vector
linear algebra
linear algebra unsolved

If $A_1,...,A_r$ are symmetric(or hermitian) real matrices then show that $A_1^2+...+A_r^2=0$ implies $A_1=...=A_r=0$.

Each $A_i^2$ is positive semidefinite (they are diagonalizable matrices with real non-negative eigenvalues), and if, say, $A_1\ne O$, then there will be a column vector $x$ s.t. $x^*A_1^2x>0$, which will yield $x^*\left(\sum A_i^2\right)x>0$, and this contradicts $\sum A_i^2=O$.


Suppose that $ P\equal{}2^m$ and $ Q\equal{}3^n$. Which of the following is equal to $ 12^{mn}$ for every pair of integers $ (m,n)$?

$ \textbf{(A)}\ P^2Q \qquad
\textbf{(B)}\ P^nQ^m \qquad
\textbf{(C)}\ P^nQ^{2m} \qquad
\textbf{(D)}\ P^{2m}Q^n \qquad
\textbf{(E)}\ P^{2n}Q^m$


number theory
greatest common divisor
number theory solved

Note that : C(1,2n)+C(3,2n)+...+C(2n-1,2n)=2^(2n-1)
Let d=gcd(C(1,2n),C(3,2n),...,C(2n-1,2n))
Hence : d|2 2n-1 thus d=2 t
on other hand : d|C(1,2n)=2n=2 u+1 . v where n =2 u . v , v is odd
Easily to prove t=u+1 from that we have : d=2 u+1

It was an Oral Examination Ecole Polytechnique 2003


function
calculus
derivative
analytic geometry
symmetry
inequalities
inequalities unsolved

Suppose $f$ is a symmetrical function and differentiable in $(e,r)$ , so:

${f(x,y)=f(y,x)}$

Or

${f(x,y,z)=f(x,z,y)=f(y,x,z)=f(y,z,x)=f(z,x,y)=f(z,y,x)}$

Etc...

It's east to see that the partial derivatives of respect variable of each rearrangement is equal

Now, Suppose $f$ has a minimum/maximum, so, to find the coordinates we solve the sistem

${\frac{\partial f(x,y,z)}{\partial x}=0}$
${\frac{\partial f(x,y,z)}{\partial y}=0}$
${\frac{\partial f(x,y,z)}{\partial z}=0}$

But, from the definition of the Symmetry of $f$, Does not matter the possition of the variable, just the value matter.

${\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial f(y,x,z)}{\partial x}=\frac{\partial f(y,z,x)}{\partial x}=\ldots }$

So, if

${f(a,b,c)}$

Is the maximum

Each combination of $a,b,c$  are the maximum.

Suppose

${u(x)=f(x,y,z)}$
${v(y)=f(x,y,z)}$
${w(z)=f(x,y,z)}$

If we make the substituition $x = y$

${u(y)=f(y,x,z)=v(y)}$

If we make the substituition $x = z$

${u(z)=f(z,y,x)=w(z)}$

So, if $a$ is the maximum of $u$, the $a$ is the maximum of $v$ and $w$.
This is from the definition of symmetry.\
But, if $a$ is a maximum on

${f(a,y,z)}$
${f(x,a,z)}$
${f(x,y,a)}$

And this combinations are equal, so $a$ is a maximum in each possition, so, for the case

${f(a,a,a)}$

$a$ is a maximum too. Analogous for the minimum

So, we can solve it

${x=y=z=a}$

${\frac{\partial f(a,a,a)}{\partial a}=0}$

we can make  $g(x) = f(x,x,x,)$

and analise $g$, the maximum, minimum, signal of derivatives, etc...

This is valid if $f$ is a high Symmetric function.

any doubts?

[quote="Thales418"]
Now, Suppose $f$ has a minimum/maximum, so, to find the coordinates we solve the sistem

${\frac{\partial f(x,y,z)}{\partial x}=0}$
${\frac{\partial f(x,y,z)}{\partial y}=0}$
${\frac{\partial f(x,y,z)}{\partial z}=0}$

[/quote]

But i have notice with you that if above system has roots- called halted points but may be the function $f$ has not extremal ?

The question is

we define  $f(x,y,z)$

and we define a condition of the variables $g(x,y,z) = 0$

So, to $f$ be symmetrical, $g$ need to be symmetrical

This proof is a little bit harder, but is intuitively

I will try to proof this after, if someone want to help

in this case, we have a function $g$, we need to analise a few things

if the conditions $f$ has a minimum/Maximum?

we need to verificate te interval

doing $x,y,z=a$

$f'$ can or cannot be equal to zero

but, if we proove that  if $g$ is symetrical so $f$ is symmetrical  $(a,a,a)$ is a solution

So, $f$ has a maximum on $f(a,a,a)$  where $a$ is solvede in $g(a,a,a)=0$

this question is a little bit harder

if someone help

[quote="Thales418"]${u(x)=f(x,y,z)}$
${v(y)=f(x,y,z)}$
${w(z)=f(x,y,z)}$[/quote]

These definitions are a nice way to deceive others and yourself. Of course, the definition of u(x) involves y and z, so the function u(x) depends on y and z, and this pretty much destroys your later argument. I could get into more details, but this would be handwaving, because your proof itself is pretty much handwaving. Just in order to directly confute your thesis that a symmetric function obtains its extrema at (x, x, x) points, here is a symmetric function:

$f\left(x,y,z\right)$
$=\left(\left(y-2z\right)^{2}+\left(2z-3x\right)^{2}+\left(3x-y\right)^{2}\right)$
$\cdot\left(\left(z-2x\right)^{2}+\left(2x-3y\right)^{2}+\left(3y-z\right)^{2}\right)$
$\cdot\left(\left(x-2y\right)^{2}+\left(2y-3z\right)^{2}+\left(3z-x\right)^{2}\right)$
$\cdot\left(\left(y-2x\right)^{2}+\left(2x-3z\right)^{2}+\left(3z-y\right)^{2}\right)$
$\cdot\left(\left(z-2y\right)^{2}+\left(2y-3x\right)^{2}+\left(3x-z\right)^{2}\right)$
$\cdot\left(\left(x-2z\right)^{2}+\left(2z-3y\right)^{2}+\left(3y-x\right)^{2}\right)$.

Guess where it obtains its minima. There are much simpler examples of such behavior - functions which are rather natural and not constructed by me as counterexamples.

  darij

Can you do an example to demonstrate your proof, Thales 418. Let's say the Schurl inequality:
$a^{3}+b^{3}+c^{3}+3abc \geq ab(a+b)+bc (b+c)+ca (c+a)$
This is a highly symmetric inequality. But please notice that the equality obtain when $a=b=c$ or $a=b$, $c=0$

when i define $u$ is so trivial that $y$ and $z$ are constants

so, $u$ depend only of $x$

if you want to figth

create a new topic

dont do this trashs on mine

[quote="Thales418"]dont do this trashs on mine[/quote]

I'm not "doing trashs on your", I'm just laughing my ... off reading your posts  :rotfl:  :rotfl:

Seriously, I have disproved your statement, but you have not reacted in any halfways constructive way. As for the functions, ok, you define u when y and z are constant, then you define v when z and x are constant and w when x and y are constant, but then you work with these functions as if x, y, z were arbitrary values. As for your polemics towards other people and me... no comment.

If you want to defend some point, we expect a minimal willingness to read others' arguments. Anything else is considered trolling.

  darij

dera daruji

are you crazy?!

if you want to learn calculus

buy a some calculus book

it is good

you will learn a loot!!
for more information

http://www.senhorcarteiro.blogspot.com

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/DoNotFeedTroll.svg/150px-DoNotFeedTroll.svg.png[/img]

  dg

Yes, i think we have only an example: Shurl's theorem that enough to prove your argument not true. There are sysmetry problems getting extreme at nonsysmetry variables:
For example: the following inequality is always true $a^{3}+b^{3}+c^{3}\leq 5abc$  $\forall a,b,c\in[1;2]$ but the sing "=" occurs at $a=2,b=c=1$

again Thales i hope you are succesful. but i like to mention that if a function is differenciable in several variables and has a min/max it must be where the partials are zero. however it doesn't mean that when the partials are zero that you found a max/min it can be a "saddle point". so in some cases this works in others it doesn't.

i know alll this guys

this conjecture is tooo much difficult to proove

there is a too much cases and etc...

and the supposition of $f$  is that $f$ is differentiable, symmetric

and i supposing that f need to be even or odd, i dont know

im trying my best to prove this

yu guys may helps although say im wrong

why do you think i solved a lot of inequality?!?!

to joke!?

no!!!

to test the strong of my conjecture

it fail in some cases

but i cannot explain  fine why it fail
 etc...

[quote="Thales418"]dera daruji
are you crazy?!
if you want to learn calculus
buy a some calculus book
[/quote]

Ok . You are the best caculus user I have ever met Thales418.
I think you should just stop speaking until you prove <<your theorem>> and don't tell the others what to do (especially darij as he is a goldmedalist and he doesn't need  suggestions of you)
Finally stop spam ( I will do it also)

But with your methods, you can't solve my problem that i have posted in previous psot

i said this before

i cannot say if a function f has a maximum in x=y=z=a

not ao symmetrical do this

but some do

i worked out in this equation

 the others, that dont do

i cant make it

but a precise definition of this i does not made

this is justa conjecture

im not saying  "all symmetrical can do"

i saying "some symmetrical..."

oh please.... ban this guy..

..well, bets are now being taken... 13 year old ?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134859

hey BB king

read


Let $P$ be a point inside triangle $ABC$; let $AP$, $BP$, and $CP$, meet sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively.  Prove that $PA\cdot PB\cdot PC \geq 8\cdot PD\cdot PE\cdot PF$.

Nice question, which I'll leave for others to solve.  Another question that I'll add just for fun:

Given triangle ABC with D on (segment) BC and E on (segment) CA.  AD and BE intersect in F.  Prove that F cannot be the midpoint of both AD and BE.

I'm also leaving both problems for others, particularly since I'm too tired to write anything like a solution, but here's a link for the original inequality:

http://www.mathlinks.ro/Forum/viewtopic.php?t=19862

(where it was given in the form $\frac{AP}{PD}\cdot\frac{BP}{PE}\cdot\frac{CP}{PF}\geq 8$).

So it's, alas, not original...

  darij

It can be done with mass points. (Or, equivalently, an appropriate combination of Ceva and Menelaus.)


real analysis
topology
logarithms
integration
function
group theory
abstract algebra

Everyone who studied Lebesgue measure surely knows the result that if $\lambda_{n}$ is the Lebesgue measure on $\mathbb{R}^{n}$ and $A \in \mathbb{R}^{n}$ then $\lambda_{n}(A) > 0 \implies A+A$ contains a ball.
[see grobber's proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47161]here[/url]]

But the converse isn't true, there are plenty of sets (like the usual Cantor middle third set) for which $A+A$ contains a ball but $\lambda_{n}(A) = 0$.

So I wanted to come to some kind of generalization of the result, like this:

Let $\mathcal{H}_{s}$ be the $s$-dimensional Hausdorff measure. Let $A \in \mathbb{R}^{n}$ and $s > 0$. If $\mathcal{H}_{s}(A) > 0$, then $A+A$ contains an $n$-dimensional ball.

But the conditions of this one are too weak. Like for example if $A$ is a line in $\mathbb{R}^{2}$ then $A+A = A$ right? So how about I will be less ambitious and look at the special case $n=1$.

Conjecture: If $A \in \mathbb{R}$ is a set with Hausdorff dimension greater than 0, then $A+A$ contains an open interval.

Does anyone have a counterexample, or any idea about how I could prove this?

:noo: Consider a thinner Cantor set, e.g. the set of all numbers whose decimal expansion has only $0$s and $1$s. Then $A+A$ has only numbers with $0,1,2$ in the decimal expansion.

:| but at least that thing is appreciably bigger. So, what about a weaker thing:

Denote the Hausdorff dimension of $A$ by $\text{dim}_{\mathcal{H}}(A)$.

Conjecture: If $A \in \mathbb{R}$ then $\text{dim}_{\mathcal{H}}(A+A) = \text{min}(1, 2\text{dim}_{\mathcal{H}}(A))$.

edit: I think this can't possibly be true (that would be too nice) but maybe something like this ... I am just trying to bumble my way to a reasonable guess.

My example has $\text{dim}_{\mathcal{H}}(A)=\frac{\log 2}{\log 10}$ and $\text{dim}_{\mathcal{H}}(A+A)\le \frac{\log 3}{\log 10}$.

Here's a modest goal: if $\text{dim}_{\mathcal{H}}(A)>0$, then $\text{dim}_{\mathcal{H}}(A+A)>\text{dim}_{\mathcal{H}}(A)$?

Argh it's so hard to estimate Hausdorff dimension from below.

So, we can easily get an upper bound of $\text{min}(\text{dim}_{\mathcal{H}}(A)+\text{dim}_{M}(A), 1)$ because there is an obvious Lipschitz map from $A\times A$ to $A+A$ and Lipschitz maps don't increase dimension and $\text{dim}_{\mathcal{H}}(A\times A) \le \text{dim}_{\mathcal{H}}(A)+\text{dim}_{M}(A)$.

We could maybe do better but I am more interested in the lower bound.

Your suggestion is a good start but not strong enough for what I wanted to do. But that's not important, we've got to start somewhere eh? So let me think about this...
The Hausdorff dimension being greater than $s$ means that there is a measure with finite $s$-energy on that set, right? But I really don't see how we're going to construct a measure on $A+A$ out of one on $A$ so that the one on $A+A$ is strictly nicer than the old one. Well how about this one, suppose $\mu$ is the measure on $A$, then
$\nu(B) = \mu(\{x | x \in A, \exists y \in A : x+y \in B\})$

So what's the $s$-energy of $\nu$?
$E_{s}(\nu) = \int \frac{d\nu(x)d\nu(y)}{|x-y|^{s}}$

Maybe I should look at measures of balls instead. God I'm so bad at this. So, if we can show that $\nu(B_{r}(x)) < cr^{s}$ then that's good.

Now I don't like $\nu$ any more, I can't really see what's going on with it. I'm going to take a break from this problem. I know this post has pretty much no actual results but maybe my ideas will be helpful to someone.... I will come back to this soon hopefully.

Unfortunately, there is no result like what you wanted. Let's generalize my example by splitting the decimal expansion of a number $x\in [0,1]$ into blocks of length $L$ and declaring that $x\in A_{L,N}$ iff the integer represented by the digits within each block is at most $N$ (here $1\le N<10^{L}/2$). Then $\dim_{\mathcal{H}}A_{L,N}=\frac{\log (N+1)}{\log 10^{L}}$ because $A_{L,N}$ is obtained by dividing intervals into $10^{L}$ parts and keeping $N+1$ of them.

The set $A_{L,N}+A_{L,N}=A_{L,2N}$ has dimension $\frac{\log (2N+1)}{\log 10^{L}}$, which is very close to the dimension of $A_{L,N}$ when $L$ is large. Thus, any estimate $\dim_{\mathcal H}(A+A)\ge (1+\epsilon)\dim_{\mathcal H}(A)$ is false.

To construct an example with $\dim_{\mathcal H}(A+A)=\dim_{\mathcal H}(A)$, you can make the length of blocks ($L$) variable and increasing to infinity, with an appropriate adjustment for $N$.

Maybe at least this much is true: if $\mathcal{H}^{s}(A)>0$ for some $0<s<1$, then $\mathcal{H}^{s}(A+A)=\infty$.  :maybe:

Ok instead of this line of reasoning, how about we turn around and try to find a sufficient condition on $A$ so that $A+A$ contains a ball? Clearly $\lambda(A) > 0$ works, but it would be nice if we can find something weaker.

Here's my first guess: $\text{dim}_\mathcal{H}(A) > \frac{1}{2}$.

Also, you destroyed all of my hopes with some very "thin" examples; I wonder if we can come up with any "thick" ones so that the Hausdorff dimension of $A$ is small, but that of $A+A$ is large. (I guess to do this we would want to find examples where the Minkowski dimension is a lot bigger than the Hausdorff dimension?)

The latest examples are not necessarily thin: the dimension of $A_{L,N}$ can be made arbitrarily close to $1$, with $A_{L,N}+A_{L,N}=A_{L,2N}$ still being a set of dimension less than $1$. Also, these sets are "regular" Cantor-type sets, for which Minkowski and Hausdorff dimensions coincide.

Okay, I think this explains why the original problem that inspired all of these investigations is hard. Probably this line of approach won't work. I am still interested in this question about sum sets, but I'm inclined to give up for now because apparently I can't even make reasonable guesses, let alone prove stuff.

I don't really understand how it would help, but maybe we can say more if $A$ is an $F_\sigma$ set? (My friend told me that making that assumption will make things easier because we can write down what $A$ is, which we can't do if it's just Borel or measurable or something. But honestly I have no idea how that helps at all with this kind of problem.)

The Cantor-type sets $A_{L,N}$ are compact, aren't they?

Sure, it's natural to ask what condition imply that $A+A$ has nonempty interior. But it's also natural to expect that the answer will involve not just the size of $A$, but also some measurement of its additive structure, such as a Fourier transform of some kind.

[quote="mlok"]also some measurement of its additive structure, such as a Fourier transform of some kind.[/quote]
Sorry I have absolutely no idea what you mean by this but it sounds like I ought to know. Can you explain a bit please?

edit: is it because $\chi_{A+A}(y) = max_{x}(\chi_{A}(x) \cdot \chi_{A}(y-x))$ or something? So it's like $L^\infty$ convolution? :|

Since the Minkowski sum is a lot like convolution, it may be helpful to look at the Fourier transform of some measure associated with $A$. If $A$ had positive Lebesgue measure, you'd be looking at the restriction of that measure to $A$, $\chi_{A}$. Indeed, $\chi_{A}\in L^{2}$ implies $\widehat{\chi}_{A}\in L^{2}$, hence $(\widehat{\chi}_{A})^{2}\in L^{1}$ and taking the inverse Fourier transform we find that $\chi_{A}*\chi_{A}$ is a nonzero continuous function, so it must be positive on some interval. Since $\chi_{A}*\chi_{A}\le \lambda(A)\chi_{A+A}$, this provides another proof of the result you started with.

Now suppose that $0<\mathcal{H}^{s}(A)<\infty$ for some $0<s<1$ --- can the Fourier transform of the restriction $\mathcal{H}^{s}_{|A}$ tell us something about $A+A$? I don't know.

Edit: I just learned of a theorem of Erdos and Volkmann (1966): for any $s\in (0,1)$ there exists an additive subgroup of $\mathbb R$ that is a Borel set of Hausdorff dimension $s$.  :!:


trigonometry algebra polynomial inequalities calculus integration function

Can somebody compute this?

$ \sum_{k\equal{}0}^{n\minus{}1} \frac{1}{1\plus{}8 \sin^2 (\frac{k \pi}{n} )}$

Find a polynomial $ P(x)$ whose roots are $ \sin^2 \left( \frac{k \pi}{n} \right)$ and use it to construct a polynomial whose roots are the terms of the desired summation.

Thanks for the quick reply. Could you get me started a bit more? I'm not that experienced I should say :)

Your hint made me construct $ P(x) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}\left(x \minus{} \sin^2\left(\frac {k\pi}{n}\right)\right)^k$ for the polynomial with roots $ \sin^2 \left( \frac {k \pi}{n} \right)$, but I'm not sure whether I'm going in the direction you had in mind.

The power of $ k$ is unnecessary.  One observation you can make is the half-angle identity $ \sin^2 \left( \frac{k \pi}{n} \right) \equal{} \frac{1 \minus{} \cos \frac{2\pi k}{n} }{2}$, so now you should try to find a polynomial $ Q(x)$ with roots $ \cos \frac{2 \pi k}{n}$.

The second observation is the existence of the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomials of the first kind[/url].

So $ Q(x)$ would be simply $ Q(x) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}\left(x \minus{} \cos\left(\frac {2\pi k}{n}\right)\right)$ ?

I read that the Chebyshev polynomials (nice observation!) of the first kind, $ T_n(x)$, have extrema ocurring at $ x \equal{} \cos\left(\frac {\pi k}{n}\right)$ which seems very relevant.
I'm not seeing exactly where we're going yet, though. I guess the idea is to relate the original summation to $ T_n(x)$, en using that at the extrema $ T_n(x)$ takes values $ \pm 1$?

Actually, it is precisely the case that $ Q(x) \equal{} T_n(x) \minus{} 1$ (why?).  Now the computation reduces to keeping track of some coefficients of $ T_n$, although now that I think about it more it might not be very easy to do so.

[quote="t0rajir0u"]Actually, it is precisely the case that $ Q(x) \equal{} T_n(x) \minus{} 1$ (why?).[/quote]
Let me define $ x_0\equiv\cos\left(\frac {2\pi k}{n}\right)$ and $ x_1\equiv \cos\left(\frac {\pi k}{n}\right)$
We have that $ T_n(x_1)\equal{}\left\{\begin{array}{cc}\ 1\mbox{ if k is even}\\0\mbox{ if k is odd}\end{array}\right.$
So $ T_n(x_0)\equal{}1$ for all $ k$, from which it follows that $ Q(x_0)\equal{}T_n(x_0)\minus{}1$.
So at least I see why it is true for $ x\equal{}x_0$ :)

Good.  Now recall that a polynomial of degree $ n$ is determined by its roots and its leading coefficient.  ($ x_1$ is totally irrelevant.)  From here, can you construct the polynomial with roots $ \frac{1}{5 \minus{} 4 \cos \frac{2 \pi k}{n} }$ (as desired in the original problem)?

[quote="t0rajir0u"](as desired in the original problem)[/quote]$ 1 + 8 \sin^2 (\frac {k \pi}{n} ) = 1 + 8(\frac {1 - \cos \frac {2\pi k}{n} }{2}) = 5 - 4\cos(\frac {2\pi k}{n})$. Yep, I see.

So my task is to transform $ Q(x) = \sum_{k = 0}^{n - 1}x - \cos(\frac {2\pi k}{n})$  to  $ R(x) = \sum_{k = 0}^{n - 1}x - \frac {1}{5 - 4 \cos( \frac {2 \pi k}{n} })$
and having done that, use the identity $ Q(x)=T_n(x)-1$.
I'm a bit bothered by the fact that the cosine has to switch from the numerator to the denominator...Let my try that tomorrow.  Thanks for the help so far!

One can do it via complex analysis .
consider  complex integral of complex function
\[ g(z) \equal{} \frac {f(z)}{\tan (nz)}
\]
on this contour (see picture pp, denote it by $ \Gamma$)
where
\[ f(z) \equal{} \frac {1}{1 \plus{} 8\sin^{2}z}
\]
it's easy to see that our function $ g(z)$ have no pole on line $ \pi/2 \plus{} ti$ and also on line $ \minus{} \pi/2 \plus{} it$ since $ g(z) \equal{} g(z \plus{} \pi)$.
note that $ \int_{\Gamma}g(z)dz \equal{} 0$ since when $ |Im z|\to \infty$ then$ |f(z)| \to 0$ and also summation  integral on lines $ \pi/2 \plus{} ti$ and $ \minus{} \pi/2 \plus{} it$ is zero.
thus we obtine
\[ \sum_{k \equal{} \minus{} n \plus{} 1}^{n \minus{} 1}f(z) \equal{} \minus{} n\sum_{Res z_{k}} g(z)
\]
where $ z_{k}$ is residue of $ f(z)$ (it's important that $ z_{k}$ is not on real line .)
in your case $ f(z)\equal{}f(\minus{}z)$

[quote="BlueVelvet"]Can somebody compute this?

$ \sum_{k \equal{} 0}^{n \minus{} 1} \frac {1}{1 \plus{} 8 \sin^2 (\frac {k \pi}{n} )}$[/quote]

I'm sure that the following files are going to appeal to you, BV...

[url=http://reflections.awesomemath.org/2008_4/andrica_piticari.pdf]1[/url]

[url=http://reflections.awesomemath.org/2008_5/article_2.pdf]2[/url]

Regards.

Wow, they sure do, thank you very much! :)

Extremal's method also seems interesting. I'll have to try that some day, and of course finish t0rajir0u's 'proposal' (don't know whether that's the right word, it's the literal Dutch translation Dutch). Hopefully the answers will be the same.


Can u make the icons near the Subjects to be different if the problem is solved and if not. Kind'a it is yellow if there is a solutiona and it is red if not... Moderators can make the icon change.  :D

When a problem is solved it is moved to solved problems! :D :D

cheers! :D


symmetry
number theory unsolved
number theory

For a positive integer $ n$, find the sum of all positive integers whose base $ 2$ representations have exactly $ n$ $ 1'$s and $ n$ $ 0'$s. (The first digit can not be $ 0$).

[hide]Clearly there are $ \binom{2n \minus{} 1}{n}$ such numbers, since we may choose any $ n$ out of the $ 2n \minus{} 1$ positions but the first.  By cyclic symmetry applied to the $ 2n \minus{} 1$ positions but the first, exactly the same number of numbers will have $ 1$ on their second position as numbers will have $ 1$ on their third position as... as numbers will have $ 1$ on their last position, with a proportion of $ n \minus{} 1$ $ 1$'s for each $ n$ $ 0$'s, or the sum is
\[ (2^0 \plus{} 2^1 \plus{} ... \plus{} 2^{2n \minus{} 2})\binom{2n \minus{} 1}{n}\frac {n \minus{} 1}{2n \minus{} 1} \plus{} 2^{2n \minus{} 1}\binom{2n \minus{} 1}{n} \equal{} 2^{2n \minus{} 1}\binom{2n \minus{} 1}{n}\frac {3n \minus{} 2}{2n \minus{} 1} \minus{} \binom{2n \minus{} 2}{n}.
\]
[/hide]


Hallo!Could you give me a hint how to do the following problem: "Let $a_{0}=0,91$ and $a_{k}=0,99\cdots 900\cdots 01$ where number of nines is $2^{k}$ and number of zeroes is $2^{k}-1$.Prove that
$\lim_{n\to \infty}(a_{0}a_{1}\cdots a_{n})=\frac{100}{111}$".


[quote="happyme"]If the length of each edge of a cube is increased by 75%, by what percent is the surface area of the original cube increased?[/quote]

[hide]The length became $\frac{7}{4}$ of the original.

Therefore the surface area should have become $\left( \frac{7}{4}\right)^{2}=\frac{49}{16}$ of the original, which is $\frac{33}{16}=\boxed{206.25 \%}$ increase.[/hide]

[quote="jli"]it gets increased by the square the side increases so $1.75^{2}-1$ which $1.0625$ or something.[/quote]

By the way, $1.75^{2}-1=2.0625$

Write hide inside brackets[] to start the part that has to be hidde, and to end it, write [/hide]


modular arithmetic
greatest common divisor
number theory
relatively prime
number theory proposed

1.Find all n such that
$ 12^n\plus{}5^n$ is a perfect square.

[quote="TTsphn"]1.Find all n such that
$ 12^n \plus{} 5^n$ is a perfect square.[/quote]

We have $ 12^n \plus{} 5^n\equal{}a^2$
If $ n$ is odd, $ 12^n \plus{} 5^n\equal{}2$ or $ 3\pmod{5}$ and $ a^2\equal{}0$, $ 1$, or $ 4\pmod{5}$. So $ n\equal{}2p$ is even.

So we have $ (12^p)^2 \plus{} (5^p)^2\equal{}a^2$ with $ p>0$ ($ p\equal{}0$ is obviously impossible).
Obviously, the three numbers $ 12^p,5^p,a$ must be coprimes and the only even number is $ 12^p$. So we have :

$ 5^p\equal{}u^2\minus{}v^2$ and $ 12^p\equal{}2uv$
$ 5^p\equal{}u^2\minus{}v^2\equal{}(u\minus{}v)(u\plus{}v)$ and so $ u\plus{}v\equal{}5^k$ and $ u\minus{}v\equal{}5^{p\minus{}k}$ and we have $ u\equal{}\frac{5^k\plus{}5^{p\minus{}k}}{2}$ and $ v\equal{}\frac{5^k\minus{}5^{p\minus{}k}}{2}$

So $ 12^p\equal{}2uv\equal{}\frac{5^{2k}\minus{}5^{2p\minus{}2k}}{2}$ and $ 2* 12^p\equal{}5^{2k}\minus{}5^{2p\minus{}2k}$

Since LHS is not divisible by $ 5$, we must have $ k\equal{}p$ and so $ 2* 12^p\equal{}5^{2p}\minus{}1$

And this last equation has obviously one unique solution : $ p\equal{}1$

So the initial equation has one unique solution $ n\equal{}2$ with : $ 12^2\plus{}5^2\equal{}13^2$

More (own problem)
Find all $ (m,n)\in N$
$ 12^m\plus{}5^n$ is a perfect square.

[quote="TTsphn"]More (own problem)
Find all $ (m,n)\in N$
$ 12^m \plus{} 5^n$ is a perfect square.[/quote][hide]
It's easy to show that m and n are even.
$ (12^{m}\plus{}i5^{n})(12^{m}\minus{}i5^{n})\equal{}a^{2}$
The gcf is $ 2\cdot 5^{n}$. Both are odd, and neither can be a multiply of 5, so they are relatively prime, so both are perfect squares.
$ (12^{m}\plus{}i5^{n})\equal{}(s\plus{}ei)^{2}$
$ 12^{m}\equal{}s^{2}\minus{}e^{2}$
$ 2se\equal{}5^{n}$
But clearly the second has no solutions. Which makes me wonder, what happened to the (1,1) case, it somehow has vanished...[/hide]
I can't find the flaw in my solution...

[quote="diophantient"]It's easy to show that m and n are even.
$ (12^{m} \plus{} i5^{n})(12^{m} \minus{} i5^{n}) \equal{} a^{2}$
The gcf is $ 2\cdot 5^{n}$. Both are odd, and neither can be a multiply of 5, so they are relatively prime, so both are perfect squares.
$ (12^{m} \plus{} i5^{n}) \equal{} (s \plus{} ei)^{2}$
$ 12^{m} \equal{} s^{2} \minus{} e^{2}$
$ 2se \equal{} 5^{n}$
But clearly the second has no solutions. Which makes me wonder, what happened to the (1,1) case, it somehow has vanished...
I can't find the flaw in my solution...[/quote]
First you must show, that m and n even (it is easy).
Second, may be $ 12^m \equal{} 2se,5^n \equal{} s^2 \minus{} e^2$ (it had solution, when $ m \equal{} n \equal{} 1 \ \minus{} \ s \equal{} 3,e \equal{} 2$ ).


vector
geometry
3D geometry
tetrahedron
analytic geometry
real analysis
real analysis unsolved

can we dissect the set $\{ \sum_{i=1}^{n}\alpha_{i}\cdot v_{i}|\alpha_{i}\in [0,1],v_{i}\in \mathbb{R}^{n}\}$ to the$n!$ sets congruent to $\{ \sum_{i=1}^{n}\alpha_{i}\cdot v_{i}|\sum_{i=1}^{n}\alpha_{i}\leq 1,v_{i}\in \mathbb{R}^{n}\}$?

Your set notation is misleading: $v_{i}$ are fixed vectors, they do not vary over $\mathbb R^{n}$. I think the answer is no in three dimensions. Let $v_{i}$ be standard basis vectors in $\mathbb R^{3}$. We are to make a unit cube out of six copies of the tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. Note that the angle between the plane $x+y+z=1$ and coordinate planes are equal to $\arctan\sqrt{2}$. The angles between coordinate planes are $\pi/2$. To build a cube, we'd have to represent $\pi$ as an integer combination of these two angles, which is clearly impossible.


geometry
3D geometry
inequalities
geometry proposed

Does every polyhedron have 3 edges being the lengths of the sides of a triangle?

No. One can consider the quadrilateral with sides 25,25,60,100

actually I think that you have the question all wrong Fedor because of a language matter.

the quadrilateral is a polygon, but the cube is a polyhedron.

the answer to the problem is yes, and the solution is quite natural and simple. suppose that there can be a polyhedron without a triangle among it's sides (err ... not to correct said, but you understand what I mean :D).

let us order it's side lengts: x_1 \leq x_2 \leq ... \leq x_n. Then we must also have x_1 + x_2 \leq x_3 , x_2 + x_3 \leq x_4, and so on xk-1 + xk \leq xk+1 for every k in the range {2, 3, ..., n-1}. Adding all of these up we obtain x_n \geq xn-2 + ... + x_2 + x_1 + x_2 (!).

But x_n is a side on [i]two[/i] of the faces of the polyhedron, one those two cannot share other sides, so we can choose one of them which hasn't got xn-1 as side. In that case the generalised inequality of the triangle applies, so x_n \leq xk_1 + xk_2 + ... + xk_i, for some k_1, ... , k_i among 1, 2, ... , n-2. But obviously this last inequality and (!) are in direct contradiction, so our assumption is false, and there exists indeed a triangle.

That's wrong Valentin :D Do you see the mistake?

ups I hurried, thanks for noticing. In fact it's the same solution, I edited it a little bit (extracting a x_{n-1} :D).


calculus
integration
function
calculus computations

hi all,
i am kind of stuck with this one.
find general soution
y"+(x-1)y'+y = 0
so i know this one is exact equation.

condition:  if  P(x)y"+Q(x)y'+R(x)y=0 and P(x)" - Q(x)'+R(x) = 0
then it is exact and can be written as [P(x)y']'+[F(x)y]' = 0
where F(x) = integral[ R(x) ]

so F(x) = x+k          (k=constant)

now plugging in  [P(x)y']'+[F(x)y]' = 0
i get
[y']'+[(x+k)y]' = 0

y"+y+xy'+k'y+ky' = 0

but y" = -(x-1)y'-y
so substitute
-xy'+y'-y+y+xy'+k'y+ky' = 0

y'+k'y+ky' = 0
 so k = - 1    k' = 0

now i am confused how am i supposed to proceed after this ?
thanks,

The null function $ \ f(x) \equiv 0 $  is solution of diff. equation.

F(x) = 0 is a solution true but how about other functions?( f(x) = 0 is generally a solution to many)

i think the method asks to solve the first order equation but i am confused which one?(i did find k and u are supposed to put that back in and end with a first order equation which leads to solution.) i was trying to use Abel's formula wronskians and try to xome up with the 2 solutions but it seems i am unable to find them..
thanks.

Your $F(x)$ from post #1 should be $x-1$. That leads to the equation

$y'+(x-1)y=C$.

If we try $C=0$, we get $y=e^{-\frac{x^2}2+x}.$

You can check this and find that it is one solution to the orginal homogeneous equation.

To find a second solution, I suggest the technique of reduction of order.

thanks, but the second solution using both techniques (reduction of order and wronskian-abel) gave me the second solution as

integral(e^(x^2/2)-x)) *  e^((x^2/2)-x)

where the the first integral is not computable in elementary functions.(not using erf functions etc)
is this correct?( i hope i made a mistake as it is really not a pleasant second solution :? )

Yes, that's correct. The second solution is not an elementary function. There's no reason we should assume it would be elementary.

Incidentally, we could fairly efficiently find power series centered at zero for two independent solutions just by assuming $y=\sum_{n=0}^{\infty}a_nx^n,$ differentiating that series, inserting it into the differential equation, and finding a recursive equation for the coefficients. As for the two independent solutions, for one, you can let $a_0=1,\, a_1=0$ and for the other, $a_0=0,\,a_1=1.$


I read the Da Vinci Code, Angels and Demons and I'm starting to read the Lost Symbol.

How much of this is actually true?  I mean, a lot of it seems like it could be, but then again, it's fiction.

What do you guys think?

It's based on actual historical fact but there are plenty of elements that are not true,  or altered.

*nods* . 'Tis good fiction, but its mostly only that. There are entire books dissecting which parts are true and what parts aren't :p

I've read the Lost Symbol, and its a really good book. I want to read The Da vinci code  but its usually out.

You should read them in order.  Doesn't really make a difference in the story but it is a series.  I found the Lost Symbol to be much slower than the other books, but at least the ending was a little better (I thought the endings for A&D and DC were terrible).


Lewis is thinking of a number. If he multiplies that number by 2, adds 4, divides by 5, multiplies by 10, finally divides by 5, he will get the number with which he started. What is the original number?

It's easier when you can read the $ \text{\LaTeX}$ better.

\begin{align*}
\frac{10 \left (\frac{2x+4}{5} \right )}{5}&=x\\
2 \left (\frac{2x+4}{5} \right )&=x\\
\frac{4x+8}{5}&=x\\
4x+8&=5x\\
x&=\boxed{8}
\end{align*}

[quote="r15s11z55y89w21"]Set up an equation.

$ 10*\frac{2x\plus{}4}{5}/5\equal{}x$.

Solving that, $ x\equal{}\fbox{8}$.[/quote]

try using \cdot instead of *.

Doesn't really matter as long as everyone knows what he's trying to say?


Let $ a,b,c,x,y,z > 0,$[size=150]such that  [/size]$ a^2 \plus{} b^2 \plus{} c^2 \equal{} x^2 \plus{} y^2 \plus{} z^2,$prove that:

$ \sqrt {3} > \frac {x^2 \plus{} y^2 \plus{} z^2}{a^2 \plus{} b^2 \plus{} c^2} > \frac {1}{\sqrt {3}}.$

BQ

[quote="xzlbq"]Let $ a,b,c,x,y,z > 0,$and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} x^2 \plus{} y^2 \plus{} z^2,$prove that:
$ \sqrt {3} > \frac {x^2 \plus{} y^2 \plus{} z^2}{a^2 \plus{} b^2 \plus{} c^2} > \frac {1}{\sqrt {3}}.$

BQ[/quote]

You should edit it :P

[quote="xzlbq"]Let $ x_i,y_i \in R^ \plus{} ,i \equal{} 1,2,...,n,n\geq 2,r \in N,f_x[n,r] \equal{} \sum_{i \equal{} 1}^n{x_i^r}, f_y[n,r] \equal{} \sum_{i \equal{} 1}^n{y_i^r}$,
such that  $ f_x[n,r] \equal{} f_y[n,r]$,prove that:

$ n^{1 \minus{} \frac {1}{r}} > \frac {f_x[n,1]}{f_y[n,1]} > n^{\frac {1}{r} \minus{} 1}.$

BQ[/quote]

if $ r \equal{} 1$, ineq get $ \equal{}$.

if $ r > 1$ ($ r$ can be reals), easy to prove that

$ f_x[n,1] > \sqrt [r]{f_x[n,r]} \equal{} \sqrt [r]{f_y[n,r]} \equal{} n^{\frac {1}{r}}\sqrt [r]{\frac {f_y[n,r]}{n}} \ge n^{\frac {1}{r}} \frac {f_y[n,1]}{n} \equal{} n^{\frac {1}{r} \minus{} 1} f_y[n,1],$

then we get the right hand, and the left hand is similar.

on this method, we can also get that when $ 0<r<1$ have

$ n^{1 \minus{} \frac {1}{r}} < \frac {f_x[n,1]}{f_y[n,1]} < n^{\frac {1}{r} \minus{} 1}.$


integration
calculus
calculus computations

Is it possible to prove that

$ \int_0^1\frac{\ln(x\plus{}1)}x\equal{}\frac{\pi^2}{12},$

applyin' double integration?

No, I cann't see any more about it! Maybe i am of low level in Calculus:(


limit
function
analytic geometry
graphing lines
slope
real analysis
real analysis solved

Prove that the following limit does not exist:

  <sub>x </sub> \lim <sub> \infty </sub> tan(x)/x.

i am not sure if i am correct but doesnt plugging in the series of tanx work just fine

[quote="galois"]... plugging in the series of tanx work..[/quote]

But we only know an expansion in power serie around 0 or around
b =/=  \pi /2 + n \pi not in infinity.

It is obvious. tan(x)/x attains arbitrary large values (for x=pi/2+pi*k-eps), and also zero values (for x=pi*k) as x goes to infinity.

Moreover, we can show that there is a sequence (xn){n>=1} which tends to 00 s.t. tan(xn)/xn tends to any real number L we want.

To show this just look at the graph of the tangent function and draw a line of slope L which passes through the origin. The line obviously intersects the graph of the tangent in infinitely many points, so we're done.

grobber's right
take
x_n=2n \pi +\pi/2-t/n(->oo), t \in R.
then your limit becomes:
lim{n->oo} (tg(x_n)/x_n) = lim{n->oo}1/(sin(t/n)*x_n)=
=lim{n->oo}(1/(sin(t/n)/(t/n)*(t/n)*x_n))=
=1/lim{n->oo}t*(2 \pi -\pi/(2n)-t/n²)=1/2t \pi.
So lim{n->oo}(tg(x_n)/x_n)=1/(2t \pi).

Right. So we have this result:

For all real a, the eqn tan(x)=ax has infinitely many solutions.

This is obviously much stronger than the fact that tan(x)/x has no limit.


Diophantine equation
number theory proposed
number theory

Let a ,b be positive intengers with $1 <a <b$ such that $ab$ divides $a^{2}+b^{2}-1$.Find all possible values of
$\frac{a^{2}+b^{2}-1}{ab}$.

1. $k\not =1$, because $k=1\Longrightarrow a=b=1$.
2. It is obviosly $a\not =b$. We have $2 =\frac{a^{2}+(a+1)^{2}-1}{a(a+1)},$
3. Let k>2 and even k=2m, then we have Pell's equation $(a+bm)^{2}-(m^{2}-1)b^{2}=1$. It had solution for all m.
4. k is odd give Pell equation $(2a+kb)^{2}-(k^{2}-4)b^{2}=4$ had solution, because $4=2^{2}$ and $k^{-}4$ is not square.


LaTeX limit calculus derivative function

The daily production cost, in pesos, for a desk manufacturer is y=0.3x squared -5.3x+546.8, where y is the daily production cost and x is the number of desks produced daily.

a.what is the daily production cost if the number of desks produced daily is 4,6,8,20 and 12?

b. What happens to the daily production cost as the company increases its daily production of desks?

c. How many desks should be the manufacturer produce to minimize the daily production cost?

d. What is the minimum production cost.

PLEASE HELP ME. Sad

Kindly show the solution and answers.

Over and out.

$ \text{\LaTeX}$'d: $ y=.3x^2-5.3x+546.8$.

[hide="a"]
Unless I'm missing something, don't you just plug in values of $ x$?
You should get the following:
530.4
525.8
523.6
560.8
526.4
[/hide]

[hide="b"]
Well... if $ x=100$, $ y$ gets pretty big.
Indeed, $ \lim_{x\to\infty}y=\infty$ (it diverges as $ x$ grows boundlessly large).
So I guess the answer is "it gets bigger."
[/hide]

[hide="c"]
This looks suspiciously like a calculus question (so this is the wrong forum), but anyway...
$ y'=.6x^2-5.3=0\implies x=5.3/.6\approx 8.83$, to which the nearest integer is $ 9$.
[/hide]

[hide="d"]
When $ x=9$, $ y=\$523.40$.
[/hide]

[quote="marcopolo"]The daily production cost, in pesos, for a desk manufacturer is y=0.3x squared -5.3x+546.8, where y is the daily production cost and x is the number of desks produced daily.[/quote]

[quote="marcopolo"]a.what is the daily production cost if the number of desks produced daily is 4,6,8,20 and 12?[/quote]

This part is easy -- just plug in each of the numbers in place of x.

[quote="marcopolo"]b. What happens to the daily production cost as the company increases its daily production of desks?[/quote]

In economics the quantity you're asking for is called the marginal cost of production and it is equal to the first derivative of the cost function, which as grn_trtle already said makes this a calculus problem that is beyond the scope of the High School Basics forum. Nevertheless, the marginal cost of production is negative if the number of desks produced per day is less than $ \frac{5.3}{0.6}$; otherwise it is positive.

[quote="marcopolo"]c. How many desks should be the manufacturer produce to minimize the daily production cost?[/quote]

The daily production cost is minimum where the marginal cost of production is zero. This occurs at $ x \equal{} \frac{5.3}{0.6}$.

[quote="marcopolo"]d. What is the minimum production cost.[/quote]

Plug the value for $ x$ found in part (c) into the formula for $ y$.


probability
symmetry
algebra
polynomial
geometry
3D geometry

A bug begins on a particular vertex of a cube. Each move it randomly and independently selects a new vertex to move to, according to pre-determined rules (given below). Find the probability that the bug will return to its starting vertex after $ n$ moves if:

$ [ \text{a} ] \blacktriangleright$ The bug may move along edges and face diagonals, but [color=red][b]not [/b][/color]space diagonals.
$ [ \text{b} ] \blacktriangleright$ The bug may move along edges and space diagonals, but [color=red][b]not [/b][/color]face diagonals.

[i]You may find [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215790]this thread[/url] to be helpful.[/i]

[hide="(a)"]
We define three sets of vertices: $ A$ containing the starting point, $ B$ containing the 6 points adjacent to $ A$ by either an edge or a face diagonal, and $ C$ being the point opposite to $ A$. Note that we don't need to distinguish between whether a point in $ B$ is connected by edge or diagonal to $ A$, because the probability that we move from $ A$ to either an edge or a diagonal is the same.

Let $ a_n, b_n, c_n$ be the probabilities that the bug is at points $ A,B,C$ at move $ n$, such that $ a_0 = 1, b_0 = c_0 = 0$. On the second move, the bug must proceed to $ B$, and on the third move, by symmetry we find that $ a_2 = c_2$. Indeed for $ n > 0$ we have $ a_n = c_n$, so we combine both into set $ A$ and halve the probability later. The only way to reach $ A$ is to take one of two segments from any point in $ B$ (out of six possible edges), and to reach $ B$ is to either leave from $ A$ or one of four segments from any point in $ B$ (again out of six); thus we have the recursions \begin{align*} a_{n+1} &= \frac{1}{3}b_n \\ b_{n+1} &= a_n + \frac 23 b_n\end{align*} By substitution we have $ a_{n+2} = \frac 23 a_{n+1} + \frac 13 a_{n}$, and using the characteristic polynomial we find that $ a_n = \frac 14 + \frac 34 \left(\frac {-1}3\right)^n$. Therefore \[ P(n) = \begin{cases}1 & n = 0 \\ \frac 18 + \frac 38 \left(\frac {-1}3\right)^n & n \ge 1\end{cases}.\][/hide][hide="(b)"]We re-define the sets of vertices as follows: Let $ A$ contain the starting point, $ B$ contain the opposite point, $ C$ containing the $ 3$ points adjacent to $ B$ by edges, and $ D$ the $ 3$ points adjacent to $ A$ by edges. Then we note that from $ A,C$ we can only move to $ B,D$ and vice versa. Also, we find that $ c_n = 3a_n$ by setting up recursions similar to above, so we find that $ P(n) = 1$ when $ n = 0$, $ P(n) = 0$ when $ n$ is odd, and $ P(n) = \frac 14$ when $ n > 1$ is even.
[/hide]

Nice :)

The intended solution (this is why I linked that thread) utilized the following two bijections:

[hide="Bijection for (a)"]
Label the four opposite pairs of vertices A, B, C, and D. From a vertex A, the bug may move to either a vertex B, C, or D, and similarly for others. We can draw an analogy between this and a tetrahedron. So we find the probability that the bug returns to the starting vertex of a tetrahedron. But since there are two point A's in the cube, etc., we must just divide by 2 (after the first move, all points with the same label are completely symmetric)[/hide]

[hide="Bijection for (b)"]
Label the starting vertex A. Label one of the vertices connected to the starting point A, and the other two such points C. Then label the other four points B and D, such that from either a point A or a point C, we may move only to either of the point B's or either of the point D's. The opposite will hold if you are starting on B or D (you must move to an A or a C).

Now, we can draw analogy between this and a square with vertices A, B, C, D, labeled clockwise. Like before, we find the probability of returning to point A in the square, and then divide by 2.[/hide]


the sum of 2 #s is 84 the first is 9 more than 4 times the second.find the #s

let two numbers be x and y.

x=4y+9

so y+4y+9=84, 5y=75, y=15

answer : 15, 69


inequalities
geometry
geometric transformation
reflection

Don't know how hard/easy this might be:

Given positive reals $x$ and $y$ and $\frac{1}{x}+\frac{1}{y} = 2$, prove that

$x+y \ge \frac{(x^{\frac{1}{3}}+y^{\frac{1}{3}})^3}{4xy}$.

Edited to hide my solution and to reflect the changes in the original inequality:

[hide]It can't be hard, it's homogeneous (or can be made so)

Turn it into $(x+y)^3\geq xy(x^{\frac 13}+y^{\frac 13})^3$

This is easy; $4(x+y)\geq (x^{\frac 13}+y^{\frac 13})^3$

and $(x+y)^2\geq 4xy$

Combine these and we're done.[/hide]

As an addendum, I was thinking of obvious ways to make this inequality stronger

Substituting $4\sqrt{x^2+y^2}$ doesn't work, but its sooooooooooo close that almost any slightly weaker term would work.

I don't quite understand blahblahblah's solution.  Power Mean seems to do everything for you:

$\sqrt{\frac{1}{xy}} \le \frac{\frac{1}{x}+\frac{1}{y}}{2} = 1 \implies xy \ge 1$

$xy(x+y) \ge x+y \implies x+y \ge \frac{4(x+y)}{4xy}$

Now note that by power mean

$\frac{x+y}{2} \ge \Big( \frac{x^{\frac{1}{3}}+y^{\frac{1}{3}}}{2} \Big)^3$

$4(x+y) \ge x^{\frac{1}{3}}+y^{\frac{1}{3}}$

Thus

$x+y \ge \frac{x^{\frac{1}{3}}+y^{\frac{1}{3}}}{4xy}$

My first step was to count degrees of both sides.  If I move the $xy$ to the other side, the LHS has degree 3, the RHS has degree 1.  Multiplying the LHS by $\textstyle\sum 1/x$ and the RHS by 2 decreases the degree by 1 each time.  So after we do it twice, we get the expression I had.

Now my first inequality follows easily from Muirhead (or Power-Mean, but for some reason I dislike that inequality)
The second one is just AM-GM


Show that any integer can be expressed as the form:

$a^{2}+b^{2}+c^{3}$

where $a,b,c \in Z$

a) Posted before.
b) Bad title ( http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 ).


I wanted to "stack" two lines under a summation sign.

I want $\sum_{k=1,k\,\mathrm{odd}}^{\infty}a_k$

only I want "k=1" and "k odd" to be on two lines, one below the other. (In the same size type, of course.)

How do I do that?

$ \sum_{ a=1 \atop b=2 } a^2b^2 $ by any chance? :)

Use substack:
[code]\sum_{\substack{k = 1 \\ k \ \text{odd}}}^\infty a_k[/code]
to get
$\sum_{\substack{k = 1 \\ k \ \text{odd}}}^\infty a_k$

Thanks. There seem to be two ways: \substack or \atop.

Using \atop:

$\sum_{k=1\atop k\,\text{odd}}^{\infty}a_k$

Using \substack:

$\sum_{\substack {k=1 \\ k\,\text{odd}}}^{\infty}a_k$

... slightly different results. Different font sizes, different vertical spacings.


algebra
function
domain
greatest common divisor

For each positive integer m > 1, let P(m) denote the greatest prime factor of m.  For how many positive integers n is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$ ?
a.) 0 b.) 1 c.) 3 d.) 4 e.) 5
Help please...

[hide]n has to be a square of a prime, $n=p^2$, (try to figure out why), and the other has to be a square of a prime too, and let $q^2=p^2+48$, $q$ must also be a prime
$p^2+48=q^2$
$48=(q-p)(q+p)$
$q=\frac{ \frac{48}{d}+d}{2}$
where $d$ is a divisor of $48$, testing,
$q=13,8,7 \implies p=11,4,1$ but the 2 solutions don't work since $8$ is not prime and 1 is not in the domain, so 1 solution $\boxed{B}$[/hide]

Wow :omighty: , does n have to be a square of a prime because if it is not then the $\sqrt{n}$ will not be a whole number...?  n then has to be a square of a prime for both equations...
That's a neat equation to test for values of q... how did you get that??

[quote="Apprentice"]That's a neat equation to test for values of q... how did you get that??[/quote]

$48=(q-p)(q+p)$ we know that both factors on the right are positive integers so they must be factors of 48, if
$(q-p)=d$, then $q+p=\frac{48}{d}$, then to solve for q, we add them together and divide by 2 to get the desired expression

It has to be the square of a prime because if it's the square of a composite number then the greatest common factor is composite, and if it's not a square then it won't be in the domain.


complex numbers
real analysis
real analysis unsolved

$ \sum_{n\equal{}0}^{\infty}x_n$ is a sequence in $ \mathbb{C}$ that converges but  $ \sum_{n\equal{}0}^{\infty}|x_n|$ does not converge. Show that there is bijection $ \pi \ : \mathbb{N} \rightarrow \mathbb{N}$ so that
\[ \sum_{n\equal{}0}^{\infty}x_{\pi(n)}\]
does not converge.

Use the fact that the sum of the positive terms must diverge.

Lydianrain is right, except remember its in the field of complex numbers, not real numbers. Anyhow, first consider real numbers.

Consider an infinite sequence $ {x_0, x_1, x_2, ...}$. This sequence has both positive and negative terms, and the series $ \displaystyle\sum_{n\equal{}0}^\infty |x_n|$ diverges. If all the negative terms of the sequence are eliminated, must this series diverge?

Lydianrain already told us yes, and here's a small demonstration. Let the mapping $ \pi$ create the sequence $ {c_0,c_1, ...}$ of all the positive terms in $ {x_n}$ and another mapping create the sequence $ {d_0, d_1, ...}$ of all $ x_n<0$. Since the sequence $ \displaystyle\sum_{n\equal{}0}^\infty x_n$ converges, and it is equal to the sum of the elements of $ {c_n}$ and $ {d_n}$, $ \displaystyle\sum_{n\equal{}0}^\infty c _n \plus{} \displaystyle\sum_{n\equal{}0}^\infty d_n \equal{} C$, where C is a constant real number. So if either sum did not diverge, neither could the other. But then the sum of all $ |x_n|$ must converge since it is the difference of two convergent series, which is against hypothesis.

Now we must extend this reasoning to complex numbers. Note that a complex number has a real and imaginary component. Addition of the imaginary components is closed and so is the addition of the real components, and the addition of the imaginary components is essentially an addition of real numbers which in the end are multiplied by $ i$. We have shown that there always exists a mapping $ \pi$ following the conditions of the problem for real numbers. If a series of complex numbers converges, $ \displaystyle\sum_{n\equal{}0}^\infty x_n$, then both its real components and its imaginary components would converge. Let's call $ a_i$ the real part of a complex number and $ b_i$ the imaginary part of a complex number $ x_i$. By hypothesis, the series $ \displaystyle\sum_{i\equal{}0}^\infty \sqrt{a_i^2\plus{}b_i^2}$ diverges. Note that if either $ \displaystyle\sum_{n\equal{}0}^\infty |a _n|$ or $ \displaystyle\sum_{n\equal{}0}^\infty |b_n|$ diverges, the modulo sum diverges because its greater than or equal to either of these. So, if the modulo sum diverges and the normal sum converges, must there exist an infinite  subset $ {f_n}$ of either $ {a_n}$ or $ {b_n}$ whose sum diverges?

Suppose the contrary, that every infinite subset of each of these must have its sum converge. Then $ \displaystyle\sum_{i\equal{}0}^\infty |a_i|$ must converge and $ \displaystyle\sum_{i\equal{}0}^\infty |b_n|$ must converge. $ \displaystyle\sum_{i\equal{}0}^\infty |a_i|\plus{}|b_i|$ converges and is greater than or equal to $ \displaystyle\sum_{i\equal{}0}^\infty \sqrt {a_i^2\plus{}b_i^2}$, so then this must also converge. So we have reached a contradiction. A subset $ {f_n}$ must exist whose sum diverges. Whether it is a sequence of real or imaginary parts does not matter, it is only important that one of these diverge for the entire sum to diverge. The mapping $ \pi$ can map $ N$ to the $ n$'s that correspond to the elements of $ {x_n}$ that correspond to the elements of $ {f_n}$, and all these sets have already been shown to never be null, and therefore such a mapping $ \pi$ always exists.

Details, details. :P

Thanks for the correction.  It's so hard to tell my Cauchy-complete fields apart sometimes...

Thanks !

It's your first post, and at the same time a great one :p


Z joins two others, A and B. Z whispers to A the sum of two natural numbers and to B the sum of the squares of the same two natural numbers. Each knows the nature but not the detail of the information being conveyed. The following conversation takes place:

B: I do not know the numbers.

A: I do not know the numbers.

B: I do not know the numbers.

A: I do not know the numbers.

B: I do not know the numbers.

A: I do not know the numbers.

B: I know the numbers.

What are the two natural numbers?

Have A Great Day!

Let $ (m,n)$ henceforth denote the two natural numbers we're looking for.  Now assume A and B are either idiot savants or brilliant number theorists.  Since B knows the sum of the squares but can't figure out the numbers, the number B got must be expressible as a sum of squares in more than one way.  i.e.
\[ (m,n)\in S_1 \equal{} \left\lbrace(a,b)\in \mathbb{N}^2: \exists (x,y)\in\mathbb{N}^2\backslash\lbrace (a,b),(b,a)\rbrace, x^2 \plus{} y^2 \equal{} a^2 \plus{} b^2\right\rbrace
\]
Now A knows the sum of the numbers and that $ (m,n)\in S_1$ (which he infers from B's claim that he doesn't know the number), but he doesn't know the numbers, thus
\[ (m,n)\in S_2 \equal{} \lbrace (a,b)\in S_1: \exists (x,y)\in S_1\backslash\lbrace (a,b),(b,a)\rbrace, x \plus{} y \equal{} a \plus{} b\rbrace
\]
Continuing on in this fashion...
\[ (m,n)\in S_3 \equal{} \lbrace (a,b)\in S_2: \exists (x,y)\in S_2\backslash\lbrace (a,b),(b,a)\rbrace,x^2 \plus{} y^2 \equal{} a^2 \plus{} b^2\rbrace
\]

\[ (m,n)\in S_4 \equal{} \lbrace (a,b)\in S_3: \exists (x,y)\in S_3\backslash\lbrace (a,b),(b,a)\rbrace,x \plus{} y \equal{} a \plus{} b\rbrace
\]

\[ (m,n)\in S_5 \equal{} \lbrace (a,b)\in S_4: \exists (x,y)\in S_4\backslash\lbrace (a,b),(b,a)\rbrace,x^2 \plus{} y^2 \equal{} a^2 \plus{} b^2\rbrace
\]

\[ (m,n)\in S_6 \equal{} \lbrace (a,b)\in S_5: \exists (x,y)\in S_5\backslash\lbrace (a,b),(b,a)\rbrace,x \plus{} y \equal{} a \plus{} b\rbrace
\]
Now B knows the answer, so it must be that
\[ (m,n)\in S_7 \equal{} \lbrace (a,b)\in S_6: \forall (x,y)\in S_6\backslash\lbrace (a,b),(b,a)\rbrace,x^2 \plus{} y^2\neq a^2 \plus{} b^2\rbrace
\]
So we can answer the question "what are these two natural numbers" if and only if $ \left \vert S_7 \right \vert \equal{} 2$.  Now if you are an idiot savant or a brilliant number theorist, you may well be able to prove that $ \left \vert S_7 \right \vert \equal{} 2$, and find those element.  I am neither, so I wrote a quickie Matlab script to find elements of $ S_7$, which returns that $ (8,9),(9,8)\in S_7$ are solutions, and that there are definitely no others within the search range $ [1,3000]^2$.

Um..could you please say that again? I didn't really get that...


linear algebra
matrix
linear algebra unsolved

Let $A= \left(\begin{array}{ccc}a & b \\ c & d \\ e & f \\ g & h\end{array}\right)$ and $B= \left(\begin{array}{cccc}i & j & l & m \\ n & o & p & q\end{array}\right)$ two inversible matrices of whole numbers. Prove or disprove that the matrix $AB$ is inversible.

Since we can take any whole numbers i thought that it depends which number you take. If somebody helps... thanx!

Sorry about my english!

The closest you can get with $A$ and $B$ by themselves is that you could ask that they be of maximal rank, which in this case would be rank 2. In any case, $\text{rank}(A)\le 2$ and $\text{rank}(B)\le 2.$ Then there's this vitally important theorem:

$\text{rank}(AB)\le \min(\text{rank}(A),\text{rank}(B)).$

Than means that $AB$ will be a $5\times 5$ matrix whose rank is at most $2.$ Thus $AB$ cannot be invertible.

It is possible - even likely - for $BA,$ which is a $2\times 2$ matrix, to be invertible.

[quote="jmerry"]$A$ and $B$ cannot be invertible, as they're not square.[/quote]
The question in about the matrix AB  :D

[quote="Kent Merryfield"]santosguzella, jmerry was reacting to the first sentence of your original post, in which you said
[quote]Let ... two inversible matrices of whole numbers.[/quote]
You must not have meant that, but you did say it.[/quote]

Oh man! Sorry.. my mistake!  :(


function
number theory proposed
number theory

Find the maximal number of functions $ f: \mathbb{N}\to\mathbb{N}$ which satisfy the conditions:
1) For any $ m,n\in\mathbb{N}$ it is true $ f\left(m \plus{} n\right) \equal{} f\left(m\right) \plus{} f\left(n\right) \plus{} 2mn$;
2) For only one of this functons $ f\left(30\right)\ge2009$;
3) None of this functions is such that $ f\left(29\right)\ge2009$.

Let $ g(n)\equal{}f(n)\minus{}n^2$, then $ g(m\plus{}n)\equal{}g(m)\plus{}g(n)\to g(n)\equal{}an\to f(n)\equal{}n^2\plus{}an$.


Let $ \minus{}1<a_0<1$ and define recursively-
$ a_n\equal{} (\frac{1\plus{}a_n\minus{}1}{2})^{\frac{1}{2}}$ , $ n>0$,
Let $ A_{n} \equal{} 4^n(1\minus{}a_n)$. Does the A sequence converge? if not - why not? if yes - to what?

I think the sequence diverges... but let me see if that's correct, I'll post again(or edit this post) when I have a full solution.

Put $ a_0\equal{}\cos \theta$ and $ a_n \equal{} \cos \left(\frac{\theta}{2^n}\right)$
apply l'hopital's rule twice; $ A_n \to 0$

[quote="TZF"]I'm going to assume you meant to write $ \boxed{ a_{n} \equal{} \sqrt {\frac {1 \plus{} a_{n \minus{} 1}}{2}}}$[/quote]
yes you are right... i guess got a check my latex after i post..sorry

[quote="ocha"]Put $ a_0 \equal{} \cos \theta$ and $ a_n \equal{} \cos \left(\frac {\theta}{2^n}\right)$
apply l'hopital's rule twice; $ A_n \to 0$[/quote]
l' wat?? well i dont know that and is there a solution without that? i used - $ \frac {sinx}{x} \rightarrow 0$
wat is it anywy?

[quote="ocha"]apply l'hopital's rule twice; $ A_n \to 0$[/quote]

Are you sure?

[hide="Evaluation of limit"]$ A_n \equal{} 4^n \left(1 \minus{} \cos \left( \frac {\theta}{2^n} \right) \right)$

$ \equal{} 4^n \left(1 \minus{} 1 \plus{} \frac {1}{2} \left( \frac {\theta}{2^n} \right)^2 \minus{} \frac {1}{24} \left( \frac {\theta}{2^n} \right)^4 \plus{} \ldots \right)$

$ \equal{} \frac {1}{2} \frac {\theta ^2}{2^{2n \minus{} 2n}} \minus{} \frac {1}{24} \frac {\theta ^4}{2^{4n \minus{} 2n}} \plus{} \ldots$

$ \to \frac {1}{2} \theta ^2$

$ \equal{} \frac {1}{2} \left( \arccos a_0 \right) ^2$[/hide]

Yeah sorry you're right
[hide="another way to evaluate the limit"]
Since $ \frac{\theta}{2^x}\to 0$ our limit is equivalent with

$ \lim_{x\to \infty}   \theta^2 \cdot \frac{1\minus{}\cos\left(\frac{\theta}{2^x}\right)}{\left(\frac{\theta}{2^x}\right)^2}$

$ \equal{} \lim_{x\to 0} \theta^2 \cdot \frac{1\minus{}\cos x}{x^2}$

multiply top and bottom with $ 1\plus{}\cos x$

$ \equal{} \lim_{x\to 0} \theta ^2 \cdot \frac{1\minus{}\cos^2x}{x^2(1\plus{}\cos x)}$

$ \equal{} \lim_{x\to 0} \theta^2 \cdot \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1\plus{}\cos x} \to \theta^2\cdot 1^2 \cdot \frac{1}{1\plus{}1}$

$ \equal{}\frac{\theta^2}{2}$

by well known limits[/hide]


I'm a 7th grader going to nationals and there were 1 or 2 other 7th graders on the state team (I can't remember which). How many 7th graders are on your state's team?

[quote="Fanatic"]*waves*

I'm a 7th grader going to Nationals. (wink wink)

I'm [b]5th[/b] place, but the coach invited me to go to Nats with them. I was 50/50 at first, but my brother said that I definitely should so I will.

Anyhow, the answer to the poll is 0-1 anyways.[/quote]

i hoped i could go to nats but i only got 10th:(

I got fourth, so I'm going.

Wohooo!!

Are there any 6th graders going?

I'm a sixth grader, I hope to get to nationals. I made it to state in Colorado.

Yeah kevin chen from Texas got 30-something

O YEAH i won states today. we have 1 seventh grader

I think we have 1 seventh grader. ME. The other three kids looked like 8th graders but I'm not 100% sure.

We have 1 7th grade on the Ohio team, she's a girl too...

There's a 6th grader on Vermont's team!  He came first at States.  I'm not sure about the grade levels of the other 3.  All boys this year.  Last 2 years we sent 1 girl.

The poll should split 0 and 1 because now we don't know the diff...

There are no seventh graders from CA :( .

there is only one 7th grader from texas and thats me  :)

EDIT: oops i see that u hav already discussed this earlier

[quote="pgpatel"][quote="xxm0rg07hxx"]Two 7th graders from AR this year.  And two from AR last year.  The two 8th graders going this year are the two 7th graders that went last year.[/quote]
Are you one of the 8th or 7th. On a side note I average .61 posts a day even though I have 2 stars!!!  :ninja:[/quote]

I'm one of the 8th.  I went last year.

The Maine team is all 8th graders this year- last year we had three seventh graders.

[quote="aznness"][quote="Klebian"][quote="biffanddoc"]That was nats right??? (although hes really god I hope he bombs at state :D :bomb:)[/quote]
Yes, as a 6th grader, he got 30-something at nats last year.[/quote]

I got 30 last year, and i was 17th. there were no 6th graders in countdown...[/quote]
30 something-th place.
29th actually.

Yeah, I'm 7th grade.

To azness: Was last year's test ridiculously hard or something? I talked to Mark and he sayed he got a 35 while Sergei Bernstein had a 39.

No seventh graders from IN, 1 last year (Nathan), 3 from HCMS! :trampoline: No sixth graders either :weightlift: We're training for next year.

P.S. Technically, I'm a sixth grader going to nats, but as a guest. :D

There aren't any 7th graders from VA, at least I think.  There were 2 eigth graders who went last year though.  I'm gonna go  to nats just to watch anyways.  Maybe I can get some pins.  They should let the top 200 people in the COUNTRY go to nats.  I think it's unfair that South Dakota gets 4 people, and so does CA.

I think that the way they do it isn't the fairest way, but definetely the most logical.

The whole point of national MATHCOUNTS is to represent your state. If there are 50 people from California and none from Puerto Rico (logical if they chose top 228), then that completely defeats the point. Also, they would HAVE to cut out the team round, and that is a huge chunk of the competition as well. If you have people with 24 going to nats in delaware and 42 not in Indiana, but have 4 people from each state, then it is still the best :)

Maybe its because Im from DE and this way benefits DELAWAREANS, but I think that would not be a good idea, since in USAMO, your not representing your state like in mathcounts at a national competition. Also, the USAMO is not a separate contest; it is just another qualifying test for the IMO team. What your saying is kinda like "every country should take the USAMO (or whatever their country has) you have to have 30 or something to make IMO. If your country has noone, then you can send a representative. and by the way, you cant place as a team"

There were 97 mathcounts participants in DE last year.

There were 98 at my chapter competition last year.

Yes this is pretty unfair, but then again GA has theoretically "smarter" people / penguins going to nationals. (I hope.)

Keep in the mind the ultimate goal isn't just to get to Nationals, it's to win Nationals (unrealistic for 227 people including me) and have fun :)
Although that's unrealisitc, do you think it would be fair for 229th in the Nation to lose the tiebreaker? Seeing as Connecticut is about an average level competition, (200+ at 4 chapters, ~90 at 1, almost 200 at state) I would still want the Top 4 from each state because if it's not, it ruins many things like someone brought up teams, sharing the experience, pin trading, and lots of fun in general.

If you wanted to be fairest, you could say 8th graders auto win tiebreakers against 7th graders, 7th against 6th as 6th and 7th graders will get more chances and probably qualify the next year, but of course, that's stupid.

PS. We're really off topic  ;)

[quote="kyyuanmathcount"]They should do a half MATHCOUNTS half USAMO thing, where you have to have at least a 34 to go or something, but your state forfeits the team round and has the top two people to go if they dont have enough 34+ people.[/quote]

I agree whole-heartedly.  :yup: We need a consolation thread or something.

I think by teams the way it's done now it is okay.  The only thing is that a state with a high population(CA,TX) could bring up to 4 individuals to nats like the schools can do at chapter.

Yeah, I really feel sorry for those from California that got 41 and KNEW they could have gone to nationals in ANY other state, and that 1st place guy in SoCal.

I would say maybe having a base number of people, and per unit of population, you get to bring another.

Don't forget, the more people going to nationals, the more $\$\$\$\$$ sponsors have to pay.

They spend atleast $\$$500 per student.

Going back to the poll, Here is an analysis of the gradewise composition of the states.
The State Department needs to post their list going to nationals.

[hide]
Total:
162-8th
56-7th
6-6th
4-not known yet.

4-8th:
AL, CA, CO, GA, IL, IN, KY, ME, MT, NM, OK, SC, SD, N. Mariana Islands, Virgin Islands

3-8th, 1-7th:
AK, DE, FL, IA, LA, MD, MI, MN, NV, NH, NY, OH, PA, RI, TN, TX, VA, WA, WV, WI, WY

2-8th, 2-7th:
AZ, AR, CT, DC, HI, ID, KS, MS, MO, NJ, NC

1-8th, 3-7th:
MA, NE

4-7th:
UT

3-8th, 1-6th:
ND, Defence Department, Peuto Rico

1-8th, 2-7th, 1-6th:
Guam

1-8th, 1-7th, 2-6th:
VT

[/hide]

[quote="kyyuanmathcount"]I would say maybe having a base number of people, and per unit of population, you get to bring another.[/quote]

How about base unit of Mathcounts participants.
WY has less population than DE but more participants.

[quote="Xantos C. Guin"]Don't forget, the more people going to nationals, the more $\$\$\$\$$ sponsors have to pay.

They spend atleast $\$$500 per student.

Going back to the poll, Here is an analysis of the gradewise composition of the states.
FL, GA,  and the State Department need to post their list going to nationals and OR needs to post the other 2 qualifiers.

[hide]
Total:
153-8th
55-7th
6-6th
14-not known yet.

4-8th:
AL, CA, CO, IL, IN, KY, ME, MT, NM, OK, SC, SD, N. Mariana Islands, Virgin Islands

3-8th, 1-7th:
AK, DE, IA, LA, MD, MI, MN, NV, NH, NY, OH, PA, RI, TN, TX, VA, WA, WV, WI, WY

2-8th, 2-7th:
AZ, AR, CT, DC, HI, ID, KS, MS, MO, NJ, NC

1-8th, 3-7th:
MA, NE

4-7th:
UT

3-8th, 1-6th:
ND, Defence Department, Peuto Rico

1-8th, 2-7th, 1-6th:
Guam

1-8th, 1-7th, 2-6th:
VT

[/hide]

[quote="kyyuanmathcount"]I would say maybe having a base number of people, and per unit of population, you get to bring another.[/quote]

How about base unit of Mathcounts participants.
WY has less population than DE but more participants.[/quote] Florida has 1 seventh grader (95% sure). ME. I am 5 percent unsure.

It is unfair, but it also kind of makes...hard to explain...

Ehehheehee: I made Nationals in Sector D-3
Person: what are you talking about?

Maybe it should be like the U.S. House of Representatives?

[quote="xxm0rg07hxx"]Ask Bill Gates to sponsor mathcounts.  That would help, don't you think?[/quote] :rotfl:  :rotfl:  :rotfl:  :rotfl: :rotfl:. It will help a lot if we tell him we're a bunch of [b]nerds[/b]. He did say be nice to nerds, someday you might be working for one.

[quote="biffanddoc"]NO! The cost is alone a factor enough to make MATHCOUNTS reject the idea. What Xantos C. Guin said about cost is quite low. First, airfare depending on where you live. If you live in CA, it is AT LEAST $\$$400 per person. The hotel rooms are probably around 300, making it another 150 per night, and for 3 nights is 450. The calculator is worth at least 100. The meal total is nearly 200. (I counted based on the guest price). That totals to 1150, not including more tests, grading, transportation to airport and back, t-shirt, etc, which probably total to about 50. That is 1200 total!!!
If you take, say 100 more people, that is 120,000 more dollars!!![/quote]

That's what I did, but I used the price from GA of 290 not from CA. Also I counted the calculator as 130. I completely forgot the T-shirt, and buses and other things.


inequalities
trigonometry
inequalities unsolved
BPSQ

Prove that in any acute triangle we have:

$\sin A + \sin B + \sin C >2.$

Its  very  easy  since  in any triangle  ABC we have $ sinA+sinB+sinC=\frac{s}{R}  $  and  since  ABC is  acute we  have  $s>2R+r$

This was discussed here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5379

  Darij

An improvement of this inequality without use of any lemmas follows from the observation

SinA +SinB +SinC > SinA)^2 + (SinB)^2 +(SinC)^2 = 2 +2CosACosBCosC>2 since the product of the cosines is positive in an acute-angled triangle.

Another proof can be based on the Jordan Inequality SinX/X >2/pi where X lies in the range0<X<pi/2.
Writing X equal to A B and C in turn and adding the three inequalities gives the result at once

[quote="Pantum"][quote="sqing"]In any acute triangle ,we have:
$\sin A + \sin B + \sin C >\sqrt{2}(\cos  A + \cos B + \cos C).$

In any acute triangle
$\cos A + \cos B + \cos C >\sqrt{2} ?$[/quote]

$\cos A + \cos B + \cos C=1+4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2} }$

but we can take  $ \sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}\rightarrow 0 }$[/quote]
In any acute triangle，have
$sin^2A+sin^2B+sin^2C\ge \((cosA+cosB+cosC)^2 .$

$2cosAcosBcosC+2\ge \((cosA+cosB+cosC)^2 .$

[quote=sqing]In any acute triangle ,we have:
$$\sin A + \sin B + \sin C >\sqrt{2}(\cos  A + \cos B + \cos C).$$
[/quote]
In any acute triangle , [url=https://math.stackexchange.com/questions/3238915/proving-that-for-an-acute-triangle-abc-sin-a-sin-b-sin-c-gt-cos-a-c]prove that[/url] $$\sin A + \sin B + \sin C >\cos  A + \cos B + \cos C.$$
[b]Solution:[/b]
$$A + B>\frac{\pi}{2}\implies \frac{\pi}{2}> A> \frac{\pi}{2}-B>0 ,\implies sin A>cosB,$$
Similarly，
$$ sin B>cosC,  sin C>cosA.$$

[quote=Mitzah]Prove that in any acute triangle we have:
$$\sin A + \sin B + \sin C >2.$$[/quote]
[url]https://artofproblemsolving.com/community/c6h172002p952771[/url]
[url]https://artofproblemsolving.com/community/c7h1657257p10506374[/url]
[url]https://artofproblemsolving.com/community/c260h1551195p10510984[/url]

[quote=sqing]In any acute triangle , [url=https://math.stackexchange.com/questions/3238915/proving-that-for-an-acute-triangle-abc-sin-a-sin-b-sin-c-gt-cos-a-c]prove that[/url] $$\sin A + \sin B + \sin C >\cos  A + \cos B + \cos C.$$
[b]Solution:[/b]
$$A + B>\frac{\pi}{2}\implies \frac{\pi}{2}> A> \frac{\pi}{2}-B>0 ,\implies sin A>cosB,$$
Similarly，
$$ sin B>cosC,  sin C>cosA.$$[/quote]
[b]Another solution :[/b]
$$sinA+sinB>1,sinB+sinC>1,sinC+sinA>1,$$
$$\sin A + \sin B + \sin C= (sinB+sinC)\cos A + (sinC+sinA)\cos B + (sinA+sinB)\cos C $$$$>\cos  A + \cos B + \cos C.$$


geometry
3D geometry
sphere
integration
calculus
calculus computations

Can someone explain how to solve this problem with basic volume techniques of Calc. I?

A round hole of radius  :sqrt: 3 feet is bored through the center of a sphere of radius 2 feet. Find the folume of the piece cut out.

Thanks.

Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve $x^2+y^2=4$ and on the left by the line $x=\sqrt{3}.$ The points where these curves intersect are $(\sqrt{3},1)$ and $(\sqrt{3},-1).$ We want the volume of the solid of revolution formed by rotating this about the $y$ axis.

Method 1 (the "shell" method):

Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position $x,$ has width $dx,$ and height $2\sqrt{4-x^2}.$ Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width $2\pi x$ (the circumference of the shell), height $2\sqrt{4-x^2},$ and thickness $dx$ so $dV=4\pi x\sqrt{4-x^2}\,dx$ and

$V=\int_{\sqrt{3}}^24\pi x\sqrt{4-x^2}\,dx.$

Method 2 (the "washer" method):

Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position $y$ and has width $dy.$ Its left endpoint is at $x=\sqrt{3}$ and its right endpoint is at $x=\sqrt{4-y^2}.$ Rotating this strip about the vertical axis gives a thin flat "washer" of thickness $dy$. The inner radius of this (annulus) is $\sqrt{3}$ and the outer radius is $\sqrt{4-y^2}.$ The volume is

$dV=\pi\left(\left(\sqrt{4-y^2}\right)^2 -\left(\sqrt{3}\right)^2\right)dy = \pi(1-y^2)dy.$ Thus

$V=\int_{-1}^1 \pi(1-y^2)dy.$

I'll leave you to do the integrals; both can be done by elementary means but this time method 2 is probably easier.


calculus
derivative
function
real analysis
real analysis unsolved

ok i need some help.......
for example u are given a graph of 3x^3-2 and thats a derivative of something u dont know what, u dont even know the equation all they give u is the graph of it.

and using that u need to determine
MINs/MAXs
critical points
points of inflection

so how do u do that??
PLEASE HELPPPPP

Find teh zeros of this equation, thus the max and min. Using second order derivatives, find the convexity and concavity of locations on the graph.

yea but u dont know the equation all is given is the graph of it..................i just wrote it as an example so u guys knows what it looks like.........
but actually ALL YOU KNOW IS THAT GRAPH

You're given a graph of a derivative. All the information you have is the sign of the derivative and its zeros. That's all you need.

A zero of the derivative is a critical point of the function. That means it could be a max or a min or an inflection point. You don't know which.

What happens to the derivative right before and after the zero tells you what happens at the point. Think about what the sign of the derivative means and see if you can figure that part out yourself.

(A point of inflection is when the derivative has a 0, but the function itself doesn't have a max or a min, it just flattens for an instant and then keeps going like it was before.)

Hmm I seem to disagree with Xevarion in this case... An inflection point is a point where a function goes from being concave up to being concave down, or vice versa.  (Right?)  The derivative of a function doesn't have to be 0 at an inflection point.

Sorry, DVO is right. What I should have said is when the derivative is 0, but it isn't a max or a min, it must be an inflection point. But the other direction is not true. An inflection point does not necessarily mean the first derivative is 0. It means the second derivative is 0, and so you also need to find those points. (How can the graph of the first derivative of a function tell you when the second derivative is 0? :) )


Hi everyone,

Could someone please help me on this problem?

I have to come up with a proof to show that a word, written in capital letters, is unchanged when reflected in a horizontal line and a vertical line. That just means that it needs to be able to be reflected across the x and y-axis, right?

So, I need to find a word that just has I's O's and H's in it, right?

Can anyone show me how to do a proof like this?

Thank you very much

[quote]is unchanged when reflected in a horizontal line and a vertical line.[/quote]
This can be understood in two ways:
(1) the word is unchanged when reflected in a horizontal line. It is also unchanged when reflected in a vertical line.
(2) the word is unchanged when it is reflected in a horizontal line, and then reflected in a vertical line.

The first interpretation leaves you with letters H, I, O, X.
The second also allows N, S, and Z.
Of course, the order of letters in the word also matters.

Please tell me this is not homework from a math class on writing proofs. :|


Let u_n= \sum1/ia  (a>0).
Find a:u_n have a limit (<> oo).

Thius really is well-known: the series is convergent if a>1.

The harmonic series is one of those typical series that defies a lot of convergence tests. Any slight change to the harmonic series and many different series result  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Are there any maximal rational subgroups in $(\mathbb{Q},+)$ ?

I guess no?  And this result is probably well known?

Looking at the primes in the denominators, one easily sees that a finitely generated subgroup can never be maximal.
But that doesn't really help.

The set ${\frac{1}{p^m} }$ with p prime and m integer apparently generates the group so I was thinking for a maximal, if there is one, of kicking a few elements out.. but to  no avail?

All comments are welcome

Notice that $(\mathbb Q,+)$ is a divisible group. That is, given any $a\in\mathbb Q$ and any positive integer $n$, the equation $nx=a$ has a solution in $\mathbb Q$. Now, a divisible (abelian, of course) group $G$ cannot have a maximal subgroup $M$. The reason is that $G/M$ would have to be cyclic, so we would have an epimorphism $\pi$ from the divisible group $G$ to the cyclic group $G/M$ generated by, say, $t\in G/M$. Take some $x\in G$ with $\pi(x)=t$. Then if $2y=x,\ y\in G$ (and such a $y$ does exist because of the divisibility of $G$), we can't define $\pi(y)\in G/M$.

a maximal subgroup is a subgroup different from the total group, that is only a subgroup of itself and the total group

that is $H$ maximal subroup of $G$ iff $H\leq G$
and $|H|<|G|$ and $|H|$ is maximum.
but if H is the only subgroup of H we have also
$|H|=p$ with p prime.

Lemma: In any group $G$, a normal subgroup $H$ is maximal if and only if $G/H$ is cyclic of prime order.

Proof: The subgroups of $G$ containing $H$ are in natural 1-1 correspondence with the subgroups of $G/H$. $H$ is maximal when there are exactly two such subgroups, which occurs when $G/H$ is cyclic with prime order.

In our case, the subgroup is abelian, so any subgroup is normal.

[quote="fredbel6"]Why would maximality in a divisible abelian group imply a cyclic factor group?

And why can't you define the image of y

I'm sorry I don't get it, divisible groups are a new concept to me[/quote]

What I wrote isn't exactly complete\correct, but it can easily be repaired.

For your first question: maximality in any abelian group actually implies that the quotient group is cyclic of prime order. This is because for any $y\in G\setminus M,\ M$ and $y$ generate $G$, which means that any non-zero $\bar y\in G/M$ is a generator of $G/M$. Obviously, this implies that $G/M$ is cyclic of prime order. Divisibility has nothing to do with this.

As for your second question, you're right: what I wrote is not exactly correct. I had an infinite cyclic group in mind. Here's the correct version (same notations as in my first post):

We now know that $G/M$ is cyclic of prime order $p$. Now, if $\pi(x)=t\ne 0\in G/M$, and $py=x$, then there's no way you can define $\pi(y)$. This is because you'd have $\pi(x)=\pi(py)=p\pi(y)=0$, contradicting the fact that the image of $x$ through $\pi$ is non-zero.


inequalities
geometry
circumcircle
perimeter
rotation
geometric transformation
geometry solved

Let $ O$ be the circumcenter of a triangle $ \triangle ABC$.  Let $ M$ and $ N$ be points on the sides $ AB$ and $ BC$, respectively, such that $ \angle MON \equal{}\angle ABC$.  Prove that the perimeter of triangle $ \triangle MBC$ is at least the length of $ AC$.

Original statement (see http://www.mathlinks.ro/Forum/viewtopic.php?t=30486 ):

Let $O$ be the circumcenter of $\triangle ABC$, $M$ and $N$ be points on sides $BA$ and $BC$ respectively, so that $2\angle MON=\angle AOC$. Prove that the perimeter of $\triangle BMN$ is at least the length of $AC$.

Rotate  triangle BMO around O until B becomes A and Let M become X. At the same way rotate(a new rotation from initial position) BNO until B becomes C and say N become Y. Then triangle XOY and MON are equal, hence XY=MN AND THE CONLUSION IS OBVIOUS.


Prove that for any polygon with perimeter Z, the greatest area is achieved with a circle with circumference Z.

This is actually a fairly difficult problem.  It is often called Steiner's Circle Problem (or isoperimetric problem), as apparently he was the first to prove it.  (Actually the correct wording would be [i]planar surface[/i] instead of polygon, but I can tell that's what you mean).  Many little axuillary proofs lead up to it, such as that for all trapezoids with common base lines and altitudes, the isosceles trapezoid is the one with the smallest perimeter.  The converse of the problem, of all plane surfaces of equal area, the circle is the one with the smallest area, is also true.

hmm thanks for the info themonster...I assume you meant of all plane surfaces of equal area, the circle is the one with the smallest perimeter?

Also, I believe i am close to a proof and will post it when it is finished so that someone can point out some probable errors :P

I would enjoy seeing anyone else's attempts at the problem as well.

Can't you say the fact that another polygon with the same perimeter but greater # of sides has a larger area? Since when a polygon is inscribed in a circle and when it has an infinite # of sides, it is a circle, so therefore the area is the greatest

But I don't know if you can call that a proof ..  :blush:

Yes, jimhu, but the thing is you have to prove that fact and thats what makes this problem so difficult

are you proving for all polygons, not just regular?

For regular polygons I think it's not bad.  For all polygons I think it's really hard (which explains why no one proved it for a long long time).

Maybe we should move this to the "Fields Medalists Playground" forum then :D.


logarithms
number theory
relatively prime
number theory unsolved

Let $ n$ be a positive number. Let $ \varphi(n)$ be the the number of positive integers less than or equal to $ n$ that are relatively prime to $ n$, and let $ s(n)$ be the sum of the digits of $ n$. Prove that set of positive integers $ n$ that satisfy the equation $ \varphi(n)\equal{}(s(n))^3$ is finite.


Determine the positive integers $ a,b,c$ with the properties:

\[ 20(a\plus{}b)\equal{}15(b\plus{}c)\equal{}12(c\plus{}a)\]

\[ a^2 b^3 c\equal{}559872\]

hello, the only two solutions are
$ a\equal{}\minus{}12,b\equal{}\minus{}6,c\equal{}\minus{}18$
$ a\equal{}12,b\equal{}6,c\equal{}18$
Sonnhard.

Except that $ a$,$ b$, and $ c$ are positive integers. So it's just the latter, $ a\equal{}12,b\equal{}6,c\equal{}18$.

How did you find that solution?? :what?:

Because $ a,b,c$ are integers, you can factorize $ 559872$ and use trial and error to figure out which values of $ a,b,c$ work out.

Take this equation.

$ 20a \plus{} 20b \equal{} 12c \plus{} 12a$

Solve for $ a$.

$ 8a \equal{} 12c \minus{} 20b$

$ a \equal{} \frac {3c}{2} \minus{} \frac {5c}{2}b$

Plug it into this equation, and solve for $ c$.

$ 15b \plus{} 15c \equal{} 12a \plus{} 12c$

$ 15b \plus{} 15c \equal{} 12\left(\frac {3c}{2} \minus{} \frac {5c}{2}b\right) \plus{} 12c$

$ 15b \plus{} 15c \equal{} 18c \minus{} 30b \plus{} 12c$

$ 15b \plus{} 15c \equal{} 30c \minus{} 30b$

$ 45b \equal{} 15c$

$ c \equal{} 3b$

Now do this step again and you should get $ a \equal{} 2b$.

Then, plug them into the final equation.

$ (2b)^2 \times b^3 \times (3b) \equal{} 559872$

$ 12b^6 \equal{} 559872$

$ b^6 \equal{} 46656$

$ b^6 \equal{} 2^63^6$

$ b \equal{} 2*3 \equal{} 6$

Plugging back into $ a \equal{} 2b$ and $ c \equal{} 3b$, we see that our only solution (for positive integers) is $ \boxed{(12,6,18)}$.

Thank you for your solutions.

$ \boxed {(12,6,18)}$ is the right answer.

The problem is from a book with exercises for 13 years old students.


[quote="thanhak5"]Let a,b,c be positive reals with $ ab \plus{} bc \plus{} ca \equal{} 3$
            Prove that
$ \frac {1}{1 \plus{} a^2(b \plus{} c)} \plus{} \frac {1}{1 \plus{} b^2(a \plus{} c)} \plus{} \frac {1}{1 \plus{} c^2(b \plus{} a)}\le \frac {1}{abc}$
 It is Rumani Inequalities but I do not find the sulotion for this problem[/quote]
see here:
http://www.mathlinks.ro/viewtopic.php?p=1154642#1154642


Hello, I am beginning a new game, for a few reasons.

1.) A lot of my recent ones died, and they really stunk.
2.) I have seen a lot of good games on this forum, and it has given me a lot of good ideas.
3.) I am bored.

Alright here are the basic rules;

A.) Only discussion is on this thread unless I say otherwise.
B.) Inactivity will result in penalties, and later replacements.
C.) All votes or descisions must be in bold, and if they are not, then I will ignore them.
D.) This game will last about 2 and a half months, so don't join if you are not going to participate, or if you are just going to quit. If this is the case, then I will not let you play in upcoming seasons of AoPS Academy
E.) I am excepting twelve people at the maximum, so there are twelve people, then please don't try to sign-up.
F.) Don't post in the thread unless you are still in the competition.
G.) Have fun.

Here is what the game is;

All the competitors are students at AoPS Academy that have very low grades and are trying to raise them up. They have joined a club after school to improve their study skills. Every week at AoPS Academy the students are given assignments. These assignments are extra-credit assignments that will raise the grades. The people that run the club have gotten tired of staying after school every day, so they decided that they would have goals for the students in the club. The last person to reach (or pass) this goal will be kicked out of the club. This will continue until there are only three people left. These three people will take the exam. (the challenges aren't actually based on academics, they are all games). The person that does the best on the exam will have their grade raised to a 100% and they will be the top student at AoPS Academy.

In the club, each point has AoPSA points. After each challenge the people who did the best will earn points based on their performance. The last person to reach a certain number of AoPS points will be kicked out of the club. If there is more than one person that is the last to reach this deadline, then they will take a pop-quiz to determine who gets kicked out of the club.

These are not the only complications of the game. We begin the game by people sending me the answers to a registration quiz. You can not post on the thread, unless you have pmed me your answers to the quiz. Based on the answers to this quiz, I will determine bullies, victims, partners, and the salesman. Everybody is a bully of another student, and a victim of another bully. Whenever your bully gets kicked out of the club, you get 25 AOPSA points. Once your current bully gets kicked out of the club, the bully of that bully becomes your new bully. Similarly, if your victim gets kicked out of the club, then your new victim will be the victim of that person. Everybody will be assigned a partner. When yoru partner gets kicked out of the club, then you lose 20 AOPSA points. If you want to prevent this from happening, then you want to make sure they do good at the challenges. When your partner is gone, you do not get a new one. The club happens to be right across the hall from the AoPS Academy store. One of the members of this club is the store owner, and chooses the prices for certain items. The items are all worth AOPSA points. The items at the store are things that will benefit you in the game. You must also not forget that when you buy items at the store, the store owner gets the AOPSA points.

Sign-ups begin now. You sign-up by pming your answers to this quiz to me. You can not post on this thread, until you have signed-up.

1. What is your favorite color?
2. What is your favorite sport?
3. What is your favorite subject in school excluding math?
4. What is your favorite math topic?
5. What do you write with? (pencil, pen, marker, etc.)
6. Why are you playing this game?

Sign-ups end two weeks from now.

Dude.  Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=800653506&t=236386]here[/url].  Specifically, at the first sentence.

[quote="Wickedestjr"]Alright I am beginning a new game. This is the last one. I promise.[/quote]

Yeah, but everybody quit for absolutely no reason.

At least don't make the rules so complicated.

W/e.  I sign up.

Umm please pm me your answers to the quiz before you post.

I signed up. Hopefully this will be fun.

It depends on how many people sign-up. If there are at least six, then it should go smoothly.

Players;
crazypianist1116

AIME15 - You need to send me your answers to the quiz for me to add you to the list.

Er hem:

[quote="Inside "Another competition", Wickedestjr"]Alright I am beginning a new game. [color=red][b]This is the last one. I promise. [/b][/color][/quote]

Too many games...

AIME15 already pointed this out, and I already replied, so your post is pointless.

dude why has nobody realized that Wickedestjr said that the other competition would be his last and he broke his promise?

[/irritating]

dude seriously just copy a game show or something. stop making games that people join and don't compete in.

[quote="Dojo"]I guess I'll be a liar.
I'm a salesman so...[/quote]

You'll find out in a few days.

Trustworthy
1. crazypianist1116
2. jjx1

Liers
1. ernie
2. SMB
3. dojo

Task 4 will begin when task three is over.

Nobody made their shots.

Line;
dojo
jjx1
SMB

Score as of task 2;
crazypianist116: 5 pts.
SuperMathBoy: 4 pts.
ernie: 3 pts.
jjx1: 0 pts.
dojo: 0 pts.

Score as of task 3;
SuperMathBoy: At least 6 points.
jjx1: At least 6 points.
dojo: At least 6 points.
ernie: 4 pts.
crazypianist1116: 3 pts.

Score as of tasks 2 & 3;
crazypianist1116: 8 pts.
dojo: At least 6 pts.
ernie: 7 pts.
jjx1: At least 6 pts.
SuperMathBoy: At least 10 pts.

Remember, the last person to get more than ten points is eliminated from the competition.

2 things, do you get a bonus for being first? Second, how does the scoring work for task 3.

1st place: 10 pts.
2nd place: 8 pts.
3rd place: 6 pts.
4th place: 4 pts.
5th place: 3 pts.

Okay, dojo is the salesman. He can charge you anything for the things in his store. Of course, nobody knows what each item does. Only I do. Dojo is in charge of making prices for each item in his store. He will get new items in his store now and then. He can't buy the items in his store. The price of each item is an amount of points. Once he makes the prices for each item, he can't change them. If you don't give a price to something then it is considered free. Oh and also, if somebody gets an item, then most of the time they will know what it does. And they are capable of trading their items with other players for their items/points.

These are the items in the store;

A golden pencil.
An orange key.
A red key.
A grape.
A broken keyboard.

Oh and since some people quit, here is the new arrangements of partners;

SuperMathBoy and jjx1
ernie and dojo
crazypianist1116

Oooh, buy the grape. :)

Shoot for a 3 too.

You missed but fortunately I have made the chance of making a three-pointer one eleventh.

jjx1 made the shot. Which means that dojo is out of this challenge.

Which means;

Score as of task 2;
crazypianist116: 5 pts.
SuperMathBoy: 4 pts.
ernie: 3 pts.
jjx1: 0 pts.
dojo: 0 pts.

Score as of task 3;
SuperMathBoy: At least 8 points.
jjx1: At least 8 points.
dojo: 6 pts.
ernie: 4 pts.
crazypianist1116: 3 pts.

Score as of tasks 2 & 3;
crazypianist1116: 8 pts.
dojo: 6 pts.
ernie: 7 pts.
jjx1: At least 8 pts.
SuperMathBoy: At least 12 pts.

Line;
SuperMathBoy
jjx1

The chance of making a three pointer is now one out of five.

Shoot for a 3.

And I bought the grape. :)

Swwwwwweeeeeeet!!!!!! I just learned that you can access AOPS from mathlinks.ro

I shoot from the best chance (I'm really not sports smart).


modular arithmetic
number theory unsolved
number theory

Give n be positive integer number . Prove that :
 $A = (2001^{720n}+2002^{1232n}-1) \vdots 60697$
 ( :maybe: )
[i]Louis Latin and Vicky[/i]

I'm guessing that the vertical dots mean divides...

So we want to show $2001^{720n}+2002^{1232n}-1 \equiv 0 \pmod{60697}$. Well, $60697=7 \cdot 13 \cdot 23 \cdot 29$ so we can break it up using CRT. Then some FLT and simplification should get the result pretty quickly.


limit
probability
real analysis
probability and stats

Is it possible to construct a sequence of random variable $ X_n$
such that

$ \lim_{n\rightarrow \infty}E[X_n] \equal{} \infty$
and $ \lim_{n\rightarrow \infty}X_n \equal{} c$ with probability 1,
where $ c$ is a constant, and $ E[\cdot]$ is the expectation.

 :huh:

Certainly we can do that. Let $ A_n$ be a nested decreasing sequence of events such that $ A_1\supset A_2\supset A_3\dots$ and $ P(A_n)\equal{}\frac1n.$ (You could just take $ [0,1]$ as a model for the probability space and let $ A_n$ be the interval from $ 0$ to $ \frac1n.$)

Now, let $ X_n$ take on the value $ c\plus{}n^2$ on $ A_n$ and $ c$ otherwise.

$ \lim_{n\to 0} X_n\equal{}c$ almost surely, but $ E(X_n)\equal{}c\plus{}n$ and tends to infinity as $ n\to\infty.$

(Some probability questions, such as this one, are in fact real analysis questions.)

@ kent merryfield
can you please suggest a book where i can find similar problems? :blush:
thanks in advance.


Prove that:
a) From $ 9$ natural numbers exit $ 5$ numbers which sum is divisible by $ 5$.
b) From $ 1999$ natural numbers exit $ 1000$ numbers which sum is divisible by $ 1000$.


here's the inequality from the inmo 2003(geometricl inequality but i'm posting it in the algebra section).
suppose there is a triangle with sides a,b,c .construct another triangle with sides a + b/2, b + c/2 , c + a/2.prove that the area of this triangle is 9/4 the area of the original triangle.and find that case in which the inequality holds.i sum how managed to mess up this question in the exam (messd up everything in fact).

Let p = (a+b+c)/2 , x = p-a , y=p-b , z=p-c . Then the inequality becomes after a few calculations :
(2x+y)(2y+z)(2z+x) >= 27 xyz , which is true by AM-GM .

Here is the solution.
Let a,b,c be the sides of triangle ABC.

a+b/2,b+c/2,c+a/2 are sides of triangle  KLM.

  \[
Set a=x+y, b=x+z , c=y+z.
Then A(ABC)= \sqrt{xyz(x+y+z)} and A(KLM)=\sqrt{\frac{3(x+y+z)(2x+y)(2y+z)(2z+x)}{16}}\geq\sqrt{\frac{3(x+y+z)27xyz}{16}}=9/4A(ABC) \]

fixed latex

Set $a=x+y, b=x+z , c=y+z.$
Then $A(ABC)= \sqrt{xyz(x+y+z)}$ and $A(KLM)=\sqrt{\frac{3(x+y+z)(2x+y)(2y+z)(2z+x)}{16}}\geq\sqrt{\frac{3(x+y+z)27xyz}{16}}=9/4A(ABC) $

You have to construct such a triangle, too. (pretty simple, just bisect each of the sides to get $\frac a2,\frac b2,\frac c2$ and copy a bunch of lengths)


Given $ a,b,c,d \in \mathbb{R}$ such that $ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$. Prove that $ abcd > a \plus{} b \plus{} c \plus{} d \plus{} 8$

we have abcd>0
so $ abcd>a^2\plus{}b^2\plus{}c^2\plus{}d^2\ge 4\sqrt{abcd}$ so $ abcd>16$
we will prove the best case :a,b,c,d>0
so $ a\plus{}b\plus{}c\plus{}d\ge 4\sqrt[4]{abcd}>8$
so $ abcd\ge a^2\plus{}b^2\plus{}c^2\plus{}d^2\ge \frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}>a\plus{}b\plus{}c\plus{}d\plus{}8$ because a+b+c+d>8
***
if a,b,c,d<0 so abcd>16>a+b+c+d+8
if a,b<0;c,d>0.let $ a\equal{}\minus{}a_{1},b\equal{}\minus{}b_{1}$ so $ a_{1},b_{1}>0$
so $ abcd\equal{}a_{1}b_{1}cd>a_{1}\plus{}b_{1}\plus{}c\plus{}d\plus{}8>a\plus{}b\plus{}c\plus{}d\plus{}8$
q.e.d

[quote="tuandokim"]we have abcd>0
so $ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\ge 4\sqrt {abcd}$ so $ abcd > 16$
we will prove the best case :a,b,c,d>0
so $ a \plus{} b \plus{} c \plus{} d\ge 4\sqrt [4]{abcd} > 8$
so $ abcd\ge a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{4} > a \plus{} b \plus{} c \plus{} d \plus{} 8$ because a+b+c+d>8
***
if a,b,c,d<0 so abcd>16>a+b+c+d+8
if a,b<0;c,d>0.let $ a \equal{} \minus{} a_{1},b \equal{} \minus{} b_{1}$ so $ a_{1},b_{1} > 0$
so $ abcd \equal{} a_{1}b_{1}cd > a_{1} \plus{} b_{1} \plus{} c \plus{} d \plus{} 8 > a \plus{} b \plus{} c \plus{} d \plus{} 8$
q.e.d[/quote]
Thanhks. goodluck to you.


calculus
derivative
function
search
calculus computations

[b](1) Can anyone tell me how can i derive the formula of
differentiation of implicit function

means  If f(x,y) =0,
Then dy/dx = -(df/dx) /(df/dy).
how can i derive this formula.
can anyone have a proof.[/b]

Try a search on the Implicit Function Theorem and the Inverse Function Theorem.

Also, those $ d$s on the right hand side of the equation should be $ \partial$s (to reference the partial derivative).


geometry
inradius
area of a triangle
Heron\u0027s formula

Can't believe how easy this was for a IIT-JEE mains problem:

Three circles externally tangent to each other ... radii 3, 4 and 5
P is the point of intersection of the tangents to the circles at the points of contact. Find the distance from P to the points of contact.

Hint:
[hide]All you need to do is find out what point P is[/hide]

Clearly the point P is the in–centre of the ΔABC, and
hence
$ r = \sqrt{ \frac{(s-a)(s-b)(s-c)}{s} }$ $ =\frac{ \Delta}{s} $

Since $P$ is the center of the triangle, you can use the $rs=A$ formula to find out the inradius of the triangle and apply Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$  to find out the area of the triangle, so:

$r(\frac{9+8+7}{2})=\sqrt{12(12-9)(12-8)(12-7)}\Longrightarrow12r=\sqrt{720} \Longrightarrow r=\boxed {\sqrt{5}}$.

[quote="frt"][quote="Iversonfan2005"]P.S. What's the latex code for the little arrows?[/quote]
[hide]
\rightarrow: $\rightarrow$

\leftarrow: $\leftarrow$

\leftrightarrow: $\leftrightarrow$

\Longrightarrow: $\Longrightarrow$

\Longleftarrow: $\Longleftarrow$

\Longleftrightarrow: $\Longleftrightarrow$
[/hide][/quote]

Thank you frt, strangely, though, it's a black arrow, not one of those other ones, but it's fine though.

Oh then you're probably talking about these... :P
[hide]
\longrightarrow: $\longrightarrow$

\longleftarrow: $\longleftarrow$
[/hide]

See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php#arrows]this[/url] for more arrows :)