus even if they miraculously find some source of energy now, it would take at least 15 years to properly test and replace the Oil infrastructure ...  [/quote]

I choose 40 years. I think that world economy won't instantly collapse, but have a steep decline, as oil becomes scarcer and scarcer(and more expensive), and that it will cause for electricity, or hydrogen power( or others), to be swept in (whether ready or not), purely because of necessity. World economy won't be totaly damaged, but it will take a while for it to recover.

EDIT: Yeah, 250th post :D

You are funny talking about electric cars. More than 50% of today's energy is produced using ... guess what? Oil (and natural gases sometimes). So no Oil, huge energy crisis.

As for nuclear plants, U238 is even scarcer than Oil, and if anything, people prefere to do Atom bombs from it. US hasn't opened a new nuclear plant in 9 years! - I think that tells a lot about the future of nuclear energy ;)

[quote="236factorial"]Okay, so basically, if the oil supply ends, we all are doomed? I'm going to suicide  :D[/quote]Basically, no we are not doomed. However our current "western" life style is doomed. Air traffic is doomed. All planes run on kerosen, which is a highly concentrated form of Oil fuel. Unless they figure out how to make a small nuclear reactor in a plane I doubt there will be any planes (other than Air Force 1 and F11s) flying around 60 years from now, which is kinda' sad.

The United States hasn't opened a new nuclear plant in the lifetime of most AoPSers (maybe not even in Richard Rusczyk's lifetime), but that's just politics. There is plenty enough nuclear "fuel" to last for long after all of our lifetimes. The United States has a domestic supply of coal that could last much more than a century, although your point is well taken that converting coal into motor vehicle fuel (which both Germany and South Africa have had to do in the past) is extremely energy-inefficient. Whether or not that is ECONOMICALLY inefficient depends on how much the price (= value) of motor vehicle transport will rise if other fuels become scarce. There are still plenty of BTUs of fossil fuels in the ground, and huge amounts of untapped solar and wind energy all around us--price incentives will decide which forms of energy will be transformed into which other forms.

tokenadult exactly my point - stuff will be [i]much[/i] more expensive. Oil is never going to end (physically speaking), because at some point it'll be much more expensive to drill it out than to sell it afterwards :)

According to [i]Discover[/i] magazine, there's actually someone who's figured out a fairly efficient way of distilling, well, basically anything of an organic nature (the current plant takes leftover, unusable material from a nearby turkey processing plant, if memory serves) into gasoline and related hydrocarbons.  Sadly, however, I don't think he's yet turned a profit and his plant is really mostly a novelty as of now.

[quote="IntrepidMath"]The earth is being bathed in energy from the sun! if we can get most of the houses on earth with solar paned roofs, we have solved the problem. We need to harness the power; it is there, we just need to utilize it.[/quote]No, I don't think so. I've read some stuff about this. Besides that fact that it's [b]extremely expensive[/b], very weather dependent (it might work in California or in some desert, but in London, New York - read blocs with  50 stories and 100 square meters worth of roof - or Vancouver - read rains a lot - that'll never work!) and [b]very[/b] inefficient.

That proves my ineptitude and ignorance. However, if the government put in more money, and the world moved towards alternative energy, that would be great. Also, we need to work (now) for more fuel efficient cars and better other sources, i.e. wind, tide, river dams, etc. Also, if we could get people biking short distances, etc. What we really need to work on are alternative forms, but as Mr. Vornicu pointed out, this is incredibly expensive. What would be nice is if we cut funding on useless programs, and increased on eduation and alternative energy. do we really need several thousand nukes hidden away? What if we could (unlikely) get the ore back out and use that as nuclear fuel? Please correct we where you can, I don't really know much about this.

Personally, I don't think we'll run out of oil for a long time.  In my opinion, someone will still be pumping oil maybe 150 years from now, just in extremely small quantities.  Instead, I think oil production will peak in about ten years, followed by a long, irreversible, painful decline.

Valentin, they actually have built a plane with a nuclear reactor, although the reactor didn't actually power the plane.  It was done sometime during the 50s or 60s or something.  It was a B-36 with a small nuclear reactor in the back.  The cockpit was shielded with a heavy lead wall, and a whole bunch of (regular) planes followed it around, including one plane with paratroopers ready to secure and clean up the reactor in case the plane crashed.  It was quite amusing.  Also, I'm sure you could make a plane fly on hydrogen, methane (which we'll also be really short of), etc.

I think you can make the supply of nuclear fuel last longer by doing reprocessing, etc., but this is illegal in the US (but allowed in some countries like France), apparently because of nonproliferation concerns.  I'm not sure why they would have such nonproliferation concerns, as the US has way more nukes than it will ever need.

[quote="IntrepidMath"]Please correct we where you can, I don't really know much about this.[/quote]Basically that's where the current problem is I think. Most of the people in the [i]world[/i] (not only developed countries like US, EU, Japan, etc.) have no idea that Oil extraction will end sometime this century. Actually most of the folks I've talked to about this recently back here seemed to think that Oil reserves are still huge, and we still have 150 years at least not to worry (which is definitely wrong, even by the most optimistic approaches).

No, I'm not saying it'll be [url=http://www.imdb.com/title/tt0082694/]Mad Max 2[/url] anytime soon, however stuff'll get quickly expensive if alternative energy doesn't start being put to good use.

Hmm...at school we read that there were 80 years of coal and 40 years of oil. Ignorance is bliss, espicially if the oil world will come crashing down in 20 years. We need to make a move towards ethenol and other "bio-oils".  I belive that they are still partially oil however. I agree with putting alternative energy to good use. The first car I buy (10 years ish from now) will be a hybrid or electric car. It would be nice to be able to harness outer space, I can think of any ways to do this efficently. I have a wind farm about 30 miles from my house, if we could have more in the sahara or antarctica(treaty problems), that would be efficient. Also, placing small tubines in rivers would help this problem.

[quote="Valentin Vornicu"][quote="IntrepidMath"]Please correct we where you can, I don't really know much about this.[/quote]Basically that's where the current problem is I think. Most of the people in the [i]world[/i] (not only developed countries like US, EU, Japan, etc.) have no idea that Oil extraction will end sometime this century. Actually most of the folks I've talked to about this recently back here seemed to think that Oil reserves are still huge, and we still have 150 years at least not to worry (which is definitely wrong, even by the most optimistic approaches).

[/quote]

People in the US also live really far away from work and drive big cars.  Really big.  No, even bigger than that.  We're talking big.  Not only that, they get tax deductions for certain really big cars, like Hummers.

On a somewhat unrelated note, they recently started work on a tokamak in France.  It's supposed to be the first fusion reactor to gain energy.

[quote="Valentin Vornicu"]Ok this is crazy :) You posted that while I was typing and I couldn't not notice afterwards that 150 years limit you suggested :D It's bit of a strech don't ya think?

BTW more and more people think that the Oil Peak has been reached sometime this year ... anyway this won't be known for sure until the decline is obvious ..[/quote]

Maybe someone might find some really small deposit of oil and pump some out.  However, for all practical purposes, oil will be over really soon.

[quote="frankthetank"]People in the US also live really far away from work and drive big cars.  Really big.  No, even bigger than that.  We're talking big.  Not only that, they get tax deductions for certain really big cars, like Hummers.[/quote]That's well known. I've met some people that told me they would never give up their 20 l/ 100km (I have no idea how much is that in galons / miles but it's a lot) Hummers, even if the gas went 100$\$$ / baril. But they are beginning to reconsider, after the Katrina and Rita phenomena. I heard at the news today that Ford has reported a 50% decrease in sales of large gabarit cars, and General Motors a 30% drop. This is all for the better, however the problem is not solved. A solution would be something like: someone invents cold fusion and it actually works! :)

I am too lazy to read everything here, but u should read the econ bloggers on oil, they have smart thigns to say.

[quote="elston"]i read a popular science article... it said that if wind power was harnessed corrected in the u.s., it could provide 70% of current energy needs[/quote]That's a good joke :)

We will never really run out of oil, but there will come a time whrn the cost of extracting oil becomes greater than the alternatives. Who determines the cost of oil? OPEC, et. al. The West is really at their mercy. (Disclosure: Canada is in OPEC)

What needs to be done is to wean ourselves off oil by gradually imposing larger taxes on petroleum products. This will speed development of alternative sources of energy, encourage conservation by reducing demand, and will as a result unburden our economies and the environment, and force oil-producing countries to grow up.

[quote="Valentin Vornicu"][quote="Drunken_Math"]I also read an internet article that said that the US has already found a substitute for fossil fuels, but they don't want the public to know.. (funny how this info was on the internet.) It also said that they'll introduce it once the oil runs out. Hmm.... :roll:[/quote]There is no source of energy that can replace Oil at the same cost. That is a simple fact of physics.

Of course they say that "they have something" and they'll introduce it when Oil runs out. If you would be the President and you would know that you would run out of Oil in 20 years, would you tell the general public the truth?! Ford Inc. would instantly go bankrupt for example. Whole economies would instantly collapse ... :o

But think about back in the 70's ('73 and '79 if I recall well). There were only a [b]couple[/b] of countries that stopped pomping Oil and there was a HUGE crisis. And we had 50% less cars and fabrics than now ...

There are alternative sources of energy but they are much more expensive than Oil. Our whole global economy is based on [b]cheap[/b] Oil so that not being available will turn things into chaos. Just think of what happened when Katrina hit and there were a [b]few[/b] rafineries not working. Imagine [b]no rafinery at all[/b] working!

In plus even if they miraculously find some source of energy now, it would take at least 15 years to properly test and replace the Oil infrastructure ...[/quote]

To paraphrase what you said:

Once we run out of oil, we're doomed.  :P

[quote="rookworm"]
What needs to be done is to wean ourselves off oil by gradually imposing larger taxes on petroleum products. This will speed development of alternative sources of energy, encourage conservation by reducing demand, and will as a result unburden our economies and the environment, and force oil-producing countries to grow up.[/quote]

Unfortunately, to raise taxes on gasoline in the United States would be politicial suicide.  Many politicians were even considering a price cap of USD 2.50/gallon after Hurricane Katrina. Americans are too used to paying <2.00/gallon.  Also, the car and oil companies have way too much power.  Did you hear of that MASSIVE subsidy that Exxon-Mobil got last year?

On the bright side, the owner of a gas station near my school stopped driving his Hummers to work (always a good job to have if you own two Hummers).

There's a documentary about this stuff called [i]The End of Suburbia.[/i]  It's quite funny.  It refers to this situation as something like "being stuck in a cul-de-sac in a cement SUV without a fill-up."

We can always do what Bush wants and drill in federally protected wildlife refuges.

If I remember correctly, I heard on the radio recently that about 50% (or something like that) of the uranium/plutonium used in nuclear reactors today in the us, are from russian nukes that have been downgraded to reactor grade. I could be mistaken though :?

EDIT: I was reffering to
[quote="IntrepidMath"]What if we could (unlikely) get the ore back out and use that as nuclear fuel? Please correct we where you can, I don't really know much about this.[/quote]

I didn't see the second page of this thread:P

[quote="xtrememath8"]Okay, heres my way-out-there solar idea: we create a giant solar panel and keep it way up in space, while connected with the ground below. Position a million of these things all over Earth; constant, reliable power.[/quote]

That would be expensive and time consuming.

[quote]Americans are too used to paying <2.00/gallon.[/quote]

i m not too sure about that. locally, we have upwards of $\$$3.00

Well, actually they've already created "green" cars - that work on a battery and an electric motor. When riding on a straight road, the car relies on the electric motor to power it. When going uphill, the battery kicks in and when going downhill, the electric motor charges the battery. Very good system, but what happens when you're going uphill all the time :?:  :help:

[quote="xtrememath8"]Okay, heres my way-out-there solar idea: we create a giant solar panel and keep it way up in space, while connected with the ground below. Position a million of these things all over Earth; constant, reliable power.[/quote]That would be a "good" idea if we there was a simple and efficient way of transferring energy instantly from one place to another (without wires :D). And of course if we had the fuel to power up all the rockets to deliver the panels in space :D

the world is so dependent on oil..but we don't even view it as a major topic. humans don't like to worry until it actually happens

There are many ideas floating about that are almost complete.  Some are the same as I've seen from professors at CalTech and MIT on TV articles.

I'm new to this board and read through the comments, I'll try to respond to some of the ideas.

We are indeed using Russian atomic fuel.  However, this is not because of a US shortage, it was to provide a subsidy to help the Russians (then the Soviets) pay for dismantling their weapons and distributing the atomic fuel.  There is no shortage of atomic fuel, be it uranium 238 or otherwise.

Kerosene is a less refined type of petroleum than gasoline, not more.

All of the plant based fuel substitutes in the US are not a part of the solution to the energy concern, as they consume more energy than they replace.  Yes, they consume energy, making money for the oil companies and sell corn, making money for the big agriculture companies the fuel generates less pollution, but only because it has less energy included, so there is lower gas mileage -- only the citizens loose.

Wind power has huge environmental problems, it tends to selectively destroy raptors, in particular the protected bald eagle.  There are many built in the US that can't be used.

Solar cells are not a solution, with today's technology as over it's lifetime, it takes much more energy to make the solar cell than the solar cell produces during its life.  (Think of it as a battery -- useful to transfer power efficiently someplace, such as when camping or out in space, but not an energy generator.)

Hydrogen is not an energy solution, contrary to many statements made by very knowledgeable folks.  The problem is that hydrogen starts out connected to something, like oxygen in water.  It takes more energy to separate hydrogen than burning it releases.  (Again, think of it as a battery.  Also, it tends to explode and burn at high temperatures in invisible fires.  (That and alcohol fires are why firemen still use natural brooms -- they are early warning systems that avoid the firemen stepping into flame fronts and keeping more of them alive.)

Coal gasification is now economically competitive.  The US has over 100 years of proven coal reserves.  However, like atomic power, we have a problem.  Do we want to allow folks to work in underground coal mines or strip mine?  So far we have said, "No."  However, if there ever was a real energy crisis, these decisions might well be reconsidered.

When talking about oil reserves, one has to know what one is talking about.  The almost universal term is proven return.  This means that it is an amount of oil that 1) they know is there, 2) that they know how to extract and 3) that can be sold for a profit at today's prices.  A known reserve is an amount of oil that meets the first two requirements.  When the price of oil increased 4 times in 1976 the record of the reported proven reserves didn't increase dramatically.  Strange, however, the US government, in a cost saving move, stopped making its own estimate of oil reserves in the mid-1930s.  The Saudi's are sitting on at least 250 years of proven reserves.  Finally, I have held in my hands a national newspaper headline that reported that the proven reserves of oil would be exhausted in less than 10 years.  The newspaper was from the mid 1930s.

Do I think that there is a great conspiracy?  No, but one has to understand how oil companies work.  If one has enough reserves to provide for one's company's estimated production for the next 10 years, how much would one spend to get the production needed for years 11 through 15?  Not much.   Plans are made, but the expensive holes have to be drilled before one can report a proven reserve.  Also, if one's exploration cost was higher than expected or the found reserves were smaller than expected the oil field will be shut in and not developed until the price comes up.  Otherwise one would have to report to one's shareholders that one lost money on an oil field.  How many executives will take this option, when they can just slow down the development of the field until inflation, or the cost of money makes the oil field profitable?  One might even just think that the reported loss might go onto another executive watch.

Next, there is a great deal of effort internationally to say that the current energy intensity of the US consumer is unsustainable.  This simply is not true.  We develop sufficient value to the rest of the world to sustain it on a financial basis, so now the attack changes to finite oil.  Why is gasoline so expensive in foreign countries, when the US also imports a great deal of its oil.  (The foreign governments tax at much higher rates, relying on the idea that oil is scarce and high prices are not the result of governmental tax policies.  If this were not true, how would Exxon and the other US oil companies be making record profits, with the US prices so much lower than the world's prices?)

Yes, the amount of anything is finite.  However, if one wants to be able to predict the future one must understand history.  For example, many of the old oil fields are being reworked yielding significant quantities of oil today from fields that our parents were told were all used up.  Some of you are ahead of me, asking how much of the oil in the fields are we currently extracting?  This sets the upper limit on the re-working potential.  The answer is, "No one knows."  Nobody put the oil in, we just know how much we have recovered from similar fields to estimate how much is in each field that is found.  Current estimates by knowledgeable folks (engineers working at oil extraction companies) are that we are currently extracting much less than half the oil that exists in the oil fields.

If one looks back to when the US started using oil as a fuel, one might find something enlightening.  Oil was discovered some decades before being economically exploited.  The difference was the civil war.  The previous whale oil based economy was replaced with oil based economy (primarily for oil lamps for light) when the US Navy impressed all the sailors and ships that were producing whale oil.  The US Army consumed much more oil for light (planning successful and unsuccessful missions) than it did before.  The price for a quart of light oil went from something like 4 cents a quart to 42 cents a quart and someplace between these two prices it was economic to produce and sell petroleum based light oil.  It will take a similar event to stimulate a shift from petroleum.

When the first OPEC oil embargo and resulting increase in prices occurred, the US responded by reducing the amount of oil it imported by twenty-five percent in less than half a decade.  Even with cars and other equipment that is more sensitive to the type of fuel it consumes than were oil lamps, the demand for oil is very dynamic.

Enough from me for now.

There are some very good concerns and thoughts, but one needs to have the appropriate background to really understand and being able to predict what is likely to occur in the future.  Even if one wanted to have a honest discussion, the local media is incapable of delivering such a complex message to its a


Given $a,b,c$ positive numbers and $a^{2}+b^{2}+c^{2}=1$
Find minimum of:
$\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}$


calculus
integration
floor function
ceiling function
combinatorics open
combinatorics

find the number of triangles with integral sides possible with the given
largest side = a (a positive integer)?

Geometric illustration would be usefull.

The answer is a(a+1)/2.

Sorry, I was wrong. I didnt count the situation (a, b, c) = (a, c, b).

So, we should act more carefull.

So we want the number of integral soutions to

$x \le y \le a$

$x + y > a$

Suppose $x + y = k$, and suppose $k \le a+1$.

Then we have \lfloor\frac{k}{2}\rfloor} possibilities.

This never actually happens but is a good place to start.

Now suppose the true case, $k > a$, then there are $\lfloor\frac{k}{2}\rfloor+a+1-k$ possibilities.

So we have $\sum_{k=a+1}^{2a} a+1-\lceil\frac{k}{2}\rceil$.

From here you can quite easily apply summation rules and such to get an answer. However, it involves case analysis with whether a is odd or even, and I'm too lazy to do that, sorry.


[quote="nhat"]i)$a,b,c >0$.Prove that:
              $\frac{ab}{a^2+b^2+3c^2}+\frac{bc}{b^2+c^2+3a^2}+\frac{ac}{c^2+a^2+3b^2} \leq \frac{3}{5}$
ii)$a,b,c >0$.Find all $k \geq 0$ that satisfies the inequality:
                $\frac{ab}{a^2+b^2+kc^2}+\frac{bc}{b^2+c^2+ka^2}+\frac{ac}{c^2+a^2+kb^2} \leq \frac{3}{k+2}$[/quote]

[b]1.[/b] I found an ugly solution for this problem, and I'm looking for a nicer one...  :?
[b]2.[/b] $k=0$ and $k \ge 2(\sqrt{3}-1)$.


For $a,b,c$ positive reals prove that

$\frac{1+a^2}{1+b}+\frac{1+b^2}{1+c}+\frac{1+c^2}{1+a}\geq 6(\sqrt{2}-1)$

Let $s=a+b+c$, so from Cauchy,

$\sum \frac{1}{1+b} \geq \frac{9}{s+3}$

$\sum \frac{a^2}{1+b} \geq \frac{s^2}{s+3}$

Add them together,

$\sum \frac{1+a^2}{1+b} \geq \frac{s^2+9}{s+3}$

Now, we will prove that $\frac{s^2+9}{s+3} \geq 6(\sqrt{2}-1)$

In fact, this is equivalent to $(s-3\sqrt{2}+3)^2 \geq 0$. $\blacksquare$


[quote](As written in the document)

Let $ n$ be a positive integer with $ k$ positive divisors. Suppose $ d_1, d_2, d_3 \ldots d_k$ are its positive divisors with $ d_1 < d_2 < \ldots < d_k$, so $ d_1 \equal{} 1, d_k \equal{} n$. Call $ n$ olympic iff for all $ 1 < i < k$,
\[ d_i \leq \frac{d_{i\plus{}1} \plus{} d_{i\plus{}2}}{2}\]
Show there is an olympic integer with 2002 distinct prime divisors.

[/quote]

Judging by the domain of the $ i$s you gave, and the fact that the definition as above makes all integers olympic, the condition was probably supposed to say $ d_i \leq \frac{d_{i\minus{}1} \plus{} d_{i\plus{}1}}{2}$. I'm going to assume that in my
[hide="Solution"]We will show by induction that we can get an olympic integer with $ m$ distinct prime divisors for all positive integers $ m$.

Base case: $ n$ any prime gives $ m \equal{} 1$; such a number is vacuously olympic.

Inductive step: Suppose we have an olympic integer $ 2n$ with $ m$ distinct prime divisors and $ k$ positive divisors. Pick a prime greater than $ 2n$, and call it $ p$. We claim that $ np$ is olympic with $ m\plus{}1$ distinct prime divisors. Since $ p > n$, $ p \nmid n$, so clearly $ np$ has $ m\plus{}1$ distinct prime divisors.

To show it is olympic, note that $ np$ has $ 2k$ divisors, and that $ d_{k\plus{}i} \equal{} pd_i$ for all $ 1 \leq i \leq k$, since $ p > n$. We already know that the condition is true for $ 1$ through $ k\minus{}1$. Since $ d_k \equal{} n$, $ d_{k\plus{}1} \equal{} p$, we have $ d_k < d_{k\plus{}1}/2$, so the condition is true for $ k$. Also since $ d_2 \geq 2d_1$, $ d_{k\plus{}2} \geq 2d_{k\plus{}1}$, so it is also true for $ k\plus{}1$.

If $ i$ is an integer between $ k\plus{}2$ to $ 2k\minus{}1$ inclusive, then we want $ d_i \leq \frac{d_{i\minus{}1} \plus{} d_{i\plus{}1}}{2}$. Dividing everything through by $ p$ ($ i \minus{} 1 > k$) shows it is equivalent to the inductive hypothesis ($ n$ is olympic) which we assume true. So $ np$ is olympic.

This is sufficient to solve the problem.[/hide]

Hi, MellowMelon
Thank you for your quick solution,
Yes, you are right and I made a typying mistake.
taozh

In order to give credit to the source, this came from this years BMO (British) 1 guide problems. I think it was about question 16?


conics ellipse geometry rectangle perimeter AMC

Rectangle $ ABCD$ has area 2006.  An ellipse with area $ 2006\pi$ passes through $ A$ and $ C$ and has foci at $ B$ and $ D$.  What is the perimeter of the rectangle?  (The area of an ellipse is $ \pi ab$, where $ 2a$ and $ 2b$ are the lengths of its axes.)

$ \textbf{(A) } \frac {16\sqrt {2006}}{\pi} \qquad \textbf{(B) } \frac {1003}4 \qquad \textbf{(C) } 8\sqrt {1003} \qquad \textbf{(D) } 6\sqrt {2006} \qquad \textbf{(E) } \frac {32\sqrt {1003}}\pi$

What do $S$, $a$, $b$ and $c$ refer to in this case?

[quote="pontios"][quote="joml88"]$2a$ and $2b$ are the lengths of its axes
[/quote]
With the standard notation:
$2a:$ the major axis
$2b:$ the minor axis
$2b:$ the distance between the focuses

$S = AB \cdot AD = 2006 \Rightarrow S$ is the area of $ABCD$[/quote]
Should be $2c$ is the distance between the foci

Yes, of course...  :blush:
it was just a typo because of Copy-Paste method  :D

I'm sorry, but I cannot edit this post anymore  :mad:

Let $ AB\equal{}CD\equal{}x$ and $ BC\equal{}DA\equal{}y$.

Area of rectangle: $ xy\equal{}2006$. (1)

Area of ellipse: $ ab\pi\equal{}2006\pi\implies ab\equal{}2006$. (2)

$ ABD$ is right: $ x^2\plus{}y^2\equal{}4c^2$. (3)

Sum of lengths to foci: $ x\plus{}y\equal{}2a$. (4)

Using (4), we get $ a\equal{}\frac{x\plus{}y}{2}$ (5)

Using (2) and (5), we get $ b\equal{}\frac{4012}{x\plus{}y}$ (6)

Using (3), we get $ c\equal{}\sqrt{\frac{x^2\plus{}y^2}{4}}$ (7)

In an ellipse, we have $ a^2\equal{}b^2\plus{}c^2$ (8)

Using (5), (6), (7), in (8), we get \[ \frac{(x\plus{}y)^2}{4}\equal{}\frac{x^2\plus{}y^2}{4}\plus{}\frac{4012^2}{(x\plus{}y)^2}\]

\[ \frac{2xy}{4}\equal{}\frac{4012^2}{(x\plus{}y)^2}\]
Plugging in (1), we get

\[ \frac{\sqrt{2*2006}}{2}\equal{}\frac{4012}{x\plus{}y}\]

\[ (x\plus{}y)\equal{}2\sqrt{4012}\]
\[ P\equal{}(2)(x\plus{}y)\equal{}8\sqrt{1003}\]

This gives $ C$ as the answer.