Find the inverse of $\left(\begin{array}{cccrrr}5&3&-2\\3&-4&3\\1&6&-4\end{array}\right)$. :)

EDIT: Can someone notify me of a strategic method to evaluate the inverse of square matrices that have dimensions greater than 2x2? And by the way, there are no inverses for non-square matrices, correct?

Masoud Zargar

The matrix questions you have been posting might probably be better suited for one of the linear algebra forums.

Oh, I thought they could be in the intermediate forums.  Can a mod please move this to the Linear Algebra Forum? Thanks. in advance.

Masoud Zargar

They are suitable for Intermediate but I think that putting them in the Linear Algebra section would be better.  For one, you will probably get responses from more knowledgeable people for the most part.
[color=red][i]
Moved from Intermediate.[/i][/color]

Only square matrices can possibly have inverses. For a square matrix, having an inverse is equivalent to many other properties: the determinant is nonzero, the columns are independent, and so on.

A general, practical algorithm for finding the inverse of a matrix: (For more elaboration, see any linear algebra textbook or many "Finite Math" textbooks.)

Form an $n\times 2n$ matrix by placing the identity matrix beside the given matrix:

$[A|I]$

Now perform Guassian row elimination on this matrix. The objective is to reach a point where the left hand half turns into the identity matrix. That is, we proceed until we see this: $[I|A^{-1}].$ At that point, the right hand half is the desired inverse.

If we cannot complete the row elimination, it would be becuase we get a row of zeros across the bottom of the left half. If that happens, $A$ was not invertible.

This algorithm requires something like $\frac43n^3$ multiplication or division operations, and a similar number of additions. If you're doing this by hand, many of the addition operations are adding fractions. With practice, it's practical for hand calculations in the $3\times3$ and $4\times4$ cases, and can be used for somewhat larger matrices if those matrices start out with a reasonable number of zeros. Beyond that, you need computer or calculator help - but this (with some additional numerical concerns) forms the basis for computer algorithms for computing inverses.*

* An experienced numerical analyst may well tell you that you don't need to find inverses of matrices: finding $LU$ factorizations is good enough.

Would you please tell me about the $LU$ factorizations?  By the way, I am aware of the Gaussian elimination method you spoke of, however, I still thank you for your time. :)

A LU-factorization is a factorization $A=LU$ where $L$ is lower triangular and $U$ is upper triangular. It can be found by Guassian elimination - or at least you could get $A=LU$ if you never interchanged rows. Row interchages introduce permutations, so in general, you get $A=PLU$ where $P$ is a permutation matrix. This is still usually referred to as an LU factorization.

Taking your matrix and not trying to do anything clever, I get the following, at least if my arithmetic is any good. (You shouldn't trust that; check.)

$\begin{bmatrix}5&3&-2\\3&-4&3\\1&6&-4\end{bmatrix} = \begin{\bmatrix}1&0&0\\ \frac35&1&0\\ \frac15&-\frac{27}{29}&1\end{bmatrix} \begin{\bmatrix}5&3&-2\\ 0&-\frac{29}5&\frac{21}5\\0&0&\frac9{29}\end{bmatrix}$

The point: suppose you need to solve an equation $Ax=b$ for $x.$ If you have computed $A^{-1},$ then you can write the answer as $x=A^{-1}b.$ Perfoming the matrix multiplication $A^{-1}b$ will take $n^2$ multiplications and about that many additions.

If on the other hand, you knew that $A=LU,$ then you could proceed as follows. First solve $Ly=b$ for $y.$ You can do this by succesive subsitition: solve the equation of the first row for $y_1$ (by a single division), then substitute that in the second row to find $y_2,$ and so on down the matrix. Once you have found $y,$ then solve $Ux=y$ by back-substition. Start with the last row, solved for $x_n$ and proceed upwards.

The total number of arithmetic operations required to solve the two triangular equations by substitutions is exactly the same $n^2$ count for the number of operations to find $A^{-1}b.$ The advantage is that $LU$ can be found with fewer operations than it takes to find $A^{-1}.$


Find height h that dish climbs when applied a momentum M in axle.

wow, that is a good problem, I've never considered a problem like that before :). I'd have to think about this (as usual ;))

I don't exactly understand what the problem says and how are the things on the picture attached to each other. If M is angluar momentum, you also need the mass of the disc or something else, and the exact configuration if the strings is not too important. Shouldn't M be torque?

YA Torque might help and in the end the strings become useless because they slacken.

The interpretation I think is that even if we remove the strings, we have a disc on an axle and when we rotate it , it goes up like a screw. Is this right?

[quote="Rushil"]This may be stupid but is M the angular momentum or what?
Also I would like to confirm whether or not the 2 attached strings become slack when the disc rises.\
Am I interpreting the problem correctly[/quote]

M Is a angular momentum applied in axle

[quote="Rushil"]can you explain problem a bit more properly[/quote]

are two lines that keep hold the dish , and you applied a momentum and make dish goes up

it's ok to solve some eqns and get the answer. But what about physical understanding. Where does the force to lift the disc come from? Am i missing something?

Well, if I'm getting the problem, then:

- the disc is rotated and gets angular momentum M
- it rotates, but the strings cause it to rise up
- therefore it spends some of its energy
- when it stops in the topmost position, all its kinetic energy is transformed into potential

It's like giving linear momentum to a ball attached to a string.

But I still don't know the purpose of giving the length of the strings, except for imposing h<r ...

[quote="kubus"]OK. Angular momentum $M$, moment of inertia of the disc $I = \frac{1}{8}mr^2$, mass $m$. Then

$\omega = M/I$

Energy conservation law:

$\frac{1}{2} I \omega ^2 = mgh$
$h = \frac{I \omega ^2}{2mg}$
$h = \frac{M^2}{2Img}$
$h = \frac{4M^2}{m^2r^2g}$[/quote]

Use geometry

[quote="kubus"]Is my solution not correct? If I understand the problem (and it is possible that I don't!), my energy solution is perfectly OK. The only thing that should be appended is, that $h$ cannot be arbitrarily large, namely, it cannot be larger than $r$ due to the geometry of the strings.

Does "using geometry" give any better insight into the nature of the problem?[/quote]
1no, is wrong
2yes

Do you understood the problem?

[quote="kubus"]In that case, I am afraid I did not understand the problem. Probably the part about applying M. What exactly is M??[/quote]

A momentum applied in axle(Force)
This can help

Anyone agree or disagree of my solution?


Let sequence $ \{a_n\}$ such that $ a_{n\plus{}1}<a_n$ with all $ n$;$ a_n \in N$ with all $ n$
Suppose that sequence $ \{(a_{n\plus{}1}\minus{}a_n)\}$ is boundedness
Prove that set of prime $ p$ such that $ p|a_n$ is unlimited

This means exits real number $ M$ such that exits $ n_1 \in N$ sauch that with all $ n \geq n_1$ we have
$ |a_{n\plus{}1}\minus{}a_n| <M$

[quote="M3LF4H"]Let sequence $ \{a_n\}$ such that $ a_{n \plus{} 1} < a_n$ with all $ n$;$ a_n \in N$ with all $ n$
Suppose that sequence $ \{(a_{n \plus{} 1} \minus{} a_n)\}$ is boundedness
Prove that set of prime $ p$ such that $ p|a_n$ is unlimited[/quote]

If $ a_n \in \mathbb{N}$, how the sequence(strictly decreasing) can be bounded? :huh:

[quote="M3LF4H"]Let sequence $ \{a_n\}$ such that $ a_{n \plus{} 1} < a_n$ with all $ n$;$ a_n \in N$ with all $ n$
Suppose that sequence $ \{(a_{n \plus{} 1} \minus{} a_n)\}$ is boundedness
Prove that set of prime $ p$ such that $ p|a_n$ is unlimited[/quote]
you want say
i) $ N\equal{}\{1,2,.....\}$
ii) $ a_{n\plus{}1}>a_n$
iii) $ \{(a_{n \plus{} 1} \minus{} a_n)\}$ is boundedness

that mean that existe $ \{b_n\}$ sequence in $ \{1,2,....,M\}$ for some $ M$
and $ a_n\equal{}\sum_{i\equal{}1}^{n}b_i$

I believe the first statement was supposed to read $ a_{n\plus{}1} > a_n$.  Anyway, if $ |a_{n\plus{}1} \minus{} a_n|$ is bounded then $ \sum \frac{1}{a_n}$ diverges; on the other hand, if the set of prime divisors of $ a_n$ is finite, say $ p_1, p_2, ... p_k$, then

$ \sum \frac{1}{a_n} \le \prod_{i\equal{}1}^{k} \frac{1}{1 \minus{} \frac{1}{p_i}}$

which converges; contradiction.

[quote="M3LF4H"]Let sequence $ \{a_n\}$ such that $ a_{n \plus{} 1} < a_n$ with all $ n$;$ a_n \in N$ with all $ n$
Suppose that sequence $ \{(a_{n \plus{} 1} \minus{} a_n)\}$ is boundedness
Prove that set of prime $ p$ such that $ p|a_n$ is unlimited[/quote]
[hide="Solution"]
Assume to the contrary and let the primes that divide the sequence be $ p_1$, $ p_2$, $ \cdots$, $ p_k$; let the difference between any two consecutive terms be at most $ m$.  Consider $ f(n) \equal{} \prod_{i \equal{} 0}^k {p_i^n}$ for any integer $ n > m$.  Notice that, for suffiently large $ n$, we can force the first $ x$ $ a_i$'s to be in the form of $ \prod_{i \equal{} 0}^k {p_i^{\alpha_i}}\bmod f(n)$ for any integer $ x$.  Furthermore, there are $ n^k$ ways to pick these numbers, so the first $ x$ numbers have $ n^k$ possibilities.  Clearly, $ x > \frac {f(n)}{m} \minus{} 1 \minus{} a_1$ since the difference between any two numbers is at most $ m$, and $ x \le n^k$.  However, this means that $ n^k > \frac {f(n)}{m} \minus{} 1 \minus{} a_1$ for all $ n$, which reaches a contradiction for big enough $ n$ as the Left Hand Side is a polynomial and the Right Hand Side is exponential.  Hence, infinitely many primes divide the sequence.  [/hide]


I had some things (clothes, water bottles, etc) exposed to sodium azide due to airbags going off in my car, is it safe to still use these things or ?  Im not familiar with the stuff, so initially my thought would be 'yes', but http://www.bt.cdc.gov/agent/sodiumazide/basics/facts.asp is a bit scary.

[quote="http://www.bt.cdc.gov/agent/sodiumazide/basics/facts.asp"]http://www.bt.cdc.gov/agent/sodiumazide/basics/facts.asp[/quote]

Sounds like you should be fine.  Unless your airbag was torn open I wouldn't worry about it too much.  You could probably call the local health department and ask just to be sure.


[quote="hien"]let a, b, c >0 so that abc = $\frac{5}{4}$
find minimum value of A = $\sqrt[4]{a^4+b^4+\frac{b^2}{c}}+\sqrt[4]{b^4+c^4+\frac{c^2}{a}}+\sqrt[4]{c^4+a^4+\frac{a^2}{b}}$[/quote]
Very easy:
Use Mincopxki:
$A\ge\sqrt[4]{2(a+b+c)^4+(\sqrt[4]{\frac{b^2}{c}}+\sqrt[4]{\frac{c^2}{a}}+\sqrt[4]{\frac{a^2}{b}})^4}$
Then use AM-GM  :D

Or just simply Cauchy it with correspondent factors.  :)


function
integration
real analysis
real analysis unsolved

that function in itself is not possible to express in terms of elementary functions, but it is   known

$ \int_0^\infty \; \frac{x^{3}}{e^{x} \minus{} 1}\;dx\equal{}\Gamma(4) \cdot\zeta(4)\equal{}\Gamma(4)\cdot  \text{Li} _ 4 (1)\equal{}\frac{\pi^4}{ 15}$

  you can probably only go as far as expressing it as follows:

$ \int \frac{x^{3}}{e^{x} \minus{} 1}\;dx\equal{}  6 \text{Li} _ 4 (e^x) \minus{}6x\; \text{Li} _3 (e^x) \plus{}3x^2 \text{Li}_2 (e^x)  \minus{}x^3\text{Li}_1 (e^x)  \minus{}\frac{1}{4}x^4$

where $ \text{Li} _ n (\theta) \equal{} \sum_{k\equal{}1}^\infty \; \frac{\theta^k}{k^n}$


Name Some that you like

I like the Geico commercials

the careerbuilder.com commercials with the monkeys at work were pretty good, but they've seem to have been left off the air now...

I think that most of the Gieco comercials are very stupid. I personally have a collection of nice comercials from around the world. I love one of Mikado, a japanease chocolate, and some others. I don't know if you remember the commercial of John West, a brand of canned food, where the guy fight with a bear, Great! Other pretty good is the one of the NHL from a few years ago ("Bowling would be better if were Hokey"), And some of pepsi vs coke in Latin America.

the one with the gecko doing the robot was pretty sweet

the WORST commercials: the "prevent taste loss" commercials. those drive me INSANE!!!

I like the geico commercials that actually have the gecko in them. I also like the aflak commercials as well as the one where the two little squirrels try to run the car off the road and then do the little hand thing. What's that advertise?

They're dumb, but funny..... and then you just see the commmercial and don't know what they advertise... or at least that's me.

THE COMMERCIAL I ABSOLUTELY CANNOT STAND IS THE PEPTO BISMOL ONE. UGGG...

[quote="elementsofmathkid"]THE COMMERCIAL I ABSOLUTELY CANNOT STAND IS THE PEPTO BISMOL ONE. UGGG...[/quote]

Nausea, Heartburn, Indegestion, Upset Stomach, DIARRHEA!!! lol.
love that one. it's soo fun. :)

also love the AFLAK commercials. lol.

:)  :)  :)

I remember when I was in 6th grade we watched this Voyage of the Mimi II video and they had I think it was Ben Afleck go into this pressure chamber or something at the end of the video. When he said Afleck it sounded like the duck on that commercial.  :)

I saw one on greatest commercials ever or something like that, and it was about some salmon company. They were advertising the freshness of their salmon and it showed these bears fishing for salmon in the stream, and this one guy coming up and trying to take the salmon. So then the bear and the guy fight, and the bear does soem martial arts on the guy. But they guy does the inevitable groin kick and the bear falls into the river and the guy takes the bear's salmon. hilarious.

You are all wrong  :P

The best commercials are those by Emerald Nuts

They are insanely hilarious.

Egyptian Navigators love Emerald Nuts!

yeah emerald nuts are pretty good
i liked a commercial (forgot what product it was pushing) that said somewhere in there "your smart! you dont need someone yelling "ORDER NOW! ORDER NOW! and also dont need it to come across the screen (which it does) and have sparkles appear (and sparkles appear) and bounce up and down (and the banner bounces up and down)

I found this[url=http://www.jengajam.com/r/Glen-Starbucks]Starbucks commercial[/url] pretty amusing

Sometimes the Enzyte commercials are funny (but always a little stupid)

I remember a commercial from the Super Bowl in 03. I think it was a Bud Light commercial with two guys with their dogs and one of the guys is shows off his dog by saying "Bud Light" and the dog runs in and gets a Bud Light. Then the other guy goes " Well Look at this, Max, Bud Light!" The dog then bites the other guy who is holding the bud light in the balls and the bud light falls in the other persons hand. It was my fav. commercial ever.

bud light makes the best ads, they are all super funny

i also like the godaddy.com one

trunkmonkey: especially the original and the theft retrieval
bud light: i havent seen too many of them, but i especially liked the alien wazzup and the guy with the dog trying to enter the no-pets bar

nothing tops the super bowl commercial with the cat that spills the spagetti on itself while the guy is holding a knife and his girlfriend walks in the door.

and the new bud light robot rocks as well as the toyota commercial about adrenalitus.

[url=http://video.luxology.com/article/113004_1/Citroen_C4.mov]Citroen C4 commercial[/url]
funny and cool

[quote="plokoon51"][quote="Scrambled"][quote="djimenez"]Hey, how could I forget... All the Capital One commercials of the last year (except may be the last two, of the new guy that always have to say no!)[/quote]
is capital one the one with an "instabranch"?[/quote]

no, thats for something else, washington mutual, or something like that???[/quote]
ok then i like washington mutual commercials

Have you guys seen the Pepsi Max "Don't worry, it's got no sugar" commercials?

There is this one when a guy catches a cab to go to the airport, the driver (in left) throws him a pepsi max can, and asks him where he wants to go. The guy says "airport" and the "driver" starts saying "tooo thhheeee aiiiirpoooort" a couple of times. The camera moves, and we see a monkey at the wheel on the other side (english car :D). The "driver" then says to the guy: don't worry, it's got no sugar. And the monkey starts driving with the guy screaming of fear in the back :D

[quote="plokoon51"][quote="elementsofmathkid"]Oh, and another insanely annoying commercial is the old Quizno's Subs commercials with the talking hamsters. I want to know what happened to them. It looks like they were mauled.  :P

We used to have this one kid who would play it on the computer in health class last year because we never did anything.....

Another senseless commercial... but they work...I've never been to Quizno's....[/quote]

speaking of Quizno commercials, the one with the talking baby is SO funny...[/quote]

I HATE that commercial! Its so "f word"ing annoying

Hey, we had forgotten the Raising Brand commercials, those on which Smith is Firing Johnson... they are very stupid, but any way a little funny  :D


algebra
polynomial
function
calculus
complex analysis
complex analysis unsolved

Show that the polynomial $z^{5}-6z+3$ has $5$ distinct complex roots, $3$ of which are real.

Seems trivial via a simple study of the function with High School tools. Did I miss any application of the argument's principle or something ?

By first-year calculus tools - on the real line:

$f(-\infty)=-\infty<0, f(0)=3>0,f(1)=-2<0,f(+\infty)=+\infty>0,$ so by the Intermediate Value Theorem, at least three real roots, in three different intervals.

$f'(x)=5x^4-6.$ We can find its roots explicity; exactly two of them, $\pm\sqrt{\frac65},$ are real. Hence by Rolle's Theorem, $f$ has at most three real roots. And we can't have any multiple real roots, because the zeros of $f'$ do not correspond to zeros of $f.$

So $f$ has exactly three real roots. By the Fundamental Theorem of Algebra, it has two more complex roots. Since it has real coefficients, they must form a conjugate pair - and thus be distinct from each other, and from the three real roots.

Did any of that involve complex analysis? Perhaps the quote of the Fundamental Theorem of Algebra. But  julien is right; it does use low-level tools.

[quote="Kent Merryfield"]Perhaps the quote of the Fundamental Theorem of Algebra.[/quote]

You don't need the Fundamental Theorem of Algebra in its strongest version. (hence no complex analysis) After finding three roots (or rather proving that they exist), the remaining polynomial is of degree $2$, whose roots can be found explicitely.


prove that there doesn't  exist positive integer x,y such that

                       x²+1=34y²

[hide]x^2 +1=34y^2, by modulo3, we get x multiple of 3 and by modulo 2 , x is odd.    .......... (2)
by mod17,  x^2  is  equiv -1                      .......(1)
hence, x can only leave remainder 3, 9, 15, [by (2)]  squaring ,none leaves -1 by mod 17

hence situation is impossible.
[/hide]
do remark as solution is quite short for a 'hard' problm.

[quote="analytic"]
hence, x can only leave remainder 3, 9, 15, [by (2)]  squaring ,none leaves -1 by mod 17
[/quote]
No. There is no contradiction by divisibility.
$ x\equal{}123$ - then $ x^2\plus{}1$ divisible by 17, $ x$ is odd and $ x$ divisible by 3.

[quote="mihai miculita"][b]HINT:[/b]
[b]I.[/b]
$ x^2 \plus{} 1 \equal{} 34y^2\Leftrightarrow x^2 \minus{} 9y^2 \equal{} 25y^2 \minus{} 1\Leftrightarrow(x \minus{} 3y).(x \plus{} 3y) \equal{} (5y \minus{} 1).(5y \plus{} 1)$
or
$ x^2 \plus{} 1 \equal{} 34y^2\Leftrightarrow x^2 \minus{} 25y^2 \equal{} 9y^2 \minus{} 1\Leftrightarrow(x \minus{} 5y).(x \plus{} 5y) \equal{} (3y \minus{} 1).(3y \plus{} 1).$
[b]II.[/b]
$ (5y \minus{} 1;5y \plus{} 1) \equal{} (5y \minus{} 1;2)\in\{1;2\}$
$ (3y \minus{} 1;3y \plus{} 1) \equal{} (3y \minus{} 1;2)\in\{1;2\}$
[b]III.[/b]
Case 1). If $ (5y \minus{} 1;5y \plus{} 1) \equal{} 1 \Rightarrow x \minus{} 3y \equal{} 5y \minus{} 1; \ x \plus{} 3y \equal{} 5y \plus{} 1\Rightarrow x \equal{} 5y\dots$
Case 2). If
$ (5y \minus{} 1;5y \plus{} 1) \equal{} 2 \Rightarrow 5y \minus{} 1 \equal{} 2h; 5y \plus{} 1 \equal{} 2k; (h;k) \equal{} 1 \Rightarrow \\
\Rightarrow 2 \equal{} (5y \plus{} 1) \minus{} (5y \minus{} 1) \equal{} 2.(k \minus{} h) \Rightarrow k \minus{} h \equal{} 1 \dots$[/quote]
I don't think it is right
Case 1). If $ (5y \minus{} 1;5y \plus{} 1) \equal{} 1 \Rightarrow x \minus{} 3y \equal{} 5y \minus{} 1; \ x \plus{} 3y \equal{} 5y \plus{} 1\Rightarrow x \equal{} 5y\dots$
if (x,y)=1 and (a,b) =1 and xy=ab
it doesn't follow that x=a and y=b
example
8.9=2.36
in case I


group theory
abstract algebra
linear algebra
matrix
calculus
integration
algebra

$P=x^n+...+a_0 \in \mathbb{Z}[X]$, irreducible. Let $a \in \mathbb{R}$ a root of $P$.
Is it true that any ideal of $\mathbb{Z}[a]$ is isomorphic to $\mathbb{Z}^n$

Yes, it's true.

First of all, we use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42888]this problem[/url] to show that every ideal, being a subgroup of $\mathbb Z[a]\cong \mathbb Z^n$ (group isomorphism), is isomorphic to $\mathbb Z^k$ for some $1\le k\le n$ (we work only with the non-trivial ideals, i.e. the null ideal does not appear here; it would correspond to $k=0$).

Let $u_1,u_2,\ldots,u_k$ be a basis for our ideal regarded as a group ($\mathbb Z$-module). We have, for all $i=\overline{1,k}$, $au_i=\sum_{j=1}^k\alpha_{ij}u_j$ for some integers $\alpha_{ij}$. Put the $au_i$'s on the RHS, and this becomes a homogenous linear system. It's matrix must have a null determinant, and if we compute this determinant we'll see that it's an integral monic polynomial computed in $a$, so it can't be $0$ unless $k=n$ (we cannot have $k<n$), and this is what we wanted to show.

Nice, grobber! I realize now that I've found the same solution, modulo the strange words you used ("module").  :) .


topology
vector
advanced fields
advanced fields unsolved

Prove that if $ S^k$ has a nonvanishing vector field, then its antipodal map is homotopic to the identity.

This is on p. 135 of Hatcher.

Sketch: let $ v(x)$ denote this vector field;
normalized so that $ |v(x)| \equal{} 1$. Then $ (cos t)x \plus{} (sin t)v(x)$ is nonzero for all $ t$, and so taking $ t$ from 0 to $ \pi$ gives a homotopy from $ x$ to $ \minus{}x$.


Sorry for my late reply, I somehow did not see it until now.

You'll have to be patient with me, since I am new to make loops.
I wished to plot points on the graphs at $ \dfrac{\pi}{6}$ intervals starting from $ \minus{}\dfrac{\pi}{2}$ (except for points that require division by zero of course).

How do I do this? Thanks!

I basically copied someone else's code for a similar graph, except with $ f(x)\equal{}x^2$   :blush:  and changed it by playing around to try to get it to work, since I couln't get it to work on my own, but I didn't really understand the code completely. Mostly, I don't know what the subinterval function is doing.

The loop you really need is
[code]
for(int i=-3;i<10;++i) if((i!=0)&&(i!=6)) dot(F(i*pi/6));
[/code]
Do you see how it works?

just out of curiosity, what exactly does the subinterval and box commands do? Why can't this work without those?

It can. In this particular program they do absolutely nothing useful. They were, probably, needed for something in the original code you modified to get yours (most likely, they were used for the graphic representation of Riemann sums there but I can only guess).


Using Second principle of finite induction, prove that

$ a^n \minus{}1\equal{}(a\minus{}1)((a^{n\minus{}1} \plus{} a^{n\minus{}2}\plus{}...\plus{}a\plus{}1)$.

Thank you.

What is the second principle of finite induction? (I googled it, but I didn't understand the explanations.)

Second principle of finite induction = strong induction, I believe.

[url=http://planetmath.org/encyclopedia/PrincipleOfFiniteInduction.html]This[/url] link states that the "second formulation" was also called strong induction... :maybe:

We only need induction to prove the statement. For our base case, we see that, for $ n\equal{}1$, $ a\minus{}1 \equal{} (a\minus{}1)(1)$.

Next, for the inductive step, we assume that $ a^{n}\minus{}1\equal{}(a\minus{}1)((a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{} \cdots \plus{}a\plus{}1)$. Clearly, $ (a\minus{}1)((a^{n}\plus{}a^{n\minus{}2}\plus{} \cdots \plus{}a\plus{}1)\equal{} (a\minus{}1)((a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{} \cdots \plus{}a\plus{}1) \plus{} (a\minus{}1)(a^n)$. Using the inductive hypothesis, we find that $ (a\minus{}1)((a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{} \cdots \plus{}a\plus{}1) \plus{} (a\minus{}1)(a^n) \equal{} a^n \minus{} 1 \plus{} (a\minus{}1)(a^n) \equal{} a^{n\plus{}1} \minus{} 1$, so our induction is complete.

That was quite nice of you.In  fact, seeing your solution, I realized that I was already using this second priniple of finite induction unwarily!

[hide]I had already done it like this, but was not sure whether this was the second principle of finite induction.[/hide]

But, this cleared my doubt.
Thanks.

Sorry, to clear up any misunderstanding that may exist, I did not use the second principle of finite induction (assuming that it is another name for strong induction).

I used only "weak" induction, where you assume that $ P(n)$ holds and prove $ P(n\plus{}1)$. I did not need to assume that the statement held for all $ P(x), 1 \le x \le n$ in order to prove $ P(n\plus{}1)$ (this is strong induction).  :)

All right.I was proved correctly,anyway.


I think it's been posted before... couldn't find it, though.

Solve $3^{x}+4^{y}=5^{z}$ in natural $x,y,z$.

If you want know solution  :wink:

http://www.kalva.demon.co.uk/short/soln/sh9117.html

So would that solution be a way to prove Pythagoras's theorem?
Or is it only used for just this question?

It has nothing to do with the Pythagorean Theorem, since the problem doesn't mention geometry anywhere.

This is  problem in Number Theory  :wink:

Of course, that brings up the interesting question of what happens for larger triples.  For example, can we find all integer solutions to

$8^{x}+15^{y}= 17^{z}$

?

Can we generalize to all primitive triples?


Basically right - we put everyone in a given class in a usergroup and give that group access to the class message board.


Can you cuss on FTW or the forum? By FTW I don't mean lobby (obviously not the lobby LANGUAGEFILTER), but I mean like in a answer or something? (sometimes i put idk or something). Also, what happens if you cuss on the forums? Is it a permanent ban or something? I tried that and got myself banned for 15 min...  :D

:roll: Is "don't be a moron" such a confusing principle that you need to hear the legalese version of it?  Try to be a good citizen and there won't be any problems.  The presence of a language filter should give you some hint about one of the things that good citizenship might entail.

[quote="akhil0422"]By FTW I don't mean lobby (obviously not the lobby LANGUAGEFILTER), but I mean like in a answer or something? (sometimes i put idk or something). [/quote]

Yeah, it's [i]possible[/i] to do that. Do you remember that guest who cursed at everyone through the answers? Additionally, some muted people talk through countdowns.


Rationalize the denominator: $ \frac1{\sqrt[4]{3}}$.

All we have to do is multiply the numerator and denominator by $ {\sqrt [4]{3^3}}$. Doing so yields $ \frac {\sqrt [4]{3^3}}{3}\equal{}\boxed{\frac{\sqrt [4]{27}}{3}}$. I think......


An item was placed on sale in January for $ 30\%$ less than its original price. A final close-out sale was offered in February, and the January sale price was reduced by $ 40\%$. What percent of the original price was the final reduced price?

We know that the price in January was $ 100\% \minus{} 30\% \equal{} 70\%$ of the original price.  The final reduced price is therefore $ 70\% \minus{} (40\%\times70\%) \equal{} 70\% \minus{} 28\% \equal{} \boxed{42}\%$ of the original price.

Well, the problem says what percent of the [b]original price[/b] which means $100\% - 42\% = 58\%$?

Please help.

Sorry, edited a lot because of poor $\LaTeX$...

:)

[hide="Solution"]
The initial price is 100%. Then, when it is reduced by 30%, it is 70%. 40% off 70% is equivalent to 60% of 70%, which is $\boxed{42}$%[/hide]


$\text{Sean }a,b,c,d\in \mathbb{R}^{+}\text{ demuestre que: }$

$\frac{bcd}{a}+\frac{acd}{b}+\frac{abd}{c}+\frac{abc}{d}\ge a\sqrt[3]{b^{2}c}+b\sqrt[3]{c^{2}d}+c\sqrt[3]{d^{2}a}+d\sqrt[3]{a^{2}b}$

Saludos  :lol:

Bueno pues la condición tal y como está lleva a que los numeros $a,b,c,d$ sean todos iguales a 1.
Por la condición de orden tenemos que:
$\\ da^{2}\ge cd^{2}\\ ab^{2}\ge cd^{2}\\ bc^{2}\ge cd^{2}\\$
Sumando todo obtenemos:
$da^{2}+ab^{2}+bc^{2}\ge 3cd^{2}$ pero la igualdad se debe dar siempre por lo que $a=b=c=d$. Además $3a^{3}=3$ por lo que tenemos que $a=1$. Entonces todos son 1 para que la condición se cumpla. :D

lo edite, ahora no necesariamente son todos 1  :D

Pues como lo veo ahora, se sigue cumpliendo que $a=b=c=d=1$ y la desigualdad es trivial.

$Tipe$

Tienes razon, creo que al ponerle muchas condiciones, fracaso el problema original; asi que ahora lo edite, y puse mi problema original.  :blush:


trigonometry
vector
calculus
geometry
Alcumus
3D geometry
sphere
\/closed

Art of Problem Solving is pleased to announce that our new [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=400][b]Precalculus[/b][/url] textbook has arrived!

[i]Precalculus[/i] covers trigonometry, complex numbers, vectors, and matrices. It includes nearly 1000 problems, ranging from routine exercises to extremely challenging problems drawn from major mathematics competitions such as the American Invitational Mathematics Exam and the USA Mathematical Olympiad. Almost half of the problems have full, detailed solutions in the text, and the rest have full solutions in the accompanying Solutions Manual.

The book is authored by Art of Problem Solving founder Richard Rusczyk.

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=400]Click here[/url] to learn more about the book and to order.

[quote="AIME15USAMO"]If we are signed up for precalc, we get "first dibs", correct?[/quote]

Yep; they're getting boxed up & shipped first.

I just wanna say... the cover looks awesome!!! I can't wait to see how awesome the book is too.

[quote="qwertythecucumber"]Can you tell how far away the Calculus book is? that's the one im determined to get.[/quote] We're hoping a couple months.

if i ordered the book today, will it be shipped this week or so?

ok, we just visited this topic, and i hate to ask, but what month would u say the calc book is gonna come out.  im testing out of calc ab ap, and i rly think that i need this book to help me score a 5 in may.  i think 2 months will be enough time for me to read it, so will the book be out by march? or earlier?

[quote="rrusczyk"][quote="$ LaTeX$"]Kinda off topic, but when is the Intermediate Geometry book coming out, and is there going to be an Intermediate Number Theory book?[/quote]

We don't have definite plans for either book.  Some of the material that was intended to go into the Intermediate Geometry book was included in the Precalculus book.[/quote]

I'm guessing that means that it won't be out for a while right?

Books ordered today will be mailed tomorrow.

Calculus book estimated ship date is February or March, 2010.

We don't have any estimated dates for other books right now.

This is not really about the precalc book, but I was wondering how you choose the designs and colors for the ball on the cover of each book.

What is covered in this book that isn't covered in a normal precalculus course?  My class last year did do all of the topics, but I don't really think I know my vectors, complex numbers or matrices very well at all...and we didn't learn stuff like $ e^{\pi\i}\minus{}1\equal{}0$

Yay the book is finally out!  :lol: I can't wait to get the book and have a nice Nerd Christmas.
Woah, the ball on the cover seems sharp compared to other AoPS covers. It is written in all books that:
[quote]Cover image designed by Vanessa Rusczyk using KaleidoTile software.[/quote]
I guess the designs are all about creativeness.

Cover design works like this: "vRusczyk, we need another cover!"  And then a cover magically appears.

In this book but not in a normal precalc book:
*Much harder and more interesting problems
*More focus: most precalc classes are grab-bags of different areas of math that weren't covered earlier (but should have been).  We have whole books for these areas of math rather than treating them as something extra to do for a couple weeks.
*More discussion of the relationship between complex numbers and trigonometry, including a couple sections specifically about roots of unity
*Applications of complex numbers and vectors to Euclidean geometry
*An approach to vectors and matrices that isn't all about memorizing methods, but rather focuses on what these mathematical objects mean, and why their operations work the way they do

Thanks for offering the book! :)

I had to take Geo Honors in school but I'm doing Alg 3 outside of school...

I've taken algebra 1/2 (same time), algebra 3, so this class should be called algebra 4 by induction. :D
Thanks, Mr. Rusczyk.

I also want it to be called Algebra 4, since it's a follow-up to the other Algebra ones. But it just can't, since it's not Algebra in particular :o It's pretty much alg/geo with trig/complex numbers.

wow! that was pretty dang fast! Between the few months (1 or 2) when I was inactive, you were just starting!

Are there any other books on the horizon after the Calculus one? Will there be an Olympiad series?

I'm sure AoPS doesn't have any plans for now, but it would be SO cool if there were Olympiad books in AoPS! Right now WOOT seems to be the only course(or book) available.

There should eventually be an advanced series... :lol: :)

We don't have any more advanced books on the immediate horizon; we are considering a brief book on mathematical writing.

Does anyone else have pages in their book that are faded out?   :|

Unfortunately, the printing process appears to have gone a little awry and left a few pages in the books somewhat faded relative to the rest of the book :(  Sorry about that!  Please let us know if they're so faded that you find them unreadable by emailing us at books@artofproblemsolving.com.

Several pages were faded out and they look somewhat awkward, but I think everything is readable and OK. :)

I only saw faded pages in chapter seven and eight.

There were about 12 - 13 total faded pages.

I wish I was good enough to do precalculus...  :(

How long does it take to work through a textbook? I only started doing math seriously this year (I didn't do math counts, math team in middle school nor actually studied anything in freshman year) and I have a ways to catch up. (I'm in 10th grade)

Ideally, I would be able to finish Intro+Intermediate C and P, Intermediate Algebra and Precalc by next year, but do you guys think I would able to pull that off? I could probably study for an hour a night... (I'm really busy, debate team captain, science olympiad, play an instrument, student govt etc... wish I had more time to do math). Currently I'm 5/7 or so done with AOPS Vol. 1 and I also own Vol. 2. Now, given time constraints, should I do the above textbook based approach or should I just stick with the AOPS books, albeit with much less depth?

Additionally, how much does of the topics of your C and P books does Alcumus cover? (I've looked at the tables of contents)

The precalc book looks awesome. There is no way my in-school class would even touch vectors.

Most important is Algebra 3. (And of course, the prerequisites for that)

[quote="bluecarneal"]What are the recommended prerequisites?[/quote]

If you have completed our Algebra 3 class or our Intermediate Algebra book, you're ready for our Precalculus book.  You may also be ready for the book if you've completed an honors Algebra 2 class in your school.  You can look at this [url=http://www.artofproblemsolving.com/Classes/Complex/PreTest.pdf]pre-test for our Precalculus class[/url] to determine if you have the required basic pre-reqs mastered.

[quote="AwesomeToad"]Dude buy it and use it later!

I did that...

But I chose priority mail for some reason[/quote]
Don't just buy it! The 1st edition is the one with the most mistake, so snowpaw can wait.

@Berzerk: Are you serious? You're delaying to buy a book just to not have to look at a few typos?

By the way, is AoPS ever going to have a book for which the book and the solutions manual together have the entire span of the sphere on its spine? Intermediate Algebra and Precalculus (oh wow, this word got highlighted by spell check) come quite close.

Also, on a very random and unrelated sidenote, I was expecting Precalculus to be a new color, since it seems that blue is for introductions and green is for intermediates and Precalculus is practically an Algebra 4, so supposedly advanced. So I was quite surprised when it came in the mail and I opened it and found it to be green.  :D Darn. I was wondering what that color would have been. Red?

Haha, but red is chosen for the classic series (Volumes 1 and 2.) Also, the online classes description shows Precalculus as an Intermediate course, and it's not as challenging as most olympiad problems. What we should [i]really[/i] wonder about is the color of the Calculus textbook, which will come out in less than a month. It would be really interesting to have pink (lol, the color of "ARML victory.") :lol:

I also think that having a book with a few typos doesn't matter at all. Yes, sometimes it affects the solution, but usually it only leads to a minor, trivial mistake which could alternatively be posted on one of the AoPS forums. Errors in the solution very rarely happens. For example, in the Intermediate Algebra textbook, most of the typos I found were small spelling/grammatical errors in a sentence, and I only saw one typo towards the end of a solution.

I ordered Precalculus last Sunday and it just came in yesterday (5 days) and I didn't even pay for priority shipping
thanks AoPS!

[quote="remy1140"]Haha, but red is chosen for the classic series (Volumes 1 and 2.)[/quote]

Oh, I thought it was orange? *goes upstairs to check the color of AoPS classic books*
Maybe I really am colorblind.

In my opinion, the calculus book should be tan...
(twisted reason: calculus in biology is kidney stone, and kidney stones are colored tan xD Okay, maybe I'm going on a tangent here, or being crazy.)

Loldude, [i]I[/i] am colorblind. Well, more of a color deficiency, but still I have trouble with shades of red and blue. It's probably more orange. (OK, I asked my parents, and they said it's orange.)

@Demosthenes: I noticed that these days, even delivery payed as regular shipping comes quite quickly. When I ordered Intermediate Algebra and Intermediate Counting and Probability, I received both within 5 days to a week. For some reason, AoPS is giving us a privilege.  :D

When I ordered a set of books (>\$100), it got lost in the mail, so they sent another set, and it was pretty fast.

[quote="rrusczyk"]Calculus book estimated ship date is February or March, 2010.
[/quote]
Sorry to be bringing this back up, but since the above post was in December, how is the Calculus book coming along now?

We expect to send it to the printer in the next week or so, which means we'll have it on the site by mid-late March.


I have a student who got perfect score on his scratch paper, but only got 23/25. He wants to pay for hand grading, if possible. Any ideas? How can he do it?

AMC 8 Teachers' Manual, page 10.
See also Item 10, Section IV, page 4 of the Teachers' Manual.

http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-amc8TM.pdf

Steve Dunbar
MAA Director, American Mathematics Competitions


Let $a,b,c$ positive such that $abc=1$, prove or disprove that $\sum_{cyc}\frac{2ab}{2+a+b+2ab}\leq 1$

[quote="campos"]this inequality is true...clear the denominators and you'll get

$\displaystyle \sum_{cyc}\frac{2ab}{2+a+b+2ab}\leq 1 \\\\\Leftrightarrow 2(ab+ac+bc)+6\leq 2(a^2+b^2+c^2)+(a^2b+a^2c+ab^2+b^2c+ac^2+bc^2)$

which is obviously true... i apologize for such an ugly solution... :blush:[/quote]
:D

How did you expand all of them (by hand? :w00t: ). My old brain no longer able to do a such expansion :D

yeah by hand... using the cyclic sums notations makes it a little faster and easier, but it's kind of boring ... one may have the hope that the remaining inequality will be a little bit easier and nice...


Let $ m$ be the number of positive integer factors of 876876, and let $ n$ be the number of positive integer factors of 678678. What is $ m\minus{}n$?

Both have repeating three digit numbers, so both have 1001 as factors.

$ 876876 \equal{} 1001\cdot 876 \equal{} 7\times 11\times 13\times 2^2\times 3\times 73$
so there are $ 3\times 2\times 2\times 2\times 2\times 2 \equal{} 96$ factors.

$ 678678\equal{}1001\cdot 678\equal{}7\times 11\times 13\times 2\times 3\times 113$
so there are $ 2\times 2\times 2\times 2\times 2\times 2\equal{}64$ factors.

96-64=32.

Is there any faster way to do this?

When I was playing someone, they apparently used a process that was much more efficient.


[size=134]I have a page of these to do and if someone can show me how to do the first one, maybe ill get the rest right  :)

Find a formula in terms of "n" (that is, a[/size][size=75]n[/size][size=134]=?) for the following sequence:
2,5,10,17,26,37... This is not an arithmetic or geometric sequence.  You need to come up with hypotheses and test them until one works.  For example, does a[/size][size=75]n[/size][size=134] = 6[/size][size=75]n [/size] [size=134]=5 work for all terms in the sequence?[/size]

I don't know if others would consider this helping with homework... but I have a few tips for sequences.

I looked at the sequence you posted and noticed that you add 3 to get from 2 to 5. Next you add 5 to get from 5 to 10, then add 7, add 9, add 11, etc.

To help you find an equation that would work here, figure out how to tweak the formula for the sum of the first n integers: n(n+1)/2.

I don't think it's so bad as long as we just do the first problem, so look at the differences:

2 5 10 17 26 37 etc.
3 5 7 9 11 etc.
2 2 2 2 etc.

Therefore, the equation is quadratic. I don't know how much math you know or are expected to discover, so I won't give you any help on finding the quadratic. See what you can do there.

Yeah...the first term is 2, then the next is 5, then 10, 17, 26.  So the differences are 3, 5, 7...I'll give you a hint.  What term would be before the 2?  What would the differences be then?

Well, the difference are consecutive odd numbers.  The difference between squares are consecutive odd numbers.  So we're dealing with square numbers.  However, none of these numbers are perfect squares...

You can use one of the formulas posted under the toolbar of this site to solve it.
The formula is: For the nth term of the sequence, the sum is n 2 You can take advantage of this.  ;)

For example, if you wanted to find the 19th term, you would add 18 2 to 2.

To find the 5672923th term, add 5672922 2 to 2.

...Well actually, the term after 2, which is 5, is 3 greater than 2.  Therefore the number before 2, 1, has to be the one in the sum.  That's the one you're adding the squares to.  Try it now.
Can't
1
1+1
1+1+3
1+1+3+5
1+1+3+5+7
Be expressed as
1
1+1
1+4
1+9
1+16
?
Now what are those numbers which we're adding to the 1? ;)


conics
parabola
vector
linear algebra
matrix
geometry
geometric transformation

Prove that the vector equation $\vec{r}=\vec{a}{S^2}+\vec{b}{S}+\vec{c}$ ($S$ is a variable) represents a parabola on a plane which has the following characteristics:

[b]I.[/b] $\vec{a}$ & $\vec{b}$ lie on this plane.

[b]II.[/b] It passes through a point that its (the point's) positional vector is $\vec{c}$.

if $\vec{c}=0$:
suppose that $\vec{a}$ and $\vec{b}$ are not collinear (the other case is trivial). Then if you take a vector $\vec{d}$ that is orthogonal to $\vec{a}$ and $\vec{b}$, then it is clear that the curve defined by $\vec{r}=\vec{a}t^2+\vec{b}t$ lies in the plane $P$ that goes through the origin and has $\vec{d}$ as orthogonal vector.
To see that the curve $C$ is a parabola, take for example an invertible matrix $A$ that sends $\vec{a}$ to $e_1=(1,0,0)$ and $\vec{b}$ to $e_1=(0,1,0)$. Then $A.C$ is clearly a parabola, hence $C$ is parabola.

For the general case, just make a translation with vector $\vec{c}$.

can you please explain a bit more?? :blush:


Find all real alphas for which
\[ [\sqrt{n + \alpha} + \sqrt{n} ] = [ \sqrt{4n+1} ]\] where n is natural.

[quote="apratimdefermat"]Wanna good problem's book. If u do    :D  :D  first solve this and i will give u the link

the problem is can u give me links to download problem books only on invarants and extremal principle..................

no, just kidding this wasn't the problem but a request . THE PROBLEM I WANT U TO SOLVE IS
:Find all real alphas for which
[code][ \sqrt{n + \alpha} + \sqrt{n} ] = [ \sqrt{4n+1} ][/code]

n is natural. :lol:  :lol:  :D
[youtube]numbers[/youtube][/quote]

$ [ \sqrt {n \plus{} \alpha} \plus{} \sqrt {n} ] \equal{} [ \sqrt {4n \plus{} 1} ]$

We 'll prove $ [ \sqrt {n \plus{} 1} \plus{} \sqrt {n} ] \equal{} [ \sqrt {n \plus{} 2} \plus{} \sqrt {n} ] \equal{} [ \sqrt {4n \plus{} 1} ]$

Because $ \sqrt {4n \plus{} 1} < \sqrt {n \plus{} 1} \plus{} \sqrt {n}$

$ \rightarrow [\sqrt {4n \plus{} 1}] \le\ [\sqrt {n \plus{} 1} \plus{} \sqrt {n}]$ (for all $ n \in N$ )

Assum $ [\sqrt {4n \plus{} 1}] < [\sqrt {n \plus{} 1} \plus{} \sqrt {n}]$

$ \rightarrow$ exist $ m$ such that :

$ [\sqrt {4n \plus{} 1}] < m \le\ [\sqrt {n \plus{} 1} \plus{} \sqrt {n}]$

$ \rightarrow 4n \plus{} 1 < m^2 < 2n \plus{} 1 \plus{} 2\sqrt {n(n \plus{} 1)} \le\ 4n \plus{} 2$

$ \rightarrow$ contradict

$ \rightarrow [\sqrt {4n \plus{} 1}] \equal{} [\sqrt {n \plus{} 1} \plus{} \sqrt {n}]$

Similar $ \alpha \equal{} 2$

So $ \alpha \in$ {$ 1;2$}

Besides, we have

[ \sqrt{n + 1} + \sqrt{n} ] =[ \sqrt{n +2} + \sqrt{n} ] = [ \sqrt{4n+2} ]  :wink:


integer sequence $x_{0},x_{1},...$ defined by $x_{0}=2,x_{1}=5$ and $x_{n+2}=5x_{n+1}-6x_{n}(n=0,1,...)$.
a)Find number of positive divisors of $x_{n+2}x_{n}-x_{n+1}^{2}$ for any $n$.
b)Prove that for all $n\geq 0$ we have $x_{n}^{2}-4.6^{n}$ is perfect square.

[hide]$x_{n}=2^{n}+3^{n}$

Here's a trick.  We build $x_{n}$ as a linear combination of other sequences which satisfy the recursive relationship.  We consider what is called the "characteristic quadratic", or $x^{2}=5x-6$, in this case (we're going to create two geometric sequences, each of which is satisfies the recursive relation, and then take a linear combination of them which has the same initial terms as $x_{n}$).  Then $x=2, 3$.  Therefore, any sequence of the form $a*2^{n}$ or $b*3^{n}$ satisfies the recursive relation.  We see $x_{n}=2^{n}+3^{n}$ produces the same initial conditions as our sequence, and satisfies the same recursive relation, so it must be the same sequence.

From this fact, we have

a) This number is equal to $2^{n}3^{n}$, which has $(n+1)^{2}$ positive divisors.

b) This number is $(2^{n}-3^{n})^{2}$.
[/hide]


limit trigonometry calculus integration function

I think this limit may be expressed as:

$\lim_{n\rightarrow \infty}\sum_{k=1}\sin(\frac{1}{2^{k}})$

Can't you use definition of integral?

[quote="Amir.S"]find:

$\lim_{n\rightarrow \infty}\sin \frac 1{2}+\sin\frac 1{2^{2}}+\dots+\sin\frac 1{2^{n}}$[/quote]
I think there is a theorem that says, for certain "well-behaved" or "elementary" functions -- I don't remember the correct term -- we have $\lim_{x \rightarrow a}f\left( g(x)\right) = f\left( \lim_{x\rightarrow a}g(x)\right).$

In our case, $f(x)=\sin x$ and $g(x) = \frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{x}}$ for $x\in\mathbb{N}.$

So, $\lim_{n\rightarrow \infty}\left(\sin \frac 1{2}+\sin\frac 1{2^{2}}+\dots+\sin\frac 1{2^{n}}\right) = \sin \left( \lim_{n\rightarrow \infty}\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}\right) = \sin 1$.

But, I doubt if my answer is correct. Anyone with the correct idea?

[quote="JBL"]Here's a reason that can't possibly be correct in general as you've stated it:  no matter what $g$ is, if $f = \sin$, you would have us converge to something between -1 and 1.  That's clearly going to be problematic.

Here's a reason it can't be correct in this case:

$\sin 1 \approx 0.8415$.
$\sin \frac{1}{2}+\sin \frac{1}{4}+\sin \frac{1}{8}\approx 0.8515$, and all other terms are positive.

On the other hand, $0 < \sin x < x$ for $\pi > x > 0$, so the sum does have to converge somewhere below 1.  The actual limit is $0.9764577364706682\ldots$, which doesn't look familiar to me.[/quote]

Do you Have any solution for it?


Prove that if $2^{p}+3^{p}=a^{n}$, with p a prime and a and n positive integers, then $n=1$.

I tried using Fermat, but I couldn't finish the proof.

Is $p$ a prime, or is it any positive integer?

[quote="Andreas"][hide][quote="FieryHydra"]$25 | (2+3)(2^{p-1}-2^{p-2}\cdot 3+...-2\cdot 3^{p-2}+3^{p-1}).$

[/quote]

Iff $p = 5$. It gives $275$ which is not a perfect power. Hence $n = 1$.[/hide][/quote]

I may be missing something, but  you still have to show that  "iff p=5" part (why you have to consider p=5 part only)
[hide="Hint"]
(One can put a less restriction on $p$, instead of being prime, you could say as long as it is not divisible by 2 and 5)[/hide]

well...

[hide]

since

$2^{p}+3^{p}= (2+3)(2^{p-1}-2^{p-2}\cdot 3+\ldots-2\cdot 3^{p-2}+3^{p-1}) = a^{n}$

take this $\bmod 5$ and we see that $a^{n}\equiv 0 \bmod 5$

now we take

$2^{p-1}-2^{p-2}\cdot 3+\ldots-2\cdot 3^{p-2}+3^{p-1}\equiv 2^{p-1}-2^{p-2}(-2)+\ldots-2 (-2)^{p-2}+(-2)^{p-1}\equiv p \cdot 2^{p-1}\bmod 5$

hence our expression is only $0 \bmod 5$ when $p=5$ so that's the only case we need to check

so $2^{p}+3^{p}= 275$ which is not a perfect power and therefore $n=1$

[/hide]

[quote="gotztahbeazn"]well...

[hide]

now we take

$2^{p-1}-2^{p-2}\cdot 3+\ldots-2\cdot 3^{p-2}+3^{p-1}\equiv 2^{p-1}-2^{p-2}(-2)+\ldots-2 (-2)^{p-2}+(-2)^{p-1}\equiv p \cdot 2^{p-1}\bmod 5$

[/hide][/quote]

Could you explain that step?

basically $3 \equiv-2 \bmod 5$

so when taking the big factor thing in $\bmod 5$, we can change all the $3$'s into $-2$'s and yea voila

Also, may  not be  (or may be :))       simpler  than above,  but one can also  consider everything modulo 25.
[hide] All calculations [b]mod 25 [/b]
$3^{3}\equiv 27 \equiv 2$
  so $8 \equiv 2^{3}\equiv 3^{9}$
and $3^{10}\equiv 8 \cdot 3 \equiv 24 \equiv-1$
(3^ (Any (and only) odd factor of 10 ) is equivalent to -1)

Since $2\equiv 3^{3}$ we have $2^{p}+3^{p}\equiv 3^{3p}+3^{p}\equiv 3^{p}(3^{2p}+1)$
this can be zero only if $2p$ is odd multiple of 10 (or p is multiple of 5)
(And of course, Only prime which is multiple of 5 is 5    .)

So  if $p$ is prime, only for p=5, the expression is 0 mod 25.

And a number which is multiple of 5 but not of 25 can not be written as $a^{n}$ where $n>1$

(And of course, you know, if p is odd  the expression is divisible  by 5)

(And you can manually check for p=2 and p=5)  [/hide]


i read in my book that in $Z[i\sqrt{5}]$ that
$(1+i\sqrt{5})|6$.
is this true ?
how is division actually defined in this ring ?
is it sufficient that $N(x)|N(y)$ to say that $x|y$ ?

$a|b$ iff there is $c$ with $b=ac$ (the definition).
Now $(1+i \sqrt 5)(1-i \sqrt 5)=6$, thus $(1+i \sqrt 5)|6$.

No, $N(a) | N(b)$ is not sufficient, for example take $a=3, b=(1+i \sqrt{5})^{2}=-4+2i \sqrt 5$.


"Every day, at noon, a ship leaves Miami for Montreal, and another leaves Montreal for Miami. Each trip lasts 6 days, exactly 144 hours. How many Montreal-to-Miami ships will the ships leaving Miami today pass at sea (not at port) from the time it departs till the time it arrives in Montreal?"

[hide]I got 5...and the answer is something else. :/ [/hide]

[hide]When the ship from Miami leaves the port, the nearest ship from Montreal is 1 day away. The two ships will meet in 1/2 a day. From here, the nearest ship from Montreal is now a day away, and the ships will meet in another 1/2 day. Thus, the ship from Miami will meet a ship from Montreal every 1/2 day, and will meet $\boxed{11}$ ships along its trip to Montreal.[/hide]

think about that there are other ships besides the first one that meet others

Did you guys actually have to draw out diagrams to understand it? I drew two, because my first one was so messy I couldn't understand it. But I drew a chart, like a multiplication table, except the sides are labelled with the day. Day 0, 1/2, 1, etc. Then I had little dots marking where the ships were located. Each day I would move a dot forward/backward one space. On the the top of the chart, at Day 6, I put Montreal, and on Day 0, Miami.

Hope that makes sense...

here is a good way to think about it, extend the line backwards 6 days, and have all the ships drawn in, but not on route to the other place, then you can see if they are at a distance of 12 days worth, the will either barely meet each other, or meet at port, so you would not count that

Day 0
S-S-S-S-S-[b]S[/b]-_-_-_-_-_-[b]S[/b]-S-S-S-S-S

bold means that that is the port, and each dash ( - ) is one day's travel time, each will move 6 times, and underscores ( _ ) mean a place they can end up at the end of the day

Wait, each [i]day[/i] the ships will move 6 times?

Here's something like the chart I made...well, the thinking behind it, at least.

Day 0,                                                 Day 1/2,                                      Day 1.....
*ship (that we're watching) at port,     ship 1/2 day's travel away,         ship 1 day away

[u]after 1/2 day[/u]

Day 1/2,                                                                  Day 1
              *ship meets ship that was 1/2 day's travel away,  ship that was 1.5 days away

Do you see how I'm thinking?

I understand the problem now, but it took a reaaaallyyy long time to figure out a suitable diagram/count/read my own writing. Are we supposed to think about this in our heads and write down an answer? Like..."hmm. Since the ship will meet another ship traveling from the opposite direction twice a day--11!" That's kind of hard...

I drew a picture too. I made lines for Miami and Montreal and 5 lines inbetween for the 5 days. Then I put dots on each of the lines, to represent the ships. I drew arrows to move the ships every 1/2 day, and found where the ship from Miami met others from Montreal.


Salut, imi trebuie o culegere cu cat mai multe tipuri de probleme de liceu rezolvate. Ar fi folositoare cat mai multe detalii despre carte cum ar fi editura si unde sa o caut.

Sunt in ultimul an de liceu, ma pregatesc sa dau la Facultatea de Automatica si Calculatoare din Timisoara.

Multumesc.

Iti recomand sa te uiti pe http://www.gil.ro/ (este cea mai buna editura de mate de la noi din tara).

[quote="alexmart"]Nu prea sunt detalii despre carti pe site-ul respectiv. E o imagine cu coperta.

O ai tu ? probleme sunt rezolvate frumos la sfarsit ? Ma refer la oricare dintre ele, cu toate ca cea de admitere mi-ar trebui mai mult.[/quote]
     Valentin are dreptate.Editura GIL e cea mai buna editura pentru matematica de la noi din tara.Poti lua legatura cu cei de la editura GIL scriindu-le pe adresa de email gil1993@zalau.astral.ro.Asa am procedat si eu cand am avut nevoie de carti. :)


geometry
incenter
inequalities
inequalities proposed

We have $A\leq B$.Geometrically it follows from $IA^{2}+IB^{2}+IC^{2}\geq \frac{a^{2}+b^{2}+c^{2}}{3}$(I the incentre of the triangle),or if you take the Ravi substitutions you will discover Schur. :lol:

Who has another way???
Can Mateivld solve this problem with Ravi susbtitutions?

[quote="VIETNAM"]a,b,c are the leng sides of triangle.Compare A and B with:
A=($\(a+b+c)^{3}$+3abc.
B=5($\ a^{2}b+b^{2}c+c^{2}a+b^{2}a+c^{2}b+a^{2}c$).[/quote]
The inequality is schur.
You use $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)$
And before we have:
$a^{3}+b^{3}+c^{3}+3abc\ge ab(a+b)+bc(c+b)+ca(a+c)$
 :D Bài này cũng đơn giản thôi nói chung cứ biến đổi về schur là xong hết  :D


inequalities geometry circumcircle inradius

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle with the circumradius $ R$, the inradius $ r$ and the $ A$- exinradius $ r_{a}$. Prove that $ \boxed{R^{2}\plus{}a^{2}\ge 5rr_{a}}$. Establish when is equality.[/color][/quote]
$ R^{2}\equal{}\left(\frac{abc}{4S}\right)^{2}\equal{}\frac{a^{2}b^{2}c^{2}}{2a^{2}b^{2}\plus{}2a^{2}c^{2}\plus{}2b^{2}c^{2}\minus{}a^{4}\minus{}b^{4}\minus{}c^{4}}.$
$ rr_{a}\equal{}\frac{s^{2}}{p(p\minus{}a)}\equal{} (p\minus{}b)(p\minus{}c) \equal{}\frac{a^{2}\minus{}(b\minus{}c)^{2}}{4}.$
Thus, $ R^{2}\plus{}a^{2}\ge 5rr_{a}\Leftrightarrow\frac{a^{2}b^{2}c^{2}}{2a^{2}b^{2}\plus{}2a^{2}c^{2}\plus{}2b^{2}c^{2}\minus{}a^{4}\minus{}b^{4}\minus{}c^{4}}\plus{}\frac{5(b\minus{}c)^{2}}{4}\geq\frac{a^{2}}{4}.$
Id est, it remains to prove that $ \frac{a^{2}b^{2}c^{2}}{2a^{2}b^{2}\plus{}2a^{2}c^{2}\plus{}2b^{2}c^{2}\minus{}a^{4}\minus{}b^{4}\minus{}c^{4}}\geq\frac{a^{2}}{4},$
 which equivalent to $ (b^{2}\plus{}c^{2}\minus{}a^{2})^{2}\geq0.$ :)
The equality holds when $ b\equal{}c$ and $ \alpha\equal{}90^{\circ}.$

Unless I expanded wrong, how is the last statement true? (Noob speaking here)

I mean, how is the rational inequality converted to $ (b^{2}\plus{}c^{2}\minus{}a^{2})^{2}$? How did you see that?

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle with the circumradius $ R$, the inradius $ r$ and the $ A$- exinradius $ r_{a}$. Prove that $ \boxed{R^{2}\plus{}a^{2}\ge 5rr_{a}}$. Establish when is equality.[/color][/quote]
[color=darkblue][b]Proof.[/b] The relations $ abc \equal{} 4RS$ and $ rr_{a}\equal{} (p\minus{}b)(p\minus{}c)$ are well-known. Therefore,

$ \left\{\begin{array}{cccccc}1\blacktriangleright & a^{2}\ge a^{2}\minus{}(b\minus{}c)^{2}&\implies & a^{2}\ge 4(p\minus{}b)(p\minus{}c) &\implies &\boxed{a^{2}\ge 4rr_{a}}\\ \\ 2\blacktriangleright &\left\|\begin{array}{c}bc\ge 2S\\ \ a\ge 2\sqrt{rr_{a}}\end{array}\right\| &\implies & abc\ge 4S\sqrt{rr_{a}}&\implies &\boxed{R^{2}\ge rr_{a}}\end{array}\right\|\bigoplus$ $ \implies$ $ \boxed{R^{2}\plus{}a^{2}\ge 5rr_{a}}$.

We'll have equality if and only if $ \left\|\begin{array}{c}bc \equal{} 2S\\ \ a^{2}\equal{} 4(p\minus{}b)(p\minus{}c)\end{array}\right\|$ $ \Longleftrightarrow$ $ A \equal{} 90^{\circ}$ and $ b \equal{} c$.

[b]Remark.[/b] $ \left\{\begin{array}{ccc}a \equal{} (p\minus{}b)\plus{}(p\minus{}c) &\implies & a\ge 2\sqrt{(p\minus{}b)(p\minus{}c)}\equal{} 2\sqrt{rr_{a}}\\ \\ bc \equal{} p(p\minus{}a)\plus{}(p\minus{}b)(p\minus{}c) &\implies & bc\ge 2\sqrt{p(p\minus{}a)(p\minus{}b)(p\minus{}c)}\equal{} 2S\end{array}\right\|$ a.s.o.

More precisely, for any positive numbers $ x,y,k$ we have $ (x\plus{}y)(k^{2}\plus{}xy)\ge 4kxy$.[/color]


calculus
derivative
abstract algebra
blogs
search
group theory
superior algebra

Let $G$ be a finite group with an abelian maximal subgroup. Prove that $G^{(3)}=\{e\}$.

PS I've spent some time trying to prove this strange result but no good approach is made so far. I had sent a somehow similar result on my blog sometime ago but I can't relate these two problems. (maybe no relation exists. :maybe: )

In other words, one must show that such a group is solvable of length at most $3$. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=337838097&t=54910]This[/url] is a weaker version, where the aim is to show merely that such a group is solvable, but a proof for the stronger version can be read between the lines :).

I'll use the notations in that other topic. Let $A$ be a maximal abelian subgroup of $G$, and let $Z$ be the center of $G$. If $Z$ is not contained in $A$, we're done: $AZ=G$, so $G/Z=AZ/Z\cong A/A\cap Z$ which is abelian, so $G$ is an extension of an abelian group by an abelian group. This means that its solvable length does not exceed $2$. On the other hand, if $A$ were to be normal, then again, the solvable length of $G$ would be at most $2$ (for $G/A$ would be a group with no proper non-trivial subgroups, hence cyclic of prime order). The problem has thus been reduced to the case when $Z\le A$ and $A$ is not normal in $G$.

Now let $g$ be an element of $G$ outside $A$. Since $A$ is maximal and non-normal, it must coincide with its normalizer. This means that $gAg^{-1}\ne A$. It follows that the centralizer of every element in $gAg^{-1}\cap A$ contains $A$ strictly, so must be equal to $G$; in other words, $gAg^{-1}\cap A\le Z$. On the other hand, $Z\le A$, so $Z\le gAg^{-1}\cap A$. Putting the two inclusions together, we get $gAg^{-1}\cap A=Z,\ \forall g\in G\setminus A$. This implies that the group $G/Z$ satisfies all the conditions imposed in the main part of the proof from that other topic: it has a non-normal maximal abelian subgroup $A/Z$, and it has trivial center (because, by the preceding remarks, $A/Z$ has precisely $[G: A]=[G/Z: A/Z]$ conjugates in $G/Z$, every two of which intersect trivially). It is shown there that such a group is an extension of an elementary abelian group by a cyclic group, so must have solvable length $\le 2$. Putting $Z$ back into the picture, we find that $G$ has solvable length $\le 3$.


geometry
inequalities
inequalities unsolved

Give triangle ABC.Prove that:
    [tex] S \leq \frac{\sqrt3}{4} \sqrt[3]{a^2b^2c^2} [/tex]
where $ S $ is area of the triangle,and $a,b,c $are three sides of it.

S^3<=3sqrt3/64 a^2b^2c^2
S^3=(1/8)a^2b^2c^2sinAsinBsinC
<=>sinAsinBsinC<=3sqrt3/8

a well known result.


analytic geometry
geometry
induction
geometry unsolved
Polygons
coordinates
China

A closed recticular polygon with 100 sides (may be concave) is given such that it's vertices have integer coordinates, it's sides are parallel to the axis and all it's sides have odd length. Prove that it's area is odd.

it's true for every polygon with $ 4(2k+1)$ sides. induction: the first case in the pic(area is ood obviously). then we can make longer one side or shorter-and it doesn't change the parity of the area cause we add even length to the side, that means even number of area. also if we change the direction of a side we add or subtract and even area because odd+odd=even. hope it's true and clear. we didn't use pick's theorem!


factorial
function
logarithms
calculus
derivative
parameterization
calculus computations

I have come across summations which involve "a"? It seems to maybe be equivalent to the natural log but I am not sure, can someone please specify what it is?

  i.e. summation of a to the power of x divided by x factorial, where the summation is taken from 0 to infinity is equivalent to e raised to the power of a?

Sorry for being so vague.[/list][/hide]

$ \sum_{n \geq 0} \frac {a^n}{n!} \equal{} e^a$ is one way to define the function exponentiation.  Here $ n$ (you used $ x$, which may have had something to do with your confusion*) is a dummy variable, with no existence outside the summation, and $ a$ can be thought of as a variable, arbitrary constant, or whatever you like.  (It happens that this sum converges for any real or complex number $ a$, so we don't have to put any limitations on it.  If we had $ \sum_{n \geq 0} a^n \equal{} \frac {1}{1 \minus{} a}$, the same would hold except that now it is only true if $ |a| < 1$.)

* Your question isn't a math question.  Rather, it's an "understanding how symbols and notation are used" question.  In particular, it suggested to me that you fell into the trap of thinking that $ x$ is always "the variable."  If so (and if not, you can just ignore this), it's important to realize that $ x$ is just another symbol like $ F, y, a, n, \pi$ or $ \frac{d}{dx}$ -- these symbols don't have inherent meaning.  $ \pi$ is just a letter of the Greek alphabet, one that happens to get used for an important mathematical constant.  $ F$ is just a capital letter in the Roman alphabet -- there is nothing inherently functional about it.  With context, we temporarily imbue certain symbols with certain meaning.  In this case, $ x$ was given the meaning of the dummy variable.

[quote="vdubz85 (in PM)"]Thanks for your response ...  I somewhat understand what you are saying but am still a bit confused.  In the same book I came across the meaning that

The derivative of "a raised to the power of x{a>0}" is "(a raised to the power x) * (ln a)"

This is why i thought 'a' had some meaning in itself?

I hope im not being too confusing but I would like to have a very clear meaning and I cant seem to find one, thank you for your time and effort.[/quote]

The quoted statement from your text, which I might write as $ \frac {d}{dx} a^x \equal{} a^x \cdot \ln a$, is a statement about a family of functions (simple exponential functions).  In particular, for each positive real number $ a$, we can define a function $ f_a(x)$ whose inputs and outputs are real numbers by the equation $ f_a(x) \equal{} a^x$.  This function is differentiable, and we can compute its derivative.  It happens that its derivative is the function $ f'_a(x) \equal{} a^x \cdot \ln a \equal{} f_a(x) \cdot \ln a$.  As in your original question, this is a theorem (in this case about derivatives of functions, in the first case about summations of series) where the object in question (here, the function that we are derivating, there the infinte series we are summing) depends on some parameter (= constant) $ a$, and the theorem tells us how to evaluate the object in question (derivative, series) for any value of $ a$.  In both cases, the "value" of the answer (derivative, sum) depends on our actual choice of $ a$, or can be interpreted as a function of the parameter $ a$.


Given $ a, b, c > 0$. Prove that:
$ \frac{a^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b^2}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{c^2}{a^2\plus{}ab\plus{}b^2} \ge$$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \frac{ab}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{bc}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{ca}{a^2\plus{}ab\plus{}b^2}$
 :)

hello, we have
$ \frac {a^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b^2}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{c^2}{a^2\plus{}ab\plus{}b^2}\minus{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\equal{}$
$ {\frac { \left( {b}^{2}\plus{}ab\plus{}{a}^{2}\plus{}bc\plus{}ca\plus{}{c}^{2} \right)  \left( {b}^{
5}c\plus{}{b}^{5}a\minus{}{b}^{4}{c}^{2}\minus{}{b}^{4}{a}^{2}\minus{}{b}^{2}{c}^{4}\minus{}{a}^{4}{b}^{
2}\plus{}{c}^{5}b\plus{}{a}^{5}b\plus{}{c}^{5}a\plus{}{a}^{5}c\minus{}{a}^{2}{c}^{4}\minus{}{c}^{2}{a}^{4}
 \right) }{ \left( {b}^{2}\plus{}bc\plus{}{c}^{2} \right)  \left( {c}^{2}\plus{}ca\plus{}{a}^{
2} \right)  \left( {a}^{2}\plus{}ab\plus{}{b}^{2} \right)  \left( ab\plus{}bc\plus{}ca
 \right) }}$
this term is nonnegative because
$ b^5c\plus{}b^5a\minus{}b^4c^2\minus{}b^4a^2\minus{}b^2c^4\minus{}a^4b^2\plus{}c^5b\plus{}a^56b\plus{}c^5a\plus{}a^5c\minus{}a^2c^4\minus{}c^2a^4\equal{}$
$ bc(b^3\minus{}c^3)(b\minus{}c)\plus{}ab(a^3\minus{}b^3)(a\minus{}b)\plus{}ac(a\minus{}c)(a^3\minus{}c^3)\geq0$
Sonnhard.

hello, your second inequality can be written as
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\minus{}\frac{ab}{b^2\plus{}bc\plus{}c^2}\minus{}\frac{bc}{c^2\plus{}ca\plus{}a^2}\minus{}\frac{ca}{a^2\plus{}ab\plus{}b^2}\equal{}$
$ {\frac { \left( {b}^{2}\plus{}ab\plus{}{a}^{2}\plus{}bc\plus{}ca\plus{}{c}^{2} \right)  \left( {b}^{
4}{a}^{2}\minus{}{b}^{3}{c}^{2}a\plus{}{b}^{2}{c}^{4}\minus{}{a}^{3}{b}^{2}c\minus{}{a}^{2}{c}^{3
}b\plus{}{c}^{2}{a}^{4} \right) }{ \left( {b}^{2}\plus{}bc\plus{}{c}^{2} \right)
 \left( {c}^{2}\plus{}ca\plus{}{a}^{2} \right)  \left( {a}^{2}\plus{}ab\plus{}{b}^{2} \right)
 \left( ab\plus{}bc\plus{}ca \right) }}$
and we have
$ b^4a^2\plus{}b^2c^4\plus{}c^2a^4\geq b^3c^2a\plus{}a^3b^2c\plus{}a^2c^3b$
because of
$ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zy$
Sonnhard.

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \frac {a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2}{a^2 \plus{} ab \plus{} b^2} \ge$$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {ab}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {bc}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} ab \plus{} b^2}$
 :)[/quote]
The first $ \frac {a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2}{a^2 \plus{} ab \plus{} b^2} \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$ is proved easily by Cauchy-Schwarz.
The second $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {ab}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {bc}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} ab \plus{} b^2}$
I have no ideas with this but from it we can prove this one
$ \frac {a\plus{}b\plus{}c}{ab \plus{} bc \plus{} ca} \ge \frac {b}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {a}{a^2 \plus{} ab \plus{} b^2}$


inequalities
algebra
polynomial
function
linear algebra
complex numbers
linear algebra unsolved

Let $ A,B,C \in M_n{\mathbb(R)}$ such that simultaneously triangularizable (commute or not commute) prove that :
  $ \det(A^6 \plus{} B^6 \plus{} C^6 \minus{} 3A^2B^2C^2) \geq 0$

no linear algebra here: $ a^6 \plus{} b^6 \plus{} c^6 \minus{} 3(abc)^2 \geq 0$ for any reals $ a,b,c$ and everything is happening on the diagonal

OK alekk,
but is this result still valid if $ A,B,C$ are simult. triang. over $ \mathbb{C}$ ?

Of course not! The inequality doesn't even make any sense, and in any case the expression on the left side of alekk's inequality is a nonconstant polynomial function in each of its variables, and the range of any nonconstant polynomial is all of $ \mathbb{C}.$

Hi Kent,
of course the matrices $ A,B,C$ are assumed to be real and the inequality makes sense; yet the Alekk's inequality doesn't work because $ a,b,c$ may be non real numbers. Moreover you cannot associate  pairwise conjugate expressions in the product giving the determinant.

Note By $ A,B,C$ [color=red]simultaneously triangularizable [/color]we have :
  $ A^{6} \plus{} B^{6} \plus{} C^{6} \minus{} 3A^{2}B^{2}C^{2} \equal{} (A^2 \plus{} B^2 \plus{} C^2)(A^4 \plus{} B^4 \plus{} C^4 \minus{} A^2B^2 \minus{} B^2C^2 \minus{} C^2A^2)$
 $ A^4 \plus{} B^4 \plus{} C^4 \minus{} A^2B^2 \minus{} B^2C^2 \minus{} C^2A^2 \equal{} (A^2 \minus{} (B^2 \plus{} C^2)/2)^{2} \plus{} 3/4(B^2 \minus{} C^2)^{2} \equal{} U^{2} \plus{} V^{2} \equal{} (U \plus{} iV)(U \minus{} iV)$
 See http://www.mathlinks.ro/viewtopic.php?t=310354
  1) $ \det(A^2 \plus{} B^2 \plus{} C^2) \geq 0$
  2)
$ \det(A^4 \plus{} B^4 \plus{} C^4 \minus{} A^2B^2 \minus{} B^2C^2 \minus{} C^2A^2) \equal{} \det(A^2 \minus{} (B^2 \plus{} C^2)/2)^{2} \plus{} 3/4(B^2 \minus{} C^2)^{2}) \equal{} \det(U^{2} \plus{} V^{2}) \equal{} |\det(U \plus{} iV)|^2 \geq 0$

If you want to write the first part of your post #6 then necessarily $ A^2B^2\equal{}B^2A^2,\cdots$.

NO. In particular $ det(A^2\minus{}B^2)\not\equal{}det(A\plus{}B)det(A\minus{}B)$.

Yes I was wrong about $ det(A^2\minus{}B^2)\equal{}det(A\minus{}B)det(A\plus{}B)$. I don't know why I wrote this !
But that is not the problem in the post #6.
 $ \det(A^{6}\plus{}B^{6}\plus{}C^{6}\minus{}3A^{2}B^{2}C^{2})\equal{}\det (A^{2}\plus{}B^{2}\plus{}C^{2})\det(A^{4}\plus{}B^{4}\plus{}C^{4}\minus{}A^{2}B^{2}\minus{}B^{2}C^{2}\minus{}C^{2}A^{2})$
is true.  But $ det(A^2\plus{}B^2\plus{}C^2)\geq{0}$ is valid (a priori) only if $ A,B,C$ pairwise commute.  See:
 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274989
Moreover the decomposition in squares of $ A^4\plus{}B^4\plus{}\cdots$ is correct if $ A,B,C$ commute. Else we can consider the  det.  Then we can write $ det(A^4\plus{}B^4\plus{}\cdots)$ as a product of $ u^2\plus{}v^2$  where $ u,v$ are complex numbers   and we can't conclude anymore.

[quote="loup blanc"]
  But $ det(A^2 \plus{} B^2 \plus{} C^2)\geq{0}$ is valid (a priori) only if $ A,B,C$ pairwise commute.  See:
 [/quote]
  We still have  :  $ \det(A^2 \plus{} B^2 \plus{} C^2) \geq 0$ If $ A,B,C \in M_n{\mathbb{(R)}}$ simultaneously triangularizable.
  I suppose that $ C \in GL_n{\mathbb{(R)}}$ (If $ C \notin GL_n{\mathbb{(R)}}$ see $ C \minus{} \lambda I_n$ here , $ \lambda \in \mathbb{R}$ and$ \lambda \notin$ Spectrum of $ C$)
 Then $ \det(A^2 \plus{} B^2 \plus{} C^2) \equal{} \det(C^2)\det((C^{ \minus{} 1}B \plus{} iC^{ \minus{} 1}A)(C^{ \minus{} 1}B \minus{} iC^{ \minus{} 1}A) \plus{} I_n) \equal{} \det(C^2)\det(D\overline{D} \plus{} I_n) \geq 0$
Here $ D \equal{} C^{ \minus{} 1}B \plus{} iC^{ \minus{} 1}A$  ... ?

Indeed it's a very pretty proof. Bravo....
It remains to show that
$ \det(A^{4}\plus{}B^{4}\plus{}C^{4}\minus{}A^{2}B^{2}\minus{}B^{2}C^{2}\minus{}C^{2}A^{2}) \geq 0$

[quote="QuyBac"]Note By $ A,B,C$ [color=red]simultaneously triangularizable [/color]we have :

  2)
$ \det(A^4 \plus{} B^4 \plus{} C^4 \minus{} A^2B^2 \minus{} B^2C^2 \minus{} C^2A^2) \equal{} \det(A^2 \minus{} (B^2 \plus{} C^2)/2)^{2} \plus{} 3/4(B^2 \minus{} C^2)^{2})$[/quote]

$ (A^2 \minus{} (B^2 \plus{} C^2)/2)^{2} \plus{} 3/4(B^2 \minus{} C^2)^{2}\equal{}A^4\plus{}B^4\plus{}C^4\minus{}A^2B^2\minus{}A^2C^2\minus{}(B^2C^2\plus{}C^2B^2)/2$

I think the condition comute is necessary.

I come back to the proof, by Quybak, of the property
$ \det(A^{4}\plus{}B^{4}\plus{}C^{4}\minus{}A^{2}B^{2}\minus{}B^{2}C^{2}\minus{}C^{2}A^{2}) \geq{0}$. Finally it's correct.
Remark about the suuri's post:
$ A^{4}\plus{}B^{4}\plus{}C^{4}\minus{}A^{2}B^{2}\minus{}A^{2}C^{2}\minus{}B^{2}C^{2}\equal{}(A^{2}\minus{}(B^{2}\plus{}C^{2})/2)^{2}\plus{}3/4(B^{2}\minus{}C^{2})^{2}$
is obviously false but the det. of these matrices are equal because the (complex) eigenvalues satisfy this relation and because all the matrices are upper triangular.


So I was thinking.... what if we hd a tournament with games where you can view stats? this would be ongoing for a period of time (say spring break) where we played a certain game and just had a fun time of politics, war, and growth. Some suggested games:

Lunar wars
Galava
Strategic Domination
some game at strategygamenetwork.com

Give your opinion of this and if we have enough people we will take a poll after narrowing down games.

plz suggest games (without any large amount of graphics) to try. This is an idea that I am posting. I will give no guarantee I will host this tourney.

We could have teams (only 3 though) with leaders. In Lunar wars we could do alliances and galava, guilds, but for strategic domination and the other game discussion will have to be done here in aops of a website for team members unknown to the rest of the participants.


abstract algebra
group theory
superior algebra
superior algebra unsolved

The problem is to show than if $ p<q$ are primes such that $ p|q\minus{}1$ then there exists a non-abelian group of order $ pq$.
I think the only way to do it is to give an example. I thought of some subgroup of group of invertible matrixes but i couldnt find any..

Such a group must be a semidrect product of $ \mathbb{Z}_p$ and $ \mathbb{Z}_q$ (this is the normal factor).

2x2 matrices( entries in integers mod q )

G= {  [x   y]   : x^p = 1 mod q , where x,y are integers mod q}
         [0   1]

By Cauchy's such an x exists in the integers mod q since p\(q-1). Its powers give p solutions and y has q possibilities. Therefore the order of G is pq and non-abelian:

[1 1]            [ x  0]
[0 1]            [0   1]

don't commute  where x is any element of order p in the integers mod q.

All such non-abelian groups are iso. to this group/example.

I apologize for not knowing how to typeset the symbols.


function
logarithms
ratio
group theory
abstract algebra
geometry
real analysis

Prove that there is a function $f: \mathbb{R}\to \mathbb{R}$, such that it takes all non-zero real values on every interval.

I think I saw such construction on the forum but I haven't found it.

$\log |f|$ will assume all real values on all intervals.   Both these functions will be everywhere discontinuous so there will not be any interesting construction.   Here is one possibility.

There is a dense set of real numbers whose base 3 representation, considered as a sequence of 0,1,2's, has some asymptotic property associated with the subsequence of digits in even-numbered positions
after the "decimal" point.  For example, we could require that almost all of the digits in the subsequence are zero.  If $x$ is a real number of this type, and the odd-numbered sequence of digits has exactly two $2$'s,  consider the first 2 as a START instruction, and of the part of the sequence after the Start, use the other 2 as a decimal point and the 0-1 digits as the binary representation of $f(x)$.   Otherwise let $f(x) = 0$.

[quote="jmerry"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31493]here[/url] for the previous construction. It maps to all reals, but making that into all nonzero reals is trivial.[/quote]

That construction is heavily dependent on Axiom of Choice.  The one in this thread is milder in various technical senses.  It can probably be simplified further.

Does it have to be intervals only? Is it possible if we consider every set that is the same cardinality as R?

That's doomed to failure; consider the preimages of the negative numbers and of the nonnegative numbers. One of those sets must have cardinality $c$.

[quote="fleeting_guest"]That construction is heavily dependent on Axiom of Choice.[/quote]
I'm not so sure. Does the existence of cosets depend on the axiom? We're not choosing a representative for each coset, just mapping the set of cosets to $\mathbb{R}$. We can avoid the axiom entirely in the cardinality arithmetic if we have cosets in general.

Also, my construction is really pretty simple to describe.

[quote] Does the existence of cosets depend on the axiom? We're not choosing a representative for each coset, just mapping the set of cosets to $\mathbb{R}$. [/quote]

Existence of the quotient is choiceless, but how do you prove existence of a surjective map $R/Q \to R$ without the axiom of choice?

Here is a simpler version of the construction in this thread:

Let $f(x)$ be the limiting ratio of the number of 1's to the number of 2's in the decimal representation of $x$.   (Define it as $0$ where the limit does not exist.)

[quote="fleeting_guest"]how do you prove existence of a surjective map $R/Q \to R$ without the axiom of choice?[/quote]
Trickery, but it doesn't quite work.
[hide="The details"]Let $G=\mathbb{Q}+\sqrt{2}\mathbb{Z}$, and $H=(\mathbb{Q}+\sqrt{2}\mathbb{Z})/\mathbb{Q}$. $H$ is a subgroup of $\mathbb{R}/\mathbb{Q}$ with countable order, so its cosets partition $\mathbb{R}/\mathbb{Q}$ into countable sets. By countability of $G$ and $H$, each coset $a+H$ in $\mathbb{R}/\mathbb{Q}$ is in bijection with each coset $b+G$ in $\mathbb{R}$. Since $\mathbb{R}/G$ and $(\mathbb{R}/\mathbb{Q})/H$ are canonically isomorphic, the coset bijections extend to a bijection between $\mathbb{R}$ and $\mathbb{R}/\mathbb{Q}$.

Incidentally, this construction gives us a map $f$ in which $f(x)-x$ is restricted to $G$ (or another similar countable subgroup).

Now that I look at it, that includes the Axiom of Choice. While we can construct bijections for any particular coset pair, we need to choose them everywhere. While the argument I based this on didn't necessarily require the axiom, that argument had a direct product structure to work with.

In general, I don't care much about invoking the Axiom of Choice- it has proven itself too useful in many areas of modern mathematics to ignore.[/hide]

[quote="jmerry"]Now that I look at it, that includes the Axiom of Choice. While we can construct bijections for any particular coset pair, we need to choose them everywhere. While the argument I based this on didn't necessarily require the axiom, that argument had a direct product structure to work with.  [/quote]

Here is a more direct-product-like version of the problem that seems to suffer the same difficulty:

consider the group $X$ of infinite 0-1 sequences mod 2.
let $Fin$ be the subgroup of sequences with only a finite number of 1's.
Is there a surjective map  $X/Fin \to X$?

This appears similar to extending the filter of finite sets to an ultrafilter, which is a use of (a slightly weaker form of) the axiom of choice.


Prove that the altitudes of the triangle intersect in 1 point .

[hide="Simple"]
Simply use Ceva's theorem
[/hide]

And how do you apply Ceva's Theorem here ?

If AD, BE and CF are altitudes, then try Ceva with noting that BD=ABcosB and so the others.

I am confusing here . Why BD= AB cosB . Please , can you give me a complete proof .

I can't give the picture but:

In the right triangle ABD, $ cosB\equal{}\frac{BD}{AB}$ so $ BD\equal{}ABcosB$
Similar in other right triangles we get:
$ CD\equal{}ACcosC,AE\equal{}ABcosA,CE\equal{}BCcosC,AF\equal{}ACcosA,BF\equal{}BCcosB$
Now use Ceva's and get the result as every cosines and every side appears equally in the numerator and the denumerator.


trigonometry ARML LaTeX function calculus integration algebra

I've never seen #2 before.  I really like it.  #1 is really cool also.  I've solved it before on the boards before.  I love trig problems :lol: .

Btw, if you know of anyone on the SC ARML team (or in SC in general) that is on AoPS or wants access to the Carolinas tell me because they can't access this forum until an admin or I give them access.

What did you get for your answer to number 1?

By the way, here are two more trig problems:
1. If $2\sin{x} + 2 = \sqrt{2}\cos{x}$, find $\sec{x} - \tan{x}$

2. Find the value of $\sin{(\pi/5)}\sin{(2\pi/5)}\sin{(3\pi/5)}\sin{(4\pi/5)}$

[Warning: This problem may be rather hard if you haven't seen anything like it before. I believe it was on a USC Math Contest once though, and it has a really nice solution].

For #1 I get [hide]$\frac 18$[/hide]

For #2 I got [hide]$-\sqrt{3}$[/hide]

These are for the first set.

You could have written out the solution...

[hide]$2 \sin{x}+2= \sqrt{2} \cos{x}$
Square: $4 \sin{x}^2+8 \sin{x} +4=2-2 \sin{x}^2$
If $ \sin{x}$=y, then $6y^2+8y+2=0$, or $3y^2+4y+1=0$
(3y+1)(y+1)=0 and y=-1 or y=-1/3. We want $1/ \cos{x}- \sin{x} / \cos{x}=(1- \sin{x})/ \cos{x}$.
Given $ \sin{x}=-1, \cos{x}=0$. This makes both $ \sec{x}$ and $ \tan{x}$ undefined, so we try $ \sin{x}=-1/3$ and $ \cos{x}=2 \sqrt{2}/3$. Therefore, the answer is $ \sqrt{2}$.[/hide]

must be nice to be able to go around "fixing" other people's posts.  :D

[quote="joml88"]Ok, to make a square root symbol in $\LaTeX$ use \sqrt{x}.  You cannot use html in the middle of $\LaTeX$.  Also, to make the trig functions look better use \sin, \cos and so on.  I fixed the square root problem for you.[/quote]

Uh... no you didn't. Now it says $ \sqrt{2sinx}$ which is wrong. I'll fix it.

Hurray! I learned $\LaTeX$!

Sry bout that.

Thanks for fixing everything...the solution looks much nicer now ;)

Here are the solutions to the first two:

[hide="Number 1"]
[Note: All angles are assumed to be in degrees in this solution]

First, use the fact that $\sin x =\cos(90-x)$ to change it into:

\[\cos 20 \cos 40 \cos 80.\]

Now, we use the fact that $\cos x=\frac{\sin 2x}{2\sin x}$ (this is an immediate consequence of double angle for sine, i.e. $\sin 2x=2\sin x \cos x$).  Thus we have:

\begin{eqnarray*}
\cos 20 \cos 40 \cos 80 &=& \frac{1}{2\sin 20}\left(\sin 40 \cos 40 \cos 80\right)\\
&=& \frac{1}{2\sin 20}\left(\frac 12\right)\left(\sin 80 \cos 80\right)\\
&=& \frac{1}{2\sin 20}\left(\frac 12\right)^2\left(\cos 160\right)\\
&=& \left(\frac 18\right) \frac{\sin 20}{\sin 20}\\
&=& \frac 18\\
\end{eqnarray*}
QED
[/hide]

[hide="Number 2"]
[again...angles are in degrees]

The equation $\tan 11x = \tan 34$ is satisfied if $11x=34+k_1\pi$.  Likewise, $\tan 19x =\tan 21$ gives us $19x=k_2\pi$.  ($k_1$ and $k_2$ are integral constants).  Playing around a little bit we find what linear combination of these two equations will give us a $5x$:

\begin{eqnarray*}
\ 2(19x &=& 21+k_2\pi)\\
-3(11x &=& 39+k_1\pi)\\
\hline \\
5x &=& -60+\pi(2k_2-3k_1)
\end{eqnarray*}

So $\tan 5x=\tan(-60 + \pi(2k_2-3k_1)).$  But $\tan(x+n\pi)=\tan(x)$  thus:
\[\tan 5x=\tan(-60)=\boxed{-\sqrt{3}}.\][/hide]

P.S. The answer to the second problem of the second set is [hide="click here"]5/16[/hide].  I will post my solution later though...

[quote="gauss202"]
2. Find the value of $\sin{(\pi/5)}\sin{(2\pi/5)}\sin{(3\pi/5)}\sin{(4\pi/5)}$[/quote]

We did a problem very similar to this one in AoPS Intermediate Trig/complex numbers class.  I highly doubt I would have found this solution on my own...but it is really nice!

[hide="Number 2"]
Those numbers involved in the problem scream out roots of unity (after we exhaust usual trig methods at least).

First we use the identity:

\[|1-e^{i\theta}|=2\sin\left(\frac{\theta}{2}\right).\]

(you can derive this by considering the distance between two roots of unity geometrically...)

So then we make our product:

\[\sin(\pi/5)\sin(2\pi/5)\sin(3\pi/5)\sin(4\pi/5)=\frac{1}{16}\prod_{k=1}^4\left|1-e^{k\pi/5}\right|.\]

If we viewed the right as a polynomial and changed all the ones into $x$'s we would see that the polynomial with all the 5th roots of unity (except 1) with $x$ set at 1!  In other words we can we view the right hand side as:

\[\frac{1}{16}(1+x+x^2+x^3+x^4)\Big|_{x=1}=\frac{5}{16}.\]
[/hide]

If you have alternate solutions (as I imagine you do) please share them!

[quote="gauss202"]A few more trigonometry problems:

5. Find all real numbers x, with $0 < x < 2\pi$, such that $\sin{x}$, $\sin{2x}$, and $\sin{3x}$ form an arithmetic sequence.

6. If $\sin^6{x} + \cos^6{x} = 2/3$, find all possible values of $\sin{2x}$.[/quote]

[hide="Number 5"]
So we have:
\[ \sin 2x = \sin x +k \\ \sin 3x = \sin 2x +k \\ \Rightarrow \sin 3x = 2\sin 2x - \sin x. \]

Now, we use the identities for $\sin 2x$ and $\sin 3x$.  We have, $\sin 2x=2\sin x \cos x$ and $\sin 3x =3\sin x - 4\sin^2 x \ \dag$.  So plugging this in, we get:

\[ 3\sin x - 4\sin^2 x = 4\sin x \cos x -\sin x\Rightarrow 4\sin^2x - 4\sin x +4\sin x \cos x =4\sin x(\sin^2 x -1+ \cos x)=0. \]

By the zero product property we have that either $\sin x=0$ or $\sin^2x+\cos x -1=0$.  $\sin$ is zero if $x=\pi$ (under the restrictions).  On the other hand we have $\sin^2 x + \cos x -1 =\cos^2 x -\cos x=\cos x(\cos x-1)=0.$  Thus $x=\pi/2, 3\pi/2.$

Thus we have $x=\boxed{\pi/2,\pi,3\pi/2}$ as answers.

$\dag$
We can derive the formula for $\sin 3x$ in a couple ways.

[u]Method 1: Complex Numbers[/u]
We use de Moivre's and match up real and imaginary parts.  $\sin 3x$ is the imaginary part of $\text{cis}^3x$.

\begin{eqnarray*} \sin 3x &=& \text{Im}(\text{cis}^3(x)\\ &=& \text{Im}(\cos^3 x+3i\cos^2 x \sin x - 3\cos x \sin^2 x -i\sin^3x\\ &=& 3\cos^2 x \sin x -\sin^3 x\\ &=& 3(1-\sin^2 x)-\sin^3x\\ &=& 3\sin x-4\sin^3 x \end{eqnarray*}

[u]Method 2[/u] The more traditional way would be to use the $\sin$ sum formula twice:

\begin{eqnarray*} \sin 3x &=& \sin(2x+x)\\ &=& \sin 2x \cos x + \sin x \cos 2x\\ &=& 2\sin \cos^2 x + \sin x (1-2\sin^2 x)\\ &=& 3\sin x - 4\sin^3 x \end{eqnarray*}
[/hide]

[hide="Number 6"]
We start with $\sin^2 x +\cos^2 x=1$ and cube both sides:

\[ \sin^6 x +3\sin^4 x \cos^2 x +3\sin^2 x \cos^4 x +\cos^6 x=(\sin^6 x+\cos^6 x)+3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)=1 \]
\[ 2/3 + \frac 34 \sin^2 2x =1 \]
\[ \Rightarrow \sin 2x=\pm 2/3 \][/hide]


inequalities
algebra
polynomial
induction
algebra unsolved

My solution is the same. Nice Aplication of Abel's Sum, NT TUAN!

Sorry I've editted it and now it's correct.


[color=blue][b]Let $ a,b,c,d$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. with $ k$ is a  given positive real number. Find the maximum value of the expression:

$ A \equal{} {\sum\limits_{cyclic} {(abc)} ^k}$[/b][/color]

I think the question is false. What is the meaning of cyclic sum of $ (abc)^k$

[color=blue][b][quote="Potla"]I think he means:
$ \sum_{cyclic}(abc)^k \equal{} (abc)^k \plus{} (bcd)^k \plus{} (acd)^k \plus{} (abd)^k$ (maybe)
 :maybe:[/quote]
That's right, thanks so much for your support.[/b][/color]


are there two integer like a and b so that the prime divisiors of a and b are the same and also the same thing for a+1 and b+1?
please post the answer whith the reason.
thank you.


integration
limit
function
probability
inequalities
calculus
real analysis

Assume $ \displaystyle\int^{ \plus{} \infty}_{ \minus{} \infty} |f(x)|^p\,dx$ exists.
Show that: $ \displaystyle\lim_{\lambda\to\ \plus{} \infty}\lambda^p\mu\{x: |f(x)| > \lambda\} \equal{} 0$.
where $ \mu X$ is Lebesque measure of the set $ X$.

It is a tool of distribution function which is created by Burkholder and Gundy, if for $ f$ is measurable let $ \alpha(\lambda)\equal{}\mu\{x: |f(x)|>\lambda\}$ then $ \|f\|_p^p\equal{}\int\limits_0^{\infty}p\lambda^{p\minus{}1}\alpha(\lambda)d\lambda$

It seems dubious that Burkholder and Gundy came up with this first. To me, this is at best an observation; I'm pretty sure people have been working with this trick for ages (e.g. in probability, they switch from expectations to integrating the survival function all the time). I could be wrong.

yes phils, I used the word "creat" may be wrong, I mean they fistly use them to attack many problems in Fourier analysis. You could take a look at some books of Fourier analysis for details  (eg. Fourier analysis on Euclidean space, of Elias M. Stein). By the way, here's their paper:
D. I. Burkholder and R. F. Gundy, "Extrapolation and interpolation of quasilinear operators on martingales". Acta Math., Vol. 124 (1970), 249-304.

pluricomplex, phils a lot of thanks!
From where can I find the solution to this problem!

Coming back to this problem (which is much simpler than all that talk of Burkholder and Gundy):

Define $ g_{\lambda}(x)=\begin{cases}\lambda^p,&|f(x)|>\lambda\\ 0,&|f(x)|\le\lambda\end{cases}.$

Note that $ g_{\lambda}\le|f|^p.$ Hence

$ \lambda^p\mu(\{|f|>\lambda\})=\int g_{\lambda}\le\int|f|^p.$

That gives is Chebyshev's inequality: $ \mu(\{|f|>\lambda\})\le\frac{\int|f|^p}{\lambda^p}.$

But now note that $ \lim_{\lambda\to\infty}g_{\lambda}(x)=0$ whenever $ |f(x)|$ is finite, which is almost everywhere. Thus $ \g_{\lambda}\to0$ almost everywhere. By the Dominated Convergence Theorem (with $ |f|^p$ as the dominating function) we have that

$ \lim_{\lambda\to\infty}\int g_{\lambda}=\lim_{\lambda\to\infty}\lambda^p\mu(\{|f|>\lambda\}) =0.$

Note also that $ g_{\lambda}\to 0$ pointwise as $ \lambda\to 0.$ We can also conclude that $ \lim_{\lambda\to0}\lambda^p\mu(\{|f|>\lambda\})=0.$

---

As for that way to compute the $ L^p$ norm - what we can show fairly directly from the definition of the integral is this:

$ \int|f|^p=\int_0^{\infty}\lambda^p(-d\alpha(\lambda))$

This is a Stieltjes integral; it should also look familiar to anyone who has studied probability theory. Stieltjes integrals are just made to be integrated by parts:

$ \int_0^{\infty}\lambda^p(-d\alpha(\lambda))=
\left.\lambda^p\alpha(\lambda)\right|_0^{\infty}+
\int_0^{\infty}p\lambda^{p-1}\alpha(\lambda)\,d\lambda$

Now the theorem we have just proved above takes care of the boundary terms:

$ \left.\lambda^p\alpha(\lambda)\right|_0^{\infty}=0-0=0.$

Kent Merryfield! A lot of thanks for your help. :)


calculus
ceiling function
floor function
Pascal\u0027s Triangle
calculus computations

i hope i have not asked this one before, no thanks to perfect_radio ......   :D
(does he/she work for google or something, 'cause i swear he/she  will find the post sooner or later, if one was made previously that is...  :P  )

find the limit

$ \left(\dfrac{2}{1}\right)^{\dfrac{1}{2^1}}\;\cdot\; \left(\dfrac{2^2}{1\cdot 3}\right)^{\dfrac{1}{2^2}}\;\cdot\;  \left(\dfrac{2^3\cdot 4}{1\cdot 3^3}\right)^{\dfrac{1}{2^3}}\;\cdot\;  \left(\dfrac{2^4\cdot 4^4}{1\cdot 3^6\cdot 5}\right)^{\dfrac{1}{2^4}}\;\cdot\;  \left(\dfrac{2^5\cdot 4^{10}\cdot 6}{1\cdot 3^{10}\cdot 5^5}\right)^{\dfrac{1}{2^5}}\;\cdot\;   \left(\dfrac{2^6\cdot 4^{20}\cdot 6^6}{1\cdot 3^{15}\cdot 5^{15}\cdot 7}\right)^{\dfrac{1}{2^6}}\;\cdot\; \ldots$

it isn't that hard really....  :oops:

looking at the $ n^{th}$ term won't help you here. :(

just a small hint: start with basic calculus, this is from a grade 8 problem (13-15 yr old students primarily)  :oops:

in every term look at the exponents of $ 1,2,3,4,5,6,7,8,9,10,\; \ldots$, they are actually numbers
taken from   pascal's triangle ( it appears to go   like this : [size=150][b]/\/\/\/\/\/ [/b][/size] for odd, and  [size=150][b]/\/\/\/\/\/\ [/b][/size] for even )

 anyway, here is what you wanted, i think....  :maybe:

find $ \textbf S\; \equal{} \; \prod_{n \equal{} 1}^\infty \left\{\; \left( \dfrac {2^{\binom{n}{1}}\cdot 4^{\binom{n}{3}}\cdot 6^{\binom{n}{5}}\cdot 8^{\binom{n}{7}}\cdot 10^{\binom{n}{9}}\cdot 12^{\binom{n}{11}}\; \cdots \quad\;\left\{2\,\lceil\dfrac{n}{2}\rceil\right\} ^{\binom{n}{2\,\lceil\dfrac{n}{2}\rceil \minus{} 1}}} {1^{\binom{n}{0}}\cdot 3^{\binom{n}{2}}\cdot 5^{\binom{n}{4}}\cdot 7^{\binom{n}{6}}\cdot 9^{\binom{n}{8}}\cdot 11^{\binom{n}{10}} \; \cdots \quad \;\left\{2\,\lfloor\dfrac{n}{2}\rfloor \plus{} 1\right\}^{\binom{n}{2\,\lfloor\dfrac{n}{2}\rfloor} }} \right)^ {\dfrac {1}{2^n} } \;\right\}$

btw/ the answer is $ \frac {\pi}{2}$ (in case you are wondering)  :D

i consider it to be similar to wallis, only   this converges to $ \frac {\pi}{2}$ much much faster.


For any point $A$ and line $k$ let us define $A_k$ as projection of $A$ onto $k$. Given two intersecting lines $k$ and $l$ and point $P$ find the locus of points $Q$ such that lines $PQ$, $P_k P_l$ and $Q_k Q_l$ intersect in one point.

i think it is not easy,even may be it is hard.because it's very strange problem. :D

the locus , i think, is the joining of the two lines-projections of P on k e l , iff P isn't on the bisector  of one of the two angles formed by k and l

The locus consists of 3 lines: $PP_{k}, PP_{l}$ and a line parallel to $P_{k}P_{l}$ through the intersection X of the lines k, l. The lines $PP_{k}, PP_{l}$ trivially belong to the locus.

Assume $XQ \parallel P_{l}P_{k},$ so that $\angle QXQ_{k}= \angle XP_{k}P_{l}.$ Since the quadrilateral $PP_{k}XP_{l}$ is cyclic, $\angle XP_{k}P_{l}= \angle XPP_{l}.$ It follows that $\angle QXQ_{k}= \angle XPP_{l}$ and similarly, $\angle QXQ_{l}= \angle XPP_{k}.$ This means that the cyclic quadrilaterals $PP_{k}XP_{l}\sim XQ_{l}QQ_{k}$ are oppositely similar. Since the non-corresponding diagonals $XQ \parallel P_{l}P_{k}$ are parallel, from the opposite similarity, the other non-corresponding diagonals $XP \parallel Q_{k}Q_{l}$ are also parallel.

Let U, V be the diagonal intersections of the quadrilaterals $PP_{k}XP_{l},\ XQ_{l}QQ_{k},$ respectively, and $\eta = \frac{XQ}{PX}= \frac{XV}{PU}$ their similarity coefficient. Let the diagonal $O_{k}Q_{l}$ meet the diagonal $P_{k}P_{l}$ at a point R. Since $XQ \parallel P_{k}P_{l},$ $\angle PUR = \angle PXQ.$ Since $Q_{k}Q_{l}\parallel PX,$ $UC = XV$ and $\frac{UR}{PU}= \frac{XQ}{PX}= \eta.$ Thus the triangles $\triangle PUR \sim \triangle PXQ$ are similar by SAS, which means that the points P, R, Q are collinear.


abstract algebra
induction
superior algebra
superior algebra unsolved

Let $ M$ be an $ R$-module of finite length($ R$ is an arbitrary ring with unit),i.e there is a finite sequence of submodules of $ M$
 $ 0 \equal{} M_{0}\subseteq M_{1}\subseteq ...\subseteq M_{n} \equal{} M$ such that $ M_{i}/M_{i \minus{} 1}$ is a simple $ R \minus{}$ module for all $ i \equal{} 1,2,..n$. Prove that $ End_{R}(M)$ is artinian.

hint:$ M$ is of finite length iff $ M$ is Artinian and Noetherian thus any set of submodules has a maximal and a minimal element.
Moreover any endomorpism $ s$:$ M\mapsto M$ is an isomorphism iff $ s$ is an epimorphism or a monomorphism(consider $ imf \subseteq imf\circ f\subseteq ... \subseteq imf\circ f\circ...f\subseteq...$.
and $ kerf \subseteq kerf\circ f\subseteq ... \subseteq kerf\circ f\circ...f\subseteq...$ which are finaly constant)
Now take a strictly decreasing sequence of ideals in $ End_{R}(M)$.
Note: in case n=1 $ End_{R}(M)$ is a divison ring by Schur's lemma since M is simple. Thus we can use induction on the length of M. It is also trivial if M has a direct summand.


Let f: (0, infinity) -> R be differentiable, and assume that f(x) + f'(x) -> 0 as x -> infinity. Show that f(x) -> 0 as x -> infinity.

I understand the basic idea behind a given solution is to estimate |f(x)| by treating f as the solution to a linear first order ODE. Is this the typical way to handle these types of problems?

[quote="Snipez"]
I understand the basic idea behind a given solution is to estimate |f(x)| by treating f as the solution to a linear first order ODE. Is this the typical way to handle these types of problems?[/quote]
Of course, $ f$ is not a solution. But you can compare it to a solution of a linear ODE (actually, to the solutions of two equations). And it is quite typical to use this idea for such kind of problems.


I am trying to find the total height and length of the bottom of a triangle.  I know part of the distance of the bottom and the angle where the hypotenuse meets the base.

I just don't know where to go from here because I have 2 unknowns.

How do I find H and X (x is the total length of the base, 5 is just part of the base)?

[img]http://img172.imageshack.us/img172/8862/mathab3.png[/img]

You cannot find H or X since the point at the top of H could be as high as you want it, and there are an infinite number of triangles for which H and X could be any one of them. SRY  :ninja:

Well, than all of the kids that got it in class and won't give me any hints or anything must have the wrong answers!

It sucks that the teacher won't help me, but will give everybody else the answer.

How can H be as high as you want it to be if S has a set height?  Isn't there some proportion or ratio of the height of S and base 5 to the base X and height H?

Just pick H to be anywhere on that upper line, and you can see that you still get a traingle.


not sure if this is right or not but:
[hide="my answer"]answer: $\frac{177}{16}$
i reversed the coefficents of the polynomial. then you have to find $(a+2)^2+...(d+2)^2$. then i did a synthetic division trick i learned outta the aops book (basically dividing the new polynomial by -2 multiple times). this created a new polynomial. i then just did the second newton sum. i'm sure if it is right to do the steps in that order and i mighta made a calcuation mistake but that's my answer[/hide]