abstract algebra
group theory
superior algebra
superior algebra solved

Suppose R is a commutative ring with unity such that characteristic of R is not 2 and the nonzero elements of R form a cyclic group under multiplication (of R).  Prove theat R is a field.

You don't need the group to be cyclic as long as the group is abelian (but it will be cyclic when finite ;) ).
Indeed, when this group is finite even commutativity isn't necessary since it can be proved.
But then all conditions for a field are fulfilled: the ring axioms, commutativity of mutliplication, existence of inverses.
Btw: also the characteristik being $\neq 2$ isn't needed.

so do i need to prove that the ring is a field and holds that the ring axioms, commutativity of mutliplication, existence of inverses.  or do i need to prove that the ring is only cyclic if the ring is finite. i'm confused a little by the post.
thanks
gap

How do you define 'field'¿
All you have to prove is the field axioms, and there are two versions I know:
1. version: it is a field iff it fulfills the ring axioms, multiplication is commutative, there's an unit and all nonzero elements have inverses.
2. version: it fulfills the ring axioms and the set of nonzero elements build an abelian group under multiplication.

The second one is as good as what you posted, you are just left with proving: every cyclic group is abelian.
For the first one: commutativity comes again from the cyclicity, the unit (which needs not extra to be given) and inverses from the multiplicative group.

I'm really lost. i don't understand the reasoning.  i feel like it is more confusing than it should be.  i understand what a field is but how do i explain the feild axom with this little info.

sorry It is suppose to be prove R is a finite field not just a field think that is where we got mixed up

So we are left to prove the finiteness:
Take any generator $\zeta$ of the multiplicative group. Then there is some $k \neq 0$ with $\zeta^k = -1$, thus $\zeta^{2k}=1$. But that means that the multiplicative group has at most order $2|k|$, so is finite and then also the whole field is finite (since only $0$ was left out).


calculus
integration
calculus computations

Evaluate $ \int_0^1 \frac{x\plus{}\sqrt{x}\minus{}1}{(\sqrt{x}\plus{}1)^2}\ dx$.

[quote="kunny"]Evaluate $ \int_0^1 \frac {x \plus{} \sqrt {x} \minus{} 1}{(\sqrt {x} \plus{} 1)^2}\ dx$.[/quote]
  let   $ t\equal{}\sqrt{x}$    we have : $ dx \equal{}2tdt$
   $ \int_0^1 \frac {x \plus{} \sqrt {x} \minus{} 1}{(\sqrt {x} \plus{} 1)^2}\ dx  \equal{} \int_0^1 2\frac{(t^3\plus{}t^2\minus{}t)dt}{(t\plus{}1)^2}  \equal{} 0$
  where$ F(t) \equal{} t^2\minus{}2t \minus{}\frac{2}{t\plus{}1}$


[hide]
$2x^2+2y^2=4(x+y)+2xy$
$(x^2-2xy+y^2)+(x^2-4x+4)+(y^2-4y+4)=8$
$(x-y)^2+(x-2)^2+(y-2)^2=8$
This allows easier analysis.  By the trivial inequality all the terms on the LHS are positive and all are also perfect squares.  There are only 3 ways to partition 8 as sums of perfect squares:
$i)$ $0 + 4 +4=8$ $\Longrightarrow$ $(4,4)$ or $(0,0)$
$ii)$ $4+4 + 0=8$ $\Longrightarrow$ $(4,2)$ or $(0,2)$
$iii)$ $4+0 +4=8$ $\Longrightarrow$ $(2,4)$ or $(2,0)$
Thus we have $\boxed{6}$ answers.
[/hide]

[hide="Other solution"]
Let $x=m$ and $y=m+p$ with $m,p \in \mathbb{Z}$
Then we have $m^2+(m+p)^2=2(m+m+p)+m(m+p)$
$\Leftrightarrow m^2+(p-4)m+p^2-2p$   $(1)$
In order for this equation to have solutions, we need $D\geq 0$
$\Leftrightarrow (p-4)^2-4(p^2-2p)\geq 0$
$\Leftrightarrow p^2\leq \frac{16}3$
$\Leftrightarrow -\frac 4{\sqrt3}\leq p\leq \frac4{\sqrt3}$
So $p$ can only be $-2,-1,0,1$ or $2$
Plugging those values in into $(1)$, we get five equations:
$m^2-6m+8=0$
$m^2-5m+3=0$
$m^2-4m=0$
$m^2-3m-1=0$
$m^2-2m=0$

Solving these for integers gives us the six solutions: $(0,0) (0,2) (2,0) (2,4) (4,2) (4,4)$[/hide]

[quote="chess64"][quote="digitalboy1"]How many ordered pairs of integers $(x,y)$ satisfy the equation $x^2+y^2=2(x+y)+xy$ ?[/quote]

[hide]
$(x+y)^2-2xy=2(x+y)+xy$
$(x+y)^2-(x+y)-(x+y)=3xy$
$(x+y)(x+y-2)=3xy$
That didn't help much... :blush:
$(x,y)=(0,0)$
Well, either $x+y$ or $x+y-2$ is divisible by 3, but not both, so $x$ and $y$ can't be multiples of $3$.
If $x+y=3$, then we have $(3)(1)=3(x)(y)$, so $x=y=1$, which is not true.
If $x+y=5$, then we have $(5)(3)=3(x)(y)$, so $xy=5$, which cannot be true.
If $x+y=6$, then we have $(6)(4)=3(x)(y)$, so $xy=8$, so $(4,2)$ and $(2,4)$.
If $x+y=8$, then we have $(8)(6)=3(x)(y)$, so $xy=16$, so $(4,4)$.
If $x+y=9$, then we have $(9)(7)=3(x)(y)$, so $xy=21$, which doesn't work.
If $x+y=11$, then we have $(11)(9)=3(x)(y)$, so $xy=33$, which doesn't work.
If $x+y=12$, then we have $(12)(10)=3(x)(y)$, so $xy=40$, so....

Okay, nevermind. I think it's time to read the other people's solutions... :lol:
[/hide][/quote]

the problem with your solution is that you can never stop checking numbers :P


To get videos and lessons, do you need to complete 60 problems or get 60 problems correct? :blush:

Before you post, please scroll down in the forum to see if your question has been answered.

The $ 60$ questions are like a diagnostic stage to gauge your level of knowledge, so as to assign you videos and problems suitable for you. Thus, it is the first thing you do in Alcumus. Of course, Alcumus is probably the best way in the world (I'm not kidding) to learn Counting & Probability, so I highly recommend that you try Alcumus for a while, since it would probably be highly beneficial to your probability knowledge and understanding. For karma information, refer to the "Karma" section of the "Instructions" tab in Alcumus. Additional information about karma can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260687]here[/url].


algebra
polynomial
absolute value
Quadratic Residues
pen

Let $a, b, c$ be integers and let $p$ be an odd prime with \[p \not\vert a \;\; \text{and}\;\; p \not\vert b^{2}-4ac.\] Show that \[\sum_{k=1}^{p}\left( \frac{ak^{2}+bk+c}{p}\right) =-\left( \frac{a}{p}\right).\]

We start by evaluating the sum $\mod p$:
$\left( \frac{a}{p}\right) \equiv a^\frac{p-1}{2}\mod p$. Write the polynomial $(ax^{2}+bx+c)^\frac{p-1}{2}= \sum_{k=0}^{p-1}a_{k}x^{k}$ with leading coefficient $a_{p-1}\equiv \left( \frac{a}{p}\right) \mod p$ and remember that $\sum_{x=1}^{p}x^{n}\equiv 0 \mod p$ for $0 \leq m < p-1$ by http://www.problem-solving.be/pen/viewtopic.php?t=676. We get:
\[\sum_{x=1}^{p}\left( \frac{ax^{2}+bx+c}{p}\right) \equiv \sum_{x=1}^{p}\sum_{k=0}^{p-1}a_{k}x^{k}= \sum_{k=0}^{p-1}a_{k}\sum_{x=1}^{p}x^{k}\equiv a_{p-1}\sum_{x=1}^{p}x^{p-1}\equiv-\left( \frac{a}{p}\right) \mod p\]

As the sum has definitely absolute value $\leq p$, we just have to exclude the case $\sum_{k=1}^{p}\left( \frac{ak^{2}+bk+c}{p}\right) = (p-1) \cdot \left( \frac{a}{p}\right)$.
If this happens, all but one of the summands have to be the same, namely $1$ or $-1$. Thus there is exactly one residue class $r \mod p$ such that $ar^{2}+br+c \equiv 0 \mod p$. Dividing out $x-r$ from $ax^{2}+bx+c$, we are left with a linear term having no root $\not\equiv r$, thus we get $ax^{2}+bx+c \equiv a(x-r)^{2}\mod p$, thus the discriminant $b^{2}-4ac$ is $0 \mod p$, which we excluded.

Similar, but I believe is shorter. Using the substitution, $ax^2+bx+c = a[(x+b/2a)^2 + (4ac-b^2)/4a^2]$, and letting $(x+b/2a)=k$, and $(4ac-b^2)/4a^2=m$, we need to prove that,
$$
\sum_{k=1}^p \left(\frac{a(k^2+m)}{p}\right)=-\left(\frac{a}{p}\right)\iff \sum_{k=1}^p \left(\frac{k^2+m}{p}\right)=-1.
$$
To prove this, we use Euler's criterion, $\left(\frac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\pmod{p}$, and write the sum above as,
$$
\sum_{k=1}^p (k^2+m)^{\frac{p-1}{2}}.
$$
Now, using the fact that, for every $1\leq \ell\leq p-2$, $p\mid 1^\ell + 2^\ell +\cdots +(p-1)^\ell$, the expression above is congruent to, $\sum_{k=1}^p k^{p-1}\equiv p-1\pmod{p}$. In particular, using the fact that Legendre's symbol takes values in the set $\{-1,1\}$, the expression above is between $-p$ and $p$, and is congruent to $-1$ in modulo $p$, hence is either $-1$, or $p-1$.

But it is easy to see that it cannot be equal to $p-1$ (if $\ell$ of them is $1$, and $p-\ell$ is $-1$, we would have, $2\ell-p=p-1$, giving $2\ell =2p-1$, a contradiction). Hence, the claim follows.

@2above there is a typo. Instead of $m$ in the fourth line it should be $n$. :P


[quote="soccerstar123"]does anyone know the area of a regular dodecagon?[/quote]

What?! This has nothing to do with the topic. Post something like this in a different thread. Man I wish I was a mod now  :(  If you're wondering, it's $\frac{3}{4}s^{2}$, where $s$ is the sidelength. I believe that formula is correct, not positive.

[size=75][color=green]I see what you mean.  I'll deal with this.

-Robin[/color][/size]


function
modular arithmetic
linear algebra
matrix
number theory
algebra unsolved
algebra

If I have a function $ f_1(x)\equal{}\frac{2x\minus{}1}{x\plus{}1}$ and $ f_{n\plus{}1}(x)\equal{}f_1(f_n(x)),n\geq1$
can it be shown that $ f_{35}\equal{}f_5$?

$ f_1(x)\equal{}\frac{2x\minus{}1}{x\plus{}1}$
$ f_2(x)\equal{}\frac{x\minus{}1}{x}$
$ f_3(x)\equal{}\frac{x\minus{}2}{2x\minus{}1}$
$ f_4(x)\equal{}\frac{1}{\minus{}x\plus{}1}$
$ f_5(x)\equal{}\frac{x\plus{}1}{\minus{}x\plus{}2}$
$ f_6(x)\equal{}x$

And, $ 35 \equiv 5 \pmod{6}$.
So, it's trivial.

[quote="Kouichi Nakagawa"]$ f_1(x) \equal{} \frac {2x \minus{} 1}{x \plus{} 1}$
$ f_2(x) \equal{} \frac {x \minus{} 1}{x}$
$ f_3(x) \equal{} \frac {x \minus{} 2}{2x \minus{} 1}$
$ f_4(x) \equal{} \frac {1}{ \minus{} x \plus{} 1}$
$ f_5(x) \equal{} \frac {x \plus{} 1}{ \minus{} x \plus{} 2}$
$ f_6(x) \equal{} x$

And, $ 35 \equiv 5 \pmod{6}$.
So, it's trivial.[/quote]

Thanks Nakagawa-san. When I compare your calculation with mine, I noticed where my mistake was.
So, it means that if I want to find $ f_{28}$, then it should have the same function with $ f_4$ right?
As $ 28\equiv4$

There is a general theory about functions $ f(x) \equal{} \frac {ax\plus{}b} {cx\plus{}d}$, called homographic, or Moebius functions, by studying the associated matrix $ F \equal{} \genfrac {(} {)} {0pt} {} {a \ b} {c \ d}$.
In a simpler vein, you may compute that $ f_6(x) \equal{} x$, so $ f_36(x) \equal{} x$. Apply $ f_1^{\minus{}1}$ to obtain $ f_5 \equal{} f_{35}$.

[quote="r0ger"]Thanks Nakagawa-san. When I compare your calculation with mine, I noticed where my mistake was.
So, it means that if I want to find $ f_{28}$, then it should have the same function with $ f_4$ right?
As $ 28\equiv4$[/quote]
Yes, that's right.

[quote="mavropnevma"]There is a general theory about functions $ f(x) \equal{} \frac {ax \plus{} b} {cx \plus{} d}$, called homographic, or Moebius functions, by studying the associated matrix $ F \equal{} \genfrac {(} {)} {0pt} {} {a \ b} {c \ d}$.
In a simpler vein, you may compute that $ f_6(x) \equal{} x$, so $ f_36(x) \equal{} x$. Apply $ f_1^{ \minus{} 1}$ to obtain $ f_5 \equal{} f_{35}$.[/quote]

Thanks for the information...

"The classical Möbius function μ(n) is an important multiplicative function in number theory and combinatorics. The German mathematician August Ferdinand Möbius introduced it in 1832. This classical Möbius function is a special case of a more general object in combinatorics"

[quote="r0ger"]
"The classical Möbius function μ(n) is an important multiplicative function in number theory and combinatorics. The German mathematician August Ferdinand Möbius introduced it in 1832. This classical Möbius function is a special case of a more general object in combinatorics"[/quote]

That's a different Mobius function, defined on partially ordered sets.


inequalities
function
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

$ (a) \ \ \ \  \sum \frac {5a(b\plus{}c)\minus{}4bc}{(b\plus{}c)^2} \ge \frac {9(a^2\plus{}b^2\plus{}c^2)}{2(ab\plus{}bc\plus{}ca)}$;

$ (b) \ \ \ \  \sum \frac {a(b\plus{}c)\minus{}bc}{(b\plus{}c)^2}\plus{} \frac 1{4} \ge \frac {a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

$ (a) \ \ \ \ \sum \frac {5a(b + c) - 4bc}{(b + c)^2} \ge \frac {9(a^2 + b^2 + c^2)}{2(ab + bc + ca)}$;

$ (b) \ \ \ \ \sum \frac {a(b + c) - bc}{(b + c)^2} + \frac 1{4} \ge \frac {a^2 + b^2 + c^2}{ab + bc + ca}$.[/quote]
Very nice inequalities, VasC. :) Here is my solutions:

a) The inequality is equivalent to
\[ 5\sum \frac{a[a(b+c)+bc]}{b+c}-4\sum \frac{bc[a(b+c)+bc]}{(b+c)^2} \ge \frac{9}{2} \sum a^2,\]
or equivalently,
\[ \frac{1}{2}\sum a^2 +abc\sum \frac{1}{b+c} -4\sum \frac{b^2c^2}{(b+c)^2} \ge 0.\]
As
\[ \sum \frac{1}{b+c} \ge \frac{9}{2(a+b+c)},\quad \mbox{and} \quad 4\sum \frac{b^2c^2}{(b+c)^2} \le \sum bc.\]
Our inequality follows from
\[{ \frac{1}{2}(a^2+b^2+c^2) +\frac{9abc}{2(a+b+c)} -(ab+bc+ca)} \ge 0.\]
Which is just Schur's Inequality of third degree. :)

b) The inequality is equivalent to
\[ \sum \frac{a[a(b+c)+bc]}{b+c} -\sum \frac{bc[a(b+c)+bc]}{(b+c)^2} +\frac{1}{4} \sum ab \ge \sum a^2,\]
or equivalently,
\[ \frac{1}{4} \sum ab \ge \sum \frac{a^2b^2}{(a+b)^2},\]
which follows from AM-GM Inequality. :)

These inequalities are really stronger than Iran 96 inequality because:

If we use a) (or b)) and my old result:
\[ \frac{a^2+b^2+c^2}{ab+bc+ca} +\frac{2bc}{(b+c)^2} +\frac{2ca}{(c+a)^2} +\frac{2ab}{(a+b)^2} \ge \frac{5}{2},\]
we will get the Iran 96 inequality. :)

Nice solution, Can_hang2007. Actually, these inequalities are particular cases of the inequality below.

If $ m\ge \minus{}1$ and $ a,b,c$ are nonnegative numbers, then

$ \ \ \ \ \sum \frac {a(b \plus{} c) \plus{} mbc}{(b \plus{} c)^2} \ge \frac {(1\minus{}m)(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)}\plus{}\frac {5m\plus{}4}{4}$.

For $ m\equal{}1$, we get Iran 96 Inequality.

I have a nice and more difficult generalization of Iran 96:

Let $ a,b,c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. If $ 1\le k\le 3$, then

$ \sum \frac 1{b^2\plus{}kbc\plus{}c^2}\minus{}\frac 3{k\plus{}2} \ge \frac {2(k\minus{}2)}{9(k\plus{}2)}(a^2\plus{}b^2\plus{}c^2\minus{}3)$.

As I posted long time ago on this forum, in my opinion an inequality is stronger then other if we can write these inequalities respectively as
$ f\ge f_1$ and $ f\ge f_2$ such that $ f_1 \ge f_2$. This opinion is the same as Spanferkel's opinion, but more practical.

If we use the symbol "$ \succ$" for "$ stronger$", then
$ (b)\succ (a) \succ (Iran)$,
because we can write these inequalities as
$ (b) \ \ \ \ f\ge f_1$,
$ (a) \ \ \ \ f\ge f_2$,
$ (I) \ \ \ \ f\ge f_3$,

with $ f_1\ge f_2 \ge f_3$, where

$ f \equal{} \sum \frac a{b \plus{} c}$,
$ f_1 \equal{} \sum \frac {bc}{(b \plus{} c)^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} \frac 1{4}$,

$ f_2 \equal{} \sum \frac {4bc}{5(b \plus{} c)^2} \plus{} \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{10(ab \plus{} bc \plus{} ca)}$,

$ f_3 \equal{} \sum \frac { \minus{} bc}{(b \plus{} c)^2} \plus{} \frac 9{4}$.

Thank you very much Vasc for the precisions.

@ Can_Hang : So I see that altogether your proof yields indeed again a proof of Iran96. But the most involved part to complete it, i.e. to show $ f_1\ge f_3$, is not much simpler than the other known proofs of Iran96.  :maybe:

[quote="spanferkel"] So I see that altogether your proof yields indeed again a proof of Iran96. But the most involved part to complete it, i.e. to show $ f_1\ge f_3$, is not much simpler than the other known proofs of Iran96.  :maybe:[/quote]
Indeed, this fact is interesting and frequent. A method can be complicated for an inequality (Iran 96), but much simpler for a stronger inequality.

[quote="spanferkel"][quote="can_hang2007"]These inequalities are really stronger than Iran 96 inequality because:

If we use a) (or b)) and my old result:
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {2bc}{(b \plus{} c)^2} \plus{} \frac {2ca}{(c \plus{} a)^2} \plus{} \frac {2ab}{(a \plus{} b)^2} \ge \frac {5}{2},
\]
we will get the Iran 96 inequality. :)[/quote]
 Yes, nice solution. But can we say they are really stronger than Iran96? I would agree with that only if your "old result" were easy to prove (and even then, the question remains: "easy" in which sense??) BTW, can you please give a link?

I suggest we should use "stronger" only in the strict sense, i.e.

$ "A\ge B"$ is stronger than $ "C\ge D"$ iff $ C \minus{} D \ge A \minus{} B\ge0$, or maybe (if suitable for writing more elegantly :)) iff  $ (C \minus{} D)\cdot P \ge A \minus{} B\ge0$ for some expression $ P$.

As Vasc has pointed out in his second post, his two inequalities are rather "particular cases of a generalization of Iran 96", i.e. in a certain sense "parallel" to Iran 96. ;)[/quote]
[quote="spanferkel"] Yes, nice solution. But can we say they are really stronger than Iran96? I would agree with that only if your "old result" were easy to prove (and even then, the question remains: "easy" in which sense??) BTW, can you please give a link?
[/quote]
I am sorry, I can't give the link for you spanferkel because I haven't posted my result anywhere. I created this inequality when trying to find the different proofs of Iran:
I rewrited the Iran Inequality as
\[ \sum \frac{a}{b\plus{}c} \plus{}\sum \frac{bc}{(b\plus{}c)^2} \ge \frac{9}{4}.\]
Now, I used the Cauchy Schwarz:
\[ \sum \frac{a}{b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} ,\]
to reduce it as
\[ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \plus{} \sum \frac{bc}{(b\plus{}c)^2} \ge \frac{9}{4}.\]
which is equivalent to
\[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{2bc}{(b\plus{}c)^2}\plus{}\frac{2ca}{(c\plus{}a)^2}\plus{}\frac{2ab}{(a\plus{}b)^2} \ge \frac{5}{2}.\]
And after checking, I proved that this inequality is true. That's why I called it as my old result.  :)

P/s: In the past, I just have a "simple definition" that "An inequality (1) is stronger than the one (2) if it can be used to solve (2)". That's all. But now I totally agree with Vasc and you, I will be more careful at my statements. Thanks anyway. :)

[quote="nguoivn"][quote="can_hang2007"]
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {2bc}{(b \plus{} c)^2} \plus{} \frac {2ca}{(c \plus{} a)^2} \plus{} \frac {2ab}{(a \plus{} b)^2} \ge \frac {5}{2}$

[/quote]

Indeed, it's Mathlinks Contest Problem.
Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \sum\ \frac {1 \plus{} a^2b^2}{(a \plus{} b)^2} \geq\ \frac {5}{2}$
We can prove it easily by SOS and I had had another proof by my lenma of SOS  :)[/quote]
I remember now that I am the author of this Contest Problem. :P  :wink:

[quote="Vasc"]Nice solution, Can_hang2007. Actually, these inequalities are particular cases of the inequality below.

If $ m\ge \minus{} 1$ and $ a,b,c$ are nonnegative numbers, then

$ \ \ \ \ \sum \frac {a(b \plus{} c) \plus{} mbc}{(b \plus{} c)^2} \ge \frac {(1 \minus{} m)(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {5m \plus{} 4}{4}$.

For $ m \equal{} 1$, we get Iran 96 Inequality.

I have a nice and more difficult generalization of Iran 96:

Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$. If $ 1\le k\le 3$, then

$ \sum \frac 1{b^2 \plus{} kbc \plus{} c^2} \minus{} \frac 3{k \plus{} 2} \ge \frac {2(k \minus{} 2)}{9(k \plus{} 2)}(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3)$.[/quote]
I will write the hints for these inequalities: :)
For the first, write the inequality as
\[ m\left( \frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)} \plus{}\sum \frac{bc}{(b\plus{}c)^2} \minus{}\frac{5}{4}\right) \plus{}\sum \frac{a}{b\plus{}c} \minus{}\frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \ge 0.\]
Using the Vasc's inequality (I am very sorry about this, VasC. I dont know that you have created it before):
\[ \frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)} \plus{}\sum \frac{bc}{(b\plus{}c)^2} \minus{}\frac{5}{4} \ge 0,\]
we can see that it suffices to prove the inequality in the case $ m\equal{}\minus{}1$ ;), which is equivalent to
\[ \sum \frac{a}{b\plus{}c} \minus{}\frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \plus{}\frac{5}{4} \minus{}\sum \frac{bc}{(b\plus{}c)^2} \minus{}\frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)} \ge 0,\]
or equivalently,
\[ \sum \frac{a(b\plus{}c)\minus{}bc}{(b\plus{}c)^2} \plus{}\frac{1}{4} \minus{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge 0,\]
which is just (b). ;) :)

For the second inequality, using convexity function, we can see that it suffices to prove the inequality in the case $ k\equal{}1$ and $ k\equal{}3$ ;) (the detailed proofs of these cases we leave to the readers) :)

[quote="can_hang2007"] Using the Vasc's inequality (I am very sorry about this, VasC. I dont know that you have created it before):
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \sum \frac {bc}{(b \plus{} c)^2} \minus{} \frac {5}{4} \ge 0,
\]
we can see that it suffices to prove ....[/quote]
The following is also interesting.

Find the smallest $ k$ such that

$ \frac {ab}{a^2 \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} a^2} \minus{} \frac 3{2} \plus{} k(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} 1) \ge 0$.

[hide]$ k \equal{} \frac {\sqrt 5 \plus{} 1}{2}$,
with equality for $ (a,b,c) \equal{} (1,1,1)$, and again for $ (a,b,c) \equal{} (\frac {\sqrt 5 \minus{} 1}{2},1,1)$.[/hide]

Put $ b\equal{}c$. Ineq $ \Longleftrightarrow$ $ (a\minus{}b)^2(\frac{k(a^2\plus{}b^2)\minus{}b^2\minus{}2ab}{(a^2\plus{}b^2)(b^2\plus{}2ab)})$. Hence $ k\geq \frac{1\plus{}\sqrt{5}}{2}$. But for $ k\equal{}\frac{1\plus{}\sqrt{5}}{2}$ I can't prove it. Maybe can be prove with SOS but I'm not sure.


Assume that the set of all positive integers is decomposed into $r$ disjoint subsets $A_{1}, A_{2}, \cdots, A_{r}$ $A_{1} \cup A_{2} \cup \cdots \cup A_{r}= \mathbb{N}$. Prove that one of them, say $A_{i}$, has the following property: There exist a positive integer $m$ such that for any $k$ one can find numbers $a_{1}, \cdots, a_{k}$ in $A_{i}$ with $0 < a_{j+1}-a_{j} \le m \; (1\le j \le k-1)$.

firstly, a bit of terminology: we will call such a sequence of integers a "k-chain, m arrow"

Lemma: If a set $ S$ satisfies this property, and $ S \equal{} A \cup B$ with $ A$ and $ B$ disjoint, then at least one of these parts must satisfy the property.

Proof: Suppose $ A$ does not satisfy the property(let's denote this property as P). Now to construct $ A$, we remove some elements from $ S$ and have the removed elements go to $ B$ and the remained elements stay in $ A$. So $ S$ does satisfy $ P$ for $ m \equal{} m_0$, but $ A$ does not. Therefore for at least some $ k \equal{} k_0$, $ A$ does [b]not[/b] satisfy the property for $ m \equal{} m_o$. However in $ S$ there exists a $ k_o K$ chain $ m_o$ narrow for all $ K$. Hence we had to remove at least $ K$ elements from this chain(to eliminate the existence of any $ k_o$ chains). Hence these $ K$ removed elements form a $ K$ chain $ k_o m_o$ narrow in $ B$. But this argument works for all $ K$ and hence $ B$ satisfies $ P$ where $ m \equal{} k_om_o$ there. $ \blacksquare$

Now this lemma is in fact equivalent to the result since we can construct our partition in a finite number of steps:

$ N \equal{} A_1 \cup S_1$

If $ A_1$ satisfies $ P$ we are done, if not $ S_1$ does by the lemma. And $ S_1 \equal{} A_2 \cup S_2$... ie and we keep applying this argument and if no $ A_i$ satisfies P then eventually we get $ S_{r \minus{} 2} \equal{} A_{r \minus{} 1} \cup A_r$ and so by the lemma at least one of these sets must satisfy $ P$


The operators "+," "-," and "x" are each used once to fill in the blanks randomly. Express as a common fraction the probability that the value of the expression formed is positive.
7 __ 8 __ 9 __ 5

7+8-9*5: Negative
7+8*9-5: Positive
7-8+9*5: Positive
7-8*9+5: Negative
7*8+9-5: Positive
7*8-9+5: Positive

Therefore the answer is 2/3.

Intuitively, you could say that the expression is negative only if the minus sign comes before the multiplication sign. You would be subtracting a large number that would, because all of the numbers are close together, "dominate" the expression. We have two ways to order the minus sign comes before the multiplication sign out of 3!=6 ways. This gives us 2/6 or 1/3, but we are using complementary counting and so we subtract from 1 to get 2/3.


algebra
polynomial
limit
complex analysis
complex analysis unsolved

Sorry, i did mistake,  there should be complement of $ \Omega$ instead of $ \Omega$.
[edited]:

Let  $ P: C\rightarrow C$  be a polynomial with  $ deg(P) > 1$  and let  $ P_n(z) \equal{} P(P(\cdots P(z)\cdots))$  be $ n^{th}$  iterate of  $ P$  and define the set $ \Omega: \equal{} \{z\in C: \displaystyle\lim_{n\to\infty}P_n(z) \equal{} \infty\}$.  Prove the following propositions:

 $ (i)$    $ C\setminus\Omega$ is compact subset of $ C$.
 $ (ii)$    $ |C\setminus\Omega| \equal{} \infty$.
$ (iii)$   Find $ C\setminus\Omega$ for $ P(z) \equal{} z^2 \minus{} 2$.

Thanks!

The first is trivially false and the second is trivially true; $ \Omega$ contains everything outside some circle, where the leading term dominates and $ |P_n(z)|$ is an increasing sequence.


function
real analysis
geometry
3D geometry
algebra
domain
geometric transformation

$ 1.$  Assume $ V$ is an open subset of $ \mathbb{R} ^k$ and $ \mu$ is a positive and finite borel measure. Is it true that the function $ \mu(V\plus{}x)$ is always continuous, lower semi-continuous or upper semi-continuous?

I found that it must be lower semi-continuous, but not necessarily continuous or upper-semicontinuous. Is it correct or should I check my argument?

$ 2.$  Assume $ T: \mathbb{R}^k\to\mathbb{R}^k$ is a linear map and $ \dim(Im(f))<k$. Can somebody give me a hint on how to prove that $ m(Im(f))\equal{}0$, where $ m$ is the Lebesgue measure? Intuitively it's clear, but I don't know how to prove it rigurously.

1. You said $ \mu$ was a positive finite Borel measure but note that you didn't say anything about it being absolutely continuous. Consider this very simple example: Let $ \mu$ be the point mass at the origin. That is, $ \mu(E)\equal{}1$ if $ 0\in E$ and $ \mu(E)\equal{}0$ otherwise. Let $ V$ be the open unit ball. Then $ \mu(V\plus{}x)\equal{}\chi_V(x),$ and that function is lower semicontinuous but not continuous.

2. As a hint, first try to prove that any bounded subset of that lower-dimensional subspace has measure zero. Just work on how many balls or cubes of a certain size it would take to cover the set.

$ Im(f)$ is the image of a map from $ \mathbb{R}^{dim(Im(f))}$ to $ \mathbb{R}^k$. Clearly, this map can be chosen Lipschitz. It follows from Lipschitz and the lower dimension of the domain that the image has measure zero

Also avoiding calculation, for any bounded piece of $ Im(f)$ infinitely many translated copies can be fitted into a bounded subset of $ R^k$. So that bounded piece has measure zero since the measure is is translation invariant.

Thanks, K & K. I did the same for the first part of 1. Nice solution for 2, the one with translations.

Here is the solution for lower semi-continuity.
Assume $ \mu(V\plus{}x)>\alpha$ and take $ \varepsilon_1>0$ such that $ \alpha<\mu(V\plus{}x)\minus{}\varepsilon_1$. If we write $ V\plus{}x\equal{}\bigcup B(x_i\plus{}x,r_i)$ where the balls are disjoint then we can take $ 0<\varepsilon_2<\varepsilon_1$ and a finite set $ F$ such that $ \mu(V\plus{}x)<\sum_{i \in F}\mu(B(x_i\plus{}x,r_i))\plus{}\varepsilon_2$. Because $ B(x_i\plus{}x,r_i)\equal{}\bigcup _{n}B(x_i\plus{}x,r_i\minus{}\frac{1}{n})$ we can take n such that $ \mu(B(x_i\plus{}x,r_i)\plus{}\frac{\varepsilon_2\minus{}\varepsilon_1}{m} <\mu(B(x_i\plus{}x,r_i\minus{}\frac{1}{n}))$ where $ m\equal{}|f|$, and we can take $ n$ good for all $ i\in F$ because $ F$ is finite.
Then \[ \alpha<\mu(V\plus{}x)\minus{}\varepsilon_1<\sum_{i \in F}\mu(B(x_i\plus{}x,r_i))\plus{}m \frac{\varepsilon_2\minus{}\varepsilon_1}{m}<\sum_{i \in F} \mu((B(x_i\plus{}x,r_i\minus{}\frac{1}{n}))<\sum_{i \in F} \mu(B(x_i\plus{}y,r_i))<\mu(V\plus{}y)\] for all $ \parallel{}x\minus{}y\parallel{}<\frac{1}{n}$. So the function is lower semi-continuous.

Let me take a crack at 1 (maybe it's Ok):

Let $ U_n \equal{} \left\{ u \in \mathbb R^k : B \left( u, \frac1{n} \right) \subset V \right\} \equal{} X \setminus \bigcup_{v \in X \setminus V} B \left( v, \frac1{n} \right)$ (so it's Borel).
We have that $ U_1 \subset U_2 \subset U_3 \subset \ldots$ and $ \bigcup_{n\equal{}1}^\infty U_n \equal{} V$, so $ \lim_{n \to \infty} \mu (U_n) \equal{} \mu (V)$.
Now, for all $ x \in \mathbb R^k$ such that $ \| x \|_2 < \frac1{n}$ we have that $ U_n \subset V \cap (V\plus{}x)$, so \[ \mu (V\plus{}x) \geq \mu (V \cap (V\plus{}x)) \geq \mu (U_n) .\]
Letting $ n \to \infty$ we get that $ \liminf_{x \to 0} \mu (V\plus{}x) \geq \mu (V)$...


A beautiful and rather difficult problem:

A finite number of stones are placed on an infinite in both directions row of cells. A cell may contain several stones. The following moves are allowed:
[i]1.[/i] Take $ 1$ stone from each of the cells $ n\minus{}1$ and $ n$ and place $ 1$ stone in cell $ n\plus{}1$.
[i]2.[/i] Take $ 2$ stones from cell $ n$ and place $ 1$ stone in each of the cells $ n\minus{}2$ and $ n\plus{}1$.

Prove that for any sequence of moves, we'll end up in impossibility of doing any of the described moves. Moreover, prove that this 'impossibility' situation is unique (doesn't depend on the order of the moves; it depends only on the initial distribution of stones in cells).
__
I'm sorry if the problem has been posted before. I didn't find it.

Is was also discussed at: [url]http://www.mathlinks.ro/viewtopic.php?t=132763[/url], starting with post #9, but I would rather prefere to continue this topic here.

note that the first move,reduces the number of stones by $ 1$,so we cant do this move infinitely many times,so from a step we would use only the second move,now from this step on,we will give each stone a weight,let $ i$ be the weight of a stone in $ i^{th}$ cell,now let $ S$ be the sum of all of our weights...
so by using the second move,we will decrease the sum by $ 1$ ($ S \minus{} 2n \plus{} (n \minus{} 2) \plus{} (n \plus{} 1) \equal{} S \minus{} 1$)...
now let there be a stone in $ k^{th}$ cell,we claim that the cells from $ k \minus{} 2$ till $ k \plus{} 1$ will never get all empty (its easy to prove,using contradiction and assuming the first step that they've become all empty)
but note that as mentioned before,the second move,in each step reduses $ S$,i.e. it reduses the smallest cell which containes a stone,but according to what we said,the union of every $ 4$ consecutive cells are nonempty,so the number of stones must be infinite,which is a contradiction,so after finitely many steps,we can't do any of the moves above,so the first part is proved...
--------------------------------------------------------
now for the second part:
according to what we said before,we can find the smallest cell which contains a stone,name it $ 1$ and from this cell,label the other cells...
now let the weight of a stone in $ n^{th}$ cell be $ F_n$ where $ F_1 \equal{} 1,F_2 \equal{} 2,F_n \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2}$
now with this kind of weight,in each step the sum of the weights ($ S$) wont change...
now there is a lemma:

----------------------------------------------

[b]lemma:every natural number $ S$ can be represented in a unique form as:

$ S \equal{} F_{i_1} \plus{} F_{i_2} \plus{} ... \plus{} F_{i_k}$

where $ i_1 < < i_2 < < i_3 < < ... < < i_k$[/b]

(the notation $ x < < y$ means $ y \minus{} x\geq 2$)

[hide="proof"]
you can prove the lemma using contradiction and these relations:

$ 1)F_1 \plus{} F_3 \plus{} ... \plus{} F_{2n \minus{} 1} \equal{} F_{2n} \minus{} 1\Rightarrow F_1 \plus{} F_3 \plus{} ... \plus{} F_{2n \minus{} 1} < F_{2n}$

$ 2)F_2 \plus{} F_4 \plus{} ... \plus{} F_{2n} \equal{} F_{2n \plus{} 1} \minus{} 1\Rightarrow F_2 \plus{} F_4 \plus{} ... \plus{} F_{2n} < F_{2n \plus{} 1}$[/hide]

-------------------------------------------

now according to this lemma,let $ S \equal{} F_{i_1} \plus{} F_{i_2}... \plus{} F_{i_k}$ be the unique representation of $ S$ as the sum of the elements of fibonacci sequence,then this means that each of the cells $ i_1,...,i_k$ containes exactly $ 1$ stone...
so the final step is unique


probability
expected value
probability and stats

Take an ordinary deck of 52 playing cards, and shuffle them so that they are sufficiently randomized. For $ 1\le i\le 51$ define a pair to occur if $ \text{card}_i \equal{} \text{card}_{i \plus{} 1}$. What is the expected value of the number of pairs in the deck? Based on a computer simulation I think the answer is 3, but I can't prove it.

the probability of getting a pair in places $ i,i\plus{}1$ is just $ 13\binom{4}{2}/\binom{52}{2}$.  the expected number of pairs is the sum of the probability of getting a pair in places $ i,i\plus{}1$ over $ 1\leq i\leq 51$.

so the answer is $ (13\binom{4}{2}/\binom{52}{2})\cdot 51 \equal{} 3$.

[b]Jimmy:[/b] What do you mean by  $ \text{card}_i \equal{} \text{card}_{i \plus{} 1}$? When we draw two cards from 52 distinct cards, how can we get two cards identical? Do I misunderstand the question?!

Edit: Sorry I got it!

blahblahblah, this is for you:

Suppose the deck has $ m$ denominations of cards, and $ k$ individual cards of each denomination, so that the whole deck contains $ mk$ cards. (For the familiar case, $ m\equal{}13$ and $ k\equal{}4.$) What now is the expected number of pairs that occur in the deck?

And if there is any kind of nice answer, does it point to another possible argument for what you said in the first place?

i'm not sure what you're getting at.  i mean, in the general setting, the probability of getting some particular pair is $ (k\minus{}1)/(mk\minus{}1)$, so the expected number of pairs is $ k\minus{}1$.  but this is the same argument...


calculus
integration
number theory unsolved
number theory

find all squares of the form $ d_n\equal{}\frac{n(n\plus{}1)(n\plus{}8)}{3}$

I know two: $ d_6\equal{}14^2$, $ d_{24}\equal{}80^2$

Hi Knoppix, your post didn't say the boundary of $ n$,
So we can choose some $ n \in \mathbb{C}$, that make $ d_{n}$ any square we want!
But that's senseless, so I'll consider now the case $ n \in \mathbb{N}$
Since $ n\plus{}8 \equiv n\plus{}2 (mod 3)$ so it is obvious that $ \frac{n(n\plus{}1)(n\plus{}8)}{3}\in \mathbb{Z}$.
We know that $ \frac{n(n\plus{}1)(n\plus{}8)}{3} \equal{} \frac{n(n^{2}\plus{}9n\plus{}8)}{3}$
We distinguish three cases,

[b][u]Case 1[/u][/b] $ 3 \mid n$,
Hence, $ n\equal{}3k$ for some $ k \in \mathbb{N}$.
Therefore, $ \frac{n(n^{2}\plus{}9n\plus{}8)}{3}\equal{} k(9k^{2}\plus{}27k\plus{}8)$
[b][u]Case 1.1[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}27k\plus{}8))\equal{}1$,
This follows, $ k\equal{}r^{2}$ for some $ r \in \mathbb{N}$.
and $ 9k^{2}\plus{}27k\plus{}8\equal{}9r^{4}\plus{}27r^{2}\plus{}8$ is a square.
Since no square is congruent to 2 (mod 3), this is a contradiction!
[b][u]Case 1.2[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}27k\plus{}8))\equal{}2$,
This follows, $ k\equal{}2r^{2}$ for some odd integer $ r \in \mathbb{N}$.
and $ (9k^{2}\plus{}27k\plus{}8)/2\equal{}18r^{4}\plus{}27r^{2}\plus{}4$ is a square.
(I stuck in this case : it doesn't contradicts because we can find some possible values like $ k\equal{}2$.  :(
And it seems to be using about the [b]elliptic curve![/b] <- I know nothing  :blush: )
[b][u]Case 1.3[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}27k\plus{}8))\equal{}4$,
This follows, $ k\equal{}4r^{2}$ for some odd integer $ r \in \mathbb{N}$.
and $ (9k^{2}\plus{}27k\plus{}8)/4\equal{}36r^{4}\plus{}27r^{2}\plus{}2$ is a square.
Since no square is congruent to 2 (mod 3), again, this is a contradiction!
[b][u]Case 1.4[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}27k\plus{}8))\equal{}8$,
(This is another case I stuck : it doesn't contradicts because we can find some possible values like $ k\equal{}8$.  :( ...)

[b][u]Case 2[/u][/b] $ 3 \mid n\plus{}1$,
Hence, $ n\plus{}1\equal{}3k$ for some $ k \in \mathbb{N}$.
Therefore, $ \frac{n(n\plus{}1)(n\plus{}8)}{3}\equal{} k(3k\minus{}1)(3k\plus{}7) \equal{} k(9k^{2}\plus{}18k\minus{}7)$
[b][u]Case 2.1[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}18k\minus{}7))\equal{}1$,
This follows, $ k\equal{}r^{2}$ for some $ r \in \mathbb{N}$.
and $ 9k^{2}\plus{}18k\minus{}7$ is a square.
Since no square is congruent to 2 (mod 3), again, this is a contradiction!
[b][u]Case 2.2[/u][/b] $ g.c.d.(k,(9k^{2}\plus{}18k\minus{}7))\equal{}7$,
This follows, $ k\equal{}7r^{2}$ for some $ r \in \mathbb{N}$.
and $ (9k^{2}\plus{}18k\minus{}7)/7 \equal{} 63r^{4}\plus{}18r^{2}\minus{}1$ is a square.
Since no square is congruent to 2 (mod 3), again, this is a contradiction!
PHEW... , I'm so happy we can go through this case with no trouble!

[b][u]Case 3[/u][/b] $ 3 \mid n\plus{}8$,
Hence, $ n\plus{}8\equal{}3k$ for some $ k \in \mathbb{N}$.
Therefore, $ \frac{n(n\plus{}1)(n\plus{}8)}{3}\equal{} k(3k\minus{}7)(3k\minus{}8) \equal{} k(9k^{2}\minus{}45k\minus{}56)$

Please help me the rest..., I'm so tired now.  :(
(especially, who can solve the case 1.2 or 1.4, please show me!  :wink: )

I have no idea of how to proceed, but I think this problem is relationed with this one (case $ n \equal{} 84a \plus{} 6$ is equivalent to say when $ 6a \plus{} 1$, $ 12a \plus{} 1$ and $ 14a \plus{} 1$ are perfect squares):

http://www.mathlinks.ro/resources.php?c=1&cid=16&year=1986&p=366557

But it is more difficult.

Epsilonist, you are not using the given information with full force (if $ xyz$ is a perfect square and $ x, y, z$ are pairwise coprimes, then $ x, y$ and  $ z$ are perfect squares), so maybe we don't need to solve your equation...

This is, of course, an elliptic curve.  As such, it's guaranteed to have finitely many integer solutions, but actually finding them is not straightforward in the general case.  One approach is to show that the curve has rank zero using a computer algebra system and then apply the [url=http://www.indopedia.org/Nagell-Lutz_theorem.html]Nagell-Lutz theorem[/url].  Unfortunately, if the curve doesn't have rank zero there are problems.

Hi,
thank you very much for your answers :)

sorry for replying 2 weeks later, but first of all, I was busy,,
and this forum is a bit too advanced for me so I'm not an 'everyday user'  ;)

$ n$ should be a positive integer there.

If I change the question:

Is the number of positive integers $ n$ such that $ d_n$ is a square, finite or not ?

could you give an elliptic proof :wink:  ?

I'm not familiar with such things...

Finite; this is a special case of [url=http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points]Siegel's theorem[/url].


complex numbers
linear algebra
linear algebra unsolved

Let $ A,B \in M_2(\mathbb{R})$. Prove that: $ \det [(AB\plus{}BA)^4\plus{}(AB\minus{}BA)^4] \ge 0$

Use this equation $ X^2 \minus{} tr(X).X \plus{} det(X) \equal{} 0$ for all $ X \in M_{2.2}(R)$

We have $ tr(AB \minus{} BA) \equal{} 0$ so $ (AB \minus{} BA)^4 \equal{} t^2I$ with $ t \equal{} |AB \minus{} BA|$

And $ det[(AB \plus{} BA)^4 \plus{} (AB \minus{} BA)^4] \equal{} det(X^2 \plus{} t^2I) \equal{} det(X \plus{} itI)det(X \minus{} itI) \geq 0$ because $ det(X \plus{} itI)$ and $ det(X \minus{} itI)$ are two conjugate complex numbers


A real number $ m$ is randomly selected such that $ 0 \le m \le 8$. What is the probability that $ m < 2$? Express your answer as a common fraction.

m=0, 1, 2, ..., 8 = 9 ways

m can only be 0 and 1 if m<2.

answer : $ \frac{2}{9}$

The only two numbers that satisfy this is 0 and 1. There are 8-0+1=9 numbers, so the probability is 2/9. The answer said 2/9. Did I do anything wrong?

EDIT: beaten to it

I think that this problem is phrased incorrectly.  If it said "m is a randomly selected integer" then the last solutions hold.

But this is saying that that [b]m is a real number[/b], not just a natural number, so to say that m=0,1,2...or 8 would be incorrect.

We can't really count "how many numbers work" and divide by "how many total numbers" there are because both answers are infinte.

However intuitively, we can think of a yardstick with a 0 mark and an 8 mark, and realize that everything below 2 is 1/4 of the 8 unit long yardstick.  So, as phrased, the answer should be $ \boxed{\frac{1}{4}}$.

However if it is a typo and "real number" should be "integer" then 2/9 is correct.

Yes, the correct answer in the game is 1/4.


linear algebra
matrix
linear algebra unsolved

$ A,B \in M_{2}(R)$. Prove that $ det((AB\minus{}BA)^4 \plus{}(AB\plus{}BA)^4) \geq 0$

$ (AB \minus{} BA)^2 \equal{} \minus{} det(AB \minus{} BA) I_{2}$ and comute with every matrix
and use $ det(Z^2 \plus{} X^2) \geq o$ if $ ZX \equal{} XZ$


This is a fun one.  Source: Canadian Math Olympiad (CMO) 2001 #1:

[b]Randy:[/b] "Hi Rachel, that's an interesting quadratic equation you have written down.  What are its roots?"
[b]Rachel: [/b]"The roots are two positive integers.  One of the roots is my age, and the other is the age of my younger brother, Jimmy."
[b]Randy: [/b]"That is very neat!  Let me see if I can figure out how old you and Jimmy are.  That shouldn't be too dificult since all your coefficients are integers.  By the way, I notice that the sum of the three coefficients is a prime number."
[b]Rachel:[/b] "Interesting.  Now figure out how old I am."
[b]Randy: [/b]"Instead, I will guess your age and substitute it for x in your quadratic equation... darn, that gives me -55, not 0.
[b]Rachel:[/b] "Oh, leave me alone!"

The problem here is:
(a) Prove that Jimmy is 2 years old.
(b) Determine Rachel's age

P.S. I love how he says "That is very neat!"  It's just funny... and now my fingers hurt.  I type too much.

I'm just gonna say.. that is *such* a cool problem! I love it..

I agree that this is an awesome problem, so I'm going to post a solution in spoiler.

[hide]Let us write our quadratic equation as (x-a)(x-b), where a and b are positive integers. Specifically, let a be Jimmy's age, and let b be Rachel's age. The sum of the coefficients is the value of (x-a)(x-b) when x=1 (can you see why?). Thus (1-a)(1-b) is prime. Yet it has a factorization. We know a<b, so then 1-a must be :pm:1. 1-a can't be 1 since that would require a=0. Thus 1-a=-1, or a=2. Thus Jimmy is 2.

Also we know that for some positive integer x, (x-2)(x-b)=-55. -55=5*-11, so let's make x-2=5 and x-b=-11. Thus x=7 and b=18. Thus Rachel is 18.[/hide]

Ahhhh, Canandian Olympiads. Love them almost as much as the Irish ones... :D

A couple of things that may or may not be important:

[hide]

Firstly, the quadratic may have a leading coefficient of k, when we come to the prime thing k must be 1 but you never know..

Secondly, when we get (x-2)(x-b)=-55, there are a few solutions. We also know that b-1 is prime from what you had before.

1 x -55 gives x=3, b=58 but 57 is not prime.

5 x -11 gives x=7, b=18, which works.

11 x -5 gives x=13, b=18, which also works.

55 x -1 gives x=57, b=58 but again 57 is not prime.

So b must be 18 (which works for two different x values.)

[/hide]

I'm removing the sticky.

This diaglogue is going to get incorporated into the Saga when I update it...


symmetry
integration
trigonometry
linear algebra
matrix
calculus
derivative

Let $A$ be a symmetric matrix. Suppose $u$ and $v$ are eigenvectors of $A$ associating with two different eigenvalues. Show that $u$ and $v$ are orthogonal.

So here is my flawed proof.

Suppose we have $Au = \lambda u$ and $Av = \mu v$ where $\lambda \neq \mu$. We now take their dot product:

$(Au)\cdot (Av) = (Av)^T(Au) = v^TA^TAu = v^TA^2u = v^T\lambda^2u = \lambda^2 (u \cdot v)$ (Remember that symmetry implies $A = A^T$),

but this is precisely equal to (\lambda u)\cdot(\mu v) and therefore we can equate the two, thus:

$(\lambda u)\cdot(\mu v) = \lambda^2 (u\cdot v)$ which then simplifies to $\lambda(\mu-\lambda)(u \cdot v) = 0$.

Here is where the problem creeps in: All is very well if $\lambda \neq 0$, but what if $\lambda = 0$?

I suspect there is one property associated to the symmetry: any of the eigenvalues cannot be $0$. Can anyone prove or refute this (that is, the problem should have stated something about the eigenvalues)?

You don't want $A^TA$ in there - one copy of $A$ is enough.

$\lambda u\cdot v=\lambda v^Tu=v^T(\lambda u)= v^TAu=v^TA^Tu$ (THAT was where we used that $A$ is symmetric).

$=(Av)^Tu=(\mu v)^Tu=\mu u\cdot v.$

So $(\lambda-\mu)u\cdot v=0,$ meaning that either $\lambda=\mu$ or $u\cdot v=0.$

Interesting challenge: emulate this proof to show that

$\int_0^{\pi}\sin mx\sin nx\,dx=0$ whenever $m\ne n.$

(The role of the "symmetric matrix" will be played by the second derivative operator.)

But is this related to Integration?

How can it be done? I can't see the relationship.

Think of $\sin mx$ as $u$ and $\sin nx$ as $v.$

Back up in post #2 I started with $\lambda u\cdot v=(Au)\cdot v,$ using the fact that $\lambda u=Au.$

Start this with $-m^2\sin mx=(\sin mx)''.$ Start with

$-m^2\int_0^{\pi}\sin mx\sin nx\,dx$ and integrate by parts twice.

It really is, at heart, the same proof.


Let $ S \equal{} \sqrt [] {9 (1) ^ 2 \plus{} (1)} \plus{} \sqrt [] {9 (2) ^ 2 \plus{} (2)} \plus{} \ldots \plus{} \sqrt [] {9 (7) ^ 2 \plus{} (7)}$. And $ A < S < A \plus{} 1$, find $ 2A \plus{} 1$.
$ A$ is integer.

Greetings  :D

hello, we got $ 85 < S < 86$ and so we have $ 2A \plus{} 1 \equal{} 171$.
Sonnhard.

Hello can you explain me as well.

$ 171$

Greetings  :D


let$x_{i} \geq 0$such that  $x_{i} \in [0;1]$ follow property :
$\sum\limits_{i=1}^{2002} {x_{i}}=2010$
 find maximal of expression :
 P=$\sum\limits_{i=1}^{2002}{x_{i}^3}$

It is nice inequality.I have long solution.


A light bulb is completely covered with a thin aluminium shell in a room maintained at $20^{\circ} C$. If the shell has an equilibrium temperature of $100^{\circ} C$, what is the wattage of the bulb?

[color=blue]Do we have the area of the shell?[/color]

Area of the shell........??? Well, it's not given......You have only the temperatures of both........they're actually enough to find the wattage of the bulb.....If no one gets it for some time, I'll later post the solution.

P.S: Think Stephen-Boltzman.

Hmmm.....I'm posting the solution since it's quite a long time and nobody else has posted anything... Here's the stephen-Boltzman law -- The total energy radiated per second by a body at a temperature $T$ to it's surroundings at a temperature $T_0$ is given by $\sigma (T^4 - T_0^4)$ where $\sigma$ is a constant which is equal to $5.669 X 10^{-8} W m^{-2} K^{-4}$

Substituting in that equation gives --

Total Energy radiated per second = $\sigma (373^4 - 293^4)$ {both temperatures need to be converted to the Kelvin scale to use that Stefen-Boltzman constant $\sigma$.}

So,  Energy radiated per second $= 5.669 X 10^-8 X (1.198682784 X 10^{10})$
      $= 6.795332703 X 10^2$
      $\approx 679 Joules/sec$

Since the system radiates around $679 Joules/sec$ the wattage of the bulb must be about $679 Watts$

I'm confused, i thought the equation of a radiator was $W = A\sigma e T^4$ where $e$ is the emissivity of the material (in the case of aluminum foil, i believe we can assume it is $1$), and $A$ is the area of the surface of the object. Since the net radiation equals the radiation of the object minus the radiation of the surroundings into the object, the equation becomes $W = A \sigma e (T_o^4 - T^4)$ where $T_o$ is the temperature of the object and $T$ is the temperature of the surroundings. (This all came from my physics book). Also, I was wondering why the equation $W = \frac{A(T_a - T_b)}{L}$ couldnt be used in this situation.

Hey psychicalchemist,..... The reason why you cant use your equation $W = \frac{A(T_a - T_b)}{L}$ is because you aren't given $A$.  :D

[quote="Djole"]Well, yes, the [i]power[/i] of a radiating body is your $W$. The emissive power $E$ is something like the intensity (or flux) of the emission and is given by $E=W/A$ where $A$ is your area. And yes, this problem is to be solved by Steffan-Boltzman's law.[/quote]

hmm...wattage most likely refers to stuff measured in watts, right? i.e. power, not emissive power...

unless someone's playing a HUGE PRANK on us all

I just saw that indeed we don't have the area of the shell... Sorry for any misunderstandings. Yes, we do need the area of the shell. Without it, the problem's not solvable.

[color=blue]I have the same idea with [b]Djole[/b]. That's why I ask [b]karthik_k[/b] for the area of the shell[/color]


function
real analysis
real analysis unsolved

Let $f: [a,b] \to \mathbb{R}$. Define the variation of $f$ as $\displaystyle V_{a}^{b}(f)=sup_{a=x_0 \leq x_1=\leq...\leq x_n=b} \sum_{i=0}^{n-1}|f(x_{i+1}-f(x_i))|$, where the supremum is taken over all the subdivisions of $[a,b]$ into the intervals $[x_0,x_1],...[x_{n-1},x_n]$. By $BV[a,b]$ denote the set of such functions $f$ for which $V_{a}^{b}(f)<\infty$.
Prove that the set $BV[a,b]$ with norm $||f||=||f||_{\infty}+V_{a}^{b}(f)$ is Banach space.

take $f_n$ a Cauchy sequence in this space.(hence $V(f_n) < C$ for a constant $C$). Then it is clear that the sequence converges uniformly towards a function $F$.

Take a subdivision $x_0<x_1<..<x_n$ of $[a;b]$ and $\epsilon>0$.
One can find $N_0$ such that for any $p>N(\epsilon)$, and any $0 \leq i \leq n$: $|F(x_i)-f_p(x_i)| \leq \epsilon$.
Hence $|F(x_i)-F(x_{i-1})| \leq 2 \epsilon + |f_p(x_i)-f_p(x_{i-1})|$.
This is why $\sum |F(x_i)-F(x_{i-1})| \leq \sum |f_p(x_i)-f_p(x_{i-1})| + n \epsilon \leq C + n \epsilon$, for any $\epsilon > 0$. Finally $V(F) < C$ i.e. $F$ is of bounded variation.

Now it suffices to show that $f_n$ converges towards $F$. And to prove this it suffices to show that $V(F-f_n)$ tends towards $0$, which is easy (almost the same as above)


LaTeX
number theory unsolved
number theory

let $m,n, a$ be positive integers. Find $gcd(a^m+1,a^n+1)$?
(  $gcd(a,b)$ is the the greatest common divisor of $a,b$  )

Is the problem "which integers $m,n,a$ give the greatest value of $g.c.d(a^m+1, a^n+1)$?" or "for any $m,n,a$ what is the greatest common divider?"

Let x(n)=a^n+1, a0=1 if a is even, else a0=2. If n|m and (m/n) is odd, then x(n)|x(m).
If m>=2n then (x(n),x(m))=(x(n),x(m-2n)). Therefore, if n|m and (m/n) is even, then (x(n),x(m))=a0.
Let d=(m,n) and m1=m/d, n1=n/d, p*m1-q*n1=1.
Case 1.  If m1 and n1 is odd, then x(d)|x(n), x(d)|x(m). One of numbers p and q is odd. Let p is odd. In this case (x(n),x(m)) is divisor  (x(pm),x(n) is divisor x(d). Therefore x(d)=(x(m),x(n).
Case 2. One of m1, n1 is even. Let n1 is even, then p and m1 is odd. We have (x(m),x(n) is divisor (x(pm),x(n)) is divisor x(d) is divisor (x(d),x(n))=a0.

Rust, here you'll learn how to use LaTeX :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

This way, it's going to be easier for us to read your proofs!

edit : see how much nicer it looks like :
[quote="Rust"]Let $x_n=a^n+1$, $a_0=1$ if $a$ is even, else $a_0=2$. If $n|m$ and $\frac mn$ is odd, then $x_n|x_m$.
If $m \ge 2n$ then $(x_n,x_m)=(x_n,x_{m-2n})$. Therefore, if $n|m$ and $\frac mn$ is even, then $(x_n,x_m)=a_0$.
Let $d=(m,n)$ and $m_1=\frac md$, $n_1=\frac nd$, $p \cdot m_1 - q \cdot n_1=1$.
Case 1.  If $m_1$ and $n_1$ is odd, then $x_d|x_n$, $x_d|x_m$. One of numbers $p$ and $q$ is odd. Let $p$ is odd. In this case $(x_n,x_m)$ is divisor $(x_{pm},x_n)$ is divisor $x_d$. Therefore $x_d=(x_m,x_n)$.
Case 2. One of $m_1$, $n_1$ is even. Let $n_1$ is even, then $p$ and $m_1$ is odd. We have $(x_m,x_n)$ is divisor $(x_{pm},x_n)$ is divisor $x_d$ is divisor $(x_d,x_n)=a_0$.[/quote]

Rust, you might like TeXmacs, it is a WySiWyG, emacs-florished editor producing output in TeX, pdf, ps, ...
http://www.texmacs.org


[quote="mr. math"]i need help with this problem and also need the formula to find the volume of pyramids.

what is the volume in cubic centimeters, of a right rectangular pyramid with a height measuring 8cm and a base measuring 3cm by 19cm?

thanks[/quote]

[hide]$\frac{(3)(19)(8)}{3}=\boxed{152}$

See if you can derive the formula now.[/hide]

[quote="mr. math"](height times area of base) divided by 3?

is this it?[/quote]

Yes, but can you prove it?

The answer is 152 because the area of any pyramid or cone is $\frac{1}{3}\cdot\text{base}\cdot\text{height}$

[quote="Totally Zealous"]The answer is 153 because the area of any pyramid or cone is $\frac{1}{3}\cdot\text{base}\cdot\text{height}$[/quote]

I think you need to check your arithmetic.  Unless you made a typo.


http://www.macon.com/mld/macon/news/politics/16609180.htm

So apparently, in Boston and 9 other cities these LED signs with the character Ignignokt from Aqua Teen Hunger Force, a Cartoon Network show, were put up in various places.  It caused the most uproar in Boston where the police took it as a bomb threat and closed roads, bridges and the T.

What do you think?  Was this an overreaction?  What about the media's reaction?

Also, what about [url=http://youtube.com/watch?v=fJkTNJ7BM9I]this[/url].

[url=http://www.schneier.com/blog/archives/2007/02/nonterrorist_em.html]Bruce Schneier[/url] had a pretty good blog post on it. Personally I think it was overreaction.[/url]

Ha, that's a really good blog post.  Also, [url=http://www.youtube.com/watch?v=J3xaZDWl51U]here[/url]'s a youtube video that show's them putting them up.

I agree that it's seems to be a huge overreaction.

I think they're kind of forced to overreact to it.

If they don't, and someone gets hurt, family members of individuals that get hurt can press charges. And then they have to deal with the whole nation thinking that they're idiots for not taking a bomb warning seriously just a few years after 9/11.

[quote="joml88"]Yes, they should overreact to an extent.  But shouldn't there be a point at which it just becomes ridiculous?[/quote]

yeah, there should be some sort of reasonable limit. in this specific case, i don't think they overacted too unreasonably.

but looking back at the article i see that boston purportedly lost \$500,000. wow  :huh: . maybe it's more unreasonable than i thought. how long did they close the highways?

the Turner company should legally be bound to pay for some of that.

maybe in the future legislators can draft a law mandating that organizations, individuals, etc. are liable to pay a certain percentage of the money that others lose due to fake terrorist or security threats.

[quote="drunner2007"]maybe in the future legislators can draft a law mandating that organizations, individuals, etc. are liable to pay a certain percentage of the money that others lose due to fake terrorist or security threats.[/quote]
What!? Citizens should be forced to pay whenever someone from a government agency does something stupid? What if they profile Middle Easterners, spend \$2000 in court costs hassling some random programmer, and expect him to pay a percentage of it? WTF is that? And what if I sneeze in the airport and someone thought I said "bomb" and they lock down the airport and like \$200,000 is lost by the airlines and airport because of my sneeze? Sneezes are involuntary! I'm not going to pay them a penny! I'll move to Europe or Canada or something!


Remember that $\frac{n(n + 1)}{2} = 1 + 2 + ... + n$. Do you see why the result follows from this?

Obviously you can use induction, but the other method is neater!

Induction seems the easiest to me.  You can also use induction for the following:

$\displaystyle\sum_{k=1}^{n}\frac{k(k+1)}{2} = \frac{1}{6}n(1+n)(2+n)$

Guys, you don't need ANY calculations or induction. But I guess I won't be able to convince you.

Ok, maybe we can use $\text{LHS}=\sum_{k=2}^{n} \binom{k}{2}=\binom{n+1}{3}$... (or maybe not?)

The triangular number trick works like this:

[hide]Rewriting k(k+1)/2 as (1+2+3+...+k):
->There are n 1's, one for each triangular number
-> n-1 2's, one for each except the first
-> n-2 3's, one for each except the first and second
etc.[/hide]

That's exactly what I meant, scorpius119.


analytic geometry
combinatorics unsolved
combinatorics

Consider a $ n \times n$ checkerboard with $ n > 1, n \in \mathbb{N}.$ How many possibilities are there to put $ 2n \minus{} 2$ identical pebbles on the checkerboard (each on a different field/place) such that no two pebbles are on the same checkerboard diagonal. Two pebbles are on the same checkerboard diagonal if the connection segment of the midpoints of the respective fields are parallel to one of the diagonals of the $ n \times n$ square.

Place the  $ n$ X $ n$ checkerboard on the coordinate plane such that the bottom left corner is on $ (1,1)$
bottom right is on $ (n,1)$ and the top corners are on $ (n,1)$ and $ (n,n)$

Consider the following $ 2n\minus{}2$ diagonals:
$ (1,1) (2,2) ... (n,n)$
$ (2,1) (1,2)$
$ (3,1) (2,2) (1,3)$
...

$ (n\minus{}1,1) (n\minus{}2,2) ... (3, n\minus{}1) (2,n)$
$ (n,1) (n\minus{}1,2) ... (2, n\minus{}1) (1,n)$
$ (n,2) (n\minus{}1,3) ... (3, n\minus{}1) (2,n)$
...
$ (n, n\minus{}2) (n\minus{}1,n\minus{}1) (n\minus{}2,n)$
$ (n,n\minus{}1) (n\minus{}1,n)$

These $ 2n\minus{}2$ diagonals span the entire board, so each diagonal must contain exactly one pebble.
But, there can be no pebble on an overlapping square.
In the $ (1,1) (2,2) ... (n,n)$ diagonal, there are two places where the pebble can go.
In the $ (2,1) (1,2)$ diagonal, there are two places where the pebble can go, and the placement of this pebble uniquely determines the placement of the pebble in the $ (n,n\minus{}1) (n\minus{}1,n)$ diagonal.
In the  $ (3,1) (2,2) (1,3)$ diagonal there are two possible placement, since the $ (2,2)$ square is on the same diagonal as the $ (1,1)$ square which already contains a pebble. Each placement uniquely determines the placement of the pebble in the $ (n, n\minus{}2) (n\minus{}1,n\minus{}1) (n\minus{}2,n)$ diagonal.

Clearly we can repeat this process for all $ n$ diagonals containing squares in the bottom row, and the placement of pebbles in the other $ (n\minus{}2)$ diagonals will be uniquely determined. So the number of ways to place the pebbles is $ 2^n$.


College of Charleston MATH MEET 2007 NC Results

Timed Math Sprints

Crossing a dime? 1st: Jenny Chen (Parrott Academy)

Sum of squares 1st: John Berman (John T. Hoggard High)

Written Test Level 1

1st Place (Out State): Jenny Chen (115) (Parrott Academy)
2nd Place (Out State): Dylan Peifer (82) (North Buncombe High)

Written Test Level 2

3rd Place (Out State): Adam Southard (93) (North Buncombe High)

Written Test Level 3
1st Place (Out State): John Berman (94) (John T. Hoggard High)
2nd Place (Out State): Drew Boyuka (67) (Charlotte Math Club)
3rd Place (Out State): Joe Lozier (65) (NC School of Science/Math)

Written Test Level 1 Team
2nd Place (Out State) Buncombe Cnty Middle Schl Math Circle
MAUHAR MARIE K, TORRENCE HANNA O, KOPELAKIS BROOKE L, PRESSLEY JOSEPH T

Written Test Level 3 Team
1st Place (Out State) NC School of Science/Math
Joe Lozier, Ray Wang, Yutong Chen, Steven Ji

2nd Place (Out State) Charlotte Math Club
Drew Boyuka, Matthew Adams, Quentin Robinson, Wisun Kim

Varsity Team Elimination
2nd NC School of Science/Math
Steven Ji, Austin Matthews, Ray Wang, Jason Zhang

Programming Competition
1st  Charlotte Math Club --Davidson, NC
2nd  Durham Academy --Durham, NC
3rd NC School of Science/Math --Durham, NC

NCSSM also won the Truths and Consequences all-day sprint. That thing was a lot of fun.

For the all-day sprint was it the first correct answer, or did they just choose one at random (because they said it had nothing to do with speed)? Or something else?


This is the question:

[img]http://i204.photobucket.com/albums/bb97/hasan4444/Capture-2.jpg[/img]

The guy who ask me this question say that he want AE as the figure (curved) not as a segment even my math teacher say that he shouldn't write like this if he want it curved like in the figure.

Note: I don't know if this is the best branch of the forum that I can put it in  :(

Well we simply need to find AE+ED+BD+AB

AE is clearly 25pi.

As BD=FC and FC=10 (its a radius), BD=10

Our sum becomes $ 25\pi\plus{}10\plus{}ED\plus{}AB$

AB=10-BC and ED=10-DC (can you see why?)

$ 25\pi\plus{}10\plus{}10\minus{}DC\plus{}10\minus{}BC\equal{}25\pi\plus{}30\minus{}(DC\plus{}BC)$

As the perimeter of the rectangle is 26, BC+DC=13.

Thus we have $ 25\pi\plus{}30\minus{}13\equal{}\boxed{25\pi\plus{}17}$

Yesterday I write this:

[img]http://i204.photobucket.com/albums/bb97/hasan4444/CircleQuestion.gif[/img]

But I'm not sure about it I think I do everything OK

$ AE\equal{}\frac{2\cdot 10\cdot \pi}{4}\equal{}5\pi$
Answer: $ 5\pi \plus{}17$

Oh haha bugi is right, i did the area/4 instead of circumference/4. the lenght of arc AB is 5pi.

Hasan you forgot AB.

LOL guys seems you forget my question?

Please enter again  :|

[quote="mathemagician1729"]I think he meant arc AE[/quote]
Yes I think he do
I think arc is like a curved segment.
I ask him do you mean AE as a segment or curved like this (arc) he say curved like this (arc)

But now what is the final answer and how?

In the question it's said: find the perimeter of AEDB. That would be a quadrilateral AEDB, except if you know that it's a curved ray.

[quote="Bugi"]You wanted the perimeter of AEDB so you need AB. You forgot it.[/quote]

Oh do you mean that when he say AEDB he means AE+ED+DB+BA

LOL I think it means AE+ED+DB only

Note for Bugi: I know he was meaning a curved ray he said it to me also

Anyway I give the answer to him already yesterday its was wrong but this mean that I learn a new thing  :)

Thanks a lot guys  :)  :lol:

[quote="hasan4444"]Yesterday I [b][u]wrote[/u][/b] this:

[img]http://i204.photobucket.com/albums/bb97/hasan4444/CircleQuestion.gif[/img]

But I'm not sure about it I think I do everything OK[/quote]

Hassan, always write the exact values, please.  :D

[hide="A Solution ... sort of"]

The perimeter of AEDB = The perimeter of the quarter-circle minus the perimeter of the triangle BCD.

The perimeter of the quarter circle = $ 2\pi r \times \frac{1}{4} \plus{} 2 r \equal{} \frac{\pi}{2} r \plus{} 2 r \equal{} \frac{\pi}{2}(10) \plus{} 2(10)$

$ \equal{} 5\pi \plus{} 20$

The tricky part is finding the perimeter of triangle BCD.

We have the perimeter of FBCD = $ 2(\ell \plus{} w) \equal{}26$

We need $ \ell \plus{} w \plus{}\sqrt{{\ell}^2 \plus{} w^2}$ = the perimeter of triangle BCD.

We are missing one necessary equation, one that tells us how $ \ell$ and $ w$ are related to one another.

Nevertheless we press on ...

$ \ell \plus{} w \equal{} 13$

$ \ell \equal{} 13 \minus{} w$

$ \ell \plus{} w \plus{}\sqrt{{\ell}^2 \plus{} w^2} \equal{} 13 \plus{}\sqrt{(13 \minus{} w)^2 \plus{} w^2}$

$ \therefore \textrm{Ans.} \equal{} 5\pi \plus{} 20 \minus{}(13 \plus{}\sqrt{(13 \minus{} w)^2 \plus{} w^2}) \equal{} 5\pi \plus{} 7 \minus{}\sqrt{(13 \minus{} w)^2 \plus{} w^2}$[/hide]

The thing is, we don't need  to find CD or CB. We need to find the length of 4 things to find the perimeter of shape AEDB.

We need AE (a quarter circle, so easily found)
DB(congruent to FC, which is a radius, so 10)
ED, and BA.

We could find CD and CB and subtrac them from 10. Or we could be tricky.

We want to find ED+AB. This is equal to (10-DC)+(10-BC), right?

10-DC+10-BC=20-(BC+CD). As the perimeter of the rectangle is 26, BC+CD=13. Thus ED+AB=7, BD=10, and AE=5pi, so 7+10+5pi=5pi+17

Mewto,

You're right ! BD is 10 ... I should have added it instead of subtracting it.  :blush:


inequalities calculus function limit logarithms

[quote="silouan"]Note that $(1-a)(1-b)\geq cd$  :wink:[/quote]
Nice solution. Your claim is easy to prove, but hard to think of  :)

[quote="silouan"]Οκ very nice .Note that
$(1-a)(1-b)\geq cd$  :wink:[/quote]

Hmm... I don't understand how this solves this inequality. Also, how is this true and how did you come up with it?  :huh:

Since $\frac{c^{2}+d^{2}}2 \geq cd$, it suffices to prove that $(1-a)(1-b) \geq \frac{c^{2}+d^{2}}2 = \frac{1-(a^{2}+b^{2})}2$
$\, \Longleftrightarrow \, \, 1-(a+b)+ab \geq \frac12-\frac{a^{2}+b^{2}}2$
$\, \Longleftrightarrow \, \, 2-2(a+b)+2ab \geq 1-(a^{2}+b^{2})$
$\, \Longleftrightarrow \, \, 1-2(a+b)+2ab+(a^{2}+b^{2}) \geq 0$
$\, \Longleftrightarrow \, \, 1-2(a+b)+(a+b)^{2}\geq 0$
$\, \Longleftrightarrow \, \, [(a+b)-1]^{2}\geq 0$, true.

Similarly, $(1-c)(1-d) \geq ab$. Now multiply  :D

I still don't know how he thought of this solution.

Very nice solution!  :)

Does anyone see a solution using Jensen's inequality?

Jensen no I think but if you  want a calculus solution there exist one .Set $a^{2}=x$ and consider the function
$f(x)=ln(\frac{1-\sqrt x}{\sqrt x})+\lambda(x-\frac{1}{4})$
and find $\lambda$ such that $f'(\frac{1}{4})=0$   :wink:

Silouan's second solution is wrong, I think.
I guessed that you wanted to find $\lambda \in \mathbb R$ such that $f(x) \geq 0$ for all $x \in \left( 0, 1 \right)$.
But
\begin{eqnarray*}\lim_{x \to 1^{-}}\left[ \ln \left( \frac{1-\sqrt x}{\sqrt x}\right)+\lambda \cdot \left( x-\frac14 \right) \right] &=& \lim_{x \to 1^{-}}\ln \left( \frac{1-\sqrt x}{\sqrt x}\right)+\lim_{x \to 1^{-}}\lambda \cdot \left( x-\frac14 \right) \\ \ &=&-\infty+\frac34 \cdot \lambda \\ \ &=&-\infty ,\end{eqnarray*}
so the inequality can't possibly hold.

Maybe you're giving too much credit to this method :maybe:
Or maybe I'm wrong :(

PS: Using Jensen's inequality is not necesarilly a calculus solution. Jan, do you have such a (Jensen) solution?

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=596705


function
combinatorics proposed
combinatorics

Ati has $ 7$ pots of flower, ordered in $ P_1,P_2,P_3,P_4,P_5,P_6,P_7$. She wants to rearrange the position of those pots to $ B_1,B_2,B_2,B_3,B_4,B_5,B_6,B_7$ such that for every positive integer $ n<7$, $ B_1,B_2,\dots,B_n$ is not the permutation of $ P_1,P_2,\dots,P_7$. In how many ways can Ati do this?

Probably the last "7" in the problem is meant to be an "$ n$" instead.  In that case, this problem asks for the number of indecomposable permutations.  These are counted by [url=http://www.research.att.com/~njas/sequences/A003319]A003319[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  One way to count these is as follows:

Every permutation can be broken up uniquely as a maximal indecomposable parts followed by some other permutation.  For example, for the permutation $ 1423756$ we have the decomposition $ 1 - 423756$, while for the permutation $ 2431756$ we have the decomposition $ 2431 - 756$ and for $ 7562431$ we have the decomposition $ 7562431 - \emptyset$.  Thus, if $ a_i$ is the number of indecomposable permutations of length $ i$, we have that $ n! = \sum_{i = 1}^n a_i \cdot (n - i)!$.  This allows us to quickly compute
\begin{align*} a_1 & = 1! = 1 \\
a_2 & = 2! - a_1 \cdot 1! = 1 \\
a_3 & = 3! - a_1 \cdot 2! - a_2 \cdot 1! = 3 \\
a_4 & = 4! - a_1 \cdot 3! - a_2 \cdot 2! - a_3 \cdot 1! = 13 \end{align*}
and so-on.

One can easily extend this analysis to get the generating-functional results mentioned in the links above.  (Note that the generating functions in question are ordinary generating functions, and so exist in a purely formal sense.)


[hide]25 ft tower

50 ft shadow at noon

length of shadow increase 10 ft/hr before 6:00

length of shadow decrease 15 ft/hr after 6:00

At 6:00 shadow inreases by (10*6) ft or 60 ft, so shadow is now 110 ft.

From 6:00 to 10:00 it is 4 hours, so shadow decreases by (15*4) ft, or 60 ft, so shadow is now 50 ft.

Shadow decreases (15/4) ft or 3.75 ft per 1/4 hour, so at 10:25, shadow is 46.25 ft.[/hide]

Just a quick question about that last problem.

[hide]I agree that the shadow length increases for 6 hrs, making the height 110 ft.  Then, the shadow length decreases for 4 hrs, shortening the height by 60 ft.  That makes the length of the shadow at 10:00, 50 ft.

But then, you said what is the length at 10:25 pm.

Every 5 minutes, or 1/12 of an hour, the shadow length is shortened by 15/12 of a foot, or 5/4 of a foot, or 1.25 feet.  After 25 minutes, or 5/12 of an hour, the shadow is shortened by 6.25 feet.  So, that makes the shadow at 10:25 pm 43.75 feet long.[/hide]

I think that's right, but I just wanted to check to see if an hour is still 60 minutes long, or if it got changed to 100 minutes.

burgie12

[quote="burgie12"]

I think that's right, but I just wanted to check to see if an hour is still 60 minutes long, or if it got changed to 100 minutes.

burgie12[/quote]

I'm not sure what do you mean by if it got changed to 100 minutes..
I'll tell whether you are right or other answer's right soon..

[tab]Duck-Billed Platypus wrote:
[quote]Shadow decreases (15/4) ft or 3.75 ft per 1/4 hour, so at 10:25[/quote]
This implies that one-quarter of an hour is 25 minutes, making one hour 100 minutes long.  Since an hour is 60 minutes long, platypus's calculations are off by a little bit.

So, I'm just checking the calculations, so some beginner doesn't get lost trying to follow the incorrect calculations.

burgie12

Silverfalcon wrote:There's a tower that has height of 25 ft and casts 50 ft shadow at noon.  If every hour before 6 o' clock, the length of shadow is increasing by 10 ft, and after 6 o' clock, the length of shadow is decreasing by 15 ft per hour, tell how hight the tower is when it's 10:25.

Enjoy this... I made this one too =)

[hide]

Actually, the question doesn't say 'How long will the shadow be?'  It says 'how hight the tower is when it's 10:25'.  I can only interpret this to be 'how high the tower is when it's 10:25'.  Since the question makes no mention of the tower changing in height, the correct answer is 25 ft.

If you take it to mean shadow, it is of course 50+(6*10)-(4+25/60)*15=43.75 feet.[/hide]

I'm pretty sure the answer should be [hide]25 ft[/hide]... No?

[quote]I'm pretty sure the answer should be 25 ft... No?[/quote]
Hehe, yeah, that was my impression.  But I'm sure it was just a typo and he/she meant the length of the shadow.


ΘΑ ΜΠΟΡΟΥΣΕ ΚΑΠΟΙΟΣ ΝΑ ΜΟΥ ΥΠΟΔΕΙΞΕΙ SITE ΓΙΑ ΤΑ ΘΕΜΑΤΑ ΤΗΣ Β ΛΥΚΕΙΟΥ Η ΚΑΙ ΓΕΝΙΚΟΤΕΡΑ ΟΛΑ ΤΑ  ΧΘΕΣΙΝΑ ΘΕΜΑΤΑ  ΕΚΤΟΣ ΑΠΟ ΤΗΣ Ε.Μ.Ε.
[url][/url]

Δες το  πρωτο ποστ http://www.mathlinks.ro/Forum/viewtopic.php?t=129363


Finally, only 5 days left, yes...

I'm looking forward to getting out of school and ... reading math books all day?!?

How does everyone feel about the summer?

I have plane curves and lie groups to look forward to  :(

I have no vacation to look forward to. :(

No vacations, AP bio notes for next year, reading AoPS books, finishing up trig, and continuing calculus!


inequalities
function
trigonometry
rearrangement inequality
33

Let $ a$, $ b$, $ c$ be positive reals. Prove that

$ \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$

For convenience, denote $a=x^2, b=y^2, c=z^2$. Squaring the inequality, multiplying by 4 and rearranging some terms yields that the inequality is equivalent to

$\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$.

But since $\sqrt{x^2+y^2},\sqrt{y^2+z^2},\sqrt{z^2+x^2}$ and $\frac{x^2y^2}{\sqrt{x^2+y^2}}$,$\frac{y^2z^2}{\sqrt{y^2+z^2}}$,$\frac{z^2x^2}{\sqrt{z^2+x^2}}$ are similarly ordered, we have by the rearrangement inequality

$\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(x^2+y^2)}}$ $= \sum \frac{8x^2y^2}{x^2+y^2}\geq \sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$,

so we're done.

Peter

[quote="Peter Scholze"]For convenience, denote $a=x^2, b=y^2, c=z^2$. Squaring the inequality, multiplying by 4 and rearranging some terms yields that the inequality is equivalent to

$\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$.
[/quote]
I think this is mistake. Peter, details please  (Squaring the inequality, multiplying by 4 and rearranging some terms yields that the inequality is equivalent to).
 Thank you.

Wow, Peter! Thank you very much! :clap2:  :omighty:

i think i have an easy solution  :roll:

we can assume that a+b+c=1 , the function x->1/√x is convexe for x positive ,we apply jensen inequality :

LHC>= 1/(√2(ab+bc+ac))
now we have to prouve:
1/(√2(ab+bc+ac))>=( √a+√b+√c)/ √2 wich equivalent to:

√(ab+bc+ac) (√a+√b+√c)=<1

But by Cauchy we have (√a+√b+√c) =<√3
And it’s easy to see that (ab+bc+ac) =<(1/3) because a²+b²+c²+2(ab+ac+bc)=1 and a²+b²+c²>=(1/3)
Then the conclusion follows!

[quote="arqady"][quote="altusi"]i think i have an easy solution  :roll:
we can assume that a+b+c=1 , the function x->1/√x is convexe for x positive ,we apply jensen inequality :
LHC>= 1/(√2(ab+bc+ac))
[/quote]
$a\cdot\frac{1}{\sqrt{a+b}}+b\cdot\frac{1}{\sqrt{b+c}}+c\cdot\frac{1}{\sqrt{c+a}}\geq\frac{1}{\sqrt{a(a+b)+b(b+c)+c(c+a)}}=$
$=\frac{1}{\sqrt{a^2+b^2+c^2+ab+ac+bc}}\leq\frac{1}{\sqrt{2(ab+ac+bc)}}.$ :)[/quote]
I think there's a problem with the ineq chain :?: THe last $\leq$ is correct, so how do you come to your conclusion?

Can we prove this using Cauchy-Shwartz

In tried and i stopped here: Prove that $\sqrt{\frac{(x+y+z)^3}{x^2+y^2+z^2+xy+yz+zx}}\ge \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$

[i](1) Cauchy-Shwarz:[/i]
$\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\right)(x\sqrt{x+y}+y\sqrt{y+z}+z\sqrt{z+x})\ge(x+y+z)^2$
[i](2) Cauchy-Shwarz:[/i]
$(x\sqrt{x+y}+y\sqrt{y+z}+z\sqrt{z+x})\le\sqrt{(x^2+y^2+z^2)((x+y)+(y+z)+(z+x))}=\sqrt{2(x^2+y^2+z^2)(x+y+z)}$
[i]From (1) and (2):[/i]
$\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\right)\ge\frac{(x+y+z)^2}{(x\sqrt{x+y}+y\sqrt{y+z}+z\sqrt{z+x})}\ge\frac{(x+y+z)^2}{\sqrt{2(x^2+y^2+z^2)(x+y+z)}}$
$=\sqrt{\frac{{(x+y+z)^3}}{{2(x^2+y^2+z^2)}}}$
[i]So it is enough to proove that[/i]
$\sqrt{\frac{{(x+y+z)^3}}{2(x^2+y^2+z^2)}}\ge\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$
[i]or equivalently we must show that[/i]
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2(x^2+y^2+z^2)\le (x^3+y^3+z^3)$
[i]But from Holder we know that[/i]
$(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}+\sqrt{z})(x^2+y^2+z^2)\ge (x^3+y^3+z^3)$
[i]And i think that the inequality is true only when x=y=z. Find my mistake[/i]

[quote="Beat"][i](1) Cauchy-Shwarz:[/i]
[color=red]$\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}\right)(x\sqrt{y+z}+y\sqrt{z+x}+z\sqrt{x+y})\ge(x+y+z)^2$
[/color][i](2) Cauchy-Shwarz:[/i]
[color=red]$(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{y+x})\le\sqrt{(x^2+y^2+z^2)((x+y)+(y+z)+(z+x))}=\sqrt{2(x^2+y^2+z^2)(x+y+z)}$
[/color][i]From (1) and (2):[/i]
[color=red]$\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{y+x}}\right)\ge\frac{(x+y+z)^2}{(x\sqrt{z+y}+y\sqrt{x+z}+z\sqrt{y+x})}\ge\frac{(x+y+z)^2}{\sqrt{2(x^2+y^2+z^2)(x+y+z)}}$
$=\sqrt{\frac{{(x+y+z)^3}}{{2(x^2+y^2+z^2)}}}$[/color]
[i]So it is enough to prove that[/i]
$\sqrt{\frac{{(x+y+z)^3}}{2(x^2+y^2+z^2)}}\ge\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$
[i]or equivalently we must show that[/i]
[color=red]$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2(x^2+y^2+z^2)\le (x+y+z)^3$
[/color][i]But from Holder we know that[/i]
[color=red]$(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}+\sqrt{z})(x^2+y^2+z^2)\ge (x+y+z)^3$[/color]
[i]And i think that the inequality is true only when x=y=z. Find my mistake[/i][/quote]
There are some mistake in your post and I've corrected it but it isn't important. Because I think Cauchy ineq  is too weak to solve this and Peter Scholze's solution is very good. So we don't need find another solution

[quote="Beat"] Prove that $\sqrt{\frac{(x+y+z)^{3}}{x^{2}+y^{2}+z^{2}+xy+yz+zx}}\ge \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$[/quote]
It's wrong, Beat. Try $x=y=1$ and $z=1.1.$ :wink:

[quote="arqady"]a>0, b>0, c>0.

then $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$[/quote]
Interesting problem but it is not a very hard problem :D We put: $x=a^{2},y=b^{2},z=c^{2}$. Squaring both side we have the inequality is equivalent to:
\[\sum_{sym}(\sqrt{\frac{x^{2}+y^{2}}{2}}-{\sqrt{\frac{2x^{2}y^{2}}{x^{2}+y^{2}})^{2}\ge 0}}\]
I remember that above inequality created in Crux Magazine :D

[quote="altusi"]we can assume that a+b+c=1[/quote]
Why can we assume that? :maybe:
What if $a+b+c=2$???
Thanks.

[quote="arqady"]a>0, b>0, c>0.

then $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$[/quote]

Since $a>0,\ b>0,\ c>0$,

$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}$

$=\frac{(\sqrt{a})^{2}}{\sqrt{a+b}}+\frac{(\sqrt{b})^{2}}{\sqrt{b+c}}+\frac{(\sqrt{c})^{2}}{\sqrt{c+a}}$

$\geq \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\geq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$

$\Longleftrightarrow \sqrt{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq \sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$.

which can be proved as follows.

Since $y=\sqrt{x}$ is an increasing and concave-up function, Jensen's theorem gives $\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\frac{a+b}{2}}$.

Similarly we have $\frac{\sqrt{b}+\sqrt{c}}{2}\geq \sqrt{\frac{b+c}{2}},\ \frac{\sqrt{c}+\sqrt{a}}{2}\geq \sqrt{\frac{c+a}{2}}$. Summing up these inequalities and multiply by $\sqrt{2}$ both sides, we obtain

$\sqrt{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq \sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$, which completes the proof.

[quote="kunny"]
Since $y=\sqrt{x}$ is an increasing and concave-up function, Jensen's theorem gives $\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\frac{a+b}{2}}$[/quote]
Jensen's theorem gives $\frac{\sqrt{a}+\sqrt{b}}{2}\leq \sqrt{\frac{a+b}{2}}.$ :wink:

Oops, I was trapped by Jensen'. :blush:
Thank you for pointing out my mistake.

kunny

???? I don't understand your post... do you have a simpler solution in mind based on this formulation? It appears like you have done nothing...

I posted just idea, but this inequality associates with cone. $\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}$.

$\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$,

so let $\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\cos^{2}\alpha ,\ \frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\cos^{2}\beta ,\ \frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\cos^{2}\gamma$, we have $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma =1$. Deirected cosine!

[quote="N.T.TUAN"][quote="FOURRIER"][quote="altusi"]we can assume that a+b+c=1[/quote]
Why can we assume that? :maybe: [/quote]
Put $x=\frac{a}{a+b+c},y=\frac{b}{a+b+c},z=\frac{c}{a+b+c}$ then $x+y+z=1$ and we need prove $(...)$.  :wink:[/quote]
Your right, Thanks :clap:
 But Can you tell me when can we apply this?
Thanks again :)

[quote="N.T.TUAN"][quote="FOURRIER"]
 But Can you tell me when can we apply this?
[/quote]
I don't know, too :P But you can see some examples use this method and you'll know answer of that question!  :wink:[/quote]

Oh I see.  :D

[quote="arqady"]Let $ a$, $ b$, $ c$ be positive reals. Prove that
$ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}}\geq\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt {2}}$[/quote]
This was proposed by Walther Janous, see:

Problem $ 1366^{*}$, [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1988, No.7, p.202.

$ \sum_{cyc}x^2\sqrt {\frac {2}{x^2 \plus{} y^2}}\geq\sum_{cyc}\frac {x^2[x^2 \plus{} y^2 \plus{} 7z^2 \plus{} 5(yz \plus{} zx \plus{} xy)]}{x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} 4(x^2y \plus{} x^2z \plus{} y^2z \plus{} y^2x \plus{} z^2x \plus{} z^2y) \minus{} 3xyz}$

$ \equal{} x \plus{} y \plus{} z$,

with equality if and only if $ x \equal{} y \equal{} z$.

[quote="silouan"][quote="Ji Chen"]This was proposed by Walther Janous, see:

Problem $ 1366^{*}$, [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1988, No.7, p.202.
[/quote]

Could you please post the Janous solution ?[/quote]
Maybe I am wrong, but I think it's the Janous's assumption only.
He didn't have a proof.

The following inequality also appeared in Crux . Prove that for a,b,c >0 and $ p\equal{}\frac{1}{3}$

$ \frac{a}{(a\plus{}b)^p}\plus{}\frac{b}{(b\plus{}c)^p}\plus{}\frac{c}{(c\plus{}a)^p}\geq\frac{a^{1\minus{}p}\plus{}b^{1\minus{}p}\plus{}c^{1\minus{}p}}{2^p}$

[quote=arqady]Let $ a$, $ b$, $ c$ be positive reals. Prove that

$$ \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$$[/quote]
Let $ a$, $ b$, $ c$ be positive reals. Prove that

$$ \frac{c+k}{\sqrt{a+b}}+\frac{a+k}{\sqrt{b+c}}+\frac{b+k}{\sqrt{c+a}}\geq 3 \sqrt{2k}$$
$$ \frac{a+k}{\sqrt{a+b}}+\frac{b+k}{\sqrt{b+c}}+\frac{c+k}{\sqrt{c+a}}\geq 3 \sqrt{2k}$$
Where $k>0.$
[hide=here]$$ \frac{a+1}{\sqrt{a+b}}+\frac{b+1}{\sqrt{b+c}}+\frac{c+1}{\sqrt{c+a}}\geq 3 \sqrt{2}$$
https://artofproblemsolving.com/community/c6h1950356p13460544[/hide]

[quote=arqady]Let $ a$, $ b$, $ c$ be positive reals. Prove that

$$ \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$$[/quote]
[size=75](Indonesia TST 2014)[/size]

Let $ a$, $ b$, $ c$ be positive reals. Prove that

$$ \frac{a^2}{\sqrt{(a^2+b^2)(a+b)}}+\frac{b^2}{\sqrt{(b^2+c^2)(b+c)}}+\frac{c^2}{\sqrt{(c^2+a^2)(c+a)}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}$$

Let $ a$, $ b$, $ c$ be positive reals. Prove that
$$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}\leq \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \sqrt{\frac{3(a+b+c)}{2}}\leq  \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}$$

Let $a,b,c$ be positive real numbers with $a+b+c=1.$ [url=https://artofproblemsolving.com/community/c6h1145950p5403768]Prove that[/url]$$\sqrt{\frac{a}{6a+1}}+\sqrt{\frac{b}{6b+1}}+\sqrt{\frac{c}{6c+1}}\leq 1$$
Let $a,b,c$ are positive real numbers. Prove that
$$ a\sqrt{\frac{b}{7a+b+c}}+b\sqrt{\frac{c}{7b+c+a}}+c\sqrt{\frac{a}{7c+a+b}} \le \frac{a+b+c}{3}$$
$$ \frac{a}{a+b}\sqrt{\frac{b}{7a+b+c}}+ \frac{b}{b+c}\sqrt{\frac{c}{7b+c+a}}+ \frac{c}{c+a}\sqrt{\frac{a}{7c+a+b}} \le \frac{1}{2} $$
$$ \frac{a+b}{a+c}\sqrt{\frac{b}{7a+b+c}}+ \frac{b+c}{b+a}\sqrt{\frac{c}{7b+c+a}}+ \frac{c+a}{c+b}\sqrt{\frac{a}{7c+a+b}} \ge 1$$
Let $ a,b ,c$ be non-negative reals. [url=https://math.stackexchange.com/questions/3280839/find-the-minimum-k-such-that-sum-cyc-fraca-sqrtab-leq-k-sqrtab?noredirect=1]Prove that[/url]
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{5}{4}\sqrt{a+b+c}$$
Let $ a,b ,c,d$ be positive reals. [url=https://math.stackexchange.com/questions/2049650/prove-that-sum-limits-cyc-fraca-sqrtab-leq-frac-sqrt334-sqrta]Prove that[/url]
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+d}}+\frac{d}{\sqrt{d+a}}\leq\frac{\sqrt{33}}{4}\sqrt{a+b+c+d}$$

Given $ a, b, c > 0$ and $ a+b+c=3$. [url=https://artofproblemsolving.com/community/c6h299971p1624568]Prove that:[/url] $$\frac{a}{\sqrt{7b^2+b+c}}+\frac{b}{\sqrt{7c^2+c+a}}+\frac{c}{\sqrt{7a^2+a+b}} \ge 1$$
[url=https://artofproblemsolving.com/community/c6h1145950p5403768]h[/url]

[quote=jokehim][b]AM-GM[/b]  Let $a,b,c>0: ab+bc+ca=3$. Prove that: $$\frac{a}{bc\sqrt{a^2+b^2}}+\frac{b}{ca\sqrt{b^2+c^2}}+\frac{c}{ab\sqrt{c^2+a^2}}\ge\frac{a+b+c}{\sqrt{2}}$$[/quote]
Because $abc\leq1$.

[quote=Peter Scholze][hide=Solution]For convenience, denote $a=x^2, b=y^2, c=z^2$. Squaring the inequality, multiplying by 4 and rearranging some terms yields that the inequality is equivalent to

$\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$.

But since $\sqrt{x^2+y^2},\sqrt{y^2+z^2},\sqrt{z^2+x^2}$ and $\frac{x^2y^2}{\sqrt{x^2+y^2}}$,$\frac{y^2z^2}{\sqrt{y^2+z^2}}$,$\frac{z^2x^2}{\sqrt{z^2+x^2}}$ are similarly ordered, we have by the rearrangement inequality

$\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(x^2+y^2)}}$ $= \sum \frac{8x^2y^2}{x^2+y^2}\geq \sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$,

so we're done.
------------------------

Sorry for giving such a short proof - I did not have the time to post the term manipulation and thus only gave the main argument. But it is quite clear that squaring and multiplying by 4 yields

$4\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)^2 \geq 2(x+y+z)^2$

$\leftrightarrow \sum \frac{4x^4}{x^2+y^2} + \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq 2\sum x^2 + 4\sum xy$

$\leftrightarrow \sum \frac{2x^4+2y^4}{x^2+y^2} + \sum \frac{2x^4-2y^4}{x^2+y^2} +$ $\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq 2\sum x^2 + 4\sum xy$

$\leftrightarrow \sum \frac{2x^4+2y^4}{x^2+y^2} + 2\sum (x^2-y^2) +$ $\sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq \sum (x^2+y^2+4xy)$

$\leftrightarrow \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq$ $\sum \frac{(x^2+y^2+4xy)(x^2+y^2) - 2x^4 - 2y^4}{x^2+y^2}$

$\leftrightarrow \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq$ $\sum \frac{-x^4+4x^3y+2x^2y^2+4xy^3-y^4}{x^2+y^2}$

$\leftrightarrow \sum \frac{8x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\geq$ $\sum \frac{8x^2y^2-(x-y)^4}{x^2+y^2}$.

Peter[/hide][/quote]

Very nice !!


function
functional equation
algebra
IMO
IMO 1994
increasing functions
david monk

Let $ S$ be the set of all real numbers strictly greater than −1. Find all functions $ f: S \to S$ satisfying the two conditions:

(a) $ f(x \plus{} f(y) \plus{} xf(y)) \equal{} y \plus{} f(x) \plus{} yf(x)$ for all $ x, y$ in $ S$;

(b) $ \frac {f(x)}{x}$ is strictly increasing on each of the two intervals $ \minus{} 1 < x < 0$ and $ 0 < x$.

A solution from Kalva:

Suppose f(a) = a. Then putting x = y = a in the relation given, we get f(b) = b, where b = 2a + a2. If -1 < a < 0, then -1 < b < a. But f(a)/a = f(b)/b. Contradiction. Similarly, if a > 0, then b > a, but f(a)/a = f(b)/b. Contradiction. So we must have a = 0.

But putting x = y in the relation given we get f(k) = k for k = x + f(x) + xf(x). Hence for any x we have x + f(x) + xf(x) = 0 and hence f(x) = -x/(x+1).

Finally, it is straightforward to check that f(x) = -x(x+1) satisfies the two conditions.

Thanks to Gerhard Woeginger for pointing out the error in the original solution and supplying this solution.

Or rather (a typo in the last line of Arne's solution)
\[f(x)=\frac{-x}{x+1}.\]

Another (trace of) solution:
- find f(0)=0
- now get $ f(f(x)) \equal{} x$
- substitute f(y) for y and get $ f(x \plus{} y \plus{} xy) \equal{} f(x) \plus{} f(y) \plus{} f(x)f(y)$
- define $ g(x) \equal{} f(x \minus{} 1) \plus{} 1$ (we conjugate f) and get the equivalent problem:

- g is from the positive reals to themselves
- $ g(ab) \equal{} g(a)g(b)$
- $ g(g(x)) \equal{} x$
- and $ \frac {g(x) \minus{} g(1)}{x \minus{} 1}$ is increasing

these conditions are no doubt very nice  :)
And from now on the problem is standard... for example, you can fix $ f(2)$ and then extend uniquely f to a dense set of points, use the last condition to get continuity, and use $ g^2\equal{}\mbox{id}$ to get $ g(x) \equal{} 1/x$ for all x.

$\frac{f(x)}{x}$ is strictly increasing on $(-1,0)$ and $(0,\infty) \Rightarrow f(x)=x$ can have atmost three solutions(atmost one in the interval $(-1,0)$,another probably in the interval $(0,\infty)$ and the third one probably $0$).Let $k$ be a fixed point of $f$ in the interval $(-1,0)$.Then $x=y=k \Rightarrow f(k^2+2k)=k^2+2k \Rightarrow k^2+2k$ is also a fixed point of $f$.Moreover $k \in (-1,0) \Rightarrow -1 \le k(k+2) <0$ forcing $2k+k^2=k \Rightarrow k=0.-1$.Contradiction!!Similarly it may be seen that there are no fixed points of $f$ in the interval $(0,\infty)$.But substituting $y=x$ in the equation we get $f(x+f(x)+xf(x))=x+f(x)+xf(x)$ and hence we are forced to conclude that $x+f(x)+xf(x)=0 \Rightarrow f(x)=-\frac{x}{x+1}$It is easy to see that this indeed satisfies the relation.

IF $ f(x + f(y) + xf(y)) \geq  y + f(x) + yf(x)$，and then how to solve it?

The function $f(x)$ can have at most 3 fixed points, one in the interval $(-1,0)$, one at $0$, and one in the interval $(0,\infty)$. Now plug in $y=x$ into the FE. We get
\[ f(x+f(x)+xf(x)) = x+f(x)+xf(x).\]This means that $x+f(x)+xf(x)$ is also a fixed point. Therefore if $x$ is a fixed point, so is $2x+x^2$. Note that $2x+x^2$ is always positive for positive $x$, so we must have $x \in (-1,0)$ and $2x+x^2 \in (0,\infty)$. However, when $x \in (-1,0)$, $2x+x^2<0$. This means that we must have a fixed point at $0$. Therefore, since $x+f(x)+xf(x)$ is a fixed point, we must have
\[ x+f(x)+xf(x) = 0\implies f(x) = -\frac{x}{x+1}.\]It is easy to check that this function satisfies the conditions.

EDIT: 1000th post!

Here is my solution for this problem
[b]Solution[/b]
$f(x + f(y) + xf(y)) = y + f(x) + yf(x)$
Let $y = x$, we have: $f(x + f(x) + xf(x)) = x + f(x) + xf(x)$
So: $x + f(x) + xf(x)$ is fixed point of $f$
Suppose $u \in S$ is fixed point of $f$, which means $f(u) = u$
Let $x = y = u$, we have: $f(u^2 + 2u) = u^2 + 2u$
If $u \ne u^2 + 2u$ then $1 = \dfrac{f(u)}{u} \ne \dfrac{f(u^2 + 2u)}{u^2 + 2u} = 1$
So: $u = u^2 + 2u$ or $u = 0, u = - 1$
It means that there is no fixed point in $(- 1; 0) \cup (0; + \infty)$
Let $x = y = 0$, we have: $f(f(0)) = f(0)$
Then: $f(0)$ is fixed point of $f$
So: $f(0) = 0$ or $x + f(x) + xf(x) = 0$
Hence: $f(x) = - \dfrac{x}{x + 1}, \forall x \in S$
Retry, we see that: $f(x) = - \dfrac{x}{x + 1}$ satisfies the problem
In conclusion, we have: $f(x) = - \dfrac{x}{x + 1}, \forall x \in S$

$P(0,y) \implies f(f(y))=y+f(0)+yf(0)$ thus $f$ is injective put$ y=0 $then$ ff(0))=f(0) \implies f(0)=0$
then $ff(x))=x \implies \frac{ff(x))}{f(x)}=\frac{x}{f(x)}$
if $x>0$ and $f(x)>x$ then $\frac{ff(x))}{f(x)}=\frac{x}{f(x)}>\frac{f(x)}{x} $then $f(x)<x$
contradiction
thus if $x>0$ then $f(x)<0$ and vice versa
$P(x,x) \implies f(x+f(x)+xf(x))=x+f(x)+xf(x) \implies x+f(x)+xf(x)=0 $
then
$f(x)=\frac{-x}{x+1}  $     $\forall x\in S$

Let $P(x,y)$ be the assertion.
$P(x,x)\implies f(x+(1+x)f(x))=x+(1+x)f(x). ~~~~~(*)$
$(b)\implies f(x)=x$ has at most one solution in the intervals $(-1,0)$ and $(0,+\infty).$
Let $f(z)=z$ for some $z\in (-1,0).$ Then $(*)\implies f(z^2+2z)=z^2+2z \implies z^2+2z=z$ since $z^2+2z=(z+1)^2-1\in (-1,0).$ Hence, $a=-1$ or $a=0,$ impossible. One can get a similar contradiction for $f(x)=x$ for $x \in (0,+\infty).$
It follows from $(*)$ that $x+(1+x)f(x)=0 \implies f(x)=\frac{-x}{1+x} ~~\forall x\in S,$ which fits. $\square$

$P(0,0)\implies f(f(0))=f(0)$ so $f$ has a fixed point.

Let $w$ be any fixed point of $f$.

$P(w,w)\implies f(w^2+2w)=w^2+2w$. Thus if $w$ is a fixed point then $w^2+2w$ is also a fixed point.

Consider if $w\neq 0$, we divide the value of $w$ into two cases.
[list]
[*] If $-1<w<0$ then $-1<w^2+2w<w<0$ which is absurd.
[*] If $w>0$ then $w^2+2w>w>0$ which is absurd.
[/list]
Hence, $w$ must be $0$.

$P(x,x)\implies f(x+(x+1)f(x))=x+(x+1)f(x)$.
$$x+(x+1)f(x)=0\implies \boxed{f(x)=-\frac{x}{x+1} \ \forall x\in\mathbb{R}_{>-1}}$$
which clearly works.

Let $P(x,y)$ denote the given assertion.

$P(0,0): f(f(0))=f(0)$.

$P(x,0): f(x+f(0)+xf(0))=f(x)$.

$P(x,f(0)): f(x+f(0)+xf(0))=f(0)+f(x)+f(0)f(x)$.

So $f(0)+f(0)f(x)=f(0)(f(x)+1)=0$ for any $x$.

Since $f(x)>-1$, we have $f(0)=0$.

$P(0,x): f(f(x))=x$.

Suppose $k$ is a fixed point of $f$.

$P(k,k): f(k^2+2k)=k^2+2k$.

Note that $\frac{f(k)}{k}=1$ and $\frac{f(k^2+2k)}{k^2+2k}=1$.

So by condition b, $k$ and $k^2+2k$ can't be both positive or both negative.

However, if $k$ is negative, then since $k>-2$, $k^2+2k = k(k+2)$ is negative also.

If $k$ is positive then $k^2+2k$ is also positive.

If $k=0$ then $k^2+2k=0$.

Thus, either $k$ and $k^2+2k$ are both positive or both negative, or $k=0$, which implies that $0$ is the only fixed point of $f$.

$P(x,x): f(x+f(x)+xf(x))=x+f(x)+xf(x)$.

So for each $x$, $x+f(x)+xf(x)$ is a fixed point, so \[x+f(x)+xf(x)=0\implies f(x)(x+1)=-x\implies \boxed{f(x)=-\frac{x}{x+1}},\]which works.

[quote=Gerhard Woeginger]
Suppose $f(a) = a.$ Then putting $x = y = a$ in the relation given, we get $f(b) = b$ where $b = 2a + a^2.$ If $ -1 < a < 0,$ then $ -1 < b < a.$ But $f(a)/a = f(b)/b.$ Contradiction. Similarly, if $a > 0,$ then $ b > a,$ but $f(a)/a = f(b)/b.$ Contradiction. So we must have $a = 0.$

But putting $x = y$ in the relation given we get $ f(k) = k$ for $k = x + f(x) + xf(x).$ Hence for any $x$ we have $x + f(x) + xf(x) = 0$ and hence $f(x) = -\tfrac{x}{x+1}.$ Finally, it is straightforward to check that $f(x) = -\tfrac{x}{x+1}$ satisfies the two conditions.[/quote]

Nice solution. Mr. Woeginger orz

EDIT: Did latex.


geometry
rectangle
calculus
derivative
calculus computations

Calculate using Lagrange Multipliers

Find the absolute minimum and absolute maximum of
$ f (x, y) \equal{} x^2 \plus{} 4y^2 \minus{} 2x^2 y \plus{} 4$ on the rectangle given by $ \minus{}1 \le x \le 1$ and $ \minus{}1 \le y \le 1$.

hello, you don't need the method of Lagrange multipliers in this case here,you must build the partial derivative
with respect to $ x$ and $ y$, then you have to solve the following equation system
$ 2x\minus{}4xy\equal{}0$
$ 8y\minus{}2x^2\equal{}0$
solving this we get the point $ (x,y)\equal{}(0,0)$, the other solutions lie not in the given rectangle. By the second derivative test we get that the point $ (0,0)$ gives the local minimum $ f(0,0)\equal{}4$. The global maximum can be attained on the sides or on the corner of the given rectangle, i have found the point $ (\minus{}1,\minus{}1)$ , it is $ f(\minus{}1,\minus{}1)\equal{}11$.
Sonnhard.

But why cant we apply method of Lagrange Multipliers/
Is this because gradient of $ g(x,y)$ cannot be found?