A particle is placed on the parabola $ y \equal{} x^2 \minus{} x \minus{} 6$ at a point $ P$ whose $ y$-coordinate is $ 6$. It is allowed to roll along the parabola until it reaches the nearest point $ Q$ whose $ y$-coordinate is $ \minus{}6$. The horizontal distance traveled by the particle (the numerical value of the difference in the $ x$-coordinates of $ P$ and $ Q$) is:

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 3\qquad
\textbf{(D)}\ 2\qquad
\textbf{(E)}\ 1$

This is just the difference of the x coordinates of P and Q. We set the y value to be 6 first.

$ x^2\minus{}x\minus{}6\equal{}6$
$ x^2\minus{}x\minus{}12\equal{}0$
$ (x\minus{}4)(x\plus{}3)\equal{}0$
$ x\equal{}4$ or $ \minus{}3$

Then, we set it equal to -6.

$ x^2\minus{}x\minus{}6\equal{}\minus{}6$
$ x^2\minus{}x\equal{}0$
$ x(x\minus{}1)\equal{}0$
$ x\equal{}0$ or $ 1$

But we know it isn't 0 this is the minimum point of the parabola ($ y\equal{}...\minus{}6$). It can't be at $ (0,\minus{}6)$, because we have a $ \minus{}x$. So it must be $ 1$.

We find that $ |4\minus{}1|\equal{}3$ and $ |\minus{}3\minus{}1|\equal{}4$. So the answer is either 3 or 4, but I don't know which. I got confused from here on.

Continuing pi, if it is on the 4x, it goes 3 to 1, and if on the -3x, it goes 3 to 0. So the answer is C.


The sequence $ u_1, u_2, u_3, ...$ is defined by $ u_1\equal{}3$ and $ u_{n\plus{}1} \equal{} 3u_{n} \minus{} 2$.

(i) Find $ u_2$ and $ u_3$ and verify that $ \frac{1}{2}(u_{4} \minus{} 1) \equal{} 27$.

(ii) Hence suggest an expression for $ u_{n}$.

(iii) Use induction to prove that your answer to part (ii) is correct.

Are there other ways of finding an expression for $ u_{n}$?

[quote="AndrewTom"]The sequence $ u_1, u_2, u_3, ...$ is defined by $ u_1 \equal{} 3$ and $ u_{n \plus{} 1} \equal{} 3u_{n} \minus{} 2$.

(i) Find $ u_2$ and $ u_3$ and verify that $ \frac {1}{2}(u_{4} \minus{} 1) \equal{} 27$.

(ii) Hence suggest an expression for $ u_{n}$.

(iii) Use induction to prove that your answer to part (ii) is correct.

Are there other ways of finding an expression for $ u_{n}$?[/quote]

$ u_{n\plus{}1}\minus{}1\equal{}3(u_n\minus{}1)$, $ u_1\minus{}1\equal{}2$

So $ u_n\minus{}1\equal{}2(3)^{n\minus{}1} \Rightarrow u_n\equal{}2(3)^{n\minus{}1}\plus{}1$

Any other methods (besides ther obvious one outlined in (i), (ii) (iii) and the nice one by stephencheng)?

You complained in an earlier thread about your poor rating.  But look at what you've done here: you've made two consecutive posts, each with essentially zero content, within half an hour of each other and very shortly after the previous response.  Most readers bothering to rate will almost certainly rate these posts as "poor" or "spam," for obvious reasons.  Since a substantial majority of your posts on this forum seem to have no content other than "Any other solutions?" it seems completely non-surprising to me that your total rating is struggling to reach above 2.

Some specific suggestions:
You should learn to make use of the "edit" button if you have comments to add to a thread and the last post is your own; you should exhibit patience in waiting to see if other people will comment; you should try to make comments that might be of interest to other readers; and you should learn how to use the "quote" button when you want to figure out how some other user has managed some particular feat of formatting, LaTeX, etc.  (Doesn't apply in this thread, but I'm not going to go around posting messages like this on every single pointless post you've made, so I might as well concentrate suggestions here.)

Joel, thank you for your unpleasant post.

Did you expect an explanation of why you have a poor rating (and how you can modify your behavior to improve it) to involve many complements?   :wink:

By the way, something like stephencheng's method works in some substantial generality: if you know the method of solving homogeneous linear recursions (like the recursion for the Fibonacci numbers, say) via characteristic equation or similar, then you can also solve equations of this sort (where "2" is replaced by any constant, or more generally any other function) by finding the homogeneous solution and a particular solution (in strong analogy with differential equations).  What stephencheng has done is shown a cute way to find a particular solution.

One place it may need modification (when we have a nonzero constant like "2" as the function in question) is if the characteristic equation is divisible by $ x \minus{} 1$ (or, if the function had as a summand $ a \cdot b^n$, if the characteristic equation were divisible by $ x \minus{} b$) -- in this case, when solving for the particular solution we act as if we were solving for the homogeneous solution but with an extra factor of $ x \minus{} 1$ (respectively, $ x \minus{} b$) tossed in for good luck.

For example, the equation $ u_{n \plus{} 1} \minus{} u_n \equal{} 0$ has solutions $ u_n \equal{} c$, but the equation $ u_{n \plus{} 1} \minus{} u_n \equal{} 2$ has solutions $ u_{n} \equal{} 2n \plus{} c$, which looks like a solution to the recursion $ u_{n \plus{} 2} \minus{} 2u_{n \plus{} 1} \plus{} u_n \equal{} 0$.


quadratics
algebra
polynomial
search
ceiling function

If you work out $ (2\plus{}\sqrt{3})^{50}$ on a computer you will find the answer

39571031999226139563162735373.999999999999999999999999999999999974728............

Explain why this answer is so close to an integer.

(Hint:Try to use the idea by constructing a suitable sequence.)

[quote="kelvin911"]If you work out $ (2 \plus{} \sqrt {3})^{50}$ on a computer you will find the answer

39571031999226139563162735373.999999999999999999999999999999999974728............

Explain why this answer is so close to an integer.

(Hint:Try to use the idea by constructing a suitable sequence.)[/quote]
because $ (2\minus{}\sqrt{3})^{50}\plus{}(2\plus{}\sqrt{3})^{50}$ is an integer and $ 2\minus{}\sqrt{3}$ is less than $ 1$.

thank you, very nice.  i feel i am dumb, can i ask if u r a undergrad or grad? or prof?

I study at 11 grade,i 'm not a professional,i just participate at math olympiads.

[hide="Discussion"] The underlying concept is that of a [i]second-order linear recurrence[/i].  Given a sequence $ a_n$ satisfying the recurrence

$ a_{n\plus{}1} \equal{} pa_n \plus{} qa_{n\minus{}1}$,

$ a_n$ has a closed form

$ a_n \equal{} A \alpha^n \plus{} B \beta^n$

where $ A, B$ are suitable constants and $ \alpha, \beta$ are roots of the quadratic polynomial $ x^2 \equal{} px \plus{} q$.  There are several different ways of proving this, and you should be able to find a few if you search for "second-order linear recurrence."

In this case, $ A \equal{} B \equal{} 1, \alpha \equal{} 2 \plus{} \sqrt{3}, \beta \equal{} 2 \minus{} \sqrt{3}$ so we have the polynomial $ x^2 \equal{} 4x \minus{} 1$ and therefore the recurrence

$ a_{n\plus{}1} \equal{} 4a_n \minus{} a_{n\minus{}1}$

with the initial terms $ a_0 \equal{} 2, a_1 \equal{} 4$.  In some cases (such as this one) we will have WLOG $ | \beta | < 1$, in which case the second term vanishes as $ n$ becomes large and we have

$ a_n \approx A \alpha^n$.

Even better, if $ B, \beta$ are both positive (as in this case), we have for all but finitely many $ n$

$ a_n \equal{} \lceil A \alpha^n \rceil$. [/hide]


8. All possible three-digit positive integers are formed using the digits 0, 1, 3, 6, 7, 8 and 9 with no digit repeated in any number. If one of these numbers is selected at random, what is the probability that it will be greater than 300 and divisible by 4? Express your answer as a common fraction.

Just checking is 7/45 the correct answer?

I dont have the answers,just found this old nats problem

[quote="Temperal"]Do numbers such as $ 036$ count?
If so, I think it would be $ \frac{25}{294}$
If not, I think it would be $ \frac{25}{252}$[/quote]

I dont think thats right.

Me neither, I have no patience. As soon as I remember how I did this in the first place, I'll correct my mistake.

[hide]The standard definition of three digit number rules out numbers such as $ 036$.

Thus, the total number of three digit numbers that can be formed from the listed digits without replacement is $ 6\times6\times5 =180$, because there are six choices for the first digit {1,3,6,7,8,9} times six choices for the second digit (the five leftovers from the previous set as well as zero) times five choices for the third digit (the five leftovers from the set used to pick the second digit.)

Now, for the number to be divisible by $ 4$ requires that its last two digits be chosen from the following eight possibilities:  $ 08, 16, 36, 60, 68, 76, 80, 96$.

So now it's pretty easy to make an organized list of possible three digit numbers satisfying the stated criteria:

3 can be the first digit combined with $ seven$ of the two-digit possibilities listed above (i.e., all but 36.)

6 can be the first digit combined with $ two$ of the two-digit possibilities listed above (i.e., only 08 and 80.)

7 can be the first digit combined with $ seven$ of the two-digit possibilities listed above (i.e., all but 76.)

8 can be the first digit combined with $ five$ of the two-digit possibilities listed above (i.e., all but 08, 68, and 80.)

9 can be the first digit combined with $ seven$ of the two-digit possibilites listed above (i.e., all but 96.)

Thus, the total number of three digit numbers greater than 300 and divisible by four is $ 7+2+7+5+7=28$

The answer I get is thus $ 28/180 =7/45$.[/hide]

I just redid it and got the same answer, working from the other direction. I see where I got things mixed up; I counted some numbers that were odd by accident somehow.

[quote="abacadaea"]how are you guys such high denominators? there are $ \binom{7}{3}$ ways to make numbers which is 35. and its asking for specific numbers so i dont think the denominator can go any higher than 35.[/quote]

There are $ \binom{7}{3}=35$ ways to pick three unordered digits from the list of seven.  However, each of those $ 35$ sets of three digits can be arranged in $ 3!=6$ orders, for a total of $ 35\times6=210$ distinct possibilities, of which $ 30$ begin with zero and must be discarded, leaving $ 180$ valid three-digit numbers.

sophia is right...you're confusing the definition of combinations with what we need.


trigonometry
inequalities
real analysis
real analysis unsolved

Let x not equal to y be distinct points in [ -pi/2, pi/2 ]. Prove that |sin(x) - sin(y)| < |x - y|.

[color=darkblue]You can use the well-known relations : $\{\begin{array}{c}\sin x-\sin y=2\cdot\sin\frac{x-y}{2}\cdot \cos\frac{x+y}{2}\\\\ |x\cdot y|=|x|\cdot |y|\\\\ x\in\mathbb R\Longrightarrow |\cos x|\le 1\\\\ x\in \mathbb R\Longrightarrow |\sin x|\le |x|\ \ (*)\end{array}$

Here is an elementary proof of the inequality $(*)\ .$ Therefore, $x\in\mathbb R$ $\Longrightarrow$

$\boxed{\{\begin{array}{cc}x\ge 0\Longrightarrow & \{\begin{array}{ccc}x>1 & \Longrightarrow & |\sin x|\le 1<x=|x|\\\\ 0\le x\le \frac{\pi}{2}& \Longrightarrow & |\sin x|=\sin x\le x=|x|\end{array}\\\\ x<0\Longrightarrow & y=-x\ge 0\Longrightarrow |\sin y|\le |y|\Longrightarrow |\sin x|\le |x|\end{array}\|\Longrightarrow |\sin x|\le |x|\ }$[/color]

We know for any $y < x$ there exists $c \in (x,y)$ such that

$\cos{c}= \frac{\sin{x}-\sin{y}}{x-y}$.

Since $|\cos{c}| < 1$ on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, we have

$\left|\frac{\sin{x}-\sin{y}}{x-y}\right| < 1 \Rightarrow |\sin{x}-\sin{y}| < |x-y|$.


A cube of side 4dm, is made up of inside 1 dm cube. How many of the 1dm cube are face-to-face with the 4dm cube?

I don't get what you mean by "face to face". Could you explain that a little?

4 of these can fit in the length; just one dimension.
for 2 dimensions, you get 4*4=16 cubes.
then, 3 dimensions, is 16*4=64 small cubes.

wouldnt the question be asking for the cubes on the outside
there would be 64(i.e 4^3)-4(i.e 2^3)=60 cubes on the outside
i didnt fully understand the question :huh:  but i think this is what it means

Yes you are correct. The center $ (4\minus{}2)^3\equal{}8$ cubes don't touch any internal face of the large cube, so take the total number of cubes and subtract 8. $ 4^3\minus{}8\equal{}64\minus{}8\equal{}\boxed{56}.$

What? you said he was correct, and you got a different answer? Or was tat a typo?

aprocks was correct in the concept, but made a calculation error. $ 2^3\equal{}8$, not $ 4$. So it's $ 56$


calculus
integration
linear algebra
matrix
vector
geometry
3D geometry

We know that when we compute a triple integration (or double integration) by changing variables, the Jacobian determinant governs the effectiveness of the substitution.
    That's:
             (*) If the Jacobian =0 or infinity or changes sign at an (or some) interior point(s) within the Original region of integration ,then the substitution isn't successful.

   My Query is:
       What Geometric Properties of the Original Region of triple Integration should have to give a Jacobian to be 0?

   More specifically, given the region G is represented by:
               0<=x+2y+2z<=6 ,  0<=2x+y+z<=3, 0<=x+y+z<=6
            Find the volume enclosed by the region G above, by making a suitable substitution.

I have tried different substitutions but all failed, Could anyone draw a Figure of the region G for me and let's see whether it has a volume or not.
(I guess it doesn't have, but the region G is too difficult to draw on a paper, I don't have matlab etc either :oops: ).

   If we then change the third constraint to 5<=x+y+z<=6, what will happen then?What geometric properties of the new region has changed compared to the old region?  :idea:

THANKS VERY MUCH FOR ANY HELP!!!  :)  :)

Notice that the boundary surfaces of the region $ G$ are given by linear equations - the equations of planes, parallel planes. $ G$ is a parallelepiped. The appropriate substitution will be a linear substitution. We want:
\begin{align*}u & = x + 2y + 2z \\
v & = 2x + y + z \\
w & = x + y + z\end{align*}
or, better, write that in matrix notation:
\[ \begin{bmatrix}u \\
v \\
w\end{bmatrix} = \begin{bmatrix}1 & 2 & 2 \\
2 & 1 & 1 \\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\
y \\
z\end{bmatrix}
\]
The Jacobian of a linear (matrix) transformation is simply the determinant of the given matrix.
\[ \frac {\partial(u,v,w)}{\partial(x,y,z)} = \begin{vmatrix}1 & 2 & 2 \\
2 & 1 & 1 \\
1 & 1 & 1\end{vmatrix} = 1 + 2 + 4 - 1 - 2 - 4 = 0
\]
And we're in trouble. The transformation in question is not invertible. Note that when we make changes of variables, we normally write then "backwards": write $ x,y,z$ in terms of $ u,v,w.$ I haven't done that here, and I can't. I'd have to invert the matrix, and the matrix is not invertible.

What does this mean? If we had a positive-volume box in $ (x,y,z)$ space and applied this transformation to it, we'd get something "squashed flat" into a plane that would have zero volume in $ (u,v,w)$ space. Since what we have would be a box of positive but finite volume in $ (u,v,w)$ space, it must come from something of infinite volume in $ (x,y,z)$ space.

Look at the three normal vectors to planes that made up the three rows of our matrix: $ \langle 1,2,2\rangle,\langle 2,1,1\rangle,$ and $ \langle 1,1,1\rangle.$ They are linearly dependent; they all lie in the same plane. Find a vector perpendicular to all of them. (Pick any two and take the cross product.) We get $ \langle 0,1, - 1\rangle.$

Pick an arbitrary point that lies inside $ G.$ Say, $ (1,0,0).$ Take the line through that point in the $ \langle 0,1, - 1\rangle$ direction. That is, let
\begin{align*}x & = 1 \\
y & = t \\
z & = - t\end{align*}
for $ t\in\mathbb{R}.$ Check this line in your inequalities. They're all still true for any real $ t.$ Hence, the whole line (along with some other lines parallel to it) lies entirely within $ G.$ $ G$ isn't a parallelepiped at all; it's a prism, and it has infinite volume.

If you make your last change, by making the last inequality $ 5\le x+y+z\le 6,$ then we still have a degenerate region but now I can't put any infinite lines in it. I can't do that because the region is in fact empty.


Hi I just want to introduce a nice game called Conquer. I would love to learn everyone's username once they sign it. I'm maniac434 on server Ruby.

It is a massive multiplayer game featuring Chinese Mythology with some sweet weapons at http://www.conqueronline.com

[quote="aznness"]I've played plenty of online games in my time, Utopia, Starkingdoms, Starfury, Runescape, but every single one of my uber-high lvl accts has been hacked..[/quote]

What's the point of hacking Starkingdoms accounts?...


$ \text{Let}$ $ U \equal{} \{z \in \mathbb{C}:$ Im $ z > 0 \}$ $ \text{and}$ $ u(z) \equal{} \frac{az\plus{}b}{cz\plus{}d}$ $ \text{with}$ $ a, b, c, d \in \mathbb{C}$.

$ \text{Prove that}$ $ a, b, c, d \in \mathbb{R}$ $ \text{upon multiplying numerator and denominator by a constant}$ $ \text{if}$ $ u: U \rightarrow U$ $ \text{is a bijective map}$.

Probably the best thing to do is to send $ U$ to $ B(0,1)$.  Then all automorphisms of $ B(0,1)$ are of the form
\[ z\to \alpha\left(\frac {z_0 \minus{} z}{1 \minus{} \overline{z_0}z}\right)
\]
for some $ \alpha$ with $ |\alpha| \equal{} 1$ and some $ z_0$ with $ |z_0| < 1$.  Finally, send $ B(0,1)$ back to $ U$.


This is a thread where you can play the 24 game.  For those of you who don't know how to play the 24 game, read the rules.  For those of you that do, read them anyway.
[color=red][b][u]THE RULES: READ THEM![/u][/b][/color]
It starts with someone(me) posting four numbers.  You must use the 4 basic operations, (addition, subtraction, multiplication, division) and as many parenthasis as  you want to get 24.  Unless specifically said in the post, you may not use exponents, roots, etc.  When you post, you must post the  solution to the previous problem, as well as four numbers of your own.

[color=red][b][u]Point Tallying:[/u][/b][/color]
When you post four numbers, you get 1 point.  For solving, you get 2 points. However, if you cannot solve it, you may request a hint.  This is minus 1 point.  If you post four numbers with exponents, roots, etc.  available, you get an extra one point.  Everytime you post, please hide the score count.  First one to 300 points wins.

That about covers it.  Enjoy!

4,4,4,4

[hide="score count"]
wrightboy10: 1
[/hide]

$ 4\times4\plus{}4\plus{}4\equal{}24$

New set $ (1,2,3,4)$.

[hide="Score count"]mszew=2+1=3?[/hide]

Atention SMF:  That is not an elligable solution, as Butler did not say you could use exponents.  In the future, do not post an answer with exponents unless the number giver says so.

Butler's:  ((2*3)+6)*2
SMF's: 7*(3/7+3)

2, 3, 8, 19

[hide="score"]
1. wrightboy10: 6
2. mszew: 3
2. Butler: 3
4. SMF: 1  (I do not count inelligable solutions.)
[/hide]

9+5+4+6

5 1 2 3

[hide="score"]
1. wrightboy10: 9
2. mszew: 9
3. Butler: 3
3.Piegraph: 3
4. SMF: 1
[/hide]

$ 10 \plus{} 6 \plus{} 6 \div 4$

DIFFERENT SOLUTION TO $ 5,6,8,10$.  You can't switch around numbers or do something cheap like that.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209924[/url]
The 105-page-long 24 game marathon in the other forum.

[quote="mszew"]((6*(6+10))/4)

5 13 7 15

[hide="How to cheat this game (spoils the fun)"][url]http://www.math.jmu.edu/~walton/24Solver.html[/url][/hide][/quote]
(15-13)(5+7)=24

22   1    2     22

(8*9)/9+16

1542

That was random...
Here is another one if that doesn't work

1451

Your not allowed to switch around the order :mad:

1*(2*(3*4))

2345

why are exponents and "!" and changing order not alowed?I hate this 24 topic.

because $ 4! \plus{} (5 \minus{} (3 \plus{} 2))$ works

Changing order [b]is[/b] allowed.  The only four other "functions" allowed are addition, subtraction, multiplication, and division.  You may add as many parentheses as you wish.

i wanted to play 24 with exponents and factorials. i didn't mean to make "spam", but i must say, exponents and factorials make a big difference
 :mad:

and btw, i wasn't repeating this topic. Why would i copy a topic that i don't like?

@eagle, that's no reason to lock it.... No one posted on it except, now....

If it doesn't need to be locked, you don't have to lock it.

[quote="SMF"][size=84][color=red]This post has been censored for extreme lack of sportsmanship. [/color][/size][/quote]

Wow. Extreme lack of sportsmanship. How is saying I don't like the only version of 24 and want a new one bad sportsmanship?

If this 24 game was enough like my other one, then I wouldn't be complaining. However, if they were different...It's like the epic fail and the epic win game. They are very alike, but still different. And I haven't seen the epic win game being locked.

My 24 game is sort of like this. It is similar in the fact of the name and objective(to get 24 with 4 numbers) but the rules are different. If you think about it, You know your an epic fail/win is about being funny and creative. Nobody is trying to warn people when they are epic failures or wins. The only difference is that the rules are different: you use fail in one and win in the other.

I am seriously wondering why my 24 game wasn't kept. You mods could have perfectly well changed my post to say it was a version of this topic. You didn't have to say it was "spam" or whatever to create another version of the topic.

bluecarneal said that there was "no need for another". The reason was because there is a long running brainteasers/puzzles topic. If that was referring to this puny 1 page topic...Then, wow. All of my reasoning is wrong and you might as well delete this post. But if there is another topic, why didn't you mods lock this topic? If you locked mine because it was similar to this one, then why not lock this being similar to this other one? If there is one. Or am I just dumb? Or is bluecarneal wrong...

[quote="piegraph"]smf, it's better than ur 24.  You are just repeating this one...[/quote]

Really. I hate one 24 game and so I copy another and make it more available.

This 24 was more serious. Anyone notice a point tally on my 24? I made my 24 to be more of a less serious, vending machine type game. And also to have more operations, with more fun. That you check up on for fun, post in a minute, just for fun. Not a real game with points.

I might have been a bit rude, but still, I am not so sure about it being totally good to lock that topic.

Tell me if my reasoning is bad, ban me from the FF, do whatever you want. But I'm just saying that my 24 is different. And I would like it unlocked.


linear algebra
matrix
integration
vector
calculus
ratio
algebra

Let $ A$ be the following matrix:

$ A\equal{}\left(\begin{array}{ccccc}1&\frac{1}2&\frac{1}3&...&\frac{1}{n}\\ \frac{1}2&\frac{1}3&\frac{1}4&...&\frac{1}{n\plus{}1}\\ ...&...&...& &...\\ \frac{1}{n}&\frac{1}{n\plus{}1}&\frac{1}{n\plus{}2}&...&\frac{1}{2n\minus{}1}\end{array}\right)$.

Show that this matrix $ A$ has an inverse and that this inverse is a matrix all of whose entries are integers.

$A$ is real symetric

The coefficients of $A$  are $\int_{0}^{1}f_{i}(x)f_{j}(x)dx$
where $f_{i}(x)=x^{i-1}$

For any vector $^{t}Y=(y_1,...,y_n)$

$^{t}YAY=\int_{0}^{1}(\sum_{k=1}^{n}y_kf_k(x)dx)^2$  is positive

$^{t}YAY=0$ iff $\sum_{k=1}^{n}y_kf_k=0$
$Y=0$

$A$ is definite positive, $A$ is invertible

$A$ is the Hilbert matrix it seems we talked about this matrix

Thanks for the reply Moubinool.  Your answer makes a lot of sense.  But this problem was found in the first chapter of a linear algebra book...in the section on invertible matrices (defining inverses of matrices).  I doubt that ideas such as positive definiteness are familiar to the reader.  Is there a very elementary solution to this problem?

You can prove it constructively by reducing it into echelon form, I think.

To row reduce the matrix on the LHS, let $v_i$ be the rows of the matrix on the LHS.  On the i-th iteration, for every $j>i$, multiply $v_j$ by $j+i-1$ and then subtract $v_i$ from that row.  This eliminates a row of nonzero entries, replacing them with zero entries.  You just need to check that the next leading nonzero entry is of the form (in lowest terms) $1/n$ for some $n$, and then we can multiply $v_{i+1}$ by $n$.

This was all very heuristic, and I haven't really experimented with it, but I think it works.

What this does to the matrix on the RHS, which is initially the identity (and hence composed only of integral entries) is that it adds integral multiples of the rows to other rows, and hence the integral entries are preserved.

Try it for the $n=4$ case: I'm pretty sure it doesn't work, and you haven't addressed the second half of the elimination, going back up. You're guaranteed to have to deal with fractional elements there.

Here are the first few inverses, to give an idea of just how these behave. Don't try anything numerical, as the Hilbert matrix is spectacularly ill-conditioned.
$[1]^{-1}=[1]$
$\begin{bmatrix}1&\frac12\\\frac12&\frac13\end{bmatrix}^{-1}= \begin{bmatrix}4&-6\\-6&12\end{bmatrix}$
$\begin{bmatrix}1&\frac12&\frac13\\\frac12&\frac13&\frac14\\\frac13 &\frac14&\frac15\end{bmatrix}^{-1}= \begin{bmatrix}9&-36&30\\-36&192&-180\\30&-180&180\end{bmatrix}$
$\begin{bmatrix}1&\frac12&\frac13&\frac14\\ \frac12&\frac13&\frac14&\frac15\\ \frac13&\frac14&\frac15&\frac16\\ \frac14&\frac15&\frac16&\frac17\end{bmatrix}^{-1}= \begin{bmatrix}16&-120&240&-140\\-120&1200&-2700&1680 \\240&-2700&6480&-4200\\-140&1680&-4200&2800\end{bmatrix}$

The lower right entry is the ratio of the determinant to the previous determinant; you can see how fast the determinant of the Hilbert matrix goes to zero.

[quote="jmerry"]Try it for the $n=4$ case: I'm pretty sure it doesn't work, [/quote]

Well, it almost works - but 'going' down, we get the entry $45/28$ in the lower right hand corner, which brings everything to a screeching halt.  Anyways, I think I'm out of my league here; this problem is unusually difficult considering the text it comes from.

$A=(\frac{1}{a_i+b_j})$ where $a_i+b_j\neq 0$

if  $A^{-1}=(h_{ij})$ exist then its coefficients are

$h_{ij}=\prod_{l=1}^{n}(a_j+b_l)\prod_{k\neq j}\frac{a_k+b_i}{a_k-a_j}\frac{1}{\prod_{h\neq i}(b_h-b_i)}$

$h_{nn}=(a_n+b_n)\prod_{j=1}^{n-1}\frac{(a_n+b_j)(a_j+b_n)}{(a_n-a_j)(b_n-b_j)}$

It says in a book of mine that the sum of the elements of the invert matrix is$n^2$? doesn`t seem to be true.

Did you make the computations? Yes, it is true and fortunately, the proof is not so difficult as the actual computation of the inverse and the sum of all elements.

$\begin{bmatrix}1&\frac12\\\frac12&\frac13\end{bmatrix}^{-1}= \begin{bmatrix}4&-6\\-6&12\end{bmatrix}$

$n=2$ and $2^2=4$ am I missing something?

And as far as I remember 12+4-6-6=4=2^2.  ;)

:oops:  :oops:
is there a way of proving it for the 11th grade level?

Yes, just think it in terms of polynomials, not matrices.


What was the answer for the problem with when you put parenthesis in

2+3+4*5

or something

how many numbers can u get?


"So you've got five kids, with the one girl in the middle," said Alan. "How old is she?"
Clem smiled. "Figure it out yourself," he replied. "They're all spaced two years apart, and the youngest boy is just half as old as the oldest."
How old was the girl ?

[i]from book "Entertaining mathematical teasers and How to solve them" by J.A.H.Hunter[/i]

[hide] Let the age of the youngest boy be $ x$. Then the eldest is $ x\plus{}8$, and $ 2x\equal{}x\plus{}8 \implies x\equal{}8$. Therefore, the girl is $ 8\plus{}4\equal{}\boxed{12}$.[/hide]

Since it said the youngest BOY in the family, then it could be any person in the family. There are many ages the middele girl can be including 8. The kids would be ages: 4,6,8,10,12. The youngest boy could be 6, and the oldest could be 12. Another situation can be 8,10,12,14,16. This situation is the youngest and the oldest, not considering the gender.

[quote="$ LaTeX$"][quote="MathBunny"]Since it said the youngest BOY in the family, then it could be any person in the family. There are many ages the middele girl can be including 8. [b]The kids would be ages: 4,6,8,10,12.[/b] The youngest boy could be 6, and the oldest could be 12. Another situation can be 8,10,12,14,16. This situation is the youngest and the oldest, not considering the gender.[/quote]

How does that work? The youngest boy in the family is not half as old as the oldest.[/quote]

the 4 year old could be a girl, so the youngest boy would be 6.

[quote="small_unicorn"]"So you've got five kids, [b]with the one girl in the middle[/b]," said Alan. "How old is she?"
Clem smiled. "Figure it out yourself," he replied. "They're all spaced two years apart, and the youngest boy is just half as old as the oldest."
How old was the girl ?

[i]from book "Entertaining mathematical teasers and How to solve them" by J.A.H.Hunter[/i][/quote]

However, read the bold part. So $ 12$ is the only answer.

[hide="Clarification"]
Let the ages of the children be $ x\minus{}4, x\minus{}2, x, x\plus{}2, x\plus{}4$.
The problem states that "the youngest boy is just half as old as the oldest", and it also states that there is only one girl.
The oldest boy must have age $ x\plus{}4$, and the youngest boy must have age $ x\minus{}4$.

$ 2(x\minus{}4)\equal{}x\plus{}4 \implies 2x\minus{}8\equal{}x\plus{}4 \implies x\equal{}12$

Thus, the girl is $ \boxed{12}$ years old.[/hide]

in 8 10 12 14 16
the middle value should be clarified.
the meaning can be misinterpreted to mean anything betweenn 8 and 16
so the answer could be 10 or 14 depending on ur meaning of "middle".
using "median" is clearerr.


Where does the website find the animated math thingy on the left hand side of the website below logout - I realized that they change, and I think they are really cool ways to demonstrate formulas.

They most likely make them themselves.

javascript + flash = random little proofs on the side

javascript to make them random
flash to make the actual animations

that's my guess...

If you rightclick on the animation, it says ``About Macromedia Flash Player 6...,'' so you can tell that they are made in Flash.

It would be real nice if there is a way to save them on the computer so we can view them in a later time.

I think if you look all around the web you will find enough examples in gif format, so you can simply save them on your disk :)

Actually, you can download the flash files:

http://www.artofproblemsolving.com/Images/Side1.swf
http://www.artofproblemsolving.com/Images/Side2.swf
...
http://www.artofproblemsolving.com/Images/Side9.swf

Click File->Save As...

We build them in house.  You can view them all at once (plus a nifty Java applet) [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]here[/url].


Suppse that $(x_n)_{n\geq 0}$ be a real sequence with conditions $x_0>0 , x_1>0$ and $\forall n\geq 0 , x_{n+2}=\frac {2+x_{n+1}}{2+x_n}$. Prove that the sequence of $(x_n)_{n\geq 0}$ is the convergent.

Manifestly the sequence is bounded :   $ 0 <  x_{n+2}=\frac {2+\frac {2+x_n}{2+x_{n-1}}}{2+x_n}=  \frac 2{2+x_n} + \frac 1{2+x_{n-1}} < \frac 32$

So the adherence points set, called $A$,  is non empty and it is bounded.  Let $I,S$ be  the inferior , superior  bound of $A$ , $I\leq S$.
For all $a\in A$ we can write : $0 < \frac {2+I}{2+S} \leq  a  \leq \frac {2+S}{2+I}$
$a=I \Longrightarrow  \frac {2+I}{2+S} \leq  I \Longrightarrow 2 -I \leq I*S $
$a=S \Longrightarrow  S \leq \frac {2+S}{2+I}  \Longrightarrow I*S  \leq 2-S $
But $2 -I \leq I*S \ \leq 2-S \Longrightarrow S \leq I$, so $I=S$ .

Finally $x_n$ is convergent because there is only one adherence point , and its limit is $1$.

 :cool:

1) an "adherence point" for a sequence $x_n$ is a point $y$ such that, for every positive real number $\epsilon$ the set $\{n\in\mathbb{N}\ |\ |x_n-y|<\epsilon\}$ isn't empty; that is, you can find a point of the sequence as near to $y$ as you want;

2) sorry, I don't speak romanian.

[quote="perfect_radio"][quote="Diogene"]So the adherence points set, called $A$,  is non empty and it is bounded.  Let $I,S$ be  the inferior , superior  bound of $A$ , $I\leq S$.[/quote]

1) What are "adherence points"?

2) How are they called in Romanian?[/quote]

$a $ is an "adherence point" for a sequence $x_n$, if there is a sub-sequence $x_{n'}$ convergent to $a$

 :cool:


IMO Shortlist
rectangle
graph theory
combinatorics
IMO
IMO 1996
modular arithmetic

We are given a positive integer $ r$ and a rectangular board $ ABCD$ with dimensions $ AB \equal{} 20, BC \equal{} 12$. The rectangle is divided into a grid of $ 20 \times 12$ unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the two squares is $ \sqrt {r}$. The task is to find a sequence of moves leading from the square with $ A$ as a vertex to the square with $ B$ as a vertex.

(a) Show that the task cannot be done if $ r$ is divisible by 2 or 3.

(b) Prove that the task is possible when $ r \equal{} 73$.

(c) Can the task be done when $ r \equal{} 97$?

Put $ A$ in the bottom left corner, and $ B$ in the bottom right corner.

Let a move be $ a$ squares to the right and $ b$ squares up. (Note that a and b can be negative).
Then, by Pythagoras' theorem, the move is permitted iff
$ r \equal{} a^{2} \plus{} b^{2}$

[b]a)[/b]
If $ 2|r$ then $ a^{2}\equiv b^{2}\mod 2$. It is clear that a move is an odd number of squares right and an odd number of square up [i]or[/i] an even number of squares right and an even number of squares up. If the board is coloured in the regular chessboard fashion, then no permitted move can go from a white square to a black square. But the bottom left square is a different colour than the bottom right square: hence we cannot move from one to the other.
If $ 3|r$ then $ a^{2} \plus{} b^{2}\equiv 0 \mod 3$ Then either $ a \equiv b \equiv 0 \mod 3$ or $ a \equiv 1 \mod 3$ and $ b \equiv 2 \mod 3$. (Or the other way round). But squares are not congruent to 2 mod 3, so $ 3|a$ and $ 3|b$. So on every good move, we move a multiple of 3 squares to the right. But the bottom right square is 19 squares to the right of the bottom left square. Since 3 does not divide 19, we cannot get from one to the other.

[b]b)[/b]
If r=73, then since $ 73 \equal{} 8^{2} \plus{} 3^{2}$ we can move from one corner of a 9x4 rectangle to the opposite corner. If we denote by ($ n,m$) the square in the $ n$th column and $ m$th row, a sequence of moves from the bottom left corner to the bottom right is:
$ (1,1) (9,4) (17,1) (20,9) (12,6) (4,3) (7,11) (15,8) (7,5) (15,2) (12,10) (20,7) (12,4) (20,1)$

[b]c)[/b]
The only way that 97 can be expressed as the sum of two squares is $ 97 \equal{} 9^{2} \plus{} 4^{2}$. Hence all permittable moves are from one corner of a 5x10 rectangle to the other.
If we colour the board such that all squares ($ n,m$) are white if:
$ n$ is odd and $ m \in \{1,2,3,4,9,10,11,12\}$ OR
$ n$ is even and $ m \in \{5,6,7,8\}$.
and black otherwise.
Clearly all permitted moves go from a white square to a white square, or a black square to a black square. But the bottom left corner is a different colour to the bottom right square, so we cannot go from one to the other.

Try to think reverse put a mark in B square and then put mark in squares which you can go through it. If you succeed put a mark in A square you are done.

[quote=viperstrike]How did you find the construction for #2?[/quote]

Try to think reverse put a mark in B square and then put mark in squares which you can go through it. If you succeed put a mark in A square you are done

Nice approach for c) is to look mod 5. If we start at 1,1 and need to go to 1,20 we notice that we need to go from 1,1 to 1,0 mod 5 . Number of moves is S.Note that both 4 and 9 are -1 mod 5 so we can add and substract 1 .From the first component we need to get from 1 to 1 so the number of +1s equals the numbers of -1s ,because of that S is even.Next look at the second component  we need to get from 1 to 0 mod 5.If x is the number of +1s and y is the number of -1 then 1+x-y=0 y=x+1 S=x+y=2x+1 but S is even contradiction!( 1+x-y=0 is because we start from 1 and need to get to 0)

For part $(a)$ note that $2\mid a^2+b^2$ implies $2\mid a+b$ and similarly $3\mid a^2+b^2$ implies $3\mid a+b$.
For part $(b)$ place $A$ in the top left, $B$ in the top right. Define a move $(x, y)$ to be moving $x$ squares to the right and $y$ squares up. Then the moves $(3, -8), (3, 8), (3, -8), (-8, 3), (8, 3), (3, -8), (-8, 3), (8, 3), (-8, 3), (-3, -8), (8, 3), (8, 3), (-3, -8), (-3, 8), (8, 3)$ work.
For part $(c)$ use a "checkerboard" but with $1\times 4$ tiles. We can check that colored squares go to colored squares, and uncolored squares go to uncolored squares. So the task is not possible.

If I understand these solutions correctly, #6 doesn't work. Since $5$ is odd, the number of $+1$s does not have to have the same parity as the number of $-1$s (for example, $1+10\cdot (+1)+5\cdot (-1)\equiv 1\mod 5$.

For c):  the 1x4 "checkerboard" is not essential, here is an explanation of the real essential thing.

Denote m=4 and n=9.
Here m and n represent "coprime numbers of different parity" (otherwise it already fails to visit every square),
and 12 takes the expression m+n-1 for this structure.

Arrange the numbers from 1 to m+n-1 in this order:
m mod (m+n), 2m mod (m+n), ..., (m+n-1)m mod (m+n).
Then two numbers differ by m or n iff they are adjacent in this sequence.

Color the m+n-1 rows in red and blue by alternating colors in this sequence.
Then starting from a white square in a red row, it can only jump to a black square in a blue row, and vice versa,
and can never reach white squares in blue rows or black squares in red rows.

But if another row is added, then they close up to a loop of odd length and is no longer bipartite.


find all triples (a,b,c) such that
        abc+ab+c=a^3

you may have a look at the olympiad section where you can post problems. This subforum is dedicated to IMO 2008 and discussions relevant to this event. (Right??)

I'm sorry??  :blush:
I think it forum was "Number Theory Unsolved Problems"
sorry :)

I assume you mean positive integers.

Note that $ a | c$, so write $ c \equal{} ad$.  Factor as

$ (ab \plus{} 1)(ad \plus{} 1) \equal{} a^3 \plus{} 1 \equal{} (a \plus{} 1)(a(a\minus{}1) \plus{} 1)$.

The question is now symmetric in $ b, d$ and we have the solutions $ (a, a\minus{}1, a), (a, 1, a(a\minus{}1))$.  The question is now how to factor the RHS into a product of two numbers congruent to $ 1 \bmod a$.

[quote="t0rajir0u"]I assume you mean positive integers.

Note that $ a | c$, so write $ c \equal{} ad$.  Factor as

$ (ab \plus{} 1)(ad \plus{} 1) \equal{} a^3 \plus{} 1 \equal{} (a \plus{} 1)(a(a \minus{} 1) \plus{} 1)$.

The question is now symmetric in $ b, d$ and we have the solutions $ (a, a \minus{} 1, a), (a, 1, a(a \minus{} 1))$.  The question is now how to factor the RHS into a product of two numbers congruent to $ 1 \bmod a$.[/quote]
Easy to check that : $ a|c$ then exist an integer k such that $ c\equal{}ka$
The equation equivalent :
$ k(ab\plus{}1)\equal{}a^2\minus{}b$
$ \Leftrightarrow a^2\minus{}kab\minus{}b\minus{}k\equal{}0$
We have $ \Delta \equal{}(kb)^2\plus{}4b\plus{}4k$
We can resume that $ b\geq k>0$
The equation have positive integer solution if $ \Delta$ is a perfect square.
It means that exist an natural  m such that $ \Delta\equal{}m^2$
$ m^2\equal{}(kb)^2\plus{}4k\plus{}4b$
$ \Leftrightarrow (m\minus{}bk)(m\plus{}bk)\equal{}4(b\plus{}k)$
So $ 4(b\plus{}k)\geq m\plus{}bk$
If $ b,k>2$ then $ m\plus{}bk>4(b\plus{}k)$
Then we consider cases :
[u]Case 1 [/u] $ k\equal{}1$
Then $ a\equal{}b\plus{}1$
[u]Case 2 [/u] $ k\equal{}2$
Then easy to check that $ b\equal{}1$ (contradiction )
So $ k\equal{}1$ or $ b\equal{}1$
If $ k\equal{}1$ then $ (a,b,c)\equal{}(m\plus{}1,m,m)$
If $ b\equal{}1$ then $ (a,b,c)\equal{}(m\plus{}1,1,m(m\plus{}1))$


In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?

$ \textbf{(A)}\ \text{All Dramps are Brafs and are Crups.}\qquad \\
\textbf{(B)}\ \text{All Brafs are Crups and are Dramps.}\qquad \\
\textbf{(C)}\ \text{All Arogs are Crups and are Dramps.}\qquad \\
\textbf{(D)}\ \text{All Crups are Arogs and are Brafs.}\qquad \\
\textbf{(E)}\ \text{All Arogs are Dramps and some Arogs may not be Crups.}$


Let $f: [0,\infty)\to \mathbb R$,$f(x)=x^{m}$ ,from $f''(x)=m(m-1)x^{m-2}\geq 0$ $\Rightarrow$ $f$ is convex,therefore from Jensen's inequality we have
\[\frac{1}{n}\sum_{k=1}^{n}f(x_{k})\geq f\left( \sum_{k=1}^{n}x_{k}\right)=\frac{1}{n^{m}}\cdot \left( \sum_{k=1}^{n}x_{k}\right)^{m}\].qed

[hide]Divide by n, then take the mth root and it follows from power mean[/hide]


MIT
college
Stanford
CMU
Harvard
Carnegie Mellon

Hello
Can anyone post a rank for best undergraduate computer science or IT programs ?
Ah, i also need a rank for under economics. Can you help me ?
Thank you very much.  :)

Did you try Googling "computer science ranking"?

Many people feel that the top programs are Carnegie Mellon, MIT, Stanford, and UC-Berkeley (listed in alphabetical order) -- it's very hard to differentiate between these four.  I have a bias toward CMU as that's where I got my B.S. in Computer Science.

I strongly dislike rankings, as it implies that one school is strictly better than another, which is rarely the case among the top schools.  Certainly there is no way that I would argue that any of the above four schools that I listed is the "best".

yea if you really want to go with rankings (i have to admit i myself am partial to rankings), google for them. USA Today usually ranks them. I"m assuming you're talking about undergrad, so for a general idea:

Computer Science - MIT, Stanford, CMU, CalTech, i.e. the hard core schools

Economics - def U Chicago. Not only do they boast lots of Nobel Prize winners as faculty, it's also relatively easier to get into, as compared to other great economic departments in schools like Harvard or something

Are you strictly looking for US schools? Some schools in UK and in Germany are quite cool as well...  ;)

Waterloo in Canada is a highly regarded school for computer science.


12 balls numbered 1 through 12 are placed in a bin.  In how many ways can 3 balls be drawn, in order, from the bin, if the first ball is replaced after it is drawn but the second ball remains outside the bin?

[quote=jxl28]For the first ball selected, we have $ 12$ choices.  For the second ball selected, we still have $ 12$ choices, since the first ball is replaced after it is selected.  For the final one, we only have $ 11$ choices, since the second ball is [i]not[/i] replaced.  So we have a total of $ 12\cdot12\cdot11 \equal{} \boxed{1584}$ total ways.[/quote]

[quote=rockets8128]For the first pick, there are $12$ possibilities.  For the second, there are again, $12$.  For the last, there are only $11$.  [hide=So multiplying all three numbers gets us the answer of]$1584$[/hide][/quote]

[quote=Samchooo][hide=Sol]The first draw allows 12 choices, Since it is replaced, the second choice also allows 12 choices. But the third choice only allows 11 because the 12th ball is out of the bin. 12*12*11=1584 [/hide][/quote]

These are all....pretty much the same.

Okay, I promise that I didn't copy off the other answers (I didn't even look at them), I just wanted to post my solution. Thanks!


log2(3)   3   -1
log2(5)   2   -2      = log2(k).  Find k.
log2(6)   1    0

the determinant of (above) = log2(k)
(log base 2 of k)

sry having trouble with LATEX. :(
log2(3)   3   -1  this is the first row of numbers
(log base 2 of 3), 3, and -1
etc

this help?

[quote="7h3.D3m0n.117"][quote="GlenthemanN"]what do you mean by "log2(3) 3 -1" is that log of 3 in base 2, multiplied by 3 and having 1 subtracted or what?[/quote]

so what you're saying-

$3(\log_{2}3)-1$?[/quote]
yes.. I have no idea what a space would denote :P

[quote="DiscreetFourierTransform"]Guys, it's clearly a matrix:
$\left| \begin{matrix}\log_{2}3 & 3 &-1 \\ \log_{2}5 & 2 &-2 \\ \log_{2}6 & 1 & 0 \end{matrix}\right| = \log_{2}(k)$

Solve for k.[/quote]
if it's a matrix then shouldn't it be:

$\begin{bmatrix}\log_{2}3 & 3 &-1 \\ \log_{2}5 & 2 &-2 \\ \log_{2}6 & 1 & 0 \end{bmatrix}= \log_{2}(k)$?

He wants the determinant, so I put the standard determinant markers on it (the absolute value bar thingys)


induction
number theory unsolved
number theory

[color=blue][size=134]Bertolt Brecht High School - PERU
Training Program - Level 0
[/size]
[size=117][b]Middle Exam[/b][/size][/color]

 Let $(2p+1)$ be a prime number, p>1 is odd.
Show for "the induction method" that for all $n \in \mathbb{Z}^+$:
$\sum_{k=1}^p k^{2^n}$ is divisible by $(2p+1)$.

                                           [i]Lima, April 18 - 2006.[/i]

Carlos Bravo
Lima - PERU

First see that $\{1,2^{2^n},...,p^{2^n}\}$ give all square residues so it is enough to prove $\sum_{k=1}^p k^2 \equiv 0 \mod 2p+1$. This can be done by observing that ($\lambda$ is a primitive root):

$2\sum_{k=1}^p k^2 \equiv \sum_{k=1}^{2p} k^2 \equiv \frac{\lambda ^{2p+1}-1}{\lambda -1} \equiv 0 \mod 2p+1$

[quote="Megus"]
This can be done by observing that ($\lambda$ is a primitive root):

$2\sum_{k=1}^p k^2 \equiv \sum_{k=1}^{2p} k^2 \equiv \frac{\lambda ^{2p+1}-1}{\lambda -1} \equiv 0 \mod 2p+1$[/quote]

Sorry, could you explain this part of your solution??

Thanks in advanced ..

Fabycv  :rotfl:


N robbers have robbed the bank. As the result of their crime they chanced to get M golden coins. Before the robbery the band has made an agreement that after the robbery i-th gangster would get Xi/Y of all money gained. However, it turned out that M may be not divisible by Y.

The problem which now should be solved by robbers is what to do with the coins. They would like to share them fairly. Let us suppose that i-th robber would get Ki coins. In this case unfairness of this fact is |Xi/Y-Ki/M|. The total unfairness is the sum of all particular unfairnesses. Your task as the leader of the gang is to spread money among robbers in such a way that the total unfairness is minimized. 1 ≤ N ≤ 1000, 1 ≤ M, Y ≤ 10000

Hint: Start by sharing out the money, rounding down for each person. Which robbers should you spread the rest of the money out to, to keep the unfairness minimal?


induction
number theory proposed
number theory

Assume that $ \alpha$ and $ \beta$ are two roots of the equation: $ x^2\minus{}x\minus{}1\equal{}0$. Let $ a_n\equal{}\frac{\alpha^n\minus{}\beta^n}{\alpha\minus{}\beta}$, $ n\equal{}1, 2, \cdots$.
(1) Prove that for any positive integer $ n$, we have $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}a_n$.
(2) Find all positive integers $ a$ and $ b$, $ a<b$, satisfying $ b \mid a_n\minus{}2na^n$ for any positive integer $ n$.

(1) is trivial - induction and $ x^{n\plus{}2}\equal{}x^nx^2\equal{}x^n(x\plus{}1)$ -- $ \alpha$ and $ \beta$ satisfy this.

(1)Nothing $\alpha+\beta=1$ and $\alpha \beta=-1$ we have
$\alpha^{n+2}-\beta^{n+2}=(\alpha+\beta)(\alpha^{n+1}+\beta^{n+1})-\alpha\beta(\alpha^n-\beta^n)=$
$=(\alpha^{n+1}-\beta^{n+1})+(\alpha^n-\beta^n)$
Dividing two sides $\alpha-\beta$ we have $a_{n+2}=a_{n+1}+a_n$

(2)By assumption we have $b|a_1-2a$ that is $b|1-2a$. But $b>a$ so $b=2a-1$. Moreover for an arbitrary
positiva integer $n$ we have
$b|a_n-2na^n, b|a_{n+1}-2(n+1)a^{n+1}, b|a_{n+2}-2(n+2)a^{n+2}$
Combining $a_{n+2}=a_{n+1}+a_n$ and $b=2a-1$ an odd number we have
$b|(n+2)a^{n+2}-(n+1)a^{n+1}-na^n$
But $gcd(a,b)=1$ so
$b|(n+2)a^2-(n+1)a-n$    (1)
Taking $n$ as $n+1$ in (1) we have
$b|(n+3)a^2-(n+2)a-(n+1)$      (2)
Subtracting the right side of (1) from tha same side of (2) we have
$b|a^2-a-1 \to 2a-1|a^2-a-1$
But $2a^2\mod a(mod2a-1)$therefore
$2a-1|-a-2, 2a-1|-2a-4$ So $2a-1|-5$, $2a-1=1$ or 5
But $2a-1=1$ implies $b=a$ a contraction. So $2a-1=5, a=3, b=5$
We will prove in the following that when $a=3$ and $b=5$ for any positive integer n  we have $b|a_n-2na^n$
that is $5|a^n-2n3^n$


search
number theory unsolved
number theory

(a) Prove $ a^2 \plus{} b^2 \minus{} 5ab \plus{} 5$ has infinitely many solutions in natural numbers.

(b) If $ a,b,\frac {(a^2 \plus{} b^2)}{(ab \minus{} 1)} \equal{} q$ are integers, show that $ q \equal{} 5$.

(c) If $ a,b, \frac{(a^2\plus{}ab\plus{}c^2)}{(ab\plus{}1)}\equal{}q$ are integers, show that $ q$ is a perfect square.

unfortunately, :(
most of the problems that u are posting right now are posted before ,
plz try a search ,thx :)


I was wondering for the individual tests, is there a restriction on how many people per team can take any given test?

For instance, at the princeton university competition, we had to enter in a maximum of 4 people per test. If more than four for one test then only the top foiur scores would count.

So, can everyone on our team for hmmt hypothetically do Algebra and Calculus and have everyone's scores count or is there a rule?

As far as I know, there is no such restriction.

I think your total score is calculated from your top 3 scores in each subject...

I am not sure exactly how, but I believe that you need to take every test.

[quote]I think your total score is calculated from your top 3 scores in each subject... [/quote]
Not according to http://web.mit.edu/hmmt/www/rules.shtml#awards

Okay wow I was not thinking of the correct thing... So I guess from that you could potentially all take the same pair of tests...

They account for difficulty variances by dividing your score by the average.

This isn't exactly on topic, but according to this: [url]http://web.mit.edu/hmmt/www/datafiles/schedules/2008.shtml[/url], there won't be a USEMO this year, why not?

It looks like they only did that last year. These mini events are identical to Harvard's last ones as well.


Just calculate $ (1\minus{}10)(1\plus{}10\plus{}100\plus{}1000\plus{}\cdots)$.

Yes but what is the rezult?
When i saw this problem i though that the rezult was inf.

Let $ S_n\equal{}1\plus{}10\plus{}10^2\plus{}10^3\plus{}\cdots 10^{n\minus{}1}$,

$ (1\minus{}10)S_n\equal{}(1\plus{}10\plus{}10^2\plus{}10^3\plus{}\cdots 10^{n\minus{}1})\minus{}(10\plus{}10^2\plus{}\cdots \plus{}10^{n\minus{}1}\plus{}10^{n})$

$ \equal{}1\minus{}10^{n}$

$ \therefore S_n\equal{}\frac{1\minus{}10^n}{1\minus{}10}\longrightarrow \plus{}\infty\ (n\rightarrow \infty).$

In short, the series is infinity geometric series.

so basically, the answer is infinity...
you're starting to confuse me... xD

assuming the terms continue indefinitely, yes, it is infinity.

However, if one were to use a different summation method which assigns a finite value to this sum, the only value would be $ \minus{}\frac{1}{9}$. In the normal summation method, this sum diverges to infinity.

[hide="I think it should be"]
1111111111111111..... depending on how many terms there are
$ \infty$ if there are infinity terms.

[/hide]

[quote="Hamster1800"]However, if one were to use a different summation method which assigns a finite value to this sum, the only value would be $ \minus{} \frac {1}{9}$. [/quote]

Specifically, we can work in the [url=http://en.wikipedia.org/wiki/P-adic_number]10-adic numbers[/url].

The original question, however, is likely to be as kunny put it, but for a finite $ n$.  Then the summation has a finite value that is easily calculated.


Βάζω τα  προβλήματα της ΔΜΟ 2006 .(αν και μάλλον θα τα έχετε ήδη δει ) αλλά καλό θα ήταν να έχουμε τις λύσεις ολοκληρωμένες εδώ στα ελληνικά .

1 .Έστω τρίγωνο $ABC$  και $I$ το έγγεντρο . Έστω σημείο $P$ στο εσωτερικό του $ABC$  ώστε $\widehat{PBA}+\widehat{PCA}=\widehat{PBC}+\widehat{PCB}$  .Δείξτε ότι $AP\geq AI$

2.Έστω Ρ ένα κανονικό 2006-γωνο .Μία διαγώνιος του Ρ ονομάζεται <<καλή>> αν τα άκρα της διαιρούν το περίβλημα του Ρ σε δύο μέρη ώστε το καθένα από αυτά να αποτελείται από περιττό  αριθμό πλευρών του Ρ . Οι πλευρές του Ρ είναι καλές .
Υποθέτουμε ότι το Ρ έχει διαιρεθεί σε τρίγωνα από 2003 διαγωνίους έτσι ώστε όχι δύο από αυτές να έχουν κοινό σημείο στο εσωτερικό του Ρ .Βρείτε το μέγιστο αριθμό ισοσκελών τριγώνων που έχουν δύο καλές πλευρές .

3.(Αγαπημένη  :D )
Βρείτε το ελάχιστο Μ ώστε
$\left| ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})\right|\leq M(a^{2}+b^{2}+c^{2})^{2}$ για όλους τους πραγματικούς a,b,c

χαιρετώ τον αγαπητό Σιλουανό κι τους υπόλοιπους φίλους μας Αμβρόσιος ρδώ είχα ένα πρόβλημα με το pc μου κι τώρα θα επανέλθω δριμύτερος ώραια μοιάζουν απόπρώτη όψη τα προβλήματα μένει να δούμε τις λύσεις για να βγάλουμε συμπέρασμά  :rotfl:  :rotfl:  :rotfl: πώς τα πήγε η ομάδα άραγε?

Παραθέτω τα υπόλοιπα τρία του διαγωνισμού που τέθηκαν σήμερα.

4.Βρείτε όλα τα ζεύγη (x,y) ακεραίων  ώστε
$1+2^{x}+2^{2x+1}=y^{2}$ .

5 . Έστω πολυώνυμο Ρ(x) βαθμού n>1 με ακεραίους συντελεστές και έστω $k$ θετικός ακέραιος.Θεωρούμε το πολυώνυμο $Q(x)=P(P(...P(x))...))$  όπου το Ρ  εμφανίζεται $k$ φορές .Αποδείξτε οτι υπάρχουν το πολύ $n$ ακέραιοι $t$ ώστε
$Q(t)=t$.

6. ορίζουμε σε κάθε πλευρά $b$ ενός κυρτού πολυγώνου Ρ το μέγιστο εμβαδό ενός τριγώνου που μία από τις πλευρές είναι η $b$ και αυτό το τρίγωνο περιέχεται στο εσωτερικό του Ρ .Δείξτε ότι το άθροισμα των εμβαδών που ορίστηκαν στις πλευρές του Ρ είναι τουλάχιστον δύο φορές μεγαλύτερο από το εμβαδό του Ρ .

Ek πρωτης οψεως φαινονται λιγο πιο δυσκολα απο τα περσινα....

Λοιπον ας τα συζητησουμε. οι ελληνες πως πηγαν?

 :)

Χθες ασχολήθηκα με τα προβλήματα 1,2,3 και έλυσα το 1 και το 3. Παραθέτω τις λύσεις μου.

1.Παρατηρούμε ότι $\widehat{BPC}=90+\frac{A}{2}=\widehat{BIC}$ άρα $BIPC$ εγγράψιμο και το το κέντρο αυτού του κύκλου έστω Κ βρίσκεται πάνω στην ΑΙ . Επίσης ΚΙ=ΚΡ άρα
$\widehat{KIP}\leq 90$ άρα $\widehat{AIP}\geq 90$  άρα
$AI\leq AP$

3.Αρκεί να βρούμε το μέγιστο της
$\frac{\sum ab(a^{2}-b^{2})}{(a^{2}+b^{2}+c^{2})^{2}}=\frac{\left|(a-b)(b-c)(c-a)(a+b+c)\right|}{(a^{2}+b^{2}+c^{2})^{2}}$ .
Θέτουμε $a-b=x$  κτλ και x+y+z=s και ζητάμε να βρούμε το μέγιστο της $\frac{9\left|xyzs\right|}{(x^{2}+y^{2}+z^{2}+s^{2})^{2}}$

$\frac{9\left|xyzs\right|}{(x^{2}+y^{2}+z^{2}+s^{2})^{2}}\leq\frac{\left|\frac{(x+y)^{2}}{4}sz\right|}{(\frac{(x+y)^{2}}{2}+z^{2}+s^{2})^{2}}=\frac{9\left|z^{3}s\right|}{(3z^{2}+2s^{2})^{2}}=\frac{9\left|z^{3}s\right|}{(z^{2}+z^{2}+z^{2}+2s^{2})^{2}}\leq \frac{9\left|z^{3}s\right|}{(4\sqrt[4]{2z^{6}s^{2}})^{2}}=\frac{9}{16\sqrt{2}}$

Ελπίζω να είναι καλές  :D

Nomizw oti tha xreiazotan ena sxolio sto telos oti to sigekrimeno $M=\frac{9}{16\sqrt{2}}$ einai kai to elaxisto DINATO. Diladi oti iparxoun times a,b,c tetoies wste na ikanopoieitai kai i isotita. Diaforetika tha mporousame na paroume san fragma to opoiodipote xwris na mas noiazei. Nomizw den fainetai ksekathara kai oti tha ithele mia kouventa :)

Filika,
Alexandros

Σιλουανέ μήπως είναι Q(t)=t στο πολυώνυμο

Γεια σε ολους.τα θεματα φαινονται ελκυστικοτατα.ειναι ολα ομορφα.xp Q(t)=t ειναι  η συνθηκη.το συγκεκριμενο ειναι πολυ ενδιαφερον...

Ναι πράγματι είναι $Q(t)=t$  έκανα ένα τυπογραφικό  :blush: .
Όσο για παραπάνω η ισότητα ισχύει όταν και οι τρεις είναι διαφορετικοί (και εκεί είναι το ωραίο και το σπάνιο)  και πράγματι αν ελέξεται την τριάδα $(a,b,c)=(1-\frac{3\sqrt 2}{2},1,1+\frac{3\sqrt 2}{2})$

η ισότητα ισχύει. :lol:

Παραθέτω τη λύση μου και στο τέταρτο πρόβλημα .
Προφανώς ο $y$  είναι περιττός έστω $y=2n+1$ και ο $x\geq0$ .
Τότε για x=0 y=2 ή y=-2 .Τώρα αν $x= 1$ δεν έχουμε λύσεις .
Αν $x\geq 2$  τότε η εξίσωση γράφεται $2^{x}(1+2^{x+1})=4n(n+1)$ (1). προφανώς ο αριθμός $\frac{4n}{2^{x}}$ ή $\frac{4(n+1)}{2^{x}}$ είναι ακέραιοι αφού οι οι παράγοντες των γινομένου στο πρώτο μέλος είναι πρώτοι μεταξύ τους .
1περ.:$4n=k2^{x}$ από όπου αντικαθιστώντας στην (1) παίρνουμε  $k^{2}\cdot 2^{x}+4k=8\cdot 2^{x}+4$   Προφανώς  $k=1$ ή  $k=2$ οι οποίες όμως δεν ικανοποιούν την αρχική .
2η περ :$4(n+1)=m2^{x}$ .Πάλι εύκολα  $m=1$ ή  $m=2$ ή $m=3$.H μόνη που μας κάνει είναι η $m=3 \Rightarrow x=4$

καλή?? :)

[quote="silouan"]
Παραθέτω τη λύση μου και στο τέταρτο πρόβλημα .
Προφανώς ο $y$  είναι περιττός έστω $y=2n+1$ και ο $x\geq0$ .
Τότε για x=0 y=2 ή y=-2 .Τώρα αν $x= 1$ δεν έχουμε λύσεις .
Αν $x\geq 2$  τότε η εξίσωση γράφεται $2^{x}(1+2^{x+1})=4n(n+1)$ (1). προφανώς ο αριθμός $\frac{4n}{2^{x}}$ ή $\frac{4(n+1)}{2^{x}}$ είναι ακέραιοι αφού οι οι παράγοντες των γινομένου στο πρώτο μέλος είναι πρώτοι μεταξύ τους .
1περ.:$4n=k2^{x}$ από όπου αντικαθιστώντας στην (1) παίρνουμε  $k^{2}\cdot 2^{x}+4k=8\cdot 2^{x}+4$   Προφανώς  $k=1$ ή  $k=2$ οι οποίες όμως δεν ικανοποιούν την αρχική .
2η περ :$4(n+1)=m2^{x}$ .Πάλι εύκολα  $m=1$ ή  $m=2$ ή $m=3$.H μόνη που μας κάνει είναι η $m=3 \Rightarrow x=4$

καλή?? :)[/quote]

Ομοια με τη λυση μου!!!!

ηταν ενα ωραιο αλλα απλο (εως πολυ) για ΙΜΟ.

Θα μου αρεσε να μπει σε αρχιμηδη.... :)

Οι χώρες που πρότειναν τα προβλήματα είναι
Problem 1 - Νότια Κορέα(Hojoo Lee για τρίτη συνεχή φορά  :o )
Problem 2 - Σερβία
Problem 3 - Ιρλανδία
Problem 4 - Αμερική
Problem 5 - Ρουμανία
Problem 6 - Σερβία

Χαρακτηριστικό είναι ότι τα δύο προβλήματα γεωμετριοσυνδιαστικής προτάθηκαν από τη Σερβία !!!!!!!!!!

Είχα ξεχάσει να βάλλω τη λύση μου στο 5ο πρόβλημα .
Αν $Q(t)=t$  όταν $P(t)=t$  το πρόβλημα γίνετα προφανές .
Θα εξετάσουμε την περίπτωση κατά την οποία
$P(x_{1})=x_{2}$ ,  $P(x_{2})=x_{3}$,...., $P(x_{n})=x_{1}$ (αυτές οι ισότητες προκύπτουν ώστε να έχουμε $Q(x_{1})=x_{1}$  χωρίς        $P(x_{1})=x_{1}$   :wink: ) kai $n\geq 3$.Υποθέτουμε ότι  $x_{1},x_{2},...,x_{n}$  διαφορετικοί μεταξύ τους .Πράγματι αν $x_{k}=x_{n}$  η περιοδικότητα απλα μειώνεται αλλά στην ουσία είναι σα να διαιρούμε το φαινόμενο σε δύο φάσεις άρα είναι το ίδιο πράγμα .Τότε θα εχουμε
$P(x)-x_{2}=(x-x_{1})P_{1}(x)$
$P(x)-x_{3}=(x-x_{2})P_{2}(x)$
....
....
$P(x)-x_{1}=(x-x_{n})P_{n}(x)$

Υποθέτουμε ότι η απόλυτη τιμή της διαφοράς $x_{1}-x_{n}$  είναι η μεγαλύτερη μεταξύ όλων των διαφορών που παρουσιάζονται στα γινόμενα του δευτέρου μέλους .Θέτουμε στην πρώτη σχέση όπου $x$  το $x_{n}$ .Τότε αυτή γίνεται $x_{1}-x_{2}=(x_{n}-x_{1})P_{1}(x)$  και αφού όλοι αυτοί είναι ακέραιοι η απόλυτη τιμή της διαφοράς $x_{1}-x_{2}$  είναι μεγαλύτεροι από την απόλυτη τιμή της διαφοράς  $x_{1}-x_{n}$ , άτοπο  άρα $n=k\leq 2$

Άρα αρκεί να λύσουμε το πρόβλημα για n=2 .Δηλαδή αν
$P(P(x))=Q(x)$ .Πάλι δεν εξετάζουμε την προφανή $P(t)=t$  και υποθέτουμε ότι $P(a)=b$  kai  $P(b)=a$
Πάλι  $P(x)-b=(x-a)P_{1}(x)$  και  $P(x)-a=(x-b)P_{2}(x)$
Βάζοντας όπου x  το b στην πρώτη παίρνουμε ότι $P_{1}(x)=-1$ kai όμοια $P_{2}(x)=-1$ .Kai exoume
$P(x)-b=a-x$  (1)
$P(x)-a=b-x$  (2)
(1)+(2) => $P(x)+x=a+b$  (3)
Έστω μια άλλη δυάδα τώρα ώστε $P(m)=n$  και $P(n)=m$
Πάλι τότε $P(x)+x=m+n$  . Άρα για κάθε τέτοιο ζεύγος θα συμβαίνει αυτό .Ή αφού $a+b=m+n$  μπορούμε την   (3) να την γράψουμε ως $a+b-m-P(m)=0$ (4) .Θέτουμε $a+b=C$  και θεωρούμε το πολυώνυμο
$F(x)=C-P(x)-x$ .Δείξαμε με την (4) κάθε ζεύγος λύσης της
$Q(t)=t$ είναι ρίζα του πολυωνύμου F(x)   :wink:
Αλλά το F(x) είναι νιοστού βαθμού (είναι οοιόβαθμο με το P(x))
άρα έχει το πολύ $n$  ρίζες .Όμως αυτές είναι τα ζεύγη λύσεων της $Q(t)=t$  (5) (κάθε ζεύγος αντιστοιχεί σε μία ρίζα της (5)) άρα η (5)  έχει το πολύ   $n$  ρίζες , το ζητούμενο......

Δεν μπορούσα να τη γράψω πιο αναλυτικά   :D


complex analysis
function
complex analysis unsolved

Show as elementary as possible (without using Picard's theorem), that there is no holomorphic function $f: \mathbb{C} \to \mathbb{H} = \{z: \Im(z) > 0\}$.

Suppose such an $f$ exists. The function $g(x)=\frac{f(x)-i}{f(x)+i}$ maps $\mathbb{C}$ to the unit disk. A bounded entire function is constant, and we have a contradiction.


Study the sequence $u_n$ defined as follows:$u_{n+1}=\frac{1}{u_n+\frac{1}{n+1}},u_0>0$.

After some tries I can accross the formula :

$u_{n+1} - 1 = (u_{n-1} - 1) \times n^2 / (n^2+2n+1+ (u_{n-1} - 1) \times n)$

From there we can prove that $u_n \rightarrow 1$


integration
calculus
calculus computations

prove that       :  equation  have 1  real  root: $ \sum_{i\equal{}0}^{n} c_ix^i\equal{}0$
  with following  condition : $ \sum_{i\equal{}0}^{n} \frac{c_i}{i\plus{}1}\equal{}0$

Let $ f(x) \equal{} \sum_{i\equal{}0}^{n} c_i x^i$.  Then we are given $ \int_{0}^{1} f(x) \, dx \equal{} 0$, so $ f(x)$ is either identically zero or changes sign.


The price of an article is cut $ 10\%$. To restore it to its former value, the new price must be increased by:

$ \textbf{(A)}\ 10\% \qquad\textbf{(B)}\ 9\% \qquad\textbf{(C)}\ 11\frac {1}{9}\% \qquad\textbf{(D)}\ 11\% \qquad\textbf{(E)}\ \text{none of these answers}$

[hide="Solution"]
Let the original price of the article be $ x$. When the price is cut by $ 10\%$, then the price is $ 9x/10$. To restore it to its former value, you must multiply by $ 10/9$ which is equivalent to increasing the price by $ \boxed{(C)11\frac{1}{9}\%}$.
[/hide]


Given a natural number n≥2.Prove that the equation $ \frac{x^n}{n!}\plus{}\frac{x^{n\minus{}1}}{(n\minus{}1)!}\plus{}...\plus{}\frac{x^2}{2!}\plus{}x\plus{}1\equal{}0$ have no rational solution.

Suppose there is a rational solution $ x$, it is easy to see $ x$ has to be an integer.
It is easy to check that the root $ x\ne\pm1$. Hence there exists prime $ p$ such that $ p|x$.
Let $ f(x)$ be the power of $ p$ in number $ x$. In other words, $ x \equal{} p^{f(x)}t$, where $ gcd(t,p) \equal{} 1$.
For all $ n > 0$, we have
$ f(n!) \equal{} \sum_{i \equal{} 1}^n[\frac n{p^i}]\le \sum_{i \equal{} 1}^n\frac n{p^i} < \frac n{p \minus{} 1}\le n\le f(x^n)$
Hence, $ f(x^n)\ge f(n!) \plus{} 1$.
Let $ g(x)$ be the LHS of the equation,
$ n!*g(x) \equal{} n! \plus{} \sum_{i \equal{} 1}^n\frac {n!}{i!}x^i \equal{} 0\Rightarrow \minus{} n! \equal{} \sum_{i \equal{} 1}^n\frac {n!}{i!}x^i$.

$ f( \minus{} n!) \equal{} f(n!)$
$ f(\frac {n!}{i!}x^i)\ge f(n!) \plus{} f(x^i) \minus{} f(i!)\ge f(n!) \plus{} 1$.

$ f(n!) < f(\sum_{i \equal{} 1}^n\frac {n!}{i!}x^i)$. Contradiction!


limit
real analysis
real analysis unsolved

Suppose $x_1,x_2,x_3,...$ is a sequence of nonnegative real numbers satisfying $x_{n+1} \leq x_n+\frac{1}{n^2}$. Prove that
$\displaystyle \lim_{n\to\infty}x_n$ exists.

$x_{n+p} - x_n \leq \sum_{k=n}^{n+p} \frac{1}{k^2}$

Take supremum limit over all $p$, and we have
$\varlimsup_{k\to\infty} x_k - x_n \leq \sum_{k=n}^\infty \frac{1}{k^2}$

Take infinum limit over all $n$, and we have
$\varlimsup_{k\to\infty} x_k - \varliminf_{n\to\infty} x_n \leq 0$

another proof:
we have
$x_{n+1}-\frac{2}{n+1} < x_n-\frac{2}{n}$
then
$ x_n-\frac{2}{n}$ is limit.
so $x_n$ is limit


1 / (2x +1)

USING FIRST PRINCIPLES.

I tried and I get totally stuck mid way.

Heres what I had so far (you know I have to write the limit stuff, but its cumbersome to write here so I won't)

____

[u]f(x+h) - f(x)[/u]
h

[u][1/ (2x +2h +1)] - [1/(2x+1)][/u]
h

____

Yeah...after that I just totally got lost. I'm supposed to factor the h at the top...but how am I going to clean up this mess?

Thanks

:/

The thing that I have trouble with is the simple math underlying the complicated math. You know, the simple rules like..probably finding the common denominator or cross multiplying. I don't know what you did to get each one at each line.

For one, how did you get the -2h? And why did you multiple both of the denominators (this may be a stupid question, but as I said, I'm not good with remembering basic math rules).

And also...how did you get to write in the solution like that? Is there some plugin in the reply box?

EDIT: Actually...now I remember...Thank you so much! But I'm still curious about how you wrote your message like that :/

[quote]Is there some plugin in the reply box?[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]Read here[/url]


I got this forwarded message yesterday:

_______________________________________________________________
FROM Arvind Khamitkar , I.A.S, Director of Medical
& Research Div,
Chennai

Dear Friends,

A few weeks ago, in a movie theatre, a person felt
something poking from her seat.
When she got up to see what it was, she found a
needle sticking out of the seat with a note attached saying

'You have just been infected by HIV'.

The Disease Control Center (in Paris) reports many similar events in many other cities recently. All tested needles were HIV Positive.

The Center also reports that needles have been found in cash dispensers at public banking machines. We ask everyone to use extreme caution when faced with this kind of situation. All public chairs/seats should be inspected with vigilance and caution before use. A careful visual inspection should be enough. In addition, they ask that each of you pass this message along to all members of your family and your friends of the potential danger.

Recently, one doctor has narrated a somewhat similar
instance that hppened to one of his patients at the Priya Cinema in Delhi . A young girl, engaged and about to be married in a couple of months, was pricked while the movie
was going on. The tag with the needle had the message

'Welcome to the World of HIV family'.

Though the doctors told her family that it takes about 6
months before the virus grows strong enough to start
damaging the system and a healthy victim could survive about 5-6 years, the girl died in 4 months, perhaps more because of the 'Shock thought'.
We all have to be careful at public places, rest God help! Just think about saving a life by forwarding this message. Please, take a few seconds of your time to pass along.

With Regards, Arvind Khamitkar ,
I.A.S, Director of Medical & Research Div, Chennai.
_______________________________________________________________

Reactions?

Just searching around a little bit for information on this turns up lots of urban legend debunking sites and no relevant news articles. The fact that your source is an email chain letter in a typical spam format doesn't help.

Also, the probability of getting HIV from an infected needle is pretty small unless you actually get injected with HIV-containing fluid. This would probably be rather difficult to arrange without the apparatus becoming very visible or very expensive (it couldn't just be a needle; it would have to be a syringe).

[url=http://www.google.com/search?as_q=&hl=en&client=safari&rls=en-us&num=10&btnG=Google+Search&as_epq=You+have+just+been+infected+by+HIV&as_oq=&as_eq=&lr=&cr=&as_ft=i&as_filetype=&as_qdr=all&as_nlo=&as_nhi=&as_occt=any&as_dt=i&as_sitesearch=&as_rights=&safe=images]This[/url] is a list of searches received after doing an advanced google search.

Like Xevarion's statement, there was little/no relevent news articles about this.

The quote below is a more accurate statement from the [url=http://www.cdc.gov/hiv/resources/qa/hoax1.htm]CDC[/url] regarding similar stories mentioned (it's at the bottom of the page of the link):
[quote="Center for Disease Control (USA)"][b]I have read stories on the Internet about people getting stuck by needles in phone booth coin returns, movie theater seats, gas pump handles, and other places. One story said that CDC reported similar incidents about improperly discarded needles and syringes. Are these stories true?[/b]

CDC has received inquiries about a variety of reports or warnings about used needles left by HIV-infected injection drug users in coin return slots of pay phones, the underside of gas pump handles, and on movie theater seats. These reports and warnings have been circulated on the Internet and by e-mail and fax. Some reports have falsely indicated that CDC "confirmed" the presence of HIV in the needles. CDC has not tested such needles nor has CDC confirmed the presence or absence of HIV in any sample related to these rumors. The majority of these reports and warnings appear to have no foundation in fact.

CDC was informed of one incident in Virginia of a needle stick from a small-gauge needle (believed to be an insulin needle) in a coin return slot of a pay phone. The incident was investigated by the local police department. Several days later, after a report of this police action appeared in the local newspaper, a needle was found in a vending machine but did not cause a needle-stick injury.

Discarded needles are sometimes found in the community outside of health care settings. These needles are believed to have been discarded by persons who use insulin or are injection drug users. Occasionally the "public" and certain groups of workers (e.g., sanitation workers or housekeeping staff) may sustain needle-stick injuries involving inappropriately discarded needles. Needle-stick injuries can transfer blood and blood-borne pathogens (e.g., hepatitis B, hepatitis C, and HIV), but the risk of transmission from discarded needles is extremely low.

CDC does not recommend testing discarded needles to assess the presence or absence of infectious agents in the needles. Management of exposed persons should be done on a case-by-case evaluation of (1) the risk of a blood-borne pathogen infection in the source and (2) the nature of the injury. Anyone who is injured from a needle stick in a community setting should contact their physician or go to an emergency room as soon as possible. The health care professional should then report the injury to the local or state health department. CDC is not aware of any cases where HIV has been transmitted by a needle-stick injury outside a health care setting.[/quote]

[quote="bubka"]I got this forwarded message yesterday:

Reactions?[/quote]

my first reaction was to laugh.

my second reaction was to think that you are not very smart.

my third reaction was to realize that this really wasn't that big of a surprise.

my fourth reaction was to think of [url=http://www.snopes.com/horrors/robbery/slasher.asp]the slasher under the car[/url].

does this cover everything?

Like Xevarion said,... this is  a hoax.

When I went to the movies a long time ago (2000?) some of my friends said he had heard of this and we laughed and I poked him with my drinking straw and then he jumped. :roll:

It could be a hoax, but I don't really have enough facts to determine that. Is it possible for these "needles" to be in seats of testing cars?

well gee thanks for your informed opinion as well

Well, this thread has evidently served its purpose of allowing two different posters to make fools of themselves. (Rhetorical question: why would you bother to make a post whose content is to note that you have nothing useful to add?)  And now, locked.


vector
combinatorics proposed
combinatorics

Given a table with $2^n * n$ 1*1 squares ( $2^n$ rows and n column). In any square we put a number in {1, -1} such that no two rows  are the same. Then we change numbers in some squares by 0. Prove that in new table we can choose some rows such that sum of all numbers in these rows equal to 0.

I’ll find some rows such that their sum (when added like vectors) is 0-all. An n-tuple is an ordered system with n elements. View all the rows as some n-tuples. Let A_n be the set of n-tuples (not necessarily rows) with entries {0,1} and B_n the set of n-tuples with entries {-1,1}. Let t be a transformation that changes any element of A_n into an element of B_n by changing all 0 into 1 and all 1 into –1. I’ll build “step by step” (i.e. inductively) a sequence of rows. I’ll start with the row which first contained only 1-s, so now containes only 0-s and 1-s. I’ll try to build the sequence such that the sum of these rows (added like vectors) contains only the elements 0 and 1. Obvioulsly, at the first step, this is true. Supose at some step, we have some rows such that the sum of these rows contains only 0s and 1s. Let s be this sum and t(s) the image of s through t. Now, take the row which was first t(s) (obviously, each row occurred exactly once). When you add this row to the sequence, the sum of all rows in the sequence keeps its property of containing only 0s and 1s (just write this down and you’ll see this easily). But eventually, during this process, a row occurs twice. Take the first time a row appears for the second time. But this means that at those times (the first appearance of the row and the second one), the sum of the rows were identical. Then, you could take all the rows in the sequence after the first appearance of this row, and you see that their sum is (0,0,…,0). Now we are done. Sorry for not using LaTeX. I prommisse I’ll learn it as fast as possible.


calculus
integration
logarithms
calculus computations

\[ If\int\limits_1^2 {e^{x^2 } } dx \equal{} \alpha ,find\int\limits_e^{e^4 } {\sqrt {\log x} } dx
\]

$ \begin{array}{l}
 I \equal{} \int\limits_e^{{e^4}} {\sqrt {\log x} dx \equal{} } \frac{1}{{\sqrt {\ln 10} }}\int\limits_e^{{e^4}} {\sqrt {\ln x} dx \equal{} } \frac{1}{{\sqrt {\ln 10} }}\int\limits_1^2 {u{{\left( {{e^{{u^2}}}} \right)}^\prime }du}  \equal{}  \\
  \equal{} \frac{1}{{\sqrt {\ln 10} }}\left( {\left[ {u{e^{{u^2}}}} \right]_1^2 \minus{} \int\limits_1^2 {{e^{{u^2}}}du} } \right) \equal{} \frac{1}{{\sqrt {\ln 10} }}\left( {2{e^4} \minus{} e \minus{} \alpha } \right) \\
 \end{array}$

Assuming $ \log : \equal{}\ln$, note that $ \alpha\plus{}\int_e^{e^4}\sqrt{\log x}\ dx\equal{}\int_1^2e^{x^2}\ dx\plus{}\int_e^{e^4}\sqrt{\log x}\ dx\equal{}2e^4\minus{}e$ (what is $ \int_a^bf\plus{}\int_{f(a)}^{f(b)}f^{\minus{}1}$ when $ f$ satisfy certain properties?).  Everything else is clear.


[quote="valeriummaximum"]f(1)=1,f(2)=2,f(3)=3,f(4)=5....

So this pattern is nothing but the fibonacci series...
for the second one it is simple P&C.. so i hope u can solve it...
[hide]
1.34
2.14400
3.377
4.3[/hide][/quote]
[hide]
Yes yes its fibinocci, easy there. But the reason why its fibi is very very interesting. Its not directly f(n) = f(n-1)  + (n-2) by obvious observation because there's more to this. Care to explain>[/hide]

One small ob which can help is that the last no can be a 1 or a 2 added so it could be like fibonacci .  :D There will be overcounting if u consider the fact that 2 1's can be added instead of a 2 so we need to consider f(n-1) for that only  :)

Consider this

4 = 2 + 2

According to the ob reason. f(5) can be got by
1) Adding 1 to f(4)
   Adding 2 to f(3)

So add 1 to the above eqn

5 = 2 + 2 + 1

but this has 3 permutations which sadly aren't accounted for in f(4)?? I didn't want to tell this point.... And this is what I meant by saying its not the obvious fibinocci. In that tile sum (INOI or the one we did in class), its ob that n can be directly derived from n-1 and n-2 indenpendtly and simultaneously. here thats not the case

[quote="Cyber-Shreyas"]According to the ob reason. f(5) can be got by
1) Adding 1 to f(4)
   Adding 2 to f(3)

So add 1 to the above eqn

5 = 2 + 2 + 1 [/quote]
Wth was that? f(5) = f(4) + 1 ?  :huh:

Okay, since I can't expect you to think at my speed. I'll explain the hidden point for you.

Remember that here, all permutations matter. Infact any combination of 5 eg: (221) will be derived partially from (4) and partially from (3).

(221) has 3 permutations

f(4) has (22) as 1 permu
so we will get only one of these from f(4)

f(3) has (21) and (12) so by inserting 2 we get the other 2 permu's from here and hence the formula f(n) = f(n-1) + f(n-2) holds good.

I meant that the fibinocci formula here is not as ob as the tiles sum.


geometry
circumcircle
parallelogram
projective geometry
geometry theorems

Yesterday, I tried solving the generalization of the butterfly 's theorem and I found two nice solutions. Here I cite the theorem first
     Let $O$ be a circle and $UV$ be its chord. Let $M$ be a point on $UV$. Call $AC$ and $BD$ two other chords of $(O)$ that pass through $M$. Let $X$ and $Y$ be the intersections of $AD$ and $BC$ with $UV$. Then $\frac{1}{MX} + \frac{1}{MY} = \frac{1}{MU} + \frac{1}{MV}$.
     Here are my two proofs :
        [b]Proof 1[/b] (projective geometry)
            Let $P$ and $Q$ be the intersections of $AD$ and $BC$, $AB$ and $CD$. Then $PQ$ is the polar of $M$ wrt $(O)$. Denote by $R$ the intersection of $UV$ and $PQ$. Then from $(RMUV) = -1$, we obtain that  $\frac{1}{MU} + \frac{1}{MV}= \frac/{2}{MR}$. On the other hand since $(PQ, PM, PD, PC) = -1$, we obtain $(RMXY) = -1$. This implies that $(RMXY) = -1$. Hence $\frac{1}{MX} + \frac{1}{MY} = \frac{2}{MR}$. And the result follows.
        [b]Proof 2[/b] (inversion proof)
              Consider the inversion center $M$ with constant $MA . MC$. The inversion will map $A, B, C, D, U, V$ to $C, D, A, B, V, U$. Therefore, $AD$ is mapped to the circumcircle of $BMC$ whose center is $O_2$, and $BC$ to the circumcircle of $MAD$, whose center is $O_2$. Therefore $X$ and $Y$ are transformed respectively to the second intersections $E$ and $F$ of $UV$ with $(O_2)$ and $(O_1)$. Let $I$ and $J$ be the projections of $O_1$ and $O_2$ on $UV$. Then $I$ and $J$ are the midpoints of $MF$ and $ME$. The problem is now equivalent to proving $ME + MF = MU + MV$ or $UF = EV$. So it is sufficient to prove that $OE = OF$.
              Since $A$ and $D$ lie on $MC$ and $MD$ and $ABCD$ is cyclic, we obtain $O_1M$ must be perpendicular to $BC$. Moreover $OO_2$ is also perpendicular to $BC$. This implies that $O_1M // OO_2$. Similarly, we obtain that $O_2M // OO_1$ and hence, $O_1MO_2O$ is a parallelogram with center $K$.
              Now since $K$ is the midpoint of $O_1O_2$, we obtain $KI = KJ$. But since $K$ is also the midpoint of $OM$, it follows that $KI = \frac{OF}{2}$ and $KJ = \frac{OE}{2}$. Hence $OE = OF$ and we 're done.

     If you have other solutions, would you mind posting it so that everybody can share it ?  ;)

[quote="treegoner"]
Then from $(RMUV) = -1$, ... $(PQ, PM, PD, PC) = -1$, we obtain $(RMXY) = -1$.
[/quote]
I'm confused. what does $( \cdots )$ mean here?

perfect radio ,treegoner made mistake by writing that:(PQ,PM,PD,PC)=-1

first i should say its P(Q,M,D,C)=-1

u can see:  http://www.mathlinks.ro/forum/viewtopic.php?t=51865

there is another solution with harminic on circles,,,

i will post it as soon as possible ;)

At the risk of being a little late, the following page contains several solutions and generalizations:

http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

AB


A prime number $p$ $>$ $3$ and positive integers $a$ , $b$ , $c$ satisfy
$a$ $+$ $b$ $+$ $c$ $=$ $p$ $+$ $1$ and the number $a^{3}$ $+$ $b^{3}$ $+$ $c^{3}$ $-$ $1$ is divisible by $p$.Show that at least one of the numbers $a$ , $b$ , $c$ equal to $1$

Very nice solution, I know that I would never have come up with that... how did you approach this problem to solve it? Is there a structured logical way to come up with your idea, I need help on general problem solving!!!
Thank you, :lol:


induction
graph theory
combinatorics unsolved
combinatorics

I have two easy yet unrelated graph theory problems here. I am pretty unsure about the solutions to my problems, so I post these problems here and maybe hopefully one kind soul would reply  :lol:

1. Let D be a digraph. Prove that if there is a u-v walk in D then there is a u-v path in D as well, where u and v are vertices in D.

2. Let G be a simple graph such that G contains only one odd cycle as a subgraph. Find the chromatic number of G and justify.

 :)

Indeed, they are easy: for 1, every walk is associated to a word which lists the vertices in order.  If the walk does not include any vertex twice, we're done.  If it does include some vertex twice, then it is of the form $ {\bf x}a{\bf y}a{\bf z}$ where $ \bf x, y, z$ are some strings and $ a$ is some vertex.  Then the walk associated to the string $ {\bf x}a{\bf z}$ is also a u-v walk, and we are done by induction (on the number of pairs of twice-visited vertices).

For number 2, the answer is $ m$ for some $ m$.  The proof is just to show that $ m\minus{}1$ is inadequate and to give a demonstration that $ m$ is adequate, for example by removing and then reintroducing a particular edge.


The stem-and-leaf plot gives the height, in inches, of thirteen
students. What is the median height in inches?
$ \begin{tabular}{c|ccccccc} 4 & 7 & 7 & 7 & 8 & 8 & 8 & 9 \\
5 & 0 & 0 & 1 & 2 & 4 & & \\
6 & 0 & & & & & & \\
\end{tabular}$

[size=59][color=red]Fixed table. -Isabella[/color][/size]

The table is unclear, however here's how to solve it:

In a stem-and-leaf plot, 5|4 8 9 for example would indicate 54, 58, and 59. To find the median height, write out the numbers and cross them off until you reach the one in the middle.

[quote="GameBot"]The stem-and-leaf plot gives the height, in inches, of thirteen
students. What is the median height in inches?
$ \begin{tabular}{c|ccccccc}
4 & 7 & 7 & 7 & 8 & 8 & 8 & 9 \\
5 & 0 & 0 & 1 & 2 & 4 & & \\
6 & 0 & & & & & & \\
\end{tabular}$[/quote]

LOl latex symbold work wonders. AS you can see, there are 6 before 49 and 6 after it, so the median is $ 49$


Find all functions f : R--> R satisfying:
 $f(x^3+x) \leq x \leq f^3(x)+f(x)$

Let $g(x)=x^3+x$, $g: \mathbb{R}\to\mathbb{R}$. Clearly, $g$ is invertible and strictly increasing. We have
\[ f(g(x))\le x, \forall x\in\mathbb{R} \] which implies
\[ g(f(g(x)))\le g(x),\forall x\in\mathbb{R} \]
Also, we have
\[ x\le g(f(x)),\forall x\in\mathbb{R} \]
so, replacing $x$ by$g(x)$ yields
\[ g(x)\le g(f(g(x))),\forall x\in\mathbb{R}. \]
We conclude that $g(f(g(x)))=g(x)$, hence $f(g(x))=x,\forall x\in\mathbb{R}$, that is, $f=g^{-1}.$


$\triangle ABC$ is an isosceles right triangle such that $AC=BC$ and $\angle C=90^\circ$. $P$ and $Q$ are on $AB$ such that $\angle PCQ=45^\circ$. Prove that $PQ^2=BP^2+AQ^2$.

Edit: $Q$ is between $Q$ and $A$.

[quote="frt"]$\triangle ABC$ is an isosceles right triangle such that $AC=BC$ and $\angle C=90^\circ$. $P$ and $Q$ are on $AB$ such that $\angle PCQ=45^\circ$. Prove that $PQ^2=BP^2+AQ^2$.[/quote]

I think this was given in some former romanian JTST (2002 maybe)

 :)

[hide]Not necessarily.  If $P$ is closer to $A$ and $Q$ is closer to $B$, then $PQ<BP$.  So $PQ^2<BP^2+AQ^2$, because $AQ$ can't be complex (the length of a line must be a positive real number).[/hide]

for the converse of this fact, see:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85445[/url]

also note that we must have Q between P and A

look down a bit to my post, i offer a concise solution that can be reversed easily

Yeah, I forgot to mention that $Q$ must be between $P$ and $A$. Sorry :oops:

Altheman, I saw your solution in the link, and that was exactly what I expected :lol:


algorithm
quadratics
MATHCOUNTS
AMC
AIME
AMC 10

does anyone know if we can use two calculators on the exam?

Probably. Why would you want to use 2 calculators?

You've never heard of a double-fisted programmer?  :P

(The only reason I can think of is to write and run some algorithm on one calculator while using the free one to do other problems... though I don't think any of the problems are efficiently solved with algorithms.)

[quote="bpms"]I personally have actually used two calculators before on mathcounts. I had one scientific that I knew backwards to forwards, and one graphing that I couldn't do stuff like fractions and what not. I used the graphing for, you guessed it, graphing.[/quote]
that's the reason. my scientific (casio) can solve quadratics, cubics, and linear equations easily. also i feel that i am more familiar with it.

i hate the buttons on texas instruments so i plan to only use that for graphing.

[quote="cincodemayo5590"][quote="bpms"]I personally have actually used two calculators before on mathcounts. I had one scientific that I knew backwards to forwards, and one graphing that I couldn't do stuff like fractions and what not. I used the graphing for, you guessed it, graphing.[/quote]

I would've used the graphing calculator as a back-scratcher.  :rotfl:

(Edit: So wait, is there [i]any[/i] restriction on the calculators allowed on the test, other than not to bring a robot screaming "DANGER WILL ROBINSON, DANGER!" ?)[/quote]
:lol:

I think that you can't use TI-92/any calculator with a qwerty keyboard.

Yeah, you can definitely use 2 calculators.  I took 3 into it last year, because I roll like that.

official word from the front of the booklet is any SAT approved calculator works... doesn't say whether you can have two.

I had three at my spot at National Mathcounts last year :rotfl: . My teachers graphing calculator that I used for graphing, my old calculator that I knew how to use, and the one that I got, because it was awesome :wink: .

[quote="life1"]Source: Scientific, Graphing, "Normal" (define, please) - which are best?
I hear many people say they take multiple calculators into tests - but some prefer scientific or graphing or different brands like casio or texas -- what are the differences, and does it matter which brand? How can I figure out which TYPE  is best for tests like AMC 10 and Mathcounts and AIME (if calcs are allowed)?[/quote]

well...
im very familiar with graphing calculators because ive never really used anything other than that [besides like stupid calculators that only add and subtract]

if your familiar with a casio scientific, they're really good for solving equations and whatnot.

and graphing calculators are great for graphing [duh]

id use both.

Just get a TI-83/84. It can do pretty much anything you'd want it to.

A TI-89 titanium (which I use) can do anything you want it to do AND things you dont want it to do.  :)

If you get one, you practically passed any algebra problems on the test.  :P


function
logarithms
inequalities
algebra
polynomial
trigonometry
linear algebra

Guys, i have 2 problems...

1) For what values of $ h$ and $ k$ is the function $ f(x) \equal{} \frac {kx \plus{} h}{hx \plus{} k}$ one-to-one?

2) Find the number of digits of N if $ \log_2 (log_3 (log_4 N)) \equal{} 2$

i have partial solution for 2) [hide] $ N \equal{} 4^{81}$. i know of a method to solve this, by using a calculator and computing for the logarithms of some numbers and use an inequality with respect to base 10 since we know that $ 10^n$ has rules for the number of zeroes and eventually its digits. is there some other way? [/hide]

[hide]2. Get the equation to the form $ \log_4 N \equal{} 81$. Now convert it to base 10 ($ \log_4 N\equal{}\frac{\log_{10} N}{\log_{10} 4}$, and the first integer greater than the $ \log_{10} N$ will be the number of digits.

1. hint: what happens if none of h and k isn't zero?[/hide]

Question 1 can be answered for the more general $ f(x) \equal{} \frac{ax \plus{} b}{cx \plus{} d}$.  Try to find an inverse function.

[quote="hsiljak"][hide]2. Get the equation to the form $ \log_4 N \equal{} 81$. Now convert it to base 10 ($ \log_4 N \equal{} \frac {\log_{10} N}{\log_{10} 4}$, and the first integer greater than the $ \log_{10} N$ will be the number of digits.

1. hint: what happens if none of h and k isn't zero?[/hide][/quote]

i seem not to understand what you said but here's what i did:

[hide]
$ \log_{10} 4^{81} \equal{} 81\log_{10} 4\approx 48.77$ so, $ 4^{81} \approx 10^{48.77}$ which means there are 48 zeros + 1 (the digit 1) = 49 digits. did you do this too? or you didnt use a calculator? [/hide]

[quote="t0rajir0u"]Question 1 can be answered for the more general $ f(x) \equal{} \frac {ax \plus{} b}{cx \plus{} d}$.  Try to find an inverse function.[/quote]

hmm, i only know that to test if a function is one-to-one, we use the horizontal line test but it needs a graph. for this case, it's difficult to graph since we are not given with the values of h and k. but isn't this one-to-one for all values of h and k since it's a rational function?

There are many rational functions that aren't 1-1, like $ x^2$ - or just the constant 0.

To show whether a function is 1-1 using the definition, check if you can find $ a,b$ [i]which aren't equal[/i] such that $ f(a) \equal{} f(b)$. If you can, it's not 1-1. If you can't (and not just because you didn't look hard enough), it is.
(This is equivalent to the "horizontal line test", as $ (a,f(a))$ and $ (b,f(b))$ both satisfy $ y \equal{} f(a)$.)

For the case of rational functions, setting $ f(a) \equal{} f(b)$ will produce a factor of $ a \minus{} b$, leaving another polynomial that can easily be checked for solutions with $ a\neq b$ if the functions are simple enough.

You could also try, as suggested above, trying to construct a left inverse (a function $ g$ such that $ g(f(x))\equal{}x$ where $ f(x)$ is defined: so a left inverse for $ x^3\plus{}1$ is $ \sqrt[3]{x\minus{}1}$ and a left inverse for $ \arctan x$ is $ \tan x$). The construction may fail for some cases, so you'll have to check them separately.

[quote="t0rajir0u"]Question 1 can be answered for the more general $ f(x) \equal{} \frac {ax \plus{} b}{cx \plus{} d}$.  Try to find an inverse function.[/quote]

torajirou...i need your help for number 1,...

[hide="Solution"] Let $ x \equal{} \frac{ay \plus{} b}{cy \plus{} d}$ and we'll try to solve for $ y$.  Well,

$ x(cy \plus{} d) \equal{} ay \plus{} b \implies$
$ y(cx \minus{} a) \equal{} b \minus{} dx \implies$
$ y \equal{} \frac{\minus{}b \plus{} dx}{a \minus{} cx}$.

This construction gives us the inverse in every case except the case that $ x \equal{} \frac{a}{c}$.  But this is only possible if

$ a(cy \plus{} d) \equal{} c(ay \plus{} b) \Leftrightarrow$
$ ad \minus{} bc \equal{} 0$.

Whenever this is not the case, $ f(x)$ is invertible.  When this is the case, $ f(x)$ is a constant. [/hide]
[hide="Motivation"] Such functions form the [url=http://en.wikipedia.org/wiki/Projective_group]projective group[/url] $ PGL(2, \mathbb{R})$ (or equivalently $ \mathbb{C}$), so composition of such functions respect matrix multiplication (up to multiplication by a constant).  The condition that $ ad \minus{} bc \neq 0$ is the condition that a $ 2 \times 2$ matrix be invertible. [/hide]

Can't you just say that $ f(x) \equal{} \frac {ax \plus{} b}{cx \plus{} d}$ is not defined for $ x \equal{} \frac { \minus{} d}{c}$, so $ f(x) \equal{} \frac {ax \plus{} b}{cx \plus{} d}$ is never one-to-one?

Here it's important to distinguish between an [url=http://en.wikipedia.org/wiki/Injective]injection[/url] ("one-to-one function") and [url=http://en.wikipedia.org/wiki/Bijective_function]bijection[/url] ("one-to-one correspondence").  When $ ad \minus{} bc \neq 0$ the above function is an injection, and if we restrict its range to $ \mathbb{R} \minus{} \left\{ \frac{a}{c} \right\}$ and its domain to $ \mathbb{R} \minus{} \left\{ \minus{} \frac{d}{c} \right\}$ it is also a bijection.

Of course, $ \mathbb{R}$ is not the natural setting in which to view a [url=http://en.wikipedia.org/wiki/M%C3%B6bius_transformation]Mobius transformation[/url].

Yes, I'm sorry. I was thinking of bijections.


[b]P is one point of intersection of two circles. Draw a line throgh P intersecting one circle at A and other at B such that AP=BP.[/b]

Draw the line through $ P$ and perpendicular to the line segment $ PM,$ where $ M$ is the midpoint of the distance between the centers of the given circles.

Kostas Vittas.


Let $ a,b,c>0$.Prove that $ 9(a^4\plus{}1)(b^4\plus{}1)(c^4\plus{}1)\geq\ 8\left((abc)^2\plus{}abc\plus{}1\right)^2$


vector
function
parameterization
analytic geometry
calculus
calculus computations

find a vector function that represents the curve of intersection of the paraboloid z = 4x^2+ y^2 and the parabolic cylinder y=x^2.

I just don't get this.

Am I suppose to just to replace x^2 with y in the first equation, set one of the variables to t and derive the rest using the equations? I'm pretty sure I'm wrong.

Also, in general, how do you find the vector representation of the intersection curve of two surfaces in space?

A vector representation of a curve involves a parameter. (One parameter for one-dimensional objects such a curves and lines. Two parameters for surfaces, which are two-dimensional.) Try to choose a variable among $x$, $y$, $z$ so that the other two variables can be expressed in terms of the chosen one, using the equations of given surfaces. Here $x$ is a good choice, because on the curve of intersection we have $y=x^{2}$ and $z=4x^{2}+y^{2}=4x^{2}+x^{4}$. Hence a vector representation: $\vec r(x)=\langle x, x^{2}, 4x^{2}+x^{4}\rangle$, or $\vec r(x)=x\vec i+x^{2}\vec j+(4x^{2}+x^{4})\vec k$ if you wish. You can rename $x$ as $t$ if this makes you or your teacher feel better.

An important part of such representation of a curve is the range of the parameter. Here $x$ can be any number; $-\infty<x<\infty$, because for each $x$ you get a point on the curve of intersection. This is not always the case. Also, sometimes none of the Cartesian coordinates works as a parameter, but some of those Greek letters do.


If $ x,y>0$ ,$ x+y=1$ prove that:
\[ \sqrt{x}+\sqrt{y}+\sqrt{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\geq 4+\sqrt{2}\]

Let the function $ f(x)=\sqrt{x}+\frac{\sqrt{2}}{\sqrt{x}}$
We have that f is convex . The result follows immediately from Jensen's inequality .

[quote="Svejk"][color=darkred]If $ x,\,y>0,\,x+y=1$ prove that:[/color]
\[ \sqrt{x}+\sqrt{y}+\sqrt{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\geq 4+\sqrt{2}\]
[/quote]
[color=blue][b]Proof.[/b] We have: [/color]\begin{eqnarray*}\sqrt{x}+\sqrt{y}+\sqrt{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)&\ge&\sqrt x+\sqrt y+\frac{4\sqrt 2}{\sqrt x+\sqrt y}\\ &=&\left[\left(\sqrt x+\sqrt y\right)+\frac{2}{\sqrt x+\sqrt y}\right]+\frac{4\sqrt 2-2}{\sqrt x+\sqrt y}\\ &\ge& 2\sqrt 2+\frac{4\sqrt 2-2}{\sqrt{2\left(x+y\right)}}=4+\sqrt 2\end{eqnarray*}


[quote="Beat"]Prove or disprove: $(x+y+z)^3\ge(\sqrt{x}+\sqrt{y}+\sqrt{z})^2(x^2+y^2+z^2)$, where $x,y,z$ are positive reals.[/quote]

Exactly the reversed inequality, $\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\left(x^2+y^2+z^2\right)\geq\left(x+y+z\right)^3$ holds, since Hölder yields

$\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(x^2+y^2+z^2\right)$
$\geq\left(\sqrt[3]{\sqrt{x}\cdot\sqrt{x}\cdot x^2}+\sqrt[3]{\sqrt{y}\cdot\sqrt{y}\cdot y^2}+\sqrt[3]{\sqrt{z}\cdot\sqrt{z}\cdot z^2}\right)^3=\left(x+y+z\right)^3$.

  Darij

Darij :( i don`t know how to prove Holder`s inequality  :(  Please help me

Hölder's inequality follows from the generalized Hölder inequality:

If n and k are two positive integers, if $a_{i,j}$ are nonnegative reals for any integers i and j such that $1\leq i\leq k$ and $1\leq j\leq n$, if $p_1$, $p_2$, ..., $p_k$ are positive reals, and if $s=\frac{1}{\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_k}}$, then

$\left(a_{1,1}^{p_1}+a_{1,2}^{p_1}+...+a_{1,n}^{p_1}\right)^{\frac{1}{p_1}}\left(a_{2,1}^{p_2}+a_{2,2}^{p_2}+...+a_{2,n}^{p_2}\right)^{\frac{1}{p_2}}...\left(a_{k,1}^{p_k}+a_{k,2}^{p_k}+...+a_{k,n}^{p_k}\right)^{\frac{1}{p_k}}$
$\geq\left(\left(a_{1,1}a_{2,1}...a_{k,1}\right)^s+\left(a_{1,2}a_{2,2}...a_{k,2}\right)^s+...+\left(a_{1,n}a_{2,n}...a_{k,n}\right)^s\right)^{\frac{1}{s}}$.

If $p_1=p_2=...=p_k=1$, then this inequality becomes the Hölder inequality.

Proofs of the generalized Hölder inequality can be found in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=111563#p111563]http://www.mathlinks.ro/Forum/viewtopic.php?t=3574 post #9[/url] and in any good book about inequalities.

  Darij


Assume that $a$, $b$, $c$ are positive reals satisfying $a+b+c \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.

Prove that

$\frac{a^{3}c}{b(a+c)}+\frac{b^{3}a}{c(a+b)}+\frac{c^{3}b}{a(b+c)}\geq \frac{3}{2}$.

Indeed!  :coolspeak:

(Zhaobin, you had a thread especially for you :P : http://www.mathlinks.ro/Forum/viewtopic.php?t=82525 )

[quote="zhaobin"]first use holder or the general of cauchy we have:
$\left(\frac{a^3c}{b(a+c)}+\frac{b^3a}{c(a+b)}+\frac{c^3b}{a(b+c)}\right)(2a+2b+2c)(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}) \ge (a+b+c)^3$
so:
$\frac{a^3c}{b(a+c)}+\frac{b^3a}{c(a+b)}+\frac{c^3b}{a(b+c)} \ge \frac{a+b+c} 2$
but we also have:$a+b+c \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3$
so the proof is over :)[/quote]

My first seeing for use holder in this case

Very beautiful ! Thanks


function
calculus
derivative
algebra
domain
algebra unsolved

While solving a problem I came across the expression below.  I need to solve for Q, however, I'm not sure if that's possible.  If it is possible to solve for Q, I'm not sure how to do it.

$ e^{ \minus{} yQ}(Q \minus{} n \minus{} \frac {1}{y}) \equal{} \frac { \minus{} a}{yb}$

Q, n, y, a, b are all positive
n greater than Q

hello, has your equation the following form:
$ e^{\minus{}yQ}(Q \minus{} n \minus{} \frac{1}{y}) \equal{} \frac{\minus{}a}{yb}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, has your equation the following form:
$ e^{ \minus{} yQ}(Q \minus{} n \minus{} \frac {1}{y}) \equal{} \frac { \minus{} a}{yb}$?
Sonnhard.[/quote]

Yes, thank you, that is exactly what I wanted.  I updated my original post now.

hello, solving this you need the LambertW function, your solution is
$ Q\equal{}\frac{ny\plus{}1\minus{}\text{LambertW}\left(\frac{ae^{ny\plus{}1}}{b}\right)}{y}$.
Sonnhard.

hello, for the LambertW function see here
http://mathworld.wolfram.com/LambertW-Function.html
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, solving this you need the LambertW function, your solution is
$ Q \equal{} \frac {ny \plus{} 1 \minus{} \text{LambertW}\left(\frac {ae^{ny \plus{} 1}}{b}\right)}{y}$.
Sonnhard.[/quote]

Ok, thanks a lot!  I never heard of the LambertW function before.

Is there a specific relationship between y and Q?  What I mean is, if you increase y (holding a, b, and n constant) will Q always increase or decrease?  Still assuming a, b, n, y, and Q are all positive.  In other words, is the partial derivative of Q with respect to y strictly positive or negative for the domain I specified (or even a subset of the domain I specified)?


[quote="tenniswimmer"]Let a=1.234, b=2.34, and c=3.4.  Round the sum a+b+c to the nearest integer.

Can someone explain to me how that would equal 160?[/quote]

Uh... it doesn't. I think you should take another look...

[quote="tenniswimmer"]i didnt put that....that was what it said the answer was when i reviewed the questions after a game...i had a completely different answer than that[/quote]

Then this is a glitch. You should have pressed "Report error", did you?

Also, who was/were your opponent(s)? There is a glitch that happens occasionally which is caused by lag that shows a different problem then the actual one, which leads to what is known as "flashies", so you can ask them what problem she/he/they recieved.

yeah i pushed report error

and i dont remember who i was playing at the time


function
search
algebra unsolved
algebra

We have a decreasing function $ f: \mathbb{R}^\plus{} \to \mathbb{R}^\plus{}$, such that $ \forall x>0, y>0$ the follow holds:
$ f(x\plus{}y)\plus{}f(f(x)\plus{}f(y))\equal{}f( f(x\plus{}f(y)) \plus{} f(y\plus{}f(x)) )$
Show that $ f(f(x))\equal{}x$ :wink:

Please don't repost problems posted before   :wink:

See here http://www.mathlinks.ro/viewtopic.php?search_id=257791748&t=137765  :P

i'm sorry, i didn't know the source and that it was posted before.. :lol:
Cheers Miranda


heres a place for Chuck Norris jokes like:

Chuck Norris counted to infinity....... twice ......
Chuck Norris doesn't worship the gods... they worship him.

Chuck Norris Vs. Bad Guys in a Blendtec.
[youtube]NdD54rG9oQA[/youtube]


When Kyle drives to work at an average speed of 48 miles per hour, he arrives 9 minutes late. When he drives to work at an average speed of 64 miles per hour, he arrives 9 minutes early. At what average speed should Kyle travel in order to arrive at work exactly on time?

[hide]
call the time he wants to get there t.  convert 9 minutes to .15 hrs

48(t+.15) = 64(t-.15) <==> t = 1.2

the total distance we want to travel is thus 48*(1.2+.15) = 64.8, and divde by the time 1.2 to get RIGHT SPEED = 54 mph.

[/hide]