trigonometry
geometry
3D geometry
function
complex numbers
algebra unsolved
algebra

Find $ x \in R$ of this equation: $ x^3\minus{}4x^2\minus{}5x\plus{}6\equal{}\sqrt[3]{7x^2\plus{}9x\minus{}4}$

Let $ u\equal{}x\plus{}1$ and $ v\equal{}\sqrt[3]{7x^2\plus{}9x\plus{}4}$, we can rewrite the equation as
$ u^3\plus{}u\equal{}v^3\plus{}v$. To find real solution, we have $ u\equal{}v$, which is
$ (x\minus{}5)(x^2\plus{}x\minus{}1)\equal{}0$. The rest is trivial.

[quote="xxp2000"]Let $ u \equal{} x \plus{} 1$ and $ v \equal{} \sqrt [3]{7x^2 \plus{} 9x \plus{} 4}$, we can rewrite the equation as
$ u^3 \plus{} u \equal{} v^3 \plus{} v$. To find real solution, we have $ u \equal{} v$, which is
$ (x \minus{} 5)(x^2 \plus{} x \minus{} 1) \equal{} 0$. The rest is trivial.[/quote]

But, $ \frac{\minus{}1\minus{}\sqrt{5}}{2}$ is not a solution.

$ P\equal{}\{x\in\mathbb{R}|x^3\minus{}4x^2\minus{}5x\plus{}6\equal{}\sqrt[3]{7x^2\plus{}9x\minus{}4}\}$, $ Q\equal{}\{x\in\mathbb{R}|(x \minus{} 5)(x^2 \plus{} x \minus{} 1) \equal{} 0\}$
So, $ P \subset Q$.

hello, your equation is equivalent with
$ (x\minus{}5)(x^2\plus{}x\minus{}1)(x^6\minus{}8x^5\plus{}7x^4\plus{}49x^3\minus{}31x^2\minus{}57x\plus{}44)\equal{}0$
$ x\equal{}5$ or $ x\equal{}\frac{1}{2}\sqrt{5}\minus{}\frac{1}{2}$ are the only solutions.
Sonnhard.

Why can you exclude $ \frac{\minus{}\sqrt{5}\minus{}1}2$?

hello, putting $ x \equal{} \frac { \minus{} 1 \minus{} \sqrt {5}}{2}$ in the left-hande side of your equation you will got $ \minus{}\frac {1}{2}\sqrt {5} \plus{} \frac {1}{2}$ and in the right-hand side we get $ \sqrt [3]{2 \minus{} \sqrt {5}}$, this is not equal.
Sonnhard.

$ \frac {1 \minus{} \sqrt {5}}2 \equal{} \minus{} \sqrt [3]{\sqrt {5} \minus{} 2} \neq \sqrt [3]{2 \minus{} \sqrt {5}}$

Do you mean $ \sqrt[3]{\minus{}1}\neq \minus{}\sqrt[3]{1}$? If that is the case, I am wasting my time here.

[quote="xxp2000"]Do you mean $ \sqrt [3]{ \minus{} 1}\neq \minus{} \sqrt [3]{1}$? If that is the case, I am wasting my time here.[/quote]

Yes.
$ \sqrt [3]{ \minus{} 1} \equal{} \frac{1}{2} \plus{} \frac{\sqrt{3}}{2}i$.
$ \forall n \in \mathbb{N}, \ a<0 \ \Rightarrow \ \sqrt[n]{a} \in \mathbb{C}$.

Even you know $ (\minus{}1)^3\equal{}\minus{}1$, you still cannot get $ \sqrt[3]{\minus{}1}\equal{}\minus{}1$.
If that is your logic,
$ \sqrt[3]{1}\equal{}\minus{}\frac12\plus{}\frac{\sqrt{3}}2i$.
$ \forall n \in \mathbb{N}$, $ a>0$, we have $ \sqrt[n]{a} \in \mathbb{C}$. There is no real solution for this equation.

That is really ridiculous!

xxp, what is the motivation for your solution? It was brilliant!

[quote="Kouichi Nakagawa"][quote="xxp2000"]Do you mean $ \sqrt [3]{ \minus{} 1}\neq \minus{} \sqrt [3]{1}$? If that is the case, I am wasting my time here.[/quote]

Yes.
$ \sqrt [3]{ \minus{} 1} \equal{} \frac {1}{2} \plus{} \frac {\sqrt {3}}{2}i$.
$ \forall n \in \mathbb{N}, \ a < 0 \ \Rightarrow \ \sqrt [n]{a} \in \mathbb{C}$.[/quote]
That's wrong AFAIK.
$ \sqrt[2k\plus{}1]{\minus{}1} \equal{} \minus{}1 \forall k \in \mathbb{N}$ by definition.

[quote="xxp2000"]Even you know $ ( \minus{} 1)^3 \equal{} \minus{} 1$, you still cannot get $ \sqrt [3]{ \minus{} 1} \equal{} \minus{} 1$.
If that is your logic,
$ \sqrt [3]{1} \equal{} \minus{} \frac12 \plus{} \frac {\sqrt {3}}2i$.
$ \forall n \in \mathbb{N}$, $ a > 0$, we have $ \sqrt [n]{a} \in \mathbb{C}$. There is no real solution for this equation.

That is really ridiculous![/quote]

$ \alpha \equal{} r e^{i\theta}$
$ (r \ge 0, \ 0\le \theta < 2\pi)$

So. $ \alpha^{n/m} \equal{} r^{n/m} \cdot e^{i (n\theta/m)} \equal{} r^{n/m} \cdot \left(\cos \frac {n\theta}{m} \plus{} i \sin \frac {n\theta}{m}\right)$.

Thus,
$ 1 \equal{} 1e^{i0}$. and $ \sqrt [3]{1} \equal{} 1 e^{i (0/3)} \equal{} 1\left(\cos \frac {0}{3} \plus{} i \sin \frac {0}{3}\right) \equal{} 1$.

[quote="Mathias_DK"][quote="Kouichi Nakagawa"][quote="xxp2000"]Do you mean $ \sqrt [3]{ \minus{} 1}\neq \minus{} \sqrt [3]{1}$? If that is the case, I am wasting my time here.[/quote]

Yes.
$ \sqrt [3]{ \minus{} 1} \equal{} \frac {1}{2} \plus{} \frac {\sqrt {3}}{2}i$.
$ \forall n \in \mathbb{N}, \ a < 0 \ \Rightarrow \ \sqrt [n]{a} \in \mathbb{C}$.[/quote]
That's wrong AFAIK.
$ \sqrt [2k \plus{} 1]{ \minus{} 1} \equal{} \minus{} 1 \forall k \in \mathbb{N}$ by definition.[/quote]

$ x^{2k\plus{}1} \equal{} \minus{}1 \ \forall k \in \mathbb{N} \Rightarrow x\equal{}\minus{}1$.

But, $ \sqrt [2k \plus{} 1]{ \minus{} 1} \equal{} \cos \frac{\pi}{2k\plus{}1} \plus{} i \sin \frac{\pi}{2k\plus{}1}$.

[quote="Kouichi Nakagawa"][/quote][quote="Mathias_DK"][quote="Kouichi Nakagawa"][quote="xxp2000"]Do you mean $ \sqrt [3]{ - 1}\neq - \sqrt [3]{1}$? If that is the case, I am wasting my time here.[/quote]

Yes.
$ \sqrt [3]{ - 1} = \frac {1}{2} + \frac {\sqrt {3}}{2}i$.
$ \forall n \in \mathbb{N}, \ a < 0 \ \Rightarrow \ \sqrt [n]{a} \in \mathbb{C}$.[/quote]
That's wrong AFAIK.
$ \sqrt [2k + 1]{ - 1} = - 1 \forall k \in \mathbb{N}$ by definition.[/quote][quote="Kouichi Nakagawa"]

$ x^{2k + 1} = - 1 \ \forall k \in \mathbb{N} \Rightarrow x = - 1$.

But, $ \sqrt [2k + 1]{ - 1} = \cos \frac {\pi}{2k + 1} + i \sin \frac {\pi}{2k + 1}$.[/quote]
I seriously doubt it. But i guess that various upon definition. As far as I'm concerned $ \sqrt[n]{a}$ is defined for $ n \in \mathbb{N}$ only if $ a \ge 0$ when $ n$ is even, and always from $ \mathbb{R} \to \mathbb{R}$.
Since $ LHS$ is real, it is most probable that thanhnam2902 uses the same definition :)

Hi every One! My idea about this nice problem: $ \sqrt [3]{ \minus{} 1} \equal{} \minus{} \sqrt [3]{1}$ is true.

The truth is this eqution have three root $ \in R$ are $ x\equal{}5, x\equal{} \frac {\minus{} \sqrt {5} \minus{}1}{2}, x\equal{} \frac {\minus{} \sqrt {5}\plus{}1}{2}$

The key word is [i]definition[/i] here.

If you take root as meaning "all numbers which, when elevated to a power, give the number", then for example we have that:

$ \sqrt{25} \equal{} \pm5$

There is no reason other than definition by which we take $ \sqrt{25}$ to mean $ 5$ only.

Similarly, everyone usually takes $ \sqrt[3]{\minus{}1}$ to mean the real cube root of -1, which is -1. There are, however, other solutions to the equation $ x^3 \plus{} 1 \equal{} 0$, namely $ \frac12 \pm \frac{\sqrt3}2i$.

[quote="MathWorld"]Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $ 9$ are $ \minus{}3$ and $ \plus{}3$, since $ (\minus{}3)^2\equal{}(\plus{}3)^2\equal{}9$. Any nonnegative real number $ x$ has a unique nonnegative square root $ r$; this is called the principal square root and is written $ r\equal{}x^{1/2}$ or $ r\equal{}\sqrt{x}$. For example, the principal square root of $ 9$ is $ \sqrt{9}\equal{}\plus{}3$, while the other square root of $ 9$ is $ \minus{}\sqrt{9}\equal{}\minus{}3$. In common usage, unless otherwise specified, "the" square root is generally taken to mean the principal square root.

by [url=http://mathworld.wolfram.com/SquareRoot.html]Square Root[/url] - MathWorld

The unique nonnegative square root of a nonnegative real number. For example, the principal square root of $ 9$ is $ 3$, although both $ \minus{}3$ and $ 3$ are square roots of $ 9$.

by [url=http://mathworld.wolfram.com/PrincipalSquareRoot.html]Principal Square Root[/url] - MathWorld[/quote]

Square roots of $ 25$ are $ \minus{}5$ and $ \plus{}5$.
But, $ \sqrt{25}$ mean principal square root of $ 25$.
So $ \sqrt{25}\equal{}5$.
By the mathematics education in Japan, We learn it carefully in this way.

[hide="mathematics education in Japan"]What is the value of square roots of $ 25$?
[hide="Answer"]$ \minus{}5$ and $ \plus{}5$.[/hide]

What is the value of $ \sqrt{25}$?
[hide="Answer"]$ 5$.[/hide][/hide]

Hmm... I looked at a different page in mathworld:

[quote="MathWorld"]The schoolbook definition of the cube root of a negative number is $ ( \minus{} x)^{\frac13} \equal{} \minus{} (x^{\frac13})$. However, extension of the cube root into the complex plane gives a branch cut along the negative real axis for the principal value of the cube root. By convention, "the" (principal) cube root is therefore a complex number with positive imaginary part.[/quote]

So, in complex numbers, the principal cube root of $ \minus{} 1$ is, indeed, $ \frac {1}{2} \plus{} \frac {3}{2}i$ (solely by definition). But we're not messing with the complex plane here, so I assume we're using the "school" definition $ \sqrt [3]{ \minus{} 1} \equal{} \minus{} 1$ unless otherwise stated.

This also explains why Sonnhard said they're different without explaining it -- his CAS probably returns the complex value for the cube root too.

[quote]This also explains why Sonnhard said they're different without explaining it -- his CAS probably returns the complex value for the cube root too.[/quote]

Most CA Systems indeed take the principal root, unless you explicitly demand real values. That's one of the reasons inexperienced students using Maple/Mathematica etc come up with strange plots of functions: Example:

[code]Plot[x^(1/3), {x, -5, 5}][/code]

in Mathematica shows only the function in first quadrant, because the internal calculation done by it shows "irreality" for negative $ x$.


MATHCOUNTS
ratio
factorial
summer program
MathPath

[quote="236factorial"]HOw long do sleep everyday? I sleep about 7 hours on school days, and 8 hours during the summer.

I saw some pretty weird answers on the do you have lunch thread, so lets see what we get here.[/quote]

School Days: 8 hours.
Summer Days: 10 hours
Sleepovers: 0 hours
Camps (1 to 7 days): Average of 2 hours
Camps (more than 7 days): Average of 6 hours

10 hours over the summer, 9 during the school year.

I think there has been about 9 days when I haven't slept for the whole night... I fall asleep the next day  :D

[quote="h78reg"]wow, you must REALLY like math. :D[/quote]

My friends say I'm obssesed with it. Everyone should be. I used to do about 30-70 problems a day. Now the problems are harder however, so I do less.  :(

yea, math problems get harder and harder. which AOPS book do you use? i'm thinking of getting art and craft of problem solving, but it seems WAY too hard.

There was a thread about this not too long ago on the forums...

For me:

This summer: 7-7.5 hrs
School: 7-9 during XC or track, less than 7 when it's not a running season.

Last night (this morning ;) ) I went to sleep at about 5:30 AM and woke up around 10.  I was up all night working on the Solution guide to the 2005 Mock AMC A.  Actually, I didn't intend to stay up that late but I was so into my work that every time I would take a break to see what time it was another hour had passed. Oh yeah, take the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46816]2005 Mock AMC A[/url]  :)

I have about 10 to 12 during vacations, unless I need to get up early for something, 10 on weekends, and 7-8 on schoolnights.

6-7 hours on school days
9-10 on weekends or vacations
0 on sleepovers  :lol:

It seems like no one sleeps on [b]sleep[/b]overs, how strange.

on school nights, 7.5-8
on weekends/summer - 10-12  (i like my sleep)

wow, 10-12 hours. i haven't slept that long since i was in kindergarten :D

i sleep about 5-6 hours during the school year and a little less over the summer. mainly because im tyring to catch up on AMC/AIME/USAMO practice that i couldnt fit in during school.

7-8 for me.  I make a point of being in bed by 1:00, which I still consider to be too late.  So it's often earlier (I think, I don't really keep track of these things).  And I'm usually up shortly after 8, or about an hour earlier when I have class at 9.

Of course, this all goes out the window when there's a problem set, or a paper, or something due the next day.

During school days,when the clock strikes 12"00(morning),i slip into my dizzy mode and doze off.7 hrs for me.Could someone tell me a nice method to stay up till like 1"30(somethin other than coffee and sugar,they don't work on me and i can't make coffee :o )

Keep yourself occupied with something active: like running on a treadmill, ping-pong, unscrewing your door hinge, etc.

[quote="tiredepartment"]Keep yourself occupied with something active: like running on a treadmill, ping-pong, unscrewing your door hinge, ...[/quote]

...swimming, playing a concerto on the piano, doing 50 stacks of math and SAT reading comprehension, go on AoPS, ...

[quote="236factorial"]Many things happen in my sleep. I sometimes can't fall asleep due to some math thing. I proved the sum of the first n squares during bettime, also some mathcount problems get solved when I'm in bed.[/quote]
the same here. i solved lot of problem while sleeping :P

It sort of sucks if you solve a famous conjecture right before you fall asleep, and then forget the proof the next morning.

Ooh ooh conspiracy theory thingy:

Every night, right before you fall asleep, you realize that we are all under hypnosis mind-control stuff and stuff, or something, but you forget it the next morning.

[quote="log0"]Does anybody have any tips for people who can't fall asleep? I usually lie in bed for one to two hours before falling asleep.[/quote]Try making yourself physically or mentally tired by the end of the day. Some people say you should clear your mind before you go to sleep - don't know if that works though. :)

I find that reading a chapter or so (or maybe even a whole book) before going to sleep helps a lot.

Actually I think I slept 15 hours on Saturday because I was sick, but other than that, I average 5 - 6 on most days, which may explain why my immune system is so crummy in the first place.  :P

I sleep, averaging speaking, about 7-8 hours per day. But I only feel myself comfortable when I sleep 10-11 hours a day...makes me think better and I generally have more ideas. That's why I love to travel; everytime I get home after a big time travelling I have excelent ideas =] even to solve math problems.

[quote="S2"]i used to sleep 20 hours a day[/quote]Wow, seriously? Not recently, I hope?


function
topology
real analysis
real analysis unsolved

$A$ a subalgebra of $C_0(\mathbb{R})$ (i.e. continuous functions that tends towards $0$ at $-\infty$ and $-\infty$).
Suppoe moreover that:
*if $x \neq y$ are two reals, there exists $f \in A$ such that $f(x) \neq f(y)$
* for any real $x$ there exists $f \in A$ such that $f(x) \neq 0$

Is $A$ dense in $C_0(\mathbb{R})$. (with the uniform norm)

It works for any locally compact Hausdorff space $X$, where $C_0(X)$ is defined as the set of continuous real functions $f$ on $X$ which vanish at infinity, i.e. for which given any $\varepsilon>0$, there is a compact $K_\varepsilon\subset X$ such that $|f|<\varepsilon$ outside $K_\varepsilon$.

Let $\tilde X$ be the one-point compactification of $X$, obtained by supplementing $X$ with a point $\infty$. The functions we're concerned with can be identified with those continuous functions on $\tilde X$ which vanish at the point $\infty$. The algebra $\tilde A$ generated by all the functions in our initial algebra $A$, together with a non-zero constant function $t$ on $\tilde X$ is easily seen to satisfy the hypotheses of the Stone-Weierstrass Theorem for compact Hausdorff spaces, so it's dense in $C(\tilde X)$. Now, every function $f$ in $C_0(X)$ can be approximated arbitrarily well by a sequence of functions $a_nt+f_n$ in our algebra $\tilde A$, where $a_n$ are real constants and $f_n$ are functions from $\tilde A$ which vanish at $\infty$ (i.e. elements of $A$). Since the function $f$ vanishes at $\infty$, we must have $a_n\to 0$, which means that the sequence $(f_n)_n\subset A$ approximates $f$ arbitrarily well, and we're done.


[quote="arkhammedos"]thanks to u all  who had help me
but where can i find some cours of olympiad
des cours dolymps  thankds[/quote]

You mean classes?

[url]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php[/url]


[quote="lyra"]I learned about the laws of logs in my honors algebra II class a little while ago.  I got all of the problems in class right, but they were pretty easy, I went looking for some harder problems, and was wondering if someone could tell me if I am solving the following correctly (they arn't that hard actually) and also if someone could give me some harder ones, or tell me where to look.   Anyway:
Rewrite as a single logarithm:
2log[size=75]x[/size][size=100]a[/size]-3log[size=75]x[/size][size=100]b[/size]+4log[size=75]x[/size][size=100]c[/size]
heres what I did:
(2-3)log[size=75]x[/size][size=100](a/b)[/size]  *division log property*
(2-3+4)(log[size=75]x[/size][size=100]((a/b)(c))[/size]=
[b]3log[size=75]x[/size][size=100]((ac)/b)[/size][/b]

is this correct, or am i nuts?

Thanks,
lyra[/quote]

you can't say $a\log{x}-b\log{y}=(a-b)\log{\frac{x}{y}}$.

You do it like this:

$2\log_x{a}-3\log_x{b}+4\log_x{c}$

$\log_x{a^2} - \log_x{b^3} + \log_x{c^4}$

$\log_x{\frac{a^2c^4}{b^3}}$


hey is there a way that i can get past mathcounts nationals tests?
I've looked everywhere online and the only ones posted are chapter and state. Could someone please give me a website or send me some recent national tests & solutins.

You can't get national mathcounts problems online. You have to buy them.

Ask your coach or your state coach if he/she haves some.

Our coach got some nationals tests from other state coaches, and today(Saturday), we got two--one we did, and the other is homework.  I may be able get those for you, but I'll have to check.

I've heard that the tests are expensive too. If your school has been in a national competition, does the school also have copies of the competitions?

My nationals team coach got some from the state coordinator.

A lot of coaches often have old tests (especially the ones who have gone to nats multiple times).  Also, google "elias saab"--his website has a lot of good practice problems for nats, albeit poorly worded.

My coach got them from other coaches.  The problem is that some still have writing on them.


trigonometry
inequalities
inequalities unsolved

Let $x,y,z>0$, in every triangle of angles $A,B,C$ we have the ineq

$\frac{\sin {A/2}}{\sqrt{y+z}}+ \frac{\sin {B/2}}{\sqrt{z+x}}+\frac{\sin {C/2}}{\sqrt{x+y}} \leq \frac{(x+y+z)}{\sqrt{(x+y)(y+z)(z+x)}} $

Hi manlio!

Unfortunately, I can't find geometrical solution :(

Let $a=y+z$, $b=x+z$, $z=x+y$. Then we need to show that $\sqrt{bc}\sin A/2+\sqrt{ac}\sin B/2+\sqrt{ab}\sin C/2 \leq (a+b+c)/2$.
Consider such triple $(A,B,C)$, s.t. $A,B,C\geq 0$, $A+B+C=\pi$ and LHS is maximal for fixed $a,b,c$.

There are two cases:
1) one of $A,B,C$ is zero. Suppose WLOG $C=0$. Then we have to prove $\sqrt{bc}\sin A/2+\sqrt{ac}\cos A/2\leq (a+b+c)/2$. But it is well known that in this case $LHS\leq \sqrt{bc+ac}$. So we can easily obtain conclusion.
2) $A,B,C>0$. Since $(A,B,C)$ is maximal triple we have $\left(\sqrt{bc}\sin t/2+\sqrt{ac}\sin \frac{\pi-t-C}{2}\right)'|_{t=A}=0$. It immediately follows that $\sqrt{\frac{b}{a}}=\frac{\cos^2 B/2}{\cos^2 A/2}$. Hence we conclude $A:B:C=\cos^2 A/2:\cos^2 B/2:\cos^2 C/2$. Thus we need to prove that
\[
\cos A/2\cos B/2\sin C/2 + \cos A/2\sin B/2\cos C/2 + \sin A/2\cos B/2\cos C/2 \leq \frac{1}{2}(\cos^2 A/2+\cos^2 B/2+\cos^2 C/2).
\]
Grouping in LHS 1st and 2nd summands, 2nd and 3rd summands, 3rd and 1st summands we see that LHS=RHS !!!

We know that $ sin{\frac{B}{2}} = cos{\frac{A+C}{2}} $ we can rewrite the inequality
$ x^2 + y^2 + z^2 \geq\ 2xycosA + 2yzcosB + 2zxcosC $

This is a very well known inequality and it has a nice geometrical solution.

I am so idiot to have not seen this substitution :(


A simple homework exercise from the introductory class a friend of mine teaches:

$ A$ is a $ 10\times 10$ matrix, $ \text{rank}(A)\equal{}3,$ and $ \text{rank}(A^2)\equal{}2.$ Is $ A$ diagonalizable? Why or why not?

[quote="mlok"]Not diagonalizable, because $ 3\cdot 2^2\ne 10$. :roll:
Do I get partial credit?[/quote]

Er, sorry, I'm sensing that this is some sort of a trick argument, but I don't really follow. How does this say anything about diagonalizability?

Sorry for my silly questions  :oops:

mlok is just joking. If you want to try answering this for yourself, please ignore mlok's post altogether.

If $ A$ is a diagonalizable matrix then $ \text{rank}(A)\equal{}\text{rank}(A^k)$ for all $ k$.

A great deal of the language of linear algebra breaks down if the set of scalars is not a field. To be honest, I am not sure what "rank" means in that case, as I don't know that I can trust that Gaussian elimination is a useful algorithm. My colleague would have nothing against having his question posed over an arbitrary field, but I doubt that he intended to include rings that aren't fields, (worse yet, rings with zero divisors.)

Rank has to be replaced with determinantal rank (i.e. there is a non-zero r-rowed minor, and all r+1-rowed minors are zero), and this works for commutative rings. There is gaussian elimination in PIDs, although it's more complicated than in a field.


is there a general formula, given two parallel lines ax+by+c=0, ax+by+d=0, for the distance between these 2 lines?

it's too late, and im tired, and i don't wanna derive it myself.

Any help would be greatly appreciated

Well, theres probably an easier way, but:

[hide]

The lines are a vertical distance of x=|(d-c)/b| apart and a horizontal distance of y=|(d-c)/a| apart. Now consider that right angled triangle with those two lengths as the sides, and we want to find that other altitude.. just equating the areas we get xy = ? sqrt(x:^2:+y:^2:), and just substitute the above in. Stuff simplifies and we get |(d-c)|/sqrt(a:^2:+b:^2:)

Going to check my working now

[/hide]


Of the following sets of data the only one that does not determine the shape of a triangle is:
(A) the ratio of two sides and the included angle
(B) the ratios of the three altitudes
(C) the ratios of the three medians
(D) the ratio of the altitude to the corresponding base
(E) two angles

[hide]D, let's say the altitude is h and the base is b. I could make h perpendicular to b at any point on b, and the third vertex is not uniquely determined[/hide]

[hide]it is D because I can draw an only-case triangle with the others[/hide]


Words are formed using letters $A,B$. Using words $x_{1},x_{2}, ..., x_{n}$ we can form a new word if we write these words next to eaxh other: $x_{1}x_{2}...x_{n}$. A word is a palindrome if it does not change after rewriting its letters in reverse order. Prove that any word with $1995$ letters $A$ and $B$ can be formed with less than $800$ palindromes.

[hide="Hint"]Prove that any string of $5$ letters can be divided into at most $2$ palindromes[/hide]
[hide="Solution"]We will first prove that any string of $5$ letters can be divided into at most $2$ palindromes. Let's look at all strings with $5$ letters that have $A$ as first letter (if the first letter is $B$ we get the same type of string by switching $A$ and $B$). We have $16$ possible strings:
$ABBBB= A+BBBB$
$ABBBA= ABBBA$
$ABBAB=ABBA+B$
$ABBAA=ABBA+A$
$ABABB=ABA+BB$
$ABABA=ABABA$
$ABAAB=A+BAAB$
$ABAAA=ABA+AA$
$AABBB=AA+BBB$
$AABBA=A+ABBA$
$AABAB=AA+BAB$
$AABAA=AABAA$
$AAABB=AAA+BB$
$AAABA=AA+ABA$
$AAAAB=AAAA+B$
$AAAAA=AAAAA$
We notice that all strings on the right side are palindromes and there are no more than $2$ strings on the right side. Therefore any string of $5$ letters can be partitioned into no more than $2$ palindromes.
We can partition the given word with $1995$ letters into $399$ strings each $5$ letters long. Each of those strings can be divided into no more than $2$ palindromes, so the whole word can be divided into at most $399*2=798<800$ palindromes. The result follows.[/hide]


I'm a high school student in India where the chemistry coursework is tough and challenging both in terms of No.of topics as well as the detail and rigour to which the topic is covered.Consequently, one needs to read advanced theory as well as practice tough problems (for IIT JEE).Unfortunately, there is an absolute dearth of such material available.I need your help here.Can you recommend to me some online resources where I can get advanced theory(especially in physical chem) and get tougher , advanced problems(similar to IChO but not the same problems - I already have those).I won't be able to get books so no point recommending them.

[b]So what do you think.Can you help me out?? [/b]:(

NOTE :- I'm refering to the kind of material available for IMO

[quote="pkothari13"]seeing as you're preparing for something like the IIT JEE, the rigor at which you are studying must be pretty intense.  i'm not sure of the exact material they cover, but try a search engine like http://www.google.com to look for specifics.

also, you might want to register at a site like http://www.chemicalforums.com/, because they cover a lottttt of chemistry from basic high school to god knows how much.  that site would be much more helpful than this in terms of chemistry. let me know if you still can't find what you're looking for  :)[/quote]

[size=200][color=red][u][i][b]Still searching but haven't found anything as yet!!!!!!!![/b][/i][/u][/color][/size]

[quote="Rushil"][size=200][color=red][u][i][b]Still searching but haven't found anything as yet!!!!!!!![/b][/i][/u][/color][/size][/quote]
[size=200][color=red][u][i][b]IF THERE ARE MORE WE WOULD HAVE TOLD YOU, NOW STOP BEING HYSTERICAL AND YELLING AROUND!!!!!!!!!!![/b][/i][/u][/color][/size]

Rushil, did you register at chemicalforums.com?? their different forums have pretty tough problems.  also, you can try asking this same question there to see if you get more responses.

[hide][u][i][b][color=red] [size=200]I LIKE BIG RED WRITING [/size] [/color][/b][/i][/u][/hide]


[hide]I'm not sure...but I did it and got 2. [/hide]

I still am not getting[hide]2[/hide]. I got [hide]1[/hide]
See edited post.

I recommend you write down every step if you are not comfortable with recursion.

[hide="LONG STEP BY STEP THIS CAN BE BORING"]Question: Find $f(1)$

Analysis of problem:

Since $1\le5$, $f(1)=1-f(f(4))$

Since $4\le5$, $f(4)=4-f(f(7))$

Since $7>5$, $f(7)=7-2=5$

$f(4)=4-f(f(7))=4-f(5)$

Since $5\le5$, $f(5)=5-f(f(8))$

Since $8>5$, $f(8)=8-2=6$

Since $6>5$, $f(6)=6-2=4$

So $f(5)=5-f(f(8))=5-f(6)=5-4=1$

So $f(4)=4-f(5)=4-1=3$

So $f(1)=1-f(f(4))=1-f(3)$

But, $f(3)=3-f(f(6))=3-f(4)=3-3=0$[/hide]

... and the answer is obvious at this point :)

Let me see what I did...hm....
[hide]$f(1)=1-f(f(1+3))$ which is $1-f(f(4))$ which is $1-f(4-f(f(7))$ which is $1-f(4-f(5))$ which is $1-f(4-5-f(f(8))$ which is $1-f(4-5-f(6))$which is $1-f(4-5-f(4))$ which is....oh well, I see what I did wrong.[/hide]

[hide="We all have different answers"]0,1,2,-4[/hide]Which is correct?????

I just plugged in the x=1 and got 1, but thats pretty cool. You can also graph this function on a Ti-84 plus.

[quote="ch1n353ch3s54a1l"]The function $f(x)$ is called a nested function, and is defined as: $f(x)=\begin{cases} x-2 \text{ for } x>5,\\ x-f(f(x+3)) \text { for } x\leq5\end{cases}.$
Find $f(1)$.[/quote]
The key to this question is organization.  Another approach would be to make a table.  Begin with a few of the functions you can solve in 1 step.  Then proceed to the case you really need to solve.

$\begin{array}{ccc} x&x-2&f(x) \\ 6&4&4 \\ 7&5&5 \\ 8&6&6 \\ \end{array}$

$\begin{array}{cccccl} x&x+3&t=f(x+3)&f(t)&f(x)=x-f(t)&\fbox{Remarks} \\ 1&4&&&&\fbox{Stuck. We need }f(4)\fbox{ to continue}\\ 4&7&5&&&\fbox{Stuck. We need }f(5)\fbox{ to continue} \\ 5&8&6&4&1&\fbox{Go back to row }x=4 \\ \\ 4&7&5&1&3&\fbox{Go back to row }x=1 \\ \\ 1&4&3&&&\fbox{Stuck again. We need }f(3) \\ 3&6&4&3&0&\fbox{Go back to row }x=1 \\ \\ 1&4&3&0&\bd{1}&\fbox{Done! }f(1) = 1 \\ \end{array}$


As far as I understand, there are several mexicans here. So, I have a few questions:
1. How much does it cost to phone from Mexico to some distant country via your local mobile operator (and, plz, name operator you're talking about)?
2. How much does 'Sambrero' and tequila cost? :) And are there any restrictions for exporting them from Mexico?
3. If anyone knows: do they plan to confiscate our (participants') mobs like in Greece, or any other sanctions?

tnx

I have some questions too.
1. When can we access Internet? How much will we have to pay for it?
2. What's the weather condition there? Is it around 30 degree?

count with me for a football match
and siuhochung, i think the weather is 30-35 degrees but rainy

[quote="Feuerbach"]
1. How much does it cost to phone from Mexico to some distant country via your local mobile operator (and, plz, name operator you're talking about)?
2. How much does 'Sambrero' and tequila cost? :) And are there any restrictions for exporting them from Mexico?

tnx[/quote]
1.- It must be arround 10 pesos per minute (1 u.s.) It depends on which country are you calling. USA and Canada are the cheapest and Asia and Oceania the most expensives.
2.- The Tequila cost from 150 pesos (14 u.s.) A sombrero must cost 50 pesos ($4.5 u.s.) but i´m not really sure. I supose you can export any of them without problem (if you don´t want to take out 300 tequilas)
[quote="siuhochung"]1. When can we access Internet? How much will we have to pay for it?
2. What's the weather condition there? Is it around 30 degree?[/quote]
1.- I supose we will. Internet usually cost form 5 to 30 pesos (.5 - 2.5 u.s.) depending on the place, the speed, and the city (yes, the city, México city is cheaper than León)
2.- The weather is tropical. Must be between 30 and 42 degree (celcius)

[quote="lightrhee"]Another question. Can you use US dollar at Mexico? :?[/quote]
Usually, you can at most of the stores, and banks, mostly at tourist cities (as Mérida) however, it will be good that you carry some pesos but you surely can live using just dollars.

Albanian team, against Portugal Team :rotfl:

I have to warn you that I have played with my national team of my age.
:weightlift: :weightlift: :weightlift: :weightlift: :weightlift: :weightlift:

[quote="Dave Vather"][quote="Albanian Eagle"]i hope it will be easy to go to the beach and come back i think its nice there!
is any one interested in a [b]team[/b] soccer match?[/quote]
Maybe we could play too. We could make a World Cup in the IMO[/quote]

Yeah! its a great idea.
I think up to now there are three countries "registered".
It would be a lot of fun.

that would be interesting
 :D :ddr:  :gathering:  :winner_first:  :winner_second:  :winner_third:  :nhl:  :surf:  :juggle:

[quote="orl"][quote="Dave Vather"][quote="Albanian Eagle"]i hope it will be easy to go to the beach and come back i think its nice there!
is any one interested in a [b]team[/b] soccer match?[/quote]
Maybe we could play too. We could make a World Cup in the IMO[/quote]

You know Germany defeated Mexico 4:3!  :D[/quote]
So?

[quote="Dave Vather"][/quote][quote="orl"][quote="Dave Vather"][quote="Albanian Eagle"]i hope it will be easy to go to the beach and come back i think its nice there!
is any one interested in a [b]team[/b] soccer match?[/quote]
Maybe we could play too. We could make a World Cup in the IMO[/quote]

You know Germany defeated Mexico 4:3!  :D[/quote][quote="Dave Vather"]
So?[/quote]

Yes.

i was hoping Mexico would win cause they played so well in the group phase and Borgetti scored to amazing goals.

[quote="ondrob"]And what about ultimate frisbee? Slovak members will bring at least one disc so I hope there will be enough people for at least one match in ultimate...[/quote]I'll join you guys after the coordination is over :D Hopefully you won't hate me :P

[quote="mahbub"]i am very sorry for my poor english.

well, what do you mean by tropical weather? Does it rain?[/quote]
Yes, it does

[quote="Armo"]Hey, guys we will play football too. And what about basketball? ;)[/quote]

Yeah, of course.

Does any f the mexicans here know if the countries have been determined the Hotels that they will stay?

[quote="son of Skanderbeg"][quote="Armo"]Hey, guys we will play football too. And what about basketball? ;)[/quote]

Yeah, of course.
[/quote]
Ok, see there. :)  :)  :)

I would like to greet everyone on this Forum, and thank for these informations. It will be pleasure to meet you all in Mexico.

BIH 8 (Bosnia and Herzegovina deputy leader and former contestant)

[quote="Tasar"]I just saw that not everybody sleeps in the same hotels. Does anybody know how far the Fiesta Americana, Holiday Inn, Conquestador and Hyatt Regency are apart?[/quote]
Conquestador, Hyatt and Holiday Inn are each in front of the others (at a corner) And the fiesta americana is one block away (i think)


function
abstract algebra
superior algebra
superior algebra solved

1. Let (G,*) be a group with the property that there exists an injective function f:G->G,s.t. f(x*f(x)) = x^2*f(x*y) for every x,y \in G. Proove that (G,*) is an abelian group.

I think there's some mistake becasue we don't even need f to be injective: take x=e and we get f(f(e))=f(y) for all y in G, so f is constant, f=k. From the initial relation we get k=kx2, so x2=e for all x in G, and it's well known that in this case G is abelian.

I'm sure this wasn't the problem.

Taking into account the level of this contest, I do not wonder if this was the problem.

I guess you're right, but I still think there's a problem because if we also take into consideration the fact that f is injective, from f(f(e))=f(y) we get y=f(e) fo all y in G, so G={e} ! I really think this isn't what they meant.

sorry
it was my mistake :(
the relation is:
f(x*f(y)) = x^2*f(y*y).
:(

Take x=e and get f(f(y))=f(y 2 ) so f(y)=y 2
Write the initial relation (xy 2 ) 2 =x 2 (y 2 ) 2  which gives xy 2 =y 2 x for all x and y, that is y 2 commutes for all y.
Now write x(yx) 2 =(xy) 2 x=x(xy) 2  use the fact that y 2 is injective and it is done :)


Να βρεθούν όλα τα $x \in Z$ ,για τα οποία ο αριθμός
$x^5+x-1$ είναι τέλειο τετράγωνο ακεραίου.

_________________

Λοιπόν παραγοντοποιώντας (σαν Θαλή 2005 Α Λυκείου ;) )
έχουμε ότι  $x^5+x-1=(x^2-x+1)(x^3+x^2-1)$
Άρα αρκεί να λύσουμε την $(x^2-x+1)(x^3+x^2-1)=m^2$
Είναι επίσης γνωστό ότι $(x^2-x+1,x^3+x^2-1)=1$ ή
$(x^2-x+1,x^3+x^2-1)=7$
Για μέγιτο κοινό διαιρέτη=1 <<εκγλωβίζουμε>>  τα τετράγωνα και τελικά προκύπτει μια  λύση η  $(x,m)=(1,1)$  και η
$(x,m)=(1,-1)$ .
Στην δεύτερη περίπτωση παίρνουμε mod 8 και καταλήγουμε σε άτοπο άρα μόνες λύσεις οι παραπάνω

HΙ Stedes!!Ωραια ασκηση,που την βρηκες?Την ελυσα αλλα βαριεμαι τωρα να γραφω την λυση.Θα στην πω απο κοντα....αντε τα λεμε αυριο....

ΡΑΛΛΗΣ :D

πιο απλα προσθαφαιρουμε το χ κυβος και παραγωντοποιουμε.  και χρησιμοποιουμε την ασκηση οτι ολοι οι πρωτοι διαιρετες του χ^2+1 ειναι της μορφης 4κ+1.Στελιο αν εχεις τη λυση πεστην αν μπορεις. :)

Καλή λυση Σιλουαν!Κωστα θα κοιταξω οταν βρω χρονο και την δικη σου λυση.Ειναι αρκετα ενδιαφερουσα.Εγω βρηκα την ιδια λυση με αυτη του Σιλουαν.Stedes εισαι ενας απο αυτους που ηθελα να εμπλουτισουν το φορουμ με διαφορες ασκησεις και το πετυχες πιστευω.Συνεχίστε ετσι παιδια! :)

για να μην το αφηνουμε ετσι, η ασκηση που χρησιμοποιω ειναι αμεση εφαρμογη του θεωρηματος του Wilson.( p-1)!=-1modp, p=πρωτος.

ναι αφου ισχυει οτι [(ρ-1/2)!]^2=-1modp για p=4κ+1 ετσι Κωστα?

σωστα  ! μαλιστα αν ειναι 1(modp) a p=4k+3. πολυ χρησιμα (πχ ΒΜΟ 98 4ο θ.) αλλα χρειαζεται καποια εξοικειωση...

Πολύ ωραίες οι λύσεις σας και διδακτικές !

Γράφω κι εγώ τη δικιά μου:

Ψάχνουμε τα $y \in \mathbb N$ ώστε να ισχύει $y^2+1=x(x^4+1)$ ,άρα $x>0$ ,δηλαδή $x\geq 1$ ,οπότε $y^2+1\geq x^4+1 \Rightarrow y\geq x^2$.
Άρα $y=x^2+a$ για κάποιο $a \in \mathbb N$ και αντικαθιστώντας στην αρχική προκύπτει $a^2+2ax^2-(x-1)(x^4+1)=0$ με διακρίνουσα $4[x^4-(x-1)(x^4+1)]$ η οποία πρέπει να είναι θετική ,όμως για $x\geq 2$ δεν είναι
αφού $(x-1)(x^4+1)\geq x^4+1>x^4$ .Συνεπώς μοναδική τιμή για το $x$ είναι το $1$ για το οποίο ισχύει το ζητούμενο και βρίσκουμε $y=1$.

Stelios...


algebra
function
domain
real analysis
calculus
calculus computations

okay, the domain is $ \mathbb R,$ hence is continuous on $ \mathbb R.$ On any compact set $ f$ will be uniformly continuous but not on the full set of $ \mathbb R.$ I was wondering how to prove that $ f$ is not uniformly continuous.

The definition of $ f$ being uniformly continuous on a set $ S:$

For all $ \epsilon>0,$ there exists $ \delta>0$ such that for all $ x,y\in S$ such that $ |x\minus{}y|<\delta,$ then $ |f(x)\minus{}f(y)|<\epsilon.$

Now let's express this as the contrapositive:

Suppose that for some $ \epsilon>0$ and for all $ \delta>0$ we can find $ x,y\in S$ such that $ |x\minus{}y|<\delta$ and $ |f(x)\minus{}f(y)|\ge\epsilon.$ Then $ f$ is not uniformly continuous on $ S.$

Take a look at $ f(x)\equal{}\sin(x^2).$ You can find lots of points at which $ f(x)\equal{}1$ and lots of points at which $ f(x)\equal{}\minus{}1.$ Now, how close together can you make such points?

don't know where you're getting at with that question, but thanks for the help anyway.

of course if you can help me a bit more that'd be great!

Given any $ \delta>0$ can you find an $ x$ and a $ y$ with $ |x\minus{}y|<\delta,$ $ \sin(x^2)\equal{}1,$ and $ \sin(y^2)\equal{}\minus{}1?$


calculus
integration
parameterization
calculus computations

Calculate: I= ∫Fdr    , F=2xi+2zj-yk over the ark AB when A(0,0,0) and B(1,1,1)
                 AB

a) If AB belongs to the curve  r=ui+u²j+ u^3k
b) If AB belongs to the straight line connecting A and B

[/hide]

Your curve is already parametrized in terms of $ u$, where $ 0\leq u\leq 1$. So get everything in terms of $ u$ and compute the single variable integral.

$ F\equal{}\langle 2u, 2u^3, \minus{}u^2\rangle$ and $ dr\equal{}\langle du, 2udu, 3u^2du\rangle$, so
$ F\cdot dr\equal{}2udu\plus{}4u^4du\minus{}3u^4du\equal{}(2u\plus{}u^4)du$

Thus, $ \int F\cdot dr\equal{}\int_0^1(2u\plus{}u^4)du$.

For (b) try to find a parametrization of a straight line from (0,0,0) to (1,1,1) and proceed as in (a).

A parametrization of a straight line from (0,0,0) to (1,1,1) can be found by this
r(t)=(1-t)(0,0,0)+t(1,1,1) , 0<t<1
r(t)= (t,t,t)
x=t, y=t, z=t
Correct?

Yes. Quick check: It starts and ends in the right place, and it is definitely linear.


Factor the polynomial completely:

$ p(x)\equal{}x^5 \minus{} 3x^4 \minus{} 5x^3 \plus{} 15x^2 \plus{} 4x \minus{} 12$.

i guess you could use the conjugate root theorem. im not sure how to test for imaginary roots, but i know that if you find one, $ a\plus{}bi$, then $ a\minus{}bi$ is also a root.

does that help in any way? :maybe:

[hide="RRT FTW"]
1 is a root.  Factor it out first.
$ x^4\minus{}2x^3\minus{}7x^2\plus{}8x\plus{}12$
2 is a root.  Factor out this next.
$ x^3\minus{}7x\minus{}6$
By Descartes' rule there is one remaining positive root.
It turns out it's 3.
$ x^2\plus{}3x\plus{}2$
Factor and get the other two roots, $ \minus{}2$ and $ \minus{}1$.

End result:
$ p(x)\equal{}(x\minus{}1)(x\minus{}3)(x\minus{}2)(x\plus{}1)(x\plus{}2)$
[/hide]

@vis

No, these are the problems from my local contest that I could not solve.


Can anyone help with explaining WHY this partial diff. eq is linear? (as said by the textbook)

u_xx + 3u_yy = x^2(y) - u_x

u_x and u_y are partial derivatives.

I thought the term (x^2)y makes this equation 2nd order nonlinear?

Thanx.

Yeah, but I thought the equation was concerned both variables x and y? Can you explain your response?

What's the definition your textbook gives of a "linear" PDE?

Isn't a 2nd order linear PDE a PDE that has the form:

$A(x,y)u_{xx} + B(x,y)u_{xy} + C(x,y)u_{yy} + D(x,y)u_x + E(x,y)u_y + F(x,y)u= G(x,y)$

You may think that for a "linear" PDE, the sum of any two solutions must also be a solution.  However, that is only true for [i]homogeneous[/i] linear PDEs, as I understand it.  It's just a matter of how you define a "linear" PDE.


a,b, and c are the roots of Polynomial $x^3+x-1=0$ find

$\frac{1+a}{1-a} + \frac{1+b}{1-b} + \frac{1+c}{1-c}$

Just repeatedly transform the polynomial, and then consider the right coefficient.

To expand a little:
Let $x=\frac{1+y}{1-y}$
Then the equation becomes
$\frac{1+y}{1-y}^3+\frac{1+y}{1-y}-1$
$(1+y)^3+(1+y)(1-y)^2-(1-y)^3$
You want to find the first and second coefficients of this polynomial.
So doing the algebra, the first = 3 and the second = -1
Therefore the sum = -1/3

I'm asking about the numerical value of
$\frac{1+a}{1-a}+ \frac{1+b}{1-b} + \frac{1+c}{1-c}$
 :lol:
the answer is $5$
$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
so using this to the given polynomial we have
$a+b+c=0$ , $ab+ac+bc=1$ , $abc=1$ and we also have
$a^3=1-a$, $b^3=1-b$, $c^3=1-c$
and then just do some basic algebra
$\frac{1+a}{1-a}+ \frac{1+b}{1-b} + \frac{1+c}{1-c}=3+\frac{2(a+b+c)-4(ab+ac+bc)+6abc}{(abc)^3} $
then subs the values above yield $3+2=5$


integration
real analysis
real analysis unsolved

[quote="fotia"]here's a question that caught my eye while flipping though my mathematics analysis book.

Given that F(x)=$
\begin{document}
\displaystyle
\mathop \int_0^{x}f(s)ds$ and $F(\infty)=1,$ prove that for any $a\geq1$
$
\begin{document}
\displaystyle
\mathop\int_0^{\infty}x^{a}dF(x) = a \int_0^{\infty}x^{a-1}(1-F(x))dx
\end{document}$, where the two sides either converge or diverge together.

this was posted on another forum, i decided to make it a topic since no one has responded yet, please check it out[/quote]

 Have you tried integrating by parts?


function
quadratics
algebra
polynomial
geometry
3D geometry
special factorizations

How do you find the power series for a function which can't be factored, say $ \frac{1}{x^2 \plus{} x \plus{}1}$?
Do I use completing the squares so it would turn out to be $ \frac{1}{(x\plus{}\frac{1}{2})^2 \plus{} \frac{3}{4}}$ and rewrite it into the form $ \frac{4}{3}(\frac{1}{1\minus{}(\minus{}\frac{4}{3})(x\plus{}\frac{1}{2})^2})$ so it would be parallel to $ \frac{1}{1\minus{}x}$? thanks!

What about $ \frac{1}{1\plus{}x(1\plus{}x)}$  :?:

it is the same you  find it's complex roots

The first function can be rewritten $ \frac{1 \minus{} x}{1 \minus{} x^3} \equal{} (1 \minus{} x)(1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} ...)$, but this isn't a technique that works most of the time.

The generic technique that you want is still partial fraction decomposition, only the roots are complex.  That is, write $ \frac{1}{x^2 \plus{} x \plus{} 1} \equal{} \frac{1}{(x \minus{} \omega)(x \minus{} \omega^2)}$ where $ \omega \equal{} e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity and work from here.

[quote="t0rajir0u"]The first function can be rewritten $ \frac {1 \minus{} x}{1 \minus{} x^3} \equal{} (1 \minus{} x)(1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} ...)$, but this isn't a technique that works most of the time.

The generic technique that you want is still partial fraction decomposition, only the roots are complex.  That is, write $ \frac {1}{x^2 \plus{} x \plus{} 1} \equal{} \frac {1}{(x \minus{} \omega)(x \minus{} \omega^2)}$ where $ \omega \equal{} e^{\frac {2 \pi i}{3} }$ is a primitive third root of unity and work from here.[/quote]

won't completing the square work? what if it's say, $ \frac{1}{x^2 \plus{} 5x \plus{} 1}$? how would you deal with that then?

Do you know how to find the complex roots of the quadratic equation $ x^2 \plus{} 5x \plus{} 1 \equal{} 0$?  ;)

The problem with completing the square is that the power series you get won't generally be centered around $ 0$ (which I assume is what you're trying to do).  For example, (and I'll change your numbers to make the calculation a little easier)

$ \frac{1}{x^2 \minus{} 2x \plus{} 2}\equal{} \frac{1}{1 \plus{} (x \minus{} 1)^2} \equal{} 1 \minus{} (x \minus{} 1)^2 \plus{} (x \minus{} 1)^4 \minus{} (x \minus{} 1)^6 \pm ...$

which is a power series centered around $ 1$ instead, and "re-centering" is no fun.

You can just use long division to get the first few terms, and to see if there's a pattern.

Divide $ 1\plus{}x\plus{}x^2$ (you want to write the terms in that order) into $ 1$ and we get

$ 1\minus{}x\plus{}x^3\minus{}x^4\plus{}x^6\minus{}x^7\plus{}\cdots$

In this case, the pattern is obvious, and it shouldn't be too hard to prove that the pattern works.

The following is essentially equivalent to long division: assume that

$ (1\plus{}x\plus{}x^2)(a_0\plus{}a_1x\plus{}a_2x^2\plus{}a_3x^3\plus{}\cdots)\equal{}1.$

Expanding this out, we get that $ a_0\equal{}1,a_1\plus{}a_0\equal{}0,$ and after that, $ a_n\plus{}a_{n\minus{}1}\plus{}a_{n\minus{}2}\equal{}0.$ This gives is a straightforward two-step recursion for the coefficients.

In this particular case, the fact that the roots of the polynomial are cube roots of $ 1$ does lead to the simple repeating pattern of the coefficients.

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

To air the idea a little further, try this one (a classic): find the power series centered at zero for $ \frac{x}{1\minus{}x\minus{}x^2}.$


Find the linear function f(x) satisfying f(-10)= 6 and the point ( 4 , -7 ) is on the graph of the inverse of f.

[quote="lordWings"][hide]You know two points of the inverse function, which is linear too. Then invert it back[/hide][/quote]

[hide]Actually you know $f$ contains (-10,6) and $f^{-1}$ contains (4,-7) $\Rightarrow$ $f$ contains (-7,4)[/hide]

so is there a formula to use to get the equation??

let's say f(x) is ax + b because it's a linear function. Then -10a + b = 6 and -7a + b = 4, and solving it we get (a,b) = (-2/3, -2/3). So the function is -2x/3 - 2/3

wait.. how do u solve that...? -10a+b=6??


$ a,b \in R$ How do you solve $ ctg(ax\minus{}b)\equal{} \frac{1}{x}\minus{}\frac{x}{4}$?

I assume $ ctg$ means cotangent ...  I'll write $ \cot$ since I am used to that.

Does this help?... $ \cot(2\theta) \equal{} \frac {\cot\theta}{2} \minus{} \frac {1}{2\cot\theta}$

hello, this equation can only be solved by an numerical way. Are the real numbers $ a$ and $ b$ given?
Sonnhard.

Thanks for the answear. No I was trying to find a genral solution.


induction
number theory solved
number theory

here is an old problem from germany82.
**
suppose that (4^n+2^n+1) is a prime number.
show that n must be a power of 3.
**

I prove that if n<>0 [mod 3],(4^n+2^n+1) is not a prime number
but i have some problem to finish the problem.


induction
floor function
linear algebra
matrix
linear algebra unsolved

prove that the maximum number of mutually commuting linearly independent complex matrices of order $ n$ is $ [\frac {n^2}{4}] \plus{} 1$

You can proceed by induction on $ n$, but the solution I have is not obvious and quite long.
The easy bit is to see that this max is at least $ \lfloor n^2/4 \rfloor \plus{}1$, just consider $ aI_n \plus{}
 \begin{pmatrix}
 0 & A\\
 0& 0 \\
 \end{pmatrix}$ with $ A$ of size $ \lfloor n/2 \rfloor \times \left( n \minus{} \lfloor n/2 \rfloor  \right)$.


superior algebra
superior algebra unsolved

maybe classical:
prove that $GL_3(\frac{Z}{2Z})$ is simple.

This is a special case of the general theorem that $PSL_n(F)$ is simple when $n\ge3$ or when $F$ has at least 5 elements. When $F=\mathbb{Z}/2\mathbb{Z}$, everything invertible has determinant 1 and there are no constant multiples to remove.

Any algebra text should have a proof of this theorem; these are the other accessible examples of nonabelian simple groups.

[quote="jmerry"]This is a special case of the general theorem that $PSL_n(F)$ is simple when $n\ge3$ or when $F$ has at least 5 elements. [/quote]

What about that "at least 5 elements" part?

shouldn't that be F must have four elements?

he needs $PSL(3,2)$ for his proof because that is the same als $GL(3,2)$

the group $PSL(2,2)$ is just $S_{3}$ and $PSL(2,3)$ is just $S_{4}$


set theory
advanced fields
advanced fields unsolved

The text I'm reading is Robert Goldblatt, "Lectures on the Hyperreals: An Introduction to Nonstandard Analysis" (which is GTM 188). I also have Applied Nonstandard Analysis by Martin Davis and Abraham Robinson's earlier book. But I think Goldblatt's goes deeper into nonstandard entities -- the others are more about applications of nonstandard analysis to standard questions.

Anyway, I'm starting to get lost in the unimaginably vast universes he creates.

"we now define a [i]universe[/i] to be any [i]strongly transitive[/i] set $\mathbb{U}$ such that
* if $a, b \in \mathbb{U}$, then $\{a, b\} \in \mathbb{U}$;
* if $A$ and $B$ are sets in $\mathbb{U}$, then $A\cup B \in \mathbb{U}$;
* if $A$ is a set in $\mathbb{U}$, then $P(A) \in \mathbb{U}$.
Such a $\mathbb{U}$ will  be called a [i]universe over[/i] $\mathbb{X}$ if $\mathbb{X}$ is a set that belongs to $\mathbb{U}$ ($\mathbb{X} \in \mathbb{U}$), and the members of $\mathbb{X}$ are regarded as [i]individuals[/i] that are not sets and have no members:
$(\forall x \in \mathbb{X}) [x \ne \emptyset \wedge (\forall y \in \mathbb{U}) (y \not\in x)]$."

$P(A)$ is the power set of $A$. Now, I know what each of these statements means. I get the impression that a "universe" is meant to be something absurdly big and contains just about everything I can think of. But here's where it starts blowing my mind:

"It is time to demonstrate that there are such things as universes. Let $\mathbb{X}$ be a set with $\mathbb{R} \subseteq \mathbb{X}$. The [i]nth cumulative power set [/i] $\mathbb{U}_n(\mathbb{X})$ of $\mathbb{X}$ is defined inductively by
$\mathbb{U}_0(\mathbb{X}) = \mathbb{X}$,
$\mathbb{U}_{n+1}(\mathbb{X}) = \mathbb{U}_n(\mathbb{X}) \cup P(\mathbb{U}_n(\mathbb{X}))$
so that
$\mathbb{U}_0(\mathbb{X}) \subseteq \mathbb{U}_1(\mathbb{X}) \subseteq ...$
The [i]superstructure[/i] over $\mathbb{X}$ is the union of all these cumulative power sets:
$\mathbb{U}(\mathbb{X}) = \bigcup_{n=0}^\infty \mathbb{U}_n(\mathbb{X})$"

What in the world is that thing? It boggles my mind. Then there is this idea of a universe embedding that I totally don't understand, and hyperfinite sets. A universe embedding is supposed to map a universe into another universe under certain conditions, so that the other universe is "larger" and contains nonstandard entities. Hyperfinite sets are hyperreal sets, almost all of whose standard sets (that make up the sequence) are finite. I don't really get why hyperfinite sets are important, and their properties are weird. I'm sorry to have such a vague "question," but I'm really getting so lost that I don't know what to ask. Can someone please try to give a bit of intuitive explanation of these basic parts of the theory? (If it is possible to have intuitive ideas about these monsters.)

Have you done any set theory?  This seems to be a somewhat typical way of defining the objects one works with.  The point is just that the superstructure includes everything you could reasonably want to include.  Everything has to behave relatively well because you constructed it to do so.  For instance, you can be sure that if you have an object in the universe, its power set is in the universe by the nature of the construction.

Roughly, the idea of all this is that you want things like universes, that have these "strongly transitive" properties.  But, if you just give me a list of axioms for the behavior of an object, there is no guarantee that such an object exists within the realm of set theory.  (You might be able, for instance, to prove results about fields with 6 elements -- but since no such fields exist, you will have wasted a lot of time.)  So the latter part is giving a construction of an object which will have the desired properties, showing that such objects exist.  You should verify for his superstructure that it actually does satisfy the axioms he lays out for a universe.

For instance, if $a, b \in \bf U$, we know by construction that there are integers $i, j$ s.t. $a \in {\bf U}_i$ and $b \in {\bf U}_j$.  But then by construction, $a, b \in {\bf U}_{\max\{i, j\}}$.  Then $\{a, b\} \subset {\bf U}_{\max\{i, j\}}$ and so $\{a, b\} \in {\bf U}_{\max\{i, j\} + 1}$ so finally $\{a, b\} \in {\bf U}$.

Ok, I think I get that part. I think my problem was I was trying to think of $\mathbb{U}$ actually was, and it's so large that there's no point. I may as well just think of it as "pretty much everything."

He does verify most of the properties and gives the rest as exercises.

But the things that confuse me are really the nonstandard ones. Let's not worry about hyperfinite sets for now, and just look at the universe embedding:

He defines a mapping between two universes, which is pretty obvious and easy to understand. Each element in $\mathbb{U}$ is mapped to an element of $\mathbb{U}'$. Sure. Then he says "A [i]nonstandard framework[/i] for a set $\mathbb{X}$ comprises a universe $\mathbb{U}$ over $\mathbb{X}$ and a map $\mathbb{U} \rightarrow^* \mathbb{U}'$ satisfying:
$^*a = a$ for all $a \in \mathbb{X}$.
$*\emptyset = \emptyset$.
[b]Transfer:[/b] an $L_\mathbb{U}$-sentence $\varphi$ is true if and only if $*\varphi$ is true. "

So it's not at all clear to me why this causes nonstandard elements. I mean, why can't $\mathbb{U}'$ just be the same as $\mathbb{U}$? Also, how do we guarantee with our mapping that these properties will hold? He seems to define these mappings and not explain how we find them.

Are we just supposed to be thinking of the standard construction of the hyperreals this whole time and verify that it is such a mapping? (It obviously is, actually.) But shouldn't these concepts be more general than some particular construction of a field that includes infinitesimals?

Ok, sorry, here is some more about the nonstandard entities. They don't just arise from the embedding, you need a bit more.

$^*A \in \mathbb{U}'$ is what a set $A \in \mathbb{U}$ gets transfered to.
Let's define $^\text{im}A$ as $\{a^* : a \in A\}$, that is, the standard elements of $^*A$. The stuff that is just copies of what was in $A$ originally. So there might be other stuff in $^*A$ if we have the right conditions.

Now, here's how we ensure getting nonstandard elements in our new universe:

"$\mathbb{U}'$ is called an [i]enlargement[/i] of $\mathbb{U}$ if there exists a universe embedding from $\mathbb{U}$ to $\mathbb{U}'$ such that:
[i]if $A \in \mathbb{U}$ is a collection of sets with the finite intersection property, then there exists an element $b \in \mathbb{U}'$ that belongs to $*B$ for every $B \in A$, i.e.,
$b \in \bigcup\{^*B : B \in A\} = \bigcup\ ^\text{im}A.$" [/i]

Ok, sure. I can see why this will create nonstandard entities. However, it is no longer obvious to me that the construction of the hyperreals is an embedding of the universe over $\mathbb{R}$ that satisfies all the properties! Also, this property is a little weird to me. I mean, isn't this basically just a definition of having nonstandard elements? How do we know such embeddings exist? Maybe he proves it later and I just haven't gotten there yet...

Also, no, I've never had a course in set theory. All I know I've been picking up as I go in this book, and it's starting to get really hard. Maybe freshmen should leave GTMs alone....  :stretcher:


If $ a \otimes b \equal{} \displaystyle{\frac{a\plus{}b}{a\minus{}b}}$ , then what does $ (6 \otimes 4) \otimes 3$ equal?

We have
\[ (6\otimes 4)\otimes 3 \Rightarrow\]

\[ (\frac {6 \plus{} 4}{6 \minus{} 4})\otimes 3 \Rightarrow\]

\[ (\frac {10}{2})\otimes 3 \Rightarrow\]

\[ 5\otimes 3 \Rightarrow\]

\[ \frac {5 \plus{} 3}{5 \minus{} 3} \Rightarrow\]

\[ \frac {8}{2} \Rightarrow\]

\[ \boxed{2}\]

All correct, except that $ \frac{8}{2}\equal{}\boxed{4}$, not 2.


there is another neat way to solve this problem..

I'm sure the administer posted this problem to solve it in different ways..

If u have noticed,

2y + 4y + 6y + 8y - y - 3y - 5y - 7y = 20.

each numbers have pairs like, (2y - y)(4y - 3y)(6y - 5y)(8y - 7y)

so they actually equals like this

y + y + y + y = 4y

which the solution would be 20 / 4 = 5

Hmm, I think shinwoo's solution is the best. It makes the problem a lot easier.

[hide]I used Shinwoo's method of solving this problem. I saw that all the negative y's were always 1 smaller than the positive y's, which means when added together, the answer is y each time. There are 4 pairs, so the answer is 20/4, which is y = 5.
(2y-y) + (4y-3y) + (6y-5y) + (8y-7y) = 20
4y = 20
y = 5[/hide]

[hide]y=5
pairing the terms like,2y-y+4y-3y and so on would make this proplem simpler to n terms[/hide]

Devendra Patel, and gr8sk8r, we are expected to write full solutions along with your answer

[quote="rrusczyk"]Find y:

2y + 4y + 6y + 8y - y - 3y - 5y - 7y = 20.[/quote]

There are several ways to group the terms on the left hand side using the associative property of addition. I prefer:

(2y + 8y) + (4y + 6y) + (-3y - 7y) +(-y - 5y) = 20

10y + 10y - 10y - 6y = 20

10y - 6y = 20

4y = 20

4y/4 = 20/4

y = 5.

notice that 2,4,6,8 are each 1 greater than 1,3,5,7 respectively
y+y+y+y=20
y=5


MATHCOUNTS logarithms AMC USA(J)MO USAMO percent probability

Find the sum of the 4536 numbers from 1,000 to 10,000 which have all their digits distinct.

If you know how to solve these, don't post them.

If you don't know how to solve, here's a hint:

Find the average of the numbers.

You should post all of these under one thread.

You could also find how many times each digit is used.

[quote="AIME15"][quote="USAMO13"]I dont get it..[/quote]

I thought you are supposedly extremely smart, and much smarter than nikeballa96, right?

Another method is to add all of the numbers up by hand, which [b]may[/b] take awhile...:P[/quote]

I bet nike couldnt do that!

[hide="solution"] Since 10,000 clearly does not work, this problem is just asking for the sum of all four digit numbers with all distinct digits.

Now consider one of these numbers, ABCD.

Each digit 1 through 9 appears 9*8*7=504 times as A (9 choices for B, 8 for C, 7 for D).
Each of those digits appears 8*8*7=448 times each as B, C, D (8 choices for A, then 8*7 for the other two letters).

Zeroes clearly do not count towards the sum, so our sum is
$ (1\plus{}2\plus{}...\plus{}9)(1000)(504)\plus{}(1\plus{}2\plus{}...\plus{}9)(100)(448)\plus{}(1\plus{}2\plus{}...\plus{}9)(10)(448)\plus{}(1\plus{}2\plus{}...\plus{}9)(1)(448)$
which is some big number I don't want to compute.[/hide]

[hide="Averages"]
The first digit can be $ 1\le k\le9$. If this is the case, the expected value for each of the last three digits is $ \frac{(0\plus{}1\plus{}2\plus{}\cdots\plus{}9)\minus{}k}{10\minus{}1}\equal{}\frac{45\minus{}k}9$, and the expected value of the four-digit number is $ 1000k\plus{}111\cdot\frac{45\minus{}k}9$. Since both $ f(k)\equal{}k$ and $ g(k)\equal{}1000k\plus{}111\cdot\frac{45\minus{}k}9$ are linear and the expected value of $ f(k)$ is $ f(k)\equal{}k\equal{}5$, the expected value of $ g(k)$ is $ 1000(5)\plus{}111\cdot\frac{45\minus{}5}9\equal{}5000\plus{}\frac{4440}9$. Now we multiply this by $ 4536\equal{}9\cdot9\cdot8\cdot7$ to get $ \boxed{24917760}$, equal to xpmath's expression.
[/hide]

[quote="math154"][hide="Averages"]
The first digit can be $ 1\le k\le9$. If this is the case, the expected value for each of the last three digits is $ \frac {(0 \plus{} 1 \plus{} 2 \plus{} \cdots \plus{} 9) \minus{} k}{10 \minus{} 1} \equal{} \frac {45 \minus{} k}9$, and the expected value of the four-digit number is $ 1000k \plus{} 111\cdot\frac {45 \minus{} k}9$. Since both $ f(k) \equal{} k$ and $ g(k) \equal{} 1000k \plus{} 111\cdot\frac {45 \minus{} k}9$ are linear and the expected value of $ f(k)$ is $ f(k) \equal{} k \equal{} 5$, the expected value of $ g(k)$ is $ 1000(5) \plus{} 111\cdot\frac {45 \minus{} 5}9 \equal{} 5000 \plus{} \frac {4440}9$. Now we multiply this by $ 4536 \equal{} 9\cdot9\cdot8\cdot7$ to get $ \boxed{24917760}$, equal to xpmath's expression.
[/hide][/quote]

bleh. beaten while typing =[.

just deleted the whole thing. usamo13, if you're gonna bet that i can't do something, at least make it hard...

you should probably put all these probs in one thread...

what are the odds of these problems appearing in nats btw?

[quote="algebra2000"]

what are the odds of these problems appearing in nats btw?[/quote]

zero. this problem is pretty heavily computational, and the other one is too hard for mc (logarithms are not in MC)

EDIT: USAMO13, did you make USAMO?

I could see it being on a State or National Team Round though.  Of course, since the problem is online, and has been discussed on a prominent mathematics forum, it now has a 0 percent chance of appearing in Nats.


geometry
inequalities
algebra
polynomial
inequalities unsolved

In any triangle ABC  show that
[tex] \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1 \geq4rR(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})  [/tex]

That's another nice inequality equivalent to [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/6[/url]:

Since the area S of triangle ABC equals $\displaystyle S=\frac{abc}{4R}$ one one hand at $\displaystyle S=\frac{1}{2}\left( a+b+c\right) r$ on the other, we get $\displaystyle \frac{abc}{4R}=\frac{1}{2}\left( a+b+c\right) r$, so that $\displaystyle \frac{2abc}{a+b+c}=4Rr$. Hence, your inequality

$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\geq 4Rr\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) $

can be rewritten as

$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\geq \frac{2abc}{a+b+c}\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) $.

But

$\displaystyle \left( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\right) -\frac{2abc}{a+b+c}\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) $
$\displaystyle =\frac{a^{3}c-a^{2}c^{2}+b^{3}a-b^{2}a^{2}+c^{3}b-c^{2}b^{2}}{\left( a+b+c\right) abc}=\frac{a^{2}c\left( a-c\right) +b^{2}a\left( b-a\right) +c^{2}b\left( c-b\right) }{\left( a+b+c\right) abc}$,

what is $\geq 0$, since $a^{2}c\left( a-c\right) +b^{2}a\left( b-a\right) +c^{2}b\left( c-b\right) \geq 0$ by the [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/6 inequality[/url] (applied to the numbers a, c, b instead of a, b, c). Thus, we have

$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\geq \frac{2abc}{a+b+c}\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) $,

and your inequality is proven.

Now I see Siuhochung (a. k. a. Soarer) has posted the same proof, but he is wrong when he says (in the last line of his proof): "I think the last inequality is true because (3,1,0) majorizes (2,2,0).". In fact, his polynomials are not symmetric and hence he cannot apply Muirhead. Instead he can say that the inequality a^2c(a-c)+b^2a(b-a)+c^2b(c-b)>=0 is true by [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/6[/url].

  Darij

Thanks darij. I'm not familiar with muirhead at all..  :blush:


limit
trigonometry
real analysis
real analysis unsolved

$ a_n \equal{} cos(\sqrt{2}n)$

$ \lim_{n \to \infty}sup\, a_n \equal{} ? \lim_{n \to \infty}inf\, a_n \equal{} ?$

[hide="Answer?"]I think the answers are 1 and -1, we can choose n such that $ \sqrt{2}n$ very close to $ \pi k$ and $ 2\pi k$, but I couldn't come up with a proof[/hide]

Recall that $ \limsup_{n\rightarrow\infty}(x_{n})\equal{}\inf_{n}y_{n}$, where $ y_{n}\equal{}\sup\{x_{n},x_{n\plus{}1},x_{n\plus{}2},\ldots\}$.
Writing out some terms might help you see the pattern:
$ a_{1}\equal{}\cos\sqrt{2}$, $ a_{2}\equal{}\cos{2\sqrt{2}}$, $ a_{3}\equal{}\cos{3\sqrt{2}}$,...  Then $ \inf_{n}\{\sup\{\cos\sqrt{2},\cos{2\sqrt{2}},\cos{3\sqrt{2}}\ldots\}\}\equal{}?$  Remember that cosine oscillates between -1 and 1.

Hmm, in my solution I assume $ cos(n\sqrt {2})$ is very close to 1 or -1 for some large n.
In your reasoning, that is,  [quote="JRav"]$ \inf_{n}\{\sup\{\cos\sqrt {2},\cos{2\sqrt {2}},\cos{3\sqrt {2}}\ldots\}\} \equal{} ?$[/quote],
I make the same assumption that for some large n  $ cos(n\sqrt {2})$ approaches 1, and so inf(sup(...)) = 1.

Still nothing changed in my mind.

by the way, I couldn't see a pattern in $ a_n$, is it periodic?

Or if you tried to explain anything else, can u expand it more, please?  :maybe:

As you take $ n\rightarrow\infty$, the sequence just continues to oscillate between -1 and 1-there is no limit.  No matter how many of the first few terms you shave off ("few" being a very arbitrary term), the supremum will always be 1 and the infimum will always be -1 because you can just keep going and you know you'll find one.  So the limit superior is 1, and the limit superior is -1, as you said.

Let $ B\equal{}1\equal{}\limsup$.  Then if that is indeed true, $ \forall\varepsilon>0$, $ \exists N\geq 1$ such that $ x_{n}<1\plus{}\varepsilon$ $ \forall n\geq N$-or only a finite number of elements in the sequence exceed the limit superior and a little extra.  So is this true?  You decide.


[quote="Orion"]How much would the Mathcount Series help me (I got in the top 30 at nats)?[/quote]

If you got in the top 30 at Nationals last year, you should be spending the bulk of this year working on harder math than MATHCOUNTS.  I would recommend working on AMCs and AIMEs most of the time, with occasional MATHCOUNTS work as the contests approach in order to hone your speed.  You know enough math for MATHCOUNTS - time to start working on harder material...

How did you get top 30 in nats if you live in Damascus?

Because it's impossible for a city in the US to be named Damascus

in Oregon, we have a city called Damascus...

There's one in Maryland too. It's not a rare name.

really? it seems pretty common, for some random reason... not many smallish capitals make as names of many cities...


floor function
ceiling function
number theory unsolved
number theory

Hi all,

Perhaps many of you know the series of problems below, but I decided to post them as I think they've not been discussed so far on this forum.

[b]Definition. [/b] The [i]Somos $k$-sequence[/i] $\{a_i\}_{i \geq 0}$ is defined by the recurrence relation
\[
a_na_{n+k} = a_{n+1}a_{n+k-1}  + a_{n+2}a_{n+k-2} + \cdots + a_{n+ \lfloor k/2 \rfloor}a_{n + \lceil  k/2 \rceil}
\]
and the initial conditions $a_0 = a_1 = \cdots = a_{k-1} = 1$.

[b]Problem 1. [/b] [i](good (and challenging) practice for olympiads)[/i] Show that all terms of the $4$-Somos sequence are integers [i]in two ways[/i]: (i) by a combinatorial interpretation of the $a_n$ (that is, construct a certain combinatorial configuration in which the number of some objects, which must be an integer, satisfies the same recurrence relation and takes the same initial values); and (ii) by a more direct, algebraic argument.

[b]Problem 2. [/b] [i](extremely challenging; not recommended to high school students :) )[/i] Show that all terms of the Somos $5$-sequence, Somos $6$-sequence, and Somos $7$-sequence are integers.

[b]Problem 3. [/b] [i](open problem)[/i] Show by a combinatorial argument that the Somos $5$-sequence consists only of integers.

[b]Problem 4. [/b] [i](open problem, I think)[/i] Find all $k$ for which all terms of the Somos $k$-sequence are integers.

--Vesselin

The problem for Somos 4-sequence was posted on forum, but no solution was proposed.

Actually, the 4-sequence problem was used in a [url=http://www.mathlinks.ro/Forum/contest/2nd%20Edition/2nd%20edition%20-%20round%202.pdf]Mathlinks Contest[/url], with its solutions [url=http://www.mathlinks.ro/Forum/contest/2nd%20Edition/2nd%20edition%20-%20round%202%20solutions.pdf]here[/url].

Unless there is something especially simple about the 4-Somos sequence
above, it is far beyond "good and challenging" to interpret it combinatorially.
I think Jim Propp and his collaborators were working on such problems for
several years before the breakthrough.

Addendum: The somos-7 has a non-integer term...


A jailer gives these 25 prisoners a chance to leave by playing his game. The prisoners have a night to discuss. The next morning, they will be locked in 25 separate rooms without a window so that none of them knows how many days he has lived in there. Every day the jailer takes one prisoner randomly out of his room to visit a room which has 2 lights with 2 switchs inside. Each switch can only be turned on or turned off and at the beginning the switchs are off and off. Once the prisoner is taken to this room, he has to turn exactly 1 switch from on to off or from off to on. He cannot make no turn and he cannot turn both. When the prisoner who is taken to that room thinks that everyone has been taken to this room at least once, he tells the jailer. If it is correct, they will be all released, but if it is wrong, they will be killed.

What is the plan?

[quote="laughinghead505"][hide="bad plan"] They all have watches on, or they all count seconds.

Told you it was bad :D [/hide][/quote]Even though they know the time, they don't know which day they should tell the jailer because it is random.

[quote="besttate"][hide]Elect a counter.

If a non-counter enters the room and switch A is on, they flip switch B. If switch A is off, and they have not yet flipped it, they flip it to on. If it is off and they have already flipped it, they flip switch B.

When the counter enters the room, if switch A is off, he flips switch B. If switch A is on, he is on, he flips it off, and adds one to his count. When he reaches 24, he alerts the warden.[/hide][/quote]Right!!! :lol:

By the way, what is the expected number of days it will take the prisoners to become free? ;)

I think we cannot know because it is random.

If the question is changed a bit to we don't know what the positions of the switches at the beginning, how should they do?

[hide]Once again, elect a counter.  Do the same plan as last time, except tell each prisoner to flip switch A the first opportunity they have AND the second one. When the counter gets to 47, he alerts the warden? I think this works.[/hide]

[quote="OHO"]I think we cannot know because it is random.[/quote]
Sure we can, expected values describe the behaviour of random.

But the jailer can choose the same prisoner everyday???

What does expected value mean? How small the number of days can be?

Nice solutions guys. I never would have got that.  :wink:

Expected value is the weighted average of all of the possible outcomes, weighted by probability. For example, the expected value of a dice roll is 3.5


Given that  $S=\{a,b,c,d,e,f,g,h,i\}$ is a set of nine elememts.
$A_1=\{a,b,c\}$, $A_2=\{d,e,f\}$ and $A_3=\{g,h,i\}$ are subsets of $S$.
And $F: S \times S \to S$ is a function satisyfing

(1) $F(A_m \times A_n)= S$ for all $m,n \in \{1,2,3 \}$,
(2) $F(\{r \}\times S)=F(S \times \{s\})=S$ for all $r,s \in S$,
(3)
$F(a,a)=F(b,h)=F(e,b)=F(g,c)=F(i,i)=a$,
$F(c,e)=F(d,b)=F(i,d)=b$,
$F(d,f)=F(f,h)=c$,
$F(a,b)=F(d,g)=F(i,h)=d$,
$F(e,d)=F(h,b)=e$,
$F(a,e)=F(e,f)=F(g,h)=f$,
$F(e,h)=F(f,d)=F(i,e)=g$,
$F(a,f)=F(c,g)=F(f,c)=h$, and
$F(a,g)=F(c,b)=F(g,f)=F(i,c)=i$.

Find the value of $F(F(g,i),F(i,g))$.

Well, I didn't visit AoPS for such a long while.
I miss the old days.
It seems that this problem is very difficult but in fact if I tell you the truth about this problem, you can solve it easily.
THE TRUTH IS THAT: It is only a disguise of a SUDOKU! :P


linear algebra
matrix
algebra
system of equations

let us conisder some  m equations which have m unknowns(or variables),

if we want to solve it by matrices(by ax=0),where a is the coefficient matrix and x is the no of unknowns and 0 is zero matrix)

for a solution other than trivial solution why is the condition that determinant of coefficient matrix(here a) to be zero?

can anybody prove?

If $ A$ is an invertible matrix or equivalently $ \det(A)\neq0$, then $ Ax\equal{}0$ implies $ A^{\minus{}1}(Ax)\equal{}0$, ie $ x\equal{}0$. Hence the system of equations has the trivial solution only.

A textbook I've used offers this under the name "The Invertible Matrix Theorem":

[quote="David Lay"]If $ A$ is an $ n\times n$ matrix, then the following are equivalent:

a) $ A$ is an invertible matrix.

b) $ A$ is row equivalent to the $ n\times n$ identity matrix.

c) $ A$ has $ n$ pivot positions. (Note: this actually refers to a row echelon matrix that is row equivalent to $ A.$)

d) The equation $ Ax\equal{}0$ has only the trivial solution.

e) The columns of $ A$ form a linearly independent set.

f) The linear transformation $ x\mapsto Ax$ is one-to-one.

g) The equation $ Ax\equal{}b$ has at least one solution for each $ b$ in $ \mathbb{R}^n.$

h) The columns of $ A$ span $ \mathbb{R}^n.$

i) The linear transformation $ x\mapsto Ax$ maps $ \mathbb{R}^n$ onto $ \mathbb{R}^n.$

j) There is an $ n\times n$ matrix $ C$ such that $ CA\equal{}I$

k) There is an $ n\times n$ matrix $ D$ such that $ AD\equal{}I.$

l) $ A^T$ is an invertible matrix.[/quote]

That theorem is proved before the author even defines determinant. Once he does introduce determinant, then he adds the theorem "A square matrix $ A$ is invertible if and only if $ \det A\ne 0.$"

All parts of this are steeped in the ideas of row reduction and Gaussian elimination, including the final connection to determinants (since we understand the effect of elementary row operations on determinants.) And although the textbook is assuming that the matrices in question are real and the multipart theorem refers to $ \mathbb{R}^n$ in several places, there's nothing in any of the proofs that wouldn't work over any field.

bkunchanapalli wants to connect item d) of this theorem to all of the other parts. There are many possible paths; one place to start would be to show that d) is equivalent to e).

ohh does ax=0 form has only a trivial solution?

according to the captain as identity matrix is not zero x should be zero matrix?is that you mean captain?

and can anyone tell me or give any link  the derivation how rank of a matrix is useful in finding solutions?

how the system is considered consistent if rank of augmented matrix = rank of  the coefficient matrix in ax=b.

i want to known how we determine whether matrix is consistent with rank,what is the theory behind it

did newton work on matrices :huh: ?just asking

anyway thanks captain and kent

bkunchanapilli: I think what you need is a textbook. This is standard material from the early chapters of many textbooks, and the most important computational method is Gaussian elimination. It wouldn't be efficient for me to try to recite such a textbook to you.


i need the step  by step of that equation... please if you can answer it now please do it... ill thank you about what you've done... ^^ because my teacher make things more complicated that's why i cant understand... thanks again...

I'm not sure how to answer this.

For the first one, you don't need the brackets: it's the same as

$ 3xy \plus{}2xyz \plus{} xy \minus{}4xyz$

$ \equal{}3xy \plus{} xy \plus{}2xyz \minus{}4xyz$

$ \equal{} 4xy \minus{}2xyz$.

It might help to think of $ xy$ as $ a$ and $ xyz$ as $ b$ so you have $ 3a \plus{} 2b \plus{}a \minus{}4b \equal{} 4a\minus{}2b$.

For the second one:

$ 3L \plus{} L^{2} \plus{}L/2 \plus{}30 \equal{} 41 \minus{} L$, (on expanding the brackets)

$ 6L \plus{} 2L^{2}\plus{}L \plus{}60 \equal{} 82 \minus{} 2L$, (on multiplying both sides by $ 2$)

$ 2L^{2}\plus{}9L\minus{}22\equal{}0$

$ L\equal{} \frac{\minus{}9\pm\sqrt{257}}{4}$ (on using the quadratic formula $ x\equal{} \frac{\minus{}b\pm\sqrt{b^{2}\minus{}4ac}}{2a}$).

I hope this is of some help to you.

oh... thank you very much... you helped me a lot... May God bless you always...

[quote="AndrewTom"]
For the second one:

$ 3L \plus{} L^{2} \plus{} L/2 \plus{} 30 \equal{} 41 \minus{} L$, (on expanding the brackets)

$ 6L \plus{} 2L^{2} \plus{} L \plus{} 60 \equal{} 82 \minus{} 2L$, (on multiplying both sides by $ 2$)

$ 2L^{2} \plus{} 9L \minus{} 22 \equal{} 0$

$ L \equal{} \frac { \minus{} 9\pm\sqrt {257}}{4}$ (on using the quadratic formula $ x \equal{} \frac { \minus{} b\pm\sqrt {b^{2} \minus{} 4ac}}{2a}$).

I hope this is of some help to you.[/quote]

what i mean in this thing is prove that $ 3L \plus{} L(L\plus{} 1/2) \plus{} 40$ is equals to $ 41 \minus{} L$... hehehe... sorry for incomplete description... thanks again...

Hi isuki,

As you've changed $ 30$ to $ 40$, $ 60$ becomes $ 80$ and the quadratic becomes

$ 2L^{2} \plus{} 9L \minus{} 2 \plus{} 0$

so that

$ L \equal{} \frac { \minus{} 9\pm\sqrt {97}}{4}$.

$ 3L \plus{} L(L \plus{} \frac {1}{2}) \plus{} 40$ is only equal to $ 41 \minus{} L$ for these two values as it is an equation and not an identity.

Hope this makes sense. If it doesn't, just ask again.

oh... so there's something wrong in the equation right? thanks for correcting me... ^^ you really helped me a lot...


algebra
polynomial
superior algebra
superior algebra unsolved

Let a a root of a third degree irreducible polynomial;b a root of a second degree irreducible polynomial.Polynomial are rational coefficients.How can we determine a primitive element for the extension field Q[a,b].The polynomial are given but not the roots.

Your question is a non sense:
$ \mathbb{Q}(a,b)$ depends on the choices of the root $ a$ and of the root $ b$.

Test small integers $ n$ till $ b \in \mathbb{Q}(a\plus{}nb)$.

1) We can use this result:
Let $ a$ be a root of $ P$ a polynomial of degree $ p$ and irreducible over $ \mathbb{Q}$.
Let $ b$ be a root of $ Q$ a polynomial of degree $ q$ and irreducible over $ \mathbb{Q}$.
If $ igcd(p,q)\equal{}1$ then $ \mathbb{Q}(a\plus{}b)\equal{}\mathbb{Q}(a,b)$.
I have forgotten the proof (it is difficult because the extensions are not necessarily Galois).
   If $ P,Q$ are given then there exist $ pq$ fields $ \mathbb{Q}(a\plus{}b)$ (not necessarily distinct) and these fields are isomorphic. Primitive elements $ a\plus{}b$ of these $ pq$ fields are solutions of a polynomial $ R$ of degree $ pq$ and irreducible over $ \mathbb{Q}$.
    Example:$ P(x)\equal{}x^3\plus{}x\plus{}1,Q(x)\equal{}x^4\plus{}x\plus{}1$; $ igcd(3,4)\equal{}1$.
Then, with Maple, $ R(c)\equal{}c^12\plus{}23\plus{}4*c^10\plus{}7*c^9\plus{}9*c^8\plus{}12*c^7\minus{}13*c^6\minus{}30*c^5\minus{}6*c^4\minus{}19*c^3\plus{}31*c^2\plus{}10*c$.
2) I have a question concerning the billwgd's post:
If $ igcd(p,q)>1$ can we find an integer $ n$ s.t. $ a\plus{}nb$ is a primitive element of $ \mathbb{Q}(a,b)$ ?


Hi

Could you help me with this one?

I need to prove that a graph which its vertices can't be colored in less than 7 colors, has at least 7 odd cycles (with different edges).

I was going on the direction that a graph like that must have a clique and than inside that clique you can find 7 triangles (sum to 21 edges) but I can't prove there's a clique.

Maybe the solution is in another direction

[quote="trioash1"]I was going on the direction that a graph like that must have a clique[/quote]

If you mean a clique with 7 elements, you're just wrong -- there exist graphs of arbitrarily high chromatic number which have no triangles, let alone $K_{7}$ subgraphs.


In triangle $ ABC$, $ AD \equal{} DE \equal{} BE \equal{} BC$, and $ m\angle ABC \equal{} 140^o$.  What is the degree measure of $ \angle C$?

[img]18586[/img]

Let $ x\equal{}m\angle C\equal{}m\angle BEC$ so that $ m\angle A\equal{}m\angle AED\equal{}180^\circ\minus{}140^\circ\minus{}x\equal{}40^\circ\minus{}x$. Thus, $ m\angle EBC\equal{}180^\circ\minus{}2x$ and $ m\angle BED\equal{}180^\circ\minus{}x\minus{}(40^\circ\minus{}x)\equal{}140^\circ\implies m\angle EBD\equal{}20^\circ$, and finally, $ m\angle B\equal{}140^\circ\equal{}m\angle EBC\plus{}m\angle EBD\equal{}200^\circ\minus{}2x\implies2x\equal{}60^\circ\implies x\equal{}\boxed{30^\circ}$.


number theory
relatively prime
number theory unsolved

Define $ f(n)$ as the sum of the positive integers less than $ n$ and prime to $ n$. Prove that $ f(m) \equal{} f(n)$ implies that $ m \equal{} n$.

Note that $ \gcd(k,n)\equal{} \gcd(n\minus{}k,n)$, so the positive integers less than $ n$ that are relatively prime to $ n$ come in pairs which add up to $ n$. The only number which could be paired up with itself, $ n/2$, is not relatively prime to $ n$ unless $ n\equal{}2$. Since there are $ \varphi(n)$ numbers less than $ n$ and relatively prime to $ n$,

$ f(n) \equal{} \frac{1}{2} n \varphi(n)$

(incidentally, note that this formula happens to work for $ n\equal{}2$ as well). Obviously, $ f(n)\equal{}1$ only for $ n\equal{}2$, so we can ignore $ n\equal{}2$ from now on and focus on $ n>2$. If there are pairs $ (m,n)$ with $ m \neq n$ and $ m \varphi(m) \equal{} n \varphi(n)$, choose among them a pair with $ mn$ as small as possible. We will derive a contradiction.

Let $ p$ be the largest prime dividing $ m$, and write $ m \equal{} m' p^k$ with $ \gcd(m',p)\equal{}1$. Similarly let $ q$ be the largest prime dividing $ n$ and write $ n \equal{} n' q^l$ with $ \gcd(n',q)\equal{}1$. We have

$ m \varphi(m) \equal{} m' p^k \varphi(m') \varphi(p^k) \equal{} m' \varphi(m')p^{2k\minus{}1}(p\minus{}1)$

$ n \varphi(n) \equal{} n' q^l \varphi(n') \varphi(q^l) \equal{} n' \varphi(n')q^{2l\minus{}1}(q\minus{}1)$.

Suppose wlog that $ p \geq q$. Then $ p$ divides $ m\varphi(m)$ exactly $ 2k\minus{}1$ times, therefore it must divide $ n \varphi(n)$ also $ 2k\minus{}1$ times, which is only possible if $ p\equal{}q$ and $ k\equal{}l$ (because $ p$ cannot possibly divide any of $ n'$, $ \varphi(n')$ or $ q\minus{}1$). But then $ m' \varphi(m') \equal{} n' \varphi(n')$ and $ m' \neq n'$, contradicting the minimality of the choice of $ (m,n)$.


limit
logarithms
real analysis
real analysis unsolved

Consider a list of values non-negative values $ (\alpha_1, \alpha_2, \ldots \alpha_k)$ such that $ \prod_{j \equal{} 1}^k (\alpha_k \plus{} 1) \equal{} z$ for some $ z > 1$.

Let $ N_k$ be the minimum value of $ \prod_{j \equal{} 1}^k p_j ^{\alpha_j}$, where $ p_n$ is the nth prime, numbered from $ p_1 \equal{} 2$.

Is $ \lim_{k \to \infty} N_k > 1$ ?

[b]Edit:[/b] [i]Sorry, I was in log mode![/i]

Erm, what do you mean? It is obvious that $ N_k$ is a non-increasing sequence and each $ N_k\ge 1$... :?

Right, thanks! I was in log mode. I've revised the problem.

Does anyone have any ideas on how to approach this?

Sorry, I'm also confused: it's obvious that we have a nondecreasing sequence of integers and that each of them is actually at least 2.  More generally, though, $ N_k$ can't ever be smaller than the smallest integer with $ z$ divisors, and this is the limiting value -- were you trying to ask something else about it?

Why must they be at least 2?

Also, why can't it be smaller than the smallest integer with z divisors? The point is that $ \alpha_i$ are not restricted to integers to allow the value to be less.

$ p^\alpha\ge 2^\alpha\equal{}(e^\alpha)^{\log 2}\ge (1\plus{}\alpha)^{\log 2}$. Hence, $ \lim N_k\ge z^{\log 2}$.

Oh! Thanks  :D

Sorry for floundering around this thread -- thanks to everyone.


Consider the following electrochemical cell:

$ \ce{Ag_{(s)}} / \ce{AgCl_{(saturated)}} \parallel{} \ce{Ag \plus{} _{(aq)}} / \ce{Ag_{(s)}}$

Given that when $ [\ce{Ag \plus{} }] \equal{} 0.01 M$ when $ E_{cell} \equal{} 0.170 V$ deduce the solubility product of $ \ce{AgCl}$.

Data:

$ T \equal{} 298.15 K$
$ E^{\circ}_{\ce{Ag \plus{} }/\ce{Ag}} \equal{} 0.80V$

Use Nernst equation:
$ E\equal{}E^0 \minus{} \frac{RT}{nF} \log{\frac{[Ag^\plus{}_{saturated}]}{[Ag^\plus{}]}}$
Then find the saturated concentration Ag+ and then $ K_{sp} \equal{}[Ag^\plus{}]^{2}$.


I was wondering what's the difference between LaTeX and BBCode??

-Swapnil

Latex is one of the bbcodes on this forum. Just like [quote] and [url] are.


Show that $ n^7\plus{}7$ is never a perfect square.

In $ \mod 4$, we know that all squares are either $ 0$ or $ 1 \mod 4$ and we see that when $ n \equiv 0 , 1 , 2 ,$or$ 3 \mod 4$, the only case that gives a possibility of a perfect square is when $ n \equiv 1$. I narrowed it down but this is still incomplete.. Does it help?

when we have 0mod4^7 as always a perfect square but when we add 7 it becomes 3mod4 which can never be a perfect square. when we have -1mod4 or 3mod4 we have 2mod4 which is never a perfect square. I was only able to disprove these two out of 4 cases.

We prove that the equation $ n^7\plus{}7$=$ x^2$ doesn't have solutions. Write it as $ n^7\plus{}2^7$=$ x^2\plus{}11^2$. [hide]Work this equation by mod4 and by primes of the form $ 4k\minus{}1$.[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=396369#396369]USA TST 2008 #4[/url]


I can't find how to latex piecewise functions, where f(x) equals a different formula for different ranges of x. Someone give an example with the large { spanning the outputs.

Use the cases environment:
\[ f(x)\equal{}\begin{cases}
1, & x<1\,;
\\
2, & 1\le x\le 2\,;
\\
3, & 2<x\,.
\end{cases}\]


geometry
geometric transformation
reflection
geometry proposed

An isosceles triangle $ ABC$ with $ AB \equal{} AC$ is given on the plane. A variable circle $ (O)$ with center $ O$ on the line $ BC$ passes through $ A$ and does not touch either of the lines $ AB$ and $ AC$. Let $ M$ and $ N$ be the second points of intersection of $ (O)$ with lines $ AB$ and $ AC$, respectively. Find the locus of the orthocenter of triangle $ AMN$.

[quote=yetti]$ F \in (O)$ is a reflection of $ A$ in $ BC.$ $ AF$ bisects $ \angle MAN$ $ \Longrightarrow$ angles of kite $ OMFN$ are fixed. Vertex $ F$ is fixed, vertex $ O$ moves on line $ BC$ $ \Longrightarrow$ its diagonal intersection $ X$ moves on a line and $ FX \perp MN.$ $ Y, Z$ are feet of perpendiculars from $ F$ to $ AB, AC,$ $ YZ \parallel BC.$ When $ M \equiv A$ coincide, then $ X \equiv Z$ coincide and when $ N \equiv A$ coincide, then $ X \equiv Y$ coincide $ \Longrightarrow$ $ YZ$ is locus of $ X.$ $ YZ$ is fixed Simson line of $ \triangle AMN$ with fixed pole $ F.$ Its orthocenter $ H$ lies on fixed Steiner line parallel to $ YZ$ at double the distance from $ F.$[/quote]
This solution is incomplete, since we should consider the cases when circle $(O)$ is tangent to $AB$ or $AC$.


let f(x,y) be a distance between (x,y) and P=(a,b)
f(x,y) = sqrt[(x-a)^2 + (y-b)^2]  <--just the usual metric
Prove that grad f is a vector which belongs to a line joining (x,y) and P,
which is "looking away from point P" and which has length equal to 1, |gradf|=1.

Use the formula of gradient.

grad(f) = (df/dx, df/dy) These are partial derivatives.
Then find the normal. It is trivial.