analytic geometry
geometry
geometric transformation
rotation
linear algebra
matrix
AMC

A particle is located on the coordinate plane at $ (5,0)$. Define a [i]move[/i] for the particle as a counterclockwise rotation of $ \pi/4$ radians about the origin followed by a translation of $ 10$ units in the positive $ x$-direction. Given that the particle's position after $ 150$ moves is $ (p,q)$, find the greatest integer less than or equal to $ |p|\plus{}|q|$.

[hide]
Let the particle's position be represented by a complex number. The transformation takes $ z$ to $ f(z) = az + b$ where $ a = \frac {\sqrt {2}}{2} + i\frac {\sqrt {2}}{2}$ and $ b = 10$. We let $ a_0 = 5$ and $ a_{n + 1} = f(a_n)$ so that we want to find $ a_{150}$.

Note that we have
\[ a_{n + 1} + \frac {b}{a - 1} = a a_n + b + \frac {b}{a - 1} = a \left(a_n + \frac {b}{a - 1}\right)
\]
Then
\[ a_{150} = a^{150} \left(a_0 + \frac {b}{a - 1}\right) - \frac {b}{a - 1} = - i \left( a_0 + \frac {b}{a - 1}\right)
\]
Since we want $ |p| + |q|$ we can find $ a_0 + \frac {b}{a - 1}$, which is
\[{ 5 + \frac {10}{\sqrt {2}}{2} - 1 + i\frac {\sqrt {2}}{2}} = 5 + \frac {5\sqrt {2} - 10 - 5i\sqrt {2}}{2 - \sqrt {2}} = 5 + (5\sqrt {2} - 10 - 5i\sqrt {2})(2 + \sqrt {2}) = - 5 - 10i\sqrt {2} - 10i
\]
We then have
\[ |p| + |q| = 5 + 10\sqrt {2} + 10 \approx 29.1
\]
so the answer is $ \boxed{029}$.

[/hide]

P.S. The computation was kind of ugly, and a few people messed up on this one.

Note: My above answer is incorrect, and it's actually 10 less. There's a computational error somewhere.

I think the answer is 029 too... but it's supposed to be 019?

I got $ 9$. Exact answer was $ 10\sqrt {2} \minus{} 5$, but I think I made a mistake and should have gotten $ 10\sqrt {2} \plus{} 5$. Basically, the thing comes out to
\[ (((5\epsilon \plus{} 10)\epsilon \plus{} 10)\epsilon \plus{} 10...) \equal{} 5\epsilon^{150} \plus{} 10 \epsilon^{149} \plus{} 10\epsilon^{149} ... \plus{} 10
\]
where $ \epsilon \equal{} e^{i\pi/4}$. Notice that
\[ 10(\epsilon^{150} \plus{} \ldots \plus{} 1) \equal{} 10(1 \plus{} \epsilon \plus{} \ldots \plus{} \epsilon^6) \equal{} \minus{} 10(\epsilon^7) \equal{} \minus{} 10( \minus{} \sqrt {2}/2 \minus{} i\sqrt {2}/2)
\]
Furthermore, $ 5\epsilon^{150} \equal{} \minus{} 5i$. So our final answer is
\[ 5\sqrt {2} \plus{} 5(\sqrt {2} \plus{} 1)
\]
So $ 19$.

Hmm, I got 029 also. I guess we'll know when official answers come out.

i did it wrong on the test i sorta assumed it was pi/2 instead of pi/4 lol.

but i think it turns out to be $ 10\sqrt 2 \plus{} 5$  which is about 19.

Grr, I half-a**ed this question, and assumed it was an octagon and periodic over 8 move intervals, thus corresponding to -2 moves. I didn't know if this was right, but I attempted to find an answer anyways, using rotation matrix to move backwards along the octagon twice, got $ (p,q) \equal{} (\minus{}5\sqrt{2}\minus{}10,\minus{}5\plus{}5\sqrt{2})$... misread the answer format and gave up thinking I was nowhere close...

Anyone with a linear algebra solution?

I didn't take the AIME 2, but it sort of looks like every 8 moves it returns to (5,0)
So the 152nd move brings it back to $ (5,0)$
now we reversify
go two inverse moves, and i get 19.  :huh:

oh yeah lol that was much easier than doing it 8 times.

[quote="CircleSquared"]
go two inverse moves, and i get 19.  :huh:[/quote]

It was 19...

pianoforte,

i got a linear algebra solution but i messed it up in the end (said the particle was going 10 units up instead of 10 units down). I am not up to tex ing so i'll explain

It's pretty easy to construct a linear transform T: R --> R that rotates plan 45 degrees...

so we construct a transofm M that replicates action of one move... so we get:
M(x) = Tx + m (where m is the column vector for the translation to the right)

applying M onto itself 150 times, you can show the answer is
$ M^{150}(x) \equal{} T^{150}(x) \plus{} T^{149}m \plus{} T^{148}m \plus{} \ldots \plus{} Tm \plus{} m$

well, since T is a rotation, we know it has period $ 8$, so yeah, the whole thing turns into jsut finding a sum of iterations of T applied onto m and then the equivalent transform of rotation point by 45 degrees 150 times... you should get 19.

[hide="My Solution"]Let $ z_{n}$ be the location of the particle on the complex plane after $ n$ moves, so that $ z_{0}\equal{}5$ and $ z_{n\plus{}1}\equal{}z_{n}\text{ cis}(\pi/4)\plus{}10$.  Now our goal is to get rid of the annoying constant $ \plus{}10$ in this recursion.  To do this, we let $ z_{n}\equal{}x_{n}\plus{}k$, substitute, and solve for the value of $ k$ (it will be a complex number) that gets rid of the constant:

$ x_{n\plus{}1}\plus{}k\equal{}(x_{n}\plus{}k)\text{ cis}(\pi/4)\plus{}10$
$ k\equal{}k\text{ cis}(\pi/4)\plus{}10$

After some boring computations that I don't want to type out, we get $ k\equal{}5\plus{}(5\plus{}5\sqrt{2})i$.  Then $ x_{0}\equal{}\minus{}(5\plus{}5\sqrt{2})i$, and

$ x_{150}\equal{}x_{0}(\text{cis}(\pi/4))^{150}\equal{}x_{0}\text{ cis}(3\pi/4)\equal{}x_{0}(\minus{}i)\equal{}\minus{}5\minus{}5\sqrt{2}$.

Then $ z_{150}\equal{}x_{150}\plus{}k\equal{}\minus{}5\minus{}5\sqrt{2}\plus{}5\plus{}(5\plus{}5\sqrt{2})i\equal{}\minus{}5\sqrt{2}\plus{}(5\plus{}5\sqrt{2})i$, so $ |p|\plus{}|q|\equal{}5\plus{}10\sqrt{2}$, which is a little greater than $ \boxed{19}$.[/hide]

[quote="pianoforte"]Anyway to prove that the path is following an octagon without actually calculating the points? +10 seemed pretty just random (correct me if I'm wrong on this).[/quote]

Take a look at my solution - I approached the problem algebraically but basically I shifted all the points by a constant complex number and got a new sequence of points that formed an octagon centered at the origin.  Therefore the original sequence of points was also an octagon. :)

If you realize that you can do this you can just do the inverse of the process twice and avoid all the ugly computations - man I wish I noticed this during the test.

[quote="matt276eagles"]
Take a look at my solution[/quote]

Ah cool  :) . I assumed that was the case and got the right coordinate by going backwards, but misread the answer format and gave up...

The first clue that tells me that it's gonna be a 8-cycle is that 5 is half of 10, which leads me to conjecture that the origin is on the midpoint of a side.  Then I drew a picture and it immediately became clear to me:  basically, everytime you rotate it by 45 degrees and translate it, you're tracing a 135 angle, and after 8 transformations a complete octagon is traced (actually, the exact word is not "traced"; the octagon was rotating with the point; in other words, it has the midpoint of one of its side on the origin and it rotates 45 counterclockwise everytime a transformation is made)  it's quite apparent that the point returns to (5,0).

You should keep in mind that the composition of any number of rotations and translations is always again a rotation or a translation. The total rotation angle is just the sum of the individual rotation angles. If this total rotation angle is 0 (modulo $ 2\pi$ of course), then the composition is a translation. Otherwise it is a rotation about the unique fixed point of the transformation.

Here, if $ P$ is the fixed point of the move, then each move just rotates about $ P$ by $ \pi/4$ counterclockwise. Since $ 150$ is $ \minus{}2$ modulo $ 8$, this means that after $ 150$ moves, the particle has just rotated $ \pi/2$ clockwise about $ P$.

To find $ P$, let $ P'$ be the image of $ P$ after rotating about the origin $ O$ by $ \pi/4$ counterclockwise. Then $ OP\equal{}OP'$, $ \angle POP'\equal{}\pi/4$, and $ P'P$ has length $ 10$ and points in the positive $ x$-direction. Clearly the positive $ y$-axis bisects this isosceles triangle, so the $ x$-coordinate of $ P$ is $ 5$. To find the $ y$-coordinate, you can either use a bit of trig or just note that if $ Q\equal{}(5,5)$, then two of the angles of $ OPQ$ are equal to $ \pi/8$, so $ PQ\equal{}OQ\equal{}5\sqrt{2}$, giving $ P\equal{}(5,5\plus{}5\sqrt{2})$.

Conveniently, $ P$ lies $ 5\plus{}5\sqrt{2}$ units directly above the particle's starting point at $ (5,0)$, so the particle ends at a point $ 5\plus{}5\sqrt{2}$ directly to the left of $ P$, which is $ (\minus{}5\sqrt{2}, 5\plus{}5\sqrt{2})$. The answer is then $ \lfloor 5\plus{}10\sqrt{2} \rfloor\equal{}\boxed{019}$.

I got a wrong understanding and a right solution.
So lucky!

[quote="Zenith Xan"][hide="Considerably messier solution without any complex numbers"]

Let $ x_n$ and $ y_n$ denote the x and y coordinates after the $ n$th rotation and shift by $ 10$ in the positive $ x$ direction, respectively.

By rotation formulas, we have:

$ x_n \equal{} x_{n \minus{} 1}\cos(\frac {\pi}{4}) \minus{} y_{n \minus{} 1}\sin(\frac {\pi}{4}) \plus{} 10 \equal{} \frac {\sqrt {2}}{2}(x_{n \minus{} 1} \minus{} y_{n \minus{} 1}) \plus{} 10$
$ y_n \equal{} x_{n \minus{} 1}\sin(\frac {\pi}{4}) \plus{} y_{n \minus{} 1}\cos(\frac {\pi}{4}) \equal{} \frac {\sqrt {2}}{2}(x_{n \minus{} 1} \plus{} y_{n \minus{} 1})$

Then, adding and subtracting these together produces:

$ x_n \plus{} y_n \equal{} \sqrt2(x_{n \minus{} 1}) \plus{} 10 \Rightarrow x_{n \plus{} 1} \plus{} y_{n \plus{} 1} \equal{} \sqrt2(\frac {\sqrt {2}}{2}(x_{n \minus{} 1} \minus{} y_{n \minus{} 1}) \plus{} 10) \plus{} 10$
$ x_n \minus{} y_n \equal{} \minus{} \sqrt {2}(y_{n \minus{} 1}) \plus{} 10 \Rightarrow x_{n \plus{} 1} \minus{} y_{n \plus{} 1} \equal{} \minus{} \sqrt {2}(\frac {\sqrt {2}}{2}(x_{n \minus{} 1} \plus{} y_{n \minus{} 1}) \plus{} 10$

Cleaning up, we're left with:

$ x_{n \plus{} 1} \plus{} y_{n \plus{} 1} \equal{} x_{n \minus{} 1} \minus{} y_{n \minus{} 1} \plus{} (\sqrt2 \plus{} 1)(10)$
$ x_{n \plus{} 1} \minus{} y_{n \plus{} 1} \equal{} \minus{} (x_{n \minus{} 1} \plus{} y_{n \minus{} 1}) \plus{} 10$

Our initial conditions are that $ x_0 \equal{} 5$ and $ y_0 \equal{} 0$

Now, we perform our recursion a few times:

$ x_0 \plus{} y_0 \equal{} 5$
$ x_0 \minus{} y_0 \equal{} 5$

$ x_2 \plus{} y_2 \equal{} 5 \plus{} (\sqrt2 \plus{} 1)(10)$
$ x_2 \minus{} y_2 \equal{} 5$

$ x_4 \plus{} y_4 \equal{} 5 \plus{} (\sqrt2 \plus{} 1)(10)$
$ x_4 \minus{} y_4 \equal{} 5 \minus{} (\sqrt2 \minus{} 1)(10)$

$ x_6 \plus{} y_6 \equal{} 5$
$ x_6 \minus{} y_6 \equal{} 5 \minus{} (\sqrt2 \plus{} 1)(10)$

$ x_8 \plus{} y_8 \equal{} 5$
$ x_8 \plus{} y_8 \equal{} 5$

We see it repeats every 8, and $ 150 \equiv 6 \pmod 8$, so:

$ |x_{150}| \plus{} |y_{150}| \equal{} |x_6| \plus{} |y_6| \equal{} |5 \minus{} (\sqrt2 \plus{} 1)(5)| \plus{} |5(\sqrt2 \plus{} 1)| \equal{} 10\sqrt2 \plus{} 5$, which is less than 20 and more than $ \boxed{19}$[/hide][/quote]

This is pretty much identical to the complex numbers solution. Complex numbers simply offer a more efficient way of encoding these transformations.

We may use the fact that the [b]rotation of a sum is equivalent to the sum of the individual rotations[/b] [i](which is true because rotation is a [url=http://en.wikipedia.org/wiki/Linear_transformation]linear map[/url])[/i].

_____________________________________________________________________________________
[hide="Very low computation"]So, let's not do any addition until the very end and write the points in polar. Each move, we add 45 degrees to the existing angles and then add the vector (10,0)

We start with (5,0)
1: (5,45)+(10,0)
2: (5,90)+(10,45)+(10,0)
3: (5,135)+(10,90)+(10,45)+(10,0)
...
8: (5,360)+(10,315)+(10,270)+...+(10,45)+(10,0) = (5,0), since pairs vectors of the form (10,a)+(10,180+a) sum to zero

So, it is clear that every 8, we return back to where we started.

Now, 150 is 6 in mod8, and we have:
(5,270)+(10,225)+(10,180)+(10,135)+(10,90)+(10,45)+(10,0)

We see that several of these cancel, and we have (5,270)+(10,90)+(10,135)

Lastly and rather trivially, we convert to rectangular, giving $ (0, \minus{} 5) \plus{} (0,10) \plus{} ( \minus{} 5 \sqrt {2}, 5 \sqrt {2}) \equal{} ( \minus{} 5 \sqrt {2},5 \plus{} 5\sqrt {2}) \implies \boxed{19}$
_____________________________________________________________________________________________________
[/hide]

When I did this, I used a long arrow to represent (10,k) and a short arrow to represent (5,k) in polar, and wrote the sums in terms of the pictures, so it was really easy to see when things canceled.

The idea here is basically the same thing that is implied by the usage of complex numbers, but more direct, and also less intimidating to many. Actually, shake(some number) expressed this approach in terms of linear algebra -- his usage of the transformation T and the column vector m is exactly the same as what is done here, except expressed in a more sophisticated way.  :wink:

[hide="the second I saw the words rotation and translation, I thought VECTORS"]
though this solution doesn't really use them..

draw this out and it's not so confusing:

I thought of this as a segment formed by the origin and the "particle" rotating about the origin in increments of $ \frac {\pi}{4}$, excluding the translation part. Let's call this particle without translation point $ A$. I then imagined a regular octagon with side length $ 10$ attached (sorry that's not very mathematical) to point $ A$. The particle would occupy one of the octagon's vertices at each phase of point $ A$'s rotation around the origin

Before any of the moves (when $ A$ is at $ (5, 0)$), the octagon would occupy quadrants $ III$ and $ IV$, with one side on the x axis and the midpoint of that side being the origin.

$ 150 \equal{} 6 \mbox{ (mod 8)}$

so after the 150th move, we have $ A$ at $ (0, \minus{} 5)$. The actual particle is located on the 6th vertice of the octagon (counting from $ A$ as $ 0\mbox{th}$, clockwise), so after using a bit of simple algebra and good ol' pythag, we find that the particle's coordinates are $ ( \minus{} 5\sqrt {2}, 5 \plus{} 5\sqrt {2})$.

$ | \minus{} 5\sqrt {2}| \plus{} |5 \plus{} 5\sqrt {2}| \equal{} 5 \plus{} 10\sqrt {2} \approx \boxed{019}$.
[/hide]

EDIT:

[quote="The Zuton Force"]We may use the fact that the [b]rotation of a sum is equivalent to the sum of the individual rotations[/b] ...[/quote]

I guess I kinda used the same principle, except I never knew it in those words :lol:


MIT
college
geometry
algorithm
geometric transformation
reflection
algebra

I read a bit about it but I was wondering if anyone have more information (preferably a personal opinion).

Thanks.

You are thrown onto the MIT campus, set up with a professor/mentor, given a cutting edge research project, and let loose. It's great fun.

[quote="t0rajir0u"]Could you be more specific about what you want to know?  Qualifications you think you might need?  A general idea of the experience?[/quote]

That would be good thanks. :)

[b]Qualifications:[/b]

RSI has something like a 5% admission rate, which is half that of MIT's undergraduate admission rate.  The admissions process is similar.  Scores are important (usually PSAT); science-related extracurriculars are important (AMC/AIME, science fairs, previous research) but not a requirement.  The actual application includes a section where you discuss 3-5 unsolved problems in your desired area of research most important to you, and I suspect that the level of expertise implied in your response is important as well.

[b]Experience:[/b]

RSI is an amazing experience.  First there's the academic part, which is satisfying as much because of the quality of your fellow students as the quality of your mentors, lecturers, and tutor.  It's exhilarating.  It took me five weeks to make any progress on my research, but when I did... I can't describe how good that felt.  Then there's the social part, which you'll likely spend more time on.  Your fellow Rickoids (that's what we call ourselves) are some of the most brilliant and talented kids in the country, and yet they still know how to have fun.  You'll become close quickly.

There are a [i]lot[/i] of planned activities (which I didn't actually expect going in).  You'll find that you'll constantly have to decide between spending your time on research and spending it exploring Boston (which is, in my opinion, the best city in the country) with your new friends: eating, shopping, watching the fireworks on the 4th of July.  Then there's frisbee.  Then there's [i]chairing[/i], which I believe I am forbidden to tell you about because I am not sure we're supposed to be doing it.  :)  You'll enjoy not (really) having a curfew.

My major problem is probably the admission process (which is why I am trying to participate in as many extracurricular competitions/Olympiad as possible).

P.S. What kind of things did you guys write about?

Well, the application requires a discussion of two fields, and I only discussed number theory / mathematics.  The questions I discussed:

- Completion of the Langlands program
- The Generalized Riemann Hypothesis
- The Collatz Conjecture
- The Bouniakowsky Conjecture

[quote="t0rajir0u"]Then there's [i]chairing[/i], which I believe I am forbidden to tell you about because I am not sure we're supposed to be doing it.  :)[/quote]

Lol, good times.

(I feel like all I do now is spam.)

[quote="Elemennop"][quote="t0rajir0u"]You'll enjoy not (really) having a curfew.[/quote]

people sitll have curfews once they get to high schooll?

wtf
?[/quote]

I meant compared to other summer programs.  After bedcheck, the counselors don't really care what you do as long as you don't wander out into Boston and get stabbed :)

[quote="K81o7"][quote="t0rajir0u"]
- Completion of the Langlands program
- The Generalized Riemann Hypothesis
- The Collatz Conjecture
- The Bouniakowsky Conjecture[/quote]

Hmm...some of those are very very nontrivial and difficult...were those what you actually worked on?[/quote]

Oh, no, no, no.  I just tried to give an idea of the kind of things that interested me.  Of course, my mentor recognized that none of those were feasible research problems, so I actually worked on elliptic curves.  :)

[quote="t0rajir0u"]...Boston (which is, in my opinion, the best city in the country) with your new friends: eating, shopping, watching the fireworks on the 4th of July.  Then there's frisbee.  Then there's [i]chairing[/i], which I believe I am forbidden to tell you about because I am not sure we're supposed to be doing it.  :)  You'll enjoy not (really) having a curfew.[/quote]

I'll second that comment about Boston.  It's hard to ever find yourself bored.

As for chairing, of all the activities which you could be doing at MIT that you're not 'supposed to be doing', I think it's the one that'll get you in the least amount of trouble =p

Btw, may I ask who your mentor was?

[quote="BanishedTraitor"]How in depth did you decscribe them?[/quote]

I don't really know enough about the GRH or the Langlands Program to describe them in any depth.  I suppose the tack I chose was somewhat unusual:  I described largely why I thought they were fascinating problems, not how much I knew about progress that had been made on them.  In fields other than mathematics, you can get a lot more in depth, but I chose some pretty difficult problems to talk about.

[quote="joml88"]Btw, may I ask who your mentor was?[/quote]

Ryan Reich (Harvard).  I owe a great debt to him :)

The last modification I made to my application was on February 6th.  I don't remember if admissions are rolling or not, but I think that information is available on the website.

[quote="Twiz"]You got your letter today, Miles? Me and another person from Georgia did not get letters yet. And I'm told they send acceptance and rejection letters at the same time. This isn't fair!
Congratulations though. I'm not all that surprised that you got in. :)[/quote]
Thanks.  I think the postmark said April 25, though, so I guess I just got fast mail.  Good luck!

Hello, abgs.  Who are you?

[quote="Boy Soprano II"]
I think the postmark said April 25, though, so I guess I just got fast mail.[/quote]

Wow. That [i]is[/i] fast  :wink:

which topics did you guys say you wanted to research?

(congrats!)

I got my acceptance letter yesterday.  I'm from Alaska.  My first and second fields were Number Theory and Combinatorics respectively.

Congratulations to all of you.  :)  The counselors for RSI 2008 are all fantastic people, so make sure you appreciate them and get to know them as well as your own fellow Rickoids!

Ugh. After reading this forum I ran outside to check my mailbox, alas the mailman has not visited my house yet today, argh.

oh well. don't think I'll make it anyways, I know 4 people that applied from my state--all of them much more academically qualified than I.

Accepted, Fields/Subfields were Math/NT and Comp. Sci/Algorithms

Congrats to those that got in!

Join the Facebook group Research Science Institute 2008, if you got accepted

Uh scratch that, join the RSI 2008 group instead.

I was rejected, but congrats to everyone who did make it!


induction
inequalities
integration
inequalities unsolved

Prove that for every $k$ natural positive number the following ineq is true

$\displaystyle 4k^2(\frac {1}{\sqrt {k}}-\frac {1}{\sqrt {k+1}})\geq (\sum_{1}^{k} (\frac {1}{\sqrt {i}}) $

I wait for your answer trembling :)

I have tried by induction, by majorization and by calculus.

But absolutely nothing !!!!!!!!!!!!!!!

There is no problem.

\[\sum_{i=1}^{k} \frac{1}{\sqrt{i}}=1+\sum_{i=2}^{k} \frac{1}{\sqrt{i}}< 1+\int_1^{k}\frac{dx}{\sqrt{x}}=2\sqrt{k}-1.\]
So you just need to show $2\sqrt{k}-1<4k^2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)$. :)

We have the obvious
$\sqrt x+\sqrt{x+1}<\sqrt{4x+2}$(1).
From here $2(\sqrt{i+\frac 12}-\sqrt{i-\frac 12})=\frac{2}{\sqrt{i+\frac 12}+\sqrt{i-\frac 12}}>\frac{1}{\sqrt i}$.
Summing we get $RHS<\sqrt 2(\sqrt{2k+1}-1)$.
We are left to prove this is smaller than LHS or
$\frac{2\sqrt 2k\sqrt{k}(\sqrt{k+1}-\sqrt{k})}{\sqrt k+\sqrt{k+1}}>\sqrt{2k+1}-1$ or ${2\sqrt 2 k\sqrt k >(\sqrt{k+1}+\sqrt{k})^2(\sqrt{2k+1}-1)}$.
Now using (1) again it suffices to prove
$2k\sqrt k>2(2k+1)\sqrt{2k+1}-2(2k+1)$ which is true.
Did  make any mistake?


calculus
integration
function
calculus computations

I'm working on a physics problem, and i got stuck on an integral.  the entire question is as follows:

----------------------------
Magnetostatics treats the "source current" (the one that sets up the field) and
the "recipient current" (the one that experiences the force) so asymmetrically that it is by no means obvious that the magnetic force between two current loops is consistent with Newton"s third law. Show, starting with the Biot-Savart law and the Lorentz force law, that the force on loop 2 due to loop 1 can be written as

$F2 = -\frac{\mu_o}{4\pi}I_1 I_2 \oint_1 \oint_2 \frac{\widehat{r}}{r^2} d\overrightarrow{l_1} \cdot d\overrightarrow{l_2}$

In this form it is clear that F2 = — F1, since i changes direction when the roles of 1 and 2 are interchanged. (If you seem to be getting an "extra" term, it will help to note that $d\overrightarrow{l_2} \cdot \widehat{r} = dr$.)

http://www.angelfire.com/planet/buzzard/5.62.jpeg
--------------------------------------

I'm okay with most of the proof - just a specific part I'm stuck on.  i got to the part where i have
$F2 = \frac{\mu_o}{4\pi}I_1 I_2 [\oint_1 \oint_2 \frac{d\overrightarrow{l_1} dr}{r^2} - \oint_1 \oint_2 \frac{\widehat{r}}{r^2} d\overrightarrow{l_1} \cdot d\overrightarrow{l_2}]$

i want to prove that the first integral is zero, so i need to show that this integral along a closed loop:

$\oint_2 (1/r^2) dr$

is equal to zero.  but i'm not sure how to prove it.  i was wondering if someone can show me a rigid proof for this.  Note that the integral is along loop 2.  Note also that the $\overrightarrow{r}$ is the distance from a point on loop 1 to a point on loop 2 - that is, $\overrightarrow {r} = (x2-x1, y2-y1, z2-z1)$, so in the integral above, where r is a scalar, then $r = [(x2-x1, y2-y1, z2-z1)]^{1/2}$.  i'm not sure how to rewrite dr in terms of x2, since r is a scaler.  should it be dr = dxdydz?  I think i'm missing something here because I'm not really that familiar with loop integrals.[img]

Does the function have an antiderivative/potential? If so, we get what we want immediately.

That assumes some things you haven't stated, but we need some for this to make sense at all.

my question is editted above to include more information


geometry
perpendicular bisector
geometry proposed

A point $ M$ is taken on the perpendicular bisector of the side $ AC$ of an acute-angled triangle $ ABC$ so that $ M$ and $ B$ are on the same side of $ AC$. If $ \angle BAC\equal{}\angle MCB$ and $ \angle ABC\plus{}\angle MBC\equal{}180^{\circ}$, find $ \angle BAC.$

It's easy to see that $ \angle A\equal{}30^o$ satifies the supposion. Assuming there exist $ \angle B'AC\neq 30^o$ and satify the supposion too.
I only solve this problem when $ \angle B'AC<30^o$ (*)(another case is similar)
Assuming that a point M' satifies the supposion.
We have $ \angle MB'B<\angle MBz\equal{}\angle ABC<\angle AB'C$ then $ 90^o\minus{}\angle MB'B>90^o\minus{}\angle AB'C$
$ \Rightarrow \angle MB'y>\angle AB'y\equal{}\angle M'B'y$
Therefore B'B' is between B'M and B'y which follow M' is between M and E.
$ \Rightarrow \angle B'AC\equal{}\angle M'CB'>\angle MCB\equal{}\angle BAC$ which is contrary with $ (*)$
So this problem only has one solution: $ \angle A\equal{}30^o$


En un examen de años anteriores de mi colegio encontré este ejercicio que no supe como hacer  :blush:

Estudiar la convergencia de la serie de término general: $e-(1+1/n)^n$

Si mal no estoy es $0$.

Para probarlo veamos lo siguiente:

sea $f(x)=ln(x)$ entonces $f'(1)=1$ pero $e^1=e^{f'(1)}=e^{\lim_{h\rightarrow 0} \frac{ln(1+h)-ln(1)}{h}}$

y $e=\lim_{h\rightarrow 0} e^{\frac{ln(1+h)-ln(1)}{h}}=\lim_{h\rightarrow 0} (1+h)^h= \lim_{n\rightarrow \infty} (1+1/n)^n$

De donde se sigue lo deseado.

En realidad no me expliqué bien a lo que se refería el problema
la serie de término general $e-(1+1/n)^n$ corresponde a la sumatoria desde i=1 hasta i=inf de $(e-(1+1/n)^n)$ con n natural... no lo expliqué mejor porque no sé usar bien todavía el Latex  :oops: Es obviamente condición necesaria que el término general tienda a cero para que la serie (o sea, toda la suma) converja... pero no es condición suficiente.
Gracias Pascual  ;) igual no es muy importante...

arriba al inicio del foro hay un icono de Latex, alli puedes encontrar un Tutorial

Respecto al problema, esta noche posteo un intento de solucion....

Me voy a "autoquemar" el ejercicio para que no se preocupen en pensar una solución (ya que no es muy divertido) y además así practico con el Latex

lo que queremos hallar es $\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^{\infty}\(e-(1+1/n)^n$

lo que hacemos es escribir $e(1-\displaystyle\frac{(1+1/n)^n}{e})=e(1-\displaystyle\frac{(1+1/n)^n}{(e^\frac{1}{n})^n})$

Ahora hacemos un poco de trampa y cambiamos por el equivalente (Taylor):
$e^\frac{1}{n}$~$1+\frac{1}{n}+\frac{1}{2n^2}$ hacemos las cuentas y nos queda $e(1-\displaystyle\frac{(2n^2+2n)^n}{2n^2+2n+1)^n})=e(1-e^{nL(1-\frac{-1}{2n^2+2n+1})\))}$

Y bueno, cambiamos el logaritmo por su equivalente:
$L(1-\frac{-1}{2n^2+2n+1})$~$\frac{-1}{2n^2+2n+1}$ entonces nos queda como término principal $e(1-e^\frac{1}{2n})$~$e.\frac{1}{2n}$ y eso es bastante fácil ver que diverge (sacás el $\frac{e}{2}$ de factor común y te queda $\frac{1}{n}$ y podés formar infinitos términos de esa suma mayores o iguales que $\frac{1}{2}$ agrupándolos de a $2^0, 2^1.... 2^n$)

Y ya me aburrí de escribir... pero queda como ejercicio para el lector ver que la serie $\displaystyle\sum_{i=1}^{\infty}\((-1)^n(e-(1+1/n)^n)$ converge... en realidad no son problemas de ingenio ni nada por el estilo... pero así se supone que deben ser los ejercicios del colegio....

Por lo menos ahora sé usar un poquito más el latex  ;)


Here are "formation" problems.
We'll start with a warmup question.
[quote="Question 1"]There are 169 stones forming a square. How many stones are there on the outer most ring? How about if there was 225 stones forming a square? Can you generalize this?[/quote]
Now for squares with holes in the middle (looks like a square donut with a square hole)
[quote="Question 2"]There are soldiers in a "protective" formation where 3 rows of 10 soldiers are formed on all sides of the person to be protected, making a square with a square hole. How many soldiers are there?[/quote]
Note that there are [b]three[/b] ways to solve this problem. Can you generalize these three ways to solve if x was the number of soldiers on each side and y was the number of rows?

Now for a harder one...
[quote="Question "]A commander is trying to form a protective formation (which is square and has a square hole in the middle) with 4 rows on each side. He knows that 80 soldiers are required to make the outer most ring. How many soldiers is he going to need in total to form this formation? Can you generalize if x was the number of soldiers required to make the outer most ring and y was the number of rows?[/quote]
Drawing a diagram and double-checking the solution is strongly recommended.
Good luck!

[hide] 13x13, 48. 15x15, 56, mxm, 4m-4.

for the second one, if you look at the center column, hold on, drawing a diagram.

for the last one, 4m-4=80, m=21, so 21^2-1= 440 soldiers are required. [/hide]

You got the first one right, though you didn't need to say [hide]1313[/hide]since the problem only asked for how many stones are in the outer ring.

As for the third one, reread the question, please. Notice that [hide]it says "'protective' formation...square hole in the middle...4 rows on each side" You got how many people required to get the side of the formation right, though.[/hide]

Keep trying...

[hide]

For the second one, if you think about it, there is a 10 by 10 square with a hole in the middle. Note that there are three rows on each side of the person, therefore 6 people. 10-6 = 4.

100-16 = 84

Third one

4(x-1) = 80

4x = 84

x= 21

21:^2:- (21-8):^2: = 21:^2:-13:^2: = (21+13)(21-13) = (34)(8) = 272

If there are an x number of people on the outer most ring,

4(y-1) = x

y = (x+4)/4

[/hide]

i dont get what the 2nd two questions are asking?  anyone from this year have any ideas  :lol:


[quote="marko avila"]hola tengo un nuevo libro se llama 'elementary number theory and its applications' de kenneth h. rosen. bueno lo que quiero saber es que temas me son utiles para la olimpiada. a continuación les pondré todos los temas que contiene:
1.the integers
2.integer represantations and operations
3.primes and greatest common divisors
4.congruences
5.applications of congruences
6.some special congruences
7mutiplicative functions
8.cryptology(este no me sirve)
9.primitive roots
10.applications of primitive roots and the order of an integer
11quadratic residues
12.decimal fractions and continued fractions
13.some nonlinear diophantine equations

eso es todo, bueno espero que alguien me pueda indicar cuales son los temas utiles para mi. muchas gracias :D[/quote]Yo honestamente pienso que, de 1 a 7 son como cosas que todo olimpico TIENE que saber, aun cuando no conozco el libro y no se que nivel tiene. Por otro lado, no esta de mas saber todo el resto, yo te recomendaria que si tenes el tiempo, le des una tratada a todo el libro.

EL  libro va de lo más básico hasta lo mas complicado. si estoy de acuerdo todo el libro me sirve. gracias por tu ayuda
 :D


limit
logarithms
calculus
calculus computations

Compute
\[ \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {\frac{2}
{3}} \right)^n }}
{{1 \minus{} \sqrt[n]{n}}}\]

$ \sqrt [n]{n} \equal{} e^{\frac {\ln n}n}\approx 1 \plus{} \frac {\ln n}n,$ so the denominator looks like $ \minus{} \frac {\ln n}n$ for large $ n.$ That goes to zero at a power-of-$ n$ rate and is overwhelmed in size by the geometric numerator. The limit is zero.


[quote="236factorial"]If the graphs of 2y+x+3=0 and 3y+ax+2=0 meet at right angles, the value of a is what number?[/quote]

[hide]Since the lines are perpendicular to each other, their slopes are the negative reciprocal of each other.

Putting the lines into slope-intercept form...

$2y+x+3=0$

$2y=-x-3$

$y=-\frac{1}{2}x-\frac{3}{2}$

Slope is $-\frac{1}{2}$

$3y+ax+2=0$

$3y=-ax-2$

$y=-\frac{a}{3}x-\frac{2}{3}$

Slope is $-\frac{a}{3}$

Now, solve for $a$:

$-\frac{a}{3}=2$

$-a=6$

$a=-6$[/hide]


Simplified result?

The answer [hide]is 2[/hide]


P lies on 8y = 15x, Q lies on 10y = 3x and the midpoint of PQ is (8,6). Find the distance PQ.

I substituted (8u/ 15 , u ) for P and (10t/3 , t) for Q. Can someone please explain to me what is incorrect about my approach?

I used (8u/5, u) and (10t/3, t)

and got  the right answer

hmm... do you think you could show me your steps? I checked my work over multiple times and I don't see any computational mistakes.

[hide="my solution"]since the midpoint of PQ is (8,6)
a) 8u/15+10t/3=16=(2*8)
b) u+t=12=(2*6)

multiply equation a) by 45
we get: 24u+150t=720...........make it equation c

then multiply equation b) by 24 (to get rid of u)
we get: 24u+24t=288............equation d

c-d=> 126t=432
t=24/7
Q(80/7 , 24/7)

vertical distance to the midpoint = 18/7
so vertical distance to point P = 36/7

horizontal distance to the midpoint = 24/7
horizontal distance to point P = 48/7

distance of  PQ =((36/7)^2+(48/7)^2)^(1/2)
= 60/7[/hide]

Thank you so much, I see what I did wrong.


A sphere of diameter of 2 cm is projected from a height of $ x_{0}$ meters above the ground and is thrown with an initial horizontal velocity of $ v_{0}$ m/s. How far from the building will the ball land? Take in account air resistance.

I know that this can be easily done if we do not care about air resistance, but how can we do this when we care about air resistance?

all u need to know is the [url=http://en.wikipedia.org/wiki/Terminal_velocity]terminal velocity[/url] of the ball

Two other things may be helpfull:

1) What is the direction of the drag force?

2) Stokes law states that, for spherical objects of radius $ r$ moving with speed $ v$ in a fluid of viscosity $ \eta$, the frictional force has intensity $ 6\pi\eta r v$. So you need to know the viscosity of air.

I think we cannot do this without approximations.
First we get expression $ v_{x}\equal{}v_{0}e^{\minus{}kt}$ where $ k\equal{}\frac{6\pi\eta r}{m}$
Similarly we get $ g\minus{}kv_{y}\equal{}ge^{\minus{}kt}$, or $ v_{y}\equal{}\frac{g(1\minus{}e^{\minus{}kt})}{k}$
Integrating, we have the equation $ x_{o}\equal{}\frac{gt}{k}\plus{}\frac{ge^{\minus{}kt}}{k^{2}}$
Now we make either of two approximations, 1:$ t<<\frac{1}{k}$ or 2:$ t>>\frac{1}{k}$
In 2: we have simple uniform velocity motion. In 1:, we get $ e^{\minus{}kt}\approx1\minus{}kt$ which means , we get a condition $ x_{o}k^{2}\equal{}g$
What is going wrong in this approach?


When $ n\equal{}2$, we see that there are $ 4$ shaded.  When $ n\equal{}3$, we see there are $ 4\times2\equal{}8$ shaded.  When $ n\equal{}4$, there are $ 4\times3\equal{}12$ shaded.  We see that for any value $ n$ (greater than $ 1$), the number of shaded is $ 4\times(n\minus{}1)$.  Therefore, our answer is $ 4\times(20\minus{}1)\equal{}80\minus{}4\equal{}\boxed{76}$

Note that the border is the whole area minus the white square inside, which has side lengths s-2 (Where s is the side of the square), making the area of the border $ s^2\minus{}(s\minus{}2)^2\equal{}s^2\minus{}(s^2\minus{}4s\plus{}4)\equal{}4s\minus{}4$.


calculus
trigonometry
function
derivative
calculus computations

Ive been stuck on this question for forever!! PLEASE HELP!

8.  You plan to estimate pi/2 to five decimal places by using Newton’s method to solve the equation cos x= 0.  Does it matter what your starting value is?  Give reasons for your answer.

You have to be within the "basin of attraction" of the particular root that you want. If you start just anywhere and try to solve $ \cos x\equal{}0,$ you might wind up with $ \frac{3\pi}2$ or $ \minus{}\frac{5\pi}2$ or some such other root of the equation.

It's also relatively inefficient to start too far away from the point you seek, although in this particular case, if you're close enough to avoid the "basin of attraction" issue, you're probably close enough for reasonable efficiency. Being inefficient means taking more steps than you need to, but it doesn't affect the final accuracy, because the arithmetic is self-correcting.

Here's a sequence of Newton approximations for $ \cos x\equal{}0$ starting at $ x\equal{}1.$ I've got my spreadsheet set to show 8 places after the decimal point.

$ 1.00000000$
$ 1.64209262$
$ 1.57067528$
$ 1.57079633$
$ 1.57079633$
$ 1.57079633$

About those last three numbers, which appear the same to 8 decimal places: for the first one, the cosine is about $ \minus{}5.9\times 10^{\minus{}13},$ while for the second one the cosine is about $ 6.1\times 10^{\minus{}17}.$ So progress was made in that step, even though you can't see it in the first 8 places. But the next line also has a cosine of about $ 6.1\times 10^{\minus{}17}.$ So we can't go any further, not in the particular floating-point system implemented in the spreadsheet I was using.

Note also: the initial guess of $ 1$ wasn't all that good a guess, and we got to your 5-decimal place approximation (it was actually accurate to about 12 decimal places) in just 3 steps. That's a feature of Newton - once you get into reasonable range, it converges fast.

Although usually not that fast. There's a coincidence here of the function in question, $ f(x)\equal{}\cos x,$ having a second derivative of zero exactly at the root in question. In more normal cases, with a nonzero second derivative, the convergence is not as fast as this.

If we try an initial guess of $ x\equal{}.5,$ we still get a 5-place approximation of $ \frac{\pi}2,$ in four steps rather than three.

If we try an initial guess of $ x\equal{}.2,$ we converge fairly quickly to about $ 4.7123,$ or $ \frac{3\pi}2.$ That's the basin of attraction issue. An initial guess of $ .1$ catapults us to a value of about $ 10.9956,$ or $ \frac{7\pi}2.$


How can you use blackboard type for well-known sets like the real numbers $ R$, the complex numbers $ C$, the rationals $ Q$, etc.? I can do $ \Re$, but that isn't really what I'm after.

\mathbb{R},\mathbb{C},\mathbb{Q},\mathbb{Z},\mathbb{N}

$ \mathbb{R},\mathbb{C},\mathbb{Q},\mathbb{Z},\mathbb{N}$

And many other letters, not that I know any conventional meanings for $ \mathbb{D}$ or $ \mathbb{X}.$

Just be sure you close off the bracket before adding a superscript or subscript.

\mathbb{R}^n becomes $ \mathbb{R}^n,$ which is probably exactly what I intended.

\mathbb{R^n} becomes $ \mathbb{R^n},$ which is probably not what I meant.

What package do you need for texcenter to recognize \mathbb? Mine isn't recognizing it.

Since stevem hasn't answered you immediately, I'll say what I think he would say: you should always, always, have \usepackage{amsmath} and \usepackage{amssymb} in every document you make. This is in one of those.

[quote="Kent Merryfield"]not that I know any conventional meanings for $ \mathbb{D}$ or $ \mathbb{X}.$[/quote]
$ \mathbb{D}$ is sometimes used to stand for the open disc in the complex plane.  Similarly $ \mathbb{H}$ is sometimes used to stand for the upper half-plane.


Does there exist a real number bigger than $1$ like $x$ such that $x\notin Z$ and:
$\left\{x\right\}+\left\{x^{3}\right\}+\left\{x^{5}\right\}+\cdots+\left\{x^{99}\right\}<\frac{1}{2^{99}}$
(where $\left\{t\right\}=t-\left\lfloor t\right\rfloor$ for every real number $t$)?


IMC 2005 Longlist is in attached pdf file.

Thank you for this Longlist, may be you have IMC 2004 or 2006 longlist :maybe:

[quote="Zubr"]may be you have IMC 2004 or 2006 longlist :maybe:[/quote]
2006 one is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160249]here[/url]. Unfortunately I do not have 2004 one. :(


they go from pretty easy to fairly hard.

1.  Find the coefficient of $ a^{3}b^{5}$ when expanding $ (a+b)^{8}$

2.  Find the coefficient of $ a^{2}b^{3}c^{4}$ when expanding $ (a+b+c)^{9}$

3.  Find the coefficient of $ x^{4}$ when expanding $ (1+x^{2})^{4}$

4.  Find the coefficient of $ x^{6}$ when expanding $ (1+x^{2}+x^{3})^{5}$

5.  Find the coefficient tof $ x^{7}$ when expanding $ (1+2x+3x^{3})^{10}$

6.  Find the coefficient of $ a^{3}b^{2}c^{2}$ when expanding $ (a+2b+3c)^{7}$

[hide]1. Using Binomial Theorem, the coefficient of $ a^{3}b^{5}$ is $ \binom{8}{5}=56$.
2. Calculating the number of times $ a^{2}b^{3}c^{4}$ appears in the expansion, we have $ \frac{9!}{2!3!4!}=1260$.
3. Using Binomial Theorem again, the coefficient of $ x^{4}$ is $ \binom{4}{2}=6$.
4. Writing out the expansion, $ (1+x^{2}+x^{3})(1+x^{2}+x^{3})(1+x^{2}+x^{3})(1+x^{2}+x^{3})(1+x^{2}+x^{3})$, $ x^{6}$ appears $ \frac{5!}{2!3!}+\frac{5!}{3!2!}=20$ times.
5. There are three cases for this one. Disregarding coefficients for now, we have 1) $ 1^{3}(xxxxxxx)$, 2) $ 1^{5}(xxxxx^{3})$, and 3) $ 1^{7}(xx^{3}x^{3})$. Adding up the number of times each appears, we have $ \frac{2^{7}*10!}{3!7!}+\frac{2^{4}*3*10!}{5!4!}+\frac{2*3^{2}*10!}{7!2!}=83320$
6. There is only one case to consider since $ 3+2+2=7$. The answer is $ \frac{2^{2}*3^{2}*7!}{3!2!2!}=7560$.
Whew.[/hide]

[hide="1"]using the binomial formula, you get $ \binom{8}{5}$ which is $ 56$.[/hide]

[hide="3"]use the binomial formula: $ \binom{8}{4}$ which equals $ 14$[/hide]

i cant get anything else can someone help me

i only could get 1 and 3 too

1.56
3.6

If i had not learned the binomial thoerm, i would not have gotten any!


inequalities
function
calculus
algebra
IMO Shortlist

Let $ n$ be a positive integer, and let $ x$ and $ y$ be a positive real number such that $ x^n \plus{} y^n \equal{} 1.$ Prove that
\[ \left(\sum^n_{k \equal{} 1} \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} \right) \cdot \left( \sum^n_{k \equal{} 1} \frac {1 \plus{} y^{2k}}{1 \plus{} y^{4k}} \right) < \frac {1}{(1 \minus{} x) \cdot (1 \minus{} y)}.
\]

[i]Author: Juhan Aru, Estonia[/i]

We have: $ \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} < x^{ \minus{} 2k}$. Then we can write:
$ \sum^n_{k \equal{} 1} \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} < \sum^n_{k \equal{} 1} x^{ \minus{} 2k} \equal{} \frac {1}{1 \minus{} x^2}\cdot (x^{ \minus{} 2n} \minus{} 1) \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {1 \minus{} x^{2n}}{x^{2n}} \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {(1 \minus{} x^{n})(1 \plus{} x^n)}{x^{2n}} \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {y^{n}(1 \plus{} x^n)}{x^{2n}}$
And:
$ \left(\sum^n_{k \equal{} 1} \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} \right) \cdot \left( \sum^n_{k \equal{} 1} \frac {1 \plus{} y^{2k}}{1 \plus{} y^{4k}} \right) < \frac {1}{1 \minus{} x^2}\cdot \frac {y^{n}(1 \plus{} x^n)}{x^{2n}}\cdot \frac {1}{1 \minus{} y^2}\cdot \frac {x^{n}(1 \plus{} y^n)}{y^{2n}} \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {1 \plus{} x^n}{x^{n}}\cdot \frac {1}{1 \minus{} y^2}\cdot \frac {1 \plus{} y^n}{y^{n}} \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {1 \plus{} x^n}{1 \minus{} y^{n}}\cdot \frac {1}{1 \minus{} y^2}\cdot \frac {1 \plus{} y^n}{1 \minus{} x^{n}} \equal{} \frac {1}{1 \minus{} x^2}\cdot \frac {1 \plus{} x^n}{1 \minus{} x^{n}}\cdot \frac {1}{1 \minus{} y^2}\cdot \frac {1 \plus{} y^n}{1 \minus{} y^{n}} < \frac {1}{(1 \minus{} x) \cdot (1 \minus{} y)}.$

This gives: $ (1 \plus{} x^n)(1 \plus{} y^n) < (1 \plus{} x)(1 \plus{} y)(1 \minus{} x^n)(1 \minus{} y^n)$ or
$ 1 \plus{} x^n \plus{} y^n \plus{} x^ny^n < (1 \plus{} x \plus{} y \plus{} xy)(1 \minus{} x^n \minus{} y^n \plus{} x^ny^n)$ or
$ 2 \plus{} x^ny^n < (1 \plus{} x \plus{} y \plus{} xy)x^ny^n$
$ 2 \plus{} x^ny^n < x^ny^n \plus{} x^ny^n (x \plus{} y \plus{} xy)$
$ 2 < x^ny^n (x \plus{} y \plus{} xy)$

This is not true :( If $ y \to 0$ we get false. Think there is some condition $ \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} < ?$ but I don't know what.

really nice solution . if y near to zero the right hand side is very great and hold that inequality , am i right? :huh:

My first case $ \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} < x^{ \minus{} 2k}$ does not give the result at the end. Inequality does not hold with it. :(

A bit late, but-

Solution) $ \frac{1\plus{}x^{2k}}{1\plus{}x^{4k}} < \frac{1}{x}$ (Equivalent to $ x^4 \minus{} x^3 \minus{} x \plus{} 1 > 0$ - equality holds only for $ x\equal{}1$ which is impossible)

Therefore, $ LHS \leq \left(\sum^{n}_{k\equal{}1}\frac{1}{x}\right)\cdot\left(\sum^{n}_{k\equal{}1}\frac{1}{y}\right)$
$ \equal{} \frac{1\minus{}x^n}{(1\minus{}x)x^n} \cdot \frac{1\minus{}y^n}{(1\minus{}y)y^n} \equal{} \frac{y^n}{(1\minus{}x)x^n} \cdot \frac{x^n}{(1\minus{}y)y^n} \equal{} RHS$

[quote="orl"]Let $ n$ be a positive integer, and let $ x$ and $ y$ be a positive real number such that $ x^n \plus{} y^n \equal{} 1.$ Prove that
\[ \left(\sum^n_{k \equal{} 1} \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} \right) \cdot \left( \sum^n_{k \equal{} 1} \frac {1 \plus{} y^{2k}}{1 \plus{} y^{4k}} \right) < \frac {1}{(1 \minus{} x) \cdot (1 \minus{} y)}.\]
[i]Author: unknown author, Estonia[/i][/quote]
$ \frac{1\plus{}x^{2k}}{1\plus{}x^{4k}} < \frac{1}{x^k}$.
So: $ \sum^n_{k \equal{} 1} \frac {1 \plus{} x^{2k}}{1 \plus{} x^{4k}} < \frac{\left ( \frac{1}{x^n} \right ) \minus{} 1}{1\minus{}x} \equal{} \frac{1}{1\minus{}x} \cdot \frac{y^n}{x^n}$
So $ LHS < \frac{1}{1\minus{}x} \cdot \frac{y^n}{x^n} \cdot \frac{1}{1\minus{}y} \cdot \frac{x^n}{y^n} \equal{} RHS$

Since $x,y>0$ and $x^n + y^n = 1$, it follows that $x,y \in (0,1)$. Hence for any $k \in \mathbb{N}$, it follows that

\[(1-x^{3k})(1-x^{k})>0\]

Rearranging this inequality yields that

\[\frac{1+x^{2k}}{1+x^{4k}} < \frac{1}{x^k}\]

Summing these inequalities for $k=1,2,\dots, n$ yields that

\[\sum^{n}_{k = 1}\frac{1+x^{2k}}{1+x^{4k}}<\sum^{n}_{k = 1}\frac{1}{x^k}=\frac{1}{x^n}\sum^{n-1}_{k = 0}x^k = \frac{1}{x^n} \cdot \frac{1-x^n}{1-x}=\frac{y^n}{x^n (1-x)}\]

Similarly, it follows that

\[\sum^{n}_{k = 1}\frac{1+y^{2k}}{1+y^{4k}}<\frac{x^n}{y^n (1-y)}\]

Multiplying these two inequalities yields that

\[ \left(\sum^{n}_{k = 1}\frac{1+x^{2k}}{1+x^{4k}}\right)\cdot\left(\sum^{n}_{k = 1}\frac{1+y^{2k}}{1+y^{4k}}\right) < \frac{1}{(1-x)\cdot (1-y)} \]

This problem is nice.

First of all we have $\frac{1+x^{2k}}{1+x^{4k}} < \frac{1+x^{2k}}{\frac{(1+x^{2k})^2}{2}}=\frac{2}{1+x^{2k}} < \frac{2}{2x^k}=\frac{1}{x^k}$

So summing over we get $\sum_{k=1}^{n}{\frac{1+x^{2k}}{1+x^{4k}} < \sum_{k=1}^{n}{\frac{1}{x^k}}=\frac{(1-x^n)}{x^n(1-x)}=\frac{y^n}{x^n(1-x)}}$

Similarly $\sum_{k=1}^{n}{\frac{1+y^{2k}}{1+y^{4k}} < \frac{x^n}{y^n(1-y)}}$

Multiplying these inequalities we obtain the result.

So we are done.

[hide]
[b]Claim:[/b] $\dfrac{1+x^{2k}}{1+x^{4k}} < \dfrac{1}{x^k}$
[b]Proof:[/b] This is equivalent to
\begin{align*} x^k (x^{2k}+1) &  < 1 + x^{2k} \\
\iff (1-x^{3k})(1-x^k) & > 0,\end{align*}
which is true and strict because $x$ strictly lies between $0$ and $1.$ $\blacksquare$

Now,
\begin{align*}
\displaystyle \sum_{k=1}^{n} \dfrac{1+x^{2k}}{1+z^{4k}}&  < \displaystyle \sum_{k=1}^{n} \dfrac{1}{x^k} \\
& = \dfrac{1-x^n}{x^n(1-x)} \end{align*}
and similarly $\displaystyle \sum_{k=1}^{n} \dfrac{1+y^{2k}}{1+y^{4k}} < \dfrac{1-y^n}{y^n(1-y)}.$ Multiplying these, we get the desired result. $\blacksquare$ [/hide]

[Hide=Sketch]

Write $$\frac{1}{1-x}=\frac{1+x+x^2+..+x^{n-1}}{1-x^n}=\frac{1+x+x^2+…+x^{n-1}}{y^n}$$,
so the RHS is now $\sum x^{-k} \sum y^{-k}$

Next note that since $x \neq 1$ , we have the following inequality :

$$\frac {1+x^{2k}}{1+x^{4k}} < \frac {2}{1+x^{2k}} < \frac {1}{x^k}$$

Adding up the stuff and multiplying , we get the desired result .

The crux of the proof is the following bound: \begin{align*}
    \frac{1 + x^{2k}}{1 + x^{4k}} < \frac{1}{x^k}.
\end{align*} It's equivalent to prove \begin{align*}
    0 < x^4 - x^3 - x + 1 = (x - 1)^2(x^2 + x + 1),
\end{align*} which is true for $x$ over the interval $(0, 1)$. Using the bound yields \begin{align*}
    \text{LHS} &< \left(\frac 1 x + \frac{1}{x^2} + \cdots + \frac{1}{x^n} \right) \cdot \left(\frac 1 y + \frac{1}{y^2} + \cdots + \frac{1}{y^n} \right) \\
    &= \frac{1}{x^n} \cdot \frac{1}{y^n} \cdot \frac{(1 - x^n)(1 - y^n)}{(1 - x)(1 - y)} \\
    &= \frac{1}{(1 - x)(1 - y)},
\end{align*} as desired.

Solved with [b]Sross314[/b].

Claim: $\frac{1+x^{2k}}{1+x^{4k}}<\frac{1}{x^k}$ for $0<x<1$ and positive integers $k$.
Proof: This inequality is equivalent to \[x^k(x^{2k}+1)<x^{4k}+1,\]which is equivalent to \[x^{4k}-x^{3k}-x^k+1>0.\]

Now we have \[x^{4k}-x^{3k}-x^k+1=(x^k-1)^2(x^{2k}+x^k+1)>0\]$\blacksquare$

It suffices to show that \[\left(\frac{1}{x}+\frac{1}{x^2}+\cdots+\frac{1}{x^n}\right)\left(\frac{1}{y}+\frac{1}{y^2}+\cdots+\frac{1}{y^n}\right)\le \frac{1}{(1-x)(1-y)} \]

We have \[\sum_{k=1}^n \frac{1}{x^k}=\frac{x^{n-1}+x^{n-2}+\cdots+x+1}{x^n}=\frac{\tfrac{x^{n}-1}{x-1}}{x^n}=\frac{x^{n}-1}{x^n(x-1)}\]

Multiplying by $\frac{y^n-1}{y^n(y-1)}$ gives \[\frac{(1-x^n)(1-y^n)}{x^ny^n(1-x)(1-y)}=\frac{1}{(1-x)(1-y)},\]so we are done.

[quote=Boy Soprano II]Since $ t \mapsto x^t$ is a strictly convex function, by Karamata's inequality, we have
\begin{align*}
x^k + x^{k+4} + x^{3k+2} + x^{3k+6} &< x^0 + x^2 + x^{4k+4} + x^{4k+6} \\
x^k(1+x^4)(1+x^{2k+2}) &< (1+x^2)(1+x^{4k+4}) \\
\frac{1+x^{2k+2}}{1+x^{4k+4}} &< \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k}
\end{align*}
for any nonnegative integer $ k$, since the sequence $ (4k+6,4k+4,2,0)$ strictly majorizes the sequence $ (3k+6,3k+2,k+4,k)$.  (One can also prove this inequality by repeated application of weighted AM-GM.)  Thus
\begin{align*}
\sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} &< \sum_{k=0}^{n-1} \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k} \\
&= \frac{1+x^2}{1+x^4} \cdot \frac{1/x^n - 1}{1/x-1} \\
&= \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} .
\end{align*}
Again by Karamata, $ (1+x^2)x = x+x^3 \le 1+x^4$, so
\[ \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} \le \frac{(1-x^n)}{x^n(1-x)} = \frac{y^n}{x^n(1-x)} , \]
since $ x^n + y^n = 1$.

It now follows that
\[ \biggl( \sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} \biggr) \cdot \biggl( \sum_{k=1}^n \frac{1+y^{2k}}{1+y^{4k}} \biggr) < \frac{y^n}{x^n(1-x)} \cdot \frac{x^n}{y^n(1-y)} = \frac{1}{(1-x)(1-y)}, \]
as desired, since all quanties in question are positive.  $ \blacksquare$[/quote]

Nice

[quote=awesomeming327.]Let $0\le n< 1.$
\[n\le 1\implies n-n^4<1-n^3\iff \frac{1+n^2}{1+n^4}<\frac{1}{n}\]
Thus, \[ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) \le \left(\sum^n_{k = 1} \frac {1}{x^{k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1}{y^{k}} \right)=\frac{1-x^n}{x^n(1-x)}\cdot \frac{1-y^n}{y^n(1-y)}=\frac{1}{1-x}\cdot \frac{1}{1-y}\] as desired.[/quote]

The inequality is strict. $n=1$ doesn't work here. Maybe it's a typo.

The problem dies to the estimate $$\frac{1+x^{2k}}{1+x^{4k}} \leq x^{-k} \iff 1+x^{4k} \geq x^k + x^{3k}.$$ Indeed this means that $$\left(\sum \frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum \frac{1+y^{2k}}{1+y^{4k}}\right) \leq x^{-1} \cdot \frac{1-x^{-n}}{1-x^{-1}} \cdot y^{-1} \cdot \frac{1-y^{-n}}{1-y^{-1}} = \frac 1{(1-x)(1-y)}.$$ Note that equality cannot hold everywhere.

Our claim is that
\[\frac{1+t^2}{1+t^4} < \frac 1t\]
for all $t \in (0,1)$, which can be proven by expanding to $(t-1)^2(t^2+t+1) > 0$. Substituting $x^k$ and $y^k$, we get
\begin{align*}
\left(\sum \frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum \frac{1+y^{2k}}{1+y^{4k}}\right) &< \left(\sum \frac {1}{x^k}\right) \left(\sum \frac {1}{y^k} \right) \\
&= \frac{1-x^n}{x^n(1-x)} \cdot \frac{1-y^n}{y^n(1-y)} = \frac{1}{(1-x)(1-y)}.~\blacksquare
\end{align*}


I have two questions on complex numbers. right now im doing aops vol 2, and i came across two equatiosn taht im not sure i understand how they got them
1) how do you know $ e^{iy}\equal{}\cos y\plus{}i \sin y$?
2) when asked to find $ i^i$ they started by saying $ i\equal{}e^{i\cdot (\frac{\pi}{2}\plus{}2\pi k)}$? how do they know taht?

1)  This is called Euler's formula, and the motivation behind it is somewhat complicated.  First, we will define the function $ \text{cis } x \equal{} \cos x \plus{} i \sin x$.  From here, we can prove the identity

$ \text{cis } nx \equal{} (\text{cis } x)^n$

which is called de Moivre's theorem (for integer $ n$).  In fact, we can extend this to rational $ n$.  The existence of this theorem suggests that $ \text{cis } x$ has exponential properties, so the first interpretation of Euler's formula is that $ e^{ix}$ is [b]notation[/b] that [b]means[/b] $ \text{cis } x$, and it is convenient notation because that function obeys the same rules as the exponential function.

2)  $ e^{i(\pi/2 \plus{} 2 \pi k)} \equal{} \cos \left( \frac{\pi}{2} \plus{} 2 \pi k \right) \plus{} i \sin \left( \frac{\pi}{2} \plus{} 2 \pi k \right) \equal{} 0 \plus{} i$.

thanks a lot torajiru. i understand demoivre's theorem and all, i just wasn't sure how they got that. thanks again!

To answer your question: how do you derive Euler's relation $ \exp{i\theta} \equal{} \cos{\theta} \plus{} i\sin{\theta}$? It is not hard: write
$ \cos{\theta} \equal{} 1 \minus{} \frac {\theta^2}{2!} \plus{} \frac {\theta^4}{4!} \minus{} \cdots$ and $ i\sin{\theta} \equal{} i\theta \plus{} \frac {(i\theta)^3}{3!} \plus{} \frac {(i\theta)^5}{5!} \plus{} \cdots$. The sum of these is
$ \sum_{k \equal{} 0}^\infty {\frac {(i\theta)^k}{k!}}$. But this is precisely $ \exp{i\theta}$.

We know that $ i\equal{}\exp{[i(\pi/2\plus{}2\pi k)]}$ by Euler's relation since $ i\equal{}\exp{[i(\pi/2\plus{}2\pi k)]}\equal{}\exp{i\pi/2}exp{2i\pi k}\equal{}\exp{i\pi/2}\equal{}i$ for all $ k$. Then $ i^i\equal{}\exp{[i \cdot i (\pi/2\plus{}2\pi k)]}\equal{}\exp{[\minus{} (\pi/2\plus{}2\pi k)]}\equal{}\minus{}i \cdot \exp{2\pi k)}\equal{}\minus{}i.$

1)  The series expansion identity isn't a valid proof without several results in complex analysis about complex power series.  And even with those, it can only serve as a motivation for the [url=http://mathworld.wolfram.com/AnalyticContinuation.html]analytic continuation[/url] of $ e^x$ to complex values (which is what we are really talking about).

2)  The last two steps are wrong.  $ e^{\minus{} \frac{\pi}{2} \plus{} 2 \pi k}$ is real.

correction to part (b): $ i^i \equal{} \exp( \minus{} \pi/2 \plus{} 2\pi k)$


probability
expected value
combinatorics proposed
combinatorics

I don't know exactly where to post this, it'll probably be a well-known problem but I don't find any information.

If you draw $ n$ numbers (uniformly) at random from $ \{1,2,...,n\}$, how many different numbers have you drawn on average?

What if you draw $ m$ numbers, is there a general formula then? Or approximate behaviour?

This is just some variation of coupon collector, isn't it?

The probability that you never draw 1 is $ \left( 1 \minus{} \frac{1}{n} \right)^m$.  So the probability that you ever do draw 1 is $ 1 \minus{} \left( 1 \minus{} \frac{1}{n} \right)^m$.  So the expected number of numbers drawn is $ n \left[1 \minus{} \left( 1 \minus{} \frac{1}{n} \right)^m\right]$.

When $ m \equal{} n$ and both are large, the expected number is about $ n \left(1 \minus{} \frac{1}{e} \right)$.  That makes sense because as $ n$ gets large, we approach a Poisson distribution.


[url=http://bildr.no/view/488550][img]http://bildr.no/thumb/488550.jpeg[/img][/url]

I can not understand the last sentence. why are they not consecutive ?

Thanks

If you didn't understand the last sentence, then there exists a previous sentence which you also didn't understand.  One cannot get uniqueness without requiring non-consecutivity because $ F_{n\plus{}2} \equal{} F_{n\plus{}1} \plus{} F_n$.

[quote="t0rajir0u"]If you didn't understand the last sentence, then there exists a previous sentence which you also didn't understand.  One cannot get uniqueness without requiring non-consecutivity because $ F_{n \plus{} 2} \equal{} F_{n \plus{} 1} \plus{} F_n$.[/quote]

Can you say more detailed?

thanks

If the terms are allowed to be consecutive, then there won't be uniqueness, since you can replace the two consecutive terms with the next Fibonacci number and obtain another sum.

[quote="mavropnevma"]This is the celebrated Zeckendorf's theorem. Enter this on Google and you'll find tons of material (wikipedia, planetmath, wolfram, etc).[/quote]

Thanks very much !


There are 12 PC's in a room. 7 people are to work. Each on only one PC.
You have 12 jobs to do.
Out of which 2 run automaticaly and don't require any user interference(Likeinstallation).
5 of them are of making C++ programs( require a person on the PC).
5 of them can run on only 7 of the PC's(There min. requirements are satisfied only by these 7 PC's)
Each program is run by one PC each.
How many ways can this be done(if it can be)?

Choose 5 of the 7 P.C.'s satisfying the minimum requiements and assign to them the 5 jobs. (7[b]P[/b]5)

Of the remaining 7 choose 5 and assign to them the jobs requiring user intervention.Then assign 5 of the 7 people to the jobs.(7[b]P[/b]5*7[b]P[/b]5)

The remaining 2 P.C.'s run the 2 automatic jobs.(2)

Thus total number of ways is 7[b]P[/b]5 * 7[b]P[/b]5 *7[b]P[/b]5 *2
                                             =2520*2520*2520*2=32006016000

Here we have assumed the P.C.'s,the jobs and the people to be distinct.

wel thanks vihang for trying the problem.
I'm happy the moderators didn't shift my problem to another forum. and thanks once again for trying it.
i have nearly forgotten the problem and these days i'm busy with chemistry so it may take me some time to analyze your solution. others will judge the solution better than me so i hope someone will shortly reply now.

Varun


A rocket is fired vertically upward. At the instant it reaches an altitude of 1120 m and a speed of 285m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 241 m/s following the explosion. The second fragment has a speed of 472 m/s and is moving east right after the explosion. What is the magnitude of the velocity of the third fragment?

This is solved through momentum.  Since there is no mass given, only the velocity component is important.  Initially, the only velocity is 285 up $ (285,0,0)$ as a vector.  One piece has velocity vector $ (241,0,0)$ and the other $ (0,472,0)$ (in taking east and up to be the positive directions).  So the final velocity of the system must be the same as well.  $ (241,0,0)\plus{}(0,472,0)\plus{}(x,y,z)\equal{}(285,0,0)\rightarrow (44,\minus{}472,0)$, which means that the piece is moving upward at 44m/s and west at 472 m/s.

So the magnitude of the velocity would just be $ \sqrt{44^2\plus{}472^2}$ ?

Yes, up and west.  Make sure you understand why that makes sense.  Since all of the momentum is initially up, the total final momentum must be up as well.  Since there is a large component in the east direction, there must be a fragment to compensate.  And since not all of the upward momentum has been attained, there has to be a fragment to add the final amount.


calculus
derivative
calculus computations

Suppose
f(x,y) = sin(xy)

Is there any direction(s) in which rate of change of f at P(sqrt{pi},sqrt{pi}) is constant?

I want to say, in all directions, because gradient computed at P is: <-sqrt{pi},-sqrt{pi}> Is this right?

Definitely not all directions; that would only happen if the gradient was zero.

What do you know about directional derivatives?


Well, yesterday my school participate in the Pascal contest. I'm not sure if anyone else did, but if so, how hard do you think it was? And what was the answer to no. 25?

I thought 25 was MUCH harder than any of the last questions on past Pascals. Even after 20 min of hacking at it, I got the wrong answer...  :( Oh well, at least I figured out the question in class.

[hide="My solution"]
John's number is just going to be a large mass of $112$s or $111$s. We will only have $5$ consecutive $1$s when $111$ and $112$ are next to each other. The first digit has to be a $1$ which would branch off to a $112$ which branches off into a $112112111$ and so on. When you observe this and divide all the numbers into $3$ "blocks", the third block is always the different one.

Every third block ends in a $2$. Every ninth block ends in a $1$. Every 27th block ends in a $2$ and it will keep switching like this. But we only want to count the ones that end in $1$ because we are looking for 5 consecutive $1$s. So we are looking for the number of whole numbers $\leq  2187$ that are divisible by $9, 81, 729$ subtract those that are divisible by $27, 243, 2187$.

$243 + 27 + 3 - 81 - 9 - 1$
$= 273 - 91$
$= 182$
$\boxed {A)182}$
[/hide]

How many people do you think got over 142?

Wow, Pascal was way too easy, which is not really good. The cutoff mark will definitely be higher. I speculate at least 134. Anyway I made a stupid mistake in question 16, didn't realize that 5+5 = 8+2.  :blush: I figured out 25 after the contest, spent too much time trying to get 24 but luckily I got it in the end.

Good luck to all the Grade 9 and below students with their Pascal results!  :)

My solution:

[hide]

We want the number of "11111" strings in our 2187 digit number.
Its easy to see that this is equal to the number of "21" strings in our 729 digit number,
equal to the number of "1" strings in our 243 digit number [not including the last digit],
equal to 162 + the number of 2 strings in our 81 digit number [not including the last digit],
equal to 162 + the number of 1 strings in our 27 digit number [not including the last digit],

If we started with 2, then we had written 112 and the leftmost digits don't match up.  So we started with 1,
implying we started with 112, implying we started with 112112111, implying we started with 112112111 112112111 112112112.
Counting the # of 1's, we have 20.  So 182 total.

[/hide]

my math teacher showed me a very nice solution today...

1st column : total # of 1's
2nd column :total # of 2's
third column: # of five consecutive 1's
beside the bracket : # of digits
(I tried to make a nicer table but it didnt work out...  :P)

[     [b]2[/b]                                 1                                                0                 ]                        3^1
[     [color=darkred]7[/color]                               [b]2[/b]                                                0                 ]                        3^2
[     [color=red]20[/color]                             [color=darkred]7[/color]                                                [b]2[/b]                 ]                        3^3
[     [color=orange]61[/color]                            [color=red]20[/color]                                               [color=darkred]7[/color]                  ]                        3^4
[     [color=brown]182[/color]                          [color=orange]61[/color]                                              [color=red]20[/color]                 ]                        3^5
[     [color=green]547[/color]                              [color=brown]182[/color]                                            [color=orange]61[/color]                 ]                        3^6
[     1640                               [color=green]547[/color]                                              [color=brown]182[/color]                ]                        3^7

and the answer is 182

Well if you put those answers, then you got 150. I'm so jealous; stupid mistakes that I always make.  :ninja:

[quote="autumnleaf"]what was #16? and was the answer to #24, 7 (B, i think)?
Thanks![/quote]

You know what? I hate question 16. It sucks when you realize after the contest that you made a mistake, thinking that there are actually 10 possibilities when 5 + 5 = 8 + 2, which implies that there are only 9 possibilities.  :( I - hate it when you're so close to a perfect score but you slip.  :oops: Ahh well, I didn't even think I was going to do adequately well in the first place.  :blush:

hey, do you remember the exact question for #16?
plz plz tell me.
maybe i did the same mistake. but i dont remember by just you telling me 8+2 = 5+5

16. An unusual die has the numbers 2, 2, 3, 3, 5, and 8 on its six faces. Two of these dice are rolled. The two numbers on the top faces are added. How many different sums are possible?
... (D) 9     (E) 10

[quote="igorsr"]Hi
Here are my Pascal answers
1 E  2 B  3 D  4 A  5 B  6 C  7 E  8 A 9 C 10 D
11 B 12 C 13 D 14 E 15 E 16 D 17 A 18 C 19 A 20 A
21 E 22 A 23 C 24 B 25 A
Any mistakes?
Thanks[/quote]

OMG. OMG. i have the different answer to you for 6. I got E. omg, #6 is supposed to be a stupid easy question. what was the question?
are you sure that all ur answers are right?
by the way, how do some of you guys remeber the exact wordings of the test, are the online now or smth?
thx.

#6 Ravindra and Hongshu made a pizza together.  Ravindra ate 2/5 of the pizza.  Hongshu ate half as much as Ravindra.  What percentage of the original pizza was left?

Also no I did not get 150.  Just checking all three contests.  I believe you calculated what they ate and not what was left.  Sorry.

A 20  B 30  C 40  D 50  E 60

People remember the questions as they kept their test paper.

Oh, I thought you got 150. I'm nervous; this contest was really simple. I'm afraid the cut-off mark will be much higher.  :( Hopefully, people were extra stressed or sleepy that day and made mistakes.  :D  :rotfl:

My answers:
1. E
2. B
3. D
4. A
5. B
6. C
7. E
8. A
9. C
10. D
11. B
12. C
13. D
14. E
15. E
16. D w00t I almost put E but realized it at the last min. Hope I erased properly.
17. A
18. C
19. A
20. A
21. E
22. A
23. C
24. B
25. I left it blank, so the max I can get is 144

Did I get my 144/150, ppl? Tell me pls.

Ah, so many people are getting close to 150. I am reluctant to believe that I'll make it on the Honour Roll this year.  :?

BTW, if anyone wants to consult the solutions and answers, the Pascal/Cayley/Fermat solutions have been posted on the website.

my answers:
EBDAB
CEACD
BCDEE
DACAA
EACBA

all the questions except for #25 are extremely easy.
for #25....
i did it in totally a wrong way and i mistakenly got 81*2=182
 i put A on the answer sheet :lol:
im still surprising about my luck...

now i might be the only gr8 who got 150


How many values of $ x$ are possible for an integer sided obtuse angled triangle with sides $ 15, 8$ and $ x$ ?

[hide="solution"]First, we must check all the possibilities for $ x$ without any restrictions.  We find that $ 8 \le x \le 22$.

Now, you have two possible equations:

$ x^2 \plus{} 8^2 < 15^2$

$ 8^2 \plus{} 15^2 < x^2$

For the first equation, we have $ x < \sqrt{161}$, or $ x < 12.688$.  So, we have $ 5$ integral solutions for this equation, $ {8,9,10,11,12}$.

For the second equation, we have $ 289 < x^2$, so $ x > 17$.  Again, we have $ 5$ integral solutions for this equation, $ {18,19,20,21,22}$.

Thus, our answer is $ 5 \plus{} 5$, or $ \boxed{10}$.[/hide]


Let $ a,$ $ b$ and $ c$ are non-negative numbers,  no two of which are zeros. Prove that:
1) $ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}\geq3,$
2) $ \sum_{cyc}\frac{a}{b^{2}\plus{}c^{2}}\geq\frac{6\sqrt3}{\sqrt{11(a^{2}\plus{}b^{2}\plus{}c^{2})\plus{}5(ab\plus{}ac\plus{}bc)}}.$

Nice inequalities, arqady.

For 1), I used Holder in the form
\[ \left(\sum_{cyc}{\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}}\right)^{4}\cdot\sum{(a\plus{}b)(c^{2}\plus{}3ab)^{2}}\geq 32 (\sum_{cyc}{a})^{5}.\]
Hence it remains to prove that
\[ \frac{32 (\sum{a})^{5}}{\sum{a^{4}b}\plus{}\sum{ab^{4}}\plus{}9\sum{a^{3}b^{2}}\plus{}9\sum{a^{2}b^{3}}\plus{}12\sum{a^{2}b^{2}c}}\geq 81,\]
which is true by simple expanding, via Muirhead + Schur.

For 2), my idea is using Cauchy-Schwarz and Vasc's inequality,

\[ 4\left(\sum_{cyc}\frac{a}{b^{2}\plus{}c^{2}}\right)^{2}\equal{} 4\left(\sum_{cyc}\frac{a^{4}}{a^{3}b^{2}\plus{}a^{3}c^{2}}\right)^{2}\geq\left(\frac{2(\sum{a^{2}})^{2}}{\sum{a^{3}b^{2}}\plus{}\sum{a^{2}b^{3}}}\right)^{2}\geq 9\left(\frac{\sum{a^{3}b}\plus{}\sum{ab^{3}}}{\sum{a^{3}b^{2}}\plus{}\sum{a^{2}b^{3}}}\right)^{2}.\]

So it remains to prove that

\[ \left(\frac{\sum{a^{3}b}\plus{}\sum{ab^{3}}}{\sum{a^{3}b^{2}}\plus{}\sum{a^{2}b^{3}}}\right)^{2}\geq\frac{48}{11\sum{a^{2}}\plus{}5\sum{ab}},\]

which by expanding feels OK, but I haven't done it all by hand.

[quote="pohoatza"]
Hence it remains to prove that
\[ \frac{32 (\sum{a})^{5}}{\sum{a^{4}b}\plus{}\sum{ab^{4}}\plus{}9\sum{a^{3}b^{2}}\plus{}9\sum{a^{2}b^{3}}\plus{}12\sum{a^{2}b^{2}c}}\geq 81,\]
which is true by simple expanding, via Muirhead + Schur.
[/quote]
It's wrong. :(  Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.$ :wink:
Id est, the first is still open.
I have proved the second. It's easy enough.

My old problem in Secrets in Inequalities is pretty easier (I posted in in Mathlinks)

\[ \sum\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}}\ge 4.\]

But I think problem of Arqady must be solvable.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers,  no two of which are zeros. Prove that:
1) $ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}\geq3,$
[/quote]
I think we can prove that $ \sum_{cyc}\frac{a}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$ and $ \sum_{cyc}\frac{ b}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$

[quote="chien than"]
I think we can prove that $ \sum_{cyc}\frac{a}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$[/quote]
No! Try $ a\equal{}2,$ $ b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="hungkhtn"]My old problem in Secrets in Inequalities is pretty easier (I posted in in Mathlinks)
\[ \sum\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}}\ge 4.\]
[/quote]
Yes, it's easy. By Holder we obtain:
$ \left(\sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}}\right)^{2}\cdot\sum_{cyc}(a\plus{}b)(c^{2}\plus{}ab)\geq8(a\plus{}b\plus{}c)^{3}.$
Thus, it remains to prove that $ (a\plus{}b\plus{}c)^{3}\geq4\sum_{cyc}(a^{2}b\plus{}a^{2}c),$ which obviously true.
But  $ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}}\geq3\sqrt2$  is wrong already.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers,  no two of which are zeros. Prove that:
1) $ \sum_{cyc}{\frac{a+b}{\sqrt{c^{2}+3ab}}\geq3,}$
[/quote]

I come again with an other solution for 1) since I've misscalculated my previous attempt.

The main idea is to use contradiction method. Denote $ x =\frac{a+b}{\sqrt{c^{2}+3ab}}$, $ y =\frac{b+c}{\sqrt{a^{2}+3bc}}$ and $ z =\frac{c+a}{\sqrt{b^{2}+3ca}}$.

The inequality rewrites as $ x+y+z\geq 3$.

Suppose $ x+y+z < 3$. Since $ 3\sqrt [3]{xyz}\leq x+y+z < 3$, it follows that $ xyz < 1$, which is equivalent with
\[ \left((a+b)(b+c)(c+a)\right)^{2}< (a^{2}+3bc)(b^{2}+3ca)(c^{2}+3ab),\]
which after expanding rewrites as
\[ \sum{a^{4}b^{2}}+\sum{a^{2}b^{4}}+6\sum{a^{3}b^{2}c}+6\sum{a^{3}bc^{2}}<\sum{a^{3}b^{3}+7\sum{a^{4}bc}+18a^{2}b^{2}c^{2},}\]
which is evidently a contradiction by plugging $ c = 0$. Hence, $ x+y+z\geq 3$.

I think your solution is wrong, pohoatza. By contradiction method, you assume that... there are three certain numbers $ (x',y',z')$ that make the inequality wrong. So the inverse inequality is assumed to be true for these numbers only, not for all. So you can not choose $ x$ or $ y$ or $ z$ equal to $ 0$. That is the reason.

But
\[\sum_{cyc}{\frac{a+b}{\sqrt{c^2+ab}}}\ge 3\sqrt{2}\frac{ab+bc+ca}{a^2+b^2+c^2}\]
is true.

We have also a stronger one.
Let $a$, $b$ and  $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a+b}{\sqrt{c^2+ab}}+\frac{a+c}{\sqrt{b^2+ac}}+\frac{b+c}{\sqrt{a^2+bc}}\geq3\sqrt2\sqrt[11]{\frac{ab+ac+bc}{a^2+b^2+c^2}}$$

[quote=Nguyenhuyen_AG][quote=hungkhtn]My old problem in Secrets in Inequalities is pretty easier (I posted in in Mathlinks)
\[ \sum\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}}\ge 4.\]
[/quote]
Consequence of
\[\frac{a\plus{}b}{\sqrt{c^{2}\plus{}ab}} \geqslant \frac{2(a^2+b^2)}{a^2+b^2+c^2}.\][/quote]

No it doesn't. Because equality holds when $a=b=c$. But if $a=b=c$, $LHS$ is not equal to $RHS$

[quote=Nguyenhuyen_AG][quote=inxang1]No it doesn't. Because equality holds when $a=b=c$. But if $a=b=c$, $LHS$ is not equal to $RHS$[/quote]
Are you sure?[/quote]

?Hmm, how do you explain about it?


Consider a jar filled with some dense liquid and on top of the liquid is water. Density of first liquid is greater than water. Now put an ice in this sysem. When the ice melts,what is the change in the level of the liquids.?

I don't think the dense liquid in the bottom changes the answer unless the ice is partially immersed in it. If not, the levels of liquid or water do not change, because before the ice melts, it displaces as much water as it weighs. When it melts, it melts just into the same amount of water (that it weighs), eaxctly filling up the initial displaced water volume.

i think you need to know whether the ice is floating only in the water or in the dense liquid too . If its only on the water then i agree with kubus that there will be no change

ummmm if its floating partially on both liquids ................ ummmm i 'll have to think :blush:

If it's in both liquids, after it melts the level of the dense liquid clearly falls, but the level of water rises. Since the ice displaces as much liquid as it weighs, after melting it has more volume than the displaced liquid, since some of the displaced liquid is the dense liquid.

Hmmmmmmmmmm......why should the ice go down such that it gets immersed in both the liquids??wont the ice just stay up with the water?

oh yeahhh..... and what if the water is really less , lesse than even the lenght of the ice OR what if the ice isnt a small cube its a quite big chunk of ice?

ohhyeahhhhhh...that condition is quite right eminem.THen its somethin what kubus said...but one thing against kubus,the level of water would rise and the level of other liquid will remain same right?cause kubus said that the level of the liquid fallss...how can it fall?

[quote="shady"]
cause kubus said that the level of the liquid fallss...how can it fall?
[/quote]
it will fall because the some part of the ice is in the denser fluid.

can someone club everything and give the proper soln???

Aint it obvious eminem?If the level of water is way above the dense liquid and the ice doesnot get immersed in both the liquids....then when ice melts,th elevel of water remains same and if the ice is submerged in both the liquids,then level of the system increases. ;)

And eminem...since when did u become the missispi spellin bee winner???u never went out of india!! :rotfl:

dont try to act a bit too wise , Mr .Shady ! It would be better if  your puny little brain( hey not to worry guys he's a friend)  would try to concentrate more on the questions rather than people's signatures . the reason why i gave that signature is.... check Out the SPELLING OF   MISISIPI which i have written    .............understood ? it was supposed to be a joke, which is obvius you couldnt understand.......hehehehhe

Dont try to wise crack me , Mr. Smarty pants!

Yo,u cant humiliate people publicly around here.anyay u little brat ,i understood that but i thought u urself wouldnt understand that.....yo mod,delete his post fore i start swearin at him...and since were u my friend @@##$&##.DOnt try to bloody mess with me choo%%#@&.

Anyway,i hope my clubbed up answer is right . :mad: (still a bit hot)

hahahahhahah :rotfl:  :rotfl:  :rotfl:  :rotfl:  :lol:  :lol:  :lol: hope this teaches you a little lesson you little pipsqueak ahahhaahahaha :lol:  :lol:  :lol:

hey mod , dont delete my post , let the whole goddamned AOPS  see what a rat fink this guy is .............hahahahahahhahha :rotfl:  :rotfl:  :rotfl:

Guys!!! Exercise some restraint. We have the PMs for personal messages. Maybe the other person doesn't likw what you are saying. Don't be expository journalists!!!

The moderator should delete this post after it has been read by the culprits


[quote="can_hang2007"]http://www.mathlinks.ro/viewtopic.php?t=114617 :)[/quote]

Thanks, !
It's very nice, interested and very hard to solve cyclic inequalities
I hope the book of (anh) Can will be publish in soon time!

thank's a lot can. it's very very nice  :lol:


Este tipo de problemas son clasicos en olimpiadas,
espero que proponan soluciones variadas a estas preguntas:

[color=blue][size=150]Preg 01.[/size].[/color]

 Sean $x,y$ enteros positivos tal que $xy +1$ divide a $x^2 + y^2$.
Mostrar que:   $\frac{x^2 + y^2}{xy +1},$
es el cuadrado de un entero.

[hide=" English Version"]Let $x,y$ positive integers, such that  $xy +1$ divides $x^2 + y^2$.
Show that:      $\frac{x^2 + y^2}{xy +1},$
is the square of an integer.[/hide]

[color=blue][size=150]Preg 02[/size].[/color]

 Mostrar que, si $x,y,z$ son enteros postivos, entonces $(xy +1)(yz +1)(zx +1)$ es un cuadrado perfecto
si y solo si $(xy+1), (yz+1), (zx+1)$ todas son cuadrados perfectos.
                                                                                [i](Kiran S. Kedlaya)[/i]

[hide="English Version"]
Show that, if $x,y,z$ are postive integers, then $(xy +1)(yz +1)(zx +1)$ is a perfect square
 if and only if $(xy+1), (yz+1), (zx+1)$ are all perfect squares.[/hide]

[hide="Buena Suerte"] Good Luck[/hide]

Carlos Bravo :)
Lima- PERU

[hide]I am not spanish or whatever, but your first question is IMO88 Q6. Search it.
Basically, infinite decent...

Bomb[/hide]


A sequence $ a_{1}a_{2}a_{3}...a_{n}$ with $ a_{i}\in {0;1}$ is called a binary string of lenght n . What is the number of binary string of length n ( $ n \geq 4$ ) in which substring 01 occurs exactly twice ?

[hide="hint"]
This is equal to the number with at least two occurrences of 01 minus the number with at least three occurrences of 01.[/hide]

[hide="Hint for an alternate solution"]
All such strings must come in the form of $ ABCBCD$, where
- A is a string of 1s with any length, even 0
- B is a string of 0s with length at least 1
- C is a string of 1s with length at least 1
- D is a string of 0s with any length, even 0
Now do balls and urns.
[/hide]

This question was recently posted and solved on the Proposed & Own forum, by a user (nmtvn) who posts almost exclusively active contest problems.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168095


Let $f: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ such that: $f(x.f(y))=x^{p}.y^{q}$ $(p,q \in Z^{+})$. Show that $q=p^{2}$


Find all real solution of: $cos(x)+cos(x.\sqrt{2})=2$

[hide]$\cos x\leq 1\Rightarrow \cos (x)+\cos (x\sqrt{2})\leq 2$.
Equality holds iff $x$ and $x\sqrt{2}$ both have cosines of 1, so that both are integral multiples of $2\pi$.
But if $x=2\pi n$, then $2\pi n\sqrt{2}$ is an integral multiple of $2\pi\Rightarrow n\sqrt{2}$ is an integer.
Since $\sqrt{2}$ irrational, $n=0$ and so $x=0$.[/hide]


[quote="abacadaea"]what does something like this:
$ f: \mathbb{N} \to \mathbb{N}\cup \{0\}$
mean?[/quote]It means you are mapping from the set  $ \mathbb{N}$ into the set $ \mathbb{N}\cup \{0\}$, the latter set is the set of natural numbers with an additional element $ 0$.

If $ f: A\to B$ than $ A$ is domen, and $ B$ is codomen of the function.

Yeah, I've tried to translate, and have done it badly. Sorry.

It means that the domain of $ f(x)$ is the natural numbers and the range of $ f(x)$ is the whole numbers.

[quote="1=2"]and the range of $ f(x)$ is the whole numbers.[/quote]Not the whole numbers, but the [i]non-negative[/i] whole numbers. Also, you say 'range' but I believe you should say 'codomain', but that may be a linguistic/defintion issue.

[quote="hsiljak"]Just a minor linguistic remark: it's [i]domain [/i]and [i]codomain[/i].[/quote]

codomain=range right?

I think the range has to be in the codomain, so not all values in the codomain must actually be achieved. I'm not sure though.

Right. A function maps into the co-domain  (the co-domain statement tells us what kind of output we can expect from the function), and onto the range (the range is the smallest set that contains all of the values that are actually achieved by a particular function over a particular domain).

For example, the function $ f(t) \equal{} (\cos t, \; \sin t)$ only maps onto the unit circle, but we could introduce it as $ f : \mathbb{R} \to \mathbb{R}^2$ because it takes inputs from the set $ \mathbb{R}$ and produces an element of set $ \mathbb{R}^2$.

Of course, we can introduce a function in a way so that the co-domain is equal to the range, but that is not always the most useful way to present a function.

Also, it is sometimes helpful to cast a function into a larger co-domain when presenting a general result. For example, if a theorem says, for all continuous $ f : \mathbb{R} \to \mathbb{R}^2$, the theorem refers to any function whose graph is a continuous curve in the plane expressed in one real parameter.

For example, $ f: \mathbb{R}\to\mathbb{R}: \ x\mapsto x^2$ has codomain $ \mathbb{R}$, and range $ \{f(x)|x\in \mbox{dom}(f)\} \equal{} \mathbb{R}_{\geq 0}$ (the set of non-negative real numbers). So in this case the range is a (proper) subset of the codomain. This implies f is not surjective.
The function $ g: \mathbb{R}\to\mathbb{R}_{\geq 0}: \ x\mapsto x^2$ has codomain $ \mathbb{R}_{\geq 0}$ and range $ \mathbb{R}_{\geq 0}$, so codomain and range are equal. This implies g is surjective.
Summarizing: $ \mbox{range}\subseteq\mbox{codomain}$


induction
combinatorics proposed
combinatorics

Given an alfabet of $n$ letters. A sequence of letters such that between any 2 identical letters there are no 2 identical letters is called a [i]word[/i].

a) Find the maximal possible length of a [i]word[/i].

b) Find the number of the [i]words[/i] of maximal length.

I'm sure I've misinterpreted at least one of the conditions in some way, but here goes:

a) Obviously there can't be four identical letters (in $a \ldots a \ldots a \ldots a$, the outer $a$'s contain the inner $a$'s), so the maximal length is not more than $3n$, but $111222\ldots nnn$ works, so it is $3n$. Is this really it?

b) We can order the letters by order of appearance in a word: for example, in $ccbcbabaa$, the first letter to appear is $c$, the second is $b$ and the third is $a$. Given a word, we can permute the kinds of letters in any way (i.e. $(c, b, a)$ to $(a, b, c)$, $(a, c, b)$, $(b, a, c)$, $(b, c, a)$ or $(c, a, b)$), so the result will be a multiple of $n!$.

Now suppose that the letters are ordered, so the first to appear is $1$, the second is $2$ and so on (thus losing the $n!$ factor). We'll show that the word begins with $11$. Indeed, it begins with $1$; if there's some letter $a$ between the initial $1$ and the next $1$, then the other two $a$'s must be after the second $1$. If they are after the third $1$ too, then there's a pair of $a$'s that contains a pair of $1$'s, while if one of them is before the third $1$, the first and last $1$'s contain two $a$'s. Thus there's no $a$ between the first two $1$'s and the word begins with $11$.

Now calculate $f(1)=1, f(2)=2, f(3)=4$ by hand. We conjecture $f(n)=2^{n-1}$. To prove it by induction, assume this is the case for $n-1$. Given a word of length $n$, remove the three $1$'s; the resulting sequence is a word, and it is a word of length $3n-3$, so it must begin with $22$. It now follows that the original word began with $11122$ or $112122$ (the third $1$ can't be after two or more $2$'s), and this is also enough for the word to be valid. Thus the number of words is, in total, $n! \times 2^{n-1}$.

Yes, that's it.  I think this question is not original, however -- it has appeared on the forum [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76254]here[/url], labeled MOP homework from 2002.


A regular polyhedron with $ f$ faces, $ v$ vertices and $ e$ edges is made by cutting a regular tetrahedron with an edge length of one unit from each of the four corners of a tetrahedron with an edge length of three units. Compute $ f\plus{}v\plus{}e$.

For every vertex on the original tetrahedron, there will be three after cutting it, so $ v \equal{} 3\cdot4 \equal{} 12$. Also, we will add a face for each cut vertex, so $ f \equal{} 4 \plus{} 4 \equal{} 8$. By Euler's formula, $ f \plus{} v \minus{} e \equal{} 2\implies e \equal{} f \plus{} v \minus{} 2\implies f \plus{} v \plus{} e \equal{} 2(f \plus{} v \minus{} 1)$
$ \equal{} 2(8 \plus{} 12 \minus{} 1) \equal{} 2\cdot19 \equal{} \boxed{38}$.


Just curious about the prince (and the princess) who attended the closing ceremony... which positions are they in? I mean, prince and princess of what? I couldn't remember very well.

Also, does anyone know about a lady called Eva Almunia, who was also one person to award medals? Which position is she in?

I think those were the actual Prince and Princess of Spain (ie when the King dies, they become King and Queen).

They are the Prince and Princess of Asturias (title given to the heir of the throne... equivalent to "Prince and Princess of Wales" in the UK)

The Prince of Asturias, Felipe de Borbón, eldest son of King Juan Carlos, is expected to be the king of Spain when his father dies. The Princess is his wife, the journalist Letizia Ortiz.

I see... and what about Eva Almunia? I remember she's something like minister of Education, but I'm not sure that of the whole Spain or not. The google search doesn't give me "certain" information. So can anybody please help?

[quote="28121941"]Eva Almunia is State Secretary of Education of the Ministry of Education, that means, more or less, she is the number 2 of the Ministery.[/quote]

You mean for the whole country, right? Not just some part of the country.

So it's like deputy minister?

Yes, I means for the whole country. The Ministery is of Education, Social Policy and Sport, and therefore there are three deputy ministers, one for each.

Hello, This my first post just for check........ :roll:


My attire is noiseless when I tread the earth,
Rest in its dwellings or ride its waters.
At times my pinions and the lofty air,
Lift me high o'er the homes of men,
And the strength of the clouds carries me far
High over the folk.
My feathers gay
Sound and make music, singing shrill,
When no longer I linger by field or flood,
But soar in the air, a wandering spirit.

[hide="A guess"]Some sort of bird?[/hide]

True.
The answer is :
[hide]wild swan[/hide]


Euler
function
inequalities
number theory proposed
number theory

Let $s < 1$ be a positive real number. Show that there exist only finitely many positive integers $n$ such that $\varphi(n) < n^s$.

Hence, or otherwise, study the convergence of the series $\sum_{n = 1}^{\infty} \frac{1}{\varphi(n)^a}$.

[quote="Arne"]Show that there exist only finitely many positive integers $n$ such that $\varphi(n) < n^s$.[/quote]

Isn't that a consequence of $\varphi(n) > n - \sqrt{n}$ ?

Your inequality $\varphi(n) > n - \sqrt{n}$ is wrong (take $n = 12$ or something).

Duh... Actually $\varphi(n) \leq n - \sqrt{n}$ for composite $n$.
I'm sorry ... :(

Can you give a little proof of that inequality ?

$\varphi(n)=n\prod_{p|n} (1-1/p) < n^s \Longrightarrow \prod (1-1/p) <\prod p^{s-1} \Longrightarrow \prod (p-1)/p^s <1$.
There are finetely number of primes p, suth that $p-1<p^s$. Let $A=\prod_{p-1<p^s} (p-1)/p^s$. There are finetely primes, suth that $A(p-1)/p^s<1$. Therefore there are finetely number of sets S prime divisors n, suth that $\prod_{p \in S} (1-1/p)< \prod_{p \in S} p^s$.  There are finetely natural a(p), suth that $n=\prod_{p \in S} p^{a(p)}, \varphi(n)/n^s=\prod_{p \in S}(1-1/p)p^{a(p)(1-s)} <1$. It proved that there are finetely n, suth that $\varphi(n)<n^s$.
Therefore  $\sum_{n = 1}^{\infty} \frac{1}{\varphi(n)^a}<\infty$ if and only if a>1.


Find $x, y \in \mathbb{Z}$ such that

$x^{2}+y^{2}+6xy+1 = 6x+2y$

[hide="solution"]Rewrite it as $x^{2}+6x(y-1)+(y-1)^{2}=0$ then

$(x+3(y-1))^{2}=8(y-1)^{2}$

So the only possibility for integer solution is $y=1$ which will give us $x=0$[/hide]


probability
modular arithmetic
complementary counting

Problem 18
2008 non-zero digits <10 are generated, find the probability that their product is divisible by 10.

Answers 23
A) $\frac{10^{2008}-9^{2008}-8^{2008}+7^{2008}}{3^{2008}}$

B) $\frac{10^{2008}-5^{2008}-8^{2008}+4^{2008}}{10^{2008}}$

C) $\frac{9^{2008}-8^{2008}-5^{2008}+4^{2008}}{9^{2008}}$

D) $\frac{9^{2008}+8^{2008}-5^{2008}+4^{2008}}{3^{2008}}$

E) $\frac{9^{2008}-8^{2008}-5^{2008}-4^{2008}}{5^{2008}}$
[hide="Solution 18"]
The numbers are divisible by ten if and only if it has an even number and a 5 in its digits.
The probability, over 2008 digits, that it has no five, is $\left(\frac{8}{9}\right)^{2008}$, and the probability that it has no even is $\left(\frac{5}{9}\right)^{2008}$

The probability that it has no 5 AND no even number is $\left(\frac{4}{9}\right)^{2008}$, so by complementary counting, and the principle of inclusion-exclusion, the probability is
$\boxed{\frac{9^{2008}-8^{2008}-5^{2008}+4^{2008}}{9^{2008}}}$, or choice (C)
[/hide]

[quote="DiscreetFourierTransform"] Problem 18
2008 non-zero integers are generated, find the probability that their product is divisible by 10. [/quote]

First, you might need to fix your LaTeX. None of it is showing up.

[hide="Now, to the problem:"]
It'll be much easier to count the complement. We need the probability that not both 2 and 5 divide the product. Thus, we can PIE it. It will be:
\[1-( ( \frac{1}{2})^{2008}+( \frac{4}{5})^{2008}-( \frac{1}{2}\cdot \frac{4}{5})^{2008}) )= \boxed{ \frac{ 10^{2008}-5^{2008}-8^{2008}+4^{2008}}{10^{2008}}}\]
[/hide]

[quote="DiscreetFourierTransform"]Yeah, I'm not sure what happened. I posted them 5 minutes before lunch was over.
I'll fix them now.

And cincodemayo, a small hint:
[hide="hint"]
NON-ZERO
[hide="big hint"]
Instead of using 10 (5 even and 5 odd), there are only 4 even and 5 odd.
[/hide]
[/hide][/quote]

What? You didn't give any restriction on the numbers other than they're in $\mathbb{Z^{+}}$. So, I just count in $\pmod 2$, $1,0;1,0;1,0 \ldots$, or in $\pmod 5$, $1,2,3,4,0;1,2,3,4,0; 1, \ldots$.

Now, were you to set the restriction of $x_{i}< 10$, then yes, the answer would be what your solution gives. But there is no restriction, thus my answer is correct.

Ooo. Correct. Ok, I'll make that distinction.


function
abstract algebra
real analysis
real analysis unsolved

Suppose that $ (x_n)$ is a sequence of real numbers. Define a sequence $ (y_n)$ by $ y_n=\frac{x_n+x_{n+1}}{2}$ $ \forall n \in %Error. "texbb" is a bad command.
{N}$
(a) Prove that $ (y_n)$ converges to a real number $ x$ if $ (x_n)$ converges to $ x$.
(b) If $ (y_n)$ converges, does $ (x_n)$ also converge?

The answer to b) is clearly no.  The function that takes $ x_n$ to $ y_n$ is linear and has a kernel spanned by $ x_n \equal{} (\minus{}1)^n$.

a) is an epsilon-delta exercise.


In a triangle $ ABC$ incircle touches the sides $ AB$, $ AC$ and $ BC$ at $ C_1$, $ B_1$ and $ A_1$ respectively. Prove that $ \sqrt {\frac {AB_1}{AB}} \plus{} \sqrt {\frac {BC_1}{BC}} \plus{} \sqrt {\frac {CA_1}{CA}}\leq\frac {3}{\sqrt {2}}$ is true.

Let $ AC_1 \equal{} AB_1 \equal{} x, BC_1 \equal{} BA_1 \equal{} y$ ,and $ CA_1 \equal{} CB_1 \equal{} z$
It's equivalent to prove that $ \sum_{cyc}\sqrt {\frac {x}{x \plus{} y}} \le \frac {3}{\sqrt {2}}$
which is well-known,
\[ \sum_{cyc}\sqrt {\frac {x}{x \plus{} y}} \equal{} \sum\frac {\sqrt {(xy \plus{} xz)(z \plus{} x)}}{\sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}} \le \frac {\sqrt {2(xy \plus{} yz \plus{} zx)}\sqrt {2(x \plus{} y \plus{} z)}}{\sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}} \equal{} 2\sqrt {1 \plus{} \frac {xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}} \le \frac {3}{\sqrt {2}}
\]

rewrite it as $ \sum \sqrt{\dfrac{1}{1\plus{}m^2}}\leq \dfrac{3}{\sqrt 2}$, where $ m,n,p$ are positive reals whose product is $ 1$... suppose wlog that $ mn\leq 1$... it's easy to prove that $ \sqrt{\dfrac{1}{1\plus{}m^2}}\plus{}\sqrt{\dfrac{1}{1\plus{}n^2}}\leq 2\sqrt{\dfrac{1}{1\plus{}mn}}$ if $ mn\leq 1$...

then, we're left to prove that $ 2\sqrt{\dfrac{1}{1\plus{}t}}\plus{}\sqrt{\dfrac{t^2}{1\plus{}t^2}}\leq \dfrac{3}{\sqrt 2}$...

after squaring it once, it follows it's equivalent to $ 7t^3\minus{}t^2\plus{}9t\plus{}1\geq 8t\sqrt{(t\plus{}1)(t^2\plus{}1)}$...

from am-gm we have that $ 4t(t^2\plus{}t\plus{}2)\geq 8t\sqrt{(t\plus{}1)(t^2\plus{}1)}$ and $ 7t^3\minus{}t^2\plus{}9t\plus{}1\geq 4t(t^2\plus{}t\plus{}2)$, so we're done  :D

[quote="campos"]rewrite it as $ \sum \sqrt {\dfrac{1}{1 \plus{} m^2}}\leq \dfrac{3}{\sqrt 2}$, where $ m,n,p$ are positive reals whose product is $ 1$... suppose wlog that $ mn\leq 1$... it's easy to prove that $ \sqrt {\dfrac{1}{1 \plus{} m^2}} \plus{} \sqrt {\dfrac{1}{1 \plus{} n^2}}\leq 2\sqrt {\dfrac{1}{1 \plus{} mn}}$ if $ mn\leq 1$...

then, we're left to prove that $ 2\sqrt {\dfrac{1}{1 \plus{} t}} \plus{} \sqrt {\dfrac{t^2}{1 \plus{} t^2}}\leq \dfrac{3}{\sqrt 2}$...

after squaring it once, it follows it's equivalent to $ 7t^3 \minus{} t^2 \plus{} 9t \plus{} 1\geq 8t\sqrt {(t \plus{} 1)(t^2 \plus{} 1)}$...

from am-gm we have that $ 4t(t^2 \plus{} t \plus{} 2)\geq 8t\sqrt {(t \plus{} 1)(t^2 \plus{} 1)}$ and $ 7t^3 \minus{} t^2 \plus{} 9t \plus{} 1\geq 4t(t^2 \plus{} t \plus{} 2)$, so we're done  :D[/quote]
very nice inequality,
put that $ a^2\equal{}\frac{x}{y}$ ... the inequality is equiavalent as:
$ \sum \frac{1}{\sqrt{1\plus{}a^2}} \leq \frac{3}{\sqrt{2}}$ with $ abc\equal{}1$ the inequality is symetric then we can assume that $ a\geq b\geq c$ the $ bc\leq1$ From Cauchy Schwarz we have:
$ \frac{1}{\sqrt{1\plus{}b^2}}\plus{}\frac{1}{\sqrt{1\plus{}c^2}} \leq \sqrt{2}.\sqrt{\frac{1}{1\plus{}b^2}\plus{}\frac{1}{1\plus{}c^2}}
\equal{}\sqrt{2}.\sqrt{1\plus{}\frac{1\minus{}b^2c^2}{b^2\plus{}1)(c^2\plus{}1}} \leq \sqrt{2}.\sqrt{1\plus{}\frac{1\minus{}b^2c^2}{(bc\plus{}1)^2}}\equal{} \frac{2}{\sqrt{1\plus{}bc}}$ and:
$ \frac{1}{\sqrt{1\plus{}a^2}} \leq \frac{\sqrt{2}}{1\plus{}a}\equal{} \frac{bc\sqrt{2}}{bc\plus{}1}$ then it's suffice to prove that:
$ \frac{bc\sqrt{2}}{bc\plus{}1} \plus{} \frac{2}{\sqrt{1\plus{}bc}} \leq \frac{3}{\sqrt{2}}$
$ \Leftrightarrow \left(\sqrt{\frac{2}{1\plus{}bc}}\minus{}1\right)^2 \geq 0$


invariant
superior algebra
superior algebra unsolved

Any two cycles in Sn are cojugate if and only if they are of the same length.

Which direction of the implication are you having difficulty with?

What does conjugation by $ \sigma$ do to a permutation $ \phi$ in cycle structure? The elements that get moved are permuted by $ \sigma$; where $ \phi$ mapped $ a$ to $ b$, $ \sigma\circ\phi\circ\sigma^{\minus{}1}$ maps $ \sigma(a)$ to $ \sigma(b)$.
You should be able to explicitly construct the permutation for the positive direction. For the negative, there are a lot of different invariants to follow.

More generally, two elements of $ \mathbf{S}_{n}$ are conjugate iff their representations as a superpositions of disjoint cycles contain the same number of cycles of each length. If the condition is fullfilled, the permutation showing the conjugacy is easy to construct, according to what [b]jmerry[/b] said. If it is not fulfilled, it is easy to show that the spectra of matrices corresponding to the permutation are different (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209600]this topic[/url]) and thus they are not conjugate.


continued fraction
calculus
calculus computations

If $ \ \left\{ a_{n}\right\}$ is the Fibonacci sequence,$ \left( a_{n \plus{} 1} \equal{} a_{n} \plus{} a_{n \minus{} 1},a_{1} \equal{} a_{2} \equal{} 1\right)$ and $ \left( r_{n} \equal{} \frac {a_{n \plus{} 1}}{a_{n}},r_{n \plus{} 1} \equal{} 1 \plus{} \frac {1}{r_{n}}\right)$
then $ \ r_{n} < \frac {1 \plus{} \sqrt {5}}{2}$ for: $ n > 2$

Please, give a proof. thanks a lot.

===========================================

thank you, i fixed this error.

[quote="IndoChina"]Hmm...what you have stated is not always true. Please check the first few terms of $ r$.[/quote]
 Example $ r_2\equal{}\frac{a_3}{a_2}\equal{}2 >\frac{1\plus{}\sqrt{5}}{2}$

$ r_n$ are the convergents of the continued fraction for $ \frac{1\plus{}\sqrt5}2$ (You can prove this by induction). So they converge to it and alternate being less than and greater than it.


Is there anyone who has solutions of the book :
Matrix Theory author: Fuzhen Zhang

If you got the book's solutions, please contact me
cerendemircan@yahoo.com

Do you have this book with solutions on PDF? :)
 Can you please send it to me to the following e-mail adress:

                         lupucezar@yahoo.com


Let $log_ax=y$ ; we have $y\geqslant 1$ to respect the definition of radical.After that the second member is positive and our inequality may be squarred in  finally

      $4y^2-6y+3\geqslant 0$
wich is true for every $y$ because the determinant $6^2-4*4*3$ is negatif
Finally $y\geqslant 1$ is the unique condition then $x\geqslant a$ is the solution of this inequation.


Suppose an experiment consists of tossing a fair coin six times. What
is the probability of observing at most two tails? Express your answer
as a common fraction.

How'd you do that? Please explain. I notice the three numbers you added were $ \dbinom{6}{1}, \dbinom{6}{2}, \dbinom{6}{3} $. But the problem stated at most two tails! ???

[hide]
You can get $0$ tails, for 1 possibility. You can get one tail, for 6 possibilities. You can get 2 tails, for 15 possibilites. There are a total of 64. The answer is 11/32. @ nowunkie, I don't see how 1 is $\dbinom{6}{1}$ or how 6 is $\dbinom{6}{2}$. They are $\dbinom{6}{0}$ and $\dbinom{6}{1}$ respectively.
[/hide]


linear algebra
matrix
abstract algebra
algebra
polynomial

Suppose I want to calculate the Jordan normal form of a matrix with entries in an algebraically closed field. The text I am reading proves the Jordan normal form with an abstract argument considering $ F[x]$-modules, and contains little info about how to actually calculate it.

From following the arguments, I draw the following conclusions. Would you please indicate if they are correct?
1) The main diagonal has the roots of the characteristic polynomial
2) The number of 1s I should put above an eigenvalue $ \lambda$ is equal to the difference between the multiplicity of $ \lambda$ as a root to the characteristic polynomial and the dimension of the nullspace $ N(A \minus{} \lambda I)$

And a question 3) How do I determine exactly the jordan blocks? Jacobson does it by looking at Smith normal forms for the $ F[x]$-modules, but this seems much more difficult for small matrices.

Is it true that $ \dim N( (A \minus{} \lambda I)^s) \minus{} \dim N( (A \minus{} \lambda I)^{s \minus{} 1})$ is the number of jordan blocks with at least size $ s$ for the eigenvalue $ \lambda$? In particular, the values of $ \dim N( (A \minus{} \lambda I)^s)$ determine the block structure completely?

C'mon, someone has to know this. I have tried googling it, but everything speaks in terms of things I do not know, since I don't know linear algebra.


The terms of an interesting sequence are defined as follows:
$ T_{1}= 1$
$ T_{2n}= T_{n}$
$ T_{2n+1}=T_{2n}+1$

a) Among the first $ 1,000,000$ terms, what is the maximum value?
b) How many times does this maximum value occur?
c) What are the values of $ n$ when $ T_{n}$ is maximum?