number theory
greatest common divisor
linear algebra
matrix
linear algebra unsolved

Let $A$ be the matrix $\left(\begin{matrix}3 &2 \\4 &2 \end{matrix}\right)$ and for positive integer $n$, define $d_{n}$ as the greatest common divisor of entries of $A^{n}-1$,where $I=\left(\begin{matrix}1 &0 \\0 &1 \end{matrix}\right)$. Prove that $d_{n}\rightarrow \infty$ as $n \rightarrow \infty$.

Did i state the problem clearly, guys?  :blush:

or nobody interest with this one??

If that's meant to be $\begin{bmatrix}3&2\\4&3\end{bmatrix}$, it was asked  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=115999]here[/url].

If it's as you wrote, it looks false.

Its seems that i didn't write any typo. But I'll try the same technique in the psot u metioned above.

i thinks its true...  and the diagonal (not main diagonal) has entries $a$ and $2a$.. the other seems unbounded...

The first 20 gcds: 1,1,9,1,1,9,1,85,27,1,1,9,1,1,9,85,1,27,1,1.

That doesn't look like a sequence that's tending to $\infty$; it's extremely likely that it's equal to 1 infinitely often.
It's also really hard to prove anything about it.

I ran the matrices mod $p$ for each prime $p<100$: the first indices for which the gcds are divisible by $p$ are 3, 8, 48, 110, 168, 8, 45, 528, 28, 30, 36, 40, 308, 2208, 2808, 1740, 3720, 66, 5040, 1332, 1248, 41, 1980, and 96. The only primes less than 100 which ever divide $A^{n}-1$ for an odd index $n$ are 3, 19, and 83.


induction
number theory solved
number theory

Prove that for natural a>=3 there are infinitely many natural numbers x such that x/a^x-1.

Let p be a prime divisor of a-1 : a - 1 = k*p , p > 1
The sequence x(n) = p^n  is a solution .
We show that by induction : manifestly it's thru for n=0 ... Suppose a^(p^n) ==1 (mod p^n)
a^(p^(n+1)) - 1 = (a^(p^(n))+1))^p - 1 = (a^(p^n)-1)*((a^(p^n))^(p-1) + ... + a^(p^n)) + 1)
But ((a^(p^n))^(p-1) + ... + a^(p^n)) + 1 == 1 + .. +1 = p (mod p^n)
So a^(p^(n+1)) - 1 is divisible by (p^n)*p=p^(n+1) ..  (q.e.d)

Thanks! It's a different type of soln from mine, and that's what I was looking for. This is mine: we define the sequence x_n recursively: we find an x_0 s.t. x_0/a^x_0-1. This is not hard to do (it can be done in many ways. Then x_(n+1)=a^x_n-1. I think this must work.

Sorry, me precedent message was in a hurry and I made a mistake..
The correct  induction initialization is :
for n=1 its manifestly thru by using the little Fermat theorem : a^p = a*(a^(p-1))==a==1 (mod p)


When using the Texnic editor, how can you type so that the equations appear in the text, not on a seperate line, without the equations being cramped? Basically, how can you type in the Texnic editor so that it appears as in this forum?
Thanks.

I usually type text in between the two equations.  I suppose you could also use tabbing.  For a detailed explanation check out [b]"tabbing"[/b] in the help index.

I don't think that's what he's talking about.

Try starting the equation with the \displaystyle command.  Without \displaystyle, it looks like this:

\[\textstyle{\sum^{\infty}_{k=1}\frac{1}{k^2}=\frac{\pi^2}{6}}\]

With \displaystyle, it looks like this:

\[\sum^{\infty}_{k=1}\frac{1}{k^2}=\frac{\pi^2}{6}\]

If you want, you can couple \displaystyle with braces, \displaystyle{math stuff}, to have just part of the equation rendered in \displaystle mode.

\displaystyle is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16057]standard[/url] on the AoPS forums.  If you want to disable it here, you can use the opposite of \displaystyle, \textstyle.

I think he is referring to the tendency of equations on the same line to crunch like [hide="this"] $y=x$      $y=x^2$ [/hide]
Rather then to be seperated like [hide="this"]$y=x$ spacing $y=x^2$[/hide]  By default spacing is insignificant in LaTeX.


Determine the smallest real number $C$ such that the inequality \[ C(x_1^{2005} +x_2^{2005} + \cdots + x_5^{2005}) \geq x_1x_2x_3x_4x_5(x_1^{125} + x_2^{125}+ \cdots + x_5^{125})^{16}  \] holds for all positive real numbers $x_1,x_2,x_3,x_4,x_5$.

[quote="Arne"]It is easy to see that $C = 5^{15}$ works by Muirhead.

Should I really give a "complete" solution here, with all symmetric sums? :D[/quote]

Could anyone post the solution mentioned above ? Thanks

Why? I'm sure you've seen that kind of solution MANY times already.

It may just be that I'm as dumb as a rock, but can you explain more? :(


[quote="April"]In triangle $ABC$: $2\cos\frac{B}{2}\cos\frac{C}{2}=\frac{1}{2}+\frac{b+c}{a}\sin\frac{A}{2}$
Find the measure of angle A[/quote]
$\to sinA/2=1/2$

Using Newton formula $\frac{b+c}{a}=\frac{\cos(\frac{B-C}{2})}{\sin (\frac{A}{2})}$ http://mathworld.wolfram.com/NewtonsFormulas.html
we get $2 \cos (\frac{B}{2}) \cos (\frac{C}{2}) = \frac12+\cos (\frac{B-C}{2})$ and this simplyfies to $\cos (\frac{B+C}{2})=\frac{1}{2}$.
So $\frac{B+C}{2}= \frac{\pi}{3}$ and $A= \frac{\pi}{3}$ :D


trigonometry
calculus
derivative
calculus computations

Hi
Im having difficulty determining how to obtain the correct answer for sine and cosine derivatives that have numerical coefficients.
An example is
y=(cos2x)^2
Using the chain rule.
u=cos2X
y=u^2
dy/du=2u
dy/dx=-sin2X

-sin2X x 2u
2(cos2X) x -sin2X
After applying the chain rule this is the answer i obtain which is incorrect.
2(cos2X)-sin2X

According to the text book the correct answer is.
-4cos 2X sin 2X
I dont understand how they end up with a 4 instead of a 2.

Any help appreciated
Thanks

[quote="Ry12"]
dy/du=2u
dy/dx=-sin2X
[/quote]
The second line should be $du/dx=-2\sin 2x$: an extra $2$ comes from the chain rule for $\cos 2x=\cos v$, $v=2x$.

how exactly does cos2X = cos v
yield an extra 2? can you please take me through all the steps.
Thanks

Edit:
I think i understand it now.
v=2x
dv/dx=2
du/dx=-sin2x
2 x -sin2x
= -2sin2x
Is this correct?


Does it help?

If so, what songs do you guys listen to?

I prefer classical violin or piano - I'm a violinist and a pianist.

Whatever I'm playing ATM, as many different recordings as I can find.

here are some links....

http://www.classical.com/album/5099706453420
http://www.classical.com/album/ATM-CD-1199  --- Gavotte by Joseph Gossec is my favorite (it was my solo last year :P)

If you don't like any of these, go to http://www.classical.com.

It helps not at all or tremendously; personally, I prefer http://www.youtube.com/watch?v=teA-BD5XSQk over most other songs.
It's rather relaxing, and best of all, it's a semi-internet meme (As in, there was an internet meme that used the program used to make this song, but this is not the internet meme).
However, it's Japanese; if you don't like it/think it's annoying, then I recommend some form of New Age music or videogame music. They help as well, just not as much, as they sometimes get on your nerves.

or go here...

http://www.youtube.com/watch?v=oeETRRsj1Ek&feature=related

 :lol:

It all depends per person; if you enjoy piano then listen to piano, if you don't then don't.

I find some music you can work very well with, other songs, like ones with humorous lyrics, are harder to work with because I find myself listening to the words themselves.

[url]http://www.classicfm.co.za[/url] and [url]http://www.classicfm.co.za/player[/url]

My favorite music station in the entire [i]world[/i]...and I live in Southern California!  :D

my parents think im crazy...but i listen to like z100..so like justin bieber, taylor swift, taiocruz(break ur heart is ,y fave song), jason derulo, shontelle, lady gaga(honestly not a huge fan)...etc.etc.etc

I always love enjoying the Classical music specially from subcontinent.

Lo fi hip hop is pretty good to listen to while studying.

lo fi hip hop!! praise,, it helps me concentrate so well, i especially like it for late night studying

Yeah, and there are always those 24/7 livestreams on youtube.

LOL one time when i was completing a writing problem on AOPS i listened to nyan cat 10 hours XDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

[quote=silverivy]Did it take 10 hours to write?  (because same)

Listening to classical music makes me stress when studying, as I'm also a violinist and get too into it.[/quote]

nope. It took about half an hour to write

The Bach Goldberg Variations is the best study music

Rap or pop is best while studying lol, but no joke, actually.

[quote=HudsonR]I like classical music, or listening to stuff like Two Steps From Hell and movie soundtracks.[/quote]

TSFH YASSSS
Which is your favorite from Skyworld?

I listen to dubstep and electrical music, but sometimes I just listen to audios
they make me feel a little less lonely

I tend to listen to either jazz or Chopin, but occasionally I can handle hearing rock while studying.

I listen mainly to rock music... Yes, the more noisy it get the more concentrated I get ;)

I prefer listening to good music that sounds nice, because it helps me concentrate better and feel nicer. Loud songs that don't really sound good but are kind of funny isn't a good choice to be honest.

I got to have a complete silence around me while I'm studying, otherwise it's really difficult for me to concentrate.

Listening to music really helps for me. Preferable instrumental rock or EDM.

I prefer listening to deep-house music or lounge and it helps me to learn smth

Update: I really like listening to youtube playlists of Chopin or Liszt, and also longer concertos like the Kissin version of the 1st Chopin concerto.

[url]https://www.youtube.com/watch?v=2WJhax7Jmxs[/url]     :-D

[quote=IceWolf10]Usually I "listen" to whatever song is stuck in my head.[/quote]

Usually when I do that it doesn't work as well as actually playing the music.

I find that piano music is really good for studying.

Tbh chillwave,retrowave and synthwave are the best genres you could listen to while studying. Just search these up on YouTube and you'll get various mixes to listen to. Also you might want to check out a bit of a channel called chill nation

I listen to either Brahms quartets or quintets, Mozart piano sonatas, or (very rarely) some Bach.

[quote=Catmate]I only listen to classical music. But when I listen while I work, it distracts me so much considering how much I am used to analyzing music and contemplating its deep layers and hidden melodies.[/quote]

lol, in some math class I did while we were doing a test, the teacher put on some music, but one of the students said they were paying too much attention to the music.

I don't usually listen to music while studying, but if I do it's usually EDM with hints of classical.

I listen to heavy metal and all kinds of rock and roll while drawing. Maybe even throw in some 80's Japanese city pop, 80's chinese pop music, 90's rap, and 90's Eurobeat. Basically anything from the 20th century. It really helps me get in the zone and focus

[quote=HZAOPS]For studying, is classical music better or music that's like more on the pop-energetic side?[/quote]

When studying, I tend to listen to classical piano and pop piano covers. You can see my Spotify playlist here: https://open.spotify.com/user/p1jxvd9euvx7ayl9dc3rnoxl2/playlist/07gdFaEKBRlnsG62XjTD2J?si=kzowCmt5SW2DpS8ZvdZ40w

[quote=HZAOPS]For studying, is classical music better or music that's like more on the pop-energetic side?[/quote]

When I'm doing math I prefer silence, but with more creative stuff like drawing, or just videogames, fast-paced, head-bobbing rap and rock help me get in the zone


$ xy\plus{}yz\plus{}zx\equal{}1$ and $ x\leq y \leq z$
then $ xz<\frac{1}{2}$

-) If $ 0\leq x\leq y\leq z$ then $ yz\geq zx\Leftrightarrow 1 \equal{} xy \plus{} yz \plus{} zx > 2zx$: OK!
-) If $ x < 0,z\geq 0$: OK!
-) If $ x \leq y \leq z < 0\Rightarrow 0<\minus{} z\leq \minus{} y\leq \minus{} x$: apply first case with $ ( \minus{} x)( \minus{} y) \plus{} ( \minus{} y)( \minus{} z) \plus{} ( \minus{} z)( \minus{} x) \equal{} 1$ then OK!

[quote="FOURRIER"]$ xy \plus{} yz \plus{} zx \equal{} 1$ and $ x\leq y \leq z$
then $ xz < \frac {1}{2}$[/quote]

Very easy
$ 2xz \leq y^2\plus{}2xz \leq xy\plus{}yz\plus{}zx\equal{}1 \Rightarrow xz < \frac {1}{2}$
Equality cannot occur because for it $ y$ must be $ 0$ and $ 1\equal{}xz \leq 0$ which is contradiction.


Can someone help with this one? Given that tr(AB) = tr(BA) and that tr((AB)^k) = tr((BA)^k), show that Tr[(AB-BA)(AB+BA)]=0.

Just expand it out; you've got $\text{tr}(ABAB)+\text{tr}(ABBA)-\text{tr}(BAAB)-\text{tr}(BABA)$. By the earlier facts, $\text{tr}(ABAB)=\text{tr}(BABA)$ and $\text{tr}(ABBA)=\text{tr}(BAAB)$.


Okay, I've fallen like a chapter behind in my linear algebra class and I realize that there is a problem that I just can't seem to find an example for. Our class text books are shitty and only provides simple examples, leaving hard examples they could have done to be problems at the end of the chapter. Finding out whether a set spans a common vector space, say R3 is easy for me. If you have (4,3,2) (6,9,2) (5,3,7), you would set them all equal to (u1,u2,u3) and use variables. c1(4,3,2) +c2(6,9,2)+c3 (5,3,7)=(u1,u2,u3). Then you simplify it and slap it into a matrix, find the determinant and you know if it spans. My question is what to do when it is c1(2,1,0,1)+c2(-1,0,1,0). The matrix you get has 4 rows and 2 columns. How would you find whether it spans R4 or not? Thank you in advance.

Don't use determinants. Use Gaussian elimination to row reduce the matrix to row echelon form.

Suppose you have $n$ vectors in $\mathbb{R}^{m}.$ Make a matrix out of these columns; it is an $m\times n$ matrix. Row reduce it, and look where the pivot elements are in the row echelon form.

If you have a pivot in every row, then these vectors span all of  $\mathbb{R}^{m}.$ If you do not have a pivot in every row (which is the same as having at least one row of zeros across the bottom), the vectors do not span $\mathbb{R}^{m}.$ In your second example, since there are only two columns, the row echelon form cannot have more than two pivots. Two vectors cannot span $\mathbb{R}^{4}.$


geometry circumcircle inradius perimeter

The sides of a triangle are three consecutive integers and its inradius is $4$. Find the circumradius.

Let the sides of the triangle be n-1, n, n+1. This means the semi perimeter of the triangle is 3n/2.
Area=rs=abc/4R
Therefore we have $4(3n/2)=(n^3-n)/(4R)$ which simplifys to $24R=n^2-1$.
Now comparing rs to Hero's Formula we have $4(3n/2)=sqrt((3n/2)(n/2)(n/2+1)(n/2-1))$
pulling out then $n/2$ and canceling gives $12=sqrt(3((n^2/4)-1))$
Squaring both sides results in $144=(3/4)n^2-3$
Multiplying both sides by 4/3 and adding 3 to both sides we obtain $195=n^2-1$

Now we can substitute this into our expression for R and we see that $24R=195$ so the circumradius is $65/8$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=57779][b]here[/b][/url].

Best regards,
sunken rock


My team got 3rd but I live in a pretty good state.

My state is the 3rd hardest?
Idk.
Anyways, Our team got 1st.
Yay...

So Mr. Frost13 (haha :) sorry coach) *will* be the MA national coach for 2008?

Anyways, good job! he's an amazing coach from what I've heard!

My old school, Smith Middle School, who has gotten 1st, 2nd, or 3rd every year for the last 7 years or so until this year, got near dead last - I'm like 'wow'!

Yeah he's a good coach.
I'm still amazed that I failed so miserably at states...

Our States is like, March 15th. The Saturday of SPRING BREAK....
We still have to meet at school to practice.
Not that I'm complaining. Our sessions are fun.

Peijin,

You didn't fail at all!!  You got thrown in to a tough situation last minute and you beat 3 of the kids that beat you at chapter.

Start practicing for AIME and IMLEM!!

No more negativity!!

Mr Frost

our team got 5th place so close.. But I made it individually so i'm ok. I look forward to seeing you all at nats.

1.East Lyme School
2.Hopkins Grammar School
3.Unknown
4.Unknown
5.Us-Fairfield Woods Middle School

This is the first year that we have ever really done MATHCOUNTS as a team.(In the last 2 years we just had one guy go each time) Our school doesn't really participate in it. I just learned about it this year and the kid and I made a team with 2 other 8th-graders.

[quote="ZhangPeijin"]My state is the 3rd hardest?[/quote]
Not necessarily...

[quote="perfect628"]Well CA is one of the hardest, if not the hardest, state, and yet this really doesn't show at nats. So you can't base state competitiveness on nats performance.[/quote]
Yeah...In CA over here (even though I don't do mathcounts anymore, high school), the nats team doesn't necessarily do well, but you need a score that is something like 1-2 points below the top score to make the countdown round.

[quote="mathcrazed"]Our team... finally... beat. Honey Creek. 1st place. I think we live in the 2nd hardest state? Based on past performances at nationals, we always end up getting 2nd or 3rd -_-.

Maybe not hard state, but we do well at nats so... hm[/quote]
Great job, btw.

And our state is really hard.  I don't know how we can rank states among level of difficulty, though, since individuals keep coming in as sixth graders and exiting as eighth graders.  We can't let our past nats rankings speak for themselves.

Worse. Besides, I went to states as an Individual.

I was there as an individual. And we were worse, regardless of how we did on the team round, because the placement of the members was 4th, 30th, 31st, and 62nd.

I was going to go as an individual, but one of our team members got sick at the last moment and I replaced him.
I think I would've done better if I wasn't on the team, but you never know.
I definitely wouldn't have gotten a big nice shiny trophy if I wasn't on the team though.
:D

TEAM ROUND IS IMPORTANT......no discussing scores in anyway.  Not even how they relate to one another!!

Hmmm....
Last time I checked we got top team round score...
Only reason we beat Diamond...


Hey,

I'm a proficient Java programmer, and I was wondering what are the benefits of learning C++? Especially in terms of academic opportunities and research, what major advantages/capabilities does C++ have over Java that would compel me to learn it?

Thanks!

C++ is faster, gives you more control (in terms of pointers, for example), and has a more useful and flexible library, as it for example allows for easy use of templates as well as conveniently allowing you to use objects as any type you want with understandable functionality (i.e. you can use bools and ints synomously). It also has functionality with OpenGL, which is not, to my knowledge, true of JAVA (though I could be mistaken -- at the very least, C++ OpenGL is much easier to use) and the code, in my opinion, is much sparser and more readable. FILE I/O is much easier, and you can use #define statements to make your code much more understandable. C++ makes it very easy to do such things as sort an array with an arbitrary comparison method, whereas to do this with JAVA using a built-in sort is very difficult. C++ allows you to index vectors and maps in the same manner as arrays, whereas JAVA only allows you to access them through awkwardly-named methods. C++ lets you do many more things in arbitrary scopes, so that for example you can use switch statements in some very novel ways to make your work easier. It also means you can do something like write 20 nested for-loops with very little effort using templates.

I'm not sure if there are any other reasons to learn C++, but if those don't convince you, then I am not sure what can.

Thanks for the response (I'm convinced)

Nonetheless, I am still curious: are there any situations (e.g. research) that would absolutely require or prefer knowledge of C++? Also, how much difference is there between C and C++?

C was the "original" language. C++ added objects to the base C language, as well as the Standard Template Library (STL) and some other goodies that I can't remember. In short, C is not object-oriented (like Java), C++ is.

High-performance situations would basically demand C++, since Java usually requires far too much memory and processor power by comparison. I'm fairly certain that about 70-80% of commercial games use C++ and OpenGL/DirectX (for Windows), since those are the most common programming requirements for games.

Other than things mentioned above, I'd say one of the disadvantages of java is from the users point of view - they have to download the entire JRE to be able to run any programs. Of course, there may be some libraries you need to include with C++ programs, but they'd be a lot smaller, and running a .exe is much easier for people.

I also think that many previously written programs are likely to be in C++, so it would be useful to know in order to understand / modify them without having to rewrite stuff from scratch. While this is true of every language, it is especially true for C++, as it has the highest level of sophistication and therefore if you know it (and php might be useful for web scripts) well, you will be able to understand almost every other language.

[quote="bus"]C/C++ might be necessary if you're writing an application for a system which doesn't have a Java virtual machine. Some examples are programmable microchips, very old computer systems or C-like scripting environments.

C is also pretty much the only option for writing operating systems, drivers and other low level code.[/quote]

I disagree. You could write it in assembly or hex.

Of course nobody [i]would[/i] but you can.


Show that for any integer $n > 10000$, there are integers $a, b$ such that $n < a^2 + b^2 < n + 3 n^{1/4}$.

Posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=292 and http://www.mathlinks.ro/Forum/viewtopic.php?t=62830 .


Probably, yes.  In general, when science texts ask you a question, they expect that you can get the measurements necessary to do the problem from their collected list of constants/measurements.


function
number theory proposed
number theory

Let $\varphi_1(n)=n\sum_{d|n}\frac{|\mu(d)|}{d}$ where $\mu$ denotes the Mobius function
Prove a) $\varphi_1(n)=\sum_{d^2|n}\mu(d)\sigma(\frac{n}{d^2})$ where the sum is over those divisors of $n$ for which $d^2|n$
b) $\sum_{n\leq x}\varphi_1(n)=\sum_{d\leq \sqrt x} \mu(d)S(\frac{x}{d^2})$, where $S(x)=\sum_{k\leq x}\sigma(k)$
($\sigma(k)$  sum of positive divisors of $k$)

anyone for this problem?
i wonder whether there is a nice way to handle the summation  rule $d^2|n$

Hi,

You could use that both functions are multiplicative and therefore you only need to check the equality on prime powers.
For example $\varphi_1(p^k)=p^k\sum_{d|p^k}{\frac{\left|\mu(d)\right|}{d}}=p^k\left(1+\frac{1}{p}\right)$ (because$\mu(p^k)=0 ~(k\geq 2)$).

Hyp


I didn't have time to build the solution PDF. I'm sorry for this fact.

But now, you can start discussing the solutions to the problems.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37208]Problems[/url] is the link for the problems.

Enjoy your discussion! :)

1.[hide]$\frac{x^2y+2xy+x^2+y^2+xy^2+x+y}{x^2y^2+xy^2+x^2y+xy}$ factors to $\frac{(x+y)(xy+x+y+1)}{xy(xy+y+x+1)}$ Simplifying and solving for $y$ gives $y=\frac{3x}{2x-3}$. Plugging in $x=\frac{15}{4}$ yields $y=\frac{5}{2}$[/hide]

3.[hide] using heron's: $\sqrt{(\frac{21}{2})(\frac{9}{2})(\frac{7}{2})(\frac{5}{2})}=\frac{21\sqrt{15}}{4}$[/hide]

4.[hide]this can be expressed as $\frac{13!}{3!3!3!}=28828800$[/hide]

5.[hide]the values for $n=1$ to $n=8$ are $3,10,35,126,4642,1716,6435,24310$ Finding the prime factorization for each number and grouping them together (this works because you're multiplying all of them) gives $2^6*3^7*5^4*7^3*11^4*13^3*17$ Then $(p_1+p_2+p_3+p_4+p_5+p_6+p_7)(abcde)=(2+3+5+7+11+13+17)(6*7*4*3*1)$ which is $29232$[/hide]

Omg, I can't believe what I was thinking for #3... I actually did the multiplication out with all the .5s and then converted the decimals number and.. woah that took me way more time than I thought it should have. Bleh, I gotta think simpler.

i did the long way for number two... i multiplied everything out, i still need to learn De Moivre's... :blush: which is why i didnt post a solution for that one...


modular arithmetic
number theory unsolved
number theory

Let $ p$ be a prime number.Show that there exists a permutation $ (x_1,x_2,....,x_{p\minus{}1})$ of the set $ \{1,2,...,p\minus{}1\}$ such that the number $ x_1,x_1x_2,x_1x_2x_3...,x_1x_2...x_{p\minus{}1}$ are distinct modulo $ p$.

[quote="KDS"]Let $ p$ be a prime number.Show that there exists a permutation $ (x_1,x_2,....,x_{p \minus{} 1})$ of the set $ \{1,2,...,p \minus{} 1\}$ such that the number $ x_1,x_1x_2,x_1x_2x_3...,x_1x_2...x_{p \minus{} 1}$ are distinct modulo $ p$.[/quote]
For $ p \neq 2$ is odd,
Let $ g$ be a primitive root mod p .
Then $ x_i \equiv g^{(p\minus{}1)\plus{}( \minus{} 1)^i(i \minus{} 1)} \pmod{p}$ ,  where $ i \in \{1,2,...,p \minus{} 1 \}$ , is a permutation satisfying the condition , done.


For a,b,c positive real numbers such that a^2+b^2+c^2=1 prove that

a/(1+bc)  +  b/(1+ac)  +  c/(1+ab)  \geq 1


The integer $A$ has 198,519,851,985 digits, and is $not$ a multiple of 3.  The sum of the digits of $A$ is $B$.  The sum of the digits of $B$ is $C$.  The sum of the digits of $C$ is $D$.  Compute the maximum possible value of $D$.

[quote="nr1337"]
 and C will be about 2 digits][/quote]

So can C be, say 98 ? and D is indeed 8 modulo 9 but  then  17 too (is 8 modulo 9)  :maybe:  (hmmmm...)  Hope that helps...

Added later:  There is a hint  above - pointing to the answer..

nr1337 wrote:[hide]
SOLUTION 1:

Let a = 9999...(198,519,851,984 nines)8.  Note that a is congruent to the sum of it's digits (mod 9).  We can guess that B will be about 13 digits, and C will be about 2 digits, so D = (A mod 9).  Since a is one less than a multiple of 9, then A :equiv: 8 mod 9.  Thus D=8.

SOLUTION 2:

Let a=99999...(198,519,851,984 nines)8.  The sum of the digits is (9*198,519,851,984+8) = 1786678667864.  The sum of the digits of that number is 80, and 8+0=8.  Thus the maximum value of D is 8.[/hide]

But how do you know that having a = 999...98 will produce the maximum D?  Couldn't you have, say, C be a string of as many 9's as possible (probably 2), or at least something other than 80?

nr1337 wrote:[hide]
SOLUTION 1:

Let a = 9999...(198,519,851,984 nines)8.  Note that a is congruent to the sum of it's digits (mod 9).  We can guess that B will be about 13 digits, and C will be about 2 digits, so D = (A mod 9).  Since a is one less than a multiple of 9, then A :equiv: 8 mod 9.  Thus D=8.

SOLUTION 2:

Let a=99999...(198,519,851,984 nines)8.  The sum of the digits is (9*198,519,851,984+8) = 1786678667864.  The sum of the digits of that number is 80, and 8+0=8.  Thus the maximum value of D is 8.[/hide]

I don't think your answer is right, becuase I tried another number for A, and got [hide]12[/hide].  [hide]I just made all of the digits 8, and got B = 1588158815880, making C=66, making D = 12.

If we change the last digit of your number for A from a  8 to a 7, C = 79, and D = 15[/hide]

[hide]n = 198,519,851,985

The maximum value of A whose digits are summed will give us a number B with its maximum number of digits.

The maximum value of A is 999...8, having n digits and not being a multiple of 3, as 999...9 would be.

So the maximum value of B is 9(n-1) + 8 = 9n -1.

However, to maximize the sum of B's digits, unless we have a number of the form 999...9, we may be able to get a value with less digits whose digit sum is greater. 1999...8 with 9n - 1 digits has the same digit sum as 999...9  with (9n - 1) - 1 = 9n - 2 digits. Yet both of these are divisible by 3, as the sum of their digits tell us. So we will subtract 1 from these numbers.

Thus if the maximum value of B is greater than 199...7, then we shall use those 9n - 1 digits to get a maximum digit sum of B, equal to C. If the maximum value of B is less than or equal 1999...7, we shall use a number with 9n - 2 digits of the form 999...8, which will maximize the sum of B's digits and give us a maximum value for C.

The maximum value of B when calculated is 9n - 1 = 1786678667864 < 1999999999997. Thus for B to have a great digit sum, we shall use the number 999999999998, which has one less digit.

The sum of this number gives us the maximum for C, which is 9*12 - 1 = 107.

From here we can see that D will be greatest if C = 99, which is not a restricted possibility as the preceding number B can be decreased to suit this. However, if C = 99, the sum of B's digits = 99, which indicates that B is divisible by three, which indicates that the sum of A's digits is divisible by 3, leading to the conclusion that A is divisible by 3. So, as before, we will subtract 1 from the highest possible value of C, which gives us C = 98.

Thus the maximum value of D is 17.[/hide]


function
integration
calculus
complex analysis
complex analysis unsolved

Let the function $ f$ be analytic in the entire complex plane and satisfy :
                $ \int_{0}^{2\pi}{|f(re^{i\phi})|d\phi} \leq r^{2009}$ for all $ r>0$
Prove that f is the zero function
 :starwars:

Something is misstated here. What about $ f(z)\equal{}\frac1{2\pi}\,z^{2009}?$

[quote="gka"]First show f is constant, using the Cauchy integral formula for f' and taking r going to 0 (it's an arguement akin to that of Louisville's thm).  Then show that this constant can only be zero.[/quote]

use the Cauchy formula, you can only prove $ |f'(0)|\equal{}0$ and nothing else.


Find all values of $k$ such that the following system of equation has the most roots:
\[ \begin{displaymath} \left\{ |x-1|+|y+1|=1 \\ x^2+y^2=k \right \end{displaymath}  \]

Whoa, there, buddy, fix up your $\LaTeX$.  It seems it should be

$|x-1| + |y+1| = 1$
$x^2 + y^2 = k$

[hide="Hint"] Graph the first on a Cartesian coordinate grid.  The meaning of the second should be clear. [/hide]

[quote="hydro"]isn't the answer
[hide]
$\frac{1}{2} \le k \le 1$[/hide]????[/quote]
I got $\frac12<k\le5$
The first equation is a square with vertices $(1,0),(2,-1),(1,-2),(0,-1)$  The maximum number of times the circle with center $(0,0)$ can intersect it is 2.  The greatest lower bound of $k$ is $\frac12$ when the circle is tangent to the nearest side.  The largest value of $k$ that works is $5$ when it goes through the two farthest vertices of the square.


MATHCOUNTS geometry AMC USA(J)MO USAMO probability conics

Don't force yourself, that'll just make it worse. Two things:

1. Keep an open mind and allow mathematics the opportunity to re-interest you. I remember I thought competition math was dry and boring when I was in 9th grade, but that's changed.

2. Let's not dismiss your analogy just yet. Do you feel like you've found a passion for something else that's interfering with your passion for mathematics? If so, what is that? Maybe there's a way for you to manage both without them interfering in the long run.

I suffered from this in 10th grade after not making USAMO in 9th, and I really regret not pushing myself. I actually think that with anything, sometimes you will not like it as much, and at that time you need to promise yourself to work on it, and keep the promise. If you truly think that math is something you like doing, then you should pick up speed after trying to revive it a bit.

Had the same problem. Try something like this:

1) Find an area of math which you like (try to solve problems, or read) and do that for some time

2) If that doesn't help - stop doing math for some time. Eventually, if you don't want to do it, then don't, nobody pushes you. Just if you are good at competitions, it doesn't mean you have to do that.

And this is maybe the best time to rest from the competition math (in fact, I don't know for USA, but here competition season is February - July).

BTW, nice analogy :D

Perhaps you are somewhat disappointed in your performance? You call yourself "moderately successful"... but perhaps you wanted more than 8th place?

I know people say it all the time, but no matter how well you do in competition, you're still good at math.

And don't you ever forget it.

Take a step back and think about the situation you're in.  You have something (math) you don't want to do, and you're trying to get the desire to do it.  But how can you ever get yourself to want to do something that you don't want to do?  That passion's not going to come back out of nowhere.

Again, take a step back.  Think about why you do math.  Maybe it's because you want to get into a good college.  Well, why do you want to get into a good college?  Play the role of the annoying kid who keeps asking, "Why?" until you find out what's really motivating you, and then figure out how math fits into this.

This may seem kind of over-the-top, but I really wish I had taken some more time to think about why I spent all the time I did preparing for math competitions throughout high school.  I graduated from high school this year, and now I realize that I was so caught up in math competitions during high school that I forgot what was really important in my life.  At times, math competition success was what was driving me more than anything else, and I wish that hadn't happened.  It's good that you're thinking about this, because math competition success should not be the driving force behind your life, in my opinion.

[quote="HilbertThm90"]Also. Math itself isn't "competition math". Maybe you can reignite the passion by looking into some other parts of math. It is probably blasphemy to say this on this site, but in general I abhor competition math. I find it boring to try to sit down and solve a problem that has been constructed in order to be solved in a nice neat way. I'm a grad student in math and have barely skimmed the surface of the surface of math. There is so much out there, that becoming uninterested in one portion means it might be time to try another.[/quote]

I'd have to agree. (I am still a high school student and have no right at all to be saying these things, but...). Solving contrived problems can get tiresome after a while, and it may or may not be a happy realization to you, after perhaps discovering that competition math is "more mathish" than school math, that competition math isn't "real math" either. I was once ashamed to study mathematics beyond what I knew very well, beyond what was tested in olympiads, because I felt that I was just memorizing, accumulating knowledge, something rather against the spirit of problem-solving. But now I know that one comes upon pretty proofs and interesting, motivated results much more often outside the realm of competition mathematics.

Ok so you may be bored of math you are familliar with. Have you tried looking at math you have  never heard, math  has vast areas in which you can study it try new problem try to find new ways to solve old problems exeperimant with theories there is sooooooooo much math has to offer.  I highly doubt you are bored with all of these suggustions if you are maybe you should take a break from math.

[quote="PerpetualMotion"]I've had a moderately successful track record at math, with 8th at state mathcounts and a few other things. Over the summer, I tried to work through Volume 1 and 2, managing to do all of the geometry of Volume 1 and skimming the surface of Volume 2. However, I can feel that I am losing the "pull" toward math and having to resort to the "push." This has left me feeling apathetic, just an aimlessly drifting ghost. Over time, I have stopped practicing. To me, math just seems dry and empty. It's like a kiss between a couple cheating on each other; their love unraveled, the statement "I love you" rend a cold lie. (Sorry for the weird analogy.) How can I find my passion again?[/quote]

I had a similar problem for a few weeks this summer. I would force myself to sit down and do math but I just didn't feel like it. So, for a week or so, I turned away from math and focused on an interest I usually shoved aside to practice more for MATHCOUNTS, the AMCs, etc.: graphics. (A lot of my friends have probably noticed my abrupt obsession with them recently.) I literally spent all day doing graphics and reading about them and learning new techniques, and basically the only math I did was the arithmetic required in daily life. By the end of the week I felt much better, and now each day I have a balance between both interests (and even found time to write too, another one of my hobbies.) Basically, what I'm trying to say is, you might feel really tired of practicing a lot, and I suggest you take a break and do something else for a while: play an instrument, read a lot, write a lot, graphics, hanging out with friends, watching TV until your brain turns to mush, being lazy all day, whatever works for you. Then, when you feel like doing math, go back to it.

I took everyone's advice and waited for an opportunity for math to re-interest me, and it came today. I was doing nothing when the thought of "Godel's Incompleteness Theorem" suddenly entered my head. Searching it up opened up the door to the realms of Provability Logic, among other things. (Btw, do you think that a proof writing program/algorithm will ever exist?)

No longer do I feel like trudging against the hours each day. It seems as though an eclipse has passed over, unleashing an invigorating light onto a barren land.

The fire is burning. The oven is hot. And I ready to move into this beautiful world.

So thanks, guys (and girls)!  :D

(And excuse my poor attempts of being poetic.)

[quote="HilbertThm90"]Also. Math itself isn't "competition math". Maybe you can reignite the passion by looking into some other parts of math. It is probably blasphemy to say this on this site, but in general I abhor competition math. I find it boring to try to sit down and solve a problem that has been constructed in order to be solved in a nice neat way. I'm a grad student in math and have barely skimmed the surface of the surface of math. There is so much out there, that becoming uninterested in one portion means it might be time to try another.[/quote]

I fully agree with HilbertThm90. My interest in mathematics began when I was in 7th grade when my teacher suggested to me to participate at a math competition. From that moment I connected math with math competitions. However, something bothered me with olympiad problems and at 10th grade I realised that this was the fact that that the require deep intution-creativity. I realised that this was not the math I like, the math I liked is the one that requires just logic and or more precisely, I realised that it's not math that I like but logic. So maybe I haven't completely solved any problems from this forum for more than 6 months but I have worked many of them until the part that creativity is required to continue (there I stop). But this realisation picked my interest for programming as huge part of it requires just logic. So now I know what I like and I think that this is what you should also do. Find what you like.

During the summer I got so bored of math that I found myself [i]having[/i] to do it, not [i]wanting[/i] to do it. I then came across a thread a lot like this one and took someones advice: just dont do math for a while. It could take a few days or a few weeks, but eventually you will want to do it again.

Push yourself

life's not about what you want, its about what your parents, teachers, peers, and society demands

sad but true

i think most of us don't feel like going to school tomorrow, but we still have to wake up at 6.

suck it up

if your parents aren't taking the responsibility, grab a belt and discipline yourself.

This is completely qualitative, but in my experience there are two kinds of math competitors: those who are self-motivated, and those who are motivated by outside forces. People can move between the groups, but I've found that by the time they get to junior or senior year, almost all the outside-motivated people have gotten sick of math. The self-motivated ones are still around.

My point is that if you are forcing yourself to do math for parents/college/peers/whatever, it's not going to end well.


Έστω $ p,q$ πρώτοι αριθμοί. Βρείτε όλους τους θετικούς ακεραίους $ x,y$ τέτοιοι ώστε $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{pq}$

Greek, it's all Greek to me, but I try the problem by imagination,

$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{pq}\Longleftrightarrow (x \minus{} pq)(y \minus{} pq) \equal{} (pq)^2.$

Since $ p,\ q$ are relatively coprime, $ (x \minus{} pq,\ y \minus{} pq) \equal{} (1,\ (pq)^2),\ (pq,\ pq),\ (p^2,\ q^2),\ (q^2,\ p^2),\ ((pq)^2,\ 1)$, yielding

$ \boxed{(x,\ y) \equal{} (pq \plus{} 1,\ pq(pq \plus{} 1)),\ (2pq,\ 2pq),\ (p(p \plus{} q),\ q(p \plus{} q)),\ (q(p \plus{} q),\ p(p \plus{} q))}$.

General....Solve in integers the equation:
$ 1/x \plus{} 1/y \equal{} 1/z$
Find all triples $ (x,y,z)$.

$ 1/x \plus{} 1/y \equal{} 1/z \equal{} > (x \minus{} z)(y \minus{} z) \equal{} z^2$
$ Let x \minus{} z \equal{} a$ and $ y \minus{} z \equal{} b$ then $ ab \equal{} z^2$
If $ (a,b) \equal{} d$ then $ dk \equal{} a$, $ dl \equal{} b$ and $ (k,l) \equal{} 1$ so we have:$ d^2kl \equal{} z^2 \equal{} > d/z$.Let $ z \equal{} dm$ so $ kl \equal{} m^2$=>$ k \equal{} e^2$ and $ l \equal{} f^2 \equal{} > z \equal{} dm \equal{} > z \equal{} def$
$ x \equal{} z \plus{} a \equal{} def \plus{} dk \equal{} def \plus{} de^2 \equal{} > x \equal{} de(e \plus{} f)$
$ y \equal{} z \plus{} b \equal{} def \plus{} dl \equal{} def \plus{} df^2 \equal{} > y \equal{} df(e \plus{} f)$
So the triples are $ (x,y,z) \equal{} (de(e \plus{} f),df(e \plus{} f),def)$ with $ (e,f) \equal{} 1$ :!:

[quote="Αρχιμήδης 6"]Solve in integers the equation:
$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \equal{} \frac {1}{z^2}$
Prove also that for $ n > 2$ there are only trivial solutions for the equation:
$ \frac {1}{x^n} \plus{} \frac {1}{y^n} \equal{} \frac {1}{z^n}$ for all integers triples $ (x,y,z)$ :wink:[/quote]

Are you sure, that there exist solutions?

$ \frac {1}{x^n} \plus{} \frac {1}{y^n} \equal{} \frac {1}{z^n} \Leftrightarrow$

$ (xz)^n \plus{} (yz)^n \equal{} (xy)^n$ which has no integer solutions for $ n > 2$, according to Fermat's last theorem.

Dear maniopas i said to prove that there are only trivial solutions not to prove that there are solutions.You are right but you dont solve the equation:
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\equal{}\frac{1}{z^2}$
So....is there any idea? :)

Οι λύσεις είναι $ x\equal{}k(m^4\minus{}n^4)$ , $ y\equal{}2kmn(m^2\plus{}n^2)$ και $ z\equal{}2kmn(m^2\minus{}n^2)$

Σωστα Σιλουανε... :)


function conics parabola algebra domain ellipse

please show work and explain how it is done so i can learn from it
thanks

1. determine the vertex of the parabola   y = 2x^2 - 12x + 12

2. determine the domain of the relation x^2 + 16y^2 = 16

3. what is the vertex of the parabola f(x) = -2 ( x + 1 )^2 + 3

4.what is the length of the major axis of the ellipse x^2/36 + y^2/81 = 1

[quote]1. determine the vertex of the parabola   $y = 2x^2 - 12x + 12$[/quote]

The parabola $y=ax^2+bx+c$ has its vertex for $x_0=-\frac{b}{2a}$

Here we have
$ \begin{array}{l} a=2 \\ b=-12\\ c=12 \end{array}  \}$

So the vertex is at $x_0=-\frac{-12}{2 \cdot 2} = \frac{12}{4}=3$.

Then apply this value to $y = 2x^2 - 12x + 12$ and you get
$y = 2\cdot 3^2 - 12\cdot 3 + 12 = 18-36+12 = -6$

[quote]2. determine the domain of the relation $x^2 + 16y^2 = 16$[/quote]
ok, actually this is an ellipse, so we can easy find the domain.
Anyway, let's see an other way

$x^2 + 16y^2 = 16 \Rightarrow$
$x^2 = 16 - 16y^2\Rightarrow$
$x^2 = 16(1 - y^2)$

The first part of this equation is $x^2\geq 0$
So it is necessary that the right part is also $\geq 0$

$16(1 - y^2)\geq 0 \Rightarrow$
$1 - y^2\geq 0 \Rightarrow$
$y^2\leq 1 \Rightarrow$

$|y|\leq 1 \Rightarrow$

$-1\leq y\leq 1$

So, the domain of $y$ is $[-1,1]$

Could you find the domain of $x$ ?

[quote]3. what is the vertex of the parabola $f(x) = -2 ( x + 1 )^2 + 3$[/quote]

The parabola $y=a(x+b)^2+c$ has a vertex at the point $(-b,c)$

Can you find the vertex now?

[quote]4.what is the length of the major axis of the ellipse $\frac{x^2}{36} + \frac{y^2}{81} = 1$[/quote]

In the ellipse $\frac{x^2}{a^2} +{} \frac{y^2}{b^2} = 1$ the length of the major and minor axes is $2a$ and $2b$ (the major axis is of course the larger of these).

For example, $a^2=36 \Rightarrow a=6\Rightarrow$ The minor axis has length $2a=12$.

Can you find the length of the major axis?


integration
trigonometry
Support
real analysis
real analysis unsolved

[color=red]Please read this carefully :[/color]

Let   $I = \int \cos^{2}xdx$ . Then

  I= $\int \cos x \cos xdx = \int (\sin x)' \cdot \cos xdx$
 ( We put    $\sin x = u , \cos xdx = du)$ ) , so

  $I= \int u'du = u+c = \sin x+c$

    WRONG !!!!!!!!!!!!!!!!!!!
[color=red]Whrere is the wrong and how can we explain the wrong place to pupils ?[/color]
  [u]Babis[/u]

It's pure nonsense. "Where is this wrong?" is the wrong question, and I would never show anything like this to one of my students. This example feels like it would inspire students to make similar mistakes rather than preventing them.

The line explaining the substitution does not match the pieces- there's only one $dx$ in the real integral.

A side note to stergiu- \cos and \sin look much better: $\int \cos^{2}x\,dx$

[quote="stergiu"]

  I= $\int cosx cosxdx = \int (sinx)' cosxdx$
 ( We put  $sinx = u , (sinx)'dx =du , cosxdx = du)$ ) , so

  $I= \int u'du = u+c = sinx+c$

   [/quote]

with your change variable
 (according we are on appropriate subset of R where it is allow to do it)

$du=\cos(x)dx$ and $\cos(x)=\pm \sqrt{1-u^{2}}$

then $I=\int \pm \sqrt{1-u^{2}}du$

To Jmerry.

 Yes , the kern of your idea is right, but - sorry - I was not clear in my writting. I have edited now. The question is
  << which part in the substutation is not right >>.

 I was given the problem today. I have an answer but I 'd like to see your opinion too.
Any way , thanks for your answer and sorry because the problem made you... a little angry :). This happens to me too , when I meet '' foolish'' questions. But I think that this question has something to teach. Of course you can support a different opinion.

[u] Babis[/u]

[quote="stergiu"]$I = \int \cos^{2}xdx=\int \cos x \cos xdx = \int (\sin x)' \cdot \cos xdx$ $\Longrightarrow$ ($\sin x = u , \cos xdx = du$) $I= \int u'du = u+c = \sin x+c$ WRONG ![/quote]
$\boxed{\ \int du=\int u'dx=u+C\ }$ and not $\int u'du=u+C\ !!!\ .$ $u=u(x)=\sin x\ ,\ u'(x)=\cos x=\cos (\arcsin u)$ $\Longrightarrow$ $\boxed{\ u'(x)du=\cos (\arcsin u)du\ }\ .$

And yet, if you wish, I=$\int u'du=\int \cos (\arcsin u) du=$ $\int \epsilon\sqrt{1-u^{2}}du=$ $\frac{\epsilon }{2}\cdot \left(\arcsin u+u\sqrt{1-u^{2}}\right)+C=$
$\frac{\epsilon }{2}\cdot\left (x+\sin x\cos x\right)+C=$ $\frac{\epsilon }{2}\cdot\left(x+\frac{1}{2}\sin 2x\right)+C$, where $\epsilon^{2}=1\ .$


hello , can we find a combinatorial interpretation of this sum ??

$ \sum_{p\equal{}0}^{n}C_{n}^{p}cos(\alpha p\plus{}\beta(n\minus{}p))$

thank's :)

That's strange question (in my humble opinion). What do you mean when you are saying combinatorial?
If you want some interpretation of the form "this is the number of ways...." then I doubt about the existence of it.
Anyway, to compute this sum just note that it equals to $ Re(\sum C\limits_n ^k (e^{ia})^k (e^{ib})^{n-k}=(e^{ia}+e^{ib})^n$ :)


[quote="Jumbler"]$x,y,z$ are positive reals
 $(\frac{x+y+z}{3})^{3}\geq \frac{(x+y)(y+z)(z+x)}{8}\geq (\frac{x+y+z}{3})(\frac{xy+yz+zx}{3})$[/quote]

Sorry but I will use expand to solve your problem.
[u]The left side:[/u], your inequality is equivalent to
$8 \sum x^{3}\geq 6xyz+3 \left (\sum xy(x+y)\right)$
It is easy b/c
$3\sum x^{3}+9xyz \geq 3 \sum xy(x+y)$
$5\sum x^{3}\geq 15 xyz$
Add both sides and done

[u]The right hand side:[/u] equavalent to
$x^{2}y+xz^{2}+x^{2}z+y^{2}z+xy^{2}+z^{2}y \geq 6xyz$
 :)

They are both also here : http://www.mathlinks.ro/Forum/viewtopic.php?t=115644
And also,both can be proven by using identities,as in the link


Is anyone in Odyssey of the Mind this year?

Is Odyssey of the mind still going?
I thought it was completely replaced by DI (destination imagination). DI's a non-profit similar type of thing that split off OM a few years ago. It's really big now. I went to the Global Finals last year if you'd like info.

1.what is that?a math competition?
2.isnt it a song of Enigma?


Find positive integers $a$ and $b$ such that

$(\sqrt[3]{a}+\sqrt[3]{b}-1)^{2}=49+20\sqrt[3]{6}$

EDIT: My previous post is deleted, since I've found the solution.

Since $\sqrt[3]{6}$ is the only irrational factor on the RHS, $\sqrt[3]{a}$ and $\sqrt[3]{b}$ can only have the form $m\sqrt[3]{6}$ or $m\sqrt[3]{6^{2}}$ with $m$ being an integer, since otherwise there would be other incommensurable terms in the expansion of LHS.

For $\sqrt[3]{a}=k\sqrt[3]{6}, \sqrt[3]{b}=l\sqrt[3]{6}$, on the LHS we'd have $(k^{2}+2kl+l^{2})\sqrt[3]{6^{2}}$, which would have to be equal to zero, hence $k=-l$ which is impossible.

For $\sqrt[3]{a}=k\sqrt[3]{6^{2}}, \sqrt[3]{b}=l\sqrt[3]{6^{2}}$, on the LHS we'd have $2(k+l)\sqrt[3]{6^{2}}$, which would have to be equal to zero, hence $k=-l$ which is impossible.

Therefore $\sqrt[3]{a}=k\sqrt[3]{6},\sqrt[3]{b}=k\sqrt[3]{6^{2}}$. That yields (after expanding)

$12kl+1+(6l^{2}-2k)\sqrt[3]{6}+(k^{2}-2l)\sqrt[3]{6^{2}}=49+20\sqrt[3]{6}$

Obviously, we have the system

$12kl+1=49$
$6l^{2}-2k=20$
$k^{2}-2l=0$

From the third equation $l={k^{2}\over 2}$, hence the first becomes $k\cdot{k^{2}\over 2}=4\iff k=2\iff l=2$. The second is also satisfied for $k=l=2$.

Therefore $\{\sqrt[3]{a},\sqrt[3]{b}\}=\{2\sqrt[3]{6},2\sqrt[3]{6^{2}}\}\iff\{a,b\}=\{48,288\}$, and the expression is $(\sqrt[3]{48}+\sqrt[3]{288}-1)^{2}=49+20\sqrt[3]{6}$

[quote="Farenhajt"]
Since $\sqrt[3]{6}$ is the only irrational factor on the RHS, $\sqrt[3]{a}$ and $\sqrt[3]{b}$ can only have the form $m\sqrt[3]{6}$ or $m\sqrt[3]{6^{2}}$ with $m$ being an integer, since otherwise there would be other incommensurable terms in the expansion of LHS.
[/quote]
This is the most difficult part of this problem (further part of solution is trivial) and it should be written more precisely.


Let p and q be given coprime positive integers. Call a subset S of nonnegative integers ``ideal'' if S satisfies the following two conditions:
 a) $0 \in S$;
 b) for all $n \in S$, $n+p \in S$ and $n+q \in S$.
 Determine the number of all ideal subsets.

I'm not sure, but may be this theorem will help

Let (p, q) = 1. Let T = {numbers of the form sp + tq with s, t are nonnegative intergers}
1) all numbers > pq - p - q belong to T
2) for a <= pq - p - q, if a in T <=> pq - p - q not in T

Have fun,

Namdung

[quote="quangtrung"][quote="Namdung"]
2) for a <= pq - p - q, if a in T <=> pq - p - q not in T
Namdung[/quote]
  how did you get it[/quote]

Sorry:

for a <= pq - p - q, if a in T <=> pq - p - q -a not in T

[quote="Namdung"]
Sorry:
for a <= pq - p - q, if a in T <=> pq - p - q -a not in T[/quote]
   I think you are still wrong
       we can choose T is the set of all  positive integer numbers ,it is satisfy


Find all real numbers $ x$ such that
\[ {\left(3\plus{}\sqrt{8}\right)}^{x/2}\plus{}{\left(3\minus{}\sqrt{8}\right)}^{x/2}\equal{} 6.\]
Explain why you have found all solutions.

Answer

Let $ f(x) \equal{} (3\plus{}\sqrt{8})^{\frac{x}{2}}\plus{}(3\minus{}\sqrt{8})^{\frac{x}{2}}$

$ x \equal{} 2$ is an obvious solution for $ f(x) \equal{} 6$

Rationalising the numerator gives:

$ f(x) \equal{} (3\plus{}\sqrt{8})^{\frac{x}{2}}\plus{}\frac{1}{(3\plus{}\sqrt{8})^{\frac{x}{2}}}$

$ f(x) \equal{} f(\minus{}x)\Rightarrow f$ is an even function, so we need only consider $ x\geq 0$

Rearranging:

$ f(x) \equal{}\frac{(3\plus{}\sqrt{8})^{x}\plus{}1}{(3\plus{}\sqrt{8})^{\frac{x}{2}}}$

As $ x$ increases from 0, the numerator of this expression increases more rapidly than the denominator. Hence for $ x\geq 0$ the function is strictly increasing, and $ f(x) \equal{} 6$ will only occur once.

Exploiting the even nature of the function gives the two solutions: $ x \equal{}\pm 2$


If $a,b$ are the legs of a right triangle and $l$ the length of the right angle bisector, prove ${\sqrt{2}\over l}={1\over a}+{1\over b}$

[hide]Call $\vartriangle ABC,\angle A = 90^{0},AB = a,AC = b$. AM is the bisector of $\angle A$.$HM \bot AB,H \in AB$
We have:$\frac{{AC}}{{AB}}= \frac{{CM}}{{BM}}\Rightarrow CM = BM.\frac{{AC}}{{AB}}\Rightarrow BC = \frac{{AC+AB}}{{AB}}.BM$
$\Rightarrow \frac{{HM}}{{AC}}= \frac{{BM}}{{BC}}= \frac{{AB}}{{AC+AB}}\Rightarrow HM = \frac{{AB.AC}}{{AB+AC}}\Rightarrow \frac{1}{{HM}}= \frac{1}{{AB}}+\frac{1}{{AC}}$
$\sqrt 2 HM = AM \Rightarrow \frac{1}{{HM}}= \frac{{\sqrt 2 }}{{AM}}$
SO THAT$\frac{{\sqrt 2 }}{l}= \frac{1}{a}+\frac{1}{b}$[/hide] :)

$\frac{1}{2}\cdot a\cdot \frac{l}{\sqrt{2}}+\frac{1}{2}\cdot b\cdot \frac{l}{\sqrt{2}}=\frac{1}{2}ab\Longleftrightarrow \frac{\sqrt{2}}{l}=\frac{1}{a}+\frac{1}{b}.$ Q.E.D.


inequalities
calculus
calculus computations

Assume $\sum_{n=0}^{\inf}\frac{1}{n^2} and \sum_{n=0}^{\inf}\frac{1}{n^3}$ both converge and $\sum_{n=0}^{\inf}\frac{1}{n^2} > \sum_{n=0}^{\inf}\frac{1}{n^3}$.... Is this true???

It seems to be but i want to konw if i am looking at it right. Like is there any cases that it would not be true???

What is this "assume" and "cases?"  Those are two convergent series, with the square series clearly larger than the cube.

Also, use \infty for $\infty$.

Actually neither of them are defined if you start the index at 0. We don't want any $\frac{1}{0}$. Start them at 1 and you have $\frac{1}{n^2} \ge \frac{1}{n^3}$ for all $n$. The same inequality for the series follows.

Yeah, certainly. You only have equality when $n=1$. For $n>1$, we have $\frac{1}{n^2} > \frac{1}{n^3}$. If $a > b$ and $c \ge b$ then $a+c > 2b$. So the series $\frac{1}{n^2}$ is strictly greater than the series $\frac{1}{n^3}$. Sorry, I should have been more clear before.


articles
geometry
geometric transformation
rotation

i came accross this article yesterday, it was pretty funny, although some of you might of already seen it because it is rether old, i like to share it with you :) :

[b]Is There a Santa Clause? A physicist point of view: [/b]

1. No known species of reindeer that can fly. BUT there are 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, this does not completely rule out flying reindeer (which only Santa has ever seen.)

2. There are 2 billion children (persons under 18) in the world. BUT since Santa doesn't (appear) to handle the Muslim, Hindu, Jewish and Buddhist children, that reduces the workload to 15% of the total-378 million according to Population Reference Bureau. At an average (census) rate of 3.5 children per household, that's 91.8 million homes. One presumes there's at least one good child in each.

3. Santa has 31 hours of Christmas to work with, thanks to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical). This works out to 822.6 visits per second. This is to say that for each Christian household with good children, Santa has 1/1000th of a second to park, hop out of the sleigh, jump down the chimney, fill the stockings, distribute the remaining presents under the tree, eat whatever snacks have been left, get back up the chimney, get back into the sleigh and move on to the next house. Assuming that each of these 91.8 million stops are evenly distributed around the earth (which, of course, we know to be false but for the purposes of our calculations we will accept),we are now talking about .78 miles per household, a total trip of 75-1/2 million miles, not counting stops to do what most of us must do at least once every 31 hours, plus feeding and etc. This means that Santa's sleigh is moving at 650 miles per second, 3,000 times the speed of sound. For purposes of comparison, the fastest man-made vehicle on earth, the Ulysses space probe, moves at a poky 27.4 miles per second-a conventional reindeer can run, tops, 15 miles per hour.

4. The payload on the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium-sized lego set (2 pounds), the sleigh is carrying 321,300 tons, not counting Santa, who is invariably described as overweight. On land, conventional reindeer can pull no more than 300 pounds. Even granting that "flying reindeer" (see point #1) could pull TEN TIMES the normal amount, we cannot do the job with eight, or even nine. We need 214,200 reindeer. This increases the payload-not even counting the weight of the sleigh-to 353,430 tons. Again, for comparison-this is four times the weight of the Queen Elizabeth.

5. 353,000 tons traveling at 650 miles per second creates enormous air resistance-this will heat the reindeer up in the same fashion as spacecrafts re-entering the earth's atmosphere. The lead pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each. In short, they will burst into flame almost instantaneously, exposing the reindeer behind them, and create deafening sonic booms in their wake. The entire reindeer team will be vaporized within 4.26 thousandths of a second. Santa, meanwhile, will be subjected to centrifugal forces 17,500.06 times greater than gravity. A 250-pound Santa (which seems ludicrously slim) would be pinned to the back of his sleigh by 4,315,015 pounds of force.

In conclusion: If Santa ever DID deliver presents on Christmas Eve, he's dead now.

:rotfl:  :rotfl:

That was really funny.

This is how a mathmetician's brain work.

In Romania Santa doesn't need cookie & milk to deliver the presents :D

Of course there is a Santa Claus!

[url=http://www.xs4all.nl/~jcdverha/scijokes/2_8.html#subindex]Here are a few rebuttals.[/url]

[quote="G-UNIT"]These arguments are completely fallacious; they contradict the fact Santa uses magic to do his bidding.  And frankly, with magic anythings possible.[/quote]
Yay for magic!  :D

Well...
We must remember that Santa is pretty fat.  Or so it's said.  So is it unreasonable to assume that he is more than 3-dimensional?
If he extends into higher dimensions, then he can curve space-time, and break the laws of physics as we know them.

Yes, I've seen the article before, and I just realized that I wrote an article that's pretty similar so I was probably influenced by it.

Speaking of

[quote]Santa [b]Clause[/b][/quote]

...

that misspelling was what first got me doubting his existence, when I got a letter with the name spelled that way (undoubtably word-processed). If Santa Claus can't even sign his own name...

[quote="amirhtlusa"]
[b]Is There a Santa Clause?[/b]
[/quote]

Yes. Tim Allen starred in it. The real question is: "Will there be another sequel to The Santa Clause?"

[url=http://www.vsubhash.com/writeups/santa_outsources2india.asp]Yes - Click Here[/url]

I think SAnta doesn't exist, because wouldn't the parents notice presents are just randomly appearing under the christamas tree???

No, he modifies their memories. Using that speel from harry potter.

so what.... that would just add more things he needed to do ata each house, Creep into the parents room and alter their memory

Yeah, but he has one of those time turner thingies, which is also from harry potter.

Now consider this. If each "good" kid left a cookie and a glass of milk (8 ounces) for santa, both weighing together about 12 ounces, and there are 91.8 million homes, we are looking at his consumption of 34425 tons of food in 31 hours. Yeah, I'd say he's dead. Or it'd be hard to miss him.

He puts them in the sack for later.

Really, what do you think he eats during the rest of the year?

year old milk?

EDIT: i just realized that the Arctic might make for a good refrigerator.

>< Santa can't be real....... theres just no proof  :P

Haven't you ever seen Miracle on 34th Street? Even the United States Postal Service -- a part of the federal government -- recognizes Santa Claus as real! :P


Find all pairs function $f: \mathbb{R}\to\mathbb{R}, g: \mathbb{R}\to\mathbb{R}$ such that:
a) $f$ is increasing(that means if $x<y$ then $f(x)<f(y)$)
b) $f(xy)=g(y)f(x)+f(y)$ for all $x,y\in\mathbb{R}$

A solution of problem is:
$g(x)=x^{m}$ where $x\geq 0$ and $g(x)=-|x|^{m}$ where $x<0$; $f(x)=t(1-g(x))$ for all $m>0>t$


algebra
polynomial
Galois Theory
number theory proposed
number theory

Prove the equivalence of the following assertions:

  a) $n$ has a multiple of the form $2^{r}-1$ with odd $r$.
  b) there are polynomials $f,g\in Z/2Z[X]$ such that $f(x)=x^{n-1}g(\frac{1}{x})$ and $f(x)g(x)=1+x+...+x^{2n-2}$ in $Z/2Z[X]$.

$n=7$ satisfies (a), but Mathematica says that $1+x+\ldots+x^{12}$ is irreducible in $\mathbb F_{2}[x]$, so this would be a counterexample. Could it be that (a) should be replaced with: "$2n-1$ has a multiple of the form $2^{r}-1$ with $r$ odd" ? I think I can prove the direction (a) $\Rightarrow$ (b) of this modified version.

Yes, sorry, I had a typo. This is the correct version.

Thank you, and sorry for the delay. I hope this is correct.

Let's restate the conditions (a) and (b) (in this new form they will be equivalent to the old (a) and (b) respectively):

(a) the order of $2$ modulo $2n-1$ is odd;

(b) for every root $\alpha$ of $p(x)=\frac{x^{2n-1}-1}{x-1}$ in an algebraic closure of $\mathbb F_{2}$, the minimal polynomials of $\alpha$ and $\frac 1\alpha$ over $\mathbb F_{2}$ are distinct.

Because $2$ does not divide $2n-1$, the prime $2$ is unramified in the cyclotomic extension $\mathbb Q(\zeta_{2n-1})$, and its residue class degree $f$ is precisely the order $f$ of $2$ modulo $2n-1$ (standard results about factorization of primes in cyclotomic extensions). At the same time, a theorem of Kummer says that (roughly speaking) the minimal polynomial $\varphi=\varphi_{2n-1}$ of $\zeta_{2n-1}$ over $\mathbb Q$ factors in $\mathbb F_{2}[x]$ "the same way" $2$ factors in $\mathbb Q(\zeta_{2n-1})$. More precisely, this means that if $\frak P_{1},\ldots,\frak P_{g}$ are the prime factors of the ideal generated by $2$ in $\mathbb Z[\zeta_{2n-1}]$ (the ring of integers of $\mathbb Q(\zeta_{2n-1})$; it's not that easy to prove that this is the case), then the reduction $\bar\varphi(x)$ of $\varphi(x)$ modulo $2$ is a product of $g$ irreducible factors $\bar\varphi_{i}(x),\ i=\overline{1,g}$, each one of degree $f$.

Now for the problem itself:

(a) $\Rightarrow$ (b)

(a) says precisely that $f$ is odd. Choose an arbitrary $i\in\overline{1,g}$, and let $\alpha_{1},\ldots,\alpha_{f}$ be the roots of $\bar\varphi_{i}$ in $\mathbb F_{2^{f}}$ (the field with $2^{f}$ elements). This means that $\alpha_{j}^{-1}$ are also conjugate over $\mathbb F_{2}$, so we cannot have $\alpha_{j}=\alpha_{k}^{-1}$ for some $j,k$ unless the inverse of [i]every[/i] $\alpha_{j}$ is among the $\alpha_{j}$. This is impossible, since $f$ is odd, so we cannot partition the $f$-element set $\{\alpha_{j}\}$ into pairs $\{\alpha_{j},\alpha_{j}^{-1}\}$.

(b) $\Rightarrow$ (a)

Suppose now that $f$ is even, and that (b) holds. The aim is to obtain a contradiction. Let $\alpha\in\mathbb F_{2^{f}}$ be a primitive $2n-1$’th root of unity. This means that it’s a root of some $\bar\varphi_{i}$. Because $\alpha$ is not conjugate to $\alpha^{-1}$ over $\mathbb F_{2}$, no power of $\alpha$ different from $1$ is conjugate to its own inverse. However, the order of the Galois group of $\mathbb F_{2^{f}}$ over $\mathbb F_{2}$ is $f$, hence even, so there is an element $\sigma$ of order $2$. We must have $\sigma(\alpha)=\alpha^{b}$ for some $b>1$, and it follows that $\alpha^{b^{2}}=\alpha\Rightarrow 2n-1|b^{2}-1\ (*)$. Now $\sigma$ sends $\widetilde\alpha=\alpha^{b-1}$ into $\alpha^{b(b-1)}=\alpha^{1-b}=\widetilde\alpha^{-1}$ due to $(*)$, contradicting the remarks above.

P.S.

I'm curious: what was the Schweitzer problem? And, more importantly, do you have access to more :)? Most Miklos Schweitzer problems are beautiful.

Beautiful solution, grobber. Unfortunately, I don't have access to many other problems from Miklos Schweitzer competition. I had a few problems on a paper. The initial problem was : Prove that there exists a subset $S$ of $\{1,2,...,n\}$ such that $0,1,2,...,n-1$ all have an odd number of representations as $x-y$ with $x,y\in S$ if and only if $2n-1$ has a multiple of the form $2\cdot 4^{k}-1$.


calculus
integration
trigonometry
function
calculus computations

show, without explicitly solving the integrals:

$\frac{\cos{(x^{2})}\;+\;\sin{(x^{2})}}{x^{2}}\;dx\;=\frac{\cos^{2}{(x)}\;+\;\sin^{2}{(x)}}{x^{2}}\;dx$
the first: because easy seen that we only must consider this integral of this function in point $0$.
the seccond: in point $0$ by Taylor we find that this two function is equal.

$\infty=\infty$ is a boring way to claim equality.

Since the first function isn't nonnegative, we do need to assert that $\int_{1}^{\infty}\frac{\cos(x^{2})+\sin(x^{2})}{x^{2}}\,dx$ converges - which it does, absolutely. That way we know that no negative values of the function are sufficient to cancel the positive infinity we get integrating from $0$ to $1.$

Kent is perfectly right, of course, but the problem still can be made sense of: prove (or disprove) that $\;\int_{0}^{\infty}\;\left(\frac{\cos{(x^{2})}\;+\;\sin{(x^{2})}}{x^{2}}\;-\;\frac{\cos^{2}{(x)}\;+\;\sin^{2}{(x)}}{x^{2}}\right)\;dx=0$ ;)

Well, that's a convergent integral. In fact, the integrand tends to a finite limit  as $x\to 0.$

It's not zero; it's negative.

Note that $\cos^{2}x+\sin^{2}x=1$ and that $\cos(x^{2})+\sin(x^{2})=\sqrt{2}\cos\left(x^{2}-\frac{\pi}4\right).$

Consider, for any $p>0,\ \int_{\sqrt{p+\frac{\pi}4}}^{\sqrt{p+\frac{9\pi}4}}\frac{\sqrt{2}\cos\left(x^{2}-\frac{\pi}4\right)-1}{x^{2}}\,dx$

Let $u=x^{2}-\frac{\pi}4.$ The integral becomes

$\int_{p}^{p+2\pi}\frac{\sqrt{2}\cos u-1}{2\left(u+\frac{\pi}4\right)^{\frac32}}\,du$

-- I've got to go now. Maybe I'll be back to finish it, or maybe someone can finish it for me. I probably want to make $p$ an integer multiple of $2\pi,$ and I probably want to integrate by parts somewhere in there.

[quote="Kent Merryfield"]
It's not zero; it's negative.
[/quote]
How do you know that, Kent? $i^{-1/2}$ seems to lie right on the bisector of the 4-th quadrant (I realize that this post looks like spam, but I have a good reason to ask it this way :))

I withdraw the claim. The method I was trying doesn't convince me and now I don't know.

[hide="Actual value"]
\begin{eqnarray*}\int_{0}^{\infty}\frac{\cos x^{2}-1}{x^{2}}\, dx & = & \int_{0}^{\infty}(\cos x^{2}-1) \cdot \frac{1}{\Gamma (2)}\left( \int_{0}^{\infty}s e^{-xs}\, ds \right) \, dx\\ & = & \int_{0}^{\infty}\int_{0}^{\infty}(\cos x^{2}-1) s e^{-xs}\, ds \, dx\\ & = & \int_{0}^{\infty}\left( \int_{0}^{\infty}(\cos x^{2}-1) e^{-xs}\, dx \right) s \, ds\\ & = & \int_{0}^{\infty}\left( \left[-(\cos x^{2}-1) \frac{e^{-xs}}{s}\right]_{0}^{\infty}-\int_{0}^{\infty}(2x \sin x^{2}) \frac{e^{-xs}}{s}\, dx \right) s \, ds\\ & = &-2 \int_{0}^{\infty}\int_{0}^{\infty}x \sin x^{2}e^{-xs}\, dx \, ds\\ & = &-2 \int_{0}^{\infty}\int_{0}^{\infty}x \sin x^{2}e^{-xs}\, ds \, dx\\ & = &-2 \int_{0}^{\infty}\sin x^{2}\, dx\\ \end{eqnarray*}

Then you can use the Fresnel integral

$\int_{0}^{\infty}\sin \left( \frac{\pi x^{2}}{2}\right) \, dx = \int_{0}^{\infty}\cos \left( \frac{\pi x^{2}}{2}\right) \, dx = \frac{1}{2}$

(You may use contour integral of $e^{iz^{2}}$ to show this result.)

to have

$\int_{0}^{\infty}\frac{\cos x^{2}-1}{x^{2}}\, dx =-\sqrt{\frac{\pi}{2}}$.

Similar argument can be applied to show that

$\int_{0}^{\infty}\frac{\sin x^{2}}{x^{2}}\, dx = \sqrt{\frac{\pi}{2}}$

So, the answer is zero.[/hide]

But this solutiion cannot be an answer to misan's problem. It 'explicitly calcluated' the value of integral.

[quote="sos440"]
(You may use contour integral of $e^{iz^{2}}$ to show this result.)
.

[/quote]
 :maybe: How?
maybe: $e^{\frac{i\pi{z^{2}}}{2}}$

[quote="sos440"]But this solutiion cannot be an answer to misan's problem. It 'explicitly calcluated' the value of integral.[/quote]

Well, to be fair, it was already pointed out that the integrals as originally written aren't convergent on $(0, \infty)$.  The evaluation of the difference of the integrands, while convergent, is a more strict condition than what would actually be needed had both sides been convergent.


Find the sum of the first ten terms of the arithmetic sequence $ 2,5,8,\dots$.

the tenth term will be 2+(9*3) or 29

The first term is 2 and there are 10 terms so 31*5 or 155.

You could just find the fifth and sixth terms, and average out, which to me is a bit faster.  So we have 8+3+3=14 and 14+3=17, so sum is (14+17)*5=155

That's legit the same thing as what mathemagician did AIME15.

No, not necessarily because AIME15 didn't solve for the tenth value, knowing the average of the first and tenth would be same was 5th and 6th.


December 24, 2008
I saw that the AMC-8 statistics were just posted.  There were only 5 states with 10 or more perfect scores.  The leader by a huge amount was California with 35 perfect scores, followed by Texas with 13, Washington with 12, Massachusetts and Virginia both with 10.

There were a total of [b]412[/b] perfect scores 205 from Taiwan (amazing!)  I think these stats show that the test was considerably easier than last year (129 perfects). There was one 5th grader that was perfect, 27 sixth graders, 103 seventh graders and 280 eighth graders.

They have not posted the list of perfect scores yet.  It will be interesting to see how this list matches up with the final National Mathcounts state list this year.

wow...500+ perfect scores...that's a lot
i wonder how much this year's test is a lot easier than last year's
if anyone remembers the problems, can you post them?

AMC 8 is posted [url=http://agmath.com/23806.html]here[/url].

WOW ONLY 35 PERFECT SCORES IN CALIFORNIA?

Should be 36...if i could add 8+9 correctly. Now I'm making excuses.  Ugh.

I really don't think that this year's test is harder than last year's.  Wondering:  How many people took the test total?

Would anyone have any idea why there are 3 answers listed for #12?

I checked the scoring distribution for NJ:
#       A          B           C           D             E         OMIT       ?
12   12.14     23.27     29.77*    18.81*    14.26*     1.56      0.18

* indicates a correct answer
? indicates unable to read, either because of a double or not completely erased blah blah blah answer.

3 *'s??

I'm SURE I got 25, not to mention my coach said I did, but

SCORE   5th Grade    6th Grade    7th Grade    8th Grade
  25           0                 0               1               4
.................................................................................
(I'm in 6th BTW)

Maybe your coach is lying :o

My friend was preparing for Certificate of Merit (piano) and didn't pass, so his teacher printed out a fake certificate...


In elementary terms, could someone explain the difference between complete combustion and incomplete combustion? How do you determine whether a reaction is complete combustion or incomplete combustion?

Complete Combustion occurs when you have sufficient (or an excess) of oxygen. We usually talk about this in terms of hydrocarbon fuels (e.g. $ \ce{CH4}$).

$ \ce{CH4} \plus{} 3 \ce{O2} \rightarrow \ce{CO2} \plus{} 2 \ce{H2O}$ Only water and carbon dioxide are formed.

Incomplete combustion occurs when oxygen becomes limited and the fuel burns to liberate a range of products including $ \ce{CO}, \ce{C}, \ce{NOx}$ etc.

$ \ce{CH4} \plus{} \frac {3}{2} \ce{O2} \rightarrow \ce{CO} \plus{} 2 \ce{H2O}$ (notice how the stoichiometric coefficient of oxygen is less that in the complete combustion).

So, just given the reactants of the equation, could you predict what kind of combustion would occur?

No, because you would also need to know the amount of each reactant (hence their stoichiometric ratio).

P.S. http://en.wikipedia.org/wiki/Combustion#Complete

If you knew their stoichiometric ratio, then you would be able to predict it? Could you provide a couple more examples of hydrocarbons combining with different ratios of oxygen?


Is there some application of LaTex for messenger ?? It would be very practical for writing math codes

It seems unlikely because MSN's servers won't have LaTeX installed.
Can you use images? If so, compose the code in AoPS's TeXer and then copy/link to the image in MSN.

good idea ... I didn't know that the outputs were pictures


My friend gave me this problem however I am struggling with it. Show that there exist infinitely many natural numbers which are not the arithmetic mean of distinct powers of two.

Ah, I think I can see what to do now. Thanks a lot and sorry if I have replied late.

If you find something, please let me know. It's interesting!


If $x = a + bi$ is a complex number such that $x^2 = 3 + 4i$ and $x^3 = 2 + 11i$ where $i = \sqrt{-1}$, then $a + b =$?

[hide="You might want to use Eulers Formula, but that would be silly"]
$x = \frac{x^3}{x^2} = \frac{2 + 11i}{3 + 4i} = \frac{ (2 + 11i)(3 - 4i) }{25} = \frac{ 50 + 25i}{25} = 2 + i$ so $a + b = \boxed{3}$
[/hide]


Let $N = 222\dots2$ ($k$ twos).  If $N$ is a multiple of 1998, find the smallest value of $k$.

[hide]Since 1998 and 3k are both divisible by 111, we first divide both by them.

200200200... (3k/3 2's) is divisible by 18
100100100... (3k/3 1's) is divisible by 9

with the divisibility by 9 rule, k=27[/hide]

I hope it is right...

Your answer is correct by what I get.  One thing though, you say k and 1998 are divisible by 111.  How is this possible if your final k value is 27?  (27*3 = 81)

sorry, I meant the number represented by k...I should have said N that was represented by 3k.


Euler
function
logarithms
number theory
totient function
real analysis
real analysis unsolved

Prove that $ \{ \frac {\varphi (n)}{n} \}_{n\equal{}1}^{\infty}$ is dense in $ [0,1]$ where $ \varphi$ is Euler 's function.

$ \frac{\varphi(n)}{n}\equal{}\prod_{p|n} \left(1\minus{}\frac{1}{p}\right)$

Define $ p(i)$ as $ i\minus{}$th prime number, e.g. $ p(1)\equal{}2 ,\; p(2)\equal{}3 ,\; p(3)\equal{}5 ,\; p(4)\equal{}7 ,\; ...$

Choose $ \alpha\in [0,1]\setminus \Bbb{Q}$ arbitarily.

Let $ i_1$ be the smallest number such that $ y_1: \equal{}\left(1\minus{}\frac{1}{p(i_1)}\right)>\alpha$ .

Let $ i_2$ be the smallest number $ >i_1$ such that $ y_2: \equal{}y_1\cdot\left(1\minus{}\frac{1}{p(i_2)}\right)>\alpha$ .

And so on.

Denote the set of the sequence $ i_1,i_2,...$ with $ I$ .

Because of $ \prod_{i\ge N} \left(1\minus{}\frac{1}{p(i)}\right)\equal{}0<\alpha$  there must be infinite many $ i_k$ with $ i_k\minus{}1\notin I$ .

Hence $ y_{k\minus{}1}\cdot \left(1\minus{}\frac{1}{p(i_k\minus{}1)}\right)<\alpha<y_{k\minus{}1}$

$ \Rightarrow \, 0<y_{k\minus{}1}\minus{}\alpha<\frac{y_{k\minus{}1}}{p(i_k\minus{}1)}<\frac{1}{p(i_k\minus{}1)} \,\Rightarrow \, y_k\to \alpha$

With $ n_k\equal{}p(i_1)\cdots p(i_k)$ we have found a sequence with $ \frac{\varphi(n_k)}{n_k}\equal{}y_k\to \alpha$.

Therefore $ \left\{\frac{\varphi(n)}{n}\right\}_{n\equal{}1}^\infty$ is dense in $ [0,1]\setminus \Bbb{Q}$ , which is dense in $ [0,1]$ .

what about $ \frac{\phi(n\plus{}1)}{\phi(n)}$ ? I asked a long time ago about this one on the forum, without answer ..

[hide="Answer"]
Dense in $ [0,\plus{}\infty)$
[/hide]
[hide="Step 1"]
If $ a,c,d$ are odd, $ b$ is even and $ gcd(a,b)\equal{}gcd(c,d)\equal{}1$, then there are infinitely many $ t$ such that $ \frac{\varphi(a\plus{}bt)}{a\plus{}bt}\approx\frac{\varphi(c\plus{}dt)}{c\plus{}dt}\approx 1$.
[hide="Reason"]
What can you say about the totient function of the numbers less than $ M$ that are not divisible by any prime less than $ 0.1\log M$?
[/hide]
[/hide]
[hide="Step 2"]
Let $ u$, $ v$ be of different parity and relatively prime. Look at the solutions of the equation $ ux\minus{}vy\equal{}1$ such that $ x$ and $ y$ are both odd. Apply observation 1 to conclude that we can ensure that $ \frac{\varphi(ux)}{\varphi(vy)}\approx \frac{\varphi(u)/u}{\varphi(v)/v}$.
[/hide]
[hide="Step 3"]
Apply the result of the original problem in this thread in an obvious way.
[/hide]


algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

We know that a polynomial of degree $n$ has at most $n$ roots in a field.

From the finite field theory, we know that the splitting field of $x^{p^n}-x$ on $F$($\mbox{char }F = p$) has the order of $p^n$. The proof says that $x^{p^n}-x$ has $p^n$ distinct roots and then it says that those roots
form a field. The conclusion follows.

My question is, why does $f(x) = x^{p^n}-x$ has $p^n$ roots? I know that its roots are distinct since $f$ and $f'$ are coprime.

Every Polynomial of degree $n$ has exactly $n$ roots in the algebraic closure of the field (counted with multiplicity), so it has $n$ different roots here.

[quote="grobber"][quote="liyi"]I know that its roots are distinct since $f$ and $f'$ are coprime.[/quote]

Isn't this the case here? :? In a field which extends $\mathbb Z_p$, which must thus have characteristic $p$, the derivative of $x^{p^n}-x$ is the constant polynomial $-1$.[/quote]

Didn't liyi write that he knows that¿


topology
function
real analysis
real analysis unsolved

There exists $ A,B \subset \mathbb{R}^n$ such that exits a contiunous bijection  from$ A$ to $ B$, and a cont. bijection from $ B$ to $ A$, and $ A,B$ are not homeomorf sets.

That's the definition of a homeomorphism, as far as I can tell.  Do you mean diffeomorphic?

The question makes sense if "bijection" is replaced by "injection" ("1-1 map" for those afraid of needles). Or by embedding (homeomorphism onto its image).

[b]Definition:[/b]
A function $ f$ between two topological spaces$ X$ and $ Y$ is called a homeomorphism if it has the following properties:
$ f$ is a bijection (1-1 and onto),
$ f$ is continuous,
the inverse function $ f^{\minus{}1}$is continuous.
[b]The problem[/b]
 In this problem exits a cont. bijection $ \varphi: A \to B$ and a cont. bijection $ \tau: B \to A$. (We don't know that $ \varphi^{\minus{}1}$and $ \tau^{\minus{}1}$ are cont)Then $ A$ and $ B$ are homeomorphic?


induction
LaTeX
logarithms
combinatorics unsolved
combinatorics

Prove the following identity for every $ n\in N$:
$ \sum_{j\plus{}h\equal{}n,j\geq h}\frac{(\minus{}1)^h2^{j\minus{}h}\binom{j}{h}}{j}\equal{}\frac{2}{n}$

(Sorry, I don't know tex yet)

Let a(n)= (j+h=n,j>=h) summation (  ((-1)^h).( 2^(j-h)).C(j,h)  )

The problem is equivalent to proving a(n) - a(n-2) =2.

By using C(n,h)= C(n-1,h)+ C (n-1,h-1), we have

a(n) = 2.a(n-1)-a(n-2).

a(1)=2 and a(2)=3.

So by induction, a(n)=n+1.

So a(n)-a(n-2)= (n+1)-(n-1) =2.

Hasan, see [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for information about writing in LaTeX.  I have to admit, however, that I fail to see a relationship between what the question asks and what you've written -- what has happened to the factor of $ \frac {1}{j}$ in the summation?

We can rewrite our sum as $ a_{n} : = \sum_{0 \leq h \leq \frac {n}{2}} \frac {( - 1)^h 2^{n - 2h} \binom{n - h}{h}}{n - h}$.  Then
\begin{align*}A(x) & : = \sum_{n \geq 1} a_n x^n \\
& = \sum_{n \geq 1} x^n \sum_{0 \leq h \leq \frac {n}{2}} \frac {( - 1)^h 2^{n - 2h} \binom{n - h}{h}}{n - h} \\
& = \sum_{h \geq 0} \sum_{\substack{n \geq 2h \\
n > 0}} x^n \frac {( - 1)^h 2^{n - 2h} \binom{n - h}{h}}{n - h} \\
& = \sum_{n \geq 1} \frac {2^n \cdot x^n}{n} + \sum_{h \geq 1} \sum_{n \geq 2h} x^n \frac {( - 1)^h 2^{n - 2h} \binom{n - h}{h}}{n - h}. \end{align*}
Now observe that for $ h > 0$ we have $ \frac {1}{n - h} \binom{n - h}{h} = \frac {1}{h} \binom{n - h - 1}{h - 1}$, so we may compute
\begin{align*} A(x) & = - \log(1 - 2x) + \sum_{h > 0} \frac { ( - 1)^h\cdot 2^{ - 2h}}{h} \sum_{n \geq 2h} (2x)^n \binom{n - h - 1}{h - 1} \\
& = - \log(1 - 2x) + \sum_{h > 0} \frac { ( - 1)^h\cdot 2^{ - 2h}}{h} \cdot \frac {(2x)^{2h}}{(1 - 2x)^h} \\
& = - \log(1 - 2x) - \log\left(1 + \frac {x^2}{1 - 2x}\right) \\
& = - 2\log(1 - x) \\
& = \sum_{n \geq 1} \frac {2x^n}{n}, \end{align*}as needed.

I've learned latex since that post, now I can write it completely:

Let $ a(n)\equal{}\sum_{j\plus{}h\equal{}n,j\geq h}(\minus{}1)^{h}2^{j\minus{}h}\binom{j}{h}.$

$ \sum_{j\plus{}h\equal{}n,j\geq h}\frac{(\minus{}1)^{h}2^{j\minus{}h}\binom{j}{h}}{j}\equal{}\frac{2}{n}\iff  \sum_{j\plus{}h\equal{}n,j\geq h}n\frac{(\minus{}1)^{h}2^{j\minus{}h}\binom{j}{h}}{j}\equal{}2\iff\sum_{j\plus{}h\equal{}n,j\geq h}\frac{j}{j}(\minus{}1)^{h}2^{j\minus{}h}\binom{j}{h}\plus{}\sum_{j\plus{}h\equal{}n,j\geq h}\frac{h}{j}(\minus{}1)^{h}2^{j\minus{}h}\binom{j}{h}\equal{}2\iff a(n)\plus{}\sum_{j\plus{}h\equal{}n,j\geq h}(\minus{}1)^{h}2^{j\minus{}h}\binom{j\minus{}1}{h\minus{}1}\equal{}2.$

Let $ r\equal{}\sum_{j\plus{}h\equal{}n,j\geq h}(\minus{}1)^{h}2^{j\minus{}h}\binom{j\minus{}1}{h\minus{}1}.$

$ r\equal{}(\minus{}1)^{0}2^{n}\binom{n\minus{}1}{\minus{}1}\plus{}(\minus{}1)^{1}2^{n\minus{}2}\binom{n\minus{}2}{0}\plus{}(\minus{}1)^{2}2^{n\minus{}4}\binom{n\minus{}3}{1}\plus{}\dots$

$ a(n\minus{}2)\equal{}(\minus{}1)^{0}2^{n\minus{}2}\binom{n\minus{}2}{0}\plus{}(\minus{}1)^{1}2^{n\minus{}4}\binom{n\minus{}3}{1}\plus{}\dots$

So $ r\equal{}\minus{}a(n\minus{}2)$ thus statement is equal to $ a(n)\minus{}a(n\minus{}2)\equal{}2.$

Now $ 2.a(n\minus{}1)\equal{}(\minus{}1)^{0}2^{n}\binom{n\minus{}1}{0}\plus{}(\minus{}1)^{1}2^{n\minus{}2}\binom{n\minus{}2}{1}\plus{}(\minus{}1)^{2}2^{n\minus{}4}\binom{n\minus{}3}{2}\plus{}\dots$

$ \minus{}a(n\minus{}2)\equal{}(\minus{}1)^{1}2^{n\minus{}2}\binom{n\minus{}2}{0}\plus{}(\minus{}1)^{2}2^{n\minus{}4}\binom{n\minus{}3}{1}\plus{}(\minus{}1)^{3}2^{n\minus{}6}\binom{n\minus{}4}{2}\plus{}\dots$

Since $ (\minus{}1)^{0}2^{n}\binom{n\minus{}1}{0}\equal{}(\minus{}1)^{0}2^{n}\equal{}(\minus{}1)^{0}2^{n}\binom{n}{0},$

$ 2a(n\minus{}1)\minus{}a(n\minus{}2)\equal{}(\minus{}1)^{0}2^{n}\binom{n}{0}\plus{}(\minus{}1)^{1}2^{n\minus{}2}(\binom{n\minus{}2}{1}\plus{}\binom{n\minus{}2}{0})\plus{}(\minus{}1)^{2}2^{n\minus{}4}(\binom{n\minus{}3}{2}\plus{}\binom{n\minus{}3}{1})\plus{}\dots\equal{}(\minus{}1)^{0}2^{n}\binom{n}{0}\plus{}(\minus{}1)^{1}2^{n\minus{}2}\binom{n\minus{}1}{1}\plus{}(\minus{}1)^{2}2^{n\minus{}4}\binom{n\minus{}2}{2}\plus{}\dots\equal{}a(n).$

$ a(1)\equal{}2$ and $ a(2)\equal{}3.$ Assume that $ a(n)\equal{}n\plus{}1$ for $ n\le k.$ $ a(k\plus{}1)\equal{}2a(k)\minus{}a(k\minus{}1)\equal{}2(k\plus{}1)\minus{}k\equal{}k\plus{}2.$ So $ a(n)\equal{}n\plus{}1.$ Thus $ a(n)\minus{}a(n\minus{}2)\equal{}(n\plus{}1)\minus{}(n\minus{}1)\equal{}2.$


Let $ a,b,c$ be positive numbers having product $ 1.$ Prove that
$ \frac{1}{a\plus{}b\plus{}1}\plus{}\frac{1}{b\plus{}c\plus{}1}\plus{}\frac{1}{c\plus{}a\plus{}1} \le \frac{1}{a\plus{}2}\plus{}\frac{1}{b\plus{}2}\plus{}\frac{1}{c\plus{}2}.$

The question is: Are there any solutions without full expansion?

[quote="rachid"][quote="can_hang2007"]Let $ a,b,c$ be positive numbers having product $ 1.$ Prove that
$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}.$

The question is: Are there any solutions without full expansion?[/quote]
I think I've found a nice proof for this inequality  :lol:

I've proved first that :

$ \frac {2}{a \plus{} 2} \minus{} \frac {1}{a \plus{} ac \plus{} 1} \geq \frac {1}{a \plus{} b \plus{} 1}$ ( It's very easy to prove ! )

It's well known ( also very easy to prove ) that $ \sum \frac {1}{a \plus{} ac \plus{} 1} \equal{} 1$,

Hence,It suffices to prove that $ \sum \frac {1}{a \plus{} 2} \leq 1$ which is easy to prove !

Done ![/quote]
Dear rachid, I think you means that $ \sum \frac {1}{a \plus{} b \plus{} 1} \leq 1$
Cuz from it and this nice result $ \frac {2}{a \plus{} 2} \minus{} \frac {1}{a \plus{} ac \plus{} 1} \geq \frac {1}{a \plus{} b \plus{} 1}$
We get $ \sum\frac {2}{a \plus{} 2}\ge 1 \plus{} \sum\frac {1}{a \plus{} b \plus{} 1}\ge \sum\frac {2}{a \plus{} b \plus{} 1}$

[quote="leedt26"]Dear rachid, I think you means that $ \sum \frac {1}{a \plus{} b \plus{} 1} \leq 1$
Cuz from it and this nice result $ \frac {2}{a \plus{} 2} \minus{} \frac {1}{a \plus{} ac \plus{} 1} \geq \frac {1}{a \plus{} b \plus{} 1}$
We get $ \sum\frac {2}{a \plus{} 2}\ge 1 \plus{} \sum\frac {1}{a \plus{} b \plus{} 1}\ge \sum\frac {2}{a \plus{} b \plus{} 1}$[/quote]

No,I meant  $ \sum \frac {1}{a \plus{} 2} \leq 1$,because $ \sum \frac {2}{a \plus{} 2} \minus{} \sum \frac {1}{a \plus{} ac \plus{} 1} \geq \sum \frac {1}{a \plus{} b \plus{} 1}$ $ \Leftrightarrow$ $ \sum \frac {2}{a \plus{} 2} \minus{} 1 \geq \sum \frac {1}{a \plus{} b \plus{} 1}$,and proving that $ \sum \frac {1}{a \plus{} 2} \leq 1$ is sufficient to conclude that

 $ \sum \frac {1}{a \plus{} 2} \geq \sum \frac {1}{a \plus{} b \plus{} 1}$.But,of course it's also sufficient to show that $ \sum \frac {1}{a \plus{} b \plus{} 1} \leq 1$,as you said,both of them are correct and easy to prove :wink:

[quote="rachid"][quote="can_hang2007"]Let $ a,b,c$ be positive numbers having product $ 1.$ Prove that
$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}.$

The question is: Are there any solutions without full expansion?[/quote]
I think I've found a nice proof for this inequality  :lol:

I've proved first that :

$ \frac {2}{a \plus{} 2} \minus{} \frac {1}{a \plus{} ac \plus{} 1} \geq \frac {1}{a \plus{} b \plus{} 1}$ ( It's very easy to prove ! )

It's well known ( also very easy to prove ) that $ \sum \frac {1}{a \plus{} ac \plus{} 1} \equal{} 1$,

Hence,It suffices to prove that $ \sum \frac {1}{a \plus{} 2} \leq 1$ which is easy to prove !

Done ![/quote]
Very nice proof, rachid. It is much nicer than mine. Thank you very much.

[quote="can_hang2007"]Let $ a,b,c$ be positive numbers having product $ 1.$ Prove that
$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}.$

The question is: Are there any solutions without full expansion?[/quote]

$ p \equal{} a \plus{} b \plus{} c\ge3,q \equal{} ab \plus{} bc \plus{} ca\ge3$
\[ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\]
\[ \Leftrightarrow 3p^2q \plus{} pq^2 \plus{} 6pq \minus{} 5p^2 \minus{} q^2 \minus{} 24p \minus{} 3q \minus{} 27\ge0\]
\[ \Leftrightarrow 3(q \minus{} 3)p^2 \plus{} (p \minus{} 3)q^2 \plus{} 4(p \minus{} 3)^2 \plus{} (q \minus{} 3)^2 \plus{} 6pq \plus{} q^2 \plus{} 3q \minus{} 72\ge0\]
...


Calculate the mass of calcium carbonate that dissolves in 1 l of pH 4 HCl.
i have an idea on how to set this problem up but im not sure how can i set this up or do this problem no answer just a ormula or something

Firstly write the reaction.

[hide="Reaction"]

$CaCO_{3}_{(s)}+2HCl_{(s)}\rightarrow CaCl_{2}_{(aq)}+CO_{2}_{(g)}+H_{2}O_{(l)}$[/hide]

Secondly what is the definition of $pH$?

[hide="pH"]

$pH=-\log_{10}[H^{+}]$[/hide]

Can you now calculate the number of mole of $HCl$ and thus deduce the number of mole of $CaCO_{3}$ which had then tell you its mass.

i wrote the formula, so then i take the anitlog of 4
then i multiply it by the molar mass CACO3

Remember for every two mole of $HCl$, you need one mole of $CaCO_{3}$.

[hide="Solution"]

Since $pH=4$ therefore $[H^{+}]= 10^{-4}M$. The solution of $HCl$ is one litre thus $n(HCl)= CV = 10^{-4}\times 1= 10^{-4}mol$. Looking at the reaction we know that the amount of $CaCO_{3}$ must be $\frac{10^{-4}}{2}mol$. Now since the molar mass of $CaCO_{3}$ is $100 gmol^{-1}$ then we have the mass of $CaCO_{3}$ is $5mg$.[/hide]

so i was right the whole time, u converted the grams to milligrams, thanks for showing me thow, i wasnt sure whether i did it right, im real appreciative

Isn't this problem more accurately proposed as "What is the mass of CaCO3 that reacts in 1 L of pH 4 HCl?"

Dissolving would call for a Ksp expression, solving for a solubility in M.

$K_{sp}$ is the solubility product and it pops up when we try to solve problems of whether precipitate had be present in a solution. I don't think that matters in this case.


function
trigonometry
logarithms
integration
real analysis
real analysis solved

Supposed that $f(x)=\sin (-\pi \ln x)\ (0<x\leqq 1)$ and $a_i \ (i=0,1,2,\cdots) ,\ a_0>a_1>a_2>\cdots\ $ be the value of $x$ which satisfies $f(x)=0$.

Let $b_n=\int^{a_n}_{a_{n+1}} f(x)dx $.

Evaluate \[\sum_{k=0}^{\infty} (|b_k|-b_k)\]

Thank you for your reply,Kent Merryfield. :)

My answer is $\frac{2\pi}{(e-1)(\pi ^2 +1)}$.

kunny

Silly mistake. $\frac{1+e^{-1}}{1-e^{-2}}$ simplifies to $\frac1{1-e^{-1}},$ not $\frac1{1+e^{-1}}.$ Fix that, and I agree with kunny.


let $ n\geq1$ an integer and $ x_{1},...,x_{n}$ are positive real numbers such that $ \sum^{n}_{i\equal{}1}\frac{1}{x_{i}}\equal{}n$
find de minimum of $ \sum^{n}_{i\equal{}1}\frac{x^{i}_{i}}{i}$.
looking for other proof.


Compute

$ cos\frac{\pi}{11} \plus{} cos\frac{3\pi}{11} \plus{} cos\frac{5\pi}{11} \plus{} cos\frac{7\pi}{11} \plus{} cos\frac{9\pi}{11}$

I believe that your latex code is typed incorrectly; I beleive that you wanted to divide by 11.  If so, consider the 11-th roots of unity.

[quote="giannis"]Compute

$ cos\frac {\pi}{11} \plus{} cos\frac {3\pi}{11} \plus{} cos\frac {5\pi}{11} \plus{} cos\frac {7\pi}{11} \plus{} cos\frac {9\pi}{11}$[/quote]
Set $ P\equal{}cos\frac {\pi}{11} \plus{} cos\frac {3\pi}{11} \plus{} cos\frac {5\pi}{11} \plus{} cos\frac {7\pi}{11} \plus{} cos\frac {9\pi}{11}$
We have $ sin{\frac{\pi}{11}} \ne 0$
We have fomurla: $ 2sin{a}cos{b}\equal{}sin{(a\plus{}b)}\plus{}sin{(a\minus{}b)}$
So:
$ 2sin{\frac{\pi}{11}}*P\equal{}sin{\frac{2\pi}{11}}\plus{}sin{\frac{4\pi}{11}}\minus{}sin{\frac{2\pi}{11}}\plus{}sin{\frac{6\pi}{11}}\minus{}sin{\frac{4\pi}{11}}\plus{}sin{\frac{8\pi}{11}}\minus{}sin{\frac{6\pi}{11}}\plus{}sin{\frac{10\pi}{11}}\minus{}sin{\frac{8\pi}{11}}\equal{}sin{\frac{10\pi}{11}}\equal{}sin{\frac{\pi}{11}}$
It follows that: $ P\equal{}\frac{1}{2}$


how do i get the polynome

$f(X)=\sum_{k=0}^{n} X^k (1+X)^{n-k}$
to a form not depending on any sum so that the new form of the
polynome has the binomial coef of the term $X^k\rightarrow C_{n+1}^k$

By using binonial theorem you can get that $f(X)=\sum_{i=0}^n\left(X^i\sum_{j=0}^i \binom{n-j}{i-j}\right)$
But this sum: $\sum_{j=0}^i \binom{n-j}{i-j}=\binom{n}{i}+\binom{n-1}{i-1}+\cdots+\binom{n-i}{0}$ can be computed by some combinatorial interpretation and you get: $\sum_{j=0}^i \binom{n-j}{i-j}=\binom{n+1}{i}$
So your polynomial is: $f(X)=\sum_{i=0}^n \binom{n+1}{i}X^i$


Let ABC be an equilateral triangle.  Find the locus of the points P in the plane with the property that PA, PB, and PC are the side lengths of a right triangle.

All points not lying on the circumcircle of ABC.
Suppose P does not lie on the circumcircle of ABC. By Ptolemy,
PA*BC+PB*AC>PC*AB < == > PA + PB > PC.
If P does lie on the circumcircle of ABC, we have equality for Ptolemy, PA+PB=PC (or PB+PC=PA, depending on which arc of the circumcircle P is situated in).

PA, PB, and PC must be the side lengths of a RIGHT triangle, however.

oh. Sorry, im such a total idiot   :blush: .
When P lies inside the triangle:
Rotate ABC by 60 degrees about A. A stays fixed, B goes to C, let C go to C' and P go to P'. <PAP'=60 and PA=PC by construction so PAP' is equilatrel. Consider the triangle <PP'C. PP'=AP,P'C=PB and PC=PC so PP'C has side lengths PA,PB,PC. if <P'PC=90,<P'CA+<PCA=90,<PCA+<PBA=90 --> <PBC+<PCB=120-90=30
--> <BPC=150 is an equivalant condition. Note <APC=150 and <APB=150 also give right angle triangles, but with different orienatations. So the locus when P lies inside the triangle are the three circles , e.g. the one passing through B and C and subtending <BPC=150.
When P lies outside the triangle, the same proof works with some modifications. (so the answer are the whole parts of the three circles described above).

oops, late by  6 minutes

[hide]Let CP>AP, BP.  Rotate the triangle ABC 60 degrees around B (P to P', C to C')

So P'PB is an equilateral triangle. If APP' is a right triangle, then the [tex]\angle APB[/tex] is 150 degrees. So the locus of points is arc C'AB of circle with center C',  Also arc BC and AC with center X and Y respectively,  such that angleAXC and angle BYC=60 will also be the locus outside the triangle[/hide]


Hi,

I discovered a relation between an infinite radical and an infinite product I can't still explain. Namely, let for m in Z, n in N :

g(n,m) = sqrt(4^m+sqrt(4^(m+1)+sqrt(4^(m+2)+....+sqrt(4^(m+n))...))

Then I noticed :

limit n-->oo (2^m+1-g(n,m))*2^(n(n+1)/2) = C*q(m)

where q(m) is rational :

q(m) = prod(k=0,m,2^k+2)   if m>=0
q(m) =1/prod(k=m,0,2^k+2) if m<0

and C is a constant expressible as an infinite product :

C=prod(k>=0,1-1/(2^k+1))= 0.20971122087...

All of this is related to the trivial identity :

(2^x+1)^2=4^x+2^(x+1)+1

But the problem defeated me.

Benoit.

why is all this related to that identity ?

Well,

(2^x+1)^2=4^x+2^(x+1)+1

gives for any n :

2^x+1=sqrt(4^x+sqrt(4^(x+1)+....+sqrt(4^(x+n)+2^(x+n+1)+1))..))

So I thought all is related to this identity.


I found a glitch in viewing who is online  Apperently 2 guest are viewing the site from the "future"(see attachment). I use internet explorer 7 and my os is windows xp

Search :)

http://www.mathlinks.ro/viewtopic.php?search_id=760438985&t=282188

Keywords: 20 November


Let p be a prime number. What's the least number of colors it takes to set markers painted in one of those colors on a p x p table, such that there are no two markers with that same color attacking each other? (Two markers are attacking each other is they are on the same horizontal, vertical or diagonal line on the square, like the queens in chess)

I'm going to conjecture p+2. I'm almost positive its on O(p), and this works for p=2,3. Not sure why it matters that its prime though.


geometry
3D geometry
sphere
geometric transformation
rotation
real analysis
real analysis theorems

Just out of sheer curiosity I wanted to see a proof of the Banach-Tarski paradox. (This may have been a bad idea in retrospect). The shortest one I could find seems to be just notes about a proof and thus has lots of gaps.

Let $ D$ be a countable subset of the unit sphere centered at the origin. Clearly we can pick an axis that does not contain any of the points of $ D$. Here is the part that seems like it should be true, but I can't seem to figure out why there exists a rotation about that axis, say $ \alpha$, such that $ D, \alpha D, \alpha^2 D, \ldots$ are all disjoint.

Here is why it "makes sense." If you union all those disjoint sets, you still have only a countable number of points, so there is "enough space" in some sense. It is hard for me to come up with a reason why you would always be able to find such a rotation.

It is precisely because it's a countable set. For any two points in the set $ a$ and $ b$, how many rotations around the chosen axis map $ a$ to $ b$? At most one. Therefore, at most countably many of the possible angles $ \alpha$ have some $ \alpha^n$ mapping $ a$ to $ b$. Take the union of these bad sets over all pairs- it's still countable, so there's something left.


linear algebra
matrix
vector
probability

Prove or disprove the following:

(a) A square matrix with linearly independent column vectors is diagonalizable.
(b) If A is diagonalizable, then there is a unique matrix P such that $ P^{\minus{}1}AP$ is a diagonal matrix.
(c) If $ v_1,v_2,$and$ v_3$ come from different eigenspaces of A, then it is impossible to express $ v_3$ as a linear combination of $ v_1$ and $ v_2$.
(d) if A is a diagonalizable and invertible, then $ A^{\minus{}1}$ is diagonalizable.

I believe the first two are false and the last two are true, but I'm not very good with proofs...

The last one is easily proved.

let PAP^-1 = D, D being a diagonal matrix. Now, det(D) is nonzero, since, none of the eigen values oa A are zero. Hence, D is invertible.

 so,
         PA^-1P^-1 = D^-1
Inverse of D is D itself, with every diagonal element replaced by its reciprocal. Hence, inverse of D is also a diagonal matrix.

(a) False. That only means that the matrix is non-singular. Take, for example, $ A \equal{} \left[\begin{array}{ccc} 2 & 1 & \minus{} 1 \\
0 & 2 & \minus{} 1 \\
\minus{} 3 & \minus{} 2 & 3 \end{array}\right]$: this matrix have a non-zero determinant (and so its column vectors are linearly independent) but it is not diagonalizable.

(b) False. In the matrix P its columns are the eigenvectors: just switch the order of these and you get another diagonalizing matrix P such that P-1AP is diagonal.

(c) True. Those vectors are associated with distint eigenvalues and so they are linearly independent, which means that any of them can be written as a linear combination of the others.

(d) True. If A is diagonalizable there is a non-singular P such that $ A \equal{} PDP^{ \minus{} 1}$, with D diagonal. Therefore, $ A^{ \minus{} 1} \equal{} PD^{ \minus{} 1}P^{ \minus{} 1}$, where $ D^{ \minus{} 1}$ is diagonal, and so $ A^{ \minus{} 1}$ is also similar to a diagonal matrix, i. e., it is also diagonalizable.

[quote="Carcul"](a) False. That only means that the matrix is non-singular. Take, for example, $ A \equal{} \left[\begin{array}{ccc} 2 & 1 & \minus{} 1 \\
0 & 2 & \minus{} 1 \\
\minus{} 3 & \minus{} 2 & 3 \end{array}\right]$: this matrix have a non-zero determinant (and so its column vectors are linearly independent) but it is not diagonalizable.
[/quote]
Hmm, did you specifically construct this counterexample, or did you just randomly come up with the matrix and by chance it does not satisfy the statement (because I don't feel like randomly inventing matrices and test them as it's a long process with tons of tedious calculations)?

Out of curiosity, what is the probability that any matrix would satisfy the statement (maybe both real and complex)?

The probability is zero in any reasonable (absolutely continuous) distribution. That matrix was specifically constructed. Of course, it would be easier to use something like $ \begin{bmatrix}1&1\\0&1\end{bmatrix}$.

That matrix was not specifically constructed, I just remembered it from some source and know it not to be diagonalizable.

Professor Jmerry, are you saying that, given a nxn matrix A the probability of A being diagonalizable is zero?

He means the probability of $ A$ [i]not[/i] being diagonalizable.  The matrices with distinct eigenvalues are dense and their complement has measure zero.  Anyway, the example jmerry gave is the simplest nontrivial example of a [url=http://en.wikipedia.org/wiki/Jordan_normal_form]Jordan block[/url], so I prefer that example.  A matrix is not diagonalizable if and only if its Jordan normal form contains a Jordan block of size larger than $ 1 \times 1$; with this information, you don't have to "guess" to construct non-diagonalizable matrices.