If each number in a set of ten numbers is increased by $ 20$, the arithmetic mean (average) of the original ten numbers:

$ \textbf{(A)}\ \text{remains the same} \qquad
\textbf{(B)}\ \text{is increased by 20} \qquad
\textbf{(C)}\ \text{is increased by 200} \\
\textbf{(D)}\ \text{is increased by 10} \qquad
\textbf{(E)}\ \text{is increased by 2}$

[hide="Solution"]Let the numbers be $ a_0, a_1, \ldots, a_9$.
Then the new mean is $ \frac{a_0\plus{}a_1\plus{}\cdots\plus{}a_9\plus{}20\cdot 10}{10}$
So it $ \boxed{\textbf{(B)}\ \text{is increased by 20}}.$[/hide]

Here is my solution
 Let $ a_1 \plus{} a_2 \plus{} ... \plus{} a_{10} \equal{} t$ then $ 20\cdot \frac {t}{10} \equal{} 2t$ or $ \boxed E$

Each number is increased by 20 so the overall sum is increased by 200.
Then the sum is x+200 where x is the original sum
The average is x/10 +200/10 which is 20 more than the original average
20 :arrow: B

@enndb0x: "Increase" means to add, not to multiply. "10 increased by 2" is 10+2=12, not 10*2=20.

ohh thnx you are right , but there're few english problems  :(


Let $ a_n$ and $ b_n$ be sequences in $ \mathbb{R}$. Show that if $ \sum{b_n}$ converges and $ \sum|a_n \minus{} a_{n \plus{} 1}| < \infty$, then $ \sum{a_nb_n}$ converges.

The hypotheses appear to be carefully chosen so that summation by parts will work.

Define $ B_n \equal{} \sum_{k \equal{} 1}^nb_k$ and $ B \equal{} \lim_{n\to\infty}B_n,$ which exists by hypothesis. Choose $ M$ so that $ |B_n|\le M$ for all $ n,$ since a convergent sequence is bounded.

Define $ A \equal{} \lim_{n\to\infty}a_n \equal{} a_1 \minus{} \sum_{k \equal{} 1}^{\infty}(a_k \minus{} a_{k \plus{} 1}).$ The series converges absolutely by hypothesis.

Now, we're ready to apply summation by parts:

$ \sum_{k \equal{} 1}^na_kb_k \equal{} a_nB_n \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}B_k(a_k \minus{} a_{k \plus{} 1})$

Take the limit as $ n\to\infty$ of both sides of this expression. The left hand side becomes the series we're interested in. On the right hand side:

$ \lim_{n\to\infty}a_nB_n \equal{} AB,$ which exists by the arguments above.

$ \lim_{n\to\infty}\sum_{k \equal{} 1}^{n \minus{} 1}B_k(a_k \minus{} a_{k \plus{} 1}) \equal{} \sum_{k \equal{} 1}^{\infty}B_k(a_k \minus{} a_{k \plus{} 1})$

This series converges absolutely by comparison to $ \sum_{k \equal{} 1}^{\infty}M|a_k \minus{} a_{k \plus{} 1}|.$

Hence, our original series converges.

This was stated for sequences in $ \mathbb{R},$ but the argument would be unchanged for sequences in $ \mathbb{C}.$


If $m_{a}$, $m_{b}$, $m_{c}$ are the medians and $r_{a}$, $r_{b}$, $r_{c}$ are the exradii of a triangle $ABC$, then prove the following inequality:

$\frac{r_{b}r_{c}}{m_{b}m_{c}}+\frac{r_{c}r_{a}}{m_{c}m_{a}}+\frac{r_{a}r_{b}}{m_{a}m_{b}}\geq 3$.

[i]Remark.[/i] Using the $\sum$ sign for cyclic sums, this inequality can be rewritten as $\sum\frac{r_{a}r_{b}}{m_{a}m_{b}}\geq 3$.

please read the source posting code at http://www.mathlinks.ro/viewtopic.php?t=3490
also in each forum at the top of the questions' list.

you should have posted "unknown, (bla bla bla whatever text you want) my prof gaved it to me .. " :D (sorry for the instrusion :) ).

Isn't this one of Hojoo Lee's problems in his new project 'Topics in inequalities'?
[url]hhttp://ultrametric.googlepages.com/tin2006new.pdf[/url]

[quote="Valentin Vornicu"]you should have posted "unknown, (bla bla bla whatever text you want) my prof gaved it to me .. " :D (sorry for the instrusion :) ).[/quote]

My prof GAVE it to me ;) It's not the first time that you are making this mistake Valentin :D

:(   I'm so sad. You don't like the inequalities I posted. As a "revenge" ( kiddin) I solved your inequality:
   Put a+b-c=z, b+c-a=x and c+a-b=y. Take also x+y+z=1.Using well-known formulae, we can "reduce" this to: \sum z/[ \sqrt 4x+(y-z)^2*  \sqrt 4y+(z-x)^2] \geq 3/4. Using AM-GM it suffices to prove that  \sum z/[4x+4y+(y-z)^2+(z-x)^2] \geq 3/8. Use the Cauchy trick to find that we can reduce to  \leq z(4x+4y+(y-z) 2 +(z-x) 2 ) \leq 8/3. This becomes immediately 8 \sum xy+2 \sum x^3- \sum x^2(y+z) \leq 8/3. But  \sum x^3=3xyz+1-3 \sum xy and  \sum x^2(y+z)= \sum xy-3xyz. And everything reduces to  \sum xy+9xyz \leq 2/3, true!

k,arne u were correct.i dint know this was ho joo lee's problem(i checkd it today).and i found a solution pretty similar to his solution .and twas ugly :D
but harazi has come up with summin good.(dude,u continue to amaze me :D )
and BTW,i luv ur inequalities ,the only problem being i feel like a dumbass
trying to solve them :D

I don't understand why I cannot visit Ho joo Lee's homepage. It simply doesn't work. My solution is not at all beautiful, but it's all I have till now. And, please, do not say so about you. I jave shown for lots of times that those huge and nasty inequalities I post have very simple ideas.

Dear Jura,

I have read carefully your proof and I have found it very nice and very interesting. Congratulations, Jura.

[quote="iura"]Hope my soln is correct.
We pass by dual principle to $x,y,z$.
Then we must prove $\sum \frac{x(x+y+z)}{m_bm_c} \geq 3$.
We prove more
$\sum \frac{x(x+y+z)}{m_b^2+m_c^2} \geq \frac32$.
Which transfroms to
$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+(x-z)^2+(y-z)^2} \geq \frac 32$ or
$\sum \frac{x^2}{y+z+((x-z)^2+(x-y)^2)\frac{x}{x+y+z}} \geq \frac 32$
Now applying the Andreescu Lemma(Cauchy) this reduces to
$(x+y+z)^2 \geq 3(xy+yz+zx)+\sum \frac{x+y}{x+y+z} (x-y)^2$ which is just
$\sum \frac{z}{x+y+z} (x-y)^2 \geq 0$ true[/quote]

Here a typo, instead of

[quote]$\sum \frac{x(x+y+z)}{m_b^2+m_c^2} \geq \frac32$.
Which transfroms to
$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+(x-z)^2+(y-z)^2} \geq \frac 32$[/quote]

you should write

[quote]$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+\frac{(x-z)^2+(y-z)^2}{4}} \geq \frac 32$[/quote]

and finally $\sum (x-y)^2(4z+x+y) \geq 0 $

Thank you very much, Jura.

This problem was proposed by Zun Shan and Ji Chen, see [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1991, No.8, p.238, Problem 1680.

Solution $I$ by Marcin E. Kuczma,  Solution $II$ by the proposers, also see Crux Math., 1992, No.8, pp.250-253.

[quote="manlio"]Dear Jura,

I have read carefully your proof and I have found it very nice and very interesting. Congratulations, Jura.

[quote="iura"]Hope my soln is correct.
We pass by dual principle to $x,y,z$.
Then we must prove $\sum \frac{x(x+y+z)}{m_{b}m_{c}}\geq 3$.
We prove more
$\sum \frac{x(x+y+z)}{m_{b}^{2}+m_{c}^{2}}\geq \frac32$.
Which transfroms to
$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+(x-z)^{2}+(y-z)^{2}}\geq \frac{3}{2}$ or
$\sum \frac{x^{2}}{y+z+((x-z)^{2}+(x-y)^{2})\frac{x}{x+y+z}}\geq \frac{3}{2}$
Now applying the Andreescu Lemma(Cauchy) this reduces to
$(x+y+z)^{2}\geq 3(xy+yz+zx)+\sum \frac{x+y}{x+y+z}(x-y)^{2}$ which is just
$\sum \frac{z}{x+y+z}(x-y)^{2}\geq 0$ true[/quote]

Here a typo, instead of

[quote]$\sum \frac{x(x+y+z)}{m_{b}^{2}+m_{c}^{2}}\geq \frac32$.
Which transfroms to
$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+(x-z)^{2}+(y-z)^{2}}\geq \frac{3}{2}$[/quote]

you should write

[quote]$\sum \frac{x(x+y+z)}{(y+z)(x+y+z)+\frac{(x-z)^{2}+(y-z)^{2}}{4}}\geq \frac{3}{2}$[/quote]

and finally $\sum (x-y)^{2}(4z+x+y) \geq 0$

Thank you very much, Jura.[/quote]

since someone revived this thread what is the dual principle


MATHCOUNTS
Pythagorean Theorem
geometry
number theory
relatively prime

Here's the question that my mathcounts coach gave me: [i]Prove that in any primitive Pythagorean triple, that at least two of the side lengths are relatively prime. [/i]

I might have worded it wrong...

maybe you mean to say that it is a base triple, because any triple, such as $ 3,4,5$ or $ 5,12,13$ can be multiplied by a scalar to create a new triple that has the property that they all have a commom factor of that scalar and are thus not relatively prime.

Hi rd5493. Yes, I think you have.

I meant to say Hi batteredbutnotdefeated: as I checked for the name rd5493 had posted in them meantime so I got it wrong.

Perhaps you can let us know what the question really is?

hi battered! your Question isn't clear !

eg. take the Pythagorean triple ( 6, 8, 10 ), no 2 side lengths are relatively prime.

Perhaps you meant [b]primitive Pythagorean triplets[/b]  :!:

He's definitely talking about primitive triples. I can prove this when I get to a computer (typing from my phone currently) but my approach would pretty much be contradiction.

Consider some factor d that divides two of the three sides. From the Pythagorean theorem, it must also divide the third. However, the triple is not primitive in that case. Thus, the original statement is proven.

to explain xpmath's post:

There are two cases to evaluate; when the two legs are the sides with the common factor, and when the hypotenuse and one leg are.

In the first case, you can represent the sides as $ nx$, $ ny$, and $ z$. So by the Pythagorean Theorem, $ n^2x^2\plus{}n^2y^2\equal{}z^2$. Factor the sides: $ n^2(x^2\plus{}y^2)\equal{}z^2$. Divide both sides by $ n^2$; $ x^2\plus{}y^2\equal{}z^2/n^2$. Take the square root of both sides; $ \sqrt{x^2\plus{}y^2}\equal{}z/n$. Since you are only evaluating this for Pythagorean Triples, $ z$ must be divisible by $ n$. Therefore, if any sides have a common factor, they all have that common factor and it's not a primative triple.

In the second case, you have sides $ x$, $ ny$, and $ nz$.
 :arrow: $ x^2\plus{}n^2y^2\equal{}n^2z^2$
 :arrow: $ n^2(z^2\minus{}y^2)\equal{}x^2$
 :arrow: $ z^2\minus{}y^2\equal{}x^2/n^2$
 :arrow: $ \sqrt{z^2\minus{}y^2}\equal{}x/n$
 :arrow: therefore, if any two sides have the common factor $ n$, all sides have $ n$ as a factor

The definition of a primitive triple is that they are relatively prime, thus it is proven. (that was simple)

Let us just say that they don't have to  be relatively prime. So let's take two even legs as one case. a^2+b^2=c^2. C is then forced to be even. even squared is always even. Let us take a   leg and the hypotenues(spelled wrong) as onther case. if c and b ar even then a is forced to be even. even squared minus even squared is always even. Now lets just say if they are two sides have common prime factor of three. a number divisible by three squared + or - a number divisible by three squared is still not a primitive (base form) pythagorean triplet. Through contradiction the sides must be relatively prime in a base form pythagorean triplet.
TEEHEE  :rotfl:

turak, you're overcomplicating things. All you need to do is let $ d$ be a divisor of any two numbers in a primitive Pythagorean triple, then take the Pythagorean Theorem $ \mod d$ to reach the contradiction.

varunrocks: I have no idea what you're saying :huh:. I don't mean to be rude or anything (perhaps English isn't your first language) but your solutions usually are hard to follow. Anyway, it looks like you just examined cases where the divisor is $ 2$ and $ 3$ but nothing else, and then immediately jumped to the conclusion that it holds for all $ d$ ($ d$ as defined previously in this post).

Well, look if they do have common factors then the difference or the sum of the squares will still contain that factor. SO it is self explanatory that the third side contains that factor right? This means it is not a primitive pythagorean triplet. So they must be relatively prime.


[quote="congthienduong"]a,b,c are 3 real numbers such that $ 1 \leq a,b,c \leq 2$.
Show that: $ (3a \plus{} 2b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\leq \frac {45}{2}$.
Have fun![/quote]

Sorry  :mad:  This problem is [b]Problems in This Issue[/b] of Vietnam Mathematics and Youth Magazine(3/2008).

Lock it please,mods  :wink:


Find all differentiable functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f'(x)\equal{}e^{f(x)}$ for all $ x\in\mathbb{R}$

hello, rewriting your equation in the form
$ \frac{df(x)}{dx}\equal{}e^{f(x)}$
$ e^{\minus{}f(x)}df(x)\equal{}dx$
integrating this we get
$ \minus{}e^{\minus{}f(x)}\equal{}x\plus{}C$
Now you can compute $ f(x)$.
Sonnhard.

If we have such an $ f$, then $ x\plus{}C\equal{}\minus{}e^{\minus{}f(x)}<0$ for all $ x\in\mathbb{R}$, which is not possible since $ x\plus{}C$ is linear. Does this mean that there is no such function?


[quote="now a ranger"]1) Find the value of digit $A$ if the five digit number $12A3B$ is divisible by both $4$ and $9$, and $A$ is unequal to $B.$

2) Find the sum of all four digit natural numbers of the form $4AB8$ which are divisible by $2,3,4,6,8,$ and $9$.

could someone explain?[/quote]

1.

[hide]1+2+3+a+b=[a number divisible by 9]
3B, the two digit number, is divisible by 4.
So b=2 or 6.
1+2+3+a+2=9
a=1

or

1+2+3+a+6=18
a=6

but a cant equal b

so a=1.[/hide]

2.
[hide]So if its divisible by 9, its divisible by 3.  Same with 8, to 4 and 2.  And with 3, 6 is taken care of (2*3).  So it must be divisible by 9 and 8.

For 9, A+B+4+8=[multiple of 9]
For 8, AB8, the three digit number, is divisible by 8.

AB8, the three digit number divisible by 8.

The smallest is 128.  then you can keep adding 40, since its the smallest divisible by 10 and 8.

So we have 128, 168, 208, 248...

So A+B+4+8=27 or 18.

A+B=15
A+B=6

So if the hundreds is

1, none work
2, 4 works
3, 3 works, but 338 is not divisible by 8.
4, 2 works, but 428 is not divisible by 8
5, 1 works, but 518 is not divisible by 8
6, 0 works, 9 works-608 is, 698 isnt.
7, 8 works, 788 isnt divisible
8, 7 works, 878 isnt divisible
9, 6 works

so our three digit numbers are 248, 608, and 968.

so our 4 digits numbers are 4248, 4608, 4968=13824.[/hide]

haha thanks.

I asked someone already, and they told me more about divisibility rules, and mods, so yay.  :lol:


linear algebra
matrix
vector
algebra
polynomial
inequalities
graph theory

Let we are given $ (p \plus{} q)\times(p \plus{} q)$ matrix $ A \equal{} \|a_{ij}\|$ and $ a_{ij} \equal{} 0$ for $ 1\leq i,j\leq p$ and $ p \plus{} 1\leq i,j\leq p \plus{} q$. Show that if
$ \lambda$ is eigenvalue of $ A$ than so is $ \minus{} \lambda$.

The hypothesis is that $ A$ has the following block structure:
\[ A\equal{}\begin{pmatrix}0&B\\C&0\end{pmatrix}.\]
Now suppose that $ \lambda$ is an eigenvalue of $ A.$ That means that there is a nonzero column vector such that

$ \begin{pmatrix}0&B\\C&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equal{}\begin{pmatrix}By\\Cx\end{pmatrix}\equal{}\lambda \begin{pmatrix}x\\y\end{pmatrix}\equal{}\begin{pmatrix}\lambda x\\\lambda y\end{pmatrix},$

which can be rewritten as $ By\equal{}\lambda x,\ Cx\equal{}\lambda y,$

Now compute

$ A\begin{pmatrix}\minus{}x\\y\end{pmatrix}\equal{}\begin{pmatrix}By\\\minus{}Cx\end{pmatrix}\equal{}\begin{pmatrix}\lambda x\\\minus{}\lambda y\end{pmatrix}\equal{}\minus{}\lambda\begin{pmatrix}\minus{}x\\y\end{pmatrix}$

Hence, if $ \lambda$ is an eigenvalue, then so also is $ \minus{}\lambda.$ Note that nothing in this rules out the possibility that $ \lambda\equal{}0$ for some $ \lambda.$

The argument used no special properties of the underlying field and is therefore true over any field. (Albeit vacuous over a field of characteristic $ 2$.)

But over every ring, even with $ 2$ non-invertible, we can still show that the characteristic polynomial of $ A$ has its all coefficients before odd powers of $ X$ equal to zero. ;)

  darij

Darij, the blocks weren't the same size. (That is, $ B$ and $ C$ weren't square.) So you need to modify your statement somewhat to accommodate, say,

$ \begin{bmatrix}0&0&1\\0&0&1\\1&1&0\end{bmatrix},$ with characteristic polynomial $ x^3\minus{}2x.$

Ehm, sorry. What I meant is:

But over every ring, even with $ 2$ non-invertible, we can still show that

- the characteristic polynomial of $ A$ has its all coefficients before odd powers of $ X$ equal to zero if $ p\plus{}q$ is even;
- the characteristic polynomial of $ A$ has its all coefficients before even powers of $ X$ equal to zero if $ p\plus{}q$ is odd.

  darij

Since the proof I posted was not strong enough to prove darij's conclusion, it would be interesting to see darij post his argument for that.

See
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202835

Ibrogimov you repeat yourselves

Here comes a sketch:

[i]Notations.[/i]

- Let $ p$ and $ q$ be two nonnegative integers. A matrix will be called [i]$ \left(p,q\right)$-antidiagonal[/i] if it is a block matrix of the form $ \left(\begin{array}{cc}0_p & B \\
C & 0_q\end{array}\right)$, where $ 0_k$ denotes the $ k\times k$ zero matrix for every $ k\in\mathbb{N}$. Of course, if an $ n\times n$ matrix is $ \left(p,q\right)$-antidiagonal, then $ p \plus{} q \equal{} n$.

- If $ A$ is an $ n\times n$ matrix, and $ i_1$, $ i_2$, ..., $ i_k$ are integers with $ 1\leq i_1 < i_2 < ... < i_k\leq n$, then we denote by $ A\left[\frac {i_1,i_2,...,i_k}{i_1,i_2,...,i_k}\right]$ the submatrix of $ A$ formed by the intersections of the columns numbered $ i_1$, $ i_2$, ..., $ i_k$ and the rows numbered $ i_1$, $ i_2$, ..., $ i_k$.

Now it is known that the characteristic polynomial of any $ n\times n$ matrix $ A$ is

$ \det\left(XI_n \minus{} A\right) \equal{} \sum_{k \equal{} 0}^n\left(\left( \minus{} 1\right)^kX^{n \minus{} k}\cdot\sum_{1\leq i_1 < i_2 < ... < i_k\leq n}\det\left(A\left[\frac {i_1,i_2,...,i_k}{i_1,i_2,...,i_k}\right]\right)\right)$.

So it remains to prove that, for any $ \left(p,q\right)$-antidiagonal $ n\times n$ matrix, we have

$ \det\left(A\left[\frac {i_1,i_2,...,i_k}{i_1,i_2,...,i_k}\right]\right) \equal{} 0$ for any integers $ i_1$, $ i_2$, ..., $ i_k$ with $ 1\leq i_1 < i_2 < ... < i_k\leq n$ for any odd $ k$.

But this is easy: The matrix $ A\left[\frac {i_1,i_2,...,i_k}{i_1,i_2,...,i_k}\right]$ itself is $ \left(p^{\prime},q^{\prime}\right)$-antidiagonal for some integers $ p^{\prime}$ and $ q^{\prime}$ with $ p^{\prime} \plus{} q^{\prime} \equal{} k$. And $ k$ is odd, so we must have $ p^{\prime}\neq q^{\prime}$. So it remains to show that if $ S$ is a $ \left(u,v\right)$-antidiagonal matrix with $ u\neq v$, then $ \det S \equal{} 0$. This is easy: Write $ S$ as a block matrix of the form $ \left(\begin{array}{cc}0_u & B \\
C & 0_v\end{array}\right)$, and conclude as you wish (e. g. by $ \det S \equal{} \sum_{\sigma\in S_n}...$ and trivial combinatorics, or prove it over fields by dimension arguments and extend to rings by algebraicity).

Please ask me if anything stays unclear (this is not really to you, Kent, as I think this was already too detailed for you).

  darij

[quote="loup blanc"]Hi Darij,
your result is in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202835[/quote]

Not completely - think of characteristic $ 2$. This is why I preferred to speak about coefficients instead of just saying that the polynomial is even resp. odd.
Of course, all arguments can be extended to arbitrary rings by the polynomial identity argument. (Nice proof for $ p\equal{}q$, by the way - I didn't think of using the Schur complement here.)

  darij

$ A$ is (in an appropriately general sense) the adjacency matrix of a bipartite graph, and bipartite graphs have the property that they do not admit closed walks of odd length, so $ \sum \lambda_i^k \equal{} 0$ whenever $ k$ is odd, from which it follows that $ \sum (\minus{}\lambda_i)^k \equal{} \sum \lambda_i^k$.  (This is not intended to be a proof of the general case.)


This should be moved to highschool basics

[quote="ntellis"](logn)^3 < n^2
please help me!!!!!! :blush:[/quote]

i concur with tinytim about the moving. Welcome to Mathlinks!

i'm not sure...i did  all $ x\in \mathbb{R}$, i think.

[hide]$ (\log n)^3 < n^2$

 if $ 10^x \equal{} n$ then the above translates to $ x^3 < 100^x$.

If we can prove this for $ x\leq 0$ and $ x>0$ then we have proof.

if $ x<0$ then $ x^3<0$ while $ 100^x>0$. and if $ x\equal{}0$ we can do that manually.

If we can prove that the expression holds for $ 0<x<1$, and that if the expression holds for $ n$ then it holds for $ n\plus{}1$, we have proof.

Part 1:
$ 0<x^3<1$ but $ 1<100^x<100$

so the expression holds for $ 0<x<1$.

Part 2.1:
If $ 0<n<1$ then

$ 0<3n<3$ and $ 0<3n^2<3$.

Also, $ 99<99\cdot 100^n<9900$.

Therefore $ 3n\plus{}3n^2\plus{}1<99$. and $ 3n\plus{}3n^2\plus{}1<99\cdot 100^x$.

$ \therefore n^3\plus{}3n\plus{}3n^2\plus{}1<100(100^n)$

$ \boxed{(n\plus{}1)^3<100^{(n\plus{}1)}}$.

Part 2.2

If $ n>1$...

$ n^3<100^n$

$ \therefore 7n^3\plus{}1<100\cdot 100^n$. We are sure about the 1 part since $ 100^n>1$ and we have sufficiently many of those.

since $ n>1, n<n^2<n^3$ therefore $ n^3\plus{}3n^2\plus{}3n\plus{}1<7n^3\plus{}1<100\cdot 100^n$, or

$ \boxed{(n\plus{}1)^3<100^{(n\plus{}1)}}$

With this proved, we are sure we have touched upon all the real numbers.... how do you do imaginary?[/hide]

Not being harsh or anything but... you haven't proved anything...  :ninja: The reason is that a proof by induction is proving the statement for positive integers, not real numbers. If you want to prove it for all real numbers, you don't use induction.
What I recommend is that:

First: prove the assertion for all positive integers $ n$ by induction.

Second: try to prove the assertion for all real numbers (assume continuity), so just give a good-enough analytical proof.

EDIT: I forgot to say that since the base of the logarithm is not specified by the OP, you can assume that the base is the standard base e or use 10. It doesn't matter, really, you can even prove it for any base $ b > 1$.

Correct me if I'm wrong, but I didn't see induction being used in the solution anyways.

Induction works like this:

Criteria:

I. The statement is true for 1.
II. If the statement is true for k, then it is also true for k+1.

If BOTH criteria are met, then the statement is true for all positive integers. (I agree with 10000th user.)


In how many ways can 10 be expressed as a sum of 5 nonnegative integers, when order is taken into account?

Display $ 10$ as $ 10$ consequent circles:
\[ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc .
\]
Now the task becomes that: in how many ways we can place $ 4$ different objects between these circles. For example,
\[ \star \bigcirc \bigcirc \bigcirc \star \bigcirc \star \bigcirc \bigcirc \bigcirc \bigcirc \star\bigcirc \bigcirc
\]
is the represantation of $ (0; 3; 1; 4; 2)$ solution. So the number of ways is $ \boxed{\binom{14}{4}}$.

Moreover, for the generalisation of the task we get $ \binom{n \plus{} k \minus{} 1}{k \minus{} 1}$ if we must find the number of solutions if $ n$ is a sum of $ k$ nonnegative integers.

Wait you put four because we need to put 4 slits for 5 integers right.

Divider argument!

We have 10 things and we wish to put dividers in them. Since we may put dividers in the same spot, we should consider doing 15 as 5 positive integers; we just subtract 1 from each positive integer to get five non-negative integers that add up to 10 and vice versa.

15 as 5 positive integers: we have 14 slits and we want to put 4 things in them $ \binom{14}{4}\equal{}1001$.

Forgive me, but what does that $ \text{H}$ stand for in $ _n\text{H}_k$?

Never seen that notation before...

Permutation:$ {}_nP_r\equal{}\frac{n!}{(n\minus{}r)!}$

Combination:$ {}_nC_r\equal{}\frac{n!}{r!(n\minus{}r)!}$

Repeated Permutation:$ {}_n\Pi_r\equal{}n^r$

Repeated Combination:$ {}_nH_r\equal{}{}_{n\plus{}r\minus{}1}C_r$

H of the initial is come from by [b]Homogeneous product[/b].


calculus
integration
search
function
probability
conditional probability
calculus computations

Hey there,
I am going through my course notes for a combinatorics class and found the example:
By using the fact that $ \int_0^1 x^i dx \equal{} \frac{1}{i\plus{}1}$ show that
a) $ \sum_{i\equal{}0}^n \frac{1}{1\plus{}i}\binom{n}{i}\equal{}\frac{1}{n\plus{}1}\{2^{n\plus{}1}\minus{}1\}$
b) $ \sum_{i\equal{}1}^n \frac{(\minus{}1)^i}{i}\binom{n}{i} \equal{} \minus{}\sum_{k\equal{}1}^n \frac{1}{k}$
a) $ \sum_{i\equal{}0}^n \frac{2^{i\plus{}1}}{1\plus{}i}\binom{n}{i}\equal{}\frac{1}{n\plus{}1}\{3^{n\plus{}1}\minus{}1\}$

So far I got $ a) \sum_{i\equal{}0}^n \frac{1}{1\plus{}i}\binom{n}{i} \equal{} \sum_{i\equal{}0}^n \int_0^1 x^i dx\binom{n}{i} \equal{} \sum_{k\equal{}0}^n \sum_{i\equal{}0}^k \binom{k}{i}\frac{1}{n\plus{}1}$ which gives the right answer, but I sort have bsed it :( Help please =)

Also advice/tips on these sort of problems would be nice =)

I don't actually see what you did there, but the relevant manipulation is $ \sum_{i \equal{} 0}^{n} \int_{0}^{1} x^i \, dx {n \choose i} \equal{} \int_{0}^{1} \left( \sum_{i \equal{} 0}^{n} {n \choose i} x^i \right) \, dx \equal{} \int_{0}^{1} (1 \plus{} x)^n \, dx$.

The other problems are essentially the same trick.  Problems of this kind tend to be of the form "take a straightforward identity and integrate it to get nontrivial denominators"; see for example the last problem [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=259311]here[/url] or the various solutions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=515511477&t=242870]to this problem[/url].  I've become quite fond of integrating a generating function in two variables with respect to one variable.

Anyway, what I think is much more interesting is trying to find a probabilistic proof of, say, the first assertion, which is that the average value of $ \frac{1}{i \plus{} 1}$ over all subsets of a set with $ n$ elements (where $ i$ is the size of the subset) is $ \frac{2}{n\plus{}1} \left( 1 \minus{} \frac{1}{2^{n\plus{}1}} \right)$.  Since most subsets are concentrated near $ i \equal{} \frac{n}{2}$, one would indeed expect this average to be close to $ \frac{1}{ \frac{n}{2} \plus{} 1}$, but then the question is how to account for the error term.

Okay, here's a go at it.

Look at $ n\plus{}1$ objects $ A\equal{}\{a_0,a_1,\cdots,a_n\}$. Choose some subset $ S$ of $ A$ containing $ a_0$ at random, and then choose an element of $ S$ randomly. What is the probability that this element is $ a_0$?

Now we twist our perspective. Choose a nonempty subset $ S$ of $ A$ at random, and a random element $ x$ of $ S$. If $ S$ happens to contain $ a_0$, what is the probability that $ x\equal{}a_0$? The overall probability that $ x\equal{}a_0$ is of course $ \frac1{n\plus{}1}$, since the procedure without checking inclusion is the same as choosing a random element directly. We have $ a_0\in S$ with probability $ \frac{2^n}{2^{n\plus{}1}\minus{}1}$ since we had to exclude the empty set, and the conditional probability we seek is $ \frac1{n\plus{}1}\cdot\frac{2^{n\plus{}1}\minus{}1}{2^n}$.

if you just want some answers..

a) $ \sum_{i\equal{}0}^n \frac{1}{1\plus{}i}\binom{n}{i}\equal{}\sum_{i\equal{}0}^n \int_0^1 x^i dx\binom{n}{i}\equal{}\int_0^1 \sum_{i\equal{}0}^n \binom{n}{i}x^i\cdot 1^{n\minus{}i}dx$
$ \equal{}\int_0^1 (1\plus{}x)^ndx \equal{}\left[\frac{(1\plus{}x)^{n\plus{}1}}{n\plus{}1}\right]_0^1\equal{}\frac{1}{n\plus{}1}\{2^{n\plus{}1}\minus{}1\}$

b)$ \sum_{i\equal{}1}^n \frac{(\minus{}1)^i}{i}\binom{n}{i} \equal{}\minus{}\sum_{i\equal{}1}^n \int_{\minus{}1}^0 x^{i\minus{}1}dx \binom{n}{i}\equal{}\minus{}\int_{\minus{}1}^0 \sum_{i\equal{}1}^n \binom{n}{i} x^{i\minus{}1}\cdot 1^{n\minus{}i}dx$
$ \equal{}\minus{}\int_{\minus{}1}^0 \frac{(1\plus{}x)^n\minus{}1}{x}\equal{} \minus{}\int_0^1 \frac{u^n\minus{}1}{u\minus{}1}\equal{}\minus{}\int_0^1 \sum_{k\equal{}0}^{n\minus{}1} u^k dx\equal{}\minus{}\sum_{k\equal{}1}^n \frac{1}{k}$

c) $ \sum_{i\equal{}0}^n \frac{2^{1\plus{}i}}{1\plus{}i}\binom{n}{i}\equal{}\sum_{i\equal{}0}^n \int_0^2 x^i dx\binom{n}{i}\equal{}\int_0^2 \sum_{i\equal{}0}^n \binom{n}{i}x^i\cdot 1^{n\minus{}i}dx$
$ \equal{}\int_0^2 (1\plus{}x)^ndx \equal{}\left[\frac{(1\plus{}x)^{n\plus{}1}}{n\plus{}1}\right]_0^2\equal{}\frac{1}{n\plus{}1}\{3^{n\plus{}1}\minus{}1\}$


If you downloaded the 2006 Chapter Round off http://www.mathcounts.org, would you please post it here?  For the last few hours, I have been trying to get into MathCounts.org, but it seems that their site is down.  So if you already downloaded those 2006 Chapter (and School) Rounds, [i][b]please please please please please[/b][/i] post them here.  I have not yet had a chance to go over them.

Thanks!

Me too, If any of you have downloaded it, PLEASE POST IT.

And an answer key? (They are different :D )


algorithm
number theory
greatest common divisor
calculus
integration
Euclidean algorithm
Diophantine equation

[quote="myc"]I
" The Euclidean algorithm tells us that there are integer solutions to the Diophantine equation am + bn = c, where c is the greatest common divisor of a and b, and no solutions for any smaller c. " (aops wiki)

but $300g+210p=N$  GCD of 300 and 210 is 30.  but $300g+210p=30$  does not give intergral solutions. $10g+7p=1$
 how does g and p have intergral solutions?  I'm confused...

Thanks[/quote]

Yes it does...for the first one $(g,p)=(-2,3)$ and then $(g,p)=(5,-7)$

Edit: Oops. too late..


trigonometry
inequalities
function
geometry
inequalities proposed

Though it looks ugly it has a nice ending! :D :D

Prove that in any triangle we have:

$ \frac{\sum \sqrt[4n]{2a \cos \frac{A}{2}}}{\sqrt[4n]{3R}}+3p^{2}+3r^{2}\leq 6R^{2}+30Rr+6\sqrt{R(R-2r)^{3}}+3$.(standard notations)

cheers! :D

Should I wonder why no one looks at these problems? Dear Lagrangia, please do not change the name inequalities into " ugly and complicated things that do not say anything ". Just my opinion!

[quote="harazi"]Dear Lagrangia, please do not change the name inequalities into " ugly and complicated things that do not say anything "[/quote]

I have changed it, because I posted the wrong inequality! sorry!

[quote="harazi"]Should I wonder why no one looks at these problems? [/quote]

BTW, I would look at these mad-man crazy ineqs! :D :D :D :D

BTW do you create all these ineqs by your own? I wonder how it is possible to find such a crazy thing. :D

[quote="harazi"]I don't find anything hard in creating such problems. All you have to do is to use one million times known inequalities and then multiply or add them.[/quote]

They are not just well-known ineqs multiplied and added ... trust me! It's a lot more!

[quote="Sasha"]BTW do you create all these ineqs by your own? [/quote]

Yes, all the ineqs which have  source : Lagrangia, are created by me! :D :D

Maybe it's true what you say, Lagrangia (though I still do not understand you!), but I strongly believe that  the following  inequalities represent a fascinating part of mathematics:
  1. Nice-looking inequalities, even easy, but with beautiful ideas.
  2. Nasty-looking inequalities (though not so nasty!) if they say something, if they refine some other inequalities, if they are the best inequalities of their type or they have splendid ideas and super-short solutions, not obtained applying a trillion times convexity and well-known inequalities, then multiplying and adding them or if they can be applyed in other mathematical arguments.
     Unfortunately, I cannot see the category to which these inequalities belong. Of course, it's just my opinion and it's not a classification nor a necessity that inequalities are only like that. To see that I'm flexible, I will create a third category: those created by Lagrangia!  :)

[quote="darij grinberg"]For the rest of you who do not know how such monsters like the inequality above are created:

[url=http://hydra.nat.uni-magdeburg.de/math4u/var/PT1.html]This site[/url] lists the parametrizations of symmetric functions of a, b, c through R, r, p (aka s).

  dg[/quote]

WOW!magnificient stuff there! the site is in German! Danke fuer es Darij!

It's actually not really in German, at least the formulas... but the notations are german: the angles of the triangle are usually called $\alpha$, $\beta$, $\gamma$, the semiperimeter is s, and the area is $\Delta$.

  dg

l think it's useless to prove such a weird inequality
does this ineq have an concrete application in a bigger problem or it is an school exercice ?
l am not able to solve it anyway

[quote="arthur"]
does this ineq have an concrete application in a bigger problem or it is an school exercice ?
[/quote]

this is a problem which is not from a bigger problem, nor from a school exercise!!

[quote="arthur"]
 think it's useless to prove such a weird inequality
[/quote]

All ineqs are weird! :)


inequalities calculus derivative algorithm

I have a TI-82+ calculator and I need to graph a polar inequality in the form (some expression) > 0.  How do I do this?

Well first get it to polar form and then graph it (use mode to change it to polar, if you didn't know that already).  The shaded part is basically just all the stuff enclosed (for most graphs)...

The problem is that it graphs stuff in the form

r = bla bla bla theta,

but I need

0 = r bla bla bla theta

It's not easy to put it into r = something form becuase of the (0.62r)^1000 in the inequality.

correction: It should be TI-83+, not 82.

(cos(theta - r) - sin(theta))(r^2 - 2r^2 cos(2theta + 2.4) + 0.9) + (0.62r)^1000 < 0.  Graphing this is supposed to give a ying-yang.   :D It works on the graphing calculator on my computer, but I can't do it on my calculator.

Your chances of getting that into a good form are slim to none, unfortunately.  TI's won't do implicit graphing.

TI-89s can graph implicitly, but the results are very bad.

[quote="zabelman"]I did some funky stuff with mathematica, but I'm not gettin a yin-yang.[/quote]

And, by the way, yin-yang is the correct spelling.

Implicit graphing is HARD in some set-ups.

Can a TI-89 do implicit differentiation?

Yes, you can do implicit differentiation with a TI-89.

Say you want to differentiate 25x^2+y^2=109.
Replace y with y(x) and type in:
[code]d(25x^2+y(x)^2=109,x)[/code]
It outputs:
[code]2*y(x)*d(y(x),x)+50*x=0[/code]
Replace d(y(x),x) with z and y(x) with y and type in:
[code]solve(2*y*z+50*x=0,z)[/code]
Make sure you solve for z, not x. The calculator outputs:
[code]z=-25*x/y[/code]
This is the answer.

I don't know if there's a simpler algorithm, but this one seems to work.


I was given this problem in my math class.
How would I attempt to solve it?

The circumference of a circle can be expressed as C = (pi)d, where d is the diameter of the circle.

Therefore, C/d = pi.

Therefore, pi is a ratio...So why is pi considered an irrational number?

Irrational numbers are those that can't be expressed as the ratio of two [b]integers.[/b]  To say that $ \pi$ is irrational is the same thing as saying that you can't construct a circle whose circumference and diameter are both integers.

Anyway, I don't know of any such proof that's appropriate for this forum.  The standard proofs I know of all require some fairly delicate calculus.

A rational number is defined as a ratio of two integers, not just two any numbers. Or else something like pi/2 would be considered rational.

I can't think right now how to solve for it, but maybe assume that pi is rational and therefore pi=p/q, where p and q are integers and go somewhere from there.

There is a proof of this [url=http://www.lrz-muenchen.de/~hr/numb/pi-irr.html]here[/url], that I don't understand at all and hope is valid :)

There's [url=http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational]two Wikipedia proofs[/url] which I don't understand at all either. :P

[quote="pascal12"]Ah, so most proofs are slightly too complicated for Middle School Forum because they require calculus.[/quote]

No, only SOME proofs require calculus. However, proofs in general are too hard for middle school forums.

[quote="t0rajir0u"][quote="moogra"]No, only SOME proofs require calculus. [/quote]
I'm reasonably certain pascal12 meant "most proofs of this result."  As far as I'm aware, [b]all[/b] proofs of this result "require calculus."  To the extent that you can't even define $ \pi$ formally without analysis, you can't really hope to escape this requirement.[/quote]

Yeah, sorry for the bad sentence. I meant most proofs for the irrationality of pi require calculus.


function
algebra proposed
algebra
Function equations

[quote="Raja Oktovin"]Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
\[ x \plus{} f(y) \equal{} af(y \plus{} f(x))\]
for all real numbers $ x$ and $ y$.
[i]Hery Susanto, Malang[/i][/quote]
Very easy.
We have $ a$ not equal $ 0$.
Easy prove that $ f$ be surjectivity
So there exist $ m$ such that $ f(m)\equal{}0$
Let $ x\equal{}m$ we have: $ f(y)\equal{}af(y)$
So $ a\equal{}1$

yes, this is too easy, but i spot some flaws in your argument. even though it is obvious to fix, too  :lol: .

1. $ a \ne 0$
2. find that $ f$ is surjective
3. find that $ f$ is injective
4. thus there is one and only one $ x_0$ such that $ f(x_0)\equal{}0$
5. find that $ x_0\plus{}f(y)\equal{}af(y)$
6. if $ a \ne 1$, $ f$ is constant, impossible
7. thus $ a\equal{}1$, example $ f(x)\equal{}x$.

whole solution
if $a=0$,easy to disprove
if $a$ isn't zero then f is surjective
it's easy to obtain
$f(af(x))=\frac{x}{a}$
so $f((a-1)f(x)\equiv 0$
hence $f(x)\equiv 0$,contradiction!
except for $a=1$
so the only solution is $a=1$

Here is my solution for this problem
[b]Solution[/b]
$x + f(y) = af(y + f(x))$
It's easy to see that $a = 0$ doesn't satisfies the problem
With $a \ne 0$: Let $y = 0$, we have: $f(f(x)) = \dfrac{x + f(0)}{a}$
So: $f$ is surjective
Then there exist $b \in \mathbb{R}$ which satisfies $f(b) = 0$
Let $x = b$, we have: $(1 - a)f(y) = b$
If $a \ne 1$ then: $f(x) = c, \forall x \in \mathbb{R}, c \in \mathbb{R}$
Retry, we see that: $f(x) = c$ doesn't satisfies the problem
If $a = 1$ then it's easy to see that $f(x) = x, \forall x \in \mathbb{R}$ satisfies the problem
In conclusion, we have: $a = 1$ satisfies the problem

$\spadesuit \color{green}{\textit{\textbf{ANS:}}}$ $a=1$.
$\clubsuit \color{red}{\textit{\textbf{Proof:}}}$ Let $P(x,y)$ denote the given assertion, we have \[P(x,f(0)): x+f(f(0))=af(f(x))\] \[P(0,f(x)): f(f(x))=af(f(x))\] so $f(f(x))=x+f(f(0))$ and \[P(x,0): x+f(0)=af(f(x))=ax+af(f(0))\] which it's now easy to see that $a$ must be $1$ in order for the equation to hold for all real $x$. In this case, $f(x)=x$ fits the given FE. $\quad \blacksquare$

Let $P(x,y)$ denote the assertion $x+f(y)=af(y+f(x))$.
$P(x,f(y))\Rightarrow x+f(f(y))=af(f(x)+f(y))$ and switching $x$ and $y$, $x+f(f(y))=y+f(f(x))$. Thus, $f(f(x))=x+f(f(0))$ and $f$ must be bijective.
$P(x,0)\Rightarrow x+f(0)=af(f(x))=ax+af(f(0))$ and letting $x$ vary, we see that $a$ must be $\boxed1$. In this case, $f(x)=-x$ works.

Here is the solution of the resulting FE:
Let $P(x,y)$ denote the assertion $x+f(y)=f(y+f(x))$.
$P(x,0)\Rightarrow f(f(x))=x+f(0)$, but, as before, $f(f(x))=x+f(f(0))$, so we get $f(f(0))=f(0)$, and using injectivity, $f(0)=0$. Then $f$ is an involution.
$P(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$ and any additive involution works.
Let $A$ and $B$ be any two supplementary subvectorspaces of the $\mathbb Q$-vectorspace $\mathbb R$. Let $a$ from $\mathbb R\to A$ and $b$ from $\mathbb R\to B$ be the two projections of any real $x$ in $(A,B)$. We get $\boxed{f(x)=a(x)-b(x)}$.

Denote the assertion by $P(x,y).$ Clearly $a\neq 0$ then $P(x,0)$ gives $x+f(0)=af(f(x)).$ Hence $f$ is bijective. If $f(x_0)=0$ then $P(x_0,x)$ implies $f(x)(a-1)=x_0$ and since $f$ is not constant, $a=1.$

[quote=Raja Oktovin]Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ x\plus{}f(y)\equal{}af(y\plus{}f(x))\] for all real numbers $ x$ and $ y$.
[i]Hery Susanto, Malang[/i][/quote]
$$x+f(y)=af(y+f(x))...(\alpha)$$
$\color{blue}\boxed{\textbf{Answer: a=1 is the only solution}}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{If a=0:}}$
$\color{red}\rule{24cm}{0.3pt}$
In $(\alpha):$
$$\Rightarrow x+f(y)=0, \forall x,y\in\mathbb{R} (\Rightarrow \Leftarrow)$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{If a=1:}}$
$\color{red}\rule{24cm}{0.3pt}$
$$\Rightarrow f(x)=x \text{ is a solution}$$
$$\Rightarrow \boxed{\text{a=1 is a solution}}$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{If a}\neq \textbf{0, 1:}}$
$\color{red}\rule{24cm}{0.3pt}$
In $(\alpha):$
$$\Rightarrow \frac{x+f(y)}{a}=f(y+f(x))$$
$$\Rightarrow \text{f is suryective}...(I)$$
By $(I):$
$$\Rightarrow \exists b/f(b)=0$$
In $(\alpha) x=b:$
$$\Rightarrow b=(a-1)f(y)$$
$$\Rightarrow f(y)=\frac{b}{a-1}$$
$$\Rightarrow \text{f is constant}$$
In $(\alpha):$
$$\Rightarrow \text{x is constant} (\Rightarrow \Leftarrow)$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{blue}\rule{24cm}{0.3pt}$
$\color{blue}\boxed{\textbf{Conclusion:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$$\Rightarrow \boxed{\textbf{a=1 is the only solution}}_\blacksquare$$
$\color{blue}\rule{24cm}{0.3pt}$

[quote=Raja Oktovin]Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ x\plus{}f(y)\equal{}af(y\plus{}f(x))\] for all real numbers $ x$ and $ y$.
[i]Hery Susanto, Malang[/i][/quote]

Clearly no sol for $a=0$, now i claim that $a=1$, assume FTSOC that $a \ne 1$, now let $P(x,y)$ the assertion of the following F.E.
Notice that the function is surjective as $a \ne 0$, by $P(c,c)$ we get $c=0$ so $f(0)=0$, now by $P(x,0)$ we get $af(f(x))=x$ so $f$ is bijective, now note that by $P(f(x),y)$ we get that.
$$f(x)+f(y)=af(y+x \cdot a^{-1}) \implies f(y+x \cdot a^{-1})=f(x+y \cdot a^{-1})$$
setting $x=0$ there and then induction we can get $f(x)=f(x \cdot a^n)$ for all $n \in \mathbb Z$, now $af(f(ax))=ax$ so $f(f(x))=x$ hence $a=1$ so $f$ is involutive and additive (any function of this kind works), thus we are done :D.

let f constant Obviously it is wrong since
$x+k = ak$
$x= k(a-1)$ implies x also constant
now let $P(x,y)$ be the assertion
$P(x,y) = x+f(y) = af(y+f(x))$
$P(x-f(y),f(y)) = x=af(y+f(x-f(y)))$
if $a = 0$ x constant contradiction so a is nonzero and f surjective
so set some b so that $f(b) = 0$ due to the surjectivity
$P(b,y) = b+f(y) = a*f(y)$
$P(b,y) = b=f(y)(a-1)$ f is non constant but b constant implies that $(a-1) = 0$ forcing $b=0$
so $a=1 , f(0)=0$


function
integration
trigonometry
calculus
calculus computations

A very old question(forgive me if it has been posted many times):It is known that many original functions of elementary functions such as $\int e^{-x^{2}}dx,\int \frac{\sin x}{x}dx$ can not be represented by elementary functions,but how to prove it?Thanks!

Seems like I am becoming a specialist in answering this particular question :(. Yes, it has been posted at least 4 times by now. I'm too lazy to look up those other posts but here is the link I gave previous times. It is still good: [url]http://math.hunter.cuny.edu/ksda/papers/churchill.pdf[/url]. If you have any questions about the article, feel free to ask them in this thread :)


calculus
derivative
function
linear algebra
matrix
integration
calculus computations

If  $ x\equal{}x\left ( u,v \right ) \ , \  y\equal{}y\left ( u,v \right )$
Could someone point out how to prove  $ \frac{\partial (x,y)}{\partial (u,v)}\cdot \frac{\partial (u,v)}{\partial (x,y)}\equal{}1$ ?

Thats a property of a Jacobian. If $ J\equal{} \frac{\partial (u,v)}{\partial(x,y)}$ and $ J'\equal{}\frac{\partial(x,y)}{\partial(u,v)}$ then, $ JJ'\equal{}1$.

Let $ u \equal{}f(x,y) , v\equal{} g(x,y)$ solving for x and y we get $ x\equal{} \phi(u,v) , y\equal{}\psi(u,v)$.

Write $ \frac{\partial u}{\partial u} \equal{} 1\equal{}\frac{\partial u}{\partial x}.\frac{\partial x}{\partial u}\plus{}\frac{\partial u}{\partial y}.\frac{\partial y}{\partial u}$

Similarly writing for $ \frac{\partial u}{\partial v}\equal{}0$ ,  $ \frac{\partial v}{\partial u}\equal{}0$ ,  $ \frac{\partial v}{\partial v}\equal{}1$ as above .

Using the definition of Jacobians , $ JJ' \equal{}\left | \begin{array}{cc}1 & 0 \\ 0 & 1 \\ \end{array}\right| \equal{}1$

I'm still a little unclear about what the expression  $ \partial (x,y)$ or $ \partial (u,v)$ mean.
Does the expression  $ \partial (x,y)$  mean $ \frac{\partial x}{\partial u}$ or $ \frac{\partial y}{\partial u}$ ?
Does the expression  $ \partial (u,v)$  mean $ \frac{\partial u}{\partial x}$ or $ \frac{\partial v}{\partial y}$ ?
Or they have another meanings ?

Consider a function $ F: R^{n} \to R^{m}$. Let the component functions be

$ y_{1}\equal{}f_{1}(x_{1},x_{2},.....,x_{n})$
$ y_{2}\equal{}f_{2}(x_{1},x_{2},.....,x_{n})$
$ \vdots$
$ y_{n}\equal{}f_{n}(x_{1},x_{2},.....,x_{n})$.

We define the Jacobian matrix to be the a matrix which consists of elements as all partial derivatives of the component function as

$ J\equal{} \begin{bmatrix} \frac{\partial y_{1}}{\partial x_{1}} 	\ldots \frac{\partial y_{1}}{\partial x_{n}} \\ \vdots \ddots \\ \frac{\partial y_{m}}{\partial x_{1}} \ldots \frac{\partial y_{m}}{\partial x_{n}}\end{bmatrix}$

Consider the change of variable $ x \equal{}x(u,v)$ , $ y\equal{}y(u,v)$ which has continuous partial derivatives on $ R^{*}$ in the $ u\minus{}v$ plane.

$ \int {\int_{R}}f(x,y).dx .dy \equal{} \int {\int_{R^{*}}}f(x(u,v),y(u,v)) \left | \frac{ \partial (u,v) }{\partial (x,y)} \right | . du.dv$


As you can see in the picture, there is a venn diagram, the notation A v B represents the entire region covered by both sets A and B.  But I don't understand this, does it mean A v B is the two circles?
Thanks.

Um, its the intersdection of the two circles. If circle a contains the numbers 1, 2, 3, 4, 5, 6, and 7 and circle b contains the numbers 2, 4, 6, 8, 10, for example, then the blue part would have: 1, 3, 5, 7, the yellow part would have 8, 10, and the green part would have 2, 4, 6, the parts where both sets intersect. This example is for sets, but there are other uses of venn diagrams as well, but they follow the same principle.

$A \cup B$ is the union of two circles, and this includes both circles, but the intersection is not to be counted twice.

However, in your picture, you have $A \cap B$, which is the intersection as superwizard188 described...

the hat looks like an A so i think of and which means the intersection

V is the opposite

you also have $\left(A \bigcup B\right)^{C}$ which is the gray area around the sets...

BTW, i think set A in the diagram contains the green part and set B also contains the green part. But the venn diagram that i see in some actual problems, 'A' usually doesn't contain the green part, for an example,if  A=14, green part is 3, then 3 doesn't include in 14.if B=10, then it also doesn't contain 3. So I am really confused about in the actual question whether A,B which both represent two number contains the central overlapped part or not.


Our school got a letter about a competition called the Problem Solving Challenge or National High School Mathematics Championships or something (the letter was quite vague).  The National competition is held on Oklahoma City on June 26,27.  No one seems to know anything about this competition, and I've never heard of it, yet the problems seem quite good.  Help?

[quote="Caroline"]Hmm.. I just went on google and found http://www.ascm.org/ , which seems like maybe what you're talking about?  I don't know; it's 2:40 AM by me, which is a little too late to actually follow up on this.  Hope that helps a little =/ If you find out anything good, let me know ;)[/quote]

This sounds interesting.  The prizes sound good, too (an intership at Lawrence Livermore, a TV, or a calculator?).  The mailing list sounds good - I thinK I'm gonna register my school.

Something gives me the impression that this contest/organization is a one-man operation, or something close to it.  But then again, you don't get sponsorship from Lawrence Livermore without connections.

I know a bit about this competition since someone from my high school won it last year. He told me that the questions are very easy, but the competitors are treated really well, and I think he enjoyed the experience. One strange thing was that there was a relay round whose time limit was an hour. (I think it's a bit like the Super Relay at ARML.) He was the third person and guessed the answer after a few minutes without getting anything passed back.

This sounds pretty cool.. Does anyone know around how many people usually attend?  (This is probably too far away for my school to go to, but maybe..)

Yes, I noticed Jon Lowd (not Lewd!)'s name on the list.  I assume he paid for travel expenses, lodging, and whatnot, right?

I wish I could go, but it'd mean traveling to Oklahoma from Hawaii.  I'll mention it to Neal, though (after all, he lives in Oklahoma), and he can take a shot at winning.

[quote="ComplexZeta"]I don't think he had to pay anything to go. And I think he said there were about 15 or 20 people there.[/quote]

Wha?  How did he get invited?

[quote="ComplexZeta"]One person from each school was allowed to go. The schools supposedly were to have participated in some contest, but we didn't take it, so we just said that he won it. He was supposed to go to that one, and I was supposed to go to the Mu Alpha Theta one, but we missed the deadline for Mu Alpha Theta, so I didn't get to go. :([/quote]

One from every school in America?  Or does the school have to be a member of the ASCM?

[quote="ComplexZeta"]He told me that the questions are very easy..[/quote]
     I saw a couple of questions from them (the school-level questions) and they seemed to be fairly good (perhaps a step down from AMC 12).  I thought that the National questions would be proportionately better.  Just how easy are they?

How should I answer that question? What does it mean to ask how easy they are? I'd usually say 11 or some other pseudorandom number.

You could relate them to other competitions.

Has anyone else qualified for this event?  I looked into it after hearing about it here, got my school registered, took their exam, and apparently qualified.   Dr. Neal sent me a letter with some information, but I'm still not sure about it - how many people participate?


calculus
integration
function
logarithms
calculus computations

Evaluate
$ \int\limits_{ \minus{} \infty }^{ \plus{} \infty }{dx\int\limits_{\sqrt [n \minus{} 1]{x}}^{b}{\frac {dy}{\left( {{y}^{n}} \plus{} c \right)}}} \equal{} 2\int\limits_{0}^{ \plus{} \infty }{dx\int\limits_{\sqrt [n \minus{} 1]{x}}^{b}{\frac {dy}{\left( {{y}^{n}} \plus{} c \right)}}}$

I'm not so sure about it.
it is clear that n is even, since $ x$ could be negtive. so will be the function $ \frac{1}{y^n\plus{}1}$.
if the equation is true, function $ f(x)\equal{}\int_{x^\frac 1{n\minus{}1}}^b \frac{1}{y^n\plus{}1} d x$ satisfy the following:
$ \int_0^\infty f(x) dx\equal{}\int_{\minus{}\infty}^0 f(x) dx$.
could this hold for n is even?

Yeah, I edit my post
Evaluate
$ \int\limits_{0}^{\plus{}\infty }{dx\int\limits_{\sqrt[2k\minus{}1]{x}}^{\plus{}\infty }{\frac{dy}{\left( {{y}^{2k}}\plus{}c \right)}}}$

$ \int_{\minus{}\infty }^{\plus{}\infty }{dx\int_{\sqrt [n\minus{}1]{x}}^{b}{\frac{dy}{\left({{y}^{n}}\plus{}c\right)}}}\equal{} 2\int_{0}^{\plus{}\infty }{dx\int_{\sqrt [n\minus{}1]{x}}^{b}{\frac{dy}{\left({{y}^{n}}\plus{}c\right)}}}$

The solution:
$ I\equal{}\int\limits_{{{a}^{n\minus{}1}}}^{{{b}^{n\minus{}1}}}{dx\int\limits_{\sqrt[n\minus{}1]{x}}^{b}{\frac{dy}{\left( {{y}^{n}}\plus{}c \right)}}}\equal{}\frac{1}{n}.\ln \left| \frac{{{b}^{n}}\plus{}c}{{{a}^{n}}\plus{}c} \right|$

$ \Rightarrow \int\limits_{\minus{}\infty }^{\plus{}\infty }{dx\int\limits_{\sqrt[2k\minus{}1]{x}}^{\plus{}\infty }{\frac{dy}{\left( {{y}^{2k}}\plus{}c \right)}}}\equal{}2\int\limits_{0}^{\plus{}\infty }{dx\int\limits_{\sqrt[n\minus{}1]{x}}^{\plus{}\infty }{\frac{dy}{\left( {{y}^{2k}}\plus{}c \right)}}}\equal{}0$


algebra
polynomial
linear algebra
linear algebra unsolved

Here is a nice problem:

If $ A,B \in \mathcal{M}_{n}(\mathbb R)$ satisfying $ B^2 \equal{} I_{n}$ and $ A^2 \equal{} AB \plus{} I_{n}$ prove that:

$ \det(A)\leq \left( \frac {1 \plus{} \sqrt {5}}{2} \right)^n$

If we use the given relation in a proper manner, I think we get that $ A$ satisfies $ p(A) \equal{} 0$, where $ p$ has roots $ \frac{\pm 1 \pm \sqrt 5}2$. It's easy from there...

What's funny about this problem is the totally unexpected official solution.
[hide]Twas something about Fibonacci, wasn't it?[/hide]

Should that be $ \mathcal{M}_2(\mathbb{R})?$ Or, alternatively, should the conclusion be that $ \det A\le \left(\frac{1\plus{}\sqrt{5}}2\right)^n\ ?$

The thing is, we can just take $ B\equal{}I$ and $ A\equal{}\left(\frac{1\plus{}\sqrt{5}}2\right)I.$

Yeah, the exponent should be $ n$. I could swear it was in place :blush: It's just one of those mechanical mistakes...

[quote="perfect_radio"]If we use the given relation in a proper manner, I think we get that $ A$ satisfies $ p(A) \equal{} 0$, where $ p$ has roots $ \frac {\pm 1 \pm \sqrt 5}2$. It's easy from there...

What's funny about this problem is the totally unexpected official solution.
[hide]Twas something about Fibonacci, wasn't it?[/hide][/quote]

I solved it using characteristic polynomial of $ A^2$ after manipulatin the initial conditions.I posted this problem after i had seen the official solution .I would have never guessed something like that. :D

[quote="perfect_radio"]If we use the given relation in a proper manner, I think we get that $ A$ satisfies $ p(A) \equal{} 0$, where $ p$ has roots $ \frac {\pm 1 \pm \sqrt 5}2$. [/quote]

$ A^4\minus{}3A^2\plus{}I\equal{}0$

$ A^2$ is diagonalisable and spectrum(A^2) is included in

3/2 + sqrt{5}/2= 1+phi  ;  3/2 - sqrt{5}/2

$ det(A^2) \leq (1\plus{} \phi )^{n}$

$ |det(A)|\leq {\phi }^{n}$


Let be $ a,b,c>0$ . Prove that :

$ \left(\frac {3a}{a \plus{} 2b \plus{} c}\right)^{\frac {3}{4}} \plus{} \left(\frac {3b}{2a \plus{} b \plus{} c}\right)^{\frac {3}{4}} \plus{} \left(\frac {3c}{4a \plus{} 4b \plus{} c}\right)^{\frac {3}{4}}\geq2$


quadratics
counting
distinguishability
algebra open
algebra

If roots of the equation $x^3 - 9x - 9 = 0$ are $a,b,-(a+b)$ then express $a$ as a quadratic expression in $b$ and vice-versa.

[quote="rohitsingh0812"]If roots of the equation $x^3 - 9x - 9 = 0$ are $a,b,-(a+b)$ satisfying $a<b<0$ then express $a$ as a quadratic expression in $b$ and vice-versa.[/quote]
[size=150]a, b satisfy $a<b<0$[/size]

May be the substitution  $x = \sqrt{12} cos(t)$ would help.

I think just a little bit of equation theory will do the trick:
[hide]$(x-a)(x-b)(x+a+b)=x^3-9x-9\\ x^3-x(a^2+ab+b^2)+(a^2b+ab^2)=x^3-9x-9\\ a^2+ab+b^2=a^2b+ab^2\\ a^2(1-b)+a(b-b^2)+b^2=0\\ a=\displaystyle\frac{b^2-b\pm\sqrt{2b^3+b^4-3b^2}}{2-2b}\\ a=\displaystyle\frac{b^2-b\pm b\sqrt{2b+b^2-3}}{2-2b}\\$

And since $a$ and $b$ are indistinguishable, they are related to each other in this manner in either direction ($a$ to $b$ or $b$ to $a$).[/hide]

[quote="gopherhole112"]
$(x-a)(x-b)(x+a+b)=x^3-9x-9\\ x^3-x(a^2+ab+b^2)+(a^2b+ab^2)=x^3-9x-9\\ a^2+ab+b^2=a^2b+ab^2$[/quote]

I think $a^2+ab+b^2=9,a^2b+ab^2=-9$
So maybe a little error.
And I think we can soluted $a,b$


[hide="Hint"]Write $ \sqrt {2} \equal{} 2\cos\frac {\pi}{2^2}$, then $ \sqrt {2 \plus{} \sqrt {2}} \equal{} 2\cos\frac {\pi}{2^3},\cdots$, etc.[/hide]


How do I make a blog on the AoPS server, there was a link floating around but when I clicked on it, it didn't do anything

Click on "Profile" in the menu at the top, and then there should be blog controls/creation somewhere.


vector
trigonometry
modular arithmetic
rotation
geometry
geometric transformation
reflection

Let be given the sequence $ (x_{i})_{i\equal{}1}^{n}$ of real numbers where $ n\geq 2$. Under what conditions on the sequence $ (x_{i})_{i\equal{}1}^{n}$ are the following vectors linearly independent: $ (sinx_{1}, ..., sinx_{n}), (cosx_{1}, ..., cosx_{n}), (1, ..., 1)$?

i know that we have to have at least two different values in the sequence, but it is not enough.

the vectors are linearly dependent iff there are constants $ a,b,c$ not all 0, such that $ a\cos x_i \plus{} b\sin x_i \equal{} c$ for all $ i$. if $ a,b,c$ are not all 0, then $ a$ and $ b$ cannot both be 0. so we may write $ \frac {a}{\sqrt {a^2 \plus{} b^2}}\cos x_i \plus{} \frac {b}{\sqrt {a^2 \plus{} b^2}}\sin x_i \equal{} \frac {c}{\sqrt {a^2 \plus{} b^2}}$. since $ \frac {a}{\sqrt {a^2 \plus{} b^2}}$ is in $ [\minus{}1,1]$, there's a $ y$ such that $ \sin y \equal{} \frac {a}{\sqrt {a^2 \plus{} b^2}}$. so we get $ \sin(y \plus{} x_i) \equal{} \frac {c}{\sqrt {a^2 \plus{} b^2}}$ for all $ i$. all this amounts to the statement that linear dependence is equivalent to the existence of a $ y$ such that $ \sin (y \plus{} x_i)$ is the same for all $ i$. that is, for all $ i$, $ y \plus{} x_i \equal{} \theta \mod{2\pi}$ or $ y \plus{} x_i \equal{} \pi \minus{} \theta \mod{2\pi}$ for some $ \theta \in [ \minus{} \pi, \pi]$. so linear independence is equivalent to having at least three different values $ \mod{2\pi}$ among the $ x_i$.

thank you for your solution. but i still have a question: why should we have at least three different values of $ x_{i}$?

my "attempt of solution" was solving $ a\cos x_{i}\plus{}b\sin x_{i}\equal{} c$ in terms of $ a, b$ and $ c$ (pretending that we've found some $ a, b, c$ and $ j$ such that $ a\cos x_{j}\plus{}b\sin x_{j}\equal{} c$ where $ i$ is different from $ j$) and then claiming that we should have at least one $ x_{k}$ which is not in the solution set of the above equation.
so "solution" would be that we have at least one value of $ x_{i}$ for each of the given sets of solutions that is not in the particular set (we get several sets of solutions due to different possible values of $ a, b$ and $ c$).

is that "good"? (somehow i have the feeling that i missed something and my attempt of solution doesn't give me exact condition on the sequence $ (x_{i})_{i\equal{}1}^{n}$ as your solution does)

if there's only one value $ \pmod{2\pi}$ among the $ x_i$, then it's clear that the three vectors are not independent. if there are exactly two different values $ \pmod{2\pi}$ among the $ x_i$, then these two values may be "rotated" around the unit circle by the same amount, until they are reflections of each other across the $ y$-axis, at which point their sines are equal. this amount will be the $ y$ from my solution, and gives us linear dependence.

on the other hand, if there are at least *three* different values $ \pmod{2\pi}$ among the $ x_i$, it's impossible to rotate them all by the *same* amount around the unit circle such that any two of them are either in the same position or reflections of each other about the $ y$-axis. this means the $ y$ from my solution does not exist, and thus we get linear independence.


MATHCOUNTS
geometric series
group theory

What is the repeating decimal for 1/99?

What is the repeating decimal for 1/98?

What is the repeating decimal for 1/97?

Challenge: Represent each of these repeating decimals of the sum of an infinite geometric series.

Another very interesting one is 1/89, although the proof that I know that it comes to what it does (without division) is fairly difficult.

s0mp,

See if you can find a rational number with neither a terminating decimal or a decimal with some repeating structure.

[quote="s0mp"]1/97 seems to be a non-repeating decimal. but how could that help? same thing with 1/47[/quote]

They repeat.  The only fractions that don't repeat are those with only powers of 2 and 5 in their simplified denominators.

[hide]1/99 is equivalent to .1/9 and a shortcut for writing fractions over 9 is to just repeat the number

Ex. 1/9 =.11111111111111111111............

Ex. 2/9 = .22222222222222222222222......

This is no different.

Since 1/9 is .1111.... then .1/9 = .0101010101......

haha thats the only one i could get

by the way. i did NOT use a calculator[/hide]

1/98 =

.0102040816326528

Whoa! Powers of two!

01
02
04
08
16
32
64
128 (which is 28, and 1 added to 64)

Interesting... when does it repeat?

I'm not sure when this particular decimal repeats.  I don't plan on counting out digits to find out.  But, viewing the powers of 2 as a geometric series, what do you notice?

[quote="Treething"]I think you mean sequence, as a series is a sum of terms.

[color=white]each term is double the other[/color][/quote]

Actually, I meant series.  Think about it why this sequence IS a series that computes to be the fraction we started with.

Ohhh...

[color=white]

(sigma notation thingy from 1 to infinity) 2(0.01^n)

[/color]

if p prime and different of 2 and 5, 1/p is k-periodic with k defined as k=min(n, 10^n=1(p)) and we see that k divide p-1

[quote="MCrawford"]Actually, it occured to me that because 1/7 has 6 repeating digits, that 1/49 has 42, and so then 1/98 must also have 42 in its repeating block (after the one non-repeating digit).[/quote]

Is there a way to figure this out without just looking at the decimal?

Let be n an integer with at least one divisor different of 2 and 5. So 1/n has k repeating decimals.
$1/n = 0,b_1b_2b_3...b_ia_1a_2...a_ka_1...a_ka_1...$
$10^k/n = b_1...b_{k},b_{k+1}...b_ia_1a_2...a_ka_1...a_ka_1...$
So $10^k / n - 1 / n = b_1...b_{k},b_{k+1}...b_i - 0,b_1b_2b_3...b_i$
$(10^k - 1)/n = a/10^i$ with a integer
So n divide 10^k - 1 since n can't divide 10^i
That's why k=min(j,10^j=1(n))

For n prime there is $C_{artin}$ approximatly 0,38 primes where k is p-1, with group theory and lagrange theorem we get that k divide p-1

I'm confused.  Your explanation doesn't work I don't think if n = 12.

right
I think that's ok if we take n non divisible by 2 or 5


quadratics
algebra
polynomial
Vieta
number theory proposed
number theory

let x and y be positive integers such that xy divides x^2+y^2+1. show that :
           (x^2+y^2+1)/(xy)=3

Let $x^2+y^2+1=kxy,k>2,k\in N.$ We have $x=\frac{ky\pm \sqrt{(k^2-4)y^2-4}}{2}$. It give $(k^2-4)y^2-4=m^2.$
If $3\not |k$ then $3|(k^2-4)\Longrightarrow 3|(m^2+4)$ contradition. Let prime $p|(k^2-4)=(k-2)(k+2)$. It give $(\frac{-4}{p})=1\Longrightarrow p=1(mod \ 4)$.
We have k=3 works, k=6,9,12 not works. First candidat, when k give not contradition is k=15: $k^2-4=221=13*17$.
I think equation $221y^2-4=m^2$ had solution, therefore you are wrong.

for k=15 the question is : is there any positive integers such as x^2+y^2+1=15xy ?????????????????????????????
I'm a new user they told me about the forum . but nobody can solve it !!!!!!

This is a basic infinite descent example. Let $S$ be a set of
solutions $(x,y)$ where $x\geq y$, and take one pair from $S$ whit
the smallest $x$.

Let $\frac{x^2+y^2+1}{xy}=k>2$, from that $x^2-x\cdot ky +y^2 +1=0.$
Now look at this like quadratic equation, put $f(x)=x^2-x\cdot ky +y^2+1$, then if one root of this equation is $x$, then by Viet's
formulas, other is $\frac{y^2+1}x=ky-x>0$. We said that $x$ is the
smallest so $ky-x\geq x\geq y$ so $f(y)\geq 0$.

$f(y)=2y^2-ky^2+1=y^2(2-k)+1\geq 0$ from that it is obvious that $k$
can only be $3$, and we can get that by, for example $x=y=1$.

bye

[quote="Nima Ahmadi Pour"]The very famous idea of swapping roots. :lol:[/quote]
Could I have a question :
what is the "idea of swapping roots"?

When you have something like $x^2+y^2-axy=b$, then you can see this as equation, either in $x$ or in $y$. Thus when you have some pair $(x,y)$ that solves it, then you can get another pair by Vietas theorem (e.g. when seeing it as equation in $x$, you have $x_1+x_2=b-y^2$).
Thats all.

Thank you ZetaX,I knew the method ,but din't know the name.


There are 190 people on the beach.  110 are wearing sunglasses, 70 are wearing bathing suits, and 95 are wearing a hat.  Everyone is wearing at least one of these items. 30 are wearing both bathing suits and sunglasses. 25 are wearing both bathing suits and a hat. 40 are wearing both sunglasses and a hat.  How many people are wearing all three items?

By PIE, $ 190 \equal{} 110\plus{}70\plus{}95 \minus{} (25\plus{}30\plus{}40) \plus{} x \implies 190 \equal{} 180\plus{}x \implies x \equal{} \boxed{10}$.

[hide][url]https://gyazo.com/193211e0d3a7b047b19ea674d46975e8[/url]
Then you must subtract the numbers that share the numbers together. Therefore the answer is $\boxed{10}$[/hide]

[hide=Alternative Solution]
We get
Only sunglasses: $110-30-40=40$.
Only bathing suits: $70-30-25=15$.
Only hats: $95-25-40=30$.
Bathing suits and sunglasses: $30$.
Bathing suits and hats: $25$
Sunglasses and hats: $40$.

So in total out of all $6$ scenarios there are $40+15+30+30+25+40=180$ people. That leaves $190-180=\boxed{10}$ people who are wearing all three items.
[/hide]


integration
real analysis
real analysis solved

Hey man come to my home I will solve it for you.You had not need to post it.Ok i wait you.

Why don't you post your solution here? :(

[quote="hardsoul"]Evaluate:
$\displaystyle\int^{1}_{0}\frac{\arctan{\frac{x}{x+1}}}{\arctan{\frac{1+2x-2x^2}{2}}}dx$[/quote]
BTW, it is a very old trick.
Prove that $f(x)+f(1-x)=1$, there $f(x)=\frac{\arctan{\frac{x}{x+1}}}{\arctan{\frac{1+2x-2x^2}{2}}}$ ;)

Пожалуй?та. ;)


trigonometry
real analysis
real analysis solved

This problem is maybe well known but I think it's quite nice.
Could someone prove this :
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/ea7f56869dbe93abec89159c27fd3f40.gif[/img]

Uh- there are uncountably many possible values and countably many integers.

Does it mean that it's wrong if x is a real number? What if it's a rational ? In this case there would be countable many  many values x.

Also if it were rationals: between -1 and 1 there are infinitely many rationals (just look at all the possible approssimations of $\frac{\sqrt{2}}{2}$) but we only have to work with integers from 0 to 359

JohnConstantine appears to be thinking in degrees, which is not the way this was intended.

It should be understood in this problem that the angle is measured in radians, not degrees. Since $\pi$ is irrational, that means that the sequence $(\sin n)$ never repeats itself and takes on (countably) infinitely many different values. The original statement above by bizz is false, for the reasons already stated.

You can say this:

Given any $x\in[-1,1]$ and any $\delta>0,$ there exists $n\in\mathbb{N}$ such that $|\sin n-x|<\delta.$ In other words, you can approximate every real number in that interval arbitrarily closely by the sine of an integer.

Said more succinctly, $\{\sin n\}$ is dense in $[-1,1].$

how do you show sin n is dense in [-1,1]

It is equivalent with the fact that $\{m-n\pi: m,n\in \mathbb{N}\}$ is dense in $[0,2\pi]$.


function
geometry
3D geometry
octahedron
tetrahedron
induction
modular arithmetic

Written by $John Kennedy, Queens College$ - he's really great!

1. It is straightforward to prove that, for $x \ge 0$, $a > 0$ and $b > 1$, the function $f(x) = ax - x^{b}$ has a maximum value when $x = (a/b)^{1/(b-1)}$.
Determine the maximum value of the function $F(\theta) = (sin\theta)(sin2\theta)$.

2. In how many ways can all of the 10 digits {0,1,2,...,9} be placed in the blanks of the integer $1$ _ $346$ _ $2$ _ $6$ _ $292$ _ $346$ _ $391$ _ $780$ _ $0$ _ $875$ _ $1$ so that when this number is divided by 3168, the remainder is 2005?

3. Evaluate the geometric mean of the proper positive divisiors of the sum of the positive divisors of 2005. (Note: a positive divisor d of a positive integer N is a proper divisor provided 1 < d < N.)

4. Eight of twelve congruent equilateral triangles are used as faces in constructing a regular octahedron. The remaining four triangles are used to construct a regular tetrahedron. If the total volume of the octahedron and tetrahedron is $2005 cm^{3}$, what is the volume of the tetrahedron? (Note: to be correct, your answer must be stated with appropriate units)

5. Determine a positive integer, not less than 100, that divides $X(n) = 13^{n} + 14^{2n-3}$ for every integer n = 2005, 2006, 2007, ... .

6. A game of [i]n-sprouts[/i] commences with n dots drawn on a page. Two players alternately make a [i]move[/i]. A move is a play consisting of drawing a line that is incident on two (not necessarily different) dots in the current figure and then inserting a new dot into the line just drawn. The rules of the game (1) prohibit any line to be drawn so that it crosses itself or any previously drawn line; and (2) limit the number of lines incident on any dot to a maximum of 3. (In graph theory terms these rules restrict any play to leave a planar graph with maximum degree 3.) What is the largest and what is the smallest total number of [u]moves[/u] that will be made during a game of [i]n-sprouts[/i]?

[edited]7. Determine the maximum value of $\sqrt[3]{4-3x+\sqrt{16-24x+9x^{2}-x^{3}}} + \sqrt[3]{4-3x-\sqrt{16-24x+9x^{2}-x^{3}}}$
 in the interval $-1 \le x \le 1$.

[added]8. Evaluate $5(_{1000}^{2005})_3 + 401(_{1005}^{2005})_3 + 2005(_{1010}^{2005})_3$, where $(_y^{x})_3$ means $(_y^{x})$ reduced modulo 3.

9. Nine points are placed on the circumference of a circle of radius 2005cm so that when all 36 chords (each connecting a pair of points) are drawn no three chords intersect at an interior point. How many triangles, each edge of which lies along one of the chords drawn, can be found in the resulting figure? (Express the answer as an integer.)

[added]10. A game of [b]triad[/b] is played on an 8 X 8 board with numbered squares and from which some squares have been removed (see figure, bolded numbered haev been removed). Two players alternately take turns in placing the [b]trio[/b] piece (see figure) so as to cover available contiguous squares on the board. None of the placed [b]trio[/b] pieces may overlap. The game ends when no play is available and the winner is the last player to place a [b]trio[/b].
Describe a play in the form (x,y,z) where x, y, and z are the square numbers occupied by placing a [b]trio[/b].
In playing [b]triad[/b] on the board illustrated what are the good moves for the first player. That is, list [u]all[/u] initial moves that will make it possible for the first player to force a win in this game.

.1...2...3...4..5...6...[b]7[/b]...8.
.9..[b]10[/b].11.12.13.14.15.[b]16[/b].
.17.18.[b]19[/b].20.21.22.23.24.
.25.26.27.28.[b]29[/b].30.31.32.
.33.34.35.[b]36[/b].37.38.39.40.
.41.42.43.44.45.[b]46[/b].47.48.
.[b]49[/b].50.51.52.53.54.[b]55[/b].56.
.[b]57[/b].[b]58[/b].59.60.61.62.63.64.

[u]the trio piece[/u]
x
xx

your interval for #7 doesn't make sense...also...can you post up solutions???

edited.

well, i thought if i posted up solutions, nobody would try?
maybe i'll post up answers first soon.

What were the time limits in this competition?  Can any school in the city participate?

15 minutes per 2 problems.

Im not exactly sure who is invited, but the number of schools participating is and will be increasing.

Thanks for the info.  To start the ball rolling, here's a solution for Problem 5.

[hide="Solution"]
Let's try some small cases: $X(2) = 183$ and $X(3) = 27 \cdot 183$.  So 183 is a candidate.  (It's true the problem said that $n \ge 2005$, but JBL once pointed out that problem writers sometimes impose such conditions just to stop folks from trying simple examples.)

At this point, we could use induction to show that 183 works, because $X$ obeys a second-order recurrence equation.  But I will try a direct approach instead:
\begin{eqnarray*}
13^n + 14^{2n - 3}
  & = & 169 \cdot 13^{n - 2} + 14 \cdot 196^{n - 2} \\
  & \equiv & 169 \cdot 13^{n-2} + 14 \cdot 13^{n-2}
                    \pmod{183} \\
  & = & 183 \cdot 13^{n-2} \\
  & \equiv & 0 \pmod{183}.
\end{eqnarray*}
So, indeed, 183 divides $X(n)$ for all integers $n \ge 2$.
[/hide]

[hide="number 4"]If you let the side of the equilateral triangle be s, the volume of the tetrahedron is $\frac{Bh}{3}=\frac{\frac{s^2\sqrt{3}}{4}\cdot\frac{s\sqrt{6}}{3}}{3}=\frac{s^3\sqrt{2}}{12}$, and the volume of the octahedron is $\frac{2Bh}{3}$ (where $B$ is the area of the square base of any pyramid that is half the octahedron)$=\frac{2s^2}{3}\cdot\frac{s}{\sqrt{2}}=\frac{s^3\sqrt{2}}{3}$.
Then $2005cm^3=\frac{s^3\sqrt{2}}{3}+\frac{s^3\sqrt{2}}{12}=\frac{5s^3\sqrt{2}}{12}$
Which means that $V_{tetrahedron}=\frac{s^3\sqrt{2}}{12}=\frac{2005}{5}=401cm^3$[/hide]
These problems are very fun but really hard compared to what i'm used to :( (except for the ones where i used calculus :D ). i really would like to see a nice way to do number 8, i couldn't come up with anything other than brute force!

yes it was hard, especially when you think about the time pressure.

I was the first place with getting 7 right, He Cui from A team was the second with getting 5 right, and no one else got more than 3.

I got 6,8, and 9 wrong.
I only put the maximum for 6 (bleh)

Have you guys seen [url=http://planetmath.org/encyclopedia/LucassTheorem.html]Lucas's Theorem[/url]?  It gives a fast way to solve Problem 8.

[quote="Stan17"]i got an answer for 7...is it[hide]2[/hide]???[/quote]

i got the same answer.  the problem writer is awesome because the numbers worked out so nicely!

[hide="my solution"]
let $a=\sqrt[3]{4-3x+\sqrt{16-24x+9x^{2}-x^{3}}}$ and $b=\sqrt[3]{4-3x-\sqrt{16-24x+9x^{2}-x^{3}}}$
$(a+b)^3=a^3+b^3+3ab(a+b)=8-6x+3x(a+b)$
$(a+b)^3-3x(a+b)+6x-8=0$
$((a+b)-2)((a+b)^2+2(a+b)-3x+4)$
Thus
$a+b=2$ or $(a+b+1)^2=3x-3$ only $1$ for $x$ in the second equation yields real values: $a+b=-1<2$ so the answer is $\boxed{2}$

coincidentally, if you were lazy and plugged in $0$ for $x$, you would also get the right answer :|

edit: actually, all values appear to yield 2 :oops_sign:
[/hide]

it is not a coincidence.

The whole thing will turn out to a [hide="constant"][b]2[/b][/hide]

[quote="Geehoon"]
3. Evaluate the geometric mean of the proper positive divisiors of the sum of the positive divisors of 2005. (Note: a positive divisor d of a positive integer N is a proper divisor provided 1 < d < N.)
[/quote]

[hide="answer"]the sum of the positive divisors of $2005$ is $1+5+401+2005=2412=2^2*3^2*67$
arrange the factors pairwise to form complementary pairs whose product equals $2412$.  There are $\frac{(2+1)(2+1)(1+1)}{2}=9$ such pairs, but only $8$ such pairs of proper divisors.
Thus the answer is $\sqrt[16]{(2412)^8}=\sqrt{2412}=\boxed{6\sqrt{67}}$
i'm assuming calculators were not allowed, so i'm not going to evaluate that. [/hide]

oops :oops: edited

calculators werent allowed, therefore, its not the answer.

the geometric root of n numbers is the nth root, not the square root, of their product

can someone explain what question 8 means (specifically that reduced modulo 3 part)?  thanks

$\binom{x}{y}_3$ means the remainder when the binomial coefficient $\binom{x}{y}$ is divided by 3.  So the result is either 0, 1, or 2, depending on the remainder.


In what order would you recommend motivated, but inexperienced (freshman/sophomore) students go through the AoPS books? Should they start with Vol. 1 and then proceed to Vol. 2, or should they follow Vol. 1 by particular subject books? Or should they by-pass Vol. 1 and go directly for the subject books?

If you've seen all of the subject books, which was your [i]favorite[/i]?

i dont support aops books frankly. there are better books on individual subjects which u shud buy in my view.

[hide="@analytic(don't take it harshly i'm just joking)"]TURN OFF YOUR COMPUTER, [b]NOW!!![/b][/hide]
for subject books on the basics, you can't beat AoPs. On the intermidiate level, AoPs books get REALLY expensive for example instead of buying intermidiate algebra i got [url: http://www.amazon.com/exec/obidos/ASIN/0817636773/artofproblems-20]this[url].(itsinthemail!)Also,purchaseProblemSolvingStrategiesbyArthurEngelonceyougettothatlevel.[/url][/hide]

Yes, I too feel that the AOPS books are ridiculously expensive..
There are cheaper alternatives available


superior algebra
superior algebra unsolved

let $ X$ be a semigroup (http://mathworld.wolfram.com/Semigroup.html)
assume that for every $ x,p,q\in X$, if $ xp\equal{}xq$ or $ px\equal{}qx$, then $ p\equal{}q$.
Prove that $ X$ has neutral element.
Is $ X$ a group or not necessary ?

I don't see why $ X$ should have a neutral element; take, for instance, the positive integers with addition. Maybe you want $ X$ to be finite? (In this case, $ X$ would be a group, because the multiplicative action of every element of $ X$ on $ X$ is injective, hence bijective, hence a permutation of a finite set, and every permutation of a finite set can be raised to some power to yield the identity.)

  darij

:oops: yes, $ X$ should be finite, sorry

but you wrote about this case :lol: and your answer works fine, thanks


geometry
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1) Given the equation:

$ x^3\minus{}4x^2\minus{}2x\plus{}10\equal{}0$

If two is added to each root in the above equation to form a new equation, find the sum of the cubes of the roots of the new equation.

How do you find the roots of such an equation? And how do you solve this question? I have no idea =)

2) Find the remainder when the sum $ 4^{45}\plus{}6^{45}$ is divided by 25.

[hide="1: Method"] First off, translate the polynomial by $ 2$; then you just need to find the sum of the cubes of the roots of some polynomial.  Second, look up Vieta's Theorem and Newton's Sums. [/hide]
[hide="2: Method"] Write as $ (5 \minus{} 1)^{45} \plus{} (5 \plus{} 1)^{45}$ and use Binomial Theorem. [/hide]

How would you use the binomial theorem for $ (5\minus{}1)^{45} \plus{} (5\plus{}1)^{45}$?

[hide]Well, all the terms with the power of $ 5$ greater than or equal to $ 2$ obviously are divisible by 25, so we can ignore those terms and focus on the last 2 of each, which are \[ \binom {45}{44} 5^1\cdot (\minus{}1)^{44} \plus{} \binom {45}{45} 5^0 \cdot (\minus{}1)^{45} \plus{} \binom {45}{44} 5^1\cdot 1^{44} \binom {45}{45} 5^0 \cdot 1^{45} \equal{} 225 \minus{} 1 \plus{} 225 \plus{} 1\] which is divisible by $ 25$, which means the remainder is $ 0$.[/hide]

Is the AOPS WIKI the right place to learn newton's sum and vieta's formula right off the bat? I don't know them.

It's as good a place as any.  The Wikipedia articles aren't bad, either (as I recall).


Let $a+b+c=3$. Determine the max value of
	\[A=9ab+10ac+22bc \]
where $a,b,c$ be the positive numbers

A general inequality:
Prove that if $a+b+c=3$ then
$A=xab+yac+(x+y)bc \le \frac{9(x+y)}{4}$ where $x,y$ are positive constants.


How many ordered n-tuples $(x_{1},x_{2},\ldots,x_{n})$ of positive integers satisfy $x_{1}x_{2}\ldots x_{n}=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{k}^{e_{k}}$, where $p_{1},p_{2},\ldots,p_{k}$ are primes and $e_{1},e_{2},\ldots,e_{k}$ are positive integers?

[quote="computer_nuke"]I don't really understand the question. Can't every inter (or product of integers) be represented as product of primes?[/quote]

I think he's asking how many ways there are to factor a number composed of $k$ primes into $n$ factors.

[quote="computer_nuke"]I don't really understand the question. Can't every inter (or product of integers) be represented as product of primes?[/quote]

how many ways can you factor a number into a product of $n$ factors. for example, with $xy=12$ we can have $(1,12),(2,6),(3,4),(4,3),(6,2),(12,1)$


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Let $a>0$ and let $g \in C^{0}([-a,a])$. Prove that there exists a unique function $u \in C^{0}([-a,a])$ such that
	\[u(x) = \frac{x}{2}u\left(\frac{x}{2}\right)+g(x)\]
	for every $x\in[-a,a]$.

isn't it a direct consequence of the Banach contraction principle ?

It would be, if $a$ were $<2$. Recall that $a>0$ is arbitrary.

yep, but if $\Phi(u)(x) = \frac{x}{2}u(\frac{x}{2})+g(x)$, then $\Phi^{n}$ is a contraction for $n$ sufficiently large, isn't it ?  :maybe:

That's right. The problem is solved.  :)


Let A={1,2,3,...,n} and
Sn = {f|f:A--->A} (1-1)
In = {(x,x)|x \in A}   ----> Identity
proof that
1.fog \in Sn, every f,g \in Sn
2. every f \in Sn, it will have g \in Sn that fog \in In

Sorry that my language is not so good because I'm in thailand, and I don't know how to use LaTeX.  :(

1) the composition of two bijections is again a bijection, because $g(f(A))=g(A)=A$ (surjective) and $g(f(x))=g(f(y))\implies f(x)=f(y)\implies x=y$ (injective).

2) all bijections have an inverse.  take that inverse.


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Two spheres touch 10 units above the ground. The smaller sphere has a radius of 8. What is the radius of the bigger sphere?

If the centers of the spheres are $C_1,C_2$ and the tangent point $D$, then drawing the perpendiculars $C_1M_1, C_2M_2, DN$ to the ground we get a right trapezoid consisting of two smaller right trapezoids: $C_1M_1=r, C_2M_2=R, C_1D=r, DC_2=R.$ Let $d=DN.$

Similarity gives $M_1N: NM_2=C_1D: DC_2=r: R$. If $M_1M_2=h$, then $M_1N=\frac{r}{r+R}h$ and $NM_2=\frac{R}{r+R}h$

Now we can express the fact that the area of the big trapezoid is the sum of the areas of the smaller trapezoids:

\begin{eqnarray*}\frac{R+r}{2}\cdot h=\frac{r+d}{2}\cdot\frac{r}{r+R}h+\frac{R+d}{2}\cdot\frac{R}{r+R}h &\implies& (R+r)^2=r(r+d)+R(R+d)\\ &\implies& 2Rr=rd+Rd\\ &\implies& R(2r-d)=rd\\ &\implies& R=\frac{rd}{2r-d}\end{eqnarray*}

For the given numeric data we get $R={40\over 3}$

it's simpler if you draw the height of the right trapezoid $C_1$ to $C_2M_2$, then use the similar triangles that you cut off, to get:

$\frac{2}{R-8}=\frac{8}{R+8}$

the same answer follows


find all $ (x,y)\in{Z}$ such that:

$ xy\equal{}20\minus{}3x\plus{}y$

thanx.

$ xy\equal{}20\minus{}3x\plus{}y,$
$ \Rightarrow xy\minus{}y\equal{}20\minus{}3x,$
$ \Rightarrow y(x\minus{}1)\equal{}20\minus{}3x,$
$ \Rightarrow y\equal{}\dfrac {20\minus{}3x}{x\minus{}1}$
$ \Rightarrow y\equal{}\dfrac {\minus{}3(x\minus{}1)}{x\minus{}1} \plus{} \dfrac{17}{x\minus{}1}$
$ \Rightarrow y\equal{}\dfrac{17}{x\minus{}1} \minus{} 3.$
Since $ x,y \in Z , \Rightarrow 17|(x\minus{}1)$, so $ x \in \{\minus{}16,0,2,18\}$

So the solutions are the couples $ (x,y)$:
$ (\minus{}16,\minus{}4);(0,\minus{}20);(2,15);(18,\minus{}2)$

Alternatively:

$ 3x \minus{} y\plus{}xy \equal{}20 \iff (x\minus{}1)(y\plus{}3)\equal{}17$


Didn't know in which forum to post this, but I try here and then I spose you'll move it if I was wrong

Let p>4,q>4 be primes. Show that 24|p^2-q^2

I can show that 8 divides the expression but how do I show that 3 does aswell?

Edit: Think I got it (used the division algorithm).

Consider this:  Since p and q are prime, they cannot divide three.  So that means that the number is either of the form 3a+1, or 3a+2.  Squaring these, you can realize that both will give a remainder of 1 being divided by 3.  Therefore, 1-1=0, thus the subtraction $p^{2}-q^{2}$ is divisible by three.

$p^{2}\equiv 1 \pmod 3$

$p^{2}-q^{2}\equiv 0 \pmod 3$


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Following http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218296&sid=f8d633c321e672de847c23ff3976d5c7
What are the maximal commutative subalgebras (for the inclusion) of $ \mathcal{M}_n(\mathbb{C})$?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1512145117&t=33640
a problem which concerns the maximal (for the dimension) commutative subalgebras of $ \mathcal{M}_n(\mathbb{C})$. But the proof is not complete.

Thanks loup blanc, I knew the result on dimensions, and I think I can prove it.
I 'm hoping to find all maximal subalgebra (up to isomorphism), not just the dimension

Hi JC_math,
can you give your proof concerning the maximal dimension ?

Hi loup blanc,
I will write it down. It's gonna take me more than just a few minutes, so it might not be today though.

Actually, for the dimension question there is a very short proof, but as usual it is absolutely not easy to find. The proof is by induction and is based on the following two observations: without loss of generality, one may assume all matrices are upper triangular. Next, associate to each matrix $ A$ the matrix formed by the lines and columns $ 2,3,...,n$, call it $ f(A)$. These $ f(A)$ form a commutative subalgebra of $ n\minus{}1$ matrices. Do the same for the function $ g(A)$ which is the $ n\minus{}1$ matrix obtained by cutting the last line and column. The commutation property ensures that the kernels of $ f$ and $ g$ are orthogonal for the standard dot product, so the sum of their dimensions is at most $ n$. On the other hand, the images of these functions are commutative subalgebras in dimension $ n\minus{}1$. One applies the rank formula to these two linear maps $ f,g$ and the induction hypothesis and the conclusion is immediate. If I'm not wrong, I think this approach would also allow findind all maximal commutative subalgebras.

Hi Gabriel,
is the correct solution $ E(n^2/4)\plus{}1$ ?
If yes I don't understand the proof by induction in your last post.

Yes, that's the answer. And for the induction step in my proof, we apply it for the images of the maps $ f,g$, which are commutative subalgebras. There is a small computation to be done, but I'll leave that to you.

OK, I got harazi's proof, which is not easy to find indeed. Here it goes, with some more details (the small computation as harazi calls it).
Let $ \mathcal{A}$ be a commutative subalgebra of $ \mathcal{M}_n)\mathbb({C})$.
We want to prove by induction that $ \mathrm{dim}\mathcal{A} \leqslant \frac{n^2}{4}+1$.
Assume it is true for $ n-1$.
We use harazi's notation from thread #6.
As he said we have $ \dim \ker f +\dim \ker g \leqslant n$
We also have $ \dim \ker f + %Error. "rank" is a bad command.
f = \dim \ker g + %Error. "rank" is a bad command.
g = \dim \mathcal{A}$.
Matrices of $ f(\mathcal{A})$ commute, and the same for those of $ g(\mathcal{A})$, so we apply the induction hypothesis to get $ %Error. "rank" is a bad command.
f \leqslant \frac{(n-1)^2}{4}+1$ and $ %Error. "rank" is a bad command.
g \leqslant \frac{(n-1)^2}{4}+1$.
We then combine the above to reach $ \dim \mathcal{A} \leqslant \frac{(n-1)^2}{4}+1 +\frac{n}{2}$.
But since $ \frac{n^2+1}{4}$ is not an integer we then have $ \dim \mathcal{A} \leqslant \frac{n^2}{4}+1$. Which we can write as
$ \dim \mathcal{A} \leqslant E\left( \frac{n^2}{4}+1 \right)$

Here is the proof I had and I said I'll write down loup blanc.

I also proceed by induction. Assume the result for $ k<n$.
If we can find $ F,G$ subspaces of $ \mathbb{C}^n$ stable by $ \mathcal{A}$ s.t. $ F\oplus G=\mathbb{C}^n$ then we can write $ \mathcal{A}$ as the product of two commutative subalgebras of $ \mathcal{M}_p(\mathbb{C})$ and $ \mathcal{M}_{n-p}(\mathbb{C})$ ($ p=\dim F$) and we are done.
So now assume that there isn't such $ F,G$. We know that for all $ M\in \mathcal{M}_{n}(\mathbb{C})$ we have $ \mathbb{C}^n = \oplus_{\lambda \in \mathrm{sp}M} \ker (M-\lambda I_n)^n$, so if $ M \in \mathcal{A}$ it can only have one eigenvalue.
We can thus write $ \mathcal{A}=\mathbb{C}I_n \oplus \mathcal{N}$, where $ \mathcal{N}$ is a commutative subalgebra of $ \mathcal{M}_{n}(\mathbb{C})$ with only nilpotent element.
We just need to show that $ \dim \mathcal{N} \leqslant \frac{n^2}{4}$ to conclude.
It is well known (most likely posted on this forum) that if $ M_1, M_2, \ldots , M_n \in \mathcal{N}$, then $ M_1 M_2 \cdots M_n =0$.
Denote $ I= \{ MX, M\in \mathcal{N}, X\in \mathbb{C}^n\}$, and $ S$ s.t. $ \mathbb{C}^n = I \oplus S$. We show that $ MX=0$ for all $ M \in \mathcal{N}$ and $ X \in S$. Let $ \varphi\sim : \sim \mathcal{N} \to \mathcal{M}_{p,n-p }(\mathbb{C})$, with $ \varphi (M)=M_{|S}$. $ \varphi$ is injective so $ \dim \mathcal{N} \leqslant p(n-p)$, hence $ \dim \mathcal{N} \leqslant \frac{n^2}{4}$.

[quote="harazi"] If I'm not wrong, I think this approach would also allow findind all maximal commutative subalgebras.[/quote]
I might be missing something obvious here, but I don't see how to conclude on the inital question form here.
Harazi??

Thanks JC_math
1) For your first proof my problem was here:
"The commutation property ensures that the kernels of  $ f$ and  $ g$ are orthogonal for the standard dot product, so the sum of their dimensions is at most $ n$".
     I found the Harazi's exposition very unfriendly and I give now some details:
$ A\in{ker(f)}$ iff $ A\equal{}\begin{pmatrix}L\\0_{n\minus{}1}\end{pmatrix}$ where $ L$ is a line.
$ B\in{ker(g)}$ iff $ B\equal{}\begin{pmatrix}0_{n\minus{}1}&C\end{pmatrix}$ where $ C$ is a column.
Thus $ ker(f)\subset\mathbb{C}^n,ker(g)\subset\mathbb{C}^n$ with $ LC\equal{}(AB)_{1n}\equal{}(BA)_{1n}\equal{}0$. $ <L,C>\equal{}LC$ is not a dot product over $ \mathbb{C}^n$ but the associated bilinear form is non degenerate.
2) For your second proof: if $ \mathcal{A}$ is undecomposable, you consider $ M\in\mathcal{A}$ and you say that $ M$ can only have one eigenvalue. Why ?

[quote="loup blanc"]
2) For your second proof: if $ \mathcal{A}$ is undecomposable, you consider $ M\in\mathcal{A}$ and you say that $ M$ can only have one eigenvalue. Why ?[/quote]
If $ M$ had $ r>1$ eigenvalues then you would write $ \mathbb{C}^n\equal{}\oplus_{i\equal{}1,\ldots , r} \ker(M\minus{}\lambda_i I_n)^{\alpha_i}$ (you can take all $ \alpha_i\equal{}n$ if you want) and I covered that case above.

OK JC_math.
If $ MN\equal{}NM$ then a $ M$-characteristic vector space is $ N$-invariant.
Another question:
"We show that  $ MX\equal{}0$ for all $ M\in\mathcal{N}$ and $ X\in{S}$. Let $ \phi : \mathcal{N} \to \mathcal{M}_{p,n\minus{}p }(\mathbb{C})$, with $ \phi(M)\equal{}M_{|S}$".
Then $ \phi\equal{}0$ ??
Do you want to say: If $ M\in\mathcal{N}$ and $ MX\equal{}0$ for all $ X\in{S}$ then $ M\equal{}0$ ?

loup blanc, I wrote it badly (and wrongly), instead of
[quote="JC_math"]We show that $ MX \equal{} 0$ for all $ M \in \mathcal{N}$ and $ X \in S$. [/quote]
I meant
We show that for all $ M \in \mathcal{N}$, $ MS \equal{} 0$ implies $ M\equal{}0$.

Then $ \varphi$ is injective.
I wanted to edit my previous mistake but it seems that I need to ask a moderator. Can someone do that? Thanks

A simple result:
If $ A\in\mathcal{M}_n(\mathbb{C})$ is a cyclic matrix (it's the case if $ A$ is a generic matrix) then $ [I,A]$ is a maximal commutative subalgebra of dimension $ n$.
Proof: $ [I,A]\equal{}\mathbb{C}[A]\equal{}\{B;AB\equal{}BA\}$ and the $ A$-minimal polynomial is of degree $ n$.
Remark: If $ A$ is not cyclic then $ [I,A]$ is a commutative subalgebra of dimension $ <n$ but is not maximal.

[quote="loup blanc"]A simple result:
If $ A\in\mathcal{M}_n(\mathbb{C})$ is a cyclic matrix (it's the case if $ A$ is a generic matrix) then $ [I,A]$ is a maximal commutative subalgebra of dimension $ n$.[/quote]
Indeed, and that's why I was surprised with
[quote="harazi"] If I'm not wrong, I think this approach would also allow findind all maximal commutative subalgebras.[/quote]

1) Another partial result:
Let $ \mathcal{A}$ be a maximal commutative subalgebra of $ \mathcal{M}_n(\mathbb{C})$ that does not contain any non zero nilpotent matrix. Then $ \mathcal{A}\equal{}[I_n,C]$ where $ C$ has $ n$ distinct eigenvalues ($ C$ is cyclic).
Proof: Let $ A\in\mathcal{A}$, $ \lambda_1,\cdots,\lambda_k$ be the distinct $ A$-eigenvalues, and $ P(x)\equal{}(x\minus{}\lambda_1)\cdots(x\minus{}\lambda_k)$. $ P(A)$ is nilpotent then is $ 0$. $ P$ has only simple roots then $ A$ is diagonalizable. Thus $ \mathcal{A}$ is a diagonalizable subalgebra. Moreover $ \mathcal{A}$ is maximal and therefore we conclude easily that it is of dimension $ n$.
Corollary: if $ \mathcal{A}$ is a maximal commutative subalgebra of $ \mathcal{M}_n(\mathbb{C})$ s.t.  $ dim(\mathcal{A})\not\equal{}n$ then $ \mathcal{A}$ does not contain any cyclic matrix but must contain a nilpotent matrix $ N$ s.t. $ N^{n\minus{}1}\equal{}0$.
2) I don't know the answer to the following question:
Can we find $ \mathcal{A}$, a maximal commutative subalgebra of $ \mathcal{M}_n(\mathbb{C})$, s.t.  $ dim(\mathcal{A})<n$ ?

[quote="loup blanc"]
2) I don't know the answer to the following question:
Can we find $ \mathcal{A}$, a maximal commutative subalgebra of $ \mathcal{M}_n(\mathbb{C})$, s.t.  $ dim(\mathcal{A}) < n$ ?[/quote]
I suspect the answer is yes for most $ n$, but at least for $ n\equal{}14$ you can find $ dim(\mathcal{A})\equal{}13$,as proved by Courter, see R. C. Courter, [i]The Dimension of Maximal Commutative Subalgebras of K n[/i], Duke Mathematical J. 32 (1965), 225-232.


Three equal circles are arranged so that each is tangential to the others.
They are enclosed in a regular hexagon so that the hexagon is tangential to the circles.
Find the radius, r, of the circles in terms of the side-length, x, of the hexagon.

$ r \equal{} \frac {\sqrt {3}\, x} {4}$

Cosine rule in $ x, x, 4r$ triangle with angles $ 120^{o}, 30^{o}$ and $ 30 ^{o}$.

Or in half the triangle, $ \cos {30^{o}} \equal{} \frac {2r} {x}$.

Let our hexagon be $ ABCDEF$ . Note that two diameters from any two circles form a segmente that is parallel and equal to segments $ \overline{AE},\overline{FD},\overline{CE}$ .

Therefore $ 4r\equal{}2\cdot\frac{x\sqrt{3}}{2}\;\therefore\;r\equal{}\frac{x\sqrt{3}}{4}$

Thanks triplebig.

I notice that my edit and your post arrived at the same time.

(I should learn to create and post diagrams.)


The first term, (1-1), equals 0, so the product will be $ \boxed{0}$.

this equation would turn out to be

$ 0*\minus{}1*\minus{}2*\minus{}3*...$

$ 0$ times anything is 0 so the solution is

[hide="Click if you need to see the answer... :D "]

$ \boxed{0}$[/hide]


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Hi.

Im sorry that is yet one more related uniform continunity question  :P .

1: I am having real hard time understanding uniform continunity, my main problem is that I never know which epislons / Deltas to pick to prove the function is uniformly continuoues.

On top of that I dont understand how the interval being given for the certain function is related to the proof whether its 0 to infinity or 0 to 1 etc...

2: For example $y=lnx$ is uniformly continuous on $(0,oo]$ because its differentiable at that interval right?
But $y=x^2$ isnt uniformly continuous on $(0,oo]$ but its still differentiable at that interval (I think). I dont understand that  :(

ANY help would be appriciated :)

Thanks.

What's the different between continue and uniform continue?

$\ln x$ is not uniformly continuous on $(0,\infty).$ It is uniformly continuous on $[1,\infty)$ but it is not uniformly continuous on $(0,1].$

The difference between continuous and uniformly continuous is in the placement of the quantifiers ($\exists$ and $\forall$).

Suppose $f$ is a function defined on a set $S.$ (I'll assume that $S\subset\mathbb{R}$ and that  $f$ is real-valued, but it would be easy to replace what I say below by general metric space language.)

We say that $f$ is continuous on $S$ iff:

$\forall\,x\in S$ and $\forall\,\epsilon>0,\ \exists\delta>0$ such that if $y\in S,\,|y-x|<\delta,$ then $|f(y)-f(x)|<\epsilon.$

The key point is that $\delta$ may depend on both $x$ and $\epsilon.$ We don't choose it until we know $x.$

We say that $f$ is uniformly continuous on $S$ iff

$\forall\,\epsilon>0,\ \exists\,\delta>0$ such that if $x,y\in S,\,|y-x|<\delta,$ then $|f(y)-f(x)|<\epsilon.$

Here, $\delta$ only depends on $\epsilon.$ For a give $\epsilon,$ the same $\delta$ works everywhere in the domain. That "one size fits all" approach is the source of the word "uniform."

Here is a chain of theorems.

$f$ has a bounded derivative on $S$ (* see note below)

$\implies$

$f$ satisfies a Lipschitz condition on $S$ - that is, $\exists\,M$ such that $|f(x)-f(y)|\le M|x-y|$ for all $x,y\in S$

$\implies$

$f$ is uniformly continuous on $S$

$\implies$

$f$ is continuous on $S.$

Each of these implications is a one-way implication with a false converse. Search for it, and you'll find where I've posted counterexamples.

* I really should have said "$f$ is continuous on $S,$ differentiable on the interior of $S,$ and its derivative is bounded on the interior of $S.$"