We have the sequences (x_n) and (y_n) n>=0 with y_n=1/2^n sum(k=0,n) (n,k)*x_k for any n natural. Show that if (x_n)n>=0 converges then (y_n) also converges and lim (n-->oo)y_n= lim (n-->oo)x_n.

where (n,k)=n!/[k!(n-k)!]

This one is trivial. First of all, we can obviously suppose that lim x_n=0. Let p>0 and k such that for all n>k we have |x_n|<p/2. Let M such that for all n>M we have nC0*|x_0|+nC1*|x_1|+...+nCk *|x_k|<p/2*2^n. Then for all n>max{k,M} we have |1/2^n*sum nCk*x_k|<=1/2^n*(nC0*|x_0|+nC1*|x_1|+...+nCk*|x_k|)+p/2<p. Thus limy_n=0.


Also, I don't know where to ask about this question, I'll try here and  moderators or administrators  should try to do something for this  technical problem:
In my Sentbox or Savebox messages there is no such operation as  "Quote" or "Edit" message, thus I can't open latex code of my messages and copy  message in latex form.
Actually we don't need such operation as "Quote". it's sufficient to    add such operation as "Edit", from this we can copy original text of our messages and  to post somewhere :)

In the savebox, there is a quote button in the top right hand corner. But I don't know about the sentbox. Also, is there a way to "unsave" a message?

EDIT: By "unsave" I mean to put back into the inbox, not delete.

In "sentbox" and "[b]savebox[/b]" there  is no quote button in the top right hand corner  :(

[quote="PowerOfPi"]Also, is there a way to "unsave" a message?[/quote]
You could delete it, but I don't know if that is "unsaving".


Given any $2n$ integers, show that there are $n$ of them whose sum is divisible by $n$.

I think everyone would think of $\mathbb{Z}_{n}$. Could you give another hint :)

$\mathbb{Z}_{n}=\{0,1,\hdots,n-1\}$, i.e., the set of all congruences modulo $n$.

Any ideas? I tried looking at the congruences mod n, but there are a lot of cases...

This is just an idea but...

What if you apply pigeon hole repeatedly: you know you need $\lfloor\frac{n}{2}\rfloor+1$ numbers to find $2$ of which have sum divisible by $n$.  Take those two numbers out of the batch and repeat with $2n-2$ numbers.

Do you think this have any chance?


I'm [i]guessing[/i] that you don't entirely understand the steps.

[hide="Explanation"]You have the sum $ S \equal{} \frac {6}{27} \plus{} \frac {30}{729} \plus{} \ldots$ (ignore the $ 1$ for now).

This is an infinite geometric series, so $ S \equal{} \frac {a}{1 \minus{} r}$, where $ a$ is the first term and $ r$ is the ratio between terms.

$ a \equal{} \frac {6}{27},\quad r \equal{} \frac {5}{27}\quad\Longrightarrow\quad S \equal{} \frac {\frac {6}{27}}{1 \minus{} \frac {5}{27}} \equal{} \frac {3}{11}$

Now, add the $ 1$ that we temporarily ignored: $ 1 \plus{} \frac {3}{11} \equal{} \frac {14}{11}$.[/hide]

i do not input \begin.

$ \begin{eqnarray}$

\begin{eqnarray}volume & = & 1 + (\frac {6}{27} + \frac {30}{729} + \cdots)\nonumber \\
& = & 1 + \frac {\frac {6}{27}}{1 - \frac {5}{27}}\nonumber \\
& = & \frac {14}{11}\nonumber\end{eqnarray}

Example Text:
This is just text but $ \sqrt {2}$ should be shown as an image and so should $ \frac {1}{2}$.

Another formula is \begin{eqnarray}\frac {43}{12} \sqrt {43}\end{eqnarray}

$ \documentclass[11pt]{article}$


For a positive integer $ n>2$, let $ \left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a sequence of $ n$ positive reals which is either non-decreasing (this means, we have $ a_{1}\leq a_{2}\leq \ldots \leq a_{n}$) or non-increasing (this means, we have $ a_{1}\geq a_{2}\geq \ldots \geq a_{n}$), and which satisfies $ a_{1}\neq a_{n}$. Let $ x$ and $ y$ be positive reals satisfying $ \frac{x}{y}\geq \frac{a_{1}-a_{2}}{a_{1}-a_{n}}$. Show that:
\[ \frac{a_{1}}{a_{2}\cdot x+a_{3}\cdot y}+\frac{a_{2}}{a_{3}\cdot x+a_{4}\cdot y}+\ldots+\frac{a_{n-1}}{a_{n}\cdot x+a_{1}\cdot y}+\frac{a_{n}}{a_{1}\cdot x+a_{2}\cdot y}\geq \frac{n}{x+y}. \]

Not increasing and not decreasing!?! Thats gotta be rearrangement or chebychev!

Very neat generalization of the ordered Shapiro inequality (discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1874 )!

For the proof, we will assume that the sequence $ \left( a_{1},\ a_{2},\ ...,\ a_{n}\right)$ is non-increasing, so that $ a_{1}\geq a_{2}\geq ...\geq a_{n}$. The case of a non-decreasing sequence is analogous and is left to the reader.

We will use cyclic indices modulo $ n$; thus, $ a_{n+1}$ will mean $ a_{1}$, and $ a_{n+2}$ will mean $ a_{2}$. Then, the inequality in question becomes

$ \sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq \frac{n}{x+y}$.

This is equivalent to

[color=green][b](1)[/b][/color] $ \left( x+y\right) \cdot \sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq n$.

Now, AM-GM yields

$ \sum_{i=1}^{n}\frac{a_{i}\cdot x+a_{i+1}\cdot y}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq n\sqrt[n]{\prod_{i=1}^{n}\frac{a_{i}\cdot x+a_{i+1}\cdot y}{a_{i+1}\cdot x+a_{i+2}\cdot y}}=n\sqrt[n]{1}=n$.

Hence, in order to prove [color=green][b](1)[/b][/color], it will be enough to show that

$ \left( x+y\right) \cdot \sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq \sum_{i=1}^{n}\frac{a_{i}\cdot x+a_{i+1}\cdot y}{ a_{i+1}\cdot x+a_{i+2}\cdot y}$.

This rewrites as

$ \left( x+y\right) \cdot \sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\sum_{i=1}^{n}\frac{a_{i}\cdot x+a_{i+1}\cdot y}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq 0$,

what is equivalent to

$ \sum_{i=1}^{n}\left( \left( x+y\right) \cdot \frac{a_{i}}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{a_{i}\cdot x+a_{i+1}\cdot y}{a_{i+1}\cdot x+a_{i+2}\cdot y}\right) \geq 0$.

This simplifies to

$ \sum_{i=1}^{n}\frac{\left( a_{i}-a_{i+1}\right) \cdot y}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq 0$.

Since $ y>0$, this is equivalent to

$ \sum_{i=1}^{n}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq 0$.

Now, we split the very last term (i. e. the term with $ i=n$) off this sum, and get

$ \sum_{i=1}^{n}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}=\sum_{i=1}^{n-1}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}+\frac{ a_{n}-a_{1}}{a_{1}\cdot x+a_{2}\cdot y}$
$ =\sum_{i=1}^{n-1}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{a_{1}-a_{n}}{a_{1}\cdot x+a_{2}\cdot y}$
$ =\sum_{i=1}^{n-1}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{\sum_{i=1}^{n-1}\left( a_{i}-a_{i+1}\right) }{a_{1}\cdot x+a_{2}\cdot y}$
$ =\sum_{i=1}^{n-1}\frac{a_{i}-a_{i+1}}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\sum_{i=1}^{n-1}\frac{a_{i}-a_{i+1}}{a_{1}\cdot x+a_{2}\cdot y}$
$ =\sum_{i=1}^{n-1}\left( a_{i}-a_{i+1}\right) \cdot \left( \frac{1}{ a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\right)$.

So we want to prove that this sum

$ \sum_{i=1}^{n-1}\left( a_{i}-a_{i+1}\right) \cdot \left( \frac{1}{ a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\right)$

is $ \geq 0$. We will do this by showing that each term of this sum is $ \geq 0$. In other words, we will show that $ \left( a_{i}-a_{i+1}\right) \cdot \left( \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\right) \geq 0$ for each $ i\in \left\{ 1,2,...,n-1\right\}$.

So pick some $ i\in \left\{ 1,2,...,n-1\right\}$. We want to prove that $ \left( a_{i}-a_{i+1}\right) \cdot \left( \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\right) \geq 0$. Since $ a_{1}\geq a_{2}\geq ...\geq a_{n}$ (here is the first time we use that the sequence $ \left( a_{1},\ a_{2},\ ...,\ a_{n}\right)$ is non-increasing), we have $ a_{i}\geq a_{i+1}$ and thus $ a_{i}-a_{i+1}\geq 0$. Now we will show that $ \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\geq 0$.

In fact, $ a_{1}\geq a_{2}\geq ...\geq a_{n}$ yields $ a_{1}\geq a_{n}$, so that $ a_{1}-a_{n}\geq 0$, and thus we can multiply inequalities by $ a_{1}-a_{n}$. Hence, the inequality $ \frac{x}{y}\geq \frac{a_{1}-a_{2}}{ a_{1}-a_{n}}$ (which is a condition of the problem) becomes

$ \left( a_{1}-a_{n}\right) \cdot x\geq \left( a_{1}-a_{2}\right) \cdot y$
$ \Longleftrightarrow$ $ a_{1}\cdot x-a_{n}\cdot x \geq a_{1}\cdot y-a_{2}\cdot y$
$ \Longleftrightarrow$ $ a_{1}\cdot x+a_{2}\cdot y\geq a_{n}\cdot x+a_{1}\cdot y$.

Now, $ a_{1}\cdot x+a_{2}\cdot y\geq a_{i+1}\cdot x+a_{i+2}\cdot y$ (in fact, since $ i\in \left\{ 1,2,...,n-1\right\}$, we must have either $ i\in \left\{ 1,2,...,n-2\right\}$ or $ i=n-1$; now, if $ i\in\left\{ 1,2,...,n-2\right\}$, then $ a_{1}\cdot x+a_{2}\cdot y\geq a_{i+1}\cdot x+a_{i+2}\cdot y$ is clear because $ a_{1}\geq a_{i+1}$ and $ a_{2}\geq a_{i+2}$ follow from $ a_{1}\geq a_{2}\geq ...\geq a_{n}$; but if $ i=n-1$, then $ a_{1}\cdot x+a_{2}\cdot y\geq a_{i+1}\cdot x+a_{i+2}\cdot y$ is equivalent to $ a_{1}\cdot x+a_{2}\cdot y\geq a_{n}\cdot x+a_{1}\cdot y$ what we have just proven). Thus, $ \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y}\geq \frac{1}{ a_{1}\cdot x+a_{2}\cdot y}$, so that $ \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y }-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\geq 0$. Multiplying this with $ a_{i}-a_{i+1}\geq 0$, we get $ \left( a_{i}-a_{i+1}\right) \cdot \left( \frac{1}{a_{i+1}\cdot x+a_{i+2}\cdot y}-\frac{1}{a_{1}\cdot x+a_{2}\cdot y}\right) \geq 0$. This completes the solution of the problem.

  Darij


Post your random thoughts here so we can all laugh at you!

AT LEAST YOU COULD HAVE TOLD ME TO KEEP MY VOLUME DOWN!!!!

i'm in the living room, and suddenly "badger, badger, badger" starts singing from my computer.  very weird.

The REAL question here is

Why do you have the volume turned all the way up?  :wink:

--shrug--

Maybe I have a annoying little brother who plays Webkinz/Club Penguin on my computer with the volume on high.

:noo:

When I was like 8 I played REAL computer games like Body Land, Math Blaster and Putt Putt Goes to the Parade. REAL GAMES, using as many as FOUR floppies.

if you think about the structure of an atom, modern model, it seems like its just a fast and small version of our solar system. our universe is a larger and slower equivalent to our solar system. Must that mean we are but in the midst of an endless model cycle?
perhaps atoms get smaller and smaller and contain miniworlds of intelligence
perhaps we are but people on an atom of greater intellegent beings.
Perhaps the universe itself is but an infinitesimally small or large level of being
either way, we are far too complex or simple to comprehend. Where does that put us? I dont want religion in here.
Now that we've realized this possibility, how can we test it?

But yet again, our microscopes can only go one atom small-tips. this means that we cant even detect what happens in single atoms, except by blowing them apart. Not a good approach, if you search for intelligent life. Blow it up. Yeah, right.

That doesn't make any sense.  Perhaps that intelligence in sub-atomic level can be broken down into simpler levels  that make up the intelligence and then somehow there is more intelligence found beneath those, maybe there is endless intelligence the smaller we look?

Dude like fractals.

WHAT IF FRACTALS BECOME SELF-AWARE?!?!?!?1

WHAT THE HECK?!?! << AND WHY IS THIS NOT CAPITALIZED?

Edit: NOW IT IS. << and for some reason now that is capitalized too. aops is so messed up right now!

If fractals became self-aware...they would undoubtedly plot for a takeover of the world. After all they have an endless amount of um intelligence.  Wait if fractals became self aware would they all be one giant thinking machine or would they have infinitely small different "brains" becoming one huge efficient fractal?

I'm thinking way too much about this.


The integers a,b and c satisfy a+b+c=0

Prove that $ 2a^4 \plus{}2b^4 \plus{}2c^4$ is a perfect square

Another way is to consider a,b, and c as being solutions of

$ x^3 \plus{} qx \minus{} abc \equal{}0$ where $ q \equal{} \sum ab$

Hence they are also solutions to

$ x^4 \plus{} qx^2 \minus{} abcx \equal{} 0$

Hence $ \sum a^4 \plus{} q\sum a^2 \minus{} abc\sum a \equal{} 0$

or $ \sum a^4 \equal{} \minus{} q\sum a^2$. But, $ \sum a^2 \equal{} \minus{} 2q$

Hence $ 2\sum a^4 \equal{} \minus{} 2q\sum a^2 \equal{} 4q^2$ which is a perfect square


Let $N$ be the greatest integer multiple of $8,$ no two of whose digits are the same. What is the remainder when $N$ is divided by $1000?$

[quote="SM4RT"]is it 120???[/quote][hide="Reply"]Yeah, the number is just 9876543120.[/hide]

yup thats what i had..hey all these questions seem familiar..

A bunch of these are from AIME's, I think...

[hide]Since mod 1000 is the same as finding the last three digits, we need to minimize the last three digits such that it is divisible by 8. Note that the last three digits of a multiple of 8 is divisible by 8. A simple check finds that 8*15 = 120 is the minimum value of the last three digits. Hence, $\boxed{120}$ is the answer. Note: the value of N is 9876543120.[/hide]

[Hide=Solution]Let's talk a bit about this number. We can say this number starts from 9876543, and the last three digits should be divisible by 8 because this number is a multiple of 8. We can say a little bit about this number, that is 9876543()()(). We now have to find largest positive number that is made of 0,1,2 and is a multiple of 8. By Guess, and work, we get it is 120, i.e. this number $ N = 9876543120 $. Now, N = 120 (mod 1000)


Sorry for being really stupid, but where is the link to make Geobra stuff? Someone said it was above the composite thingy, and what is the composite thingy?


[b]Find a stictly  increasing sequence $(x_{n})_{n\geq 0}$ of real numbers such that $x_{0}=2\; ,\; x_{1}= r \; \; , r$ being rational $> 2$,and  $(k+1)x_{k+2}=(k+2)x_{k}\; \; ,\; \; \forall k\in \{0,1,... \}.$
The question seems to be very simple, but I hav'nt found any such sequence.
If possible, please  give an example .

I'm afraid that the only value of $r$ for which this sequence is strictly increasing is $r=\pi$. However, pi is not a rational number.

I didn't try to prove this, I only wrote a program to compute the first couple of digits of $r$ but I don't think that the proof will be very complicated.

[quote="bus"]I'm afraid that the only value of $r$ for which this sequence is strictly increasing is $r=\pi$. However, pi is not a rational number.[/quote]
Right ! $r=\pi \not \in{\mathbb Q}.$ Please prove it !

Found it :).

[url]http://en.wikipedia.org/wiki/Wallis_product[/url]

Transforming the original question into question about convergence of Wallis' product is straightforward.


function
algebra
polynomial
rational function
complex analysis
complex analysis theorems

Hi There,

I'm still wondering about exponential decay in the real axis of subexponential meromorphic functions in twice a cut plane. I came across the Herglotz decomposition, that states the following:

If f(z) is meromorphic in twice a cut plane, hermitian analytic (f(z)=f(z*)*), polynomial bounded (|f|<z^n), has a finite number of poles and Imf(x) has a well defined limit as x->+-infty, then f(z) = R(z)H(z), with R(z) a rational function (P/Q) and H a herglotz function, e.g. ImH>0 for Imz>0, and C/|z| < |H(z)|. Which would be exactly what I need!
Except for the fact that I want Imf-> A, with 0<A<1, so I don't quite run into a contradiction, unless (Imf-A) has well defined limits, but what if it doesn't? What if Imf has well defined limit but  Imf(z)-A does not?

Looking at representations of Herglotz functions, unless A=0, I don't see any contradiction! But there must be a way to show this is IMPOSSIBLE!

thanks!
any good reference would also help :)

I think I'm slowing getting around the problem. I'm wondering if something like the minimal modulus theorem also applies to twice cut analytic functions. Let's say without poles nor zeros. I mean, is there a general theorem that states that if |f| < e^(|z|^p) then
log|f| > -r^(p+eps), for a function that may have branch points in the real axis? Under which conditions is this true?

I have in mind a function F(z)=e^{-sqrt{z}) which does not obey the theorem since in fact p~epsilon (unless I'm missing on something F(z) is polynomial bounded in the upper half plane), and it has no zeros nor poles in the upper half plane. It has a branch point at z=0, it also has infinite oscillations for ImF(-s+i0) over the cut. I could also imagine a function sqrt{z}(e^{-sqrt{z}}-A) to avoid the oscillations, so I'm not sure exactly how to play this game, maybe there is a constraint in p? I've noticed it doesn't work if p>1/2, so that's probably the way to go... any clue?

It's a hard problem eh? ;)


Hello,

as I have said a couple of times, apart from maths, I am a die hard movie fanatic.  I also really like Asian movies and Japanese animation is my cup of tea!

As these are movies (series) too, I think this thread belongs here (as it IS a form of art)?

I was wondering which mangas are really really good?
I have had the opportunity to enjoy :

Akira (the movie is very well done, but can't capture the whole comic book series)

Dragonball Z (you kinda outgrow this one, but it still is influential and worth watching)

Neon Genesis Evangelion (clever and thought provoking)

Rurouni Kenshin : interesting view on Japanese history, and a spectacular second saga, the four first ovas are an independent masterpiece

Elfenlied : intelligent and exciting, but I don't recommend watching this with a very full stomach or if you can't stand blood.  You have to take a  lot of nudity in account too.

So please, share your thoughts!

I can only think of three Rurouni Kenshin OAV's...  What's the fourth?

fredbel6, have you seen Trigun?

I'm sort of looking for another good anime series to watch...  There's a lot of unoriginal and immature anime out there, and I'm having trouble finding one that suits my tastes...  Does anyone have a favorite series or movie that they'd like to share?  Here's a few that I like that fredbel6 didn't mention:
The Twelve Kingdoms
Visions of Escaflowne
Fushigi Yuugi
Haibane Renmei
Trigun
Princess Mononoke, Nausicaa, etc. (anything Miyazaki can't be that bad :))

could you describe trigun (genre, number of episodes, basic plot) without spoilers?

Well, actually there is one nice kenshin movie consisting of four ova's , hence possible confusion

I just mean the whole movie depicting kenshin's youth during the bakumatsu, and the tomoe relationship (four parts of half an hour each)

there is also a ridiculous movie of kenshin situated in the same timeline as the episodes, which is horrible
and there is also a seisouhen, consisting of two half hour parts, with some reanimated scenes from the episodes, and some things with enishi...

OK, I know what you're talking about now.

Thematically, Trigun is similar to Rurouni Kenshin in many ways, but of course the plot and characters are completely different.  There are 26 episodes, and I've copied this basic plot description from imdb.com:
"Trigun takes place in the distant future on a deserted planet. Vash the Stampede is a gunfighter with a legend so ruthless that he has a $\$\$$60,000,000,000 bounty on his head. Entire towns evacuate at the rumor of his arrival. However, the real Vash the Stampede is not the same man that rumor portrays him to be. The enigmatic and conflicted lead character in Trigun is actually more heroic in nature; as well as a complete and utter idiot at times."
But believe me, it's a lot cooler than it may seem from that description.

if you were to describe the matureness of the series, where would you place it?

I'd say now : Dragonball < Kenshin (the episodes) < Neon Genesis Evangelion< Elfen Lied

(ending with most mature, on right)

I would be mostly interested in something quite serious (although some comic relief is always welcome)

I forgot to mention Appleseed
It has animation that made my jaw drop : the movements are done by placing sensors on real actors, and the city where most of the action takes place is gorgeous
However, they could've bothered to write a decent story, everything happens so quick and there are so many twists, emotions seem forced...

I'd say it's about as mature as Kenshin, but, unlike Kenshin, it is thankfully without many fillers (Kenshin's third season (as well as much of its first season) is completely unnecessary filler, but the second season makes up for it all).  Anyway, I definitely recommend Trigun if you liked Kenshin.

Is Love Hina any good? I've heard people talk about it?

yes, it's good, but I think in this case manga is better than the anime series.

have you seen or read naruto, bleach or full metal alchemist?

Naruto!!!!!!!!!!!  heh, that's probably the most overrated and yet one of the most bestest manga/anime ever made (though the US dub messed up the voices).  I love it still!!!!

Anything Studio Ghibli is also good (like Grave of Fireflies which is a must watch) and I personally really like Tsubasa Chronicles since the characters are so lovable.

If you like CGs then definitely watch the Final Fantasy Movies.  Other mangas/animes I've read/watched which have not been listed before and I think are pretty good are Negima, Saiyuki, DNAngel, ID_entity (which is a Korean manga), Chrono Crusade, and Fruits basket.  I also want to make a special note for the first manga i have ever read (when I was like 7) which I fell in love with: Cardcaptor Sakura.  It's really cute, but kiddy-ish.

FMA is amazing :)
Not a big fan of Bleach but I think Naruto is pretty good too :)

The Earl and the Fairy
The manga is A-MAZING!
:):):):):)

FULLMETAL ALCHEMIST. THE BEST EVER. READ. IT.

Ahahaha the last posts before this one are on March and April 22, this one is on May 22.

Anyways, Death Note is awesome, couldn't stop watching/reading it, and Hetalia never fails to make me laugh. Also, I like Skip Beat and Ouran High School Host Club for shojo, and Fullmetal Alchemist and Soul Eater have been recommended to me.

POKEMON!!!
being a former religious pokemon player, the anime brings back memories.

Well, well. Looking for manga? I am an otaku genius, and will recommend some great ones.
Movies:
Spirited Away-this movie always touches me. I watched it when I was about 5, and still watch it. Try to keep yourself from crying!
Wolf Children Ame and Yuki-I watched this recently, loved it. FEELS.
My Neighbor Totoro-Watch it, its fantastic
Princess Mononoke-I loved it!! A childhood fave.

Manga/Anime:
Horror:
Shingeki no Kyojin-A hot anime right now, really popular. Its gory but so interesting!
Sankarea-a zombie manga that is cute and fluffy, but still a bit gory at times.
High School of the Dead-A bit ecchi, but my favorite Zombie Apocalypse manga
Death NOte-My first manga, really logical. Makes you think.

Romance/Slice of Life/School:
My all time favorite is Inuyasha. I love it so much! Its really old though.
Love Hina-an old classic manga that I like to call the original Harem/Romance manga.
Nisekoi-This manga will have you laughing so hard, I loved it! Its still in the making.
Orange Marmalade-My favorite manwha.
Ranma 1/2-The funniest gender-bender.
Yamada and the Seven Witches-This one is good
Horimiya-So kawaii!!
Kamisama Hajimenshita-A really great one.
Skip Beat!-Awesome manga and anime.
Hetalia-SO hilarious.
Ouran-Genderbender hilarity.
Fruits Basket-Adorable and awesome art.
Kaichou wa Maid Sama-This one is really funny, but a bit draggy at times.
Kenshin is awesome, FYI. SO is fullmetal.

Shounen:
Fairy Tail
Naruto
Bleach
One Piece

^These ones I don't like as much. They're typical for boys, and very popular, but I don't like them as much.

most of the good Japanese manga have already been mentioned...

Korean manga (Manhwa):
Tower of God - fantasy, fighting [also full color]
The Breaker - martial arts, somewhat romance/harem

Chinese manga (Manhua):
Feng Shen Ji [TL: The Legend and the Hero] - just plain epic lol [full color too, and the illustrations are pretty good!]

all three are still in progress

@LIONBLAZE
wow you read a lot of manga
do you read any light novels?
-ex. Haruhi Suzumiya, SAO, Shakugan no Shana, etc. (Not my personal favorites, but those are some of the more popular/famous ones)

Yes, I am an expert of anime:D I have watched Haruhi Suzumiya, and plan to read it, and I have read Tower of God.
Oh, I can recommend a few more:
Kuroko no Basket
Deadman Wonderland
Ao no Exorcist
Cradle of Monsters
Cage of Eden-a bit ecchi, but superb!

I have read/watched 183 anime/mangas, not considering the ongoing ones like FREE! and Attack on Titan :)

Meh.
I've heard of a lot of these manga, but I'm probably never going to find time to read them all.
My preferred medium is light novels, but the translation for light novels is much slower than the translation for anime/manga because there's a lot more text (obviously)

There's a bunch of really good ones that aren't adapted into anime or manga yet.
Have you heard of baka-tsuki? That's probably the largest collection of LN translations online.
Some of my favorite works:
Tabi ni Deyou [standalone]
Iris on Rainy Days [standalone]
Mahouka Koukou no Rettousei series
Tasogare-iro no Uta Tsukai series
Sayonara Piano Sonata series [TL by cetranslation.blogspot.com]
Hyouketsu Kyoukai no Eden series [V 1-3 TL by cetranslation, v4 TL by sakurahonyaku.wordpress.com]
Kikou Shoujo wa Kizutsukanai series [TL by nanodesutranslations.wordpress.com]
Yahari Ore no Seishun Love Come wa Machigatteiru (Oregairu) series [TL by nanodesu]

Fate/Zero... usually I don't like horror, but I have to admit that this series was masterfully written. The baka-tsuki translation disgraces it though.
There's an anime for this, too, as well as for Oregairu. Kikou Shoujo wa Kizutsukanai (Unbreakable Machine Doll) is getting an anime in October of this year, or so I've heard. Mild ecchi, but still pretty good.

But my favorite LN series by far is Utsuro no Hako to Zero no Maria
Not many people have heard of it, which makes me kind of sad, but I admit you would need rather unique tastes to enjoy that series.
It's kind of tough to come up with a description of it. The only genre that it definitely falls under is probably psychological. It's also technically a romance, but not the kind you usually think of. The author also makes wonderful use of limited first person perspective.

Yes, yes, yes, no, yes , no, no, yes.

Guys! Watch Angel Beats. Now!

Okay. Angel Beats is amazing. While you are at it, so are Clannad (YESYESYES) and Kanon.

Angel Beats made me cry. Clannad did too.

MIRAI NIKKI IS AMAZING.
Elfen Lied, Higurashi, Sword Art Online.


trigonometry
limit
calculus
calculus computations

if f(0)=0

g(0)=1

f'(x)=g(x)

g'(x)=f(x)

then value of lim x--pi/4   f(tan x)+f(cot x)

kenn4000, where you went wrong was in mistaking the differential equations $ f''\equal{}f$ and $ f''\equal{}\minus{}f.$ The equation $ f''\equal{}\minus{}f$ has $ \cos x$ and $ \sin x$ as solution, but the equation $ f''\equal{}f$ has $ \cosh x$ and $ \sinh x$ as solutions.

But $ 2\sinh 1\equal{}e\minus{}e^{\minus{}1},$ not $ e.$

Perhaps ayush_2008 intended the final limit to have one term that was $ f$ and one term that was $ g$ rather than having them both be $ f.$

Hello
[hide="hint"]
the general form of f(t) =$ C_1e^{t} \plus{} C_2e^{\minus{}t}$
$ f(0)$=$ C_1 \plus{}C_2$=0
$ f'(0)$=$ C_1 \minus{} C_2$=1
thus our eq. is $ \frac{1}{2}[e^t\minus{}e^{\minus{}t}]$
[/hide]
Thank u.

Hello
[hide="reply"]
$ y''\equal{}y$
The general solution of this equation depends on the character of the roots of the characteristic equation that is $ r^2\equal{}1$ thus u have $ r\equal{}\minus{}1$  and $ \plus{}1$
now if we have distinct roots suppose p and q
the general eq is in the form $ C_1e^{px} \plus{}C_2e^{qx}$
in ur ques
the eq. is $ C_1e^{t} \plus{} C_2e^{\minus{}t}$

[/hide]
Thank u.


If 6 blops = 1 glop and 5 glops = 7 flops, how many blops are equal to 21 flops?

Work backwards on this problem.
If 5 glops=7 flops, then 15 glops equals 21 flops.
Then, if 6 blops=1 glop, then 90 blops=15 glops=21 flops.
The answer is 90.

I did it slightly differently.

If 6 glops are the same as 1 glop, we find that:

$ \frac{6 \text{ blops}}{1 \text{ glop}} \equal{} \frac{x \text{ blops}}{5 \text{ glops}}$

30 blops are equal to 5 glops, so 30 blops = 7 flops. We want to find how many blops are equal to 21 flops, so $ 30 \times 3 \equal{} \boxed{90 \text{ blops}}$.


counting
distinguishability
search
articles
number theory
Diophantine equation

suppose you have a rectangular tray of brownies, and a bunch of people who have strong opinions about the relative goodness of edge brownies and center brownies.

how many rows and columns can you cut the brownies into so there are the same amount of edge brownies as center brownies?

(example: if you cut it into 4 rows and 5 columns, there are 14 edge brownies (because corner ones count as edge brownies) and 6 center brownies.)

[hide]the only ones I found are $5 \times 12$ and $6 \times 8$[/hide]

[hide]Let there be $r$ rows and $c$ columns.  We have $2(r-2)(c-2)=cr$, or $2cr-4c-4r+8=cr$.  It's just rearranging the terms to get a Diophantine equation:
$8-4r=4c-cr$
$8-4r=c(4-r)$
$c=\frac{8-4r}{4-r}$
We know that $r,c\ge3$.  The reason is common sense.  If this were not so, there'd be no center pieces.  Scanning for positive values of $c$ as we go through $r$, we find acceptable values of 5, 6, 8, and 12.  Possible values are $5\times12$ and $6\times8$.[/hide]

[hide="Solution"] Let $r$ be the number of rows and $c$ be the number of columns. Now each row has 2 edge brownies and each column has 2 edge brownie, but the 4 corner brownies were counted twice. So the number of total edge brownies is:
$2r+2c-4$.
The number of center brownies is just the total number of brownies minus the edge brownies:
$rc-(2r+2c-4) = rc-2r-2c+4$.
Since we want to have the same number of edge brownies as center brownies:
$2r+2c-4=rc-2r-2c+4$
$rc-4r-4c+8=0$
$rc-4r-4c+16=8$
$(r-4)(c-4)=8$
Because $r$ and $c$ are indistinguishable (just turn the brownie tray sideways), we have these 2 solutions:
$r-4=1, c-4=8$, so $r=5, c=12$
$r-4=2, c-4=4$, so $r=6, c=8$
[/hide]

[hide]m is row's, n is column's. The number of edge brownies=total brownies-center brownies=center brownies, or
mn=2(m-2)(n-2)
mn=2mn-4m-4n+8
0=mn-4m-4n+8
8=(m-4)(n-4) from Simon's Favorite Factoring Trick.
We get our solutions 5X12, and 6X8.[/hide]

variation of the problem: what if you wanted there to be $k$ times as many center brownies as edge brownies?

[hide="Variation Solution"] Let $r$ be the number of rows and $c$ be the number of columns. As before, the number of edge brownies is $2r+2c-4$ and the number of center brownies is $rc-2r-2c+4$. If we want there to be $k$ times as many center brownies as edge brownies, then we are looking for integer solutions to the equation
$k(2r+2c-4)=rc-2r-2c+4$
$rc-2(k+1)r-2(k+1)c+4(k+1)=0$
$rc-2(k+1)r-2(k+1)c+4(k+1)^{2}=4(k+1)^{2}-4(k+1)$
$(r-2(k+1))(c-2(k+1))=4k(k+1)$
From there you can just find the prime factors of $4k(k+1)$ and find the solutions in $r$ and $c$.[/hide]

right.  and we know there's always multiple solutions, because $4k(k+1)$ has a bunch of factors.

[quote="bpms"]Simon's favorite factoring trick is the key.[/quote]
What is that?

(see the [url=http://www.artofproblemsolving.com/Wiki/index.php/Special:Search?search=simon%27s+favorite+factoring+trick]aops wiki article about it[/url].)  it's a way to solve equtions whose solutions are integers, like for example:

$xy-y-x-9=0$
$xy-y-x+1=10$
so you factor it:
$(x-1)(y-1)=10$

and then you use the factors of 10 to find possible values for x and y. it's pretty cool.  :D


A simpler approach: We want $ x_1\plus{}x_2>x_3$, with $ x_1\plus{}x_2\plus{}x_3\equal{}2n\plus{}1$. This is equivalent to $ x_1\plus{}x_2\ge n\plus{}1$, or $ x_3\le n$. Let $ z_i\equal{}n\minus{}x_i$; we have $ 0\le z_i$ for each $ i$, and $ z_1\plus{}z_2\plus{}z_3\equal{}n\minus{}1$. That's $ \binom{n\plus{}1}{2}$ options.

Where did you go wrong? Your $ y_i$ must be even, so you're consistently overcounting.

Jmerry: do you have the problem archvied somewhere?
Jrb599: This is a warning.

  darij


You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 7 seconds. The depth of the well is ________ feet.

Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.

set up and solve a quadratic equation

:| EDITTED EDIT: To make sure that my answers are not taken as true :oops: redone AGAIN.
[hide]
To start off you need a "correction" in the measure of time since that includes the time it took for the sound to travel up the well.
$t-\frac{d}{1100}$
$t=7$
Substitute
$7-\frac{d}{1100}$
Now this time component becomes the "new" time, $T$, in
$d=\frac{gT^{2}}{2}$
Where
$g=32$, THE CORRECT VALUE [u]FINALLY[/u], thanks rnwang2
Substitute
$d=16\left(7-\frac{d}{1100}\right)^{2}$
Simply, steps missing, and then use quadratic formula getting the answers of
$d \approx 656.088,90368.912$
With the second be discarded the answer is $656.088ft$, which is [u]FINALLY[/u] correct  :wink: .

EDIT: $g\neq 9.8$ making the previous answers wrong
[/hide]

Im having trouble following this above post

Thanks, that answer looks alot closer to what i got

:|  :blush:
I used the $9.80 \frac{m}{s^{2}}$
I did not read that is was in feet.
Sorry.

I redid the problem.  :)  Your answer is invalid though because it is not $16$ it is $8$, because the formula for distance, $d$, is:
$d=\frac{gt^{2}}{2}$
Where
$g$ is gravity, in this case $32 \frac{ft}{s^{2}}$, EDIT: now it is right
$t$ is time

[quote="AstroPhys"]I redid the problem.  :)  Your answer is invalid though because it is not $16$ it is $8$, because the formula for distance, $d$, is:
$d=\frac{gt^{2}}{2}$
Where
$g$ is gravity, in this case $16 \frac{ft}{s^{2}}$
$t$ is time[/quote]
Err...no. The acceleration due to gravity is 32 ft/s^2.

:blush:  :rotfl:
Ignore me. Sorry Again, I am use to metric.
 :wink:

umm...you're given the speed of sound in feet per second, so you can convert to meters per second.

1100 ft/s = 335.28067 m/s.

solving this time, we get 199.97 m which is suspiciously close to 200 m.  hmm....


vector
linear algebra
matrix
algebra
polynomial
abstract algebra
group theory

$p$ is a prime number and $1 \leq n \leq p$. Show that $Aut((Z/pZ)^{n})$ has no element of order $p^{2}$.

$V=(\mathbb Z/p\mathbb Z)^n$ is a vector space of dimension $n$ over $\mathbb Z/p\mathbb Z$, and every automorphism is a bijective linear transformation from $V$ to itself. We must thus prove that there are no $n\times n$ matrices $A$ over $\mathbb Z/p\mathbb Z$ s.t. $A^{p^2}=I_n$, but $A^k\ne I_n,\ \forall k<p^2$.

Suppose there is such a matrix $A$. It's minimal polynomial is a divisor of $x^{p^2}-1=(x-1)^{p^2}$, so it's of the form $(x-1)^k,\ k\le n\le p$. This means that the matrix $A$ also satisfies the equation $x^p-1=(x-1)^p=0$, or, in other words, $A^p=I_n$, contrary to the assumption that $p^2$ is the order of $A$.

P.S.

The "Solved Problems" section is for problems to which a solution has been posted. This was not the case when you posted this problem :).

Edit:

Indeed, Julien, I meant "minimal polynomial". I've corrected that now. Thank you.

Thanks Grobber. Do you mean "characteristic polynomial" or "minimal polynomial" ?

By the way my approach was different:

If $f$ were such an automorphism, then the decomposition of $f$ (viewed as a permutation) into disjoint cycles would contain a cycle $(g,f(g),\dots,f^{(p^{2}-1)}(g))$ of order $p^{2}$. But the previous family is free with $p^{2}$ elements in the free abelian group $(Z/pZ)^n$, which is absurd. Is that correct ? I suspect it is not, for your proof uses the fact that $n \leq p$, while mine only uses $n < p^{2}$, but I can't see my mistake.

Here is an element of order $p^2$ in $GL_{p+1}(\mathbb{F}_p)$

$A=\begin{bmatrix}1&1&0&\cdots&0\\0&1&1&\cdots&0\\0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{bmatrix}$
$A^p=\begin{bmatrix}1&\binom p1&\binom p2&\cdots&\binom pp\\0&1&\binom p1&\cdots&\binom p{p-1}\\0&0&1&\cdots&\binom p{p-2}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{bmatrix}= \begin{bmatrix}1&0&0&\cdots&1\\0&1&0&\cdots&0\\0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{bmatrix}$
$A^{kp}=\begin{bmatrix}1&0&0&\cdots&k\\0&1&0&\cdots&0\\0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{bmatrix}$
$A^{p^2}=I$

This does show that your argument can't work. The key point is that the elements of that cycle don't have to be independent.


A right angled triangle $ ABC$ with $ \angle A = 90$ is inscribed in a circle. The incircle of $ \triangle ABC$ touches side

 $ AB$ at $ D$. If $ N$ is the point where the line joining the midpoints of sides $ AB,AC$ meets the arc $ AC$  and $ M$ is

the middle of the arc $ BC$ which does not contain $ A$ , prove that $ MD = MN$ .

[b] Note [/b]

[i] I have arrived to this problem , trying  to solve a problem from an old issue of CRUX ( 1992).[/i]

[quote="vittasko"][quote="mr.danh"]I think M is the midpoint of the arch BC, which [b]contain[/b] A.[/quote]
I think somethink is not correct in the statement of the problem, because the relation $ MD \equal{} MN$ doesn't hods, for $ M$ as the midpoint of the arc $ BC,$ either contains or not contains the point $ A$ ( except the particular case of $ AB \equal{} AC$ ).

Kostas Vittas.[/quote]

 Dear Kostas,  really sorry  :( .The figure was very bad  and I have really considered wrong point. Point $ N$ is  the point where the line joining the midpoints of $ AB,AC$ meets arc $ AC$ ( or arc  $ AB$)  of the circle .

  I apologize ! Hopefully it is OK now , because I have also spent much time with this problem.

 [u]Babis [/u]

[quote="stergiu"] [color=darkred]The incircle of the $ A$ - right triangle $ ABC$ touches the side $ [AB]$ at the point $ D$. Denote

the point $ N$ where the line joining the midpoints of the sides $ [AB]$ , $ [AC]$ meets the arc $ AC$ of the circumcircle $ w \equal{} C(O)$

and the middle $ M$ of the arc $ BC$ which does not contain $ A$ . Prove that $ MD \equal{} MN$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] In a $ A$ - right triangle $ ABC$ exist the relations $ S \equal{}\frac {bc}{2}\equal{} p(p \minus{} a) \equal{} (p \minus{} b)(p \minus{} c)$ . Apply the [b]generalized Pythagoras' theorem[/b] in $ \triangle MBD$ :

$ MD^2 \equal{} BM^2 \plus{} BD^2 \minus{} 2\cdot BM\cdot BD\cdot\cos\widehat {DBM} \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} 2\cdot\frac {a\sqrt 2}{2}\cdot (p \minus{} b)\cdot \cos (B \plus{} 45^{\circ}) \equal{}$

$ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} a(p \minus{} b)(\cos B \minus{} \sin B) \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} a(p \minus{} b)\left(\frac ca \minus{} \frac ba\right) \equal{}$

$ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \plus{} (b \minus{} c)(p \minus{} b) \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)(p \minus{} c)$ $ \implies$ $ \boxed {\ MD^2 \equal{} \frac {a^2 \plus{} bc}{2}\ }$ . Denote $ \{M,M'\} \equal{} MO\cap w$ .

 In the $ C$ - right triangle $ MCM'$ we have  $ MN^2 \equal{} a\cdot \left(\frac a2 \plus{} \frac h2\right) \equal{} \frac {a^2 \plus{} ah}{2}$ $ \implies$ $ \boxed {\ MN^2 \equal{} \frac {a^2 \plus{} bc}{2}\ }$ . In conclusion, $ MD \equal{} MN$ .[/color]

[color=darkblue][b][u]Proof.[/u][/b] In a triangle $ ABC$ exist the relations $ S \equal{} p(p \minus{} a) \equal{} (p \minus{} b)(p \minus{} c)$ . Apply the [b]generalized Pythagoras' theorem[/b] in $ \triangle MBD$ :

$ MD^2 \equal{} BM^2 \plus{} BD^2 \minus{} 2\cdot BM\cdot BD\cdot\cos\widehat {DBM} \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} 2\cdot\frac {a\sqrt 2}{2}\cdot (p \minus{} b)\cdot \cos (B \plus{} 45^{\circ}) \equal{}$

$ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} a(p \minus{} b)(\cos B \minus{} \sin B) \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \minus{} a(p \minus{} b)\left(\frac ca \minus{} \frac ba\right) \equal{}$

$ \frac {a^2}{2} \plus{} (p \minus{} b)^2 \plus{} (b \minus{} c)(p \minus{} b) \equal{}$ $ \frac {a^2}{2} \plus{} (p \minus{} b)(p \minus{} c)$ $ \implies$ $ \boxed {\ MD^2 \equal{} \frac {a^2 \plus{} bc}{2}\ }$ . Denote $ \{M,M'\} \equal{} MO\cap w$ .

 In the $ C$ - right triangle $ MCM'$ we have  $ MN^2 \equal{} a\cdot \left(\frac a2 \plus{} \frac h2\right) \equal{} \frac {a^2 \plus{} ah}{2}$ $ \implies$ $ \boxed {\ MN^2 \equal{} \frac {a^2 \plus{} bc}{2}\ }$ . In conclusion, $ MD \equal{} MN$ .[/color][/quote]

Thank Virgil Nicula. It's nice! :P


AMC
AMC 12
probability
geometry
trapezoid

A point $P$ is randomly selected from the rectangular region with vertices $(0,0),\, (2,0),\, (2,1),\, (0,1)$. What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$?

[hide="Hint"] Draw the locus of points equidistant from the origin and $(3, 1)$.  [/hide]

hey thanks for the hint.

i found the area of the larger trapezoid and divided that by the area of the rectangle. the answer is $\frac{3}{4}$.


trigonometry
algebra
polynomial
rational numbers

Find all rational numbers $k$ such that $0 \le k \le \frac{1}{2}$ and $\cos k \pi$ is rational.

One can do this problem using Chebyshef polynomials and some computation.
But here's a proof using the brutal fork of abstract algebra  :grin:

The primitive $n$-th roots of unity have algebraic degree $\varphi(n)$ (showing this requires us to prove the irreducibility if the cyclotomic polynomials, but I will skip it now).
Let $k= \frac mn$ with $m,n$ coprime integers and let $\zeta= e^{\frac{2\pi m i}{n}}$ (a primitive $n$-th root of unity).
Since $c:=2\cdot \cos \left( \frac{m}{n} \right) = \zeta + \zeta^{-1}$, we get that $\zeta$ is a root of $x^2-2c x +1=0$.
Especially, the dimension $[\mathbb{Q}[\zeta]:\mathbb{Q}[c]]$ is $1$ or $2$. If $n>2$, then $\zeta$ is nonreal, but $c$ is, giving that the degree is in fact $2$.
Thus the algebraic degree of $c$ is $\frac{\varphi(n)}2$ if $n \geq 3$, thus $c$ is rational iff $n=1,2,3,4,6$.

So, what is your final answer to the question? For those denominators you have an algebraic number, but the question is when it is rational... or did I misunderstand you?

Read the last line again ;)
At the beginning, I showed that it's algebraic and then computed it's algebraic degree.

In $ c: \equal{} 2\cdot \cos \left( \frac {m}{n} \right)$ you are probably missing a $ \pi$  there, but I still don't get it, you say $ \cos\left(\frac {1}{6}\pi\right)\equal{}\frac{\sqrt{3}}{2}$ is rational?

Sorry, let me kill the little errors:

The primitive $ n$-th roots of unity have algebraic degree $ \varphi(n)$ (showing this requires us to prove the irreducibility if the cyclotomic polynomials, but I will skip it now).
Let $ k \equal{} \frac {2m}n$ with $ m,n$ coprime integers and let $ \zeta \equal{} e^{\frac {2\pi m i}{n}}$ (a primitive $ n$-th root of unity).
Since $ c: \equal{} 2\cdot \cos \left( \frac {2m}{n} \pi \right) \equal{} \zeta \plus{} \zeta^{ \minus{} 1}$, we get that $ \zeta$ is a root of $ x^2 \minus{} 2c x \plus{} 1 \equal{} 0$.
Especially, the dimension $ [\mathbb{Q}[\zeta]: \mathbb{Q}[c]]$ is $ 1$ or $ 2$. If $ n > 2$, then $ \zeta$ is nonreal, but $ c$ is real, giving that the degree is in fact $ 2$.
Thus the algebraic degree of $ c$ is $ \frac {\varphi(n)}2$ if $ n \geq 3$, thus $ c$ is rational iff $ n \equal{} 1,2,3,4,6$.

Another proof:

Every $ \zeta \equal{} e^{\frac {2\pi m i}{n}}$ is an algebraic integer (it is a root of $ x^n\minus{}1$). But then $ c: \equal{}2\cdot \cos \left( \frac {2m}{n} \pi \right) \equal{} \zeta \plus{} \zeta^{ \minus{} 1}$ is an algebraic integer, too. Thus $ c$ is rational iff it is an integer (as a rational is an algebraic integer iff it is an integer).
Especially if $ c$ is rational,then $ c \in \{ \minus{}2 ,\minus{}1 , 0 , 1 , 2 \}$ as $ \minus{}1\leq \cos(x) \leq 1$ for all $ x$. Now checking these cases we get the same as before.


Euler
trigonometry
algebra
polynomial
complex analysis
complex analysis theorems

I'm looking for the proof of the formula
$\sin(x)=x\prod_{n=1}^{+\infty}(1-\frac{x^2}{n^2\pi^2})$

The most elementary approach is to note that $\frac{\sin x}{n\sin(x/n)}$ is a polynomial in $\sin(x/n)$ whose roots are easy to find. Writing the factorization for this polynomial and accurately passing to the limit, you'll get the desired formula.

First idea: if $f(z)=\sin z$ then $\frac{f'(z)}{f(z)}=\cot z.$

Let $g(z)=z\prod_{n=1}^{+\infty}(1-\frac{z^2}{n^2\pi^2})$

Write $\frac{g'(z)}{g(z)}$ as an infinite sum.

Show that that sum is periodic.

Show that $\frac{g'(z)}{g(z)}-\cot z$ has removable singularities, is periodic, and is bounded on a vertical strip of one-period width - hence bounded, hence, by Liouville's Theorem, constant.

Now all you have to do is find the constant.

The polynomial $z^n-1$  possesses the roots $z=\sqrt[n]{1}=e^{2ik\pi/n}\;k=0\pm1\pm2\pm3\cdots$.  Now a factorizazion of $z^n-1$ brings us to:
\[ z^n-1=(z-1)\prod_{k=1}^{(n-1)/2}\left(z-e^{2ik\pi/n}\right)\left(z-e^{-2ik\pi/n}\right)  \]
\[ =(z-1)\prod_{k=1}^{(n-1)/2}\left(z^2-2z\cos \frac{2k\pi}n+1\right)\,\,with\,n\,\,odd  \]
By replacing $z\rightarrow z/a$ and multiplying by $a^n$ we obtain a slightly more general result:
\[ z^n-a^n=(z-a)\prod_{k=1}^{(n-1)/2}\left(z^2-2az\cos \frac{2k\pi}n+a^2\right)  \]

We now insert $z=1+x/N,\,\,\,a=1-x/N$   and put $n=N$. This gives
\[ \left(1+x/N\right)^N-\left(1-x/N\right)^N=\frac{2x}N\prod_{k=1}^{(N-1)/2}\left(2+\frac{2x^2}{N^2}-2\left(1-\frac{x^2}{N^2}\right)\cos\frac{2k\pi}N\right)  \].
After algebraic manipulation and for large N, we get
\[ \frac{e^x-e^{-x}}2=x(1+x^2/\pi^2)(1+x^2/4\pi^2)(1+x^2/9\pi^2)\cdots  \].
Replacing x by ix, we obtain the Euler formula.


[quote="nguyenvanhuyen"]Given  $ a,b,c > 0$ and  $ abc \equal{} 8$. Prove that
$ \frac {a}{\sqrt {(a^3 \plus{} 1)(b^3 \plus{} 1)}}\plus{}\frac {b}{\sqrt {(b^3 \plus{} 1)(c^3 \plus{} 1)}}\plus{}\frac {c}{\sqrt {(c^3 \plus{} 1)(a^3 \plus{} 1)}} \ge \frac {4}{3}$[/quote]

isn't it $ \frac{2}{3}$, instead of $ \frac{4}{3}$??

it can prove by one lemma...
(1+a^3)^0.5 <= 1+0.5a^2


geometry
parallelogram
analytic geometry
function
trigonometry
trapezoid
ratio

[i]Lemma 1[/i]. Let $D$ be such a point that $ABDC$ is a parallelogram, then point $P$ lies on the angle bisector of $\angle{BDC}$.

[i]Lemma 2.[/i] The bisectors of $\angle{BAC}$ and $\angle{BDC}$ are parallel.

Therefore points $M,P,D$ are collinear and hence $MC=DC=AB$.

My solution uses barycentric coordinates: (these kind of solutions are usually messy, but this one came out nicely)

Let $BK=CL=t$. Then $P(t: c-t: b-t)$.

The equation of $AD$ (the interior bisector) is $0x+cy-bz=0$ and its point at infinity is $(b+c: -b : -c)$.

Let $N$ s.t. $CN=BA$ (properly oriented).

The equation of $PN$ is $(c-t)(x(b-c)+yb-zc)=0$. The case $c=t$ is trivial, so $c \neq t$ and the equation of $PN$ is $(b-c)x+yb-zc=0$ and its point at infinity is $(b+c: -b: -c)$.

Thus, $AD \| PN$ as they have the same point at infinity.

Conclusion: $M=N$, so $CM=AB$.

Sailor's solution is very nice, but I don't think I'll ever understand geometry in order to come up with synthetic solutions like that. ( :( )

I have a strange solution.

WLOG $AB\geq AC$

Let $M$ be a point on ray $CA$ such that $AB=CM$, $l$ is a line parallel through $M$ to angle bisectore of $\angle BAC$

Denote $N, L$ be the intersections of line $l$ and $AB, BC$ respectively.

Let's think a point $X$ on edge $ML$. We'll call $P, Q$ as the intersection of $BX, AC$ and $CX, AB$ respectively.

And Let $S(X)= CP$, $R(X) = BQ$.

It's obvious that $S(X), R(X)$ is the 2-dimension functions of $X$.

cuz $S(X)=R(X)$ for $X=L, N$, we can know that two functions are same.

Therefore, edge $ML$ is the trace of $P$.

$Q.E.D.$

Use Menalaus twice and you are done.

Bomb

Could anyone post the solution with Menelaos ?Thank you

[quote="Valentin Vornicu"]Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK \equal{} CL$ and let $ P \equal{} CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM \equal{} AB$.[/quote]
My solution is based on the method of working backward and using Menalaus Theorem.

Let the angle bisector of $ A$ meet $ BC$ at $ Y$, $ CK \cap AY \equal{} X$. We take $ M'$ on $ CA$ such that $ AB \equal{} CM'$ and let $ M'P \cap BC \equal{} Z, BZ \cap CA \equal{} L'$. Now $ \angle ZM'C \equal{} \frac A2$.

So, by sin law we get, $ \frac {\sin \frac A2}{\sin \angle M'ZC} \equal{} \frac {CZ}{CM'} \equal{} \frac {BY}{AB} \Longrightarrow BY \equal{} ZC \iff BZ \equal{} YC$

Using Menelaus's Theorem on $ \triangle ABY$ and transversal $ KXC$ we get,
\[ \frac {AK}{KB} \cdot \frac {BC}{CY} \cdot \frac {YX}{XA} \equal{} \minus{} 1
\]
Using Menelaus's Theorem on $ \triangle CMZ$ and transversal $ LPB$ we get,
\[ \frac {CL'}{M'L'} \cdot \frac {M'P}{PZ} \cdot \frac {ZB}{BC} \equal{} \minus{} 1
\]
Also, $ M'Z \parallel AY \Longrightarrow \frac {M'P}{PZ} \equal{} \frac {AX}{XY}$

From all these relations we get,
\[ \frac {AK}{KB} \equal{} \frac {M'L'}{L'C} \iff \frac {AB}{KB} \equal{} \frac {M'C}{L'C} \iff KB \equal{} L'C
\]
So, $ L' \equiv L \iff M' \equiv M$ and we are done. :)

$ QED$

Take $D$ on $(BA$, order $B-A-D$ so that  $AD = AC$. Obviously, $PM||CD$. Apply Menelaos in triangle $\triangle ACK$ with transversal $LPB$ and get
$\frac {KP}{PC}=\frac{b-k}{c} \ (\ 1\ )$, where $b, c$ and $k$ are the lengths of $AC, AB$ and $BK$ respectively.

Next, suppose, w.l.o.g. $AB > AC$. The parallel from $P$ to $CC'$ intersects $AB$ at $M'$ and $\frac{KM'}{M'C'} = \frac{KP}{PC} ( 2 )$. Since $AC' = AC = b$,
$KC' = b + c - k$. From (1) and (2) we get $KM' = b - k$, that is, $BM' = b = AC' = AC ( 3 )$, hence $C'X = AB ( 4 )$. Now, extend $PM'$ till it intersects $AC$ at $M$; $MM'CC'$ is an isosceles trapezoid, hence the problem is solved.

If $AB < AC$ we get the solution in a similar way.

Best regards,
sunken rock

Let $ T$ the intersection point of $ AB$ and $ PM$. Obviously $ AT\equal{}AM$.

Apply Menelaus theorem in triangle $ ABL$ with transversal $ KPC$.
$ \frac{CL}{CA}.\frac{KA}{KB}.\frac{PB}{PL}\equal{}1$ and finally $ \frac{KA}{CA}\equal{}\frac{PL}{PB}$.

Apply again Menelaus theorem in triangle $ ABL$ with transversal $ TMP$.
$ \frac{PL}{PB}.\frac{TB}{TA}.\frac{MA}{ML}\equal{}1$ and finally $ \frac{PL}{PB}\equal{}\frac{ML}{TB}$.

So $ \frac{KA}{CA}\equal{}\frac{LM}{TB}$ which is equivalent to $ \frac{KA}{CL\plus{}LM\plus{}MA}\equal{}\frac{LM}{TA\plus{}AK\plus{}KB}$
or $ \frac{KA}{KB\plus{}LM\plus{}TA}\equal{}\frac{LM}{TA\plus{}KA\plus{}KB}$.

After a few calculations we have that $ (KA\minus{}LM)(KA\plus{}LM\plus{}TA\plus{}KB)\equal{}0$ which means $ KA\equal{}LM$.

I have solution with projective way:
let's take $PM \cap AB =N  ,PM\cap BC=Q$ by projecting AB to PM with centre C
cross ratios R(M,N,P,Q)=R(A,N,K,B) then PM to AC with centre B R(MNPQ)=R(MALC) so R(A,N,K,B)=R(MALC) then
$\frac{ \frac{MA}{MC}}{\frac{LA}{LC}}=\frac{\frac{AN}{AB}}{\frac{KN}{KB}}$ since we know that MA=AN and BK=LC  we find that MC*LA=AB*KN the rest of solution is obvious hence MC-LA=BA-KN

Hi
Use this : If we make $D$ such that $ABDC$ is a paralellogram then the locus of $P$ will be the angle bisector of $BDC$.
 :wink:


geometry
parallelogram
geometry unsolved

We are given any quadrilateral. We mark the midpoint of each side and name them as A,B,C and D. We have to find a point P inside this quad so that when we join P with A,B,C and D the quadrilateral will be divided into four equal areas. What is the location of P ?

I thought I'd answered this one.  Oh, well, it must have been elsewhere.

P will be the point of intersection when you join mid-points of the opposite sides.  This can be shown since those mid-points form a parallelogram.  Draw the parallelogram as well as the above lines and you'll see that the outer triangles so formed are each 1/8 of the quadrilateral [similar triangle relationship], and the lines joining mid-points are diagonals of that parallelogram, dividing it also into equal parts of 1/8 of the quadrilateral.


\[
x_1 x_2 ..x_n \left( {\frac{1}{{x_1 }} + \frac{1}{{x_2 }} + .. + \frac{1}{{x_n }} - n + f(n)} \right) \le f(n)
\]
where $x_i ,i = 1,2,..,n$ are positive reals and $x_1+...+x_n=n$

My question is, for what kind of f(n) is this inequality true?

For f(n)=3 it is vasc's http://www.mathlinks.ro/Forum/topic-21613.html
for $f(n) = \left( {\frac{n}{{n - 1}}} \right)^{n - 1} $ it is harazi's http://www.mathlinks.ro/Forum/topic-21612.html

I found that for f(n)=k where k>=3 the inequality is valid.
Any other interesting f(n)?

Fantastic and super hard open question, but very very very difficult. ;)

For any $f(n)$ such that
$\displaystyle f(n) \ge \left( {\frac{n}{{n - 1}}} \right)^{n - 1}$.


Proof:

[hide]
Consider some converging geometric series with first term $a$ and common ratio $r$, where $-1<r<1$ and r is not 0 (in this case, $a=r=\frac{1}{2}$). Then the sum of the first n terms of the series, $S(n)$, can be generally expressed:

$S(n)=a+ar+ar^2...+ar^{(n-1)}$ Let this be Equation 1.

Multiply by $r$:

$rS(n)=ar+ar^2+ar^3...+ar^n$ Let this be Equation 2.

Subtract Equation 2 from Equation 1:

$S(n)-rS(n)=a-ar^n$

Isolating S(n) and a little factoring gives:

$(1-r)S(n)=a-ar^n$, $S(n)=\frac{a(1-r^n)}{1-r}$

This is the general formula for the sum of a converging series to the nth term. Now, the sum of the series to the infinite nth term can be reduced: since r is between -1 and 1, $r^\text{infinity}=0$. So the equation is:

$S(n)=\frac{a(1-0)}{1-r}=\frac{a}{1-r}$.

Now finally for your specific question. Since $a=r=\frac{1}{2}$, the sum of the infinite series is: $\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$.

And the proof is complete.

[/hide]

By the way, I learned that from AoPS volume I. Those books are AMAZING.

[hide]
$S=\frac 12+\frac 14+\frac 18+...$
$2S=1+\frac 12+\frac 14+\frac 18+...$
$2S-S=1, S=1$
Q.E.D.
[/hide]

Shouldn't proofs be in the Getting Started forum?

Use the sum of a geometric/infinite series equation :D which is what bthings_2000 proved

[quote="ajai"]Prove that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...=1$[/quote]

You can't really prove that this is true, you can only show this.  I know this is weird, because show and prove are the same thing, but this is true.

Oh, that just seems so obvious.  I should've noticed that as well.

[quote="mathnerd314"]
Shouldn't proofs be in the Getting Started forum?[/quote]
They are not allowed in the Getting Started forum, except really, really easy ones.


algebra
polynomial
ARML
calculus
integration
Vieta
modular arithmetic

Polynomial, I think, is difficult enough that it deserves its own thread and not belong under "Algebra" (which should be posted when I feel like doing algebra problems  :rotfl: ).  And I know I'm doing this on my own, I would still appreciate if anyone says thanks to me for putting up the problems by category.  :)

1. (AMC 12A 2004 #23) A polynomial

$ \displaystyle P(x) \equal{} c_{2004} x^{2004} \plus{} c_{2003} x^{2003} \plus{} \cdots \plus{} c_1 x \plus{} c_0$

has real coefficients with $ c_{2004} \neq 0$ and 2004 distinct complex zeros $ z_k \equal{} a_k \plus{} b_k i, 1 \leq k \leq 2004$ with $ a_k$ and $ b_k$ real, $ a_1 \equal{} b_1 \equal{} 0,$ and

$ \displaystyle \sum^{2004}_{k \equal{} 1} a_k \equal{} \sum^{2004}_{k \equal{} 1} b_k$

Which of the following can be a nonzero number?
A) $ c_0$
B) $ c_{2003}$
C) $ b_2 b_3 \cdots b_{2004}$
D) $ \sum^{2004}_{k \equal{} 1} a_k$
E) $ \sum^{2004}_{k \equal{} 1} c_k$

2. (AMC 12B 2004 #23) The polynomial $ x^3 \minus{} 2004 x^2 \plus{} mx \plus{} n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two.  How many values of $ n$ are possible?
A) 250,000
B) 250,250
C) 250,500
D) 250,750
E) 251,000

3. (AMC 12A 2003 #21) The graph of the polynomial

$ \displaystyle P(x) \equal{} x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e$

has five distinct $ x \minus{}$intercepts, one of which is at (0,0). Which of the following cannot be zero?
A) $ a$
B) $ b$
C) $ c$
D) $ d$
E) $ e$

4. (AHSME 1982) Suppose $ x \equal{} a \plus{} bi$ is a solution of the polynomial equation

$ \displaystyle c_4 z^4 \plus{} i c_3 z^3 \plus{} c_2 z^2 \plus{} i c_1 z \plus{} c_0 \equal{} 0,$

where $ c_0, c_1, c_2, c_3, c_4, a,$ and $ b$ are real constants and $ i^2 \equal{} \minus{} 1.$ Show that $ \minus{} a \plus{} bi$ is also a solution.

5. (AHSME 1981 #30) If $ a,b,c,d$ are the solutions of the equation $ x^4 \minus{} bx \minus{} 3 \equal{} 0,$ find the polynomial with leading coefficient of 3 whose roots are

$ \displaystyle \frac {a \plus{} b \plus{} c}{d^2}, \frac {a \plus{} b \plus{} d}{c^2}, \frac {a \plus{} c \plus{} d}{b^2}, \frac {b \plus{} c \plus{} d}{a^2}$

A) $ 3x^4 \plus{} bx \plus{} 1 \equal{} 0$
B) $ 3x^4 \minus{} bx \plus{} 1 \equal{} 0$
C) $ 3x^4 \plus{} bx^3 \minus{} 1 \equal{} 0$
D) $ 3x^4 \minus{} bx^3 \minus{} 1 \equal{} 0$
E) None of these

6. (NYSML 1985 I4) Let $ P(x) \equal{} 0$ be the polynomial of equation of least possible degree, with rational coefficients, having $ \sqrt [3] {7} \plus{} \sqrt [3] {49}$ as a root. Compute the product of all of the roots of $ P(x) \equal{} 0.$

7. (ARML I8) Let $ P(x)$ be a fifth-degree polynomial equation with integer coefficients that has at least one integral root. If $ P(2) \equal{} 13$ and $ P(10) \equal{} 5,$ compute a value of $ x$ that [u]must[/u] satisfy $ P(x) \equal{} 0.$

8. (ARML 1997 T9) Let $ P(x)$ be a polynomial whose degree is 1996. If $ P(n) \equal{} \frac {1}{n}$ for $ n \equal{} 1,2,3,...,1997,$ compute the value of $ P(1998)$.

Is there a typo on 8 by any chance?

Its suppose to be P(n) right?

Then you can easily construct a degree 1997 polynomial and you are basically done.

[hide="3"]
$ d$ is the sum of the products of each set of four of the roots.  All but one of these products includes $ 0$ and is $ 0$.  The other cannot be $ 0$ if the roots are distinct.  So $ d$ is nonzero.
[/hide]

[hide="1"]
Suppose we factor out the $ c_{2004}$ and instead we now have the real coefficients $ d_{2003}$ etc.

(A) must be 0 because $ d_0 =$ the product of the roots, and $ z_1 = 0$, thus implying that $ c_0 = 0$.
(B) must be 0 because $ d_{2003} =$ the negative of the sum of the roots, and because all of the coefficients are real, for any complex root its conjugate must also be a root. Thus $ \displaystyle\sum_{k = 1}^{2004} b_k = 0$, implying that $ \displaystyle\sum_{k = 1}^{2004} a_k = 0$, and the sum of all of the roots must be $ 0$.
(C) There are an odd number of terms being multiplied, but each complex root must be paired with a conjugate, giving an even number of roots. Thus at least one of the roots must not have an imaginary part, and this product is $ 0$.
(D) Must be zero, see (B).

Thus (E) is the answer. $ \sum_{k = 1}^{2004} c_k = 0$ only if $ P(1) = 0$, which is not implied by the problem conditions.
[/hide]

[hide="2"]
Sorry, this is a bit hard to follow. I can clarify if necessary.

Let the roots be $ r,s,r + s$. Then
\[ (x - r)(x - s)(x - (r + s)) = x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0
\]
, and by matching coefficients, $ 2(r + s) = 2004 \Longrightarrow r + s = 1002$. Then our polynomial looks like $ x^3 - 2004x^2 + (rs + 1002^2)x - 1002rs$, and we need the number of possible products $ rs = r(1002 - r)$.

Since $ r > 0$ and $ r(1002 - r) > 0$, it follows that $ 0 < r(1002 - r) < 501^2 = 251001$, with the endpoints not achievable because the other roots are not integers. Now let $ m = r(1002 - r) \Longrightarrow - r^2 + 1002r - m = 0$. Because $ r$ cannot be an integer, by Vieta’s we have that the sum of the roots of $ r$ is $ 1002$ and their product is $ m$. There are exactly $ 500$ occurrences in which $ m$ then must be an integer, so the answer is $ 251000 - 500 = \mathrm{(C)}$.
[/hide]

[hide="3"]
Let the roots be $ x_{1,2,\cdots,5}$, and let $ x_1 = 0$. By Vieta’s, $ d = \sum x_1x_2x_3x_4 = (0) + (0) + (0) + (0) + x_2x_3x_4x_5$. Also, all of the roots are distinct, so none of $ x_{2, 3, 4, 5}$ can be zero, and so $ d$ cannot be zero $ \Rightarrow \mathrm{(D)}$.

The others can be found to be zero through trial and error.
[/hide]

[hide="5"]
By Vieta’s, $ a + b + c + d = 0$, and so each of the new roots are actually just $ - \frac {1}{a}, - \frac {1}{b}, - \frac {1}{c}, - \frac {1}{d}$. Thus the polynomial in question is $ 3(x + 1/a)(x + 1/b)(x + 1/c)(x + 1/d) = 0$. Expanding, (there should be an easier way to find the polynomial instead of just brute-force comparing the terms, but I can't think of it)
\begin{align*} & 3x^4 + 3\left(\frac 1a + \frac 1b + \frac 1c + \frac 1d\right)x^3 \\
& + 3\left(\frac 1{ab} + \frac 1{bc} + \frac 1{cd} + \frac 1{ac} + \frac 1{bd} + \frac 1 {ad}\right)x^2 \\
& + 3\left(\frac 1{abc} + \frac 1{abd} + \frac 1{acd} + \frac {1}{bcd}\right)x + \frac {3}{abcd} = 0\end{align*}
Collecting,
\[ 3x^4 + \frac {3 \displaystyle \sum abc}{abcd}x^3 + \frac {3 \displaystyle \sum ab}{abcd}x^2 + \frac {3 \displaystyle \sum a}{abcd}x + \frac {3}{abcd} = 0
\]
By Vieta's on the original polynomial, we have $ abcd = - 3, \displaystyle \sum abc = b, \sum ab = 0, \sum a = 0$. Thus, the polynomial becomes
\[ 3x^4 - bx^3 - 1 = 0
\]
And the answer is $ \mathrm{(D)}$.
[/hide]

Edit: Oh, and thank you Silverfalcon for taking the time to gather these problems.  :)

[hide="6"]$ P(x)$ cannot be of degree 2.  So it must be of degree 3 or more.
Let $ x\equal{}\sqrt[3]{7}\plus{}\sqrt[3]{49}$.  Therefore $ x^3\equal{}(\sqrt[3]{7}\plus{}\sqrt[3]{49})^3\equal{}7\plus{}3\cdot7^{\frac{5}{3}}\plus{}3\cdot7^{\frac{4}{3}}\plus{}49\equal{}56\plus{}21(\sqrt[3]{7}\plus{}\sqrt[3]{49})\equal{}56\plus{}21x$.  Solving for 0 yields $ x^3\minus{}21x\minus{}56\equal{}0$.  So the product of the roots is $ \boxed{56}$.[/hide]

[hide="5"]From viete's, we know that $ a \plus{} b \plus{} c \plus{} d \equal{} 0$, so $ a \plus{} b \plus{} c \equal{} \minus{} d$, so the first root of the new polynomial is actually equal to $ \minus{} \frac {1}{d}$. Using similar logic, we deduce that the roots of the new polynomial are $ \minus{} \frac {1}{a}$, $ \minus{} \frac {1}{b}$, $ \minus{} \frac {1}{c}$, and $ \minus{} \frac {1}{d}$. The new polynomial can be written as $ 3(x \plus{} \frac {1}{a})(x \plus{} \frac {1}{b})(x \plus{} \frac {1}{c})(x \plus{} \frac {1}{d})$. Expanding and manipulating the coefficients a little yields $ 3(x^4 \plus{} (\frac {abc \plus{} abd \plus{} acd \plus{} bcd}{abcd})x^3 \plus{} (\frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{abcd})x^2 \plus{} (\frac {a \plus{} b \plus{} c \plus{} d}{abcd})x \plus{} \frac {1}{abcd})$
You really don't have to expand that entire thing if you realize what's going on, but anyways using viete's, we now only need to plug some stuff in, and we determine that D gives the correct polynomial.[/hide]
okay someone edited their post and did this problem while I was doing it, but I'll leave it up anyways.

[quote="Twiz"]You really don't have to expand that entire thing if you realize what's going on, but anyways using viete's, we now only need to plug some stuff in, and we determine that D gives the correct polynomial.[/quote]

But one of the choices is None of the above, so wouldn't you have to apply vieta's to all of the coefficients anyways? (which would then be basically the same thing as expanding it out). I'm pretty sure there are cleaner algebraic manipulations then repeated applications of Vieta.

Edit: Wow, I can't read. I thought you were trying to quote my solution ..  :oops:

Guys, you guys are thinking way too hard for Number 5 with brute-forcing.

Here, do this problem.

Prove that for any polynomial $ P(x)$ and $ Q(x)$ where $ Q(x) \equal{} P(1/x),$ the coefficients of $ Q(x)$ are same as $ P(x)$ but in [i]reverse order[/i] of $ P(x)$.

You mean $ Q(x) \equal{} x^n P \left( \frac{1}{x} \right)$.

Consider $ y^n P \left( \frac {x}{y} \right)$.  ;)

Blah, you're right. That's what I meant. I guess doing math problems straight for two days make me think with typos.  :D

[quote="Silverfalcon"]7. (ARML I8) Let $ P(x)$ be a fifth-degree polynomial equation with integer coefficients that has at least one integral root. If $ P(2) \equal{} 13$ and $ P(10) \equal{} 5,$ compute a value of $ x$ that [u]must[/u] satisfy $ P(x) \equal{} 0.$
[/quote]
[hide="7, maybe?"]
For a polynomial $ P$ with integer coefficients and integers $ a,b$, then $ a \minus{} b|P(a) \minus{} P(b)$...so $ \frac {13}{2 \minus{} x}$ is an integer, $ \frac {5}{10 \minus{} x}$ is an integer, and finding solutions, $ \boxed{x \equal{} 15}$...does this work...? Because I don't really use the fact that $ P$ is fifth degree, and when it says $ P$ has at least one integral root, I just said it was $ x$? Because it just says compute a value of $ x$...but I guess I haven't really shown $ x\equal{}15$ must satisfy $ P(x)\equal{}0$...maybe that's where the degree 5 comes into play?[/hide]

[quote="ch1n353ch3s54a1l"][quote="Silverfalcon"]7. (ARML I8) Let $ P(x)$ be a fifth-degree polynomial equation with integer coefficients that has at least one integral root. If $ P(2) \equal{} 13$ and $ P(10) \equal{} 5,$ compute a value of $ x$ that [u]must[/u] satisfy $ P(x) \equal{} 0.$
[/quote]
[hide="7, maybe?"]
For a polynomial $ P$ with integer coefficients and integers $ a,b$, then $ a \minus{} b|P(a) \minus{} P(b)$...so $ \frac {13}{2 \minus{} x}$ is an integer, $ \frac {5}{10 \minus{} x}$ is an integer, and finding solutions, $ \boxed{x \equal{} 15}$...does this work...? Because I don't really use the fact that $ P$ is fifth degree, and when it says $ P$ has at least one integral root, I just said it was $ x$? Because it just says compute a value of $ x$...but I guess I haven't really shown $ x \equal{} 15$ must satisfy $ P(x) \equal{} 0$...maybe that's where the degree 5 comes into play?[/hide][/quote]

[hide="Comments"]
You use the fact that it's 5th degree to make sure that you can have a real value $ x$ that satisfies $ P(x) \equal{} 0.$ Your solution looks correct. You have shown that 15 satisfies $ P(x) \equal{} 0$ because of $ a,b \in \mathbb{Z} \rightarrow a \minus{} b|P(a) \minus{} P(b)$ and $ \frac {13}{2 \minus{} 15}$, $ \frac {5}{10 \minus{} 15}$ are integral.
[/hide]

I think you do need to use that $ P$ is $ 5$th degree in another way. As the method stands now, anything that is $ 15 \pmod{65}$ works. These alternate solutions need to be eliminated somehow. The word "must" is underlined for a reason.

EDIT: Sorry, mixed up the numerator and denominator.

Er... no.  $ \frac{13}{2 \minus{} x}$ and $ \frac{5}{10 \minus{} x}$ are simultaneously integer precisely when $ x \equal{} 15$ and for no other integer value of $ x$.  The fifth-degree condition is unnecessary; note that $ P(x) \equal{} 15 \minus{} x$ satisfies the problem conditions.

8 mabey?
[hide]

Let $ Q(n) \equal{} nP(n) \minus{} 1$. Therefore, $ Q(n)$ has zeros at $ 1, 2, 3, \ldots, 1997$ and has degree 1997. Therefore, $ Q(n) \equal{} (n \minus{} 1)(n \minus{} 2)(n \minus{} 3)\cdots(n \minus{} 1997)$, and by substitution and solving for $ P(n)$:

$ (n \minus{} 1)(n \minus{} 2)\cdots(n \minus{} 1997) \equal{} nP(n) \minus{} 1 \Rightarrow$

$ \displaystyle P(n) \equal{} \frac {(n \minus{} 1)(n \minus{} 2)\cdots(n \minus{} 1997) \plus{} 1}{n}$

Then simply solving for $ P(1998)$:

$ \displaystyle P(1998) \equal{} \frac {(1997)(1996)\cdots(1) \plus{} 1}{1998} \Rightarrow$
$ \displaystyle P(1998) \equal{} \frac {1997! \plus{} 1}{1998}$
[/hide]

Your current expression for $ P(n)$ is not a polynomial.  The second step should be $ Q(n) \equal{} a(n\minus{}1)(n\minus{}2)...(n\minus{}1997)$.  For what value of $ a$ is $ P(n)$ a polynomial?

[quote="Silverfalcon"]

4. (AHSME 1982) Suppose $ x \equal{} a \plus{} bi$ is a solution of the polynomial equation

$ \displaystyle c_4 z^4 \plus{} i c_3 z^3 \plus{} c_2 z^2 \plus{} i c_1 z \plus{} c_0 \equal{} 0,$

where $ c_0, c_1, c_2, c_3, c_4, a,$ and $ b$ are real constants and $ i^2 \equal{} \minus{} 1.$ Show that $ \minus{} a \plus{} bi$ is also a solution.
[/quote]

[hide="4"]
[hide="hint"]
The $ i$s are only coefficients of the odd degree terms.
[/hide]
[hide="solution"]
Let $ C(z) \equal{} c_4 z^4 \plus{} i c_3 z^3 \plus{} c_2 z^2 \plus{} i c_1 z \plus{} c_0$; then $ C'(z) \equal{} C(i \cdot z) \equal{} c_4 \cdot z^4 \plus{} c_3 z^3 \minus{} c_2 z^2 \minus{} c_1 z \plus{} c_0 \equal{} 0$ is a polynomial equation of $ z$ with real coefficients. If a root of a polynomial with real coefficients is complex, then its complex conjugate must also be a root. As $ C(a \plus{} bi) \equal{} C(i(b \minus{} ai)) \equal{} C'(b \minus{} ai) \equal{} 0$, it follows that $ C'(\overline{b \minus{} ai}) \equal{} C(i(b \plus{} ai)) \equal{} \boxed{C( \minus{} a \plus{} bi) \equal{} 0}$.
[/hide][/hide]

[quote="t0rajir0u"]Your current expression for $ P(n)$ is not a polynomial.  The second step should be $ Q(n) \equal{} a(n \minus{} 1)(n \minus{} 2)...(n \minus{} 1997)$.  For what value of $ a$ is $ P(n)$ a polynomial?[/quote]

[hide]So, a=1/1997, and P(1998)=(1996!+1)/1998?[/hide]

Nope, but getting closer.  What's the constant term of $ Q(n)$?

The constant term of $ Q(n)$ must be $ \minus{}1$, since $ Q(n) \equal{} n \cdot P(n) \minus{} 1$ and the only $ n$-free term is $ \minus{}1$.

Thus $ a \cdot (\minus{}1997!) \equal{} \minus{}1 \Rightarrow a \equal{} \frac {1}{1997!}$.

Now, $ Q(1998) \equal{} 1998 \cdot P(1998) \minus{} 1 \equal{} \frac {1997!}{1997!} \equal{} 1 \Rightarrow P(1998) \equal{} \frac {2}{1998} \equal{} \boxed {\frac {1}{999}}$.

[hide="5 using Vieta's formulas"] From [url=https://en.wikipedia.org/wiki/Vieta%27s_formulas] Vieta's formulas[/url], we know $a+b+c+d=0 \Rightarrow a+b+c=-d$ so  let the polynomial $P(x)=3(x^4+c_3x^3+c_2x^2+c_1x+c_0)$ be the one we want to find.
From the original polynomial, we know that:
$\Sigma a=0$
$\Sigma ab=0$
$\Sigma abc=b$
$abcd=-3$
so,
$-c_3=\Sigma - \frac{1}{a} \Rightarrow (abcd)(-c_3)=-(\Sigma  bcd) \Rightarrow c_3=-\frac{b}{3} $, $(\binom{4}{1}=\binom{4}{3})$
Similarly,
$c_2=-\Sigma \frac{1}{ab} \Rightarrow (abcd)(c_2)=-\Sigma cd=0$

$-c_3=-\Sigma \frac{1}{abc} \Rightarrow (abcd)(c_3)=\Sigma  d=0 $
$c_4=\frac{1}{abcd}=-\frac{1}{3}$
So the wanted polynomial is: $3(x^4-\frac{m}{3}x^3-\frac{1}{3})$


geometry
parallelogram
geometry proposed

Points $ P$ and $ Q$ do not lie on the diagonal $ AC$ of a parallelogram $ ABCD$ and satisfy $ \angle ABP\equal{}\angle ADP, \angle CBQ\equal{}\angle CDQ.$ Prove that $ \angle PAQ\equal{}\angle PCQ.$

Construct a line which passes through P and parallel to AD, it cuts AB and CD at X, Y, similarly for ZT.
$ \Delta DZP\sim \Delta BXP \Rightarrow \frac{PZ}{PX}\equal{}\frac{DZ}{BX} \Rightarrow \frac{PZ}{AZ}\equal{}\frac{PY}{CY}$
$ \Rightarrow \Delta AZP\sim \Delta CYP \Rightarrow \angle PAD\equal{}\angle PCD$
Similarly, $ \angle QAD\equal{}\angle QCD$
$ \Rightarrow \angle PAQ\equal{}\angle PCQ$


Find all number triples $(x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is 2.

[hide]We have \begin{eqnarray} x+yz &=& 2\\y+xz &=& 2\\z+xy &=& 2\end{eqnarray} Substracting $(1)$ from $(2)$ gives $x(z-1)=y(z-1)\iff z=1\vee x=y$. If $z=1$, then $x+y=2,\quad xy=1\implies x=y=z=1$. If $x=y$, we have $x+xz=2,\quad z+x^2=2$, and solving we get $x=y=1\vee -2$ and $z=1\vee -2$. Finally, $(x,y,z)=\boxed {(1,1,1),(-2,-2,-2)}$.[/hide]

[hide]
Given: $x+yz=2$ (A), $y+zx=2$ (B), $z+xy=2$ (C)

Subtracting (B) from (A), we get $(x-y)+z(y-x)=0\Longleftrightarrow (x-y)(1-z)=0$.

If $x\ne y$, then $z=1$, which gives $x+y=2$ from (A) and $xy=1$ from (C).
Solving the system, we get $x=1$ and $y=1$, which contradicts $x\ne y$.

If $x=y$, then we have $x+xz=2$ from (A) and $z+x^2=2$ from (C).
Combining these two equation gives $x+x(2-x^2)=2\Longleftrightarrow x^3-3x+2=0\Longleftrightarrow (x-1)^2(x+2)=0$
If $x=1$, then we have $y=1$ and $z=1$.
If $x=-2$, then we have $y=-2$ and $z=-2$.

Hence the answers are $(1,1,1)$ and $(-2,-2,-2)$.
[/hide]

Edit: chess64 already posted his solution while I was writing my solution.... :P


I'm not too sure about this...

Are these two statements true or false?

1. Vapor pressure of a liquid increases as the surface area increases.

2. Liquids with higher surface tension have higher viscosities.

Here's what I think, but I'm not too sure:

[hide]1. False, I would think the vapor pressure would stay the same, in equilibrium.  Also, there is no mention of temperature increase or decrease.

2. True, because the intermolecular forces holding the molecules in a liquid with high viscosity also will make the surface tension strong.[/hide]

Thanks!

I think it would be [hide]false, true, too![/hide]

Vapor Pressure of a liquid is a property of the liquid and doesn't change as the surface area changes.  However, if we can make this untrue if we talk about a liquid mixture like water with a surfactant like soap, that is present in just enough concentration that the mycells are formed and the surface of the water is coated.  If we then increase the surface area, (moving the surface lower in Florence flask) the surface will be partially uncoated with the surfactant and the effective vapor pressure will increase.  A greater change in occur in the speed of the vapor evolving from the liquid.

The first answer is correct unless there is a significant difference in the surface tension, for a material such as a soap.


Who here is planning to go to GHP? (If you made it  :) )

Me!

Is everyone here on the Georgia ARML team? I nearly made it this year... next year definately.

[quote="drunner2007"]yeah, what happens if you made ghp and arml?[/quote]Nothing, I should think.  There shouldn't be a conflict, as GHP doesn't start until well after ARML (see link in code below), and ARML's only for a weekend, anyway.[code]http://admin.doe.k12.ga.us/gadoe/blogs/ghpblog.nsf/2/CBYE-6G7RVC/$\$$File/2006GHPCalendar.pdf[/code]

Yeah I was worried earlier that I was going to have scheduling conflicts with the practices. But I didn't realize that it was during school. Now I know.  :D But yeah, as Boy Soprano said, you shouldn't have any scheduling conflicts... GHP starts around June 11th-ish and we get back from ARML on 6/6/06... my birthday (a good thing, or a bad thing?  :? )

@ drunner2007:

Mark, I presume?


Let a,b,c be non-negative real numbers.Prove that:
     $ \frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y}\ge\frac{x\plus{}y}{x\plus{}y\plus{}2z}\plus{}\frac{y\plus{}z}{y\plus{}z\plus{}2x}\plus{}\frac{z\plus{}x}{z\plus{}x\plus{}2y}$    (R.M.T., Marin Chirciu)

Let $ x\plus{}y\plus{}z\equal{}a ,      xy\plus{}yz\plus{}xz\equal{}b   ,        xyz\equal{}c            \Longrightarrow$

The inequality equivalents to $ 2a^5\plus{}3bc\plus{}11a^2c \geq ab^2\plus{}7a^3b$

By $ Schur's$ inequality $ a^3\plus{}9c \geq 4ab$

$ c\geq \frac{4ab\minus{}a^3}{9}$ and $ 2a^5\plus{}c(3b\plus{}11a^2) \geq 2a^5\plus{}(3b\plus{}11a^2).\frac{4ab\minus{}a^3}{9} \geq ab^2\plus{}7a^3b \Longleftrightarrow$

$ (a^2\minus{}3b)(7a^2\minus{} b)\geq0$ which is obviously true.  :wink:

I think this one is more nice,
let put $ a\equal{}x\plus{}y$ ,$ b\equal{}y\plus{}z$, $ c\equal{}z\plus{}x$
then the inequality becomes,
$ \sum\frac{a\plus{}b}{2c} \geq \sum\frac{a}{b\plus{}c} \plus{} \frac{3}{2}$
we have with Cauchy-Scwarz,
$ \frac{1}{a\plus{}b}\leq\frac{1}{4}\left(\frac{1}{b}\plus{}\frac{1}{c}\right)$
$ \Leftrightarrow \frac{a}{b\plus{}c}\leq\frac{1}{4}\left(\frac{a}{b}\plus{}\frac{a}{c}\right)$
then $ \sum\frac{a}{b\plus{}c}\leq\frac{1}{4}\left(\sum \frac{a\plus{}b}{c}\right)$
then it's suffice to prove that,
$ \sum \frac{a\plus{}b}{c} \geq \frac{3}{2}$ wich is Nessbit.

Add 1 to each of the fractions to get

$ \sum\frac{x\plus{}y\plus{}z}{y\plus{}z}\ge\sum\frac{2x\plus{}2y\plus{}2z}{x\plus{}y\plus{}2z}$

Divide by $ x\plus{}y\plus{}z$ to get

$ \sum\frac{1}{y\plus{}z}\ge\sum\frac{2}{x\plus{}y\plus{}2z}$

We have by AM-HM $ \frac{1}{2}\left(\frac{1}{y\plus{}z}\plus{}\frac{1}{x\plus{}z}\right)\ge\frac{2}{x\plus{}y\plus{}2z}$.  Cycling the variables and adding the three inequalities yields the desired inequality.

[quote="baleanu"]Let a,b,c be non-negative real numbers.Prove that:
     $ \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y}\ge\frac {x \plus{} y}{x \plus{} y \plus{} 2z} \plus{} \frac {y \plus{} z}{y \plus{} z \plus{} 2x} \plus{} \frac {z \plus{} x}{z \plus{} x \plus{} 2y}$    (R.M.T., Marin Chirciu)[/quote]

using well-known ineq
\[ \frac {1}{a} \plus{} \frac {1}{b} \ge \frac {4}{{a \plus{} b}}
\]
,we have
\[ \begin{array}{l} \frac {{x \plus{} y}}{{(x \plus{} z)(y \plus{} z)}} \le \frac {{x \plus{} y}}{4}(\frac {1}{{x \plus{} z}} \plus{} \frac {1}{{y \plus{} z}}) \equal{} \frac {1}{4}(\frac {x}{{x \plus{} z}} \plus{} \frac {y}{{x \plus{} z}} \plus{} \frac {x}{{y \plus{} z}} \plus{} \frac {y}{{y \plus{} z}}) \\
... \\
\end{array}
\]
Writing the analoguous ineq ,and adding them up .and using cyclic ineq ,we have  (Q.E.D) :D

We have $ \frac{1}{2}\left(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x}\right)\equal{}\frac{1}{2}\left(\frac {x^2}{xy \plus{} xz} \plus{} \frac {y^2}{yz \plus{} xy}\right)$  and  $ \frac{1}{2}\left(\frac {x^2}{xy \plus{} xz} \plus{} \frac {y^2}{yz \plus{} xy}\right)\ge\frac{1}{2}\frac{(x\plus{}y)^2}{2xy\plus{}yz\plus{}zx}$ (Andreescu). but $ \frac{1}{2}\frac{(x\plus{}y)^2}{2xy\plus{}yz\plus{}zx}\ge\frac{x\plus{}y}{x\plus{}y\plus{}2z}$.  (is obviously true).Cycling the variables and adding the three inequalities yields the desired inequality.


geometry
3D geometry
analytic geometry
calculus
integration
calculus computations

Find the volume of the solid that lies between the paraboloids z = x^2 + y^2 and z = 4x^2 + 4y^2, and
between the planes z = 1 and z = 4.

z=x^2+y^2 is a cone.
Same for z=4x^2 +4y^2, just smaller.

So it's asked for the volume between two cones from z=1 to z=4

Do it in cylindrical coordinate:
1=< z =<4
0=< theta=<2pi
z/r =< r =< z  (when z=x^2+y^2=r, r=z ;  when z=4x^2 +4y^2=4r, r=z/4)

So the Vol is equal to the triple integral of 1, where
dr ranges from z/4 to z
dz ranges from 1 to 4
d(theta) ranges from 0 to 2pi.

For final answer: it should be 45pi/2

Thanks for your reply :)

But from your volume equation, shouldn't the final answer be 45pi/4 instead?

I tried to do it another way but came up with 45pi/8.

Here's how I did it:

Volume inside the large parabolid (from z = 1 to z = 4) = double integral of (4-r^2)*r drd(theta) from 0 to 2, from 0 to 2pi - double integral of (1-r^2)*r drd(theta) from 0 to 1 and from 0 to 2pi.

Volume inside the small parabolid (from z = 1 to z = 4) = double integral of (4-4*r^2)*r drd(theta) from 0 to 1, from 0 to 2pi - double integral of (1-4*r^2)*r drd(theta) from 0 to 1/2 and from 0 to 2pi.

I substracted the small volume from the large volume and arrived at 45pi/8, did I do anything wrong?

you are actually correct.
I ran over my method. I've made a couple of mistakes!
In my way of doing this, r should range from root(z)/2 to root(z)
and I missed the r (det of the polar tranformation) in my integrand.

I hv 45pi/8 as well!


number theory
relatively prime
algebra unsolved
algebra

1)Let a,b element of Z+
    if a+b=12  calculate  of maximum value of (1/a)+(4/b).Please don't use derivative.

2)Show that
      i)p and q relatively prime iff p.q and p+q relatively prime
      ii)p and q relatively prime iff  p.q and p-q relatively prime.

This isn't really advanced level...

[quote="mathmaniack"]1)Let a,b element of Z+
    if a+b=12  calculate  of maximum value of (1/a)+(4/b).Please don't use derivative.[/quote]
for the maximum, b should be minimal (just work it out) so $(a,b)=(11,1)$

[quote="mathmaniack"]2)Show that
      i)p and q relatively prime iff p.q and p+q relatively prime
      ii)p and q relatively prime iff  p.q and p-q relatively prime.[/quote]
suppose they had a common factor, then both $p,q$ would contain the common factor. contradiction, proven.

Are you sure for the  first problem?I found 0,75  for a=4,b=8 using derivative.
I am looking for a solution without derivative.

Well, then you missed in your derivates. Since $a=11$ and $b=1$ results in $\frac{45}{11}$.

In fact, Z+ also contains 0 so principially we can say a=12, b=0 and the thing skyrockets to $+\infty$.

But honestly, I think you meant [b]minimum[/b] value :D


algebra
polynomial
vector
linear algebra

for a polynomial of degree 2, f(x) = a + bx + cx^2

I see that the monomials 1, x, and x^2 span P2 but how do I verify that they're linearly independent?

Could someone also explain in simple terms what exactly a linear space is? Its my understanding that its a set of equations that can only be scaled or added. I'm using the 3rd edition Bretscher book for a summer lin alg class. While I'm on the topic of asking a lot of questions can someone link me to a thread or tutorial that explains how to use a code for math symbols in this forum.

Thanks, your help is much appreciated.

A set of vectors $ \{v_1,v_2,\dots,v_i\}$ is linearly independent if the equation $ a_1v_1 \plus{} a_2v_2 \plus{} \dots \plus{} a_iv_i \equal{} 0$ does not have a nontrivial solution. That is a solution other than $ a_1 \equal{} a_2 \equal{} \dots \equal{} a_i \equal{} 0$.

So you try to solve $ a_1\plus{}a_2x\plus{}a_3x^2\equal{}0$. (you should find that it does not have any nontrivial solutions)


calculus
integration
algebra
partial fractions
calculus computations

hello, make the substitution $ \sqrt{2x^2\plus{}5x\plus{}7}\equal{}x\sqrt{2}\plus{}t$.
Sonnhard.


Salutare! Am o intrebare legata de o problema cu un caracter foarte general..aflati n nr natural astfel incat $1+2+3+...+n$ sa fie patrat perfect.pt $n=8$ se verifica!

Poti determina $n$ prin intermediul unei ecuatii Pell. http://www.mathlinks.ro/Forum/viewtopic.php?t=75328 Aici gasesti mai spre sfarsit rezolvarea la ce vrei tu insa folosesc matrici. Sper ca esti familiar cu matricile. Daca nu citeste cate ceva despre solutiile generale ale ecuatiilor de tip Pell($x^2-dy^2=1$) ;) .

Imi pare rau,dar nu cunosc astfel de ecuatii. la nivel de clasa a 8a se poate rezolva?

Oricum ai lua-o tot la ecuatii de tip Pell ajungi.

[quote="andyciup"]la nivel de clasa a 8a se poate rezolva?[/quote]Raspuns simplu: nu.


Hello,

Let $ \alpha$, $ \beta$ and $ \gamma$ be the angles of a triangle with ($ \alpha$;$ \beta$)<$ \frac{\pi}{2}$, and $ sin^2\alpha\plus{}sin^2\beta\equal{}sin\gamma$.
Prove that $ \gamma\equal{}\frac{\pi}{2}$

$ \Longleftrightarrow \sin \alpha \cdot \sin \alpha \plus{} \sin \beta \cdot \sin \beta \equal{} \sin \alpha \cdot \cos \beta \plus{} \sin \beta \cdot \cos \alpha$

but $ \alpha \plus{} \beta < \frac{\pi}{2} \ \Longrightarrow \ LHS < RHS$ and $ \alpha \plus{} \beta > \frac{\pi}{2} \ \Longrightarrow \ LHS > RHS$

and we have $ \alpha \plus{} \beta \equal{} \frac{\pi}{2} \ \Longrightarrow \ LHS \equal{} RHS$ then $ \gamma \equal{} \frac{\pi}{2} \ \Longleftrightarrow \ LHS \equal{} RHS$


modular arithmetic
superior algebra
superior algebra unsolved

Show that $ x^2\plus{}x\plus{}1$ is irreducible over the field $ \mathbb{Z}_{2}$.

Can anyone check my work?

Work: If $ x^2\plus{}x\plus{}1$ was reducible, then there is some $ a\in \mathbb{Z}_{2}$ such that $ x\minus{}a|x^2\plus{}x\plus{}1$. This means that there is some $ a$ such that $ a^2\plus{}a\plus{}1\equiv 0\pmod{2}$. However, no such $ a$ exists, as seen by plugging in $ a\equal{}0$ and $ a\equal{}1$. So $ x^2\plus{}x\plus{}1$ is irreducible over the field $ \mathbb{Z}_{2}$.


Hi,

This is bit stupid idea but I'm wondering is it possible to have a forum that just holds out the homeworks. Not that it's going to answer them but:

[b]Help people understanding the concept[/b]

[b]Teach them to learn how to solve some questions in case they don't know[/b]

[b]If they're doing extra work problems, like from math books, the homework forum would be good place to hold them.[/b]

I'm registered to the science forum and I found it very nice thing.

:)

Just suggestion!

disagree, cos any homework can be put into the existing categories, if you add another 'homework' forum then it would be chaos.
P.S.which science forum do you go to? sciforums or SFN?


It is discussed at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=279363[/url] starting at post #16.


integration
calculus
function
trigonometry
calculus computations

[hide="Solution"]Use $ u$-substitution, and let $ u \equal{} x^2\plus{}1$, Therefore, $ du \equal{} 2x dx$, and $ u\minus{}1 \equal{} x^2$. $ \displaystyle\frac{x^3dx}{(x^2\plus{}1)^3} \equal{} \displaystyle\frac{u\minus{}1du}{2u^3} \equal{} \displaystyle\frac{du}{2u^2} \minus{} \displaystyle\frac{du}{2u^3}$. The integral of the latter function is $ \minus{}\displaystyle\frac{1}{2u} \plus{} \displaystyle\frac{1}{4u^2}\plus{}C \equal{} \displaystyle\frac{1\minus{}2u}{4u^2} \plus{} C$, and we substitute to obtain $ \boxed{\displaystyle\frac{\minus{}2x^2\minus{}1}{4(x^2\plus{}1)^2}\plus{}C}$.[/hide]

Thanks. These two answers are exactly the same: what happens if we use the substitution $ x \equal{} \tan \theta$?

$ \int \frac{x^{3}}{(x^{2}\plus{}1)^{3}}\ dx$

$ \equal{} \int \frac {\tan^{3}\theta\sec^{2}\theta}{\sec^{6}\theta}\ d\theta$

$ \equal{}\int \sin^{3}\theta \cos \theta \ d\theta$

$ \equal{} \frac {\sin^{4}\theta}{4}$.

Subtracting the other answer, we get $ \frac{1}{4}$, that is they differ by a constant (the constant of integration).


algorithm
ratio
function
rotation
trigonometry
geometry
geometric transformation

*edited*

I'm doing a survey to see how successful the algorithm was in inducing pleasure, basically im trying to measure it for the write up. Feel free to have a go (requires audio)…

http://www.thedivide.co.uk/study.html

I would much appreciate it ! :D

Heres a pick of it running as VSTI in SX2...

http://www.thedivide.co.uk/iiiath/IIIATH.JPG

Daniel

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

the old post....

Hi, nice forum

I’m basically stuck and a little stressed out with a joint university project/personal interest of mine and I am posting here because I don’t know where else to. Sorry if its in the wrong section  :ninja:

My situation is, I have created a midi sequencer in a piece of software called max MSP and what I’m doing is an experiment to see if you can generate aesthetically pleasing music through pure mathematical functions. I am in the middle of doing a literature report and I have looked into geometrical aspects and curiousities in art and how they influence aesthetice beauty (fib, golden ratio, etc)

The thing is I have already made midi sequencer and basically I used a formula that was already in an example patch. What I did is added a few things to it, I’m pretty good with logic and I know what the formula does. I just cant explain in proper mathematical terms (I know this would be classed as a function or something)

The way the system works is I have a counter going from 0 to inf. Every number outputted from the counter goes into a C like expression box which is then manipulated by this formula and sent out as a new number…

int($f2 x pow ($f1/$f3\3) x sin (8 x atan(1) x $f4 x $f1/$f3) + $f5)
Each $f in that formula is connected to variable faders which output various numbers….

$f1 = input value, it counts up from zero (from counter)
$f2 = Range
$f3 = Rotates??
$f4 = Root note (0-127) ( basically I can transpose the whole thing via a midi kebd.

If say my input from the counter counts up from 0-1000 I would end up with a midi sequence like this...

http://www.thedivide.co.uk/MSP/MSP%20MATH%20BANG%203.JPG

if i change the parameters...

http://www.thedivide.co.uk/MSP/MSP%20MATH%20BANG%201.JPG

or if i switch the sine for tan I get

http://www.thedivide.co.uk/MSP/MSP%20MATH%20BANG%205%20SHATTERED%20CHOAS.JPG

Basically my problem is im not brilliant with maths (I wish I was tho), I did try to figure bits of it out

For example..

(44 x pow (530/1000\3) x sin (8 x atan(1) x 480 x 530/1000) + 48) = 51

8 x atan(1) = 360, so thats got something to do with rotation

as 530 counts up the output number modulates probably as a result of x sin (blah). I dont know where to start when it comes to explaining all this to accademic standard, although im sitting a music tech degree, it would be great if I can explain it in the correct terms

http://www.thedivide.co.uk/MSP/Examples.JPG

And if you want to hear this thing played live (not pure maths, im looping and transposing the $f4 value)...

http://www.thedivide.co.uk/iiiath/IIIATHMSP.mp3

http://www.thedivide.co.uk/iiiath/IIIATHMSP2.mp3

Any help would be much appreciated

Dan

OK, so you want to explain.... what all the parameters do, or why did you choose such function, or what precisely?

And while you're answering this here's a couple of technical questions: does pow(a/b\c) mean $\left(\frac{a}{b}\right)^c$? f1 is the counter, that's clear, f2 and f3 and f5 are some constants I suppose? But what is f4? I don't know too much about midis - is that the number of the instrument that plays that part?

(And by the way, a very interesting idea :).)

Hi

Firsly thanks for your reply and bash at this, I am glad you like the idea.

the way this works is its a number thats changed into another number as it counts. That number then becomes the value of the note. Your standard midi keybourd has 0-127 notes 0 Been the very bottom C and 127 Been the very top C (I think, mite not be C but you get the idea). $f5 is the transposition value of the whole furmular, if you seen the pics the notes seem to modulate over a center point. That center point value is$f5. Im not certain with your arithmetic as Im only really getting my head around this myself (you could be right)

I did make a mistake in the previous post tho, maybe this will clear it up slightly, its actually...

$f1 = input value, it counts up from zero (from counter)$f2 = Range (see below)
$f3 = Frequency?$f4 = Rotation?
$f5 = Root note (0-127) eg C,C\#,D\#,E,F
I did wonder if im on the right lines with this... int($f2 x pow ($f1/$f3\3) x sin (8 x atan(1) x $f4 x$f1/$f3) +$f5)

atan(1) = 45 degress
8 x 45 = 360

what i dont undertsand is why its not just...

sin (360 x $f4 x$f1/$f3) +$f5)

 :|

ok that pow bit I think means...

pow(a/b\c) =

a/b to the power of (or squared by?) c

As the \ does...

"If the function requires a comma between arguments, the comma must be preceded by a backslash (\) so that Max will not be confused by it. For example: pow($i1\,2).

Arithmetic operators understood by expr are: +, -, *, /, %. ther operators ~ (one’s complement), ^ (bitwise exclusive or), &, &&, |, ||, and ! (not).

C language functions understood by expr are: abs, min, max, sin, cos, tan, asin, acos, atan, atan2, sinh, cosh, tanh, int (convert to integer), float (convert to float), pow, sqrt, fact (factorial), exp (power of e to x), log10 (log), ln or log (natural log), and random. Additional functions can be added by means of external code resources placed in Max’s startup folder."

OK, now I understand, the mistake you made in the first post confused me :).

To atan(1) - its value is not 45 but $\pi/4$ - it's in radians and that's why you can't write 360 instead of 8 x atan(1). (You could write 2 x 3.14159.... but atan(1) is probably more convenient.)

I've tried to plot the function you describe here but it didn't look too much like your pictures, could you post some values of f2 .. f5 you used plus pictures (if possible)?

At the min I have the formula down to...

o = (h x (i/f)^3) x sin(2 x pi x t x i/f) + n

i = input
h = height range
f = frequency
t = rotation
n = root note
o = output

so if

$f1 = 1,2,3,4,5,6,7,8,9,10,11, to 1000
$f2 = 44
$f3 = 1000
$f4 = 480
$f5 = 54
=
http://www.thedivide.co.uk/MSP/MATH%20WITHOUT%20DELTA.JPG
however if I take the counter input (same one used to generate the notation) and wire it into a new expression block with this formula...
(
$f2/pow($f3\,($f1-$f4)%125/124.))

http://www.thedivide.co.uk/MSP/MATH%20WITH%20DELTA.JPG

That basically slows and speeds up the timing making it look more interesting

Now I have the formula just dont know the correct mathematical terminology for it.

Any ideas would be killah

[quote="The Divide"]sure I will do, leave it to me, what are you plotting them on btw?[/quote]Oh, I wrote my own program for plotting generic things...

[quote="The Divide"]however if I take the counter input (same one used to generate the notation) and wire it into a new expression block with this formula...

(f2/pow(f3\,(f1-f4)%125/124.))

http://www.thedivide.co.uk/MSP/MATH%20WITH%20DELTA.JPG

That basically slows and speeds up the timing making it look more interesting[/quote]Er, I don't really understand this part. Is that some brand new formula? The graph you're showing here is definitely not of this function, could you explain a little more what you're doing here?

As to the former one, your function is basically a product of sine and a cubic function. I don't know about any special name for this kind of function - it's just plain sin(x) but it's amplitude increases with time - the amplitude is equal to the value of your cubic function. f1 is time, f2 and f3 affect the amplitude which is equal to $\frac{f_2}{f_3^3}\times f_1^3$ (cubic function multiplied by constant f2 / f3^3; the amplitude increases with time), f4 is the frequency of the sine - frequency = 1 / period = $\frac{f_4}{f_3}$ (so every time f1 increases by 1 the sine advances by f4/f3 of its period) and f5 is the root note.

It's interesting to notice how insanely high is the frequency you use. If you draw a line instead of points, you'd notice that between every two of your points there's about one half of a sine wave that you skipped out. It's like drawing a sin(x) function and taking values of sin(2 radians), sin(4 radians), sin(6 radians)... if you plot only these values, they won't resemble a sine function but they may, it depends on the frequency. Your frequency is close to 1/2 (but not equal, that's a crucial part) so the resulting function looks like a double sine wave.

To sum it up, it's a sine with variable (increasing) amplitude and a very high frequency. (This frequency is close to 1/2 per increment of f1 or per sample - it's close to 1/2 of the sampling frequency. This creates an illusion of a double sine wave of much smaller frequency.)

man, that really helped. Im gonna crack on with this now  :ninja:

thanks

Dan

No problem. If you ever have any further questions, feel free to ask. :trampoline:

Wow, some interesting results. I like q16c the most although I also think that your composition still lacks something... what is it? Theme, diversity, complexity? It might be interesting to try one more level of randomness - random parts of your composition changing seamlessly into other random parts of your composition... and by the way, you can also ask in the Music & Art forum, there should be some more competent people.


This is basically a Linguistics version of the Science Olympiads, and it will be held for the first time this year.

There will be a live competition in Pittsburgh, Philadelphia, and Boston, but if enough people sign up, there will be an Internet Competition too.

And this will also decide who will represent the US at the ILO, International Linguistics Olympiad.

There's a lot more information at the website, [url]http://www.namclo.org[/url]

Hey - I was just about to post this.  I'm one of the organizers for the Boston test.  If anyone's in the Massachusetts area is interested, feel free to contact me.  We may even come to your school and give a little presentation/problem solving class.

Here are some sample problems:
http://www.eecs.umich.edu/~radev/namclo/

This year it's more computational than in previous years.

- Catherine

It would be nice (and I would be more likely to participate) if this were more like the ILO in difficulty and content.

It would be cool if we had a language/linguistics forum in arts/sciences. I think there is a relatively high correlation between math and linguistics.


ratio
geometric sequence
arithmetic sequence

[quote="espark52"]can someone tell me what it means when a term of an arithmatic sequence "tends to zero?"[/quote]

You mean a geometric sequence, since a non-constant arithmetic sequence always hits 0.

mathgeniuse^ln(x) is correct.

Basically, all it means that as you keep equating more and more terms, they get smaller and smaller and approach $0$.

Let the common ratio of the sequence be $r$. If $|r|<1$, then the terms tend to $0$. If $|r|> 1$, then the terms approach infinity.

Try seeing what happens when $|r|=1$ ($r=\pm 1$).

to see what happens....

[hide="in here"]If r=1, then all of the terms are m m m m m m m m m m m m m m m m m m m m m m m m m m m m m ..........

If r=-1 then the terms are m and -m alternatingly.[/hide]


function
integration
Putnam
algebra
polynomial
real analysis
real analysis unsolved

Let $f(x)$ be a continuous function on $[0, 1]$ such that
\[ \int_0^1 {x^k f (x)} = 0 \]
for $k = 0, 1, \dots n - 1$ and
\[ \int_0^1 {x^n f (x)} = 1 \]
Prove that maximum of $|f (x)|$ on $[0, 1]$ is greater than $2^n (n + 1)$.

Let $M$ be the maximum value of $|f(x)|$ on $0\leq x\leq 1$, we are to prove

\[ \left|\int_0^1 x^n f(x)\ dx\right|\leq \frac{M}{2^n(n+1)}.  \]

Let's calculate $\int_0^1 (x-t)^n\ f(x)\ dx.$

Kunny's idea is correct.  This is an old Putnam problem, I believe.

$\int_0^1x^nf(x)dx=\int_0^1(x-t)^nf(x)dx\le M\int_0^1(x-t)^ndx=\frac{M((1-t)^{n+1} +t^{n+1})}{n+1} = \frac{M}{2^{n}(n+1)}$

 for $t=\frac 1 2$.

Very nice problem! But, can someone give example of such function?

[quote="titeo"][quote="kunny"]Is $P_n(x)$ Legendre's polynomial?[/quote]
No, it isn't. --> http://en.wikipedia.org/wiki/Legendre_polynomials
But I built them by depending on Legendre's polynomial.[/quote]

Thank you, titeo. :)

kunny


The perimeters of two regular polygons with the same number of sides are together, as is 6 to 15. Knowing that the far-side of the second polygon is 60V2 cm, calculate the far side of the first polygon.

Portuguese question:

Os perímetros de dois polígonos regulares com o mesmo número de lados estão entre si como 6 está para 15. Sabendo que a medida do lado do segundo polígono é 60V2 cm, calcule a medida do lado do primeiro polígono.

Hi luis. In the Portuguese version, you say "a medida do lado", so in English it would be "the length of a side" (and not the far side).

Anyway as the perimeters are in the ratio of $ 6$ to $ 15$ = $ 3$ to $ 5$ $ \equal{}36\sqrt{2}$ to $ 60\sqrt{2}$, the length is $ 36\sqrt{2}$.

All the best.

I think Andrew meant $ 2 : 5$. The answer is $ 24\sqrt{2}$.