limit
induction
function
calculus
calculus computations

Prove that $\displaystyle \lim_{n \to \infty} \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{n}}}}$ is finite. Furthermore, letting the limit be $\ell$, prove that \[ \left| x_n - \ell \right| < \frac{1}{2^{n-1} \cdot \sqrt{n!}} ,  \] where $x_n$ is the ugly square root thingy.

I only did the first part.

[hide]say $a$ is a number greater than all the terms of the sequence.
define the sequence $a_1=a$; $a_{i+1}=a_i^2-1$.
we see that $\sqrt{1+\ldots}<a_1 \Leftrightarrow \sqrt{2+\ldots}<a_2 \Leftrightarrow \ldots \Leftrightarrow \sqrt{n}<a_n$ for an arbitrary $n$.
We take $a=2$. To prove that any term of our sequence reduces to proving $n<a_n^2$.
But by induction we can prove that $a_n>n$, $\forall n$ (the induction step: $a_{n+1}=a_n^2-n>n^2-n>n+1$)

So, is it correct?[/hide]

But the second part seems horrible by all means

The second part is just false as written. Indeed, let us consider the function $F_n(x)=\sqrt{1+\sqrt{2+\dots+\sqrt{n+x}}}$ on the interval $0\le x\le\sqrt{n+1}$. We have $F'(x)=\frac1{2^n} \frac1{\sqrt{1+\sqrt{2+\dots+\sqrt{n+x}}}} \frac1{\sqrt{2+\sqrt{3+\dots+\sqrt{n+x}}}}\dots\frac1{\sqrt{n+x}}$. It is not hard to show by induction going from $k=n$ to $k=1$ that $\sqrt{k+\sqrt{k+1+\dots+\sqrt{n+x}}}\le \sqrt k+1$. Hence, the difference between $x_{n+1}$ and $x_n$ is at least $\frac{\sqrt{n+1}}{2^n}\frac 1{(\sqrt 1+1)(\sqrt 2+1)\dots(\sqrt n+1)}$, which is much larger than $\frac 1{2^{n-1}n!}$. It is not hard to see that this approach gives the estimate $|x_n-\ell|\le \frac{\sqrt{n+1}+1}{2^{n}\sqrt{n!}}$. I am quite inclined to believe that using some clever trick one may improve it to $\frac 1{2^{n-1}\sqrt{n!}}$. So, maybe, this square root was just lost when typing the problem  :?.

[quote="fedja"]Hence, the difference between $x_{n+1}$ and $x_n$ is at least $\frac{\sqrt{n+1}}{2^n}\frac 1{(\sqrt 1+1)(\sqrt 2+1)\dots(\sqrt n+1)}$, which is much larger than $\frac 1{2^{n-1}n!}$.[/quote]

For starters, how did you deduce this?  :?

it follows from $\sqrt_{n} +1 < n$ for $n \geq 3$. just like telescoping.

I have estimate that when $ n \minus{} > \plus{} oo$ we have :
$ x_{n \plus{} 1} \minus{} x_{n} \equal{} \frac {1 \plus{} o(1)}{2^{n}\sqrt {(n \minus{} 1)!}}\exp( \minus{} \sum_{1\leq k < n}(\frac {1}{2\sqrt {k}} \plus{} O(\frac {1}{k^{3/2}})))$
thus
$ x_{n \plus{} 1} \minus{} x_{n}$~$ \frac {const}{2^{n}e^{\sqrt {n}}\sqrt {(n \minus{} 1)!}}$
hence
$ l \minus{} x_{n}$~$ x_{n \minus{} 1} \minus{} x_{n}$~$ \frac {const}{2^{n}e^{\sqrt {n}}\sqrt {(n \minus{} 1)!}}$
also we can rewrite it as
$ (l \minus{} x_{n})^{1/n}$~$ \frac {1}{2}\sqrt {\frac {e}{n}}$


calculus
integration
rotation
function
calculus computations

I'm trying to use the disc method to find the volume of an region bounded by $y = \frac{3}{x+1}$ from $x =0$ to $x=8$ about the $x$ axis.

I know that the first thing we need to do is draw a picture to figure out what the radius $R(x)$ is, and then integrate with respect to $x$ from there. However, I tired drawing the picture, but I wasn't able to get the correct R(x).

The integral I set up was:

$V = \int_{0}^{8}\pi [R(x)]^{2}\ dx$, where $R(x) = \frac{3}{x+1}$

But that doesn't work out. Can anyone tell me what I'm doing wrong, or at least what the integral is supposed to be?

That integral looks right to me.  Perhaps you made an error while integrating?

Hmm, okay - this is what I did:

$V = \int_{0}^{8}\pi \left[\frac{3}{x+1}\right]^{2}\ dx$

Re-writing this integral, we obtain:

$\pi \int_{0}^{8}\left(\frac{9}{{(x+1)}^{2}}\right)\ dx$, which is equal to: $9\pi \int_{0}^{8}\frac{dx}{{(x+1)}^{2}}$

We now let $u = x+1$, and $du = dx$. Thus,

$9\pi \int_{0}^{8}\frac{dx}{{(x+1)}^{2}}= 9\pi \int_{0}^{8}\frac{du}{{u}^{2}}=-\frac{1}{u}$

Substituting $u = x+1$ back into this answer, we get $-\frac{1}{x+1}$ to be evaluated from 0 to 8, we get $\frac{8}{9}$, and since we took a $9\pi$ out of the integral, the final answer is $8\pi$.

Is this correct? If so that's kinda weird - first time around I kept getting a different answer.

That looks right to me.  The following lines contain a minor error:

[quote="Mario_KaRt"]We now let $u = x+1$, and $du = dx$. Thus,

$9\pi \int_{0}^{8}\frac{dx}{{(x+1)}^{2}}= 9\pi \int_{0}^{8}\frac{du}{{u}^{2}}=-\frac{1}{u}$

Substituting $u = x+1$ back into this answer, we get $-\frac{1}{x+1}$ to be evaluated from 0 to 8, we get $\frac{8}{9}$, and since we took a $9\pi$ out of the integral, the final answer is $8\pi$. [/quote]

When you change variables, you also have to change the limits of integration.  So if I were doing this, I would have written:
$9\pi \int_{0}^{8}\frac{dx}{{(x+1)}^{2}}= 9\pi \int_{1}^{9}\frac{du}{{u}^{2}}=-\frac{9\pi}{u}\Big |_{1}^{9}$
and I've already taken care of the "substituting back in for $u$" bit (and I get the right answer).  (Also, stragically, I think it's never a good idea to do things like drop a constant -- you write an equals sign, but the two things aren't equal, and often you'll find yourself forgetting to put the constant back in or whatnot.)

Ah I understand - so whenever I change from x to u or something like that, always change the limits of integration?

Another question - this time to do with the shell method. Mainly, given a region - lets say it'll be bounded by $y= 4-x^{2}$ and $y=0$, rotated around the $x$-axis.

The formula for the shell method is $V = \int_{a}^{b}\pi ([R(x)]^{2}-[r(x)^{2}])\ dx$. Given our problem, do we set $4-x^{2}= 0$ to find the limits of integration, which would be $-2$ and $2$?

Also, what's the difference between the r(x) and R(x)? Sorry for so many questions - but i'm trying to see what, if any differences are there between these two methods of finding volumes.

Yes, always change the limits when you change the subject of the integral  :lol:

In this case, really what you are to find is the volume after $y=4-x^{2}$ is rotated about the $x$ axis between $x=2$ and $x=0$ multiplied by 2, as the function is symetrical.

So $V=2\pi\int^{2}_{0}{(4-x^{2})^{2}}\,dx$

$=2\pi\int^{2}_{0}{16-8x^{2}+x^{4}}\,dx$

Sort that all out, and we get $V=\frac{512\pi}{15}$ (I hope!)

In terms of the different methods, I don't think there is much difference at all except for the method.  I just always use $V=\pi\int{y^{2}}\,dx$  :lol:

[quote="Mario_KaRt"]Ah okay - so basically there isn't any real way to determine which method would be easier to use :lol:

Thanks for the help though![/quote]

Kent Merryfield offered some thoughts on this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102670]here[/url].

Another way to set up the original integral is
$\int_{0}^{8}\left( \int_{-\frac{3}{x+1}}^{\frac{3}{x+1}}2\sqrt{\frac{9}{(x+1)^{2}}-y^{2}}dy \right) dx$

Not that this is particularly easy to evaluate.

Two thoughts.

coffeym said "always change the the limits".

I probably do a majority of the time. But here's a way I sometimes write things in class:

$9\pi \int_{0}^{8}\frac{dx}{{(x+1)}^{2}}= 9\pi \int_{?}^{?}\frac{du}{{u}^{2}}=-\frac{9\pi}{u}\Big |_{?}^{?}=-\frac{9\pi}{x+1}\Big |_{0}^{8}$

The essential meaning of those question marks: yes, the limits of integration change when you make the substitution. But if I've decided that I want to come back to the original variable in the end, I don't really have to figure out what the new limits are. It would be incorrect to write 0 and 8 there, and I won't write down something that's incorrect. It would be correct to write down 1 and 9, but to know that would be to to do work we don't need to. Hence the question marks: incomplete information, to be sure, but not incorrect.

But, as I said, I do change the limits to what they are in the new variable a majority of the time.

The other issue is coffeym saying, "I just always use .." and then naming one of two possible methods.

You shouldn't restrict yourself like that. If there are two methods, there are going to be problems in which it makes little difference which one you use, there are going to be problems which are intractable whichever way you go, but there are also going to be problems in which the first method is much easier than the second, and problems in which the second problem is much easier than the first.

You don't always know up front which is the better method; you'll only know for sure in hindsight. You have to be willing to try both just to see which one is better.


[quote="orangefronted"]The lengths of the medians of a triangle with sides 5,6 and 7 are a, b, and c. Find the area of the triangle with sides a, b, and c.[/quote]

area of triangle=4/3x[s(s-m1)(s-m2)(s-m3)]^1/2
 s=(m1+m2+m3)/2

now it is cakewalk

You could also find $ a, b$ and $ c$ by using my answer here: http://www.mathlinks.ro/viewtopic.php?t=327833

@umesh123

Can you show me how u arrived at the ratio 4/3?


Let $ (I_a),(I_b),(I_c)$ be excircle of triangle $ ABC$Let $ P$ be a point in the triangle and $ A_1,B_1,C_1$ be intersection of $ AP,BP,CP$ with $ BC,CA,AB$ respectively.Let $ A_2,B_2,C_2$ lie on  $ (I_a),(I_b),(I_c)$ respectively such that $ A_1A_2,B_1B_2,C_1C_2$ touch  $ (I_a),(I_b),(I_c)$ respectively($ A_1 ,B_1,C_1$  not $ \in$ $ BC,CA,AB$.Prove that $ AA_2,BB_2,CC_2$ are concurent.

[quote="jayme"]Dear Hoangclub and Mathlinkers,
your problem is an extraversion of the one when the three excircles become the incircle. This last problem has been solved on Mathlinks but where?
Sincerely
Jean-Louis[/quote]
Dear Jean-Louis,

Do you mean [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=797172#797172]this one[/url]? This problem has many solutions. And if I am not wrong, then our Moderator, Cosmin Pohoata, has a paper included 3 solutions for this problem.

Best regards

Dear April,
yes.
I saw this problem in   Papelier G., Rapports anharmoniques, proposition 63, Exercices de Géométrie Moderne, Paris (1927) Rééditions J. Gabay;
                                 Ehrmann J.-P., Message Hyacinthos # 6966.
                                 Danneels E.,    Transformation based on the incircle, Message Hyacinthos # 8716.
Darij Grinberg gave a nice proof in     Variations of the Steinbart Theorem", http://www.cip.ifi.lmu.de/~grinberg/
For the Steinbart theorem you can also see my article on my website :
http://perso.orange.fr/jl.ayme     vol. 3  Les points de Steinbart et de Rabinowitz.

"if I am not wrong, then our Moderator, Cosmin Pohoata, has a paper included 3 solutions for this problem": I woulb be interested in the paper of Cosmin...
Sincerely
Jean-Louis

Sorry to say this, but this problem was NOT solved in my article and is NOT an extraversion of an easier problem.

  darij


calculus
derivative
real analysis
real analysis unsolved

Hello everyone,

I am having some trouble completing the following two problems and would appreciate help. My work is shown below.

Thank you!

---

1. Find the points on the lemniscate, with the equation $ 2(x^2 \plus{} y^2)^2 \equal{} 25(x^2 \minus{} y^2)$, where the tangent is horizontal.

2. Prove the following using implicit differentiation:

http://img411.imageshack.us/my.php?image=inversefunctionnn1.jpg
---

1. Since the tangent is horizontal when the derivative = 0:

$ d/dx$ of lemniscate equation: $ 8x^2y \frac {dy}{dx} \plus{} 8y^3 \frac {dy}{dx} \plus{} 50y \frac {dy}{dx} \minus{} 50x \plus{} 8x^3 \plus{} 8xy^2$.

However, when I set $ \frac {dy}{dx}$ = 0 and solve for y, the x values do not match the ones given by my textbook as the answers.

$ y^2 \equal{} \frac { \minus{} 50 \plus{} 8x}{8}$

2. I denoted $ (f^{ \minus{} 1})(X)$ as y to get:

$ y' \equal{} \frac {1}{f'(y)}$

However, when I diffentiated both sides, I got a second derivative which seems to complicate the theorem.

hello, 1.) using implicit differentiation i got
$ 2\cdot2(x^2+y^2)\cdot(2x+2y \cdot y^')=25(2x-2y\cdoty^')$
Sonnhard.

hello, from my first derivative  we have
$ y^' = \frac {25x - 4x^3 - 4xy^2}{4x^2y + 4y^3 + 25y}$. Then you have two equations to find points where the tangent is horizontal
$ 25x - 4x^3 - 4xy^2 = 0$ and $ 2(x^2 + y^2)^2 = 25(x^2 - y^2)$
Sonnhard.

[quote="catalan"]$ y^2 \equal{} \frac { \minus{} 50 \plus{} 8x}{8}$[/quote]
You appear to have lost a minus sign somewhere -- $ y^2$ should be the negative of this.  Then substitute back into the original equation and solve.

hello to your second problem, we have $ f(g(x))=x$, differentiating this we get $ f^'(g(x)) \cdot g^'(x)=1$.
Sonnhard.

Thank you very much to everyone for the responses!


projective geometry
geometry proposed
geometry

See http://www.mathlinks.ro/Forum/viewtopic.php?t=18426 (and feel free to post your proof, since I don't particularly like using the Pascal permutation theorem)...

  darij

Dear Darij Grinberg:
I think it works with the method which I solve http://www.mathlinks.ro/Forum/viewtopic.php?t=166387
but I can not solve (IV).

Your (IV) is my [b](4)[/b] and follows via Desargues from my [b](6)[/b], which, if I remember correctly, is a consequence of Pappos or Pascal.

[b]EDIT:[/b] I didn't remember correctly. Actually, the proof of [b](6)[/b] uses [b](5)[/b]:
The triangles A'XB and YB'A are perspective (since A'Y, XB', BA concur at C'). Thus, by Desargues, the points A'X /\ YB', XB /\ B'A and BA' /\ AY are collinear. In other words, the points Q, ZX /\ CA, YZ /\ BC are collinear. Similarly, Q', ZX /\ CA, YZ /\ BC are collinear. From here on it's clear how to proceed.

  darij


In the rectangle below, line segment $ MN$ separates the rectangle into 2 sections. What is the largest number of sections into which the rectangle can be separated when 4 line segments (including $ MN$) are drawn through the rectangle?

[asy]size(3cm,3cm);

pair A,B,C,D,M,N;

A=(0,0);
B=(1.5,0);
C=(1.5,1);
D=(0,1);

draw (A--B--C--D--A);

M=(0.8,0);
N=(1.2,1);

draw(M--N);

label("M",M,S);
label("N",N,NNE);[/asy]

The largest number of sections happens only when the Nth line cuts through all of the previously drawn lines.
Then it makes N more sections than N-1 lines would make.
The first line makes 2 sections.
Thus, 4 line segments would make $ 2\plus{}2\plus{}3\plus{}4 \equal{} \boxed{11} sections maximum.$


[i]Μιας και δεν είδα πουθενά παρόμοιο topic σκέφτηκα να κάνω ένα...[/i]

1) Διπλασιάστε ένα δοσμένο ευθύγραμμο τμήμα [b]μόνο[/b] με τη χρήση διαβήτη.

2) Βρείτε το μέσο ενός ευθυγράμμου τμήματος [b]μόνο[/b] με τη χρήση διαβήτη.

3) Βρείτε το κέντρο ενός δοσμένου κύκλου [b]μόνο[/b] με τη χρήση διαβήτη. (Αυτή η κατασκευή είναι και η πιο δύσκολη...)

[i]Σημείωση[/i]: Όλα τα παραπάνω βασίζονται σε ένα θεώρημα κάποιου Ιταλού (μου διαφεύγει το όνομα), που απέδειξε πως ό,τι γίνεται με κανόνα γίνεται και με διαβήτη.

[i]PS[/i]: Όποιος έχει να προτείνει και άλλες κατασκευές μπορεί να τις βάλει εδώ!

[quote="mostel"]
[i]Σημείωση[/i]: Όλα τα παραπάνω βασίζονται σε ένα θεώρημα κάποιου Ιταλού (μου διαφεύγει το όνομα), που απέδειξε πως ό,τι γίνεται με κανόνα γίνεται και με διαβήτη.
[/quote]

Lorenzo Mascheroni (1750-1800)

Το πρώτο είναι πολύ απλό.

1. Έστω $AB$ το δοσμένο ευθ . τμήμα . Τότε φτιάχνουμε το σχήμα..

Πολύ σωστά Ramanujan  :D

Σχηματίζεται ένα κανονικό εξάγωνο αν το παρατηρήσετε και έτσι αποδεικνύεται και η κατασκευή!

Yπάρχουν πάρα πολλές γεωμετρικές κατασκευές που μου έρχονται αυτή τη στιγμή στο μυαλό,αλλά δυστυχώς στις μέρες μας ελάχιστοι ασχολούνται μ'αυτές...Λοιπόν η τελευταία κατασκευή που θέτεις ως πρόβλημα αναφέρεται και ως [b]Κατασκευή του Ναπολέοντα (του Βοναπάρτη)[/b] και την έχω προτείνει κι εγώ παλιότερα.Υπάρχουν πολλές λύσεις-με χρήση αντιστροφής βέβαια το πρόβλημα <<απλουστεύεται>>!Για περισσότερα:
[b]Τitu Andreescu,Mathematical Olympiad Challenges  σελ21,Προβλημα 9[/b](στο πίσω μέρος υπάρχει η λύση με αντιστροφή) ή  σε οποιδήποτε άλλο βιβλίο Γεωμετρίας.

[b]Νικος Ράπανος[/b]

Το αγαπημένο μου κεφάλαιο στη γεωμετρία είναι οι κατασκευές [u]με κανόνα και διαβήτη[/u].
Μόνο με διαβήτη είναι αρκετά πολύπλοκες, αλλά παρουσιάζουν μεγάλο ενδιαφέρον

http://www.mathlinks.ro/Forum/viewtopic.php?t=24888
http://www.mathlinks.ro/Forum/viewtopic.php?t=53829
http://www.mathlinks.ro/Forum/viewtopic.php?t=74230


[img]http://www.cs.toronto.edu/~moraes/optical-illusion-wheels-circles-rotating.png[/img]
Pretty nice eh?
:o

[quote="distracted523"]Hmm, the picture's a bit large.

When you picked Greek for your sig, I got this:

Βετ υοθ ψαν'τ τρανσλατε τηισ. Ιν υοθρ φαψε σθψκερσ.
Vet yoth psan't translate tiis. In yothr fapse sthpskers.

So I'm guessing it's really more of a bad substitution cipher:
Bet you can't translate this. In your face suckers.

 :roll:[/quote]
Dang you got me.
I'll have to change it in another language.

If you focus on one specific point on the image, it stops moving.

[quote="distracted523"]1st part Russian/Cyrillic-cipher:
Иуе нщг сфтэе екфтыдфеу ершы тщц ур, Щцтув!
Iue n(shch)g sftee ekftydfeu ershy t(shch)(ts) ur, (Shch)(ts)tuv!

Too lazy to decipher this...not as obvious as the first one...

2nd part Greek-cipher:
Στιλλ ψαν'τ τρανσλατε τηισ σθψκερσ! Δ
Still psan't translate sthpskers tiis!
Still can't translate suckers this!

Now your grammar is messed up too...[/quote]

Actually, you messed up, my second to last word is 4 letters, so it can't be "suckers"

That's the point of the optical illusion, I've seen it before. Also, to make sure it's a genuine optical illusion, you can just stare at one spot to see that it isn't just an animated picture.

[quote="ZhangPeijin"]

Actually, you messed up, my second to last word is 4 letters, so it can't be "suckers"[/quote]

Eh, okay, it's "this suckers!" right?

I didn't notice the movement until it was pointed out.
__________________

[quote="modeler"][quote="ZhangPeijin"]
The illusion is caused by movement, so if you don't move your head and keep your eyes still, the picture becomes still.[/quote]

It does not work for me. I guess the illusion is due to the difference between strengths of signals from the red, green and blue cones in the retina (yellow = red + green), nothing about the movement of eyes. As the light sensitive cells is more concentrated at the centre, the point you are focusing on will not cause illusion.[/quote]
Do the same thing in black and white, it works too.
It doesn't work as well since I sort of resized it, so bad quality.

@ZhangPeijin: Is your avatar a hidden cube?

Yes it's hidden because you can't see it.

[quote="ZhangPeijin"][quote="7h3.D3m0n.117"]Watch the offtopic conversation...

@randomdragoon: huh, looking at one particular spot [i]does[/i] make it seem still.[/quote]
The illusion is caused by movement, so if you don't move your head and keep your eyes still, the picture becomes still.[/quote]
But your eyes move involuntarily all the time... you can't really control it.

[quote="modeler"]I get the answer :D

Note the blue and yellow are complementary, (The sum of their R values is 255, the same for G and B) and so do the two gray colors with different lightness.

Now let's see how our brain detect moving objects

A still object should look like this:
000444000
where numbers are the color intensity and the 4s represent an object.

When the object is moving leftward, you should be able to see this
003441000
where the 3 and 1 is caused by motion blur. Since the sum of intensity remains constant, the color in front of and behind the object should be complementary (when a black object is moving in a white background or vice versa)

Then take a look at ZhangPeijin's snakes. It is just like the above case.
Therefore your brain will mistake the image as moving :D[/quote]
Hmm, I feel confused...


Another question from moi. Page 343, vol 2, exercise 23-21...
Let the divisors of (p-1) be d1, d2, d3.... Prove that if we have a primitive root g (mod p), then for each dI, there is an element with period dI.

THe answer provided just seems to contradict the very definition of a primitive root.

Answer provided: Since g has period (p-1), the numbers g^((p-1)/d1),
g^((p-1)/d2).... have periods d1, d2.. respectively. THese numbers taken to the respective powers all  yield g^(p-1). Now we introduce a new number c, for which g^(((p-1)/dI)^c) is congruent to 1 (mod p) for c <dI. This is where I get tripped up. Doesnt c have to be a multiple of the specific dI? If g^((p-1)/dI) is congruent to 1, then for it stilll to be a multiple of 1 when taken to a power, doesnt the power have to be a multiple of dI? I dont see how c < dI....
Then it ends by saying that g^c((p-1)/dI) is congruent to 1 for c(p-1)/dI<(p-1), violating the assertion that g has period p-1. Thus the given numbers have the desired periods....
But if it violates the main rule given of g being a primitive root, then how can this be true?!

Please clear this up... And to Gauss, thanx for pointing out where i went wrong in the fermat's theorem problem.

You're not as far from understanding it as you think.  In fact, your comment

[quote]But if it violates the main rule given of g being a primitive root, then how can this be true?! [/quote]

is exactly the point.

The proof is by contradiction - we assume that c < dI, then show that this leads us to a contradiction of our given that g is a primitive root.  Thus, we conclude that no such c<dI can exist such that g^(((p-1)/dI)^c) :equiv: 1 (mod p).  From this we conclude that the given numbers have the desired periods.  (Had we stated more explicitly that this was a proof by contradiction, this would have been much clearer...  Sigh.)

(You note that the only way to get g^(((p-1)/dI)^c) :equiv: 1 (mod p) is if c is a multiple of dI - the proof by contradiction in the solutions is essentially how we prove this fact.  We assume that c is less than dI then show that this assumption contradicts a given fact.  We conclude c is not less than dI; we can do something very similar to show that c must be a multiple of dI.  As you know, it's kind of intuitively clear - contradiction gives us a rigorous way to show it.  This is often true of contradiction; it's a very useful tool to formally prove something that seems 'obvious' (once you get past the notation in this case).)


probability
induction
real analysis
algebra proposed
algebra

A is a non-empty set of positive integers. There are positive integers b1, b2, ... , bn, c1, c2, ... , cn such that each set biA + ci = {bia + ci : a belongs to A} is a subset of A, and the n sets biA + ci are pairwise disjoint. Prove that 1/b1 + 1/b2 + ... + 1/bn ≤ 1.[/tex]

it was proposed from Iran with really nice idea behind,i let every one enjoy with this such a hard problem.

I'm not sure about this, but I think it's worth mentioning:

Choose a random number form $A$. $A$ is infinite, and I think it can be shown that the probability that the number belongs to $b_iA+c_i$ is $\ge\frac 1{b_i}$. By adding up the probabilities, we get the probability that the number belongs to $\bigcup(b_iA+c_i)$ (this is where we use the fact that those sets are pairwise disjoint). This probabilityy is, of course, $\le 1$, which means that $\sum \frac 1{b_i}$ is also $\le 1$.

Think something like this could work?

Here is my solution :

   We suppose 1/b1+...+1/bn>1 and seek a contradiction.

  let a be the smallest number in A. Let f(x) be the number of elements of A which are :le: x .
If x:ge:a , then f(x):ge:f((x-c1)/b1)+...+ f((x-cn)/bn)+1  (#). There are indeed at least the elements of the form bi*y+ci, which are all distinct, and a.

Part 1 :   let :varepsilon:  be min( 1/(bi*a+ci) ,  1/(c1/b1+....+cn/bn) ) >0

We prove by induction on [x] that f(x) :ge:  :varepsilon:x for all x:ge: a

for x :le: max(bi*a+ci) , it's clear because f(x) :ge: 1 :ge: x/(bi*a+ci) :ge:  :varepsilon: x
if  x > max(bi*a+ci) , we use (#) and induction :
   f(x) :ge:  :varepsilon: ((x-c1)/b1+...+(x-cn)/bn)+1 :ge:  :varepsilon: x by the choice of epsilon, and because we had supposed 1/b1+...+1/bn>1.

Part 2:

let l=liminf( f(x)/x). By part 1, l  :ge:  :varepsilon: , and thus l <> 0.

Take d>0 and let y be such that for x>y, f(x)/x>l*(1-d) (y exists by definition of d). Let then (xk) be a sequence of numbers with f(xk)/xk-->l, those numbers being chosen big enough so that for all k and i, (xk-ai)/bi>y.

By (#), f(xk):ge:f((xk-c1)/b1)+...+ f((xk-cn)/bn)+1
  And   f(xk)/xk :ge: l(1-d)(((xk-c1)/b1)+...+((xk-cn)/bn))/xk, dividing by xk, and using f(x)/x>l*(1-d) for x>y.
As xk tends to infinity, we get  l :ge:  l(1-d)(1/b1+...+1/bn)
Since l>0, 1:ge:  (1-d)(1/b1+...+1/bn)
And as d tends to 0,  we get (1/b1+...+1/bn) :le: 1, which is the contradiction sought...

I tried to find examples of such subsets A of IN . You can take A=IN, b1=...=bn=n, and ci=i. But does there exist one with, for example,n=2, b1=2 and b2=3 ?


The numerator of a fraction is 3 less than its denominator. When the numerator and denominator of the fraction are both increased by 1, the resulting fraction is equivalent to $ \frac23$. What was the original fraction?

We have $ \frac{x\minus{}3}{x}$ as the original, and then $ \frac{x\minus{}2}{x\plus{}1}\equal{}\frac{2}{3}\Rightarrow 3x\minus{}6\equal{}2x\plus{}2\Rightarrow x\equal{}8$.  Hence the original is $ \frac{5}{8}$.


A referendum failed by a vote of 36 No's and
24 Yes's. To make this pie chart of this result, what
is the lesser measure in degrees of angle $ ACB$?

[asy]import graph;
draw(Circle((0,0),20));
draw((0,0)--(0,20));
draw((0,0)--(12,-16));
label("$A$",(0,20),N);
label("$B$",(12,-16),SE);
label("$C$",(0,0),W);
label("YES",(10,13),S);
label("NO",(-10,13),S);
label("36",(-10,10),S);
label("24",(10,10),S);[/asy]

Hint: It's a sector!

Click to reveal solution.

[hide](24/60) x 360 = 144 degrees.[/hide]
________________________________
Mathcounts roxs! ~ pi man  :read:


Hi, I'm learning the rigorous defintions of functions and its properties.  Today I'm stuck on the concepts of images.  I was wondering if anyone here could pelase help me with this.  Thank You!

Question:  Let $f: A\rightarrow B$ be a function.  Let $W$ be a subset of $B$.  Prove that $f(f^{-1}(W))\subseteq W$.

I'm pretty sure that this is above High School Basics.

To Intermediate it goes!  :ninja:

Let $y\in W$ then $f^{-1}(y)=x\iff y=f(x)$ so $f(f^{-1})=f(x)=y\in W$ so $f(f^{-1}(W)\subseteq W$
p.s : I don't really understand the point of this question

2.- Give an example of a $f$ satisfying $f\left(f^{-1}(W)\right)\subsetneq W$.

3.- $f\left(f^{-1}(W)\right)=W$ when $f$ is what type of function?


Find $ k$ such that 9 is the solution for $ 2x\plus{}k\equal{}3(x\minus{}1)$.

Plug in $ x\equal{}9$. Then solve for $ k$.

$ 18 \plus{} k \equal{} 3(8)$

$ \boxed{k \equal{} 6}$


Well, I found thi sproblem in another thread but I just want to post it as its own seperate problem.

Find the exact value of $4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right)$

THis is what I did, I dont know the right answer but I want to check if its right.

[hide]Well for that call $4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right)$=x, so we find tan x.  So basically it becomes tan (4a-b)=tan x, where a=$4\tan^{-1}\left(\frac{1}{5}\right)$  and b=$\tan^{-1}\left(\frac{1}{239}\right)$.  Now you can apply the tangent sum formula to make it become (tan4a-tanb)/(1+tan4atanb)=tan x.  Now you know tanb=1/239, and from multiple double angle formulas, you find tan4a=120/119.  Now plug that back into the equation and you get (120/119-1/239)/(1+1/239*120/119)=tan x.  This now simplifies to tan x=1, so [b]x=45+(180)k for any integer k[/b][/hide]

There should be only one answer knowing that the tangent inverse is a function...

[quote="Art of Owna"]So its only 45?[/quote]
I believe it's only $\frac{\pi}{4}$, yes.

Oh in btw, I found the problem on
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=816124&sid=ec10a375d0cc8f9450cb0e8f4b45e5be#816124

rwang's post (second one)..I just moved it so we wont have two problems in one thread.

[quote="Art of Owna"][quote="rnwang2"]The answer is pi/4.

Anyway, I hope this was more fun than the pointless distance formula problem.  :roll:[/quote]
wait, I can get all the hard parts of that problem until I get to the final step of tanx=1, why is 45 degrees the only answer (sorry if this is a stupid quetsion, for inverses do you put the value that is closest to 0)?[/quote]
because the tangent ratio technically only exists in the interval -90 < x < 90 in degrees.. vertical asymptotes are on the values -90 and 90.

[quote="GlenthemanN"][quote="Art of Owna"][quote="rnwang2"]The answer is pi/4.

Anyway, I hope this was more fun than the pointless distance formula problem.  :roll:[/quote]
wait, I can get all the hard parts of that problem until I get to the final step of tanx=1, why is 45 degrees the only answer (sorry if this is a stupid quetsion, for inverses do you put the value that is closest to 0)?[/quote]
because the tangent ratio technically only exists in the interval -90 < x < 90[/quote]

oh thanks.

[quote="rnwang2"]Technically I should have written $\text{Tan}^{-1}$, which denotes the [i]principal value[/i].[/quote]

Ya I was just thinking the same thing..because I wanted to be sure and I was reading over ch.2 of art of problemsolving vol2 and it said there was a difference between $\text{Tan}^{-1}$ and $\text{tan}^{-1}$.

So if a problme asked like $\text{Tan}^{-1}(1)$ would that just be 45
but if its like tan x=1, would x be 45+180n?

[quote="Art of Owna"]So if a problme asked like $\text{Tan}^{-1}(1)$ would that just be 45
but if its like tan x=1, would x be 45+180n?[/quote]
Yes.

$y=\text{Tan}^{-1}x\iff\tan y=x$ and $-\frac{\pi}{2}<y<\frac{\pi}{2}$.


When the value of $ y$ is doubled and then this increased value is
divided by 5, the result is 10. What is the value of $ y$?

All you have to do is reverse everything that has been done to the original number so you get 25 like it shows below;

10*5=50    50/2=25     25 is the answer


prove that:
    $a/(pb+qc) + b/(pc+qd) + c/(pd +qa) + d/(pa+qb)$ $\geq$ $4/(p+q)$
                                             for positive numbers a,b,c,dp,q satisfying p $\geq$q
Does the inequality hold for$p<q$

$\frac{a}{pb+qc} + \frac{b}{pc+qd} + \frac{c}{pd +qa} + \frac{d}{pa+qb} \geq \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2}{p(a+b+c+d)+q(a+b+c+d)}$ (caushy)

So, I should prove that $(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2 \geq 4(a+b+c+d)$

But, $(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2 = a+b+c+d+2(\sqrt{ab}+\sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}+\sqrt{cd})$

So,  $(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2 \leq 4(a+b+c+d)$...

Can anybody help me?

At first, let's make sure the inequality doesn't hold for p < q: Just take a = 1, b = 2, c = 1, d = 2, p = 1, q = 2 for a counterexample.

Now, we are going to prove the inequality for $p\geq q$. In other words, we are going to show:

[color=blue][b]Theorem 1.[/b] If a, b, c, d, p, q are six positive reals such that $p\geq q$, then

$\frac{a}{pb+qc}+\frac{b}{pc+qd}+\frac{c}{pd+qa}+\frac{d}{pa+qb}\geq\frac{4}{p+q}$.[/color]

[i]Proof.[/i] We have

$\frac{a}{pb+qc}+\frac{b}{pc+qd}+\frac{c}{pd+qa}+\frac{d}{pa+qb}$
$=\frac{a^2}{pab+qac}+\frac{b^2}{pbc+qbd}+\frac{c^2}{pcd+qac}+\frac{d^2}{pda+qbd}$.

Now, after [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in Engel form[/url],

$\frac{a^2}{pab+qac}+\frac{b^2}{pbc+qbd}+\frac{c^2}{pcd+qac}+\frac{d^2}{pda+qbd}$
$\geq\frac{\left(a+b+c+d\right)^2}{\left(pab+qac\right)+\left(pbc+qbd\right)+\left(pcd+qac\right)+\left(pda+qbd\right)}$
$=\frac{\left(a+b+c+d\right)^2}{p\left(ab+bc+cd+da\right)+2q\left(ac+bd\right)}$.

Now,

$\left(p+q\right)\left(a+b+c+d\right)^2-4\left(p\left(ab+bc+cd+da\right)+2q\left(ac+bd\right)\right)$
$=\left(p-q\right)\left(a-b+c-d\right)^2+2q\left(\left(a-c\right)^2+\left(b-d\right)^2\right)\geq 0$

(here we used $p-q\geq 0$, since $p\geq q$), so that $\left(p+q\right)\left(a+b+c+d\right)^2\geq 4\left(p\left(ab+bc+cd+da\right)+2q\left(ac+bd\right)\right)$, and thus

$\frac{\left(a+b+c+d\right)^2}{p\left(ab+bc+cd+da\right)+2q\left(ac+bd\right)}\geq\frac{4}{p+q}$.

Combining the above chain of inequalities, we arrive at

$\frac{a}{pb+qc}+\frac{b}{pc+qd}+\frac{c}{pd+qa}+\frac{d}{pa+qb}\geq\frac{4}{p+q}$,

and Theorem 1 is proven.

  darij


Let $ P\in\mathbb{Q}[X].$ $ \omega\in\mathbb{C}$ is a root of $ P$ with multiplicty $ >\frac{1}{2}degP$.
Prouve that  $ \omega\in\mathbb{Q}.$

If $ \omega \not \in Q$, then exist at least one root $ \omega'$ with suth multiplicity. It give contadition.

Let $ m$ be the multiplicity of $ \omega$. Note that $ 2m > deg(P)$

If $ \omega \in \mathbb{C} \minus{} \mathbb{Q}$, then $ \omega$ must be a non-rational algebraic number.

This means that $ \omega$ must be a root (multiplicity $ 1$) of some polynomial $ Q \in \mathbb{Q}[x]$ with $ deg(Q) \ge 2$ such that $ Q^m \mid P$.

However, this requires, $ deg(Q^m) \equal{} m \cdot deg(Q) < deg(P)$, a contradiction since $ deg(Q) \ge 2$.

Therefore, $ \omega \in \mathbb{Q}$.


Hmm...

The sum of the angle measures in a quadrilateral is $ 360$ degrees. $ \angle AFE$ and $ \angle ADE$ are both right angles, and $ \angle FED$ is an interior angle of octagon $ ABCDEFGH$, so its measure is $ 135$ degrees (the interior angle measure of regular octagons). Thus, the answer is $ 360\minus{}90\minus{}90\minus{}135\equal{}45$.

Although you could probably guess it looking at the picture  :lol: .

If you draw a line between G and D it's kind of visually clear that you form a 45-45-90 triangle with DAF as a 45.

But textangle's solution brings up that formula which everyone should know:  a regular polygon of n sides has interior angles of $ \frac {(n \minus{} 2)180}{n}$ degrees.

In this case $ \frac {(8 \minus{} 2)180}{8} \equal{} 135$


inequalities calculus function derivative

find minimum of \[ a^2 + 4ab + 8b^2 + 16c^2  \]
if \[ abc = 1  \] and $a,b,c>0$
i edited the "z" mistake... thank you durt

[quote="yookon88"]there's a problem... if you aply AMGM this way you get a different answer
for example: \[ a^2 + ab + 3ab + 8b^2 + c^2 + 15c^2 \geq 6\sqrt[6]{360(xyz)^4}\approx 16.003  \]
which is smaller than \[ 3\cdot2^{\frac{17}{6}}\approx 21.3815  \][/quote]

But wait, the equality case is only when all terms on the LHS are equal, correct? If so then your example does not have an equality case, since equating different terms gives different ordered triples $(a,b,c)$... then LHS>RHS always and its not the minimum

[quote="cincodemayo5590"]
But wait, the equality case is only when all terms on the LHS are equal, correct? If so then your example does not have an equality case, since equating different terms gives different ordered triples $(a,b,c)$... then LHS>RHS always and its not the minimum[/quote]

You're right about the equality case...
[quote="cincodemayo5590"]I got a minimum of a liiiitle smaller...

By AM-GM we have $a^2+4ab+8b^2+16c^2 \geq 4 \sqrt[4]{512a^3b^3c^2}= 16 \sqrt[4]{2ab}=16\sqrt[4]{\frac{2}{c}}$

Then minimum will be at the equality case
$a^2=4ab=8b^2=16c^2,$ so $4ab=\frac{4}{c}=16c^2$
$c^3=\frac{1}{4} \Rightarrow c=\frac{1}{\sqrt[3]4}$. Then, plugging this back in we get the minimum to be $\boxed{ \sqrt[12]{2^{53}} \approx 21.357}$.
[/quote]
I think there is a smaller minimum than yours

[quote="yookon88"][quote="cincodemayo5590"]
But wait, the equality case is only when all terms on the LHS are equal, correct? If so then your example does not have an equality case, since equating different terms gives different ordered triples $(a,b,c)$... then LHS>RHS always and its not the minimum[/quote]

You're right about the equality case...
[quote="cincodemayo5590"]I got a minimum of a liiiitle smaller...

[hide]By AM-GM we have $a^2+4ab+8b^2+16c^2 \geq 4 \sqrt[4]{512a^3b^3c^2}= 16 \sqrt[4]{2ab}=16\sqrt[4]{\frac{2}{c}}$

Then minimum will be at the equality case
$a^2=4ab=8b^2=16c^2,$ so $4ab=\frac{4}{c}=16c^2$
$c^3=\frac{1}{4} \Rightarrow c=\frac{1}{\sqrt[3]4}$. Then, plugging this back in we get the minimum to be $\boxed{ \sqrt[12]{2^{53}} \approx 21.357}$. [/hide]
[/quote]
I think there is a smaller minimum than yours[/quote]

ok, well then i dont think the solution to this will involve splitting up terms further, becuase this will either give an inequality without an equality case, or as durt did, not give the lowest possible minimum...

maybe it involves factoring.... im not sure, but im trying things to increase or decrease the number of terms... ill get back to ya

[quote="cincodemayo5590"][hide]
I got a minimum of a liiiitle smaller...

By AM-GM we have $a^2+4ab+8b^2+16c^2 \geq 4 \sqrt[4]{512a^3b^3c^2}= 16 \sqrt[4]{2ab}=16\sqrt[4]{\frac{2}{c}}$

Then minimum will be at the equality case
$a^2=4ab=8b^2=16c^2$, so $4ab=\frac{4}{c}=16c^2$
$c^3=\frac{1}{4} \Rightarrow c=\frac{1}{\sqrt[3]4}$. Then, plugging this back in we get the minimum to be $\boxed{ \sqrt[12]{2^{53}} \approx 21.357}$.[/hide]

oh wait, this probably isnt the smallest answer b/c of different ways to apply Am-Gm..[/quote]

the equality case doesn't work here because you have \[ a^2=4ab \leftrightarrow a=4b  \] and \[ a^2=8b^2 \leftrightarrow a=2\sqrt{2}b  \]

the same problem with durt's solution... that explains shyong's result

[quote="paladin8"][quote="cincodemayo5590"][quote="durt"]$a^2 = n a b$
$a = n b$
$a^2 = n^2 b^2$
So, in the inequality, the coefficient of $b^2$ must be the square of that  of $a b$. I'm still trying to figure out what to do with the $c^2$.[/quote]

ok i think i have it...
So, to get the $b^2$ coefficient as the square of the $ab$ coefficient, we need to cut the LHS into $a^2+2ab+2ab+4b^2+4b^2+16c^2$. Then, what we can do is cut the $c^2$ term into enough terms so that when applied in AM-GM, the power of $c$ matches at least one of the other terms. But, i think we need to make sure that $a^2=2ab=4b^2=nc^2=k$.
So... actually, $n=16$ works:
$a^2=16c^2 \Rightarrow c=\frac{a}{4}$ and $a^2= 4b^2 \Rightarrow a=\frac{b}{2}$... so $16c^2=16(\frac{a}{4})(\frac{b}{2})=2ab$. I checked other values for the $c^2$ coefficient and they dont work.

But... this leaves us without two powers of variables that match up. If we were to do AM-GM now it would be $a^2+2ab+2ab+4b^2+4b^2+16c^2 \geq 6\sqrt[6]{1024 a^4 b^6 c^2} = 12 \sqrt[6]{16 a^2 b^4}= 12\sqrt[6]{\frac{16 b^2}{c^2}}= 12\sqrt[3]{\frac{4b}{c}}$. Then, becuase we want the equality case, $4b^2=16c^2 \Rightarrow \frac{b}{c}=2 \Rightarrow MIN=12\sqrt[3]{8}=\boxed{24}$.

This occurs at $6k=24 \Rightarrow \boxed{(a,b,c)=(2,1, \frac{1}{2})}$. (such pretty numbers!)

But.. this is smaller than our original minimum. Darn. Does anyone see a place where i messed up, or another way to do this?[/quote]

The problem is if you apply AM-GM and the RHS minimum has variables, you don't know for sure that the minimum actually occurs at the equality case. You only know that at the equality case, both sides are equal (since the RHS is changing).[/quote]

ok.. so, is there another way to do this other than with Mathematica?    :P

Yes, there is - using calculus (I haven't found a better way).  We want to find critical points of the function

$f(a,b) = a^2 + 4ab + 8b^2 + \frac{16}{a^2 b^2}$.

We take the partial derivatives with respect to $a$ and $b$ and set them equal to zero:

$2a + 4b - \frac{32}{a^3 b^2} = 0$

$4a + 16b - \frac{32}{a^2 b^3} = 0$.

These simplify to

$a^2 + 2ab - \frac{16}{a^2 b^2} = 0$

$8b^2 + 2ab - \frac{16}{a^2 b^2} = 0$.

Subtracting the second equation from the first yields

$a^2 = 8b^2$.

Substituting this into either equation and solving for $b$ yields

$b = \left(\frac{2 - \sqrt{2}}{4}\right)^{1/6}$

which gives us

$a = (128 (2 - \sqrt{2}))^{1/6}$.

Plugging these into $f(a,b)$ and simplifying gives

$f(a,b) = 12 (3 + 2 \sqrt{2})^{1/3} \approx 21.5956$, which seems by consensus to be our minimum value.  I don't see how such a messy number could turn up from a simpler method...

[quote="shyong"]thats cool zanttrang ! :D
Once we know the equality case , it is trivial apply of AM-GM

$a^2+8b^2\geq 4\sqrt{2}ab$

So
$a^2+4ab+8b^2+16c^2\\\geq 4ab(\sqrt{2}+1)+16c^2\\=\frac{2(\sqrt{2}+1)}{c}+\frac{2(\sqrt{2}+1)}{c}+16c^2\\\geq 3\sqrt[3]{\left((2(\sqrt{2}+1))^2\cdot 16\right)}\\=12\sqrt[3]{3+2\sqrt{2}}$

 :)[/quote]
Oh, excellent - that is a better way to proceed once we've found $a^2 = 8b^2$.  The question now is how we could deduce that without calculus...

I guess we can know that $a^2=8b^2$ because there can be only one equality between $a$ and $b$ or several equalities of the same kind... anyway... i think you were allowed to solve it with calculus though i don't like calculus...
I found that this question has been posted before so I apologize. No need to continue this thread.

[quote="luimichael"]How comes this question was a school olympiad problem?
What did the examiners expect the students to do ?[/quote]

Well,  one can be expected to find  minimum value of  ($2A/x+Bx^2$) (see note 1)
and, of course, can also be expected to exploit the knowledge that  minimum value of a square is zero ..(complete the square etc..)

So  (as Shyong above has shown)   you write the expression as:
$a^2+8 b^2 +4 a b+16 c^2 = ( a - \sqrt 8 b)^2 + 2 \sqrt 8 ab + 4 ab + 16 c^2$

This is nothing but  of the form $(square \ term)+ 2A/c + Bc^2$   (with $B=16$ , and  $A= \sqrt 8 + 2$ (see note 1)

[hide="note1"]
AM-GM gives:

$(A/x+A/x+Bx^2) \ge 3 \sqrt[3]{A^2B}$[/hide]

[quote="yookon88"]there's a problem... if you aply AMGM this way you get a different answer
for example: \[ a^2 + ab + 3ab + 8b^2 + c^2 + 15c^2 \geq 6\sqrt[6]{360(abc)^4}\approx 16.003  \]
which is smaller than \[ 3\cdot2^{\frac{17}{6}}\approx 21.3815  \][/quote]
"="<=>a=b=3b(false)

[quote="shyong"]I check from mathematica that the minimum is occur at $21.5956$ where $a=2.05349$ ,$b=0.726017$ ,$c=0.670752$ .

Well , I cant see anyway how to apply AM-GM to have such equality case .... :([/quote]
2.05349*0.726017*0.670752=1.00000312827539616


You probably heard a lot about the swine flu from the news it's everywhere about how we need to be prepared for the worst.  I don't think it very serious but I am curious about other peoples thoughts.

It's serious because flu is always serious.
[quote="Wikipedia"]Although the incidence of influenza can vary widely between years, approximately 36,000 deaths and more than 200,000 hospitalizations are directly associated with influenza every year in the United States.[/quote]
The government warning was... maybe twice that, about what we would expect for a novel strain of similar virulence to the the usual seasonal strains but without most of the population already immune from previous cases.
What we can do about it:
- Take routine precautions to slow down transmission, and make it less likely that everybody gets sick at once.
- Get people at high risk for complications (such as myself due to asthma) vaccinated, to significantly reduce the number of serious cases and the burden on the medical system.

Its been proven that Swine Flu is worse at infecting cells than Influenza

I'm actually worried about the vaccinations. If I remember correctly, the original bird flu vaccinations killed healthy 20-year-olds.

Perhaps not quite as much of a threat as it was originally made out to be but it is definitely a problem. :(

I had it 3 weeks ago and I was completely better in 3 days but again I was on meds.

2bh no. i think the media is making this more 'serious' than it is.

I had swine flu, I spent 4 days of my week off of school playing video games, if you remember when they said you had to stay home from school if you had it.  Though, for three days i had a pretty high fever

And, no medications

The swine flu is being blown up by the media.  But no one wants this to become the next pandemic, so they have to make a big issue on it.

According to a government run website, there are 556 deaths and a little over 8,500 hospitalized cases in the US.  Not a lot compared to our population.

The main reason swine flu is being blown up by the media is because it is very possible that it can mutate into something more dangerous. Plus, in fall, when many people go to school, anyone with the virus will spread it around very quickly. Otherwise, it's nothing really serious. For instance, Sabien5 said he had it and he was perfectly fine (fever going away by itself without medication is pretty good.)

Furthermore, those deaths are mainly young children/babies, elderly people, and people who already suffer from some sort of complication.

[quote="Treeswitheyes"]According to a government run website[/quote]  Your post could have been actually informative if it had included the following link: http://www.cdc.gov/h1n1flu/

[quote="isabella2296"]Furthermore, those deaths are mainly young children/babies, elderly people, and people who already suffer from some sort of complication.[/quote]  That's true of seasonal flu, but it's notably [i]false[/i] of the swine flu -- the incidences of infection and death are centered somewhere on those aged 5-64: http://www.cdc.gov/h1n1flu/surveillanceqa.htm

This thread suffers badly from a preponderance of assertions lacking any semblance of source or evidence.  Other similar comments following this one run a substantial risk of just being deleted as worthless.

[quote="isabella2296"]The main reason swine flu is being blown up by the media is because it is very possible that it can mutate into something more dangerous.[/quote] I think you're giving the media too much credit. Imo, they blew it up so much because a good scare makes them get more viewers which equals more money.


IMO Shortlist
modular arithmetic
invariant
combinatorics

There are $ n$ markers, each with one side white and the other side black. In the beginning, these $ n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $ n \minus{} 1$ is not divisible by $ 3$.

[i]Proposed by Dusan Dukic, Serbia[/i]

One implication is easy by induction on $n$.

We say that a positive integer $n\ge 3$ is "good" if starting with $n$ white tokens, we can reach a state with only two tokens. We can prove that if $n$ is good, then $n+3$ is good as well: if the tokens are numbered $1,2,\ldots,n$ from left to right, say, then we take tokens $2,4,3$ (in this order), and we're left with the same situation, except that with $n-3$ tokens instead of $n$. Since it's obvious that $3$ and $5$ are good, it follows that all $n$ for which $3\not|n-1$ are good.

Conversely, we have to prove that if $3|n-1$, then $n$ is not good.

Denote white by $a$ and black by $b$. We denote a sequence of tokens by the corresponding word in $a$ and $b$ (and we disregard the spaces that are left after eliminating some of the tokens). The rules for elimination tell us that $aaa=bb,aab=ba,baa=ab,bab=aa$. We write these in condensed form as $a^3=b^2,a^2b=ba,ba^2=ab,bab=a^2$. These look just like some relations in a group, don't they? :). In fact, they're satisfied by the elements $a=(123),b=(12)$ of the group $S_3$ of permutations of a set with three elements, so if we show that in this new setting, with $a,b$ as above, if $3|n-1$ then $a^n$ is not equal to a word in two letters among $a,b$, then we're done. This is simple: notice that $3|n-1\Rightarrow a^n=a\not\in\{a^2,b^2,ab,ba\}$.

My congrats: it is a very cool solution.

I have solved it with old tricks: colouring, parity and stuff like that... It is a bit different, so it may interest some of you.

[size=75]I assign 0 to black coins and +1 or -1 to white ones. I choose the sign with this rule: the sign is + iff the number of black coins to the left is even. In other words, I start from left with + signs and every time I encounter a black coin I change the sign.

It is easy to see that the sum of the coins can only stay the same, increase by 3, or decrease by 3. The sum mod 3 is therefore invariant.

Moreover, the black coins are always an even number. Thus, when there are only 3 coins left, you will always have 3 whites (sum = +3) or 1 white + 2 blacks. In this case, the game can be solved only if the white one is in the middle. The sum is then -1.

The sum starts at +n, thus if n=1 (mod 3), then the sum is always 1 mod 3. But when we are with 3 coins left, the sum can be only 3 or -1 and never 1 mod 3. [/size][]

[color=blue]EDIT: Well, my proof can be simplified further: I have considered what happens when there are 3 coins left. Silly my... why not 2 coins left?

[size=75]In that case, they can be either both white (sum = +2) or both black (sum = 0). It can never be 1 mod 3.[/color][][/size]

[quote="grobber"]$a^{3}=b^{2},a^{2}b=ba,ba^{2}=ab,bab=a^{2}$. These look just like some relations in a group, don't they? :). In fact, they're satisfied by the elements $a=(123),b=(12)$ of the group $S_{3}$ of permutations of a set with three elements, [/quote]

What does this mean? I don't really understand. can somebody help?  :blush:

I found still another solution:

In my solution, we remember the holes that occur. Now we take away coins until, for the first time, we take a coin next to a hole. as we were able to take it away, it must have been reversed two times. this implies that, in fact, both coins next to it are already taken away.
now, the first steps can be commuted arbitrarily. therefore we could as well take at first the three coins that lie next to each other(the middle one at the end) and then proceed in the old way. but taking these three coins leaves us with $n-3$ white coins. this reduces the problem from $n$ to $n-3$ coins and we can proceed by induction.

Interesting... people came up with all three of these solutions at MOP! Not being particularly clever at combinatorics, I worked out a bunch of cases until I found the second solution :). There is yet another slightly messier solution where you use strong induction. It involves considering how you could have arrived at any given position with 3 markers, but I forget exactly how it goes.

let 1 denote white, 0 black. 11 is already in the desired state, 111 turns to 00 in one move. these be the base cases. then notice that in three moves 11...1 becomes 11...1 with 3 fewer 1's, so by induction we've covered all $ n \equiv 2,3\pmod{3}$. for the converse, find an invariant (this is the more interesting part ;)).

oh, it is'nt easy. Can you post you solution more details?

yo!, I found the invariant of this problem but I can't prove that when there are 2 markers on a row, they are in same color. Who can help me??

what do you mean by two markers in a row are the same color? because certainly it can happen that not all markers are the same color, and then there'll certainly be two adjacent markers of different colors. do you maybe mean two markers left at the end?

:( , my English is quite bad. but I think when there are [b]only[/b] 2 markers on a row (when the game ends), they are the same color. Who can help me prove that? Thanks a lot!

okay, so it's what i thought... to prove that isn't hard... let whites be +1, let blacks be -1. then every time we make a move, the product of the cards stays the same. at the beginning it's +1, so at the end, we must have two + or two - :)

[hide="Ugly Solution"]It is easy to see that $n$ not congruent to $1$ modulo $3$ can win.

Now the hard part. Notice the outermost markers do not matter, so just write an 'x' there. We wish to get to the configuration xx. In the first move, you will get something like: x111....1110011...11x where 0=black, 1=white. Then, if we never select the markers that right now have a 0 in them, we will never achieve only 2 markers, we will always have at least 4 (the x's and the 0's). So we must select them at some point. I will demonstrate this is equivalent to having x11...11x (where we have $n-3$ markers) and finish by induction.

So first, call the two markers that are now 0 X and Y. We must select them in the future. So say at some point we have the configuration B00B', where B is a block that ends with a 1, and B' is another block of numbers. Then we select the final marker of B (because X and Y must become 1 at some point). Now we have (B-1)10B'. WLOG marker X will not change until we select it. Suppose we perform the set of operations O on B-1 and O' on B' (the order doesn't matter). Then we select marker X. It is easy to see that this is the same as selecting marker X and then doing O and O'. In both cases, we get O(B)O'(B'). And this is the same as first, when we have x1..1001..1x, selecting the marker before X, and then selecting X, and then doing O and O'. So this configuration could have been achieved with the initial configuration x1..1x with $n-3$ markers.

The case x011...1x is similar, so we are done by induction.

So we are done by induction.
[/hide]
[hide="One-liner"]There is a nicer solution I have found. We define S(C), where C is a configuration, to be the number of 1's that have an even number of 0s to the left minus the number of 1's that have an odd number of 0s to the left. Trivially S is invariant mod 3 so we are done.[/hide]

[hide="Remarkable fact"]It is worth noting that the order of operations doesn't matter. [/hide]

Indeed, Juan's one-liner is basically equivalent to grobber's dihedral group solution.

[hide="This is a frustrating problem: Sketch of Solution"]
Let $W$ denote a white marker and let $B$ denote a black marker. We will record a configuration of markers in the obvious manner; e.g. $WBWWB$ denotes a row of 5 markers, colored white, black, white, white, black in that order from left to right.
We will write $W^i$ to denote a row of $i$ white markers. Similarly define $B^i$.

Note that the total number of markers decreases by 1 at each step.
Note that the number of black markers is always even at each step.

Thus, the problem boils down to showing that if $n \equiv 1 \pmod{3}$, then there is no way to get from $W^n$ to $K$, where $K \in \{ BB, WW \}$.

We will take advantage of the fact that the number of black markers is always even. Given a configuration of markers, let the black markers be $B_1, B_2, B_3, \cdots, B_{2k}$ from left to right. Define the [i]score[/i] of the configuration as $d(B_1, B_2) + d(B_3, B_4) + d(B_5, B_6) + \cdots + d(B_{2k-1}, B_{2k})$, where $d(X,Y)$ is defined as the distance between markers $X$ and $Y$.

Note that the score of the configuration always increases by 1 (mod 3). Define $c$ to be the number of markers remaining. Define $s$ to be the score of the current configuration. Thus $c+s$ is invariant modulo 3.

Suppose for the sake of contradiction that we can get from $W^n$ to $K$, where $K \in \{ BB,WW \}$. At the end, $c+s$ is equal to either 2 or 3. However, at the beginning $c+s$ is equal to $n$, which is congruent to 1 (mod 3). We have our desired contradiction, since $n$ is not congruent to 2 nor 3 (mod 3). QED

this took me 2-3 hours to find :(
[/hide]

Label the markers $1, \cdots , n$. It's easy to see by induction that we can get two markers in the end if $n \equiv 0, 2\pmod{3}$ if we remove markesr $2, 4, 3$ in this order.

Now we will prove via induction that for $n \equiv 1 \pmod{3}$ it would be impossible to get down to two markers, for $n \equiv 0 \pmod{3}$ we would get two black markers and for $n \equiv 2 \pmod{3}$ we would get two white markers. The base cases $n = 2, 3, 4$ are easy to see.

If for $n$ we can get down to two markers,those two markers must be markers $1, n$ and also there must be some sequence of steps for which we can get down to three markers. Let the marker be $i$. Thus for $n$ we consider two subproblems before getting down to three, one for the markers $1, 2 \cdots i$, the other for markers $i, i+1, \cdots n$ with the constraint that we can't remove markers $1, i, n$. This is valid since flippings in the former subproblem can't affect any in the latter subproblem and vice versa with the exception of $i$. So we just keep track of the color of $i$.

Note that the two subproblem each have size less than $n$ which sum to $n+1$. Now, we do case work:
[list]
[*]$n \equiv 0 \pmod{3}$.
The subproblems are of size $0 \pmod{3}$ and $1 \pmod{3}$ or both size $2 \pmod{3}$.
By induction, the former case is impossible and in the latter case $1$, $i$ and $n$ are white after the subproblem.
Then when we have three markers and flip $i$, the first and last markers become black as desired.
[/*]
[*] $n \equiv 1 \pmod{3}$.
Suppose we can get down to three markers at some point for contradiction.
The subproblems are of size $0 \pmod{3}$ and $2 \pmod{3}$ or both size $1 \pmod{3}$.
By induction, the latter case is impossible and in the former case we would have $1$, $i$, and $n$ be black, black, white respectively, so that we can't flip $i$.[/*]
[*]$n \equiv 2 \pmod{3}$.
The subproblems are of size $1 \pmod{3}$ and $2 \pmod{3}$ or both size $0 \pmod{3}$.
By induction, the former case is impossible and in the later case $1$, $i$, and $n$ are black, white, black respectively and after flipping $i$ we get two white end markers. [/*]
[/list]

[quote=mathadventurer][quote=JuanOrtiz]
[hide="One-liner"]There is a nicer solution I have found. We define S(C), where C is a configuration, to be the number of 1's that have an even number of 0s to the left minus the number of 1's that have an odd number of 0s to the left. Trivially S is invariant mod 3 so we are done.[/hide][/quote]
Can someone explain how to come up with this invariance?[/quote]
Bump !

[quote=pandadude]It is very hard to come up with this invariance without first thinking of the group $D_6$, however if you don't know group theory, just look at the equivalences
bwb=ww
www=bb
bww=wb
wwb=bw
Then you should be able to construct something similar to the group $D_6$, and set up some invariance.[/quote]
Can you explain it with full details, I am just learning group theory and I don’t understand. Thank you!

The dihedral group of order $2n$ is constructed by two elements, $r$ and $s$, which correspond to transformations of a regular $n$-gon. In particular, $r$ corresponds to a rotational symmetry of the $n$-gon that sends vertices to adjacent vertices, and $s$ to a rotation.
We will consider $D_6$, the symmetries of an equilateral triangle.
Interpret the state of the game as a expression by writing $r$ when we see a white marker, and $s$ when we see a black marker.
So when we start, our expression is $r\cdot r\cdot r\dots = r^n$.
When ever we make a valid move, we end up with an expression that is equal to the original in $D_6$. This is because we are replacing $r^3$ with $s^2$, $r^2s$ with $sr$ and so on, and $r^3=s^2, r^2s = sr$ in $D_6$.
Therefore, when we start with $n\equiv 1\pmod 3$ markers, our expression has the value $r^{3k+1} = r$, as the order of $r$ is $3$.
But since none of $rr, rs, sr, ss$ evaluate to $r$, it is impossible that we have two markers appearing at any point.

First, we will show that if $3\nmid n-1$, it is possible. Consider the sequence of moves $$OOOOO\to XXOO\to XOX\to OO$$
This shows that if there are 5 white markers at the end, we can remove 3 of them. Using this repeatedly, if $n\equiv 2\pmod 3$, we can get it down to 2 white markers, and if $n\equiv 0\pmod 3$, we can get it to 3, after which one move gets us down to 2 markers.

Now, we will show $3|n-1$ is impossible. Since we always flip two markers at once, the number of black markers remains even. Now, if we have a state where black markers are at positions $x_1,x_2,\ldots, x_{2k}$, and there are $m$ markers left, I define the leekiness of the arrangement to be $$m-x_1+x_2-x_3+\ldots+x_{2k}\pmod 3$$. It is not hard to show, through casework, that the leekiness is indeed an invariant. Suppose $n\equiv 1\pmod 3$. Then, its initial leekiness is $1$. So, if we get it down to 2 markers, the arrangement should still have a leekiness of 1. However, there are only 2 possible arrangements of markers at 2 markers, namely $XX$ and $OO$, and these have leekinesses of 0 and 2 respectively. Thus, $n\equiv 1\pmod 3$ is impossible.

This is essentially the same group theory solution, but I believe with a lot more motivation. In particular, the approach is direct and doesn't involve magically noticing the dihedral group of order $6$. Therefore, the part about showing $n\equiv 1\pmod{3}$ can't be done will read less like a concise solution, and more like how I came up with the solution.

Firstly, we kill the easy part by noting that
\[WWWWW\to BBWW\to BWB\to WW,\]
so we can reduce $n$ whites to $n-3$ whites for $n\ge 5$. This shows that we can resolve $3\nmid n-1$ case.

Now we'll show that if a string of $n$ whites can be reduced to a string of length $2$, then $3\nmid n-1$. After playing with this problem for a bit, one realizes that this is essentially word manipulation (in the group theory sense) with relations
\begin{align*}
    w^3&=b^2 \\
    w^2b&=bw \\
    bw^2&=wb \\
    bwb&=w^2.
\end{align*}
Indeed, let $G$ be the group with generators $b$ and $w$ with the above relations. We see that if we can reduce a string $A$ to a string $B$ using the operation of the problem, then $A$ and $B$ are equal in the group $G$. Note that the reverse implication is technically not logically true, but it's irrelelvant for this direction of the proof, since we'll just show that $w^n$ is not equal to any of $w^2,wb,bw,b^2$ in the group $G$ for $n\equiv 1\pmod{3}$.

Our goal now is to determine $G$. Combining the third and fourth relations, we have
\[b^2wb=b(bwb)=bw^2=wb,\]
so $b^2=\mathrm{id}$. We also have
\[(wb)^2=wb(bw^2)=w^3=b^2=\mathrm{id}\]
and
\[(bw)^2=(bwb)w=w^3=b^2=\mathrm{id}\]
Thus, $\boxed{b^2=(wb)^2=(bw)^2=w^3=\mathrm{id}}$. It's not hard to see using these and the original relations that any word of length four or more can be reduced to a word of length at most $3$. Furthermore, it's easy to check that all words can be reduced to some word in the set
\[\{\mathrm{id},b,w,w^2,bw,wb\}.\]
We're technically not done, since it is theoretically possible that one of these words could be further reduced through some tricky maniuplation. However it's very unlikely that using the relations given that we'd be able to reduce any of these words further, so to show that this is indeed the case, all we have to do is present a group of order $6$ that satisfies our relations. We would like $bw\ne wb$, since otherwise, our group collapses to a much smaller size. There is exactly one nonabelian group of order $6$, and that is the symmetric group on $3$ elements (or the dihedral group of order $6$). Thankfully this works just fine with say $w=(123)$ and $b=(12)$, so in fact, our group $G$ consists of the $6$ distinct elements
\[\{\mathrm{id},b,w,w^2,bw,wb\}.\]
Now, if we could show that $w^n\not\in\{b^2,w^2,bw,wb\}$ for $n\equiv 1\pmod{3}$, we'd be done. Indeed, we see that $w^3=w$, and
\[w\not\in\{b^2,w^2,bw,wb\}=\{\mathrm{id},w^2,bw,wb\}.\]
This completes this direction of the solution, so we're done.

as we can see 4 is not good and 7 is also not good so we assume like in induction that 3(k-1) +1 is also not good now we prove that 3k+1 is also not good  now if we take the 3nth white and then 3n -2 white  and the white which we get in the middle of the two black markers we get a 3(k-1) +1 sequence of white markers. which by our hypothesis is not possible.Now if we take some other arrangemets in the starting so after q moves where q>3 and less than 3n-1 we see they are the arrangements of the previous 3k +1s  hence not possible

[color=#f00][b]Inductive Construction: [/b][/color]Let $k$ be a positive integer. We provide an inductive construction for $n = 3k+2$ and $n=3k$.

Note that the base cases $n=2$ and $n=3$ are trivial. Denote a white marker by $W$, and a black marker by $B$. We represent the row of markers as a string of length $n$. Note that the original state is $$WWWWW \dots $$ We will prove that we can reduce the number of white markers by $3$ and still keep all the markers white. But this is achievable by removing the $1$st, $2$nd, and $2$nd markers from the left, in that order: $$WWWWW \dots \to BBWW \dots \to BWB \dots \to WW \dots$$ completing the induction.

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[color=#f00][b]Proof of Impossibility: [/b][/color]Let $k$ be a positive integer. We will show that $n=3k+1$ is impossible.

We set up an invariant. For each marker located $x$ places from the left, consider the function

$$ f(x) = \begin{cases} 1 & \text{if marker $x$ is white and the number of black markers to the left of $x$ is even} \\ 0 & \text{if marker $x$ is black} \\ -1 & \text{if marker $x$ is white and the number of black markers to the left of $x$ is odd} \end{cases}$$

Let $S$ be the sum of $f(i)$ for all $1 \leq i \leq n$. Note that $S \equiv 1\pmod 3$. The key observation is that after each step, $S$ is invariant$\pmod 3$.

-----
Suppose that marker $i$ was removed. Note that there are only three values of $f(i)$ that are changed: $f(i-1), f(i),$ and $f(i+1).$ Moreover, note that since $f(i-1),f(i)$ and $f(i), f(i+1)$ have opposite parities, the sum $S$ is decreased by $3 f(i)$, and so stays the same$\pmod 3$.

-----
Now suppose for the sake of contradiction that we were able to reach a state with $2$ markers. Note that the total number of black markers is always even, so in the end the two markers must either be both black or both white. Now $S \equiv 0 \pmod 3$ or $S \equiv 2 \pmod 3$, and we have our desired contradiction. $\blacksquare$

Let $P(n)=1$ if $n$ fulfills the condition and $P(n)=0$ otherwise.

The proof that $P(n)=1$ if $3\nmid n-1$ is quite simple. Note that if $n>4,$ we can remove three white pieces from the row. To do this, we utilize the following diagram, where $X$ indicates a black piece and $O$ indicates a white piece. So given any consecutive $5$ white pieces, we can do the following steps:
\[\cdots OOOOO\cdots\]
\[\cdots XXOO\cdots\]
\[\cdots XOX\cdots\]
\[\cdots OO\cdots\]
which means we now can induct from $n$ to $n+3.$ Obviously $n=2$ just works, and $n=3$ goes from $OOO$ to $XX,$ so that also works.

Now we show that $P(n)=0$ if $3\mid n-1.$ To do this, define the inverse operation as well. Note that this is an associative operation, so we just need to think about the operation $A^n$ for some integer $n,$ where $A$ is the original element (remove and reverse) and $A^{-1}$ is the inverse. So we define the infinite group of all orderings that can be achieved from $4$ white markers, $A,$ and $A^{-1}.$

Let $Q(n)$ be the function on the original operation and its inverse for $n\geq 2.$ Then $P(n)=1$ implies $Q(n)=1,$ and also note that $Q(n)=Q(n+3)$ because we can reverse our steps to add three white pieces to the row. Thus
\[Q(4)=Q(7)=\cdots.\]
Now we just show that $Q(4)=0$ by showing that we cannot get from $4$ white pieces to $2$ black pieces, $2$ white pieces, or $1$ black and $1$ white piece.

We obviously can't have $1$ black and $1$ white piece because it cannot even reach the state of all white pieces. This is because the parity of the black pieces is invariant.

Now it is obvious we can get from $2$ black or $2$ white to all white, since we went from all white to those cases. Also note that the $2$ black case is equivalent to the $3$ white case.

Now we show that it isn't possible if we start with $n\equiv 1\pmod{3}$ whites. Let the minimum number of consecutive white pieces we can remove or insert into one place be $k.$ We clearly know that $k\mid 3,$ so we just have to show $k\neq 1.$ Since each operation changes the number of pieces in the row by $1,$ and the group operation is associative, we just need to show that $A$ does not just remove a white piece and $A^{-1}$ does not just add a white piece, which is obvious.

Clearly $P(1)=0,$ so we don't even need to consider that case in our main proof.

Group theory  :maybe: .

Firstly, if $n=2,3$, then it is clearly achievable. For higher $n$ that are still $\equiv 2,3 \pmod{3}$, the series of moves
\[WWWWW\cdots \mapsto BBWW\cdots \mapsto BWB\cdots \mapsto WW\cdots\]
is valid since $n\geq 5$. This removes 3 '$W$'s, which allows us to induct downwards. $\square$.

If $n\equiv 1\pmod{3}$, it is impossible. We consider the dihedral group $D_6$, where we parse the list of white and black markers as a composition of actions on an equilateral triangle; white represents clockwise rotation and black represents swapping the bottom pair. By inspection, this satisfies
\[WWW=BB\]
\[WWB=BW\]
\[BWW=WB\]
\[BWB=WW\]
Thus, each move preserves the total transformation. Finally, since $W^{3k+1} = W \notin \{WW,BW,WB,BB\}$, we can never get to a state with two markers and we are done. $\blacksquare$.

We say a marker is a $valence$ marker if it's one of the two outer markers and a $core$ marker otherwise. We say that we operate on a marker if we remove it and let the $singleton$ be the last marker operated on. Notice that if a marker ends up black it was flipped an even under of times and white otherwise. the construction for $n=3k$ and $n=3k-1$ are easy to find.

we claim that if $n \equiv 0,1,2$ (mod $3$) then the valence markers are both black, different colors, and both white respectively. We will proceed with induction. Indeed, notice that this result clearly holds for $n=2,3,4$.  So, assume it holds for all $n<k$. This gives three cases.

Case 1: $n \equiv 0$ (mod $3$)

Notice that the singleton splits the set of markers into two groups. Since we will not remove the singleton until the end, it acts as a sort of sub-valence marker as it will not be removed until the very end. Notice that the singleton must end with its white side facing up to be removed on the last move.  Let the black marker be in position $N$ from the left.

i. If $N \equiv 0,1$ (mod $3$) then this splits this case into two sub-cases of $a \equiv 0$ and $1 \equiv 2$ (mod $3$). Notice that the $b \equiv 0$ (mod $3$) leaves the singleton flipped an even number of times and the $a \equiv 1$ (mod $3$) flipped the singleton the number of times opposite to one of the valence markers. It is known that the singleton is flipped an odd number of times since it must end up on the white side. Thus, the two valence markers were flipped an even number of times.

ii. If $N \equiv 2$ (mod 3) then this splits into two sub-cases of $a \equiv 2$ and $b \equiv 2$. Each of these cases flips the singleton an odd number of times. Flipping the singleton an even number of times in total leaves it black. Contradiction. So this case is impossible.

Case 2: $n \equiv 1$ (mod $3$)

i. If $N \equiv 1,2$ (mod $3$) then the two sub-cases are  $a \equiv 2$ and $b \equiv 0$. $b \equiv 2$ makes one valence marker white and the $a \equiv 0$ case makes the other valence marker black. This also makes the singleton white.

ii. if $N \equiv 0$ (mod $3$) then we get that $a \equiv b \equiv 1$ if the two valence markers are the same color then we flipped the singleton an even number of times. This is contradiction. Thus, the two valence markers are different colors.

Case 3: $n \equiv 2$ (mod $3$)

i. If $N \equiv 2,0$ then the two subcases are $a \equiv 1$ and $b \equiv 2$. For the singleton to be white, the valence markers are clearly both white

ii. If $N \equiv 1$ then the subcase is $a \equiv b \equiv 2$ leaves the singleton black. Contradiction.

This completes the induction. $\square$

Thus, if we assume that in $n=3k+1$ then we end with one black marker and one white marker. But, the number of black markers is clearly always even. Thus, this is impossible. $\blacksquare$

if: we can always delete 3 markers at the end if there are 5 or more at the start.
only if: suppose that a counter $x$ exists, starting from 0; from left to right, each instance of a B marker increases $x$ by $1$, and each instance of a W marker negates $x$. Then, the final value of $x$ is invariant mod 3.

I pretty much knew immediately that there would be an invariant, and this 2-liner still took me close to 3 hours. It also seems to be isomorphic to other "nicer" solutions.

[hide=solution] First, note that the two markers remaining have to be the outermost markers, as they never get removed. Therefore, we have to be able to remove all the other markers, and so we need to show that given $n$ markers, all white side up at first, if we do the same algorithm but can remove the markers on the sides as well, then it is possible to remove all markers unless $n+1$ is a multiple of $3$ (if $n=1$ in the original game then it is impossible as there is only one marker, and if $n=2$ then there are already exactly two markers). Number the positions of the markers from $1$ to $k$ (increasing from left to right), where $k$ is the number of markers left.\\
Now, we provide an inductive construction for when $3 \nmid n+1$.\\
$Base$:
If $n=1$, then we can just remove the marker (then, a leftmost and a rightmost marker have been removed an odd number of times), and if $n=3$, then we remove the leftmost marker, followed by the rightmost marker. Then, the remaining marker is white, so we can remove it. Note that for $n=3$, the leftmost marker has been removed twice, and the rightmost marker has been removed twice, and we cannot remove marker $2$ first as then both of the remaining markers are black, so in this case we must have that the leftmost and rightmost markers are both removed an even number of times.\\
$Step$:
We can remove marker $3$, followed by marker $1$. Then marker $4$ is black and marker $2$ is white. Therefore, removing marker $2$ makes marker $4$ white. Then we have $n-3$ markers remaining, so we are done by the inductive hypothesis.\\
\\
Now, we will also prove that it is impossible if $3 \mid n+1$ by induction, by at the same time proving that if $3 \mid n$, then the leftmost marker and the rightmost marker both get removed an even number of times, while if $3 \mid n-1$ then the leftmost marker and the rightmost marker both get removed an odd number of times.\\
$Base$: n=2 \\
WLOG we remove marker $1$ first. Then marker $2$ is black, and so can't be removed.\\
$Step$ $1$: $n=3k$.\\
Let the position of the last marker to be removed be $k$. Then the markers to the left of $k$ and to the right of $k$ don't affect each other, as $k$ doesn't get removed until the end.
Therefore, we cannot have that $k \equiv 0 \pmod 3$, as then it is impossible to remove all the markers to the left of $k$, and we also cannot have that $k \equiv 1 \pmod 3$, as then it is impossible to remove all of the markers to the right of $k$. Therefore, we must have that $k \equiv 2 \pmod 3$, so there are $1 \pmod 3$ markers to the left of $k$ and $1 \mod 3$ markers to the right of $k$. Therefore, by the inductive hypothesis the leftmost marker to the right of $k$ gets removed an odd number of times, and so does the leftmost marker to the right of $k$, so $k$ changes colour an even number of times and so is white. However, the leftmost marker and rightmost marker have both been removed an odd number of times. Therefore, when $k$ is removed, since when $k$ is removed it is both the leftmost and rightmost marker, in total the leftmost and rightmost markers both get removed an even number of times, as required.\\
$Step$ $2$: $n = 3k+1$.\\
In the same way as before, we must have that $k \equiv 1 \pmod 3$, and so in a similar way to before $k$ ends up being white, but the leftmost and rightmost markers have both been removed an even number of times, so as before we must have that the leftmost and rightmost markers both get removed an odd number of times.\\
$Step$ $3$: $n = 3k+2$.\\
Again, let $k$ be the last marker to be removed. We cannot have that $k \equiv 0 \pmod 3$, so we must have that on one side of $k$ there are $0 \pmod 3$ marbles and on the other side $1 \pmod 3$ marbles. Therefore, $k$ will change colour an odd number of times, and so will be black, a contradiction, and so we are done. [/hide]

Let the [i]characteristic sequence[/i] of a state of marker parity be determined in the following way: start from the left. The first number of your sequence will be the number of white markers before the first black marker. The second number will be the same number when the first black marker and the ones to the left of it are removed. For example, the characteristic sequence of BBWWBBWBBW will be $(0,0,2,0,1,0,1)$.

Note that the parity of the number of black markers is invariant, so the number of terms in the sequence is odd. If the characteristic sequence is $(a_1,a_2,\dots,a_{2k+1})$ then let the characteristic index of the sequence be $a_1-a_2+a_3-\dots +a_{2k+1}$.

Suppose we flip a WWW. Then, the characteristic sequence goes from $(\dots, a+b+3 , \dots)$ to $(\dots,a,0,b,\dots)$.

Suppose we flip a BWW. Then, the characteristic sequence goes from $(\dots,a,b,\dots)$ to $(\dots, a+1,b-2,\dots)$.

Suppose we flip a WWB. Then, the characterstic sequence goes from $(\dots,a,b,\dots)$ to $(\dots,a-2,b+1,\dots)$.

Suppose we flip a BWB. Then, the characterstic sequence goes from $(\dots,a,1,b,\dots)$ to $(\dots,a+b+2,\dots)$.

In each case, it is easy to check that the index $\pmod 3$ is invariant. This index $\pmod 3$ always starts the same as $n$. Note that the only possible endings are WW or BB since the number of black markers is always even. Clearly, the index $\pmod 3$ must be $0$ or $2$. Now, to prove that this is sufficient: $\dots$WWWWW $\to$ $\dots$WWBB $\to$ $\dots$BWB $\dots$WW and we are done by induction.

You are the dihedral group in the shape of my heart.  :love:

[b][color=pink]Claim:[/color][/b] This is impossible for $3 \mid n - 1$.
[i]Proof.[/i] Let $b, w$ be elements of $D_3$ such that $b$ is a reflection and $w$ is a rotation. Then $bwb = ww, bww = wb, wwb = bw, www = bb$.
This then implies that $w^n = w \in \{ww, wb, bw, bb\}$ which is false. $\blacksquare$

[b][color=pink]Claim:[/color][/b] This is possible for all other $n$.
[i]Proof.[/i] Note that we can easily construct the cases $n = 2$ and $n = 3$.
Then repeating the pattern $wwwww \to bbww \to bwb \to ww$ which trims off $3$ of the $w$s finishes. $\blacksquare$


geometry
circumcircle
inequalities proposed
inequalities

If $a, b, c$ are the side lenghts of a triangle, $R$ is circumradius of the triangle and $S$ its area such that $a^{2}+b^{2}+c^{2}=3$. Show that

           \[6R^{2}+S^{2}\geq 3.\]

i thought using the fact that the sum of the squares of the sides is at most
$9R^{2}$[/list][/code]


inequalities
function
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
\[ \sqrt{\frac{b\plus{}c}{2a^2\plus{}bc}}\plus{}\sqrt{\frac{c\plus{}a}{2b^2\plus{}ca}}\plus{}\sqrt{\frac{a\plus{}b}{2c^2\plus{}ab}} \ge 4.\]
:)

[quote="nguoivn"]Using Honde:
$ VT^2 \sum\ (b \plus{} c)^2(2a^2 \plus{} bc) \geq\ 8$
We only need to prove: $ \sum\ (b \plus{} c)^2(2a^2 \plus{} bc) \leq\ \frac {1}{2}$
It's easily to finish  :)[/quote]
That's it! :) As you see, the problem is so easy but nice.

Well I have simplified the given inequality in the following way:
That is By Power mean inequality for the positive reals we have
$ (\sum a^2 \plus{}9) {(\sum a^2 \plus{} 3)}^ \minus{} 1 \le 9$.

So our given inequality becomes equivalent to proving that the given LHS is $ \le$ the acquired expression.
But I do not know to finish it off.
PS I have already got a solution but brute force and very lengthy . So I thought if this method has a good finish, it would be very elegant. Can some one complete it for me .???

[quote="nguoivn"]Yes, it's so njce.
And here is my gift for the last year:
Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 2.$ Prove that:
 $ \frac {11 \minus{} a}{a^2 \plus{} 3} \plus{} \frac {11 \minus{} b}{b^2 \plus{} 3} \plus{} \frac {11 \minus{} c}{c^2 \plus{} 3} \leq\ 9$
PS: It exists an easy proof with Am-Gm
 :)[/quote]

Let assume $ 0 \leq a \leq b \leq c \leq 2$. Consider function
\[ f(x) \equal{} \frac {11 \minus{} x}{x^2 \plus{} 3}
\]
There are two properties of this function
1) $ f(x)$ is a decreasing function on interval $ \frac {2}{3} \leq x \leq 2$ . Because
\[ f'(x) \equal{} \frac {x^2 \minus{} 22x \minus{} 3}{(x^2 \plus{} 3)^2}
\]
The function $ g(x) \equal{} x^2 \minus{} 22x \minus{} 3 \leq 0$ for $ 11 \minus{} 2\sqrt {31} \leq x \leq 11 \plus{} 2\sqrt {31}$ or $ \minus{} 0.13552873\leq x \leq 22.13552873$

2) For $ x \leq \frac {7}{5}$ , we have
\[ \frac {11 \minus{} x}{x^2 \plus{} 3} \leq \frac {123}{31} \minus{} \frac {45}{31}x
\]
since
\[ LHS \minus{} RHS \equal{} \frac {1}{31} \frac {(5x \minus{} 7)(3x \minus{} 2)^2}{x^2 \plus{} 3} \leq 0
\]
and also $ f(x) \leq \frac {123}{31}$

Consider our inequality
If $ 0 \leq a \leq \frac {2}{3} \leq b \leq c \leq \frac {7}{5}$           applied second property
If $ 0\leq a\leq \frac {2}{3}\leq b \leq \frac {7}{5} \leq c \leq 2$ then
\[ f(a) \plus{} f(b) \plus{} f(c) \leq \frac {123}{31} \plus{} f(\frac {2}{3}) \plus{} f(\frac {7}{5}) \leq \frac {123}{31} \plus{} 3 \plus{} \frac {60}{31} < 9
\]
If $ 0\leq a\leq \frac {2}{3}\leq \frac {7}{5} \leq b \leq c \leq 2$  then
\[ f(a) \plus{} f(b) \plus{} f(c) \leq \frac {123}{31} \plus{} f(\frac {7}{5}) \plus{} f(\frac {7}{5}) \leq \frac {123}{31} \plus{} 2\frac {60}{31} < 9
\]
:)

[Edit]
Later, I realize that $ f(x)$ is a decreasing function on interval $ [0..2]$ so $ f(a,b,c) \leq f(a,b,b) \equal{} f(a,\frac{2\minus{}a}{2},\frac{2\minus{}a}{2}) \leq 9$
Because
\[ 9\minus{} f(a,\frac{2\minus{}a}{2},\frac{2\minus{}a}{2})  \equal{} \frac{(a^2\minus{}3a\plus{}4)(3a\minus{}2)^2}{(a^2\plus{}3)(a^2\minus{}4a\plus{}16) }\geq 0
\]

You can see the traditional proof here (of can_hang and me)  :)
http://www.maths.vn/forums/showthread.php?t=17157


logarithms
function
algebra
functional equation
algebra proposed

Find all $ f: R\minus{}>R$ such that
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ for all $ x,y \in R$  and $ f(x)f( \frac{1}{x})\equal{}1$ for all $ x \in R*$

if $ a \not \equal{} 0$,
$ \begin{array}{llll}
f(x\plus{}y) \equal{} f(x) \plus{} f(y), & f'(0)\equal{}a & \iff & f(x) \equal{} ax \\
f(x\plus{}y) \equal{} f(x) f(y), & f'(0)\equal{}a & \iff & f(x) \equal{} e^{ax} \\
f(xy) \equal{} f(x) \plus{} f(y), & f'(1)\equal{}a & \iff & f(x) \equal{} a \log |x| \\
f(xy) \equal{} f(x) f(y), & f'(1)\equal{}a & \iff & f(x) \equal{} x^a \\
\end{array}$

But, even the continuity is not assumed...

Pierre.

The statements are not correct, since $ f$ does not need to be continuous.

The additive property $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ only implies $ \forall r\in\mathbb{R},q\in\mathbb{Q}: f(qr)\equal{}q\cdot f(r)$ and there exist examples for discontinuous respectively non-linear functions over $ \mathbb{R}$ fullfilling this property. The question is, how much structure will the additional property add. Obviously, the multiplicative property implies $ \forall q\in\mathbb{Q}^\plus{}: f(\sqrt{q})^2\equal{}q$, in particular $ f(1)\equal{}\pm 1$. I assume that we can construct a non-linear function using a Hamel basis approach. This will require some kind of multiplication table for a Hamel basis over the rationals in order to control the multiplicative functional property.

[quote="Kouichi Nakagawa"]My post (functional equation) assumed that $ f(x)$ was differentiable in each domain.
I forgot to attach this 1 sentence.
sorry :([/quote]

It is then sufficient to be differentiable at $ 0$.Then we take(means the [b]definition[/b] of f at x , that $ f(x) \equal{} ax , f'(0) \equal{} a$.The secont condition gives the functions $ f(x) \equal{} x , f(x) \equal{} \minus{}x$.

actually, it will be sufficient to have continuity at one point to conclude linearity

P.S. a non-linear function over $ \mathbb{Q}(\sqrt{2})$, which fulfills the properties, is $ f(a\plus{}b\sqrt{2})\equal{}a\minus{}b\sqrt{2}$

[quote="ThangTongHopK41"]Find all $ f: R \minus{} > R$ such that
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ for all $ x,y \in R$  and $ f(x)f( \frac {1}{x}) \equal{} 1$ for all $ x \in R*$[/quote]

We only need prove exist a constant c such that $ |f(x)|<c,\forall x\in [0,1]$.
It is quite easy from AM-GM inequalities.
So $ f(x)=ax$
So $ a^2=1$ . It mean that ${ a\in\{-1,1}$

sometimes, it pays to ponder more thoroughly over it: For $ 0\not\equal{}xh(x\plus{}h)$ we get
$ \begin{array}{lcl} 0 & \equal{} & \frac{1}{f(x)}\minus{}\frac{1}{f(x\plus{}h)}\minus{}f\left(\frac{1}{x}\minus{}\frac{1}{x\plus{}h}\right)\equal{}\frac{f(h)}{f(x)^2\plus{}f(x)f(h)}\minus{}f\left(\frac{h}{x^2\plus{}xh}\right)\\
 & \equal{} & \frac{f(h)}{f(x)^2\plus{}f(x)f(h)}\minus{}\frac{1}{f\left(\frac{x^2}{h}\right)\plus{}f(x)}\end{array}$

which implies $ \frac{f(x)^2}{f(h)}\equal{}f\left(\frac{x^2}{h}\right)$, i.e. for all $ y,h>0$ we get: $ f(hy)\equal{}f(h)f(y)$. Using the earlier found symmetries of $ f$ and the knowledge of its zeroes, we get for abritrary $ y,h$ the property $ f(hy)\equal{}f(h)f(y)$


function
logarithms
ratio
real analysis
real analysis unsolved

Calculate
\[ \sum_{n \equal{} 1}^{\infty}{\frac {\left(\frac {3 \minus{} \sqrt {5}}{2}\right)^n}{n^3}}
\]
[hide="hint"]
If $ \phi (x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^3}{n^3}},\; |x|\leq\frac {1}{2},$
then $ \phi\left(\frac {x}{x \minus{} 1}\right) \plus{} \phi (x) \plus{} \phi (1 \minus{} x) \minus{} \phi (1) \equal{} \frac {\pi^2}{6}\ln(1 \minus{} x) \plus{} \frac {1}{6}\ln^2 (1 \minus{} x)\left(\ln(1 \minus{} x) \minus{} 3\ln x\right)$
[/hide]

Let $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^n}{n^3}}$. So, it converges evenly. You will have:
$ f'(x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^{n \minus{} 1}}{n^2}}$.
$ xf'(x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^{n}}{n^2}}$.
$ xf''(x) \plus{} f'(x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^{n \minus{} 1}}{n}}$.
$ x^2f''(x) \plus{} xf'(x) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {x^{n}}{n}}$.
$ (x^2f''(x) \plus{} xf'(x))' \equal{} \sum_{n \equal{} 1}^{\infty}{x^{n \minus{} 1}} \equal{} \frac {1}{1 \minus{} x}$.
$ x^2f'''(x)\plus{}3xf''(x)\plus{}f'(x)\equal{}\frac{1}{1\minus{}x}$.
So, remember that $ f(0)\equal{}f'(0) \equal{} 0$ and add some condition for solving this differential equation.
Is it clear?

The only little (but rather annoying) problem is that the solution is not an elementary function (if it were, we would know $ \zeta(3)$ explicitly) ;). So, no, this is not clear.

Yes, we should not solve this differential equation, it has no solution in elementar functions as fedja says. Solution based on the [b]hint[/b] and on fact that $ x \equal{} \frac {3 \minus{} \sqrt {5}}{2}.$ Try to go this way.

Let $ z$ be a positive number.

$ \text{Li}_1(\minus{}z)\minus{}\text{Li}_1\left(\minus{}\frac{1}{z}\right)\equal{}\minus{}\ln z$

Divide both sides by $ z$ and integrate with respect to $ z$ from $ 1$ to another $ z$ to obtain

$ \text{Li}_2(\minus{}z)\plus{}\text{Li}_2\left(\minus{}\frac{1}{z}\right)\equal{}\minus{}\frac12 \ln^2 z\minus{}\frac{\pi^2}{6}$ .

Do this again to get

$ \text{Li}_3(\minus{}z)\minus{}\text{Li}_3\left(\minus{}\frac{1}{z}\right)\equal{}\minus{}\frac16 \ln^3 z\minus{}\frac{\pi^2}{6}\ln z$ .

In particular for $ z\equal{}\phi\equal{}\frac{1\plus{}\sqrt{5}}{2}$ (golden ratio)

$ \text{Li}_3(\minus{}\phi)\minus{}\text{Li}_3\left(\minus{}\frac{1}{\phi}\right)\equal{}\minus{}\frac16 \ln^3\phi\minus{}\frac{\pi^2}{6}\ln \phi \quad (*)$ .

Now we need the trilogarithm identity in your hint, already shown [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=226482]here[/url], namely

$ \text{Li}_3(z)\plus{}\text{Li}_3(1\minus{}z)\plus{}\text{Li}_3\left(1\minus{}\frac{1}{z}\right)\equal{}\zeta(3)\plus{}\frac16 \ln^3 z\plus{}\frac{\pi^2}{6} \ln z\minus{}\frac12 \ln^2 (z)\, \ln(1\minus{}z)$ .

For $ z\equal{}\frac{1}{\phi^2}$ we obtain

$ \text{Li}_3\left(\frac{1}{\phi^2}\right)\plus{}\underbrace{\text{Li}_3\left(1\minus{}\frac{1}{\phi^2}\right)}_{\text{Li}_3\left(\frac{1}{\phi}\right)}\plus{}\underbrace{\text{Li}_3\left(1\minus{}\phi^2\right)}_{\text{Li}_3(\minus{}\phi)}\equal{}\zeta(3)\plus{}\frac23 \ln^3 \phi\minus{}\frac{\pi^2}{3} \ln \phi$ .

And when we substitute $ \text{Li}_3(\minus{}\phi)$ by formula $ (*)$ then

$ \text{Li}_3\left(\frac{1}{\phi^2}\right)\plus{}\underbrace{\text{Li}_3\left(\frac{1}{\phi}\right)\plus{}\text{Li}_3\left(\minus{}\frac{1}{\phi}\right)}_{\frac14 \text{Li}_3\left(\frac{1}{\phi^2}\right)}\equal{}\zeta(3)\plus{}\frac56 \ln^3 \phi\minus{}\frac{\pi^2}{6} \ln \phi$ .

Thus $ \text{Li}_3\left(\frac{1}{\phi^2}\right)\equal{}\frac45 \zeta(3)\plus{}\frac23 \ln^3\phi\minus{}\frac{2\pi^2}{15} \ln\phi$ .

Here is mine:
Let $ x \equal{} \frac {3 \minus{} \sqrt {5}}{2}.$ Then $ (x \minus{} 1)^2 \equal{} x$ and $ \frac {x}{x \minus{} 1} \equal{} x \minus{} 1.$ After substitution $ v \equal{} 1 \minus{} x$ our hint will be $ \phi (x) \plus{} \phi (v) \plus{} \phi ( \minus{} v) \equal{} \phi (1) \plus{} \frac {\pi^2}{6}\ln v \minus{} \frac {5}{6}\ln^3 v.$
Using representation of $ \phi (x)$ as series we get: $ \phi (v) \plus{} \phi ( \minus{} v) \equal{} \frac {2}{2^3}\phi (v^2) \equal{} \frac {1}{4}\phi (x).$
Now using $ v \equal{} \frac {\sqrt {5} \minus{} 1}{2}$ and $ \phi (1) \equal{} \zeta (3)$ we get $ \sum_{n \equal{} 1}^{\infty}{\frac {\left(\frac {3 \minus{} \sqrt {5}}{2}\right)^n}{n^3}} \equal{} \frac {4}{5}\zeta(3) \plus{} \frac {2\pi^2}{15}\ln\left(\frac {\sqrt {5} \minus{} 1}{2}\right) \minus{} \frac {2}{3}\ln^3\left(\frac {\sqrt {5} \minus{} 1}{2}\right).$


Hello, when I use the command line form of Asymptote in linux (command asy), the output is in eps format and it looks exactly the way I want it to. But when I put outformat = "pdf" in config.asy, it says a bunch of error messages:
[code]
/usr/local/share/asymptote/plain_shipout.asy: 70.3: runtime: )

! LaTeX Error: File `ifpdf.sty' not found.

Type X to quit or <RETURN> to proceed,
or enter new name. (Default extension: sty)

Enter file name: ! Improper alphabetic constant.
<to be read again>
                   \spacefactor
l.16 \RequirePackage
                    {ifdraft}
! Improper \spacefactor.
<recently read> \spacefactor

l.16 \RequirePackage
                    {ifdraft}
! Missing $ inserted.
<inserted text>
                $
l.16 \RequirePackage
                    {ifdraft}

! LaTeX Error: Missing \begin{document}.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...

l.16 \RequirePackage
                    {ifdraft}
(/usr/share/texmf-texlive/tex/latex/tools/.tex File ignored)
! Improper alphabetic constant.
<to be read again>
                   \spacefactor
l.16 \RequirePackage
                    {ifdraft}
! Improper \spacefactor.
<recently read> \spacefactor

l.16 \RequirePackage
                    {ifdraft}
(/usr/share/texmf-texlive/tex/latex/ifdraft/ifdraft.sty
[/code]

I already installed ifdraft.sty and ifpdf.sty. It definitely has something to do with the package three because all 2D images render perfectly. One quirk though: the eps file looks exactly like what I want, so is there any way to configure Kile so that it embeds an EPS file rather than a pdf file that is output by Asymptote?

Thanks!

It'd help if you posted the code which gave the error, as well. :)

As for embedding eps: use latex instead of pdflatex. This will output a .dvi file on which you'll need to run dvi2ps to convert the .dvi to a .ps file. The necessary buttons should be in the toolbar.


geometry
circumcircle
trigonometry
parallelogram
Pythagorean Theorem
area of a triangle
geometry proposed

Since $O_9$ is the midpoint of the segment $OH$ between the circumcenter $O$ and orthocenter $H$, the segment $AO_9$ is a median of the triangle $\triangle AOH$. The length of this median is given by the equation

$AO_9^2 = \frac 1 4 (2AO^2 + 2AH^2 - OH^2)$

Let $a = BC$, $b = CA$, $c = AB$ be the triangle sides, $\alpha = \angle BAC$, $R  = OA$ the circumradius and $\triangle = \frac 1 2 bc\ \sin \alpha$ the triangle area. Then

$OH^2 = 9R^2 - (a^2 + b^2 + c^2)$

Using the cosine theorem for the triangle $\triangle ABC$,

$OH^2 = 9R^2 - 2(a^2 + bc\ \cos \alpha) =  9R^2 - 2(a^2 + 2 \triangle \cot \alpha)$

Let $D$ be the intersection of the line $CO$ with the circumcircle $(O)$ other than the vertex $C$. The angles $\angle CAD = \angle CBD = 90^o$ are the both right. $AD$ is parallel to the altitude from the vertex $B$ (both are perpendicular to $CA$) and $BD$ is parallel to the altitude from the vertex $A$ (both are perpendicular to $BC$). Hence, $AHBD$ is a parallelogram and $AH = BD$. By Pythagorean theorem for the right angle triangle $\triangle BCD$,

$BD^2 + BC^2 = 4R^2$

$AH^2 + a^2 = 4R^2$

Substituting for $AO^2$, $AH^2$ and $OH^2$ into the equation for the median $AO_9$,

$AO_9^2 = \frac 1 4 [2R^2 + 2(4R^2 - a^2) - 9R^2 + 2(a^2 + 2 \triangle \cot \alpha)] = \triangle \cot \alpha + \frac{R^2}{4}$

[quote="Leon"]Let $ O_9$, $ R$, $ \Delta$ denote the center of the ninepoint circle, the circumradius and the area of the triangle $ \triangle ABC$ respectively. Show that:
\[ AO_9 \equal{} \sqrt {\Delta \cot A \plus{} \frac {{R^2 }}{4}}
\]
[/quote]

There is also a synthetic proof:

Let E be the midpoint of the side CA of triangle ABC, and let Y be the foot of the altitude issuing from the vertex B. It is well-known that the area of a triangle equals the halved length of a side, multiplied with the corresponding altitude; thus, the area of triangle ABC equals the halved length of the side CA, multiplied with the corresponding altitude BY. In other words, $ \Delta \equal{} \frac12\cdot CA\cdot BY$. Since the point E is the midpoint of the side CA, we have $ AE \equal{} \frac12\cdot CA$, so that

$ AE\cdot BY \equal{} \frac12\cdot CA\cdot BY \equal{} \Delta$.

Now, in the right-angled triangle AYB, we have $ AY \equal{} BY\cdot\cot\measuredangle YAB \equal{} BY\cdot\cot A$. Thus,

$ AE\cdot AY \equal{} AE\cdot \left(BY\cdot\cot A\right) \equal{} \left(AE\cdot BY\right)\cdot\cot A \equal{} \Delta\cdot\cot A$.

But the points E and Y lie on the nine-point circle of triangle ABC (since E is the midpoint of a side and Y is the foot of an altitude of this triangle). Hence, $ AE\cdot AY$ is the power of the point A with respect to the nine-point circle of triangle ABC. On the other hand, the power of the point A with respect to the nine-point circle of triangle ABC equals $ AO_9\ ^2 \minus{} \left(\frac {R}{2}\right)^2$, since $ O_9$ is the center and $ \frac {R}{2}$ is the radius of the nine-point circle of triangle ABC. Hence, we have

$ AO_9\ ^2 \minus{} \left(\frac {R}{2}\right)^2 \equal{} AE\cdot AY$.

And since $ AE\cdot AY \equal{} \Delta\cdot\cot A$, this becomes

$ AO_9\ ^2 \minus{} \left(\frac {R}{2}\right)^2 \equal{} \Delta\cdot\cot A$, and thus

$ AO_9\ ^2 \equal{} \Delta\cdot\cot A \plus{} \left(\frac {R}{2}\right)^2 \equal{} \Delta\cdot\cot A \plus{} \frac {R^2}{4}$,

so that $ AO_9 \equal{} \sqrt {\Delta\cdot\cot A \plus{} \frac {R^2}{4}}$.

QED.

  darij


abstract algebra
linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

Let $ H$ be an additive, finite abelian group with generators $ x,y,z$ satisfying the relations

\[ 10y\plus{}5z\equal{}5x\plus{}10z\equal{}10x\plus{}5y\equal{}0.\]
Express $ H$ as a product of finite cyclic groups.

I represented the above system in matrix form and, by column and row additions and interchanges, reduced it to

\[ \begin{pmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 25\end{pmatrix},\]
so that $ H\simeq\mathbb{Z}_{5}\oplus\mathbb{Z}_{5}\oplus\mathbb{Z}_{25}$.  Is this the correct method?

It's the right method, but I think the answer is a little off...

Thank you for the reply.  Perhaps my original matrix representation was off.  I wrote it as

\[ \begin{pmatrix}5 & 0 & 10 \\ 0 & 10 & 5 \\ 10 & 5 & 0\end{pmatrix}.\]
Should the matrix be different?

The original one is correct, but the derived one isn't. Their determinants don't even agree (that $ 25$ should be a $ 45$).


calculus
integration
analytic geometry
trigonometry
geometry
logarithms
calculus computations

Let $ p,\ q\ (p < q)$ be the $ x$ coordinates of curves $ x^2 \plus{} y^2 \equal{} 1,\ y\geq 0$ and $ y \equal{} \frac {1}{4x}.$

(1)  Find $ \alpha ,\ \beta$ such that $ \cos \frac {\alpha}{2} \equal{} p,\ \cos \frac {\beta }{2} \equal{} q\ (0 < \alpha < \pi ,\ 0 < \beta < \pi).$

(2) Find the area bounded by the curves.

To simplify $ x,y >0$

The intersection of the 2 curves are the points

$ \left\{\frac{1}{4} \left(\minus{}\sqrt{2}\plus{}\sqrt{6}\right),\frac{1}{4}
   \left(\sqrt{2}\plus{}\sqrt{6}\right)\right\}\land \left\{\frac{1}{4}
   \left(\sqrt{2}\plus{}\sqrt{6}\right),\frac{1}{4} \left(\minus{}\sqrt{2}\plus{}\sqrt{6}\right)\right\}$

The area is the double of:

$ \int_{\frac{1}{4} \left(\sqrt{6}\minus{}\sqrt{2}\right)}^{\frac{1}{4} \left(\sqrt{2}\plus{}\sqrt{6}\right)}
   \left(\sqrt{1\minus{}x^2}\minus{}\frac{1}{4 x}\right) \, dx\equal{}\frac{1}{4} \log \left(2\minus{}\sqrt{3}\right)\plus{}\frac{\pi }{6}$

Note: To evaluate this integral is a little bit troublesome, but not impossible (in five lines can be donne), just use the fact that $ \cos ^{\minus{}1}\left(\frac{1}{4} \left(\minus{}\sqrt{2}\plus{}\sqrt{6}\right)\right)\minus{}\cos ^{\minus{}1}\left(\frac{1}{4}
   \left(\sqrt{2}\plus{}\sqrt{6}\right)\right)\equal{}\frac{\pi }{3}$


trigonometry
inequalities unsolved
inequalities

[quote="N.T.TUAN"][hide]Try setting $ x=\tan\frac{A}{2},...$, where $ A,B,C$ are angles of a triangle![/hide][/quote]

Yes its the idea.

But how can we proove this:
$ tg(A/2)=\frac{1}{tg(\frac{B+C}{2})}$ $ \Rightarrow$ $ A+B+C=\Pi$

[quote="N.T.TUAN"]$ \tan\frac{A}{2}=\tan (\frac{\pi}{2}-\frac{B+C}{2})$, now note that $ 0<A,B,C<\pi$.[/quote]

Yes, thanks.


integration
limit
calculus
real analysis
real analysis unsolved

For $p\in N^{*}$ define the sequence $a_{n}=\int_{0}^{1}x^{pn}\left(1+x^{p}\right)^{n}\ dx$ , where $n\in N^{*}\ .$ Prove that $\lim_{n\to\infty}\frac{na_{n}}{2^{n}}=\frac{2}{3p}\ .$ Study the cases $p\in \overline{1,2}\ .$

Here p and n are positive real.
 y=x^p implies a_n=1/p*I(n,1/p-1) with I(n,c)=
int(y^n*(1+y)^n*y^c,y=0..1).
1°) We show that I(n,0)~(2^(n+1))/(3n) .
u=y*(1+y) implies I(n,0)=int((u^n)/sqrt(1+4u),u=0..2).
It is easy to show that I(n,0)-int((u^n)/3,u=0..2)=
O((2^n)/n^2 ) and int((u^n)/3,u=0..2)~(2^(n+1))/(3n).
2°) We show that I(n,c)~I(n,0).
Let a in ]0,1[. We suppose for example c>0.
It is enough to show that
J(n,c)=int(y^n*(1+y)^n*(y^c-1),y=a..1)=o((2^n)/n).
J(n,c)<=int((1+y)^n*(y^c-1),y=a..1).
We conclude with an integration by parts: the bracket is in
O((1+a)^n) and the integral is in O((2^n)/n^2).

[quote="loup blanc"]Here p and n are positive real.
 $y=x^{p}\Rightarrow a_{n}=\frac{1}{p}I\left( n,\frac{1}{p}-1\right)$ with $I(n,c)= \int_{0}^{1}y^{n}(1+y)^{n}y^{c}\ dy$
1°) We show that $I(n,0)\sim \frac{2^{n+1}}{3n}$ .
$u=y(1+y) \Rightarrow I(n,0)= \int_{0}^{2}\frac{u^{n}}{\sqrt{1+4u}}\ du$.
It is easy to show that $I(n,0)-\int_{0}^{2}\frac{u^{n}}{\sqrt{1+4u}}\ du = \mathcal{O}\left( \frac{2^{n}}{n^{2}}\right)$ and $\int_{0}^{2}\frac{u^{n}}{3}\ du \sim \frac{2^{n+1}}{3n}$.
2°) We show that $I(n,c)\sim I(n,0)$.
Let a in ]0,1[. We suppose for example c>0.
It is enough to show that
$J(n,c)=\int_{a}^{1}y^{n}(1+y)^{n}(y^{c}-1)\ dy=o\left( \frac{2^{n}}{n}\right)$.
$J(n,c)\leq \int_{a}^{1}(1+y)^{n}(y^{c}-1)\ dy$.
We conclude with an integration by parts: the bracket is in
$\mathcal{O}\left( (1+a)^{n}\right)$ and the integral is in $\mathcal{O}\left( \frac{2^{n}}{n^{2}}\right)$.[/quote]
I hope I translated it faithfully, (I speak maple).

  --Ben