function
absolute value
complex analysis
complex analysis unsolved

Hey all. I'm having trouble with the following problem:

Let f(z) be an entire function. Let a be an arbitrary real-valued constant.

Show that if Re(f(z)) is less than or equal to a for all z, then f is constant.
Show that if Re(f(z)) is greater than or equal to a for all z, then f is constant.

I'm pretty sure I need to use Liouville's Theorem, but I don't see how Re(f(z))<=a implies |f(z)|<=a.

Thanks!

[hide="Hint"]$ \text{Re}f(z)$ is a harmonic function on the plane and use the fact that any harmonic function which is bounded from above (or below) is constant.  :wink: [/hide]

If you need to use Liouville's Theorem then apply it to the function $ g(z)\equal{}Exp[f(z)]\equal{}Exp[u\plus{}iv]$.   Absolute value of $ g(z)$ then is $ |Exp(u)|$ but that's bounded since u is by statement of theorem.  Therefore $ |Exp(u)|\leq M$.  Now, using Liouville's theorem, we can say:

$ |g'(z_0)|\leq \frac{M}{R}$ for ANY R and $ z_0$ so $ |g'|$ tends to 0 for R large enough which implies g(z)=C.  Only way for that to happen is if u and v are constants.  Same dif for the other one instead use Exp[-f].

I think an interesting problem would be:  describe the intrinsic geometric property of analytic functions which causes this behavior.


calculus
derivative
calculus computations

can someone help me derive .5(x radical (1-x^2) + arcsinx)

In other words, evaluate $\frac{d}{dx}\ \frac{1}{2}(x\sqrt{1-x^2} + \arcsin x)$.

Masoud Zargar


Fie $a, b, c$ trei numere reale strict pozitive astfel incat $ab+bc+ca=1$. Sa se arate ca \[{\sqrt [3]{\frac{1}{a}+6b}+\sqrt [3]{\frac{1}[b}+6c}+\sqrt [3]{\frac{1}{c}+6a}\leq\frac{1}{abc}.\]

Nu stiu cat de corecta e solutia de mai jos, dar..macar am incercat.
[i]Solutie[/i]
  Cu Inegalitatea Mediilor(MA-MG) obtinem relativ usor inegalitatea $\sqrt [3]{\frac{1}{a}+6b}\leq\frac{\frac{1}{a}+6b+2}{3}$ si analoagele.Adunandu-le, aflam ca $\ LHS\leq\frac{1}{3}\left(\sum\frac{1}{a}+6\sum a+6\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\sum a+2\leq\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\sum a+2(a^{2}+b^{2}+c^{2})$.Aici am folosit ca $\ 1=ab+bc+ca\leq\ a^{2}+b^{2}+c^{2}$.Pentru a arata ca $\ LHS\leq\frac{1}{abc}$ e necesar ca inegalitatea $\ (a+b+c)(a^{2}+b^{2}+c^{2})\leq\frac{1}{3abc}$ sa fie adevarata, adica $\ a^{2}bc+ab^{2}c+abc^{2}+a^{3}bc+ab^{3}c+abc^{3}-\frac{1}{3}\leq\ 0$.Sa notam membrul stang al inegalitatii cu $\ f(a,b,c)$.Trebuie sa demonstram ca $\ f(a,b,c)\leq\ 0$, ceea ce e echivalent cu $\Delta_{f(a,b,c)}\geq\ 0$.Dar discriminantul functiei $\ f(a,b,c)$ este $\Delta=(b^{2}c+bc^{2}+b^{3}c+bc^{3})^{2}+\frac{4}{3}(abc+bc)$, care este evident pozitiv, caci $\ a,b,c>0$.Cu asta problema e rezolvata. :)

Aplicam $\sum \sqrt[3]{\frac{1}{a}+6b}\leq\sqrt[3]{9\sum\frac{1}{a}+54\sum a}$.Acuma $\frac{1}{abc}\geq\sqrt[3]{\frac{27}{abc}}$ si se reduce la $3abc\sum a\leq 1$ :D

Bine mai VLadutz mai! :wink:
Vezi ca se mai pot da cel putin inca 3-4 solutii. :)

am gasit si eu o solutie
sper ca e bine si ca n`am incurcat semne :)
$LHS=\sum\frac{\sqrt[3]{\left(\frac{1}{a}+6b\right)\cdot 3\sqrt{3}\cdot 3\sqrt{3}}}{3}$.
aplicand inegalitatea mediilor, obtinem ca
$LHS\leq\sum\frac{1}{9}\left(18\sqrt{3}+\sum\frac{1}{a}+6\sum a\right)$.
apoi aducem la acelasi numitor in suma obtinuta, si vom obtine ca
$LHS\leq\frac{1}{9abc}\left(18\sqrt{3}abc+1+6abc\sum a\right)$.
mai ramane sa aratam ca
$18\sqrt{3}abc+1+6abc\sum a\leq 9$.
cum $3abc\sum a\leq 1$, si $abc\leq\frac{1}{3\sqrt 3}$, problema este rezolvata.
avem egalitate atunci cand $a=b=c=\frac{\sqrt{3}}{3}$.


does taking the old practice test + mock tets help? im sure they help to some degree, but what about after that? are there tests/books that will help to improve on one's aime/amc score?

old and mock tests are DEFINITELY helpful. you should take advantage of having such wonderful resources... if you run out,
then the book i would recommend is AoPS book1. Or if u think u know everything on book1, which i doubt (perusing through your topics of questions), then you can buy book2.

ty for your info. do you know where i might be able to find past test of amc/aime?

for AIME: http://www.kalva.demon.co.uk

for AMC: go to google and type "Harold Reiter"
              he should have AMC's on his website.

Actually, in my experience, all but 1 or 2 problems can be solved using techniques in only Volume 1. You don't need to even do Volume 2 for a score of 13 or so.

Practice. Can you do all the problems on the old contests? If not, stop asking for new sources and learn how to do all the old problems.

And if you want to buy the paper versions of the AIMEs, AMC 8/10/12s etc, based of their supply, you can order from the official AMC website:(just browse around the site) [url]http://www.unl.edu/amc/[/url]


μπορεί κανείς αναλυτικά να περιγράψει πως γίνεται ο παρακάτω πολλαπλασιαμός μεταθέσεων;

(1,4,5,6)(2,1,5)

εχω μπερδευτεί βασικά γιατί οι δυο κύκλοι έχουν κοινά στοιχεία και το βιβλίο δίνει το αποτέλεσμα χωρίς να εξηγεί τίποτα.

Σύνθεση συναρτήσεων είναι βασικά...
Πρώτα ενεργεί ο κύκλος από "δεξιά" και μετά ο άλλος.
ας πούμε το 1 πάει στο 5 με την δεύτερη μετάθεση και το 5 με την πρώτη πάει στο 6 άρα το 1 πάει στο 6κτλ.

Ευχαριστώ πολύ για την απάντηση σου. Με βοήθησε πολύ. Κάτι τέτοιο δεν αναφέρεται στο βιβλίο και δεν ήξερα πως γίνεται ο πολλαπλασιασμός. Τώρα κατάλαβα πλήρως.


calculus
algebra
polynomial
limit
function
calculus computations

I had essentially this problem assigned as homework for my Calculus III class for today. It proved a tad bit more challenging than they really appreciated.

Suppose $ f(x)\equal{}\begin{cases}0&x\le 0\\ P(x)&0<x<1\\ 1&x\ge 1\end{cases}\equal{}$

subject to the curve $ y\equal{}f(x)$ in $ \mathbb{R}^2$

(1) is continuous,

(2) has continuous unit tangent direction, and

(3) has continuous curvature, with

(4) P(x) being a polynomial.

Find the polynomial $ P(x)$ of lowest possible degree that satisfies these conditions.

[color=green][The problem in the textbook was phrased in terms of designing a section of railroad track. The textbook also specified the degree of the polynomial; I'm upping the stakes slightly by asking you to determine the lowest possible degree.][/color]

$ \lim_{x\to 0} P(x)\equal{}\lim_{x\to 0} P'(x)\equal{}0 \to P(x)\equal{}x^2Q(x)$
$ \lim_{x\to 1} P(x)\equal{}1\to \lim_{x\to 1} Q(x)\equal{}1\to Q(1)\equal{}1$
$ \lim_{x\to 1} P'(x)\equal{}\lim_{x\to 1}  x^2Q'(x)\plus{}2xQ(x)\equal{}0\to Q'(1)\equal{}\minus{}2$

letting $ Q(x)\equal{}Q(1)\plus{}Q'(1)(x\minus{}1)\plus{}Q''(1)\frac{(x\minus{}1)^2}{2}\plus{} ...$
we can obtain a minimal solution by assuming
$ Q(x)\equal{}1\minus{}2(x\minus{}1)\equal{}3\minus{}2x \to P(x)\equal{}3x^2\minus{}2x^3$

but i didnt check (2) and (3)

If the function behaves like $ cx^2$ near zero, then the curvature is [i]not[/i] continuous. Hence I do not accept that as a solution.


function
logarithms
floor function
algebra proposed
algebra

find all functions $ f: \mathbb{R}\equal{}\equal{}\equal{}>\mathbb{R}$ such that:
$ (\forall x\in \mathbb{R})$  $ f(3x)\equal{}2f(x)$

 :wink:

[quote="greatestmaths"]find all functions $ f: \mathbb{R} \equal{} \equal{} \equal{} > \mathbb{R}$ such that:
$ (\forall x\in \mathbb{R})$  $ f(3x) \equal{} 2f(x)$

 :wink:[/quote]

This is a very well known problem.

A general solution is :

Let $ h_1(x)$ and $ h_2(x)$ any two functions defined for $ x\in[0,1)$.
Then :
$ f(0) \equal{} 0$

$ f(x) \equal{} x^{\frac {\ln(2)}{\ln(3)}}h_1(\{\frac {\ln(x)}{\ln(3)}\})$ $ \forall x > 0$

$ f(x) \equal{} ( \minus{} x)^{\frac {\ln(2)}{\ln(3)}}h_2(\{\frac {\ln( \minus{} x)}{\ln(3)}\})$ $ \forall x < 0$

Where $ \{a\}\equal{}a\minus{}\lfloor a\rfloor$ is the fractional part of $ a$

[quote="greatestmaths"]would you do it without  $ In$[/quote]

No, I'm sorry.

My proposal is a general solution :

1) every function built as I suggest is a solution
2) any solution can be put in the form I suggest.

But I don't know any general solution not involving logarithms (although it maybe exists).

I'm interested in your own solution.

observe that  $ f(x)\equal{}2^{n}f(x/3^{n})$ :lol:

[quote="greatestmaths"]observe that  $ f(x) \equal{} 2^{n}f(x/3^{n})$ :lol:[/quote]

And then what ?
This leads nowhere.
You just can conclude that $ f(x/3^{n})$ may be as few as you want but you can conclude nothing because you have no information about continuity.

And $ f(0)\equal{}0$ is an immediate conclusion from $ f(3\times 0)\equal{}2f(0)$

So, could you give us, as I requested in my previous post, a general solution without using logs ?

Thanks in advance.


Some good problems:

1. There are a white, b black, and c red chips on a table. In one step, you may choose two chips of different colors and replace each by a chip of the third color. Find conditions for all chips to become the same color.

2. n numbers are written on a blackboard. In one step you may erase any two of the numbers, say a and b, and write instead (a+b)/4. Repeat until there is one number left. Prove that, initially, if there were n ones on the board, at the end, a number, which is not less than 1/n will remain.

First question

[hide]WLOG, assume we will be left with all whites. Consider the three possible moves:

(type 1): (a,b,c) -> (a+2,b-1,c-1)

(type 2): (a,b,c) -> (a-1,b+2,c-1)

(type 3): (a,b,c) -> (a-1,b-1,c+2)

In each, the value of b-c remains unchanged mod 3. If the end position of b=c=0 were reachible, then that invariant value of b-c would be 0, meaning we MUST have b-c divisible by 3 if the desired result were to be reachible.

On the other hand, if we start with a position (a,b,c) where b :equiv: c mod 3, then I will prove that we can reach (a+b+c,0,0). WLOG, we may assume b>=c. First, apply type 1 moves as many times as possible to obtain (a+2c,b-c,0). If b-c = 0, we're done. If not, I will give a process to change from position (x,y,0) to (x+3,y-3,0), which must eventually take us to (a+b+c,0,0) since b-c is divisible by 3.

If we start with the position (x,y,0) where x>0, then we make a type 3 move to get to (x-1,y-1,2), and then 2 type 1 moves to get to (x+3,y-3,0). This is the desired sequence of moves, and it is valid as long as a+2c>=1. If it weren't, then we would be at the position (0,b,0) and we'd already be done.

To recap, we start at (a,b,c), move to (a+2c,b-c,0) through repeated type 1 moves, and then decrease the #black by 3 with the above process until this number goes to 0, which it must because b-c :equiv: 0 mod 3.[/hide]

Sorry if it's unclear or repetitive. I'm tired.

The second question:

[hide]We will proceed by induction.  Consider the final step: we have two numbers, a and b, which are converted to (a + b)/4.  Now, the a comes from some number of 1's -- let us say m of them.  Then b comes from (n - m) 1's.  We have m in {1, 2, ..., n - 1}.  So, by inductive hypothesis, a :ge: 1/m, b :ge: 1/(n - m).  Thus, our final number (1/m + 1/(n - m))/4 = n/(4*m*(n - m))

Now, (n - 2m)^2 :ge: 0.

n^2 - 4mn + 4m^2 :ge: 0

n^2 :ge: 4mn - 4m^2

n/(4mn - 4m^2) :ge: 1/n.

Q.E.D.[/hide]


[quote="7h3.D3m0n.117"][quote="ckck"]Find the distance between the point(2002,2003) and the line 3x-2y=0.[/quote]

since the line isn't horizontal or vertical can't the distance change depending on the point chosen on the line?[/quote]
I think they usually mean the shortest distance.

[quote="7h3.D3m0n.117"][quote="ckck"]Find the distance between the point(2002,2003) and the line 3x-2y=0.[/quote]

since the line isn't horizontal or vertical can't the distance change depending on the point chosen on the line?[/quote]

The distance between the line and the point is the length of the segment from the point which is perpendicular to the line.  So rearranging the equation

[hide]
$y=3x/2$

So the slope will be the negative reciprocal, or $-2/3$.  Now we can find the y-intercept.

$2003=2002(-2/3)+b$

Then solve for b, find the intersection between the two lines, then use the distance formula to find the distance.[/hide]

What Aegor said works, and that's how I would solve it, but there is also a formula:
Distance from point $(x_{1},y_{1})$ to line $Ax+By+C=0$: $d=\frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$

Yea, what anyone said is how i did it.  off '03 hoover, right?  :)

and chris, im pretty sure you knew this? maybe not.

aegors way works, but when you get some weird slopes, it can be a tad confusing, at least for me.

I tend to forget formulas, so I do problems like that with Aegor's way. With formulas (or at least that one), I can't really tell what I'm doing.

Oh.  I don't know all of the tips and tricks, so I guess I learned something new today (i.e., the formula).  I try not to use things until I derive them myself, so thanks for letting me know.


recently we have created tmf the MATH forum. i know what your thinking.. but we aops! yea but does aops have the game battleships? or presidental paintball?? and soon to come chess checkers and board games!!!!! we also have a chatroom where you can block but cant get muted!!! so join now and maybe you could become a mod or even an ADMIN!!! w00t!  :D  :D  :!:

[hide="website url"][url]http://themathforum.forumotion.com/forum.htm[/url][/hide]

sorry this site isn't as good as AOPS...

not yet but we are accelerating fast and soon it WILL be better s we need no worries of for the win yet

those two games "advertised"... I can get those games + thousands more at miniclip. For centralized math I go to AOPS. Also confusing is math and math is AOPS

james you dont HAVE to join but dont insult the website because someone is trying to make a good website thats like aops. some people need time to make their websites different. all in all dont say aops is better until TMF has had time to improve. if you dont want to join then fine but dont make the makers of TMF sad because you dont like their website

ha. I jjust signed up.

So far, it's okay for a starting website.

I just joined with the same username.

BLEH

I WILL NOT BETRAY AOPS!!!!

or maybe i will in the future...

meh...


probability that 2 heads show

2/4= 1/2

probability that 1 is heads,
after guess and check i got 1/2

1/2 * 1/2= 1/4

[hide]

Probability of:

Probability when dime is heads and one other coin is heads:

$(\frac{1}{2})^{4}*3 = \frac{3}{16}$

Probability when dime is heads and two other coins are heads:

$(\frac{1}{2})^{4}*3 = \frac{3}{16}$

Probability when all are heads:

$(\frac{1}{2})^{4}= \frac{1}{16}$

Add them all together and you get $\frac{7}{16}$.
[/hide]

[quote="ProtestanT"][quote="mathmatatics"]A penny, a nickel, a dime, and a quarter are tossed simultaneously. What is the probability that at least two heads show and one of the heads showing is the dime? Express your answer as a common fraction.[/quote]
[hide]
probability of dime = heads is $\frac{1}{2}$
probability of no heads in the other three coins is $\frac{1}{8}$, so at least 1 heads in the other three coins is $\frac{7}{8}$
therefore the answer is $(\frac{1}{2})(\frac{7}{8})$
simplified, it is $\frac{7}{16}$
[/hide][/quote]

ahh.. i SO need to remember these shortcuts.. ;D

[quote="mathmatatics"]Sorry, but that is incorrect.[/quote]
oh, [b][i][u]at least[/u][/i][/b]


How does one view the posts saved as favourites? Ijust can't find the icon for it. Could someone send me how to access the favourites.

Feature is not yet re-installed. Have a bit of patience, it will be back.


algebra
polynomial
number theory unsolved
number theory

Let $ a,b\in\mathbb{Z}$ .Define $ a_{0},a_{1},a_{2},...$ by $ a_{0}=a,a_{1}=b,a_{2}=2b-a+2,a_{n+3}=3a_{n+2}-3a_{n+1}+a_{n}$ .Find the general term of the seuqence and find $ a,b$ so that $ a_{n}$ is perfect square for $ n\ge1998$ :wink:

We're talking over a linear difference equation. Then consider the associated characteristic polynomial
\[ p(\lambda) = \lambda^{3}-3\lambda^{2}+3\lambda-1 = (\lambda-1)^{3}. \]
Since $ p(\lambda) = 0$ iff $ \lambda = 1$, theory entails $ a_{n}= c_{0}+c_{1}n+c_{2}n^{2}$, for any $ n \in \mathbb{N}_{\;\!0}$, where $ c_{0}, c_{1}, c_{2}\in \mathbb{C}$ are absolute constants to calculate by imposing $ a_{0}= a$, $ a_{1}= b$ and $ a_{2}= 2b-a+2$ (the initial conditions). Just a few counts to find that $ c_{0}= a$, $ c_{1}= b-a-1$ and $ c_{2}= 1$, so yielding $ a_{n}= a+(b-a-1)\cdot n+n^{2}$, for any $ n \in \mathbb{N}_{\;\!0}$ (here $ \mathbb{N}_{\;\!0}= \{0, 1, \ldots\}$).

By the Eyler`s theorem we have:
$ a_{n}=C_{1}+C_{2}n+C_{3}n^{2}$
By the beginning facts we have:
$ a_{n}=a+(b-a-1)n+n^{2}$
Perfect sqare will be for example if $ b-a-1=2$ and $ a=1$, so: if $ a=1$ and $ b=4$.
Then: $ a_{n}=(n+1)^{2}$

[quote="s.tringali"]Well, I see I missed some terms in my calculations... :roll: I'm going to edit my previous post.[/quote]
 :P   :oops:

[quote="s.tringali"]Sometimes I'm near to think that I can't count at all... :blush:[/quote]
I`m joking...)))


factorial
number theory unsolved
number theory

The equation  $ 1!\plus{}2!\plus{}...\plus{}x!\equal{}y^n$ where $ n\geq 2$ and x and y and n are positive integers has solution?

OK, I'll tell ya this, boy  :D :

[hide="Solution"]
First, if $ x\equal{}1$, for $ (x, y, n)$ we obviously have solution $ (1, 1, t)$, $ t$ being a positive integer larger than $ 1$ .
If $ x\equal{}2$, we have $ 3\equal{}y^n$, which has no solutions, obviously.
Now, let's look at $ x>2$.
Obviously, $ y$ is odd and $ y$ is divisible by $ 3$ ($ 1!\plus{}2!\equal{}3$). So, $ y\equal{}6y_1\plus{}3$. Now, we have $ 1!\plus{}2!\plus{}...\plus{}x!\equal{}(6y_1\plus{}3)^n$. If $ n\geq 3$, we have the right side which is divisible by $ 27$. If $ x\geq 9$, the left side gives $ 1\plus{}2\plus{}6\plus{}24\plus{}12\plus{}18\plus{}18\plus{}9\equiv 9(mod 27)$. That's because $ 9!$ is divisible by $ 27$. That's a contradiction and we can now say that either $ x\leq 8$ or $ n<3$.
Let's see what if $ x\leq 8$. Then, if $ n\geq 3$, again, $ y^n$ is divisible by $ 27$. However, by looking $ (mod 27)$, we can say that $ x\equal{}7$ is the only possibility. Now, by doing a little summing, we can get that $ 1!\plus{}2!\plus{}...\plus{}7!\equal{}5913$. That equals $ 73\cdot 3^4$, and, because of $ 73$, there are obviously no solutions.
Now, we can say that $ n<3$. So, $ n\equal{}2$.
If $ x\geq 5$, we can see that $ 1!\plus{}2!\plus{}...\plus{}x!\equiv 3(mod 10)$. So, we have $ 10k\plus{}3\equal{}y^2$. However, there are no solutions: $ r^2$ can only end in $ 0$, $ 1$, $ 4$, $ 6$, $ 5$ and $ 9$. Therefore, $ x<5$. We already saw about $ x$ being equal to $ 1$ or $ 2$, now we have only two other cases.
First, $ x\equal{}3$. We have $ 9\equal{}y^2$, and we have the solution $ (3, 3, 2)$. If $ x\equal{}4$, we have $ 33\equal{}y^2$, which gives no solutions.

So, we have checked all the cases and reached the conclusion that there are infinitely many solutions $ (x, y, n)$ of form $ (1, 1, t)$, $ t\geq 2$, $ t$ is an integer. There is only one other solution and that is $ (3, 3, 2)$. Mission accomplished! :D [size=75](Looks like I didn't need $ y$ being odd after all... weird.)[/size]
[/hide]

Hope it helps!


geometry
rhombus
perimeter
Pythagorean Theorem

The diagonals of a rhombus have lengths 12 and 16 units. Find the number of units in the perimeter of the rhombus.

[asy]import geometry_dev;
import olympiad;
pair A=(0,0), B=(0,12), C=(-8,6), D=(8,6);
draw(A--D--B--C--cycle);
draw(A--B);
draw(C--D);
markrightangle(C, (0,6), B, 5);
label("$6$",(0,9), E);
label("$6$", (0,3), E);
label("$8$", (-4,6), N);
label("$8$", (4,6), N);[/asy]
The diagonals of any rhombus perpendicularly bisect each other.  In this rhombus, the diagonals make four right triangles with legs of $ 6$ and $ 8$ units.  The hypotenuse of each of these right triangles is a side of the rhombus.  Using the Pythagorean Theorem, we find that each side of the rhombus is $ \sqrt {6^2 + 8^2} = 10$ units.  The perimeter of the rhombus is thus $ 4\cdot 10 = \boxed{40}$ units.

Excuse me, but you have an error in your $Latex.$  :stink:

[quote=jxl28][asy]
pair A=(0,0), B=(0,12), C=(-8,6), D=(8,6);
draw(A--D--B--C--cycle);
draw(A--B);
draw(C--D);
draw(rightanglemark(C, (0,6), B, 20));
label("$6$",(0,9), E);
label("$6$", (0,3), E);
label("$8$", (-4,6), N);
label("$8$", (4,6), N);[/asy]
The diagonals of any rhombus perpendicularly bisect each other.  In this rhombus, the diagonals make four right triangles with legs of $ 6$ and $ 8$ units.  The hypotenuse of each of these right triangles is a side of the rhombus.  Using the Pythagorean Theorem, we find that each side of the rhombus is $ \sqrt {6^2 + 8^2} = 10$ units.  The perimeter of the rhombus is thus $ 4\cdot 10 = \boxed{40}$ units.[/quote]

Fixed to the best of my abilities.


I have a chemistry problem I need help with:

A sample of a gaseous binary compound of Boron and Chlorine weighing $ 2.842 g$ occupies $ 0.153 L$. This sample is decomposed to give $ 0.664 g$ solid boron and enough gaseous chlorine $ Cl_{2}$ to occupy $ 0.688 L$ at the same temperature and pressure. Determine the molecular formula of the compound.

Can anyone help?

[hide="hint"]
let the molecular formula of the compound be $ BCl_{x}$ then
$ BCl_{x}$ $ \rightarrow$  $ B$ $ \plus{}$  $ \frac{x}{2}Cl_{2}$ hence $ 1$ mole of the compound gives $ 1$ mole of boron and $ \frac{x}{2}$ moles of chlorine.hence calculate the moles of boron chloride from the number of moles of boron.hence equate this by clalculating the molar mass of boron in variable $ x$

This is done assuming quantitative yield[/hide]

This is a very nice problem. We are to assume quantitative yield, as Pardesi stated, and also ideal gas behaviour.
However Pardesi, your reasoning is only partially correct. Why the boron chloride must have molecular formula $ BCl_{x}$, and not the more general $ B_{x}Cl_{y}$?

that's because i assume $ Cl$ to have an oxidation number of $ \minus{}1$
well we could have well assumed $ B_{x}Cl_{y}$ and then on used the volume of $ Cl_{2}$ produced also

Chlorine can have several oxidation numbers. Anyway, according to my calculations I get

[hide="Answer"]The molecular formula is $ B_{9}Cl_{9}$, which could not be reached by assuming it to be $ BCl_{x}$. [/hide]

Yes, it is, although a $ B_{4}Cl_{4}$ compound in known. However, note that we always get nine chlorines for the molecular formula, and $ BCl_{9}$ is naturally an impossible compound. Unless, of course, I am missing something.


absolute value
number theory unsolved
number theory

Let $ (m,n)$ be a pair of positive integers.

(a) Prove that the set of all positive integers can be partitioned into three pairwise disjoint nonempty subsets such that none of them has two numbers with absolute value of their difference equal to either $ m$ or $ n$.

(b) Find all pairs $ (m,n)$ such that the set of all positive integers cannot be partitioned into two pairwise disjoint nonempty subsets satisfying the above condition.


Okay, thanks.

I tried to post the picture earlier, but had problems...

By the way, when I click on [url=http://www.artofproblemsolving.com/Forum/faq.php?mode=attach&f=144]Allowed Extensions and Sizes[/url], not only does the page look weird (the Bookstore, Online School, etc. links are a couple of inches to the right of where they should be), it says that no matter what kind of attachment I want to make, the maximum upload size is 0 bytes (I'm using the AoPS skin).

Yeah, that's a confirmed bug. Post splitted :)

Actually it's not a bug. There is no maximum file size defined :)


[quote=G-UNIT]Hi, I was just wondering if anyone was particapating in the ELMACON ( it was formerly called PIMS) contest.  Since this is a regional based contest I am not expecting a great deal, if any, replies.....[/quote]

I am


Prove that for all $ n\in\mathbb{N}$,there always exist $ n$ distinct natural numbers,such that they and arithmetic means of any number of them can be represented as $ r^{s}$,where $ r,s\in\mathbb{Z},r,s\geq 2$.

Balkan MO 2000
http://www.mathlinks.ro/Forum/viewtopic.php?p=495308#495308


Given iscoceles triangle ABC, construct P in its interior so <PBC=40, <PCB=30 and <BAC=80. Find APC Im looking for a trig solution

[hide="answer"]100[/hide]

Solution[hide]
Let $\angle CAP=x$. All angles are in degrees. From $\frac{AP}{AC}=\frac{AP}{AB}$ with sine theoreme we have $\frac{\sin20}{\sin(x+20)}=\frac{\sin10}{\sin(90+x)}$ or $\frac{2\sin10 \cos10}{\sin(x+20)}=\frac{\sin10}{\cos x}$. This simplyfies to $2 \cos x \cos10 = \sin(x+20)$
$\cos(x-10)+\cos(x+10) = \cos(x-70)$
$\cos(x+10) = \cos(x-70)-\cos(x-10)=\sin(x-40)$
And the solution is $x=60$, so $\angle APC = 100$ [/hide]

[quote="D.P.L"]Given iscoceles triangle $ABC$, construct$P$ in its interior so $\angle PBC=40, \angle PCB=30$and $\angle BAC=80$. Find $\angle APC$ . Im looking for a trig solution.

[hide="answer"]90[/hide][/quote]

 [color=red]If you want to have a synthetic solution , you can read this HINT :[/color]
The bisector of angle $A$ meets $CP$ at $D$.Then

 $\angle BDC = 120, \angle PBA = \angle PBD = 10, \angle PDB = 60, \angle ADB = \angle ADC = 120$ ,

 so  $DP, BP$   are bisector lines in triangle $BAD$.

But then $AP$ is also bisector of  $\angle BAD = 40$ , thus

 $\angle PAD = \frac{1}{2}\angle BAD = 20$ and finally $\angle APC = 180-20-60 = 100$

 Babis


AMC USA(J)MO USAMO FTW AIME Gauss geometry

Pick two or more AoPS users.

Next person who posts: say what you get when you cross the two.

dragon96
james4l

A smart not-so-smart person who smiles while saying "I fail".

isabella2296
AIME15
Mewto55555
miller4math
EcstaticPotter

--syntax error--
or, Memzville

jjx1
Mewto55555

KABOOM.

Hexahedron
MysticTerminator
stevenmeow
variable
Lazarus.

Good luck!

Aren't they all 1337 MC Nats Champions?

r15s11z55y89w21
ch1n353ch3s534a1l
5849206328x

al2929387fhw73fqn73chq

gauss1181
suma_milli
guineapiggies

BLAGIDIBLAGIDIBLAH

Though these two would never cross, I wonder:
mewto55555
isabella2296

Hey you put in only one person...

AMCDirector
rrusczyk

Hmmm.

Ribbit.
Onewholovesfrogs4795
Frogslikepie

Many many frogs.

@SuperPawn: you're supposed to wait for the next person to respond.
Thus I don't need to respond to anything.

Mewto5555
lucario55555
SicilianFury
RomanianGenius
rg
nt811581147
manifester

edit:SicilianMewcario manifest ntrg genius 811581147555555555

I just want to see what happens when failure and success combine!
asterday1(ME! I got 21st at state(PA))
abacadaea(he got 2nd!(PA))

a guy who gets $ \frac{23}{2}$th place at PA states

Fanfan
izzy

:O

SOMEONE VERY BEAST

[newest user]
[second newest user]

AIME15's: 1L1k3H4rv357P4rk
previous: Bergen_County_1337

dragon96
nikeballa96
isabella2296

Ack! The year of the Nerds!

AIME15
r15s11z55y89w21
15!!!
cyao15
cgyao15

AIME15!!!11cyao55cgyao89w21

Do the entered things have to be AoPS people?
If not: apples and oranges
If yes: dragon96 and eragon97

Deragon96.5

mz94
coolts
lotsofmath
AIME15

An very smart turtle who does lots of math and is amazing at skateboarding.

dragon96
mz94
AIME15

Amazing Dragon.

gauss1181
ThinkFlow
cheesymath
suma_milli

Gauss thinks that cheesy math is suma_milli.

superquark1123
leafiness0
ak909090

Quarkiness90

ak909090
bowei
james4l
james320
dragon96
AIME15
r15s11z55y89w21
Nerd_of_the_Ages
lifeisacircle
infinite_sigma
vahalla

[quote="mathelete"]A singularity[/quote]

I don't even want to know.

[Just an interjectory post]

[quote="mathelete"]A singularity

isabella2296
AIME15[/quote]

FTW system failure

Dojo
ak909090
bowei
james4l
james320
dragon96
AIME15
r15s11z55y89w21
Nerd_of_the_Ages
lifeisacircle
infinite_sigma
vahalla
mathelete
gauss1181
ThinkFlow
cheesymath
suma_milli

Memory overload (too many people)

lifeisacircle
leafiness0
donghyun
james4l

(Note something)

Bay Area pwnage.

gauss1181
ReseesPieces
HolyMushroomspwnthywrld17
LHM
dragonlaird7

Er,

[quote="dragon96"]
lifeisacircle
leafiness0
donghyun
james4l
[/quote]

is Hopkins pwnage, not SFBA pwnage.

[quote="gauss1181"]
gauss1181
ReseesPieces
HolyMushroomspwnthywrld17
LHM
dragonlaird7
[/quote]

Heaven for dragon96 :D

Hm...

AIME15
jerry125
4zn.prid3l337
midnightgladius186
mcfail
glaser06

CA pwnage then??? hmm im not sure of ur exact locations tho

not_trig
CatalystOfNostalgia
kyyuanmathcount
SplashD
b-flat
probability1.01
Ubemaya
serialk11r


[quote="kunny"]For $ A \equal{} \left( \begin{array}{cc} 3a^2 \minus{} 3 & 3a \\
2a^2 \minus{} a \minus{} 2 & 2a \minus{} 1 \end{array} \right),\ B \equal{} \left( \begin{array}{cc} \minus{} 1 & \minus{} 3a \minus{} 2 \\
a & 3a^2 \plus{} 2a \minus{} 3 \end{array} \right)$, Find $ AB.$[/quote]

$ AB\equal{} \left( \begin{array}{cc} 3 & 6 \\
2 & 7 \end{array} \right)$


To be magical when I grow up.

I want to be just like Harry Potter :) .

biggest dream? Well, short-term is to qualify for MOP this year.

It would be nice, in the math world, to get over 25 at the Mathcounts state competition next year.

to be happy. happy:
1. have a successful career that challenges me.
2. have a family that loves me and that i love in return.
3. have enough free time to devote to the care of others and the world around me.
4. have enough money to know where my next meal is coming from.

I would like to have a problem created by me on the IMO paper :)

To always just be happy. What more could anyone ask for?

I know 'happy' can be the wrong word to use here, as everyone here hopes to be happy after they've achieved theyre goal.  So, to those wondering, to  me, being happy is being happy w/ what you have. So, happy could mean living frugally - but I could be happy w/ that. Really then, my goal is to learn to be happy w/ anything that I end up with.

[quote="h78reg"]well, anyone who likes math's biggest dream is an IMO gold medal, though that's just a dream and not reality................. :rotfl:[/quote]

No, I don't dream about an IMO gold medal... I dream about a perfect paper :D

[quote]No, I don't dream about an IMO gold medal... I dream about a perfect paper [/quote]

well, that pretty much tops the list. the only things better than that are beating IMO medal records or receiving a Fields Medal. :D

[quote="h78reg"]
 but, i dont think we will live to see the end of poverty. :([/quote]

I don't think anyone will ever live to see the end of poverty.

well, that depends on how you define poverty. of course, the world will always become more developed and life will get better. however, life in some countries will always be worse than life in other countries.

[quote="h78reg"][quote]Or how about solving a millenium problem? The Riemann hypothesis. I am interested in either becoming the Chairman of the Federal Reserve or becoming the CFO or CEO of one of the top 100 companies in the world. How about manking the people in countries of poverty live like the average person in the US? I want to see that achieved.[/quote]

well, to be a CEO of a company that large, you must have an MBA and usually work with the company for 30 years, or be a renowned chief executive. as for the countries in poverty living like the average person in the US, keep in mind that the average person makes aroung $30,000 a year in the US, which isn't that much. but, i dont think we will live to see the end of poverty. :([/quote]

Or you can be like the founders of google :D

I think you dont need 30 years to become an executive all the time.

I think something should be done about the countries in poverty. Their political and economic infrastrucutre needs change and if the regime is not willing to go through that change, other countries should intervene to help, but not just by giving loans like the IMF but promoting the right infrastrucutre needed for the company to succeed. I am I read in a book that 3 billion people live on $2 a day, so that is like less then $800 a year, compare that to the not so loarge $30,000. We have to act.

[quote]
I think you dont need 30 years to become an executive all the time.

I think something should be done about the countries in poverty. Their political and economic infrastrucutre needs change and if the regime is not willing to go through that change, other countries should intervene to help, but not just by giving loans like the IMF but promoting the right infrastrucutre needed for the company to succeed. I am I read in a book that 3 billion people live on 800 a year, compare that to the not so loarge $30,000. We have to act.
[/quote]

forcing your way into a country won't solve anything. the people in those countries must first have the initiative to rise up and show the international community that they want a change in their regime.

Come on, how can people in a country like say India rise up and say they have problems? To who will they rise? It is obvious that they need change. Can any normal person who doesnt even want to be rich will be willing to live for $2 a day? Come on, its obvious they dont.

just look at Iraq, many people aren't happy that the US came there. theoretically, now they get suicide attacks from people who want the US to leave. millions of people are not comfortable with the idea of a country just barging into their territory.

Hi, I'm new, and one of my bigger dreams is to be a good math student. I'm not exactly a math prodigy and it's something I struggle with, but I'm really getting interested in it.

My biggest dream this year probably is to make it into the USAMO... :blush:

My god I cant believe no one has mentioned the greatest problem of all!

[size=200][b]GRAND UNIFICATION .A.K.A THE THEORY OF EVERYTHING[/b][/size]

  Now thats the just about the biggest dream you can have.

g-unit, hasn't that already been mentioned?

My dream is to grow up, get married, have a family, lead a comfortable existance, give my children a leg up in life like my parents gave me. Any more than this is just icing on top.

For all the math kids out there, is qualifiying for the USAMO really your biggest dream? Don't get me wrong, it's an admirable and achievable goal, but the biggest dream?

[quote="tetrahedr0n"]For all the math kids out there, is qualifiying for the USAMO really your biggest dream? Don't get me wrong, it's an admirable and achievable goal, but the biggest dream?[/quote]
well, it's probably the biggest short term goal for some of us. i'm sure that college, good job, happiness, family, etc. are on everybody's list.
[quote="PenguinIntegral"]To die with more money than god.[/quote]
um, i'd say that god doesn't have any money so you've pretty much accomplished your goal

G-Unit, I was also livin in Canada before. Anyway, for mathematicians the Riemann Hypothesis is a bigger problem then the theory of everything.

Whoever it was that mentioned Iraq, I am not saying that the war is right, a lot of the Iraqi people were brainwashed anyway so they dont even understand everything.

Like I said, I am pretty sure that any normal person in any country will want to earn more then $2 per day, try to get some survey, I cant be wrong, imagine if someone says that they prefer to live on $2 a day then live under the normal conditions in the US. Is that really possible?

i imagine living on 2 dollars a day won't make you rich, but, it doesn't mean you're poor either. here in the US, its impossible to live on 2 dollars a day because there are always going to be jobs open at a McDonalds or something like that which pay 40 dollars a day at least.(minimum wage)because of deflation, 2 dollars a day may not mean that you're that poor in other countries. you may still be below average in terms of your income, but you won't always be poor or in poverty. maybe someone knows of a global salary survey or something like that?

What r u talkin about? I watched movies about how people work in African and Asian countries and how they live, thats pretty bad. Trust me, $2 is nothing, nothing regardless of where it is.

[quote="krustyteklown"]Since its biggest, and I've been reading up on physics recently.

How about, find the "general theory" in physics? :D[/quote]

Well, the general theory would be the unification of String Theory into M-Theory, also know as the theory of everything.

Speaking about physics, my goal (you may consider them as dreams, but nothing is a dream if it really is a dream...it/they can be accomplished if they are true things that you desire to accomplish) is to become a great physicist and mathematician who discovers great things, such as the mathematics of the infinitude of singularities, wormholes, and at what speed do you travel even being under great gravitational pressure of blackholes...I have lots more scientific thoughts/goals than the ones mentioned.

Masoud Zargar

hm, from what i've read, which isn't a lot

isn't string theory by itself sufficient? even predicting gravity, etc.

i thought that we just didn't have the instruments to see these strings, and that some of the models don't have complete solutions yet
i didn't quite catch anything about unifying string theory with anything...
(we should prob talk about this in a PM tho, instead of clogging up forum space)

String theory, that has been unified to M-Theory, can predict many things.  Many physicicts are currently working on the theory of everything; this reminds of the time when Einstein and many otehr scientists were seeking for teh theory of relativity.  The curernt theories are exiguous in terms of predicting nature to teh maximum extent.  It is true that we currently do not have the material to physically observe strings, but I, including many theorists, rely on mental experiments (logic) ,and hence, mathematical reasoning is sufficient for them.

Masoud Zargar

I'm very surprised that nobody of you has mentioned the Nobel Prize yet.

Well, that's actually not what I'm dreaming of - my currently biggest wish is a medal at the IMO (and when this is achieved, I'll find myself a new 'biggest dream' like a perfect paper or something :lol: )

My dream is to be a rock star, I'm surprised that hasn't been mentioned (I want to be a celebrity 0_o)

Too bad it won't happen though.

[quote="math92"]my biggest dream is to beat Roger Federer in tennis.[/quote]

Heheh,u interested in tennis too?HOW old r u right now?I was sooooooooo into tennis at one time but then the stupid education syshtem in india doesnt let u do anythin extra,not even related to ur career, not even from a young age of 13,its so tough,not math and science,the othert useless subjects like social,hindi,biology,etc.

Hey even i feel like becomin a rock sta,for now,i wanna get out of india( :rotfl:  :rotfl: ),do my 11th and 12th ,gain knowledge,maybe some olympiad.then hopefully get into IIT,not dedcide then......
Well it still is my dream to b the hansomest and intelligent person on this earth,too bad one of them cant be true.. ;)


I'm interested in taking the AMC 12 in February, but no one at my school has ever taken it. How should I go about registering? On the AMC website, it appears registering is done at the group level, not at the individual level. Should I try to see if one of the math teachers at my school can register, or should I look into taking the test at another high school?

Does anyone have any experience with this?

Thanks!

I am having the same trouble, I am trying to recruit more people at my school to take it so my teachers will sign us up, if not I will just take it at a near by highschool. Really it just depends if there are any teachers that are willing to take the time to give a test to just a few students.

I'm a homeschooling parent, and I run into this issue every year. I used to try to interest my friendly local public school district in administering the test, but what I do now (as a parent) is order a test kit (ten tests), communicate to AMC headquarters about the mailing address to send the tests to (for example, a local after-school tutoring center) and then announce to other parents that they can join us in having their children take the test. By asking each parent to pay part of my cost, I get enough funds to pay a proctor and arrange space. Last year, both state champions in the AMC 8 in our state and both top scorers on the AMC 10 in our state tested through such arrangements. Use parent power to overcome bureaucratic barriers.

Our school district is pretty unwilling to do anything math related, sports only.
Everything needs to be done on an individual level, among the similar minded families.
But I was pretty successful with registering my son for individual Mathcounts last year. He did pretty well, this year he recruited a team for Mathcounts (school is not getting involved though but as long as they do not object - it's as good as it gets).
The local high school answer to AMC question was that they do not give the exam due to no interest in it. When I said there are at least 3 8-graders and 1 9-grader interested, they agreed to find a teacher to proctor the exam on Feb. 6 (but the registration and costs will be taken care of by parents). Which is as good as it gets too.
For date B I am taking my son and his friends to one of the colleges (there are 3 within 150 miles from us). All three colleges said out of state students can take their AMCs there.

So now my son will take AMC 10 on date A in our local high school and AMC 12 on date B at some college.

You are definitely on the right track recruiting the other students. When you have at least a few kids, you can always disagree with the first argument school districts give you - no interest.
Of course I had to point out directly that all we are asking is  75 minutes of a teacher's time and a classroom. The costs, paperwork, preparation etc. will be taken care of by the parents.

(It is very sad, unfair, aggrevating etc. that school districts use our taxes on sports and not on academics and then act as if they do us favors by allowing us to arrange math competitions for our kids. But this is a different story. I understand that I cannot change the school district, just trying to do what has to be done for my son and his buddies).

Do not give up, good luck!

PM me if you have more questions.


inequalities
function
algebra unsolved
algebra

Show that $8x^4 - 16x^3 + 16x^2 - 8x + k = 0$ has at least one real root for all real $k$. Find the sum of the non-real roots.

Are the terms of the problem correct ?

for $k = 2$ we get $2 \times (2x^2 - 2x + 1)^2$ which has no real roots ???  :mad:  :(

[quote="Fermat -Euler"]Show that $ 8x^4 \minus{} 16x^3 \plus{} 16x^2 \minus{} 8x \plus{} k \equal{} 0$ has at least one real root for all real $ k$.
Find the sum of the non-real roots.[/quote]

It seems that the real problem said : Show that $ 8x^4 \minus{} 16x^3 \plus{} 16x^2 \minus{} 8x \plus{} k \equal{} 0$ has at least one [b][u]non-[/u][/b]real root for all real $ k$.

:(

I'm assuming we are allowed to post:

We have $ 8x(x^3 \minus{} 1) \minus{} 16x^2(x \minus{} 1) \equal{} \minus{} k$
This reduces to $ (x^2 \minus{} x)(x^2 \minus{} x \plus{} 1) \equal{} \minus{} k/8$
Letting $ J \equal{} \minus{} k/8$ (J is  a real obviously), we have
$ (x^2 \minus{} x \plus{} 1/2)^2 \minus{} 1/4 \equal{} J$
$ x^2 \minus{} x \plus{} 1/2 \equal{} \sqrt {J \plus{} 1/4}$
$ x \equal{} 1/2\pm\sqrt { \minus{} 1/4\pm\sqrt {J \plus{} 1/4}}$
If we take the minus sign of the inner square root we will get the square root of either a negative no. (J >= -1/4)
or a complex number (J < -1/4); either way, x is a complex number.
I know my proof is lacking in clearness and conciseness; might've made some mistakes.
EDIT: And it is trivial to derive the latter part from this.

Consider $f(x) = 8x^4 - 16x^3 + 16x^2 -8x + k = 0 $
Consider $f'(x) = 8(4x^3 - 6x^2 + 4x -1 ) $
If $a,b,c$ are roots of $f'(x)$, $\sum a = \frac{3}{2}$, $\sum ab = 1$ , $abc = \frac{1}{4}$, so we see that the inequality $(\sum ab)^2 \ge 3abc ( \sum a) $ is not satisfied. This implies there are 2 non real roots for $f'(x)$. This therefore implies there are atleast 2 non real roots for $f(x)$ by Rolle's theorem.

Another idea:It's easy to get that the sum of the squares of the roots is $0$.This is not possible if all the roots are real as in that case the equation will be $x^4=0$

My solution is like this...
We can write $8x^4-16x^3+16x^2-8x+k=0$ as $\frac{1}{2} (2x-1)^{2}((2x-1)^2+2)+k-\frac{3}{2}=0$
Note that this function is symmetric about $x=\frac{1}{2}$ and also strictly increasing in the interval $[\frac{1}{2},\infty)$ and strictly decreasing in the interval $(-\infty,\frac{1}{2}]$ So there are atmost two roots in this two intervals. Hence, there exist atleast two non-real roots of this equation. As this function is symmetric about $x=\frac{1}{2}$so this real roots should be in the form $\frac{1}{2}+p$ and $\frac{1}{2}-p$ for some non-negative real number $p$. So the sum of the real roots is $1$ or $0$(when there is no real roots. Sum of all four roots is $2$. So, the sum of the non-real roots is $1$ or $2$.

The graph is as shown.
So no matter what k is it will have at max 2 real roots...

Differentiate the equation twice. The second derivative has no real root. Hence [b]f'(x)[/b] won't have a minima or maxima. Since [b]f'[/b] is an odd degree polynomial with no points of extrema, it will have only one root. Hence, [b]f(x)[/b] will have a minima and no other extremum points. Thus, [b]f(x)[/b] will have at most two roots and thus we are done.

How do you realise that the equation is symmetric about 1/2? Once you have the value, it is easy to prove it is symmetric but how do you find the value 1/2?


[quote="Treething"]For reading an easy problem that Treething posts, Puloc gets one IQ point smarter. For reading a hard problem that somebody else posts, Puloc loses four IQ points. If Puloc had an IQ of 73 this morning and now has an IQ of 100, what is the minimum number of problems that he could have read today?

Assume that his IQ was not influenced by outside sources.
Assume that he read at least one USAMO question today.
Assume that dogs are cute.
If USAMO questions are easy, dogs are not cute.
Assume that problems are only easy or hard.
Assume that Treething only posts easy questions.
Assume that other people only post hard questions.[/quote]

Does reading a USAMO question count as reading a problem someone else posts?

[hide]Puloc starts with an IQ of 73 (It's the "c" in his name), and then reads a hard USAMO problem, bringing his IQ down to 69. He then reads 31 Treething problems and brings his IQ up to 100.
31 + 1 = 32, our answer.[/hide]

I think this is right...
[hide]Let's say the first thing he reads is a USAMO question so his IQ becomes 69. By reading 31 more Treething questions his IQ is now 100. So 1+31=32.
ANSWER: (32)[/hide]

[quote="Treething"]looks right and that's what I got, so I'm assuming we're both right.[/quote]

[hide]Well I got 32 as well, so now all three of us are right.  Plus four if you want to include the new kid.[/hide]


A sequence is defined by $ x_1 \equal{} 1, x_2 \equal{} 4$ and $ x_{n\plus{}2} \equal{} 4x_{n\plus{}1}\minus{}x_n$ for $ n\geq 1$. Find all natural numbers $ m$ such that the number $ 3.x_n^2 \plus{} m$ is a perfect square for all natural numbers n.


This is another questions that was recommended to do for our own benefit, and I think I sort of have it figured out. But once again any help would be greatly appreciated.  The way I tried to approach this question was by finding a subset of three vectors that are linearly independent, is that the right way to do this.

The Question is:
$\mbox{The Vectors}$
$x_1=\ \left(\begin{array}{cc}1\\2\\2\end{array}\right)$,   $x_2=\ \left(\begin{array}{cc}2\\5\\4\end{array}\right)$,   $x_3=\ \left(\begin{array}{cc}1\\3\\2\end{array}\right)$,   $x_4=\ \left(\begin{array}{cc}2\\7\\4\end{array}\right)$,   $x_5=\ \left(\begin{array}{cc}1\\1\\0\end{array}\right)$

$\mbox{span } R^3\mbox{. Pare down the set }\{x_1\mbox{, }x_2\mbox{, }x_3\mbox{, }x_4\mbox{, }x_5\} \mbox{ to form a basis for }R^3$.

Since any three of $x_1,x_2,x_3,x_4$ are linearly dependent (all have the form $\begin{bmatrix}a\\b\\2a\end{bmatrix}$, we need $x_5$ and two of the first four. Any two of the first four will do.

A generic procedural answer to the question:

Form these column vectors into a matrix. (In your case it will be a $3\times5$ matrix.) Perform row elimination until you arrive at a row echelon form. Note the columns in which pivots occur. Then use the original vectors (before the row elimination) that correspond to the those columns.

The property that jmerry observed concerning the first four vectors will manifest itself during the row elimination. A step in which twice the first row is subtracted from the third row will result in the first four elements of the third row being zero, and a pivot occurring on the third row in the fifth column.

Also: in the statement of the problem, you were told that these vectors span $\mathbb{R}^3.$ You wouldn't have to be told that; the appearance of a pivot in every row (three pivots) would tell you that.


Jack starts at the summit of Mt. McKinley (elevation, 20,320 feet) and travels down the mountain at a rate of 4 feet per second. Jill starts at the bottom at the same time and travels up the mountain at a rate of 1 foot per second. At what elevation, in feet, will they meet?

they get closer to each other at a rate of 5 feet per second, of which 1 foot is traveled by jill upwards.
we find the distance jill has traveled to find their meeting elevation.
$ \frac15\cdot20320\equal{}4064$
4064.


i logged in after all...............

by the way i got these question.........it came in RMO 2006 india

1) let ABC be an acute angled triangle and let D,E,F be the feet of perpendiculars from A,B,C respectively to BC,CA,AB let the perpendiculars from F to CB ,CA,AD ,BE, meet them in P,Q,M,N respectively.....prove that P,Q,M,N are collinear

2) find the least possible value of a+b where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b

3)

Pls don't [url=http://www.mathlinks.ro/viewtopic.php?t=236806]double post[/url].
and try to post one problem in a topic unless they are closely related by content.

Apply [url=http://en.wikipedia.org/wiki/Simson%27s_line]Simpson's line theorem[/url] on no.1.

[hide="Solution"]Denote $ H$ the orthocenter of $ \Delta ABC$
Clearly $ AEHF$ concyclic. Hence $ f$ is a point on the circumcircle of $ \Delta AEH$
By Simpson's line theorem $ QMN$ collinear.
Similarly it can be proved that $ PMN$ collinear.
Hence Result. QED.[/hide]

Try to find one divisibility condition instead of two in no.2.

[hide="Solution"]From the first condition: $ 143|13a\plus{}169b$ and from the second $ 143|11a\plus{}121b$
So $ 143|(13a\plus{}169b)\minus{}(11a\plus{}121b)\equal{}2a\plus{}48b$. Hence $ 143|a\plus{}24b$
Clearly the positive pair $ (a,b)$ satisfying $ 143|a\plus{}24b$ for which $ a\plus{}b$ is least is $ (23,5)$.
And that pair satisfies the two original conditions. Hence $ \min(a\plus{}b)\equal{}23\plus{}5\equal{}28$[/hide]

Akash is Simpson's line theorem a well known one or should we write it as a lemma?
It can otherwise be done by proving congruences by construction....

OMG Everything need be proven ??? I think it is quite well known.
Check out the proof given at the end of the Wikipedia link I gave. It's that simple.
BTW, I want to know your proof proving congruences by construction.
Would you be so kind to write it up here?


Let $ a,b,c\ge0$ such that $ abc\equal{}1$.Prove that
$ ab^3\plus{}bc^3\plus{}ca^3\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2\ge a\plus{}b\plus{}c\plus{}\sqrt{3(a^2\plus{}b^2\plus{}c^2)}$

This is straightforward. The inequality is equivalent to

\[ \sum \frac{b^2}{c}\plus{}\sum \frac{bc}{a}\ge \sum a\plus{}\sqrt{(a^2\plus{}b^2\plus{}c^2)},\]

which is true since

\[ \sum \frac{b^2}{c} \ge \sum a\]

and

\[ \sum \frac{bc}{a}\ge \sqrt{3(a^2\plus{}b^2\plus{}c^2)},\]

[quote="hungkhtn"]This is straightforward. The inequality is equivalent to
\[ \sum \frac {b^2}{c} \plus{} \sum \frac {bc}{a}\ge \sum a \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)},
\]
which is true since
\[ \sum \frac {b^2}{c} \ge \sum a
\]
and
\[ \sum \frac {bc}{a}\ge \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)},
\]
[/quote]

Thank you,your solution is very simple,hungkhtn!
We have stronger following result:
$ (ab\plus{}bc\plus{}ca)^2\ge 6\plus{}\sqrt{3(a^2\plus{}b^2\plus{}c^2)}$
Can your method use for it?


Assume ai>0 for i=1,..,5. Prove that

  \sum (cyclic) (a1a2a3/a4a5)^4  \geq  \sum (cyclic)a1a2^2a3

It is Murhead for (8,8,8,0,0) > (6,5,5,4,4).
Of course, it can be done by using AM-GM inequality.

Namdung

I don't think Muirhead works as the sums are cyclic, not symmetric.

$\frac{1}{4} \cdot \frac{a_1a_2a_3}{a_4a_5}^4 + \frac{1}{4} \cdot \frac{a_2a_3a_4}{a_5a_1}^4 + \frac{1}{8} \cdot \frac{a_3a_4a_5}{a_1a_2}^4 + \frac{1}{8} \cdot \frac{a_4a_5a_1}{a_2a_3}^4 + \frac{1}{4} \cdot \frac{a_5a_1a_2}{a_3a_4}^4 \ge a_1a_2^2a_3$ by AM-GM, and we add the similar inequalities.

[quote="manlio"]Assume $a,b,c,d,e>0$. Prove that
$\left(\frac{abc}{de}\right)^{4}+\left(\frac{bcd}{ea}\right)^{4}+\left(\frac{cde}{ab}\right)^{4}+\left(\frac{dea}{bc}\right)^{4}+\left(\frac{eab}{cd}\right)^{4}$
$\geq ab^{2}c+bc^{2}d+cd^{2}e+de^{2}a+ea^{2}b$.[/quote]
I think the following stronger inequality

$\left(\frac{abc}{de}+\frac{bcd}{ea}+\frac{cde}{ab}+\frac{dea}{bc}+\frac{eab}{cd}\right)^{24}$

$\geq5^{23}(a^{6}b^{12}c^{6}+b^{6}c^{12}d^{6}+c^{6}d^{12}e^{6}+d^{6}e^{12}a^{6}+e^{6}a^{12}b^{6})$.

I haven't solution to this inequality.


vector
trigonometry
calculus
calculus computations

My vector skills are weak and I'm just wondering if I have the right idea for this problem:
Here's the Prob:
Suppose that $ \bold{v}$ is a unit vector in  that makes a  $ 75^\circ$ angle with the
positive x-axis and a $ 60^\circ$ angle with the negative z-axis. Write $ \bold{v}$ in its com-
ponent form.

My current idea -- which I immediately thought when I saw the problem -- is to find the components of the vectors in individual planes and then somehow combine, but I'm currently clueless of how?

Yeah, my gut call works, I suggest
HINT+SOLUTION
finding ways to satisfy the angular properties of each axis individually by focusing on its plane with the y-axis, then using simple trig and finding the resultant


The following problem is similar to the IMO2003Problem2,And also interesting :
Find all pairs of $(a,b)\in Z^{2}$ such that $\frac{a^{2}}{2a^{2}b-b^{3}+1}$is a postive integer

It is obviosly, that $c=\frac{a^{2}}{2a^{2}b+b^{3}-1}$ is integer if and only if b=0, then $c=-a^{2}\le 0$.

[quote="Rust"]It is obviosly, that $c=\frac{a^{2}}{a^{2}b+b^{3}-1}$ is integer if and only if b=0, then $c=-2a^{2}\le 0$.[/quote]
Sorry,My problem is $c=\frac{a^{2}}{2a^{2}b+b^{3}-1}$

I'm really sorry, :blush: ,rust.I made a typo in the first post :blush: ,It should be :
Find all pairs of $(a,b)\in Z^{2}$ such that $\frac{a^{2}}{2a^{2}b-b^{3}+1}$is a postive integer.(I have edited the my post.)

It is easy to chek, that for any a we have no more one natural b, were $|2a^{2}-b^{2}|<a$.
Consider Pele equation $x^{2}-2y^{2}=\pm 1$, with solution $x_{n}+y_{n}\sqrt 2=(1\sqrt 2 )^{n}$.  It is not difficult to check, that $a=y_{n},b=x_{n}$ work.

Yes,the result you get is same as mine,though I can't understand your solution clearly.


arithmetic sequence
number theory proposed
number theory

Prove that : one can not $ x ,y \in Z+ $ such that : $ x^2-xy+y^2 $ and $ x^2+xy+y^2 $ are perfect squares

I understand your idea but I don't know how I can apply it to solve this problem  :(

I'm sure there's a more direct appraoch, but I was just looking at another post mentioning squares in arithmetic progression...  Perhaps the method used to prove that $a^2,b^2,c^2$ and $d^2$ cannot be consecutive terms of an arithmetic progression could also be used to show that $a^2, b^2, n, c^2$ and $d^2$ cannot be consecutive terms of an arithmetic progression, where $n$ is some integer (or, to use more of what was given, where $n=x^2+y^2$).

I haven't actually seen the proof I'm talking about though, so I'm really only guessing.

[quote="marko avila"]the two numbers are  between  $(x-1)^2$ and $(x+1)^2$ so there could only be one perfect square[/quote]
  It seems no valid here !

This is something I got :
  Put $x^2-xy+y^2=u^2 (1)$
       $x^2+xy+y^2=v^2 (2)$
  We can suppose $x \geq y$
 From $(1)$ and $(2)$ , we get : $y+\sqrt{4u^2-3y^2}=\sqrt{4v^2-3y^2}-y$
  This means $4y^4-2y^2(u^2+v^2)+(u^2-v^2)^2 =0$

  Can you help me finish it ?

[quote="romano"]  This means $4y^4-2y^2(u^2+v^2)+(u^2-v^2)^2 =0$
    Can you help me finish it ?[/quote]

Put  $y^2=a$  and then you will have a second degree equation and you will try to find the integer roots of this. This could help I think

I have an idea :
  We will prove that :
 Lemma : $x^2-xy+y^2$ is a perfect square $\iff x=y$ or $3x=8y$ or $5x=8y$
 WLOG , suppose $(x,y)=1$
 So $(x,u)=(y,u)=1$
   We have : $x(x-y)=(u-y)(u+y) (1)$
  Suppose $x \geq y >1$
  From (1) , we get $x >u >y >1$
  We have $(x,x-y) =1$ and $(u-y,u+y) =1$ or $2$

  Case 1: $(u-y,u+y)=1$
    So $(u+y)$ is divisible by $x$ and $x-y$ is divisible by $u-y$
 On the other hand , we have $2x > u+y \geq x$
  So $u+y =x$ , contradict with $x^2-xy+y^2=u^2$

  This means $x=y=1$

   The original problem can be solved by using this lemma

I have a big mistake but no problem , I can correct it
 $(u-y, u+y) = 1$ or $2$
 and we have $x^2-xy+y^2$ is a perfect square $\iff x=y$ or $3x=8y$ or $5x=8y$


calculus
integration
trigonometry
calculus computations

[quote="AndrewTom"]$ \int_{0}^{\frac {\pi}{2}} \cos x\ \sin x\ e^{\cos x}\ dx.$[/quote]
$ \int_{0}^{\frac {\pi}{2}} \cos x\ \sin x\ e^{\cos x}\ dx.\equal{}\minus{}\int_{0}^{\frac {\pi}{2}} \cos x d e^{\cos x}\equal{}\minus{}\cos x e^{\cos x}|_{0}^\frac {\pi}{2}\plus{}
\int_{0}^{\frac {\pi}{2}} e^{\cos x}d\cos x\equal{}2\minus{}e$

Let $ z \equal{} cosx$.
Then $ \minus{}dz \equal{} sinx dx$.
Also $ z(\frac{\pi}{2}) \equal{} 0$, and $ z(0)\equal{}1$.

$ \int_{0}^{\frac{\pi}{2}}\cos x\ \sin x\ e^{\cos x}\ dx \equal{} \int_{0}^{{1}}\ z e^{z}\ dz \equal{} (ze^z \minus{} e^z)|_{0}^1 \equal{} 1$

In Deng Qunyi's solution, when putting in the limits, it should be

$ (0 \minus{} \minus{} e) \plus{} [e^{\cos x}]_{0}^{\frac {\pi}{2}} \equal{} (0 \plus{} e) \minus{} (1 \minus{} e) \equal{} 1$,

giving the same answer as MrMagician.

I wanna try my integration by parts.

$ \int \cos x \sin x e^{\cos x}dx\equal{}\minus{}e^{\cos x}*\cos x\minus{}\int \sin x e^{\cos x}dx$
$ \equal{}\minus{}e^{\cos x}*\cos x\minus{}(\minus{}e^{\cos x}\plus{}C_1)\equal{}\minus{}e^{\cos x}*\cos x\plus{}e^{\cos x}\plus{}C\equal{}e^{\cos x}(1\minus{}\cos x)\plus{}C$.

$ \int_0^{\frac{\pi}{2}}\equal{}[e^{\cos x}(1\minus{}\cos x)]_0^{\frac{\pi}{2}}\equal{}e^0*1\minus{}e^1*0\equal{}1$.


If $ \displaystyle\lim_{n\to\infty}a_n\equal{}0$ then the sequence $ b_n$ given by $ b_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$ is a convergent sequence.

Can the question be rephrased to:

Is $ \lim_{n\to \infty} a_n\equal{}0$ sufficient and necessary for the series $ \sum_{n\equal{}1}^{\infty}a_n$ to converge?

The answer is: it is necessary, but not sufficient. Example, $ a_n\equal{}\frac{1}{n}$.


Let $M$ be a set in the plane with area greater than 1. Show that $M$ contains two distinct points $(x_1, y_1)$ and $(x_2,y_2)$ such that $x_2 - x_1$ and $y_2-y_1$ are integers.

Obviously we can restrict our attention to $[0,1]^2$ because $x - y \in Z \Leftrightarrow {x} - {y} \in Z$.

Now if a point is "doubled" then clearly those 2 points can be used.

Else, no point is doubled and $(1,1), (0,0)$ can be used.


I want some way of expressing the following y values in terms of x, using any type of math expressions (eg. summations, factorials, addition whatever!).

x:-1-|-2-|-3-|--4--|--5--
y: 1  |  3  | 15 | 105 | 945

As you can see there is an easy pattern - y(n)= (2n - 1).y(n-1) (or the y values are being multiplied by 1, 3, 5, 7... respectively to get the next y value) but can y be expressed in terms of x alone?

After exploring a bit i found some interesting things about the pattern, none of which have helped in the solving of this problem. For instance the equations for each y value (individually) are y1=2x-1  y2=(2x-1)(2x-3)  y3=(2x-1)(2x-3)(2x-5)
              OR y2=2x-1       OR y3=(2x-1)(2x-3)
However this one has stumped me for a while.

Cheers

You should do a wikipedia search about "double factorial".
$ y=(2x-1)!!=\frac{(2x)!}{2^{x}x!}$


calculus
integration
real analysis
function
algebra
polynomial

I have read in several places that the generalized Riemann integral is superior to the Lebesgue integral (it includes it as a special case), it has a simple treatment, .. etc.
(see the article of R.G. Bartle "Return to Riemann Integral" http://www.jstor.org/stable/pdfplus/2974874.pdf )

so why didn't the mathematical community put the Lebesgue integral aside and shift toward the more general (and simpler) integral?

Regards,

One answer I've seen is that the Henstock-Kurzweil integral doesn't generalize well to spaces other than $ \mathbb{R}$.  The Lebesgue integral, on the other hand, opens up the world of measure theory, which is widely applicable.

I don't really know anything about it, but the fact that the FTC holds for all differentiable functions has a weird feel. Since there are differentiable functions that are not of bounded variation, this means that the Kurzweil-integral can behave very badly in comparison to the usual integral. I cannot access JSTOR article (doesn't seem to be on MathSci), but I would like to know how a lot of fundamental theorems work in this more general setting. Is there a "space of all integrable functions" with a reasonable metric?

There are of course important improper integrals whose properties depend upon a lot of cancellation, but theorems about such are usually quite deep anyway, so the question is how much a more general framework for integrals helps.

[quote="Kalle"]Is there a "space of all integrable functions" with a reasonable metric?[/quote]
That's the big issue right there - that and the convergence theorems. Any metric space has an abstract completion. Take, say, $ C[a,b]$ with the 1-norm: that's a metric space that isn't complete. Is there any way to envision its completion as a set of functions - or at least, something close enough to being functions that we can think of them that way? The answer to this is precisely the Lebesgue space $ L^1.$ (Whose members aren't functions, but they're close enough.)

Of course, to prove that - to prove the completeness of $ L^1$ - we must make direct and powerful use of the Monotone Convergence Theorem and the Dominated Convergence Theorem.

The advantage of the Lebesgue integral doesn't really lie in its ability to integrate a relatively large set of functions. The advantage lies with those convergence theorems - MCT, Fatou, DCT.

That's one of the issues raised in [url=http://www.math.vanderbilt.edu/~schectex/ccc/gauge/]this webpage[/url]:

[quote]Here is another drawback to the gauge theory: The Lebesgue theory yields the Banach space $ L^1[a, b]$ of all Lebesgue-integrable functions. It's a very nice space, with a nice metric and nice convergence theorems. The gauge theory yields a slightly larger vector space, which we might call $ G[a, b]$. But the metric and the convergence theorems for this space are much less nice. That's not surprising: this space contains not only the nice functions of $ L^1[a, b]$, but also some additional functions that are rather nasty. If we don't really need to work with those nasty functions, we might prefer to stick to the setting of $ L^1[a, b]$, which has nicer worktools (i.e., the metric, the convergence theorems, etc.). [/quote]

[quote]That's not surprising: this space contains not only the nice functions of L^1[a, b], but also some additional functions that are rather nasty.[/quote]
Here's another way to say the same thing (and the same thing that I was saying earlier): there are all sorts of spaces of "nice" functions - step functions, continuous functions, $ C^{\infty},$ polynomials, trig polynomials, and so on - that are [i]dense[/i] in $ L^1.$ They can't be dense in that other space.

Thanks for the replies.
I quote this analogy from t0rajir0u's link which I see it meaningful:
[quote]This analogy may be helpful: The gauge integrable functions are like convergent series; then the Lebesgue integrable functions are like absolutely convergent series. The absolutely convergent series are easier to work with, and yield a tidier theory. The convergent series are more general, but only occasionally do we need to work with a series that is conditionally convergent.  [/quote]

Generalized Riemann may be superior [b]on the real line[/b].  I like it.

Have you looked into it for the plane?  Nowhere near the simplicity.

Lebesgue theory has natural extensions of vast generality.  Not just the plane, but manifolds, fractals, function-spaces, abstract probability, and so on.  Generalized Riemann, not.


Let x,y two real numbers with 1 \leq x2 - xy + y2  \leq 2. Show that:
1) 2/9 \leq x4 + y4 \leq 8;

2) x2n + y2n \geq 2/3n, (V) n natural number, n \geq 3.

$(x-y)^4 \ge 0 \Rightarrow x^4 + 6x^2y^2 + y^4 \ge 4x^3y + 4xy^3 \Rightarrow x^4 + y^4 \le 2(x^4 - 2x^3y + 3x^2y^2 - 2xy^3 + y^4) = 2(x^2-xy+y^2)^2 \le 8$

Equality for $(x,y) = (\pm \sqrt{2}, \pm \sqrt{2})$.

$(x-y)^4 + 6(x^2-y^2)^2 \ge 0 \Rightarrow 7x^4 + 7y^2 \ge 4x^3y + 6x^2y^2 + 4xy^3 \Rightarrow 9x^4 + 9y^4 \ge 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4 = 2(x^2+xy+y^2)^2$. If $x, y < 0$, replace $x, y$ with $-x, -y$ and all conditions will remain the same, so in this case and for $x, y > 0$, note that $2(x^2+xy+y^2)^2 \ge 2(x^2-xy+y^2)^2 = 2$. If one of $x, y$ is positive, WLOG x, replace $y$ with $-y$; the only thing changed will be $x^2-xy+y^2$ to $x^2+xy+y^2$. In either case, $x^4+y^4 \ge \frac{2}{9}$.

Equality for $(x,y) = (\pm \frac{1}{\sqrt{3}}, \mp \frac{1}{\sqrt{3}})$.


I realized the other day that I know almost nothing about American politics or even general politics for that matter.  Where is a good place on the web to learn about this or any other useful subjects for the real-world / college.

The best way to start learning about politics is to talk to people who already know something about it (although there are potential pitfalls, of course, mostly having to do with those who believe they Know things that are either opinion or just wrong).  An alternative method is to read the newspaper daily, especially one of the bigger national papers if you're interested in political news at the federal level.  I go with the New York Times, myself.  (Television news, especially cable, has really cruddy coverage if you want to figure out what's going on.  C-SPAN is far better in terms of accuracy but worse in terms of being incredibly boring most of the time.)  If you have a local NPR affilliate (check [url=http://www.npr.com]their website[/url]), find out when they run their news programs -- they tend to be more in-depth and well-thought-out than TV news, with good national and international segments.  Also, political blogs are an excellent resource, but you have to know how to sift the good stuff from the bad (and both from the plain-old crazy).

Another worthwhile thing to consider is to take to google and find out everything you can about your own representative, senators, state representatives, town councilpeople, etc.

Go to nytimes.com or any other very very well-known and prestiged news source and start reading some articles there (maybe look in international and US to start, but don't limit yourself). Radio programs are also good ideas.

For me I just read the paper and watch the news everyday. If something interists me, then I will google it to find out more. Its also good to see two sides to the story, ex. alleged muderers view, and the accusing officers view. This way, using both forms of imfermation, you may be able to recunstuct a better story about what has happened.

Well, I get information by reading the Economist, and listening to the radio on my way to mathcounts I almost always talk to dad about news. That's how I first got to know that Saddam was killed.

[quote="Basketball4Life"]Thanks for all the responses. The only issue I'm having is that I am outside the US for the year so my only source basically is the internet.[/quote]

No, in most cases fortunately  it's not. Start reading the news, I've spent long periods out of my country and it was never a problem to be informed about what happened in the world. You don't have to read the nytimes. If you are in a not very unusual country, surely there is a nation wide newspaper that treats abundantly the international issues. If you are in some unusual country where there isn't - this I've only seen in Africa actually - then you'll probably have to go to the internet but in such unusual countries there is a great deal of regional politics who tends to go unnoticed in the western world that you might consider interesting.

Ursula

Wikipedia is your friend.

From there you can get links to probably good sites.

Reading newspapers is not enough.  Neither was my education in school.  We spent several lessons nagging about how great the Belgian system was back in 1830 etc.. but never discussed anything about the present system.
I had already voted twice (on a federal and regional level) before I found out how many representatives there are in both parliaments and (more importantly) that we have electoral districts.
And I did watch television and read newspapers.
Sometimes the news can spend days and days about topics without properly explaining them, always giving "non-news" (basically that nothing has changed yet) without and then all of a sudden the topic disappears!

The internet is way better.  Wikipedia is good but be careful about sensitive topics.  But if you wanna know stuff like who is my representative(s) etc... Wikipedia is indeed reliable.


Yeah, um, I don't really understand recursion. I know what it means, but I'm not really sure how to use it in advanced problems. Could anyone explain?

And by advanced, I mean like this question:

[hide="Recursion question"]
I have 2048 letters numbered 1 to 2048. I have to address every single one. Originally they're stacked in order with #1 on top. To make the task a bit less mind-numbing, I address every other one, starting with #1. When I address a letter, I put it in my outbox. The ones I skip I stack as I skip them (so #2 is on the bottom of the stack after my first pass). After I finish my first pass, I have 1024 letters which are not addressed; #2048 is on top, #2 is on the bottom. I then repeat my procedure until there's only one letter left. What is that letter?
[/hide]

By the way, that is quoted word by word.

Recursion in mathematics and computer science is a method of defining functions in which the function being defined is applied within its own definition. The term is also used more generally to describe a process of repeating objects in a self-similar way. The Fibonacci sequence and factorial are the examples of recursion.


Euler
function
ratio
modular arithmetic
quadratics
number theory unsolved
number theory

how can one prove that: $\frac{\phi(n+1)}{\phi(n)}$ is dense in $R^{+}$
($\phi$ is the Euler function)

On the other hand, I propose to solve a more simple problem:

Prove that $\frac{\varphi(n+1)}{\varphi(n)}$ takes arbitrary small and arbitrary large values.

 ;)

First of all, in both cases we can look at the ratio $f(n)=\frac{\frac{\varphi(n+1)}{n+1}}{\frac{\varphi(n)}{n}}$. Take $n=p^t-1$, where $p$ is a prime, and $t$ is chosen s.t. $n=p^t-1$ has as many different prime factors as we want. Computing $f(n)$ in these conditions will show that we can make it arbitrarily large. Something similar should work for the other case (arbitrarily small), but it needs a tiny bit more work.

Unless I'm wrong, of course... :)

Yes, you are right, it was exactly my idea :)

As for the second part, I think my approach is a bit too complicated, but..

We take $n=p^t$ so that $n+1=p^t+1$ has as many prime factors as we want. We can accomplish this by taking $p=3$, for example, and considering a list of primes $p_1,p_2,\ldots,p_k>3$ s.t. $p_i\equiv 3\pmod 4,p_i\equiv 1\pmod 3$. We then have $\left(\frac p{p_i}\right)=-1$ (by Quadratic Reciprocity), so we can take $t=\prod_{i=1}^k\frac{p_i-1}2$ to get $p^t\equiv -1\pmod{p_i},\ \forall i\in{1,k}$.

I know there must be easier ways, but this is what I was thinking of when I posted my previous message.

It is a not sufficient argument, since you don't know (?) that $\prod_{i=1}^k \left(1-\frac{1}{p_i}\right)$ can be an arbitrary small value. Am I right?

Oh, darn! I'll have to rethink the whole thing now...

I used Dirichlet's Theorem in my last argument without mentioning it, so I'll use it here too :). Again, I'm sure there are easier ways.

There is a prime $p$ s.t. $p_1,p_2,\ldots,p_k|p+1$, where $(p_n)_{n\ge 1}$ is the sequence of the primes. Take $n=p$.

Is it Ok now?

Have any one a complete solution of first problem?


Hi. I'm taking the SAT for the first time this year as an 8th grader.

I can get an 800 math, 650-700 writing, and 570-630 critical reading.

Is this something to be ashamed of as an 8th grader? Because a few of my friends can do 2260 total.

No need to be ashamed!  As an 8th grader, if you don't like your score, there are still four years to work on it.

There is nothing to be ashamed of. Also, your 8th grade SAT score is completely inconsequential.

If you invest time into improving your reading, writing, speaking, and overall command over the language, chances are you'll do a lot better when it actually matters.

[quote="monkeygirl13"]If this was like your real score to apply into college you may not be able to get into Princeton[/quote]

Building onto this point, my eighth grade SAT score was in the 1800s or so (CR ~ 500). Guess what college I am attending.

The SATs are generally unimportant, especially those that you take during middle school, and if you score in the ranges you indicated now, by the time you take them as the real thing (and with a bit of work), your score will probably be just as high as those of your friends indicated (who, by the way, aren't a very good comparison for a normal score).

Maybe by 10th grade you will get in the range your friends indicated.

I answered half of the english questions (both critical reading and writing) but I got an 8 on the actual essay (out of 12).  I also NEVER studied whatsoever but I ended up with a 720 Math, 500 Reading, and 430 Writing.  A 720 on math is the 99.999999999999999 percentile for a 7th grader.  Even my reading score is way above average because average is a 510 for a 12th grader.  Just thought I would share how I did (you guys did much better though -- GREAT JOB!).

To the OP: As the post above states, your score amongst your age group is already amongst the very highest percentiles (I could check with my CTY data if you're interested).  Even competing with people years above you, I believe that score is in the 97th or above percentile, it is definitely nothing to be ashamed about.

There's a reason that the SAT erases scores from tests taken in middle school (unless explicitly required not to), people change significantly in the approximately three or four years from first taking the SAT in MS (if they do at all) to finally taking them junior year, don't stress about your score compared to your peers.

[quote="monkeygirl13"]If this was like your real score to apply into college you may not be able to get into Princeton or another good school like that but you beat out about half of all SAT takers..[/quote]
More like 90% of all SAT takers.  I remember I got 710 Math, which it said was 91st percentile, 630 CR(83rd percentile), and 600 W(81st percentile).  I hate the math curve.  4 questions wrong = 710(if you left the 4 blank, 720).  The CR and Writing curves are easier but the test is significantly harder.
It's definitely nothing to be ashamed of. Especially the 800 Math.  I always make 2-4 stupid mistakes on Math, usually in Algebra.

[quote="monkeygirl13"]If this was like your real score to apply into college you may not be able to get into Princeton or another good school like that but you beat out about half of all SAT takers. Considering that you have 4 years to bring up your CR and writing scores you'll do great when you take the SAT in high school.[/quote]

Actually, you can most definitely get into Princeton or even Harvard with that sort of score. Princeton does not really look at SAT scores that much. 2100 to 2200 range is fine for them. CR is all vocabulary so I suggest ypu imrpive your vocab. Study root and stems, they help you the most. Writing is just experience. 4 yeard of high school will cover that for you. I think you will get like a perfect in high school with what you are doing now.

http://www.princeton.edu/admission/applyingforadmission/admission_statistics/
2200 won't do much for you as far as getting you in or keeping you out if you want to get into Princeton etc.


yay
it is what now

tell me if you want to join

yes beat ra to making this topic
 :lol:

dude budi edit your post to match ra's...when he posts

rules for "what now"
1. There are some number of players
2. One of them is "the ding" (usually it is dingzhou)
3. The remaining players either say "what what", "what now", "now what", or "now now"
4. If all but one of them says the same thing (like there are five players, one is the ding, three of the remaining four say "what now", and the last says "now what", then the last guy is the person who does not say the same thing), then the one of them which did not say the same thing gets a point
5. If anything else happens when the players other than "the ding" say "what what", "what now", "now what", or "now now", then "the ding" gets a point
6. When one player gets three points, then that person wins
this is a game we made up at nats, and it is totally awesome
somehow though not many people were interested in playing it other than dingzhou
i wonder why

it is normally played with 4 players or else "the ding" will win way too much

i'll join (tho i still dont really get it)

one of my blog entry's has the directions...
srry im not sure wut the URL is so search me on the memberlist, click blog and find the RLY FUN GAME
in this post, 16 ppl can join.
the DING will be randomly chosen
in the first round, the 16 ppl will be broken into 4 groups of 4
these groups will face off and the 4 winners *1 from each group* will advanced to the final game
more details l8r

I'm joining, but maybe we should have multiple dings to even this out a bit...

/in
even tough i don't get what i'm supposed to do  :D

[quote="Arvind_sn"]IM IN CUZ I PLAYED DIS AT NATS 24 7!!![/quote]

nats wasn't 7 days lawlz

/in if there's room.

Played this at Nats with Arvin as well.

I will play, but how does everyone say "what now", "now what", "what what", and "now what" at the same time?

Just use darn PMs and post what you decided

I really don't care either way. Both PM's or FTW are fine.

But FTW might have some problems with timing and people not being able to make it.

For some reason, I think its going to take forever to do this game on FTW or PM... Most people have already given up.

you are supposed to pm me 9 responses not just 1

are we close to having all of the entires in yet? Or has everyone given up on this game?

Wow, this game is fun, and addictive.

I played it with iYOA, superbeast, and permutations. We rotated who was Ding. superbeast won 2 out 4, so he won. He even won 3-0-0-0  when he wasn't Ding.

Can I play next tournament?

With the looks of how well this one's progressing, I don't think they'll be a next one.

[quote="12markkram34"]Hey I remember this game from the Denver zoo...it made no sense over there but now it seems perfectly logical![/quote]

oh i saw u 12markkram34! in front of the penguin place, i was one of the 4 idiots dancing around while it was snowing....

I remember watching Arvind_sn play this game on Stevenmeow's calculator.

Good times...


Let $a,b,c$ be non-negative numbers. Prove that

$a\sqrt{a^2+3bc}+b\sqrt{b^2+3ca}+c\sqrt{c^2+3ab} \ge 2(ab+bc+ca)$.

I will thank Valentin as Vornicu-Schur inequality solves it..  :D

My solution:
\[
\begin{gathered}
  \sum {a\sqrt {a^2  + 3bc} }  \geqslant 2\left( {\sum {ab} } \right) \hfill \\
   \Leftrightarrow \sum {a\left( {\sqrt {a^2  + 3bc}  - b - c} \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {a\frac{{a^2  + 3bc - (b + c)^2 }}
{{\sqrt {a^2  + 3bc}  + b + c}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {a\frac{{a^2  + bc - b^2  - c^2 }}
{{\sqrt {a^2  + 3bc}  + b + c}}}  \geqslant 0 \hfill \\
\end{gathered}
\]

Denote \[
f(a) = \frac{1}
{{\sqrt {a^2  + 3bc}  + b + c}}
\]
If $a \ge b$, then
\[
\begin{gathered}
  \frac{1}
{{\sqrt {a^2  + 3bc}  + b + c}} \geqslant \frac{1}
{{\sqrt {b^2  + 3ca}  + c + a}} \hfill \\
   \Leftrightarrow \sqrt {b^2  + 3ca}  + c + a \geqslant \sqrt {a^2  + 3bc}  + b + c \hfill \\
   \Leftrightarrow \sqrt {b^2  + 3ca}  + a \geqslant \sqrt {a^2  + 3bc}  + b \hfill \\
   \Leftrightarrow a^2  + b^2  + 3ac + 2a\sqrt {b^2  + 3ca}  \geqslant a^2  + b^2  + 3bc + 2b\sqrt {a^2  + 3bc}  \hfill \\
   \Leftrightarrow a^2  + b^2  + 3ac + 2\sqrt {a^2 b^2  + 3ca^3 }  \geqslant a^2  + b^2  + 3bc + 2\sqrt {a^2 b^2  + 3b^3 c}  \hfill \\
   \hfill \\
\end{gathered}
\]
So $f(a) \ge f(b)$. That means f is increasing function.
Now, we have to prove
\[
\sum {a\frac{{a^2  + bc - b^2  - c^2 }}
{{\sqrt {a^2  + 3bc}  + b + c}}}  \geqslant 0
\]
or,
\[
\sum {af(a)\left( {a^2  + bc - b^2  - c^2 } \right)}  \geqslant 0
\]
 by vornicu-schur, \[
\sum {af(a)\left( {a^2  + bc - ab - ac} \right)}  \geqslant 0
\]
Now we only have to prove,
\[
\begin{gathered}
  \sum {af(a)\left( {ab + ac - b^2  - c^2 } \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {f(a)\left( {ab(a - b) + ac(a - c)} \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {ab\left( {a - b} \right)\left( {f(a) - f(b)} \right)}  \geqslant 0 \hfill \\
\end{gathered}
\]
which is true.

[quote="siuhochung"]by vornicu-schur,
$\sum {af(a)\left( {a^2  + bc - ab - ac} \right)}  \geqslant 0$[/quote]
Sorry but I didn't understand that  :?
and what it is "vornicu-schur" ???

[quote="Valentin Vornicu"][quote="siuhochung"]I will thank Valentin as Vornicu-Schur inequality solves it..  :D[/quote] :blush:  :blush:  :P  :P[/quote]
 :D  :D  :D

i really like your solution ,Sihochung  :)

I think that this inequality is right $f(a,b,c) \geq f(a, \sqrt{bc}, \sqrt{bc})$, here $f(a, b,c) = LHS - RHS$, $a \leq b \leq c$, but I have no proof of that. :(

hi  :D
I think that i have another sollution :

by jensen inequality :

$ w_1f(x_1)\plus{}w_2f(x_2)\plus{}w_3f(x_3)\, \succeq \, (w_1\plus{}w_2\plus{}w_3)f(\frac{w_1x_1\plus{}w_2x_2\plus{}w_3x_3}{w_1\plus{}w_2\plus{}w_3})$

with :

$ f(x)\equal{}\sqrt{x} \,\, and \,\, w_1\equal{}a \,\, and \,\, w_2\equal{}b \,\, and \,\, w_3\equal{}c \,\, and \,\, x_1\equal{}a^2\plus{}3bc \,\, and \,\, \,\,\,\,x_2\equal{}b^2\plus{}3ac \,\, and \,\, x_3\equal{}c^2\plus{}3ba$

so we have :

$ a\sqrt{a^2\plus{}3bc}\plus{}b\sqrt{b^2\plus{}3a}\plus{}c\sqrt{c^2\plus{}3ba}\,\succeq\,\sqrt{abc}\sqrt{a^3\plus{}b^3\plus{}c^3\plus{}9abc}$

now by shur we have:

$ a^t(a\minus{}b)(a\minus{}c)\plus{}b^t(b\minus{}a)(b\minus{}c)\plus{}c^t(c\minus{}a)(c\minus{}b)\,\succeq\,0$

put $ t\equal{}2$:

$ a^4\plus{}b^4\plus{}c^4\plus{}bca^2\plus{}cab^2\plus{}abc^2\,\succeq\, a^3c\plus{}a^3b\plus{}b^3a\plus{}b^3c\plus{}c^3a\plus{}c^3b$

so :

$ (a\plus{}b\plus{}c)(a^3\plus{}b^3\plus{}c^3\plus{}9abc)\,\succeq\,2(a^3c\plus{}a^3b\plus{}b^3a\plus{}b^3c\plus{}c^3a\plus{}c^3b)\plus{}8(bca^2\plus{}cab^2\plus{}bac^2)$

and we have:

$ 2(a^3c\plus{}a^3b\plus{}b^3a\plus{}b^3c\plus{}c^3a\plus{}c^3b)\plus{}8(bca^2\plus{}cab^2\plus{}bac^2)\,\succeq\,4(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2bca^2\plus{}2acb^2\plus{}2bac^2)$

because ;

$ a^3c\plus{}c^3a\,\succeq\,2a^2b^2$

then :

$ (a\plus{}b\plus{}c)(a^3\plus{}b^3\plus{}c^3\plus{}9abc)\,\succeq\,4(ab\plus{}ac\plus{}bc)^2$

$ \longleftrightarrow\,\sqrt{a\plus{}b\plus{}c}\sqrt{a^3\plus{}b^3\plus{}c^3\plus{}9abc}\,\succeq\,2(ab\plus{}ac\plus{}bc)$

so:

$ a\sqrt{a^2\plus{}3bc}\plus{}b\sqrt{b^2\plus{}3ca}\plus{}c\sqrt{c^2\plus{}3ab}\,\succeq\,2(ab\plus{}ac\plus{}bc)$

[quote="memath"]By jensen inequality :

$ w_1f(x_1) \plus{} w_2f(x_2) \plus{} w_3f(x_3)\, \succeq \, (w_1 \plus{} w_2 \plus{} w_3)f(\frac {w_1x_1 \plus{} w_2x_2 \plus{} w_3x_3}{w_1 \plus{} w_2 \plus{} w_3})$

with :

$ f(x) \equal{} \sqrt {x} \,\, and \,\, w_1 \equal{} a \,\, and \,\, w_2 \equal{} b \,\, and \,\, w_3 \equal{} c \,\, and \,\, x_1 \equal{} a^2 \plus{} 3bc \,\, and \,\, \,\,\,\,x_2 \equal{} b^2 \plus{} 3ac \,\, and \,\, x_3 \equal{} c^2 \plus{} 3ba$

so we have :

$ a\sqrt {a^2 \plus{} 3bc} \plus{} b\sqrt {b^2 \plus{} 3a} \plus{} c\sqrt {c^2 \plus{} 3ba}\,\succeq\,\sqrt {abc}\sqrt {a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc}$[/quote]
It is wrong, since $ f(x)\equal{}\sqrt x$ is not convex.

[quote="Vasc"]Let $ a,b,c$ be non-negative numbers. Prove that

$ a\sqrt {a^2 \plus{} 3bc} \plus{} b\sqrt {b^2 \plus{} 3ca} \plus{} c\sqrt {c^2 \plus{} 3ab} \ge 2(ab \plus{} bc \plus{} ca)$.[/quote]
My solution :)

can you write your solution here . i dont have M.Word

Nice your solution Can_hang.
I attached it in PDF for Memath.


If you have a bucket of 10 coins, and you flip each one of them, what is the probability that 6 or more of the coins will land on heads?
What about 15 coins, getting 11 or more coins being heads up?

[quote="ashwinrk_jain"]$\frac{5}{1024}$
in the case of 15 the answer is
$\frac{5}{2^{15}}$[/quote]
Are you sure?
For 10, having 6 be heads: $\frac{\binom{10}{6}}{\binom{2}{1}^{10}}=\frac{210}{1024}=\frac{105}{512}$ And so on....


After applying inversion of pole $B$, if, in general, we denote the image of $X$ through this inversion by $X'$, the statement is "reduced" (:)) to $\frac{\frac 1{BM_1'}-\frac 1{BM_2'}}{\frac 1{BM_2'}-\frac 1{BM_3'}}=\frac{\frac 1{BK_1'}-\frac 1{BK_2'}}{\frac 1{BK_2'}-\frac 1{BK_3'}}$. This looks nastier, but after completing the computations, it's easily seens to be equivalent to the fact that the four lines ($AB$ and the three images of the circles through the inversion) determine the same cross-ratio on $BM_1',BK_1'$, which is obvious because the four lines are concurrent at $A'$.

[quote="orl"]Three circular arcs $w_1,w_2,w_3$ with common endpoints $A$ and $B$ are on the same side of the line $AB$; $w_2$ lies between $w_1$ and $w_3$. Two rays emanating from $B$ intersect these arcs at $M_1,M_2,M_3$ and $K_1,K_2,K_3$, respectively. Prove that $\frac{M_1M_2}{M_2M_3}  = \frac{K_1K_2}{K_2K_3} $.[/quote]

This problem was already discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=14550 , so I didn't want to open a new thread for it.

  Darij


topology
real analysis
real analysis unsolved

Conside the collection of subsets of [b]C[i]n[/i][/b] which are the zero set of some finite collection of polynomials. Show that this collection of subsets form the closed sets of a topology on [b]C[i]n[/i][/b].[b]

In the case n=1. Show that C with this topology is not hausdorff.

How can I do this problem without involving any abstract
algebra???

[/b]

The fact that they are the closed sets of a topology is obvious.
In order to show that this topology is not Hausdorff, ask yourself the question: if $O$ and $U$ are two nonempty open sets for this topology, can their intersection be empty ?

Is the topology not Hausdorff for any $n$?

It's never Hausdorff, because two nonempty open sets will never be disjoints.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that $A_n$ is the only subgroup of $S_n$ with index of 2.

Otherwise, that group would have a subgroup of index $2$, formed by the even permutations in it, and that group would have index $2$ in $A_n$ as well, which means that $A_n$ has a normal subgroup. This is impossible for $n\ge 5$ (since then $A_n$ is simple), and you can check it manually for smaller $n$'s.

Another approach:

Suppose $H$ is a normal subgroup of $S_n$ with index of 2. We claim that $H$ should not contain any transpositions. (If it contains one, then it can contain all transpositions from the normality. Since $S_n$ is generated by transpositions, $H=S_n$, and it is a contradiction). Thus $S=H\cup H$, where $'H'=aH$ for any transposition $a$.
Thus $H=(H')^2$.

Now let a permutation $\sigma = \tau_1\tau_2 \cdots \tau_m\in H$, where $\tau_i$ are transpositions. Then
$\sigma H = \tau_1 H \tau_2 H \cdots \tau_m H = (H')^m$
and it is $H$ or $H'$ depending on the parity of $m$.


What is the minimum possible value of $\frac{a}{2b}+\frac{b}{4c}+\frac{c}{8a}$, where a, b, and c are positive real numbers?

Just use simple AM-GM

We have $\frac{a}{2b}+\frac{b}{4c}+\frac{c}{9a}\geq 3 \cdot\sqrt[3]{\frac{abc}{2b \cdot 4c \cdot 8a}}=\boxed{\frac{3}{4}}$


1. the decimal expansion of 1/7 has a six-digit block of repeating digits. what is the maximum length of the block of repeating digits in the decimal expansion of 1/n? explain.

2. are there more rational or irrational numbers? why?

3. is the sum of two sine curves always another sine curve? prove it if it is.

[hide="1"] Let $n = ab$ where $gcd(b, 10) = 1$.  It's not hard to prove that the length of the repeating block is $ord_{b}(10)$, which divides $\lambda(b)$, which divides $\varphi(b)$, which is less than or equal to $b-1$, which is less than or equal to $n-1$.  This is because

$0.\overline{d_{1}d_{2}d_{3}}= \frac{abc...}{999...}= \frac{1}{b}$

So $\frac{999...}{b}= d_{1}d_{2}d_{3}$.  In other words, if the length of the repeating block is $l$, then $b | 10^{l}-1$.  [/hide]
[hide="2"] This is well-known.  [url=http://mathworld.wolfram.com/CantorDiagonalMethod.html]Cantor's diagonal argument[/url] is one way of showing that the cardinality of the irrationals is larger than the cardinality of the rationals (which are countable). [/hide]
[hide="3"] If and only if they have the same frequency.  In that case,

\begin{eqnarray*}A \sin (\omega t+\phi)+B \sin (\omega t+\varphi) &=& (A \cos \phi+B \cos \varphi) \sin (\omega t)+(A \sin \phi+B \sin \varphi) \cos (\omega t) \\ &=& \sqrt{ (A \cos \phi+B \cos \varphi)^{2}+(A \sin \phi+B \sin \varphi)^{2}}\cos (\omega t-\arctan \left( \frac{A \cos \phi+B \cos \varphi}{A \sin \phi+B \sin \varphi}\right) ) \\ &=& \sqrt{ A^{2}+B^{2}+2AB \cos (\phi-\varphi) }\cos (\omega t-\arctan \left( \frac{A \cos \phi+B \cos \varphi}{A \sin \phi+B \sin \varphi}\right) ) \\ \end{eqnarray*}  [/hide]


inequalities
trigonometry
function
geometry proposed
geometry

[quote="gemath"]Given triangle $ABC$ prove that
$\sin^{3}A+\sin^{3}B+\sin^{3}C\le \frac{9\sqrt{3}}{8}$[/quote]

Aplying Jensens's inequality for de function $f: (0,\pi)\to R$, $f(x)=sin^{3}x$ we got $f\left(\frac{x+y+z}{3}\right)\ge \frac{f(x)+f(y)+f(z)}{3}$. So $sin^{3}\left(\frac{A+B+C}{3}\right)=\frac{3\sqrt{3}}{8}\ge \frac{sin^{3}A+sin^{3}B+sin^{3}C}{3}$.

[quote="gemath"]But how can we show $f(x)$ is convex ? :maybe:[/quote]
That's very hard to show, because $f(x)$ is not convex, nor concave on the given interval.  :wink:


$ a_1\equal{}2,a_2 \equal{} 8, a_n \equal{} 4a_{n \minus{} 1} \minus{} a_{n \minus{} 2}, n  \geq 3$, define

$ \sum_{n \equal{} 1}^{\infty} arccot{ a_n^2}$

$ arccota_n^2 \equal{} \lambda( arccot b_n \minus{} arccot b_{n \plus{} 1})$

the  result  is  $ \frac {\pi}{12}$,  but I don't  know  how  to  get  it.

are  there  any  other  method?

$ a_0 \equal{} 0$,

$ arccota_n^2 \equal{} arccot(\frac {a_n^2 \minus{} a_{n \minus{} 1}^2}{2} \plus{} \frac {a_n^2 \plus{} a_{n \minus{} 1}^2}{4}\sqrt {3}) \minus{} arccot(\frac {a_{n \plus{} 1}^2 \minus{} a_{n}^2}{2} \plus{} \frac {a_{n \plus{} 1}^2 \plus{} a_{n}^2}{4}\sqrt {3})$


Fredrico paid $ \$1.44$ to make 24 copies at the copy center. At this rate, how many dollars would it cost him to make 50 copies?

Each copy costs the same, so we have two choices here - proportions, or finding the cost of one copy and multiplying by 50. In different cases, one of these two methods is faster. Because of the easy numbers, both of these are sufficiently quick.

If we were using proportions, we would have \[ \frac{144}{24} \equal{} \frac{x}{50}\] Notice that I used cents to avoid decimals. We cross-multiply to get $ 50 \cdot 144 \equal{} 24x$. Dividing by 24 gives us $ \frac{50 \cdot 144}{24} \equal{} x \implies 25 \times 12 \equal{} x \implies 300 \equal{} x$. Therefore, it costs 3 dollars.

Alternatively, we would have seen that each copy costs $ 144 \div 24 \equal{} 6$ cents. Since Fredrico needs 50 copies, this is $ 50 \times 6 \equal{} 300$ cents, or 3 dollars.


operation x*y is defined by x*y=2x+y
 find x if 4*x=(2x)*3

[hide="My Solution"]

4*x=8+x

2x*3=4x+3

x+8=4x+3.  Solve to get

$x=\frac{5}{3}$[/hide]


Find all pairs $(x,y)$ integers, such that:

$9^{x}-7^{y}=2$

$7^{y}= 9^{x}-2 \equiv 3 (mod 4) \Rightarrow y \equiv 1 (mod 2) \Rightarrow 7^{y}\equiv 2 (mod 5)$ or $3 (mod 5)$
$9^{x}= 7^{y}+2 \equiv 4 (mod 5)$ or $(mod 5)$, but $9^{x}$ can only be $\equiv 4 (mod 5) \Rightarrow x \equiv 1 (mod 2) \Rightarrow 9^{x}\equiv 9 (mod 10)$
$7^{y}= 9^{x}-2 \equiv 7 (mod 10) \Rightarrow y \equiv 1 (mod 4) \Rightarrow 7^{y}\equiv 7 (mod 100)$
$9^{x}= 7^{y}+2 \equiv 9 (mod 100)$

[Sorry... Unfinished solution....]  :blush:

Thus, $(x,y) = (1,1)$

It can't be periodic after $31$:
$9$ is invertible $\mod 100$, thus $9$ will reoccure.
In fact, the sequence is wrong, the correct one is $9,81,29,61,49,41,69,21,89,1,9,...$

PS: you wrote $9^{y}$ instead of $9^{x}$ in the last lines.

$9^{x}=2+7^{y}$. If y=0 no solution, if y=1 then x=1 is solution.
If y>1 then $9^{x}=2(mod 7^{y})$ if and only if $x=1(mod 6*7^{y-1})$. Ig x>1 it give $9^{x}>2+7^{y}$.

Is correct the solution of RdeepMath91??  Its confused!!!

Please Zetax, help me!!!
 :oops:

His solution is correct till the point where he considers the explicitely given sequence which he calculated wrong.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=48431 , the method mentioned in the second post (by maxal) works here, too.
This method works indeed very often (see the rest of that thread for more examples).

mmm.... do you can explain the solution of my problem??
I dont understand to RDeepMath91!!!
Please, i see that technique is very interesting...


Demonstrate that is possible to color a trivalent graph (every vertex has degree 3) With 2 colors so that one vertice has maximun one adjacent vertices with the same color.

Bye, Daniele!

Are you allowed to color every vertex the same color?

Okay, that's better.  Actually, I have seen this problem before, so I'll leave it for others.


I write with a normal pencil. Please specify the brand of the pencils you use. I use Executive Choice (its the whole sale company; my dad got them from work, they write very nicely)

mechanical. BIC. the typical ones. or the korean mechanical pencils. :)

but otherwise, pen. pilot g-2. or korean ones. steadtler pens are the best though. but expensive.

i used to write with pens but they are very unreliable. But, i still use some fountain pens. got em from india, a lot cheaper, in the USA they cost like $ \$300$!

[hide="PhD Ultra Mechanical Pencils"]
[img]http://www.staples.com/sbd/img/cat/enl/s0048752_enl.jpg[/img]
[/hide]

Perfectly weighted and balanced for spinning.  :) Costs anywhere from 5-9 USD, depending on the store.

Pen. The sole exceptions are the AMC (only the real) and the AIME (real and practice). Of course, I sometimes use pencils outside of mathematics, too.

For pens, I like BIC Atlantis (black or blue) for regular use.

For pencils, I like Ticonderoga or Staples regular pencils. I can't stand mechanical pencils.

I just use whatever pecncils I can find around. Usually, they are mechanical pencils. I don't like using pens because I tend to make write-oes (based off typoes, except in writing), so I can't erase my mistakes. On some contests, I tend to bubble in 'C' when I got 'B' as the answer.

i used to use pens but i find them very annoying because they write all blotty unless they're one of the VERY good kind.


$ ABC$ is an arbitrary triangle , the external bisectors of $ < BAC, < ABC, < BCA$ hit the straightlines $ BC,CA,AB$ at $ U,V,W$ respectively. Prove that if $ P$ is a radom point chosen on the line $ UVW$ then one of the distances from $ P$ to the sidelines of the triangle $ ABC$ equals the sum of the other two distances

[quote="shoki"]sth like this is also true if we consider the line passing through two of the intersections of the internal bisectors with their opposite side,(and obviously,this line passes also through $ U$ or $ W$ or $ V$ depending which two sides we consider)[/quote]

Yes shoki it is also true for the three lines conecting the feet of the internal bisectors , mi program says it is truly too :D


Find the positive integer solutions of the equation
$2\sqrt{x}+3\sqrt{y}=10\sqrt{10}$

Justify your answer

$(40,40)$ any others? my solution's ugly. you don't wanna see it. neways here's the solution:
squaring we get $4x+9y+12\sqrt{xy}=1000$ so $xy=k^{2}$=>$x=\frac{k^{2}}{y}$
substituting this into the original equation, $4k^{2}+9y^{2}+12xy=1000y=(2k+3y)^{2}$
so $2k+3y=10\sqrt{10y}$=>$y=10t^{2}$
now we have $2k+30t^{2}=100t$ solving, $t=\frac{50\pm\sqrt{625-15k}}{30}$
so $625-15k=e^{2}$=>$15k=(25+e)(25-e)$ from here we get $e=5,10,20$
=> $k=40,35,15$
=> $t=1,2,3$
=> $y=10,40,90$
=> $x=160,40$
=> $(x,y)=(160,10),(40,40)$
but (160,10) doesn't satisfy the equation( i don't know why)
hence the result follows.$\blacksquare$.

ps. i've skipped the details because then the solution will get pretty ugly. :D

Another approach...  :)
Let $(x,y) = d, x = dp, y =dq$
By squaring we get $4x+3y+12\sqrt{xy}= 1000 \Rightarrow xy$ is a square. Since $(p,q) = 1$, then$p = A^{2}, q = B^{2}$ where $A,B \in \mathbb{N}$
The equation becomes $(2A+3B)\sqrt{d}= 10\sqrt{10}\Rightarrow d = 10k^{2}\Rightarrow k(2A+3B) = 10$
$2A+3B \geq 5 \Rightarrow k \leq 2$
i) $k = 2 \Rightarrow 2A+3B = 5 \Rightarrow d = 40, A = B = 1 \Rightarrow x=y=40$
ii) $k = 1 \Rightarrow 2A+3B = 10 \Rightarrow d = 10, A = B = 2 \Rightarrow x=y=40$


[quote="anirudh"][quote="G-UNIT"]I picked four integers.  Now in all possible ways, I pick 3 of them and every time I add those 3 numbers.  Finally, I add all these sums.  The result is 51.  The product of my 4 numbers is 216.  What are my numbers?[/quote]

How do you do this problem[/quote]
just take a peak at my post :D :roll: