Find all function $f: \mathbb R^{+} \to \mathbb R^{+}$ such that for every real  $ x>0$ and $ z \in (0,1)$, we have \[(1- z)  f(x)= f\left( \frac{(1- z)f(xz)}{z}\right).\]

[quote="hoangclub"]Find all function satisfying: $ f$:$ R^ \plus{}$-->$ R^ \plus{}$
           with every $ x$>0 and $ z$ $ \in$ (0;1) then (1-$ z$) $ f(x)$=$ f$($ f(xz)$(1-$ z$)/$ z$).[/quote]

Let $ f(x)\equal{}xg(x)$. The equation becomes $ (1\minus{}z)xg(x)\equal{}xzg(xz)\frac{1\minus{}z}{z}g(xzg(xz)\frac{1\minus{}z}{z})$ and so $ g(x)\equal{}g(xz)g(g(xz)(x\minus{}xz))$
Let $ X\equal{}x(1\minus{}z)$ and $ Y\equal{}xz$. The equation becomes $ g(X\plus{}Y)\equal{}g(Y)g(Xg(Y))$

So we have to find all functions $ g(x)$ : $ \mathbb R^\plus{}\to\mathbb R^\plus{}$ such that assertion $ P(x,y)$ : $ g(x\plus{}y)\equal{}g(y)g(xg(y))$ is true $ \forall x>0,y>0$

If $ \exists a$ such that $ g(a)>1$, then $ P(\frac{a}{g(a)\minus{}1},a)$ $ \implies$ $ g(\frac{ag(a)}{g(a)\minus{}1})\equal{}g(a)g(\frac{ag(a)}{g(a)\minus{}1})$ and so $ g(a)\equal{}1$

So $ g(x)\leq 1$ $ \forall x$ and so $ g(x\plus{}y)\leq g(y)$ and so $ g(x)$ is non decreasing.

If $ g(a)\equal{}g(b)$ for some $ a>b$ then $ P(a\minus{}b,b)$ $ \implies$ $ g(a)\equal{}g(b)g((a\minus{}b)g(b))$ and so $ \exists u$ such that $ g(u)\equal{}1$.
Then $ P(x,u)$ $ \implies$ $ g(x\plus{}u)\equal{}g(x)$ and so $ g(x)$ is constant (since non decreasing) and so $ g(x)\equal{}1$ $ \forall x$

If $ g(x)$ is not the constant function, then it's injective.

Then $ P(\frac{x}{g(y)},y)$ $ \implies$ $ g(\frac{x}{g(y)}\plus{}y)\equal{}g(x)g(y)$ and $ P(\frac{y}{g(x)},x)$ $ \implies$ $ g(\frac{y}{g(x)}\plus{}x)\equal{}g(y)g(x)$

So $ g(\frac{x}{g(y)}\plus{}y)\equal{}g(\frac{y}{g(x)}\plus{}x)$ and, since $ g(x)$ is injective : $ \frac{x}{g(y)}\plus{}y\equal{}\frac{y}{g(x)}\plus{}x$

And so $ \frac{x}{\frac{1}{g(x)}\minus{}1}$ $ \equal{}\frac{y}{\frac{1}{g(y)}\minus{}1}$ and so $ \exists c$ such that $ \frac{x}{\frac{1}{g(x)}\minus{}1}\equal{}\frac{1}{c}$

And so $ g(x)\equal{}\frac{1}{1\plus{}cx}$

And so the solutions of the initial equation :
$ f(x)\equal{}x$

$ f(x)\equal{}\frac{x}{1\plus{}cx}$

And it's easy to check back that these two solutions fit the initial requirements.


Find a power series, centered at 2, for the function:
$ f(x) \equal{} \frac{1}{3x\minus{}2}$

[hide="Solution"]
Translate left by $ 2$ to get $ \frac{1}{3x\plus{}4}\equal{}\frac 14\cdot\frac{1}{1\minus{}(\minus{}\frac{3x}4)}\equal{}\frac 14\sum_{k\equal{}0}^{\infty}\left(\minus{}\frac{3x}4\right)^k$

Now translate it back to the right:

$ \boxed{\frac 14\sum_{k\equal{}0}^{\infty}\left(\minus{}\frac{3x\minus{}6}{4}\right)^k}$
[/hide]


[quote="flavian"]can somebody give me a very short solution to this inequality?

If x,y,z are positive numbers  (x+y+z)^4   >=    27(  (xy)^2+(xz)^2+(yz)^2   ).[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=87328 :)

[quote="flavian"]can somebody give me a very short solution to this inequality?

If x,y,z are positive numbers  $(x+y+z)^4 \geq 27( (xy)^2+(xz)^2+(yz)^2$   ).[/quote]


[quote="Hong Quy"]A,B,C phạm tội phản bội tổ quốc bị nhà vua tuyên án tử hình nhưng cả 3 đều là người cực kì tài giỏi, thông minh nên giết đi nhà vua lại thấy tiệc Nhà vua ra 1 tình huống nếu bất kì người nào trả lời đúng( chỉ 1 người đại diện trả lời và chỉ trả lời 1 lần ) trong thời hạn 15' thì sẽ tha tội chết cho cả 3.
 Tihf huống như sau :
 1 tên lính cầm trong túi bên trong có 4 cái nón( hai xanh 2 đỏ ) và đội lên đầu mỗi người 1 cái nón( giả định tên lính cố tình đội A : nón đỏ ; B : nón xanh ; C : nón đỏ ) Biết rằng mỗi người đều không biết trên đầu mình đội mux gì , B thây được nón của A ; C thấy được nón của A và B.
 Hỏi ai là người có thể đoán đúng nhất cái nón còn lại trong túi của tên lính và thời gian trả lời là phút thứ mấy ?[/quote]

Đầu tiên là suy nghĩ của C : Nếu anh ta thấy 2 cái mũ đỏ(hoặc 2 cái mũ xanh) thì anh ta có thể đoán ngay màu mũ của mình. Nhưng do anh ta không nói gì nên sau khoảng 5->10 phút, anh B sẽ biết được màu mũ của A và B sẽ là 1 xanh 1 đỏ. B nhìn thấy màu mũ của A nên anh ta nói ngay được màu mũ của mình. :D

minhf hoi ai doan duoc ten linh cam mu j ma` ? chu khong hoi B co the doasn minh doi mu j ?

Dù C đội nón xanh hay đỏ thì C cũng không thể đoán được vì hai người kia đội hai nón khác màu. Sau khi B nhìn thấy A nón đỏ, B sẽ biết mình đội nón xanh. Nhưng lại lhông biết C đội nón gì, nên cũng không thể đoán được. A lại càng không vì chỉ biét mình và B đội nón khác màu. KHÔNG AI CÓ THỂ ĐOÁN ĐƯỢC. vua ác ghê!!!

À không. Sau khoảng 12 phút sẽ có một người biết là không thể đoán được và đoán bừa một màu bất kì. Màu đó có thể đúng hoặc sai và nếu may mắn họ sẽ thoát chết. :D


[quote="ch1n353ch3s54a1l"][quote="ragnarok23"]James bond is trying to sabotage a missile that is made up of a right cylinder with a hemisphere on top of it. Due to cameras, security guards and whatnot 007 must revolve around the cylinder once in the shortest length possible such that when he reaches the top of the cylinder he is above the same place he started. Then he must climb to the top of the hemisphere to the exact center and sabotage the missile. Then he goes down the same way he came back up. If the cylinder has radius $7\pi$ units and height $2\sqrt{5}$ units, and the hemisphere has the same radius as the cylinder, then how long is the total distance up and down the missile silo? Express your answer in terms of $\pi$.

I hope my wording isn't too bad....no pic, sorry, I don't have a scanner[/quote]

^^
[hide]The circumference of the base of the cylinder is $14\pi^2$. Expand the cylinder to get a length of the diagonal of the rectangle with sides $14\pi^2$ and $2\sqrt{5}$ which is ...$2\sqrt{49\pi^4+5}$. The hemisphere has radius $7\pi$ so then he travels $1/4$ the circumference, or $\frac{7\pi^2}{2}$ and add them up and multiply by two to get...$\boxed{4\sqrt{49\pi^4+5}+7\pi^2}$...unusual number[/hide][/quote]
Correct! :D
You know, we should take all of these problems made by us and make a mock MathCounts test. :)


limit
logarithms
trigonometry
calculus
calculus computations

Bring these indeterminated forms of limits to other known forms and solve them. (For example $\mathop{\lim }_{x \to 0}\frac{{\ln (1+x)}}{x}= 1$.)
$1)$ $\mathop{\lim }_{x \to 3}\frac{{2x}}{{x-3}}\ln (\frac{x}{3})$

$2)$ $\mathop{\lim }_{x \to e}\frac{{x-e}}{{\ln x-\ln e}}$

$3)$ $\mathop{\lim }_{x \to 0^{+}}[\ln \sin^{3}x-\ln (x^{4}+ex^{3})]$

math_music, your solution is correct but the final answer should be $\ln e^{-1}=-1$ and not $e^{-1}.$ You forgot something at the last step of your solution.

yes, i was typing when saw your message. I will pay more atention next time. Thanks FieryHydra :oops:

[hide="2) solution"]
$\lim_{x \to \text{e}}\frac{x-\text{e}}{\ln x-\ln \text{e}}$

Let $t=x-\text{e}$, when $x\to \text{e}\, \Rightarrow \, t\to 0$

$\lim_{t \to 0}\frac{t}{\ln{(t+\text{e})}}=\lim_{t \to 0}\frac{\ln \text{e}^{t}}{\ln{(t+\text{e})}}=\lim_{t \to 0}\log_{t+\text{e}}\text{e}^{t}=\lim_{t \to 0}t\log_{t+\text{e}}\text{e}=0$
[/hide]
correct me please

sorry, forgot $-\ln \text{e}$ in the last post. Using L'hospital rule i got
${\lim_{x \to \text{e}}\frac{x-\text{e}}{\ln x-\ln \text{e}}}=\lim_{x \to \text{e}}\frac{1}{\frac{1}{x}}=\text{e}$
ps: i do  have a problem with atention...


You could also use
81^0.25 = 81^1/4 = fourth root of 81 = 3
(3^4 = 81)


Let $x=\frac{\pi}{2}-\theta$, we have $I=J$ and $I+J=\frac{\pi}{2}$, yielding $I=J=\frac{\pi}{4}$.


geometry
rhombus
Gauss
cyclic quadrilateral
geometry proposed
U sviti matematyky

In convex quadrilateral $ ABCD$, let $ E\equal{}AB\cap DC, F\equal{}BC\cap AD$. The angle bisectors of $ \angle AED,\angle BFA$ intersect at $ K$. Given that $ \angle EKF\equal{}90^{\circ}$, prove that

$ [AKB]\plus{}[CKD]\equal{}[BKC]\plus{}[AKD]$, where

$ [...]$ denotes the area of triangle $ ...$

Hello!

The bisector of the angle $ \angle BEC$ intersect the lines $ BC$, $ DA$ at the points $ N$, $ Q$ respectively.
The bisector of the angle $ \angle BFA$ intersect the lines $ AB$, $ CD$ at the points $ M$, $ P$ respectively.
Point $ X$ is the midpoint of $ AC$ and point $ Y$ is the midpoint of $ BD$.
I prove your problem in few steps:
1.The quadrilateral $ ABCD$ is cyclic.
2.The quadrilateral $ MNPQ$ is a rhombus and the point $ K$ is it's center.
3.$ K\in XY$
4.$ \left[AKD\right] \plus{} \left[BKC\right] \equal{} \left[AKB\right] \plus{} \left[CKD\right] \equal{} \frac {\left[ABCD\right]}{2}$.

1.$ \angle EKF \equal{} 180^{\circ} \minus{} \angle KEF \minus{} \angle KFE \equal{} 180^{\circ} \minus{} \angle KEP \minus{} \left(\angle PEF \plus{} \angle QFE\right) \minus{}$
$ \minus{}\angle KFQ \equal{} 180^{\circ} \minus{} \frac {180^{\circ} \minus{} \angle B \minus{} \angle C}{2} \minus{} \left(180^{\circ} \minus{} \angle D\right) \minus{} \frac {180^{\circ} \minus{} \angle A \minus{} \angle B}{2} \equal{}$
$ \equal{} ... \equal{} \frac {\angle B \plus{} \angle D}{2}\Rightarrow \angle EKF \equal{} \frac {\angle B \plus{} \angle D}{2}$. Since $ \angle EKF \equal{} 90^{\circ}\Rightarrow ABCD$ is a cyclic quadrilateral.

2.$ \frac {AQ}{QD} \equal{} \frac {AE}{ED} \equal{} \frac {sin\left(180^{\circ} \minus{} \angle D\right)}{sin\left(180^{\circ} \minus{} \angle A\right)} \equal{} \frac {sin\angle B}{sin\angle C}$. Similarly $ \frac {AM}{MB} \equal{} \frac {sin\angle B}{sin\angle C}$, so $ \frac {AQ}{QD} \equal{} \frac {AM}{MB}\Rightarrow MQ\parallel{}BD$.     (1)
Similarly $ NP\parallel{}BD$.     (2)
(1),(2)$ \Rightarrow MQ\parallel{}NP$.     (3)
Similarly $ MN\parallel{}PQ$.     (4)
(3), (4), $ MP\bot NQ$, $ \left\{K\right\} \equal{} MP\cap NQ\Rightarrow MNPQ$ is a rhombus and $ K$ is it's center.

3.$ \left\{V\right\} \equal{} BX\cap MN$ and $ \left\{W\right\} \equal{} DX\cap PQ$
$ MN\parallel{}AC$ and $ X$ is the midpoint of $ AC$ $ \Rightarrow V$ is the midpoint of $ MN$     (5)
$ PQ\parallel{}AC$ and $ X$ is the midpoint of $ AC$ $ \Rightarrow W$ is the midpoint of $ PQ$     (6)
(5), (6), $ MNPQ$-rhombus $ \Rightarrow K$ is the midpoint of $ VW$. But $ VW\parallel{}MQ\parallel{}BD$, so $ K\in XY$.
The point $ K$ lies on the Newton-Gauss line.

4.Construct $ AA'\bot XY$ and $ CC'\bot XY$, $ A',C'\in XY$. Since $ AX \equal{} XC$ $ \Rightarrow AA' \equal{} CC'\Rightarrow \left[AYK\right] \equal{} \left[CYK\right]$     (7)
$ KY$ is a median of triangle $ KBD$ $ \Rightarrow \left[DYK\right] \equal{} \left[BYK\right]$     (8)
(7), (8) $ \Rightarrow \left[AYK\right] \plus{} \left[DYK\right] \equal{} \left[BYK\right] \plus{} \left[CYK\right]\Rightarrow$
$ \Rightarrow \left[AYD\right] \minus{} \left[AKD\right] \equal{} \left[BKC\right] \minus{} \left[BYC\right]\Rightarrow$
$ \Rightarrow \left[AKD\right] \plus{} \left[BKC\right] \equal{} \left[AYD\right] \plus{} \left[BYC\right] \equal{} \frac {\left[ABCD\right]}{2}\Rightarrow$
$ \Rightarrow \left[AKD\right] \plus{} \left[BKC\right] \equal{} \frac {\left[ABCD\right]}{2}\Rightarrow \left[AKB\right] \plus{} \left[CKD\right] \equal{} \frac {\left[ABCD\right]}{2}$.
So $ \left[AKD\right] \plus{} \left[BKC\right] \equal{} \left[AKB\right] \plus{} \left[CKD\right] \equal{} \frac {\left[ABCD\right]}{2}$.  :)

This problem is coming from the past.

The angle bisectors of $ \angle E,\ \angle F,$ are perpendicular each other and their intersection point lies on the line connecting the midpoints $ M,\ N,$ of the diagonals $ AC,\ BD$ of the given cyclic quadrilateral $ ABCD.$

Every point $ K,$ lies on the line segment $ MN,$ has the property as the problem states ([b][size=100]Leon Anne[/size][/b]).

Kostas Vittas.


[quote="filomen"]If

$ a,b,c > 0$

then

$ \sum\frac {a^2}{b \plus{} c}\geq\frac {3}{4}(\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \sqrt [4]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}})$[/quote]

We have : $ \sum\frac {a^2}{b \plus{} c} \geq \frac{3}{2} \sqrt [4]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}$

And : $ \sum\frac {a^2}{b \plus{} c} \geq \frac{3}{2} (\frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2})$

[quote="Materazzi"][quote="filomen"]If

$ a,b,c > 0$

then

$ \sum\frac {a^2}{b \plus{} c}\geq\frac {3}{4}(\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \sqrt [4]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}})$[/quote]

We have : $ \sum\frac {a^2}{b \plus{} c} \geq \frac {3}{2} \sqrt [4]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}$

And : $ \sum\frac {a^2}{b \plus{} c} \geq \frac {3}{2} (\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2})$[/quote]

Dear Materazzi, can you post full proff? thank you very much  :(


For any polynomial $ p \in \mathbb R[x]$, prove that there exist polynomials $ p_1(x),\ldots,p_{100}(x)$ such that
\[ p(x)\equal{}(p_1(x))^3\plus{}\cdots\plus{}(p_{100}(x))^3.\]

[quote="MJ GEO"]for all $ p(x)$ prove that there are polynmial $ p1(x),...,p100(x)$ that $ p(x) \equal{} (p1(x))^3 \plus{} ... \plus{} (p100(x))^3$[/quote]

$ \left(\frac{x}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{1\minus{}x}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{x\minus{}3}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{2\minus{}x}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{\sqrt[3]{18}}{\sqrt[3]{12}}\right)^3$ $ \equal{}x$

And so :

$ P(x)\equal{}\left(\frac{P(x)}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{1\minus{}P(x)}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{P(x)\minus{}3}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{2\minus{}P(x)}{\sqrt[3]{12}}\right)^3$ $ \plus{}\left(\frac{\sqrt[3]{18}}{\sqrt[3]{12}}\right)^3$

And just take the $ 95$ other polynomials as $ 0$

[quote="MJ GEO"]how did you fid that strange solution.[/quote]

I tried to eliminate $ x^3$ between $ (ax\plus{}b)^3\plus{}(cx\plus{}d)^3$ and so $ c\equal{}\minus{}a$. I took then :

$ (x\plus{}c)^3\plus{}(d\minus{}x)^3\equal{}3x^2(c\plus{}d)\plus{}3x(c^2\minus{}d^2)\plus{}c^3\plus{}d^3$

Hence a first sum : $ x^3\plus{}(1\minus{}x)^3\equal{}3x^2\minus{}3x\plus{}1$

Then, I need to find another couple to eliminate $ 3x^2$, so such that $ c\plus{}d\equal{}\minus{}1$

Hence a second sum : $ (x\minus{}3)^3\plus{}(2\minus{}x)^3\equal{}\minus{}3x^2\plus{}15x\minus{}19$

Summing : $ x^3\plus{}(1\minus{}x)^3\plus{}(x\minus{}3)^3\plus{}(2\minus{}x)^3\equal{}12x\minus{}18$

And so $ x^3\plus{}(1\minus{}x)^3\plus{}(x\minus{}3)^3\plus{}(2\minus{}x)^3\plus{}18\equal{}12x$

Hence the result


conics
parabola
function
analytic geometry
calculus
calculus computations

Express the distance between the point (3, 0) and the point P (x, y) of the parabola y = x2 as a function of x.

$P(x,y)=(x,x^{2})$
let $D(x)$ be the distance from the point $(3,0)$ so the point w/ the desired x coordinate

$D(x)=\sqrt{(x-3)^{2}+(x^{2}-0)^{2}}=\sqrt{(x-3)^{2}+x^{4}}=\sqrt{x^{4}+x^{2}-6x+9}$


[color=darkblue]For $a>0$ and $m\in N^{*}$ ascertain the range of $\left(a+x^{m}\right)\left(a+y^{m}\right)\ ,$ where $x>0\ ,\ y>0\ ,\ x+y=1\ .$

[b]Remark.[/b] For $m\in \overline{2,3}$ see http://www.mathlinks.ro/Forum/viewtopic.php?t=115900[/color]

[quote="now a ranger"]what is the symbol that looks like an e?[/quote]
It means "in". In other words $m$ is a natural number.

"$a\in A$" means "the element $a$ belongs to the set $A$ and "$n\in\overline{1,4}$" means $n\in\{1,2,3,4\}\ .$


Between which two consecutive integers does $S$ lie? \[ S=\sqrt{18+\sqrt{18^2+\sqrt{18^3+\cdots}}} \]

(Don't you just hate it when you change your answer from right to wrong? :roll:)

[hide]
Write out the expressions for $s^2$ and $s^4$, from that,

$s^4=2(18^2)+36(s^2-18)+s^2-18$
substitute $a$ for $s^2$

$a^2-37a+18=0$
$a=34.939...$ We only take the positive one since $a$ represents a square.
$s=5.911...$
$5, 6$

[/hide]

Adunakhor: answer is not correct. Use another method...

Thanks for pointing that out, a simple check should have shown me the answer was too small. Now to find what I did wrong  :?

umm...

[hide]

we have:

$s^2 - 18 = \sqrt{18^2 + \sqrt{18^3 + \ldots}} = s\sqrt{18}$ (i think)

using quadratic and taking positive solution you get:

$s = \frac{3\sqrt{2}(\sqrt{5}+1)}{2}$

which is between $6$ and $7$

is that right...?

[/hide]

Well, $S*\sqrt{18}=\sqrt{18+S}$, so $18*S^2=18+S$, and we solve that to get $S=1.03$ ?
And s is between 0 and 1?
Weird, what did I do wrong? :?

EDIT: oops, bad first equation

[b]@ pieterminate and gotztahbeazn:[/b]
$\sqrt{18}S=\sqrt{18}\sqrt{18+\sqrt{18^2+\sqrt{18^3+...}}}=\sqrt{18(18+\sqrt{18^2+\sqrt{18^3+...}})}$
$=\sqrt{18^2+\sqrt{18^2(18^2+\sqrt{18^3+...})}}=\sqrt{18^2+\sqrt{18^4+\sqrt{18^7+\sqrt{18^{12}+...}}}}$

Basically, the general exponent is $2^{n-1}+n$

[hide="What I would do at a competition"]Clearly, the sum is bigger than $6=\sqrt{18+18}=\sqrt{18+\sqrt{18^2+\sqrt{0}}}$

What about 7?
$7=\sqrt{49}=\sqrt{18+31}=\sqrt{18+\sqrt{18^2+637}}=\sqrt{18+\sqrt{18^2+\sqrt{18^3+399937}}}$ etc

So clearly the sum is smaller than $7$

The sum lies between $6$ and $7$[/hide]

[hide=" Similar to pieterminates method"]
$\frac{S}{\sqrt{18}}=\sqrt{1+\sqrt{1+\sqrt{\frac{1}{18}+\sqrt{\frac{1}{18^4}+...}}}}$
Where the general exponent is $n-2^{n-1}$
The exponent gets very negative very quickly, so we can say:
$\frac{S}{\sqrt{18}}$ is going to be slightly bigger than $\sqrt{1+\sqrt{1}}=\sqrt{2}$
So $S$ will be slightly bigger than $\sqrt{2}\sqrt{18}=6$ and our bounds are $6$ and $7$
[/hide]

Hint:
[hide]$\phi = \sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$[/hide]

Yeah, I messed up on my first equation :oops:
So,
$S=\sqrt{18+S*\sqrt{18}}$, so $S^2=18+S*\sqrt{18}$, and we solve that to get $S=6.8647......$
So it's between 6 and 7.
Yay for an actual numerical!  :lol:

actually i think tahts the same thing i had which is wrong... as zuton force pointed out

Nice solutions...I had 6-7 but then I changed it :huh:

answer is 6 & 7 as many of u correctly noted

to see its greater than 6, show that S > sqrt(18+sqrt(18+sqrt(18)+...

to show that its less than 7 show that S < sqrt(18 + sqrt(18^2 + sqrt(18^4 + ...


What is the largest number of distinct positive integers such that no four of them have sum divisible by 4?

[hide="something that may be of use..."]
If there are evenly many odd addends, then their sum is even.
[/hide]

I got
[hide]$6$

3 numbers in the form of $4k$ and 3 more numbers in the form of $4k+1$.

Example: $1, 4, 5, 8, 9, 12$
[/hide]
I don't know if it's right...


Prove that the points K, L, C
from the picture: http://www.picturerack.com/ims/guest/7625.gif
lie on a stright line.

for the power of C wrt the circumcircle of ABDE we have $ \overline{CE} \cdot \overline{CD} \equal{} \overline{CA} \cdot \overline{CB}$ so C stay on the Radical Line of the other 2 circle (that is KL).


topology
group theory
abstract algebra
search
superior algebra
superior algebra unsolved

If we have a binary structure $ (A,*)$ which is not group, Can be extend it o be group?

Sorry , if my post already here, but I searched about this question here and I couldn't find?

I appreciate if tell me where I can find it if it's exist

 :)

What does "extend" mean in this context?  Please be precise.

we know that in Topology every incomplete metric space can be isometrically embedded in a complete metric space.

I've just wonder if we could "extend" the structure to be group such as embedding process !!

If we can... :huh:

Let $ F$ be the free group with alphabet $ A$
and $ N$ be the smallest normal subgroup of $ F$ generated by the words $ (a*b)b^{\minus{}1}a^{\minus{}1}$ for all $ a,b\in A$.
Then $ \Phi: A\to F/N,a\mapsto aN$ is a homomorphism
and $ *$ can be extended to some group operation iff $ \Phi$ is injective.

In fact if $ \Phi$ is injective, then we can identify $ A$ with $ \Phi(A)\subseteq F/N$,
in which case $ *$ becomes the group operation of $ F/N$ restricted to $ A$.

Conversely if we have some group $ G$ with $ A\le G$, then
using the universal property we find some group homo $ \Psi: F\to G, a\mapsto a$ for all $ a\in A$.
Then $ \ker \Psi$ contains all words $ (a*b)b^{\minus{}1}a^{\minus{}1}$, so $ N\le \ker \Psi$ and we get
the group homo $ \overline{\Psi}: F/N\to G, xN\mapsto \Psi(x)$.
Then $ \overline{\Psi}(\Phi(a))\equal{}a$ for all $ a\in A$ gives $ \Phi$ injective.

So the question becomes under which conditions for $ *$ we have $ \Phi$ injective.
Clearly we need $ *$ associative, i.e. $ (a*b)*c\equal{}a*(b*c)$ for all $ a,b,c\in A$.
Also we need, that for each $ a,b\in A$ the equations $ a*x\equal{}b$ and $ y*a\equal{}b$ each have at most one solution $ x,y\in A$.

But are those conditions enough to get $ \Phi$ injective? :huh: I don't know.

Nice work dude :wink:

I think we need to add the condition that $ \Phi$ must be well-defined ??? :huh:

The map $ \Phi: A\to F/N,a\mapsto aN$ is well-defined!
It explicitly says to where each element $ a\in A$ is mapped to
and the elements $ aN\in F/N$ exist by definition of $ F$.

Sorry , I didn't notice that...

I don't see we need another condition... :maybe:

Yes we do. :P
If we have $ a,e\in A$ with $ a*e\equal{}a$ or $ e*a\equal{}a$
then in fact $ b*e\equal{}b$ and $ e*b\equal{}b$ for all $ b\in A$. (i.e. $ e$ is the neutral element)
I think we need that too.

[quote="olorin"]Yes we do. :P
If we have $ a,e\in A$ with $ a*e \equal{} a$ or $ e*a \equal{} a$
then in fact $ b*e \equal{} b$ and $ e*b \equal{} b$ for all $ b\in A$. (i.e. $ e$ is the neutral element)
I think we need that too.[/quote]

Excellent  :P  :D

According to the Wikipedia article, a [url=http://en.wikipedia.org/wiki/Semigroup]semigroup[/url] can be embedded into a group only if it has the cancellation property.

The problem with the analogy to an embedding of metric spaces is that, intuitively, a metric space can be made complete just by adding more elements to it; this does not affect the properties of the metric.  On the other hand, your original question is about a magma, and the binary operation on a magma is not required (in general) to be associative or to have the cancellation property, and neither problem can be solved simply by enlarging the underlying set.

O.k.

Is there a general way for any set? or we depend on binary operation ?

 :D

For every cardinality there is a group. Use search function.


Sean $x,y,z >0$ que satisfacen, $xyz=x+y+z+2$
demostrar que:
                $xyz(x-1)(y-1)(z-1)\leq 8$

A ver si esto no es muy estúpido; desde luego, aunque esté bien, bruto es un rato.
La condición inicial equivale a $\sum \frac{1}{x+1}=1$ así que podemos hacer el cambio $x=\frac{b+c}{a},\;\;y=\frac{a+c}{b}\;\;z=\frac{a+b}{c}$ con $a,b,c>0$.
Nos queda que hay que probar:
$\frac{(b+c)(a+c)(a+b)(-a+b+c)(a-b+c)(a+b-c)}{(abc)^{2}}\leq 8$.
Después de una larga y lasciva serie de manipulaciones algebraicas, si mis cálculos son correctos, pasando por:
$(\sum a^{2}b+2abc)(\sum a^{2}b-\sum a^{3}-2abc)\leq 8(abc)^{2}$
llegamos a que la desigualdad inicial es equivalente a:
$\sum a^{5}b+6(abc)^{2}\geq 2\sum a^{3}b^{3}+\sum a^{3}b^{2}c$.
Pero de la desigualdad de Schur, $\sum a(a-b)(a-c)\geq 0 \Leftrightarrow \sum a^{3}+3abc\geq \sum a^{2}b \Leftrightarrow$
$\Leftrightarrow 6(abc)^{2}\geq 2\sum a^{3}b^{2}c-2\sum a^{4}bc$.
Esto quiere decir que bastaría con probar:
$\sum a^{5}b+\sum a^{3}b^{2}c\geq 2\sum a^{3}b^{3}+2\sum a^{4}bc$
Pero como por reordenamiento $2\sum a^{3}b^{3}\leq \sum a^{4}b^{2}$, habríamos acabado si lográsemos demostrar:
$\sum a^{5}b+\sum a^{3}b^{2}c\geq \sum a^{4}b^{2}+2\sum a^{4}bc \Leftrightarrow \sum a^{3}(b+c)(a-b)(a-c)\geq 0$, que es otra versión de Schur; el hecho clave en su demostración es que $a\geq b\geq c\Rightarrow a^{3}(b+c)\geq b^{3}(a+c)\geq c^{3}(a+b)$.
Se que es muy esquemático y probablemente esté mal, pero bueno, igual le sacáis algún partido.

[quote="Mimoide"]A ver si esto no es muy estúpido; desde luego, aunque esté bien, bruto es un rato.
La condición inicial equivale a $\sum \frac{1}{x+1}=1$ así que podemos hacer el cambio $x=\frac{b+c}{a},\;\;y=\frac{a+c}{b}\;\;z=\frac{a+b}{c}$ con $a,b,c>0$.
Nos queda que hay que probar:
$\frac{(b+c)(a+c)(a+b)(-a+b+c)(a-b+c)(a+b-c)}{(abc)^{2}}\leq 8$.
.[/quote]bueno, creo que desps de llegar a eso, notamos que $a,b,c$ tiene que ser los lados de un triangulo, por que sino, en el lado izquerido seria negativo, y la desigualdad seria obia. desps lo que sigue es usar un poco de desigualdades geometricas y sale, en estos momentos estoy corto de tiempo, desps posteo la solucion,


What is $ 7 \frac{5}{7} \%$ of the epxression $ 17.7 \minus{} 2 \frac{1}{9} \plus{}2.3$? Express your answer as a decimal to the nearest hundredth.


The product of four different integers, exactly three of which are odd, is 7!. The sum of the
four integers is 63. What is the largest of the four integers?

Since $ 7!\equal{}7*5*3*6*4*2\equal{}7*5*3*48$ We easily see that largest integer among 4 of them is $ 48$.


Let $n \in \mathbb N^\ast$ and $\mathcal P(A)$ be the power set of $A = \left\{ 1, 2, \ldots, n \right\}$. A function is called [i]ascending[/i] if $f : \mathcal P(A) \to \mathcal P(A)$ and, for all $X,Y \subset A$, we have that $X \subset Y \, \Longrightarrow \, f(X) \subset f(Y)$.
(a) Prove that if $f$ is an ascending function, then there is $B \subset A$ such that $f(B) = B$.
(b) Prove that for all $k \in \overline{1,2^{n}}$ there is an ascending function $f_{k}$ such that the equation $f_{k}(X) = X$, $X \subset A$, has exactly $k$ solutions.

My question is if $f$ need to be bijective and if the notation is $X,Y \subseteq A$ instead of $X,Y \subset A$

Yes, I think that $\subset$ is supposed to mean $\subseteq$ (else the problem doesn't have much sense).


$21$ distinct points are selected on a circle, prove that at least $100$ arcs with endpoints in these points, substend an interior angle of at most $120$ degrees.

Something similar has been posted before, I believe. Not exactly this problem, but the idea was similar.

Connect two points with a segment iff the angle ($\pi$) they subtend is $>120^{\circ}$. We cannot have triangles, so this graph has at most $\frac {21^2}4$ edges, i.e. at most $110$. This means that there are at least $\frac{21\cdot 20}2-110=100$ pairs of points which determine good arcs ($\le 120^{\circ}$).

View the points as the vertices of a graph,where two vertices are joined by an edge if their corresponding points subtend an arc of at least 120 degrees. Cleary this graph contains no three cycles, so by Turan´s Theorem,it contains at most [21^2/4] = 110 edges, leaving ("21 choose 2" - 110) = 100 pairs of points taht subtend an arc of at most 120 degrees. Equality is achieved by placing 20 points near one endpoint of  a diameter and the other 21 points near the other endpoint of the diameter.  :D

[quote="grobber"]Something similar has been posted before, I believe. Not exactly this problem, but the idea was similar.

Connect two points with a segment iff the angle ($\pi$) they subtend is $>120^{\circ}$. We cannot have triangles, so this graph has at most $\frac {21^2}4$ edges, i.e. at most $110$. This means that there are at least $\frac{21\cdot 20}2-110=100$ pairs of points which determine good arcs ($<120^{\circ}$).[/quote]
Yes, it was problem about $90^\circ$.


Let a, b be the positive integer number such that  $ \frac{ab\plus{}1}{a\plus{}b} <\frac{3}{2}$
Find the minimum value of $ P\equal{}\frac{a^3b^3\plus{}1}{a^3\plus{}b^3}$


Please see the attached.

There is this large horizontal space that I can't minimize when I go online. It started acting like this a few days ago, when I accidentally do something that my taskbar go from horizontal to vertical. Although I fixed the taskbar, the firefox remain unmovable (I can't drag the border up or down). Please help me!!! :blush:

I'm not sure if this will work, but you could try clicking on view, then clicking on status bar.

Thanks but the space is still there even though the status bar and the "find" bar are closed.

On the picture you posted, I can see an up arrow on the left side. What happens if you click on it?

It moved the page further down, blocking some of the space. Anyways, I downloaded a new browser - Opera and it worked perfectly. I think I should delete Firefox then reinstall it :D .

Or you could just use Opera....

Just a suggestion.


Find all positive intigers x,y
$ x!\plus{}y!\equal{}x^y$

Already posted. See [url=http://www.mathlinks.ro/viewtopic.php?p=933240#933240]here[/url].  :wink:

if $ x > y$  $ \implies$  $ x! \plus{} y! \equal{} y!((y \plus{} 1)...x \plus{} 1) \equal{} x^{y}$

$ ((y \plus{} 1)...x \plus{} 1,x) \equal{} 1,(y \plus{} 1)...x \plus{} 1|x^{y}$  contradiction!

if $ y > x$  $ \implies$  $ x!((x \plus{} 1)...y \plus{} 1) \equal{} x^{y}$

$ \implies$  $ (x \minus{} 1)|x^{y}$  if $ x \minus{} 1 > 1$  contradiction!  $ \implies$  $ x \minus{} 1 \equal{} 1$ $ \implies$  $ x \equal{} 2$  $ \implies$ $ y! \plus{} 2 \equal{} 2^{y}$

$ 2^{k}\parallel{}y!$ , $ k > y$  $ \implies$  $ 2^{y}\parallel{}2$  $ \implies$  $ y \equal{} 1$ contradiction!

$ y > k$  $ \implies$  $ 2^{k}\parallel{}2$  $ \implies$  $ k \equal{} 1$  $ \implies$  $ y \equal{} 1,2,3$

$ \implies$  $ y \equal{} 3$  because $ x\not \equal{} y$

if $ y \equal{} x$  $ \implies$  $ 2x! \equal{} x^{x}$ ,  $ x \minus{} 1|x^{x}$

if $ x \minus{} 1 > 1$  contradiction!

$ x \minus{} 1 \equal{} 1$  $ \implies$  $ x \equal{} 2$

$ \{(2,2),(2,3)\}$


calculus
integration
continued fraction
real analysis
real analysis unsolved

Proove that:

\[ \int_{1}^{\infty }\frac{1}{x^{n}\plus{}\frac{1}{x^{n}\plus{}\frac{1}{x^{n}\plus{}\frac{1}{x^{n}\plus{}\ldots }}}}\, dx\equal{}\frac{n\plus{}i (2 i)^{1/n}(n\plus{}1) B_{\minus{}4}\left(\minus{}\frac{n\plus{}1}{2 n},\frac{3}{2}\right)}{2 n (n\plus{}1)}\]

For all Real Number $ n > 1$

That continued fraction is the positive root of the equation $ \frac{1}{x^{n}\plus{}t}\equal{}t$, or $ t^{2}\plus{}x^{n}t\equal{}1$; this root is $ t\equal{}\frac{1}2\left(\sqrt{x^{n}\plus{}4}\minus{}x^{\frac{n}{2}}\right)\equal{}\frac{2}{\sqrt{x^{n}\plus{}4}\plus{}x^{\frac{n}{2}}}$

Since I have no idea what the RHS means, I'm not going to try to go any farther.


superior algebra
superior algebra unsolved

We have "Every Commutative ring with unity has a maximal ideal."We can prove this using Zorn's lemma.
Can you constuct an example for the result does not hold if we drop unity?

by
Savithri.K.P.
MSc Student
India

[quote="ZetaX"]At the moment, I don't see where in the proof we need that the ring has a unity...[/quote]

In the proof that I know, proof which uses Zorn, there is a step when you take a totally ordered set of ideals and you want to prove that there is a PROPER ideal of the ring that contains all these ideals. The solution there is given by taking the union of all the ideals in the totally ordered set and proving that this is an ideal and that it does not containt $1$, hence is a proper ideal.


Let a,b,c,d reals such that (a^2+1)(b^2+1)(c^2+1)(d^2+1)=16. Prove that ab+bc+cd+da+ac+bd \leq 5+abcd.

my friend found that if we let
a=tgx
b=tgy
c=tgz
d=tgt
then we have prod( cosx) = 1/4
another hand he used
cos(x+y+z+t)=...
your ineq will be cosX \leq 1.


inequalities
trigonometry
quadratics
function
algebra
geometry proposed
geometry

Let $ ABC$ be an acute triangle. Prove that:
$ \sum{\sin{\frac {A}{2}}}\geq \frac {5}{4} \plus{} \frac {r}{2R}$.

There is a stronger problem propose by Jack Garfunkel in Crux problem 987 of Crux
With any acute trtangle prove that
$\sum\sin\frac{A}{2}\ge \frac{4}{3}(1+\frac{r}{4R})\ge\frac{5}{4}+\frac{r}{2R}$
I only known an ugly solution for it .

Substituting $\frac{r}{R}= \sum \cos A-1$, we have to show that

$F(A, B, C) = \sum \sin \frac{A}{2}-\frac{1}{3}\sum \cos A \ge 1$

Find extrema of $F(A, B, C),$ subject to the constraint $G(A, B, C) = \sum A-\pi = 0,$ using Lagrange multiplier $\lambda.$ Define $H(A, B, C) = F(A, B, C)-\lambda G(A, B, C)$ and put $\frac{\partial H}{\partial A}= \frac{\partial H}{\partial B}= \frac{\partial H}{\partial C}= 0.$ This leads to

$3\lambda = \frac{3}{2}\cos \frac{A}{2}+\sin A = \frac{3}{2}\cos \frac{B}{2}+\sin B = \frac{3}{2}\cos \frac{C}{2}+\sin C$

$f(A) = \frac{3}{2}\cos \frac{A}{2}+\sin A$ is concave down for $A \in (0, \pi/2),$ because $f''(A) < 0$ in this interval. Thus a horizontal line meets $f(A)$ at most at 2 points and the $\triangle ABC$ has to be at least isosceles, WLOG $A = B,\ C = \pi-2A.$ Then

$\frac{3}{2}\cos \frac{A}{2}+\sin A = \frac{3}{2}\cos \left(\frac{\pi}{2}-A\right)+\sin(\pi-2A) = \frac{3}{2}\sin A+\sin 2A$

$\frac{3}{2}\cos \frac{A}{2}= \frac{1}{2}\sin A+\sin 2A = \sin \frac{A}{2}\cos \frac{A}{2}(1+4\cos A)$

$16 \sin^{3}\frac{A}{2}-10 \sin \frac{A}{2}+3 = 0$

This cubic has trivial root $\sin\frac{A}{2}= \frac{1}{2}.$ Factoring it out, the remaining quadratic equation $8 \sin^{2}\frac{A}{2}+4 \sin \frac{A}{2}-3 = 0$ has roots $\sin\frac{A}{2}= \frac{-1 \pm \sqrt{7}}{4},$ only the positive one acceptable. For the isosceles $\triangle ABC,$

$F(A, A, \pi-2A) = 2 \sin \frac{A}{2}+\cos A-\frac{2 \cos A-\cos 2A}{3}= 2 \sin \frac{A}{2}+\frac{\cos A+\cos 2A}{3}$

If $\sin \frac{A}{2}= \frac{1}{2},$ then $\cos A = \frac{1}{2},\ \cos 2A =-\frac{1}{2}$ and the above expression is equal to 1. If $\sin \frac{A}{2}= \frac{\sqrt{7}-1}{4},$ then $\cos A = \frac{\sqrt{7}}{4},\ \cos 2A =-\frac{1}{8}$ and the above expression is greater than 1:

$\frac{\sqrt{7}-1}{2}+\frac{2\sqrt{7}-1}{24}= \frac{7 \sqrt{7}}{12}-\frac{13}{24}> 1$

On the boundary of $\sum A = \pi$ for an acute $\triangle ABC,$ we have WLOG $C = \pi/2 = A+B.$ For $A = 0,\ B = \pi/2,$ we get $F\left(0, \frac{\pi}{2}, \frac{\pi}{2}\right) = \sqrt{2}-\frac{1}{3}> 1.$ Similarly as before, the boundary extrem is at

$3\lambda = \frac{3}{2}\cos \frac{A}{2}+\sin A = \frac{3}{2}\cos \frac{B}{2}+\sin B$

Since this function is not symmetric WRT $\pi/4,$ it must be $A = B = \pi/4$ and

$F\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2}\right) = \sqrt{2-\sqrt{2}}+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{3}= \sqrt{2-\sqrt{2}}+\frac{\sqrt{2}}{6}> 1$

As a result, the absolute minimum equal to 1 is at $\sin \frac{A}{2}= \sin \frac{B}{2}= \frac{1}{2},$ $A = B = C = \frac{\pi}{6},$ when the $\triangle ABC$ is equilateral.

See and [url=http://www.mathlinks.ro/viewtopic.php?p=1077118#1077118]here[/url] where I applied the [b]Popoviciu's inequality[/b] to the concave function $ \cos$ on $ \left(0,\frac {\pi}{2}\right)$ !

You can prove analogously $ \sum\sin A\equal{}\frac pR\le\boxed {\ \frac {p}{2R}\plus{}\frac {3\sqrt 3}{4}\le\sum\cos\frac A2\ }\le\frac {3\sqrt 3}{2}$ .


If p(n) be the nth prime number prove that:
p(n+1) ^2<p(1).p(2).p(3)...p(n) :roll:

This is trivially true. Bertrand's Postulate states that $\exists$ a prime in the interval $(n, 2n) \ \forall \ n >1$, thus $\forall \ n \in \mathbb{Z}^+$, we have $p_n < p_{n+1} < 2p_n \Rightarrow {p_n}^2 < {p_{n+1}}^2 < 4 {p_n}^2$. Using also the fact that $\forall \ n \geq 2$, $p_{n-1} < p_n < 2p_{n-1}$, we have ${p_{n+1}}^2 < 8p_{n-1}p_n$. Since $\prod_{i=1}^{n-2}p_i \geq 8 \forall \ n>4$, we have ${p_{n+1}}^2 < \prod_{i=1}^n p_i \forall \ n>4$. It lefts to check the finite cases $n = 1,2,3,4$, which upon checking, we find that $n = 4$ also works, thus proving the assertion for $n>3$.

Of course it is true but it must solve with primary solution.(without using Bertrand's postulate)  :?

Let $a=p_1p_2...p_n-1$
If $a$ is not a prime then:
It exists a prime number dividing $a$ and less than $\sqrt{a}$, this prime number is not $p_1,...,p_n$ so it's $p_{n+k}$ with $k>0$. Now : $p_{n+1}^2\leq p_{n+k}^2\leq p_1...p_n-1 < p_1...p_n$
If $a$ is prime, I don't know.


complex analysis
algebra
function
domain

What does holomorphic determination means?

It is more commonly called a holomorphic branch. Let $\Omega$ be a domain and let $f$ be a holomorphic function defined in a neighborhood of some point of $\Omega$. Suppose that $f$ has analytic continuation over any curve in $\Omega$. We say that $f$ has a holomorphic branch in $\Omega$ if the result of analytic continuation depends only on the endpoints of the curve, but not on its shape.
[url=http://en.wikipedia.org/wiki/Analytic_continuation]Analytic continuation[/url]

This thread should have been a part of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153074]another one[/url].


function
floor function
irrational number
algebra solved
algebra

a.)  Prove that there are two numbers $a,b \in \mathbb{R}$ that

\[ c_n = a \cdot n + b - \lfloor a \cdot n + b \rfloor, n = 1,2, \ldots \]

generates a non-constant sequence with all its elements $c_n \neq 0$.

b.) Prove that for two chosen $a,b \in \mathbb{Q}$ the sequence $c_n$ attains a minimum.

[quote="grobber"]The first one is, of course, clear (take $a=b=$any irrational number), and there's something I do not understand regarding the second part: $a,b$ chosen for this part should also satisfy the condition from [b]a)[/b]?[/quote]

Yes, part a.) is assumed.  :)

And by "$c_n$ attains a minimum" you mean that there is an $n_0$ for which $c_n\ge c_{n_0},\ \forall n\ge 1$?

Strange... Maybe Orlando mean "doesn't attain minimum"?

You can show that the sequence $c_n$ just attains finitely many values. Thus the sequence must have a minimum. My first argument is your turn.  :P

But we can take $a=\frac 13,b=\frac 12$, can't we? The sequence has only finitely many values, but more than $1$ (so it's non-constant), meaning that it has a minimum.

On the other hand, if $a$ is irrational, there's no minimum.

orl, do we have to prove a certain result (which one ?:?) for all $a,b$ which satisfy [b]a)[b/], or do we have to show that we can choose $a,b$ s.t. a certain property (again, which one?) holds? This is very unclear, you know.

[You guys posted these replies while I was typing :)]

b.) Prove that for each two $(a,b) \in \mathbb{Q}^2$ the sequence defined in a.) attains a minimum.  :)

You have just made it trivial, haven't you? :D

Hmm. I thought you want to have some easy problems as there are still some old tricky problems on the forum.  :D

[quote="orl"]b.) Prove that for each two $(a,b) \in \mathbb{Q}^2$ ...[/quote]
Of course, it is a small unimportant condition...  :lol:  :lol:  :lol:


limit
logarithms
number theory unsolved
number theory

Show that there exist two consecutive perfect squares such that there are at least $ 1000$ primes between them.

[b]Theory[/b]
$ \lim_{n\to\infty}\frac {\pi(x).\ln{x}}{x} \equal{} 1$
Now suppose that for each n then between $ x^2,(x \plus{} 1)^2$ has at most k-1 prime then we have
$ \pi(n^2)\leq (k \minus{} 1)n$
Therefore $ \frac {\pi{(n^2)}.\ln{n^2}}{n^2} < \frac {(k \minus{} 1)\ln(n^2)}{n}$
It gives contradiction when $ n\to \plus{} \infty$

[quote="TTsphn"][b]Theory[/b]
$ \lim_{n\to\infty}\frac {\pi(x).\ln{x}}{x} = 1$
Now suppose that for each n then between $ x^2,(x + 1)^2$ has at most k-1 prime then we have
$ \pi(n^2)\leq (k - 1)n$
Therefore $ \frac {\pi{(n^2)}.\ln{n^2}}{n^2} < \frac {(k - 1)\ln(n^2)}{n}$
It gives contradiction when $ n\to + \infty$[/quote]
nice  :) ,
so if i understand the solution $ \pi(x)=\#\{p\ prime|\ p\le x\}$?


Substitute $a=\frac{1}{b}$ and multiply both sides by $(a^{2}+3)(1+3a^{2})$.
The given inequality is equivalent to \[(a-1)^{2}(3a^{2}-2a+3)\geq 0\]which is obviously true.

Equality occurs if and only if $a=b=1$. $\blacksquare$

[quote="ionbursuc"]Generalization 2
Let  $a>0,b>0,x>0,y>0,k>0,r\in \left[ 0,1 \right]$ and $ab=xy=1.$ Prove that
$\frac{{{a}^{k+r}}}{{{a}^{2k+2}}+1+2x{{a}^{k}}}+\frac{{{b}^{k+r}}}{{{b}^{2k+2}}+1+2y{{b}^{k}}}\le \frac{1}{2}.$[/quote]
The same method to kill it.
Not worth it to write...

[quote="sqing"]
The same method to kill it.
Not worth it to write...[/quote]
Are you sure?

Sorry.
Thanks.
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geqslant  \frac{1}{3}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2
\geqslant \frac{a}{b}+\frac{b}{c}+\frac{c}{a}.$

[quote=hamup1][quote=sqing]Let $a,b,c>0$ . Prove that\[\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geqslant  \frac{a}{b}+\frac{b}{c}+\frac{c}{a}.\][/quote]
Multiply both sides by $a^2b^2c^2$ to get $$\sum a^4c^2 \geq \sum a^3bc^2.$$ Since $(4,0,2)\succ (3,1,2)$, the result follows from Muirhead's divided by 2. $\blacksquare$[/quote]
 We can not use Murhead here. The inequality is not symmetric.

hello, consider $f(a)=\left( 1+a \right) ^{-1}+ \left( 1+\sqrt {-{a}^{2}+2} \right) ^{-1}+{
\frac {1+a\sqrt {-{a}^{2}+2}}{1+a+\sqrt {-{a}^{2}+2}}}
$ and show that for $a=1$ we get $f(1)=\frac{5}{3}$, where $(1,f(1))$ is the global minimum point of $f(a)$, for $0\le a\le \sqrt{2}$.
Sonnhard.

$S= \frac{a}{a^{2}+3}+\frac{b}{b^{2}+3}\leqslant\frac{1}{2}.$

By $AM-GM$ we have that $a^2+1 \ge 2*\sqrt{a^2*1}=2a$ same do with b and c and replace in inequality.
Now we have:
$S\le \frac{a}{2a+2}+\frac {b}{2b+2}=\frac{4ab+2a+2b}{8ab+4a+4b}=\frac{2}{4} \frac{a+b}{a+b}=\frac{1}{2}$

[quote=VadimM][quote=sqing]Let $a\ge 0,$ $b\ge 0$ and $a^2+b^2=2.$ Prove that\[\frac{1}{1+a}+\frac{1}{1+b}+\frac{1+ab}{1+a+b}\geqslant\frac{5}{3}.\][/quote]

let $ a+b=x, ab=y $ Hence $ x^2=2+2y$ and $y\leq 1$This inequality is equivalent to:$ 3y^2-y(3+2x)+2x=3(y-1)(y-\frac{2}{3}x)\geq 0 $
Prove that : $ 3y\leq 2x\iff $
$9y^2\leq 4(2+2y)=8+8y\iff $
$9(y-\frac{8+\sqrt{352}}{18})(y-\frac{8-\sqrt{352}}{18})\leq 0$ This is true.[/quote]
Thanks.
\[\frac{1}{1+a}+\frac{1}{1+b}+\frac{1+ab}{1+a+b}\geqslant\frac{5}{3}\iff \frac{2}{a+b}+\frac{(a+b)^2}{2(a+b+1)}\geqslant\frac{5}{3}\]


For positive real number a,b,c satisfy: $ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}  \geq \frac{27}{8}$
Prove that:   $ \sqrt{abc}  \leq \frac{a\plus{}b\plus{}c}{3}$

Using the condition it is enough to prove :

$ \sqrt{\frac{8(ab\plus{}bc\plus{}ca)}{27}}\le \frac{a\plus{}b\plus{}c}{3}$ , true .

[quote="bigzero"]i don't know when equality occurs[/quote]
It doesn't.

yaa, please don't [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=297028]double post[/url].

[quote="bigzero"]i don't know when equality occurs[/quote]

I doesn't occurs .


Hi, I proved this inequality and I found it interesting..
If x,y,z,a,b,c are positive and x,y,z, a,b,c are similarly sorted, then
\[
x(ka - b)(ka - c) + y(kb - c)(kb - a) + z(kc - a)(kc - b) \geqslant (k - 1)^2 \left( {a^2 x + b^2 y + c^2 z} \right)
\]
where $k \ge 1$
Put k=1 we obtain a special case of vornicu-schur inequality.
I would like to see if there's any application of this inequality.

[quote="siuhochung"]If x,y,z,a,b,c are positive and x,y,z, a,b,c are similarly sorted, then
\[
x(ka - b)(ka - c) + y(kb - c)(kb - a) + z(kc - a)(kc - b) \geqslant (k - 1)^2 \left( {a^2 x + b^2 y + c^2 z} \right)
\][/quote]

Hmm, you seem to assume $k\geq 1$ here, else the inequality is wrong (take, for instance, a = 1, b = 2, c = 2, $k=\frac12$). With $k\geq 1$, the problem becomes very easy: By the Vornicu-Schur inequality,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$,

and $k\geq 1$ yields $k-1\geq 0$, so that

$\left(x\left(ka-b\right)\left(ka-c\right)+y\left(kb-c\right)\left(kb-a\right)+z\left(kc-a\right)\left(kc-b\right)\right)-\left(k-1\right)^2\left(a^2x+b^2y+c^2z\right)$
$=x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)$
$+\left(k-1\right)\left(\left(b+c\right)\left(b-c\right)^2+\left(c+a\right)\left(c-a\right)^2+\left(a+b\right)\left(a-b\right)^2\right)\geq 0$,

and thus

$x\left(ka-b\right)\left(ka-c\right)+y\left(kb-c\right)\left(kb-a\right)+z\left(kc-a\right)\left(kc-b\right)\geq \left(k-1\right)^2\left(a^2x+b^2y+c^2z\right)$,

qed.. Nice inequality although.

  Darij

Thanks for pointing out my mistake.  :lol:

[quote="Reyes"]Siuhochung, i think they should have :$x\leq y\leq z $ (or $ x\geq y\geq z$) and $ a\geq b\geq c $ .[/quote]

I wrote x,y,z,a,b,c are similarly sorted..

Sorry, I made some idiot calculation mistakes in my post #2. Still, the problem requires $k\geq 1$. We have the following correct version of the problem:

[color=blue][b]Theorem 1.[/b] Let a, b, c, x, y, z be six nonnegative reals such that $a\geq b\geq c$ and $x\geq y\geq z$. Also, let k be a real number satisfying $k\geq 1$. Then,

$x\left(ka-b\right)\left(ka-c\right)+y\left(kb-c\right)\left(kb-a\right)+z\left(kc-a\right)\left(kc-b\right)\geq \left(k-1\right)^2\left(a^2x+b^2y+c^2z\right)$.[/color]

[i]Proof of Theorem 1.[/i] Since $a\geq b$ and $x\geq y$, we have $ax\geq by$, so that $ax-by\geq 0$. Together with $a-b\geq 0$ (what follows from $a\geq b$), this yields $\left(a-b\right)\left(ax-by\right)\geq 0$. Similarly, $by-cz\geq 0$ and $b-c\geq 0$, and thus $\left(b-c\right)\left(by-cz\right)\geq 0$. Similarly, $cz-ax\leq 0$ and $c-a\leq 0$, and thus $\left(c-a\right)\left(cz-ax\right)\geq 0$.

By the Vornicu-Schur inequality,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$,

and $k\geq 1$ yields $k-1\geq 0$, so that

$\left(x\left(ka-b\right)\left(ka-c\right)+y\left(kb-c\right)\left(kb-a\right)+z\left(kc-a\right)\left(kc-b\right)\right)-\left(k-1\right)^2\left(a^2x+b^2y+c^2z\right)$
$=x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)$
$+\left(k-1\right)\left(\left(b-c\right)\left(by-cz\right)+\left(c-a\right)\left(cz-ax\right)+\left(a-b\right)\left(ax-by\right)\right)\geq 0$,

and thus

$x\left(ka-b\right)\left(ka-c\right)+y\left(kb-c\right)\left(kb-a\right)+z\left(kc-a\right)\left(kc-b\right)\geq \left(k-1\right)^2\left(a^2x+b^2y+c^2z\right)$,

what proves Theorem 1.

By the way, I don't think that a similar assertion can be made about the case when $a\geq b\geq c$ and $x\leq y\leq z$.

  Darij


linear algebra
matrix
superior algebra
superior algebra solved

A,B \in M_2(R) from wich at least one is invertible
Show that I_2+AB , I_2+BA have the same determinant
--------------------------------------------------------------
Here a general result
A matrix with p-rows, n-colomns
B matrix with n-rows, p-columns
coefficients of the matrix is in R or C
then det(XI_n - BA) = X^(n-p).det(XI_p - AB)
Proof:
We will consider matrix define by blocs

J=
(I_n 0)
(A I_p)

L=
((I_n-BA) B)
( 0 XI_p)  where X \in R or C

M=
(XI_n  B)
(0   (XI_p-AB))

If you compute LJ and JM you get
((XI_n) B)
((XA) (XI_p))

det(LJ)=det(JM)
J is lower triangular with diagonal only 1, det(J)=1
detL=detM
detL=X^p.det(XI_n -BA)
detM=X^n.det(XI_p - AB)

For the problem X=-1, n=p=2

consider that there exists A^(-1) then we have that: det(I_n+AB)=det(A*A^(-1)+AB)=det[A(A^(-1)+B]=det(A)*det(A^(-1)+B)=det(A^(-1)+B)det(A)=det(I_n+BA)

cheers! :D :D

[quote="Lagrangia"] det(I_n+AB)=det(I_n+BA)
[/quote]
yes, but this equality is also true if A,B are not invertible.


We have a pyramid with the base a $9$-gon. The lateral edges and the diagonals of the base are colored in red and blue. Prove that there is at least a monochromatic triangle formed by all these edges.

$\star \, \, \star \, \, \star$

Maybe they did not want to pronounce the word "graph", I guess.


show :D

$ \sum_{n\equal{}1}^\infty \frac{1}{16^{n}}\left(\;\zeta(2n)\minus{}1\;\right)\;\equal{}\;\boxed{  \minus{} \frac{\pi}{8}\plus{}  \frac{13}{30} }$

That looks like a double sum to me: $ \sum_{n\equal{}1}^{\infty}\frac1{4^{2n}}\sum_{k\equal{}2}^{\infty}\frac1{k^{2n}}$

What do you do when you see an iterated sum/integral? Reverse the order. In this case, everything's positive and justified by Tonelli.
$ \sum_{k\equal{}2}^{\infty}\sum_{n\equal{}1}^{\infty}\frac1{(4k)^{2n}}\equal{} \sum_{k\equal{}2}^{\infty}\frac{\frac1{(4k)^2}}{1\minus{}\frac1{(4k)^2}}$
$ \equal{}\sum_{k\equal{}2}^{\infty}\frac{1}{(4k)^2\minus{}1}\equal{} \frac12\sum_{k\equal{}2}^{\infty}\frac1{4k\minus{}1}\minus{}\frac1{4k\plus{}1}$
$ \equal{}\frac12\left(\frac17\minus{}\frac19\plus{}\frac1{11}\minus{}\frac1{13}\plus{}\cdots\right)$

It is well-known that $ 1\minus{}\frac13\plus{}\frac15\minus{}\frac17\plus{}\frac19\minus{}\frac1{11}\plus{}\cdots\equal{}\frac{\pi}{4}$ (arctangent power series), so our sum is $ \frac12\left((1\minus{}\frac13\plus{}\frac15)\minus{}\frac{\pi}{4}\right)\equal{} \frac12\left(\frac{13}{15}\minus{}\frac{\pi}{4}\right)\equal{}\frac{13}{30}\minus{}\frac{\pi}{8}$.


Given three non-negative numbers such that $xy+yz+zx=1$, prove that
\[ \frac9{2+(x+y+z)^2}\leq\frac1{1+x^2+yz}+\frac1{1+y^2+zx}+\frac1{1+z^2+xy}\leq\frac95. \]

[quote="mathpk"]It's same as the one you pose before. Use AM-HM to prove LHS. Inequality.[/quote]
Yes, for the first inequality, it is not difficult at all. We use AM-GM. But for the second one, I have not solved it.


topology
function
geometry
3D geometry
sphere
group theory
advanced fields

Can someone explain to me what a quotient topology is in an intuitive manner. I know that it is the unique topology on a set $ Y$ for which the surjective function $ q: X\rightarrow Y$ from a topological space $ X$ is an identification map, however I am failing to see the consquences of this. Any help is greatly appreciated, thank you.

Quotient constructions in general are there to satisfy a certain universal mapping property.

We can say that $ q : X \to Y$ is a quotient map if it is continuous, and for any continuous map $ f : X \to Z$ such that $ q(x) \equal{} q(y) \implies f(x) \equal{} f(y)$, then there is a unique continuous map $ g : Y \to Z$ such that $ f \equal{} g \circ q$. This is how I usually think about quotient maps; the idea agrees with that of a quotient map in group theory, for example, as well.

(Note: You made need to add in surjectivity as a condition as well... but I find this extension property of quotient maps to be its useful 'defining' property.)

You can show that giving $ Y$ the quotient topology with respect to some equivalence relation $ \sim$ on $ X$ makes $ q : X \to Y$ a quotient map.

Man your fast. I think I got it. Thanks.

While Nukular gives the general category theoretic picture of a quotient, I'm not sure how intuitive it is (if you're fine with it then thats cool too). But the basic intuition behind it is you want the quotient space to look the same as what you're starting out with, except you collapse a subset into a point. For example, if you take the disc $ D$ and you quotient out by its boundary, is the same thing as collapsing the boundary to a point, thus making the disc into a sphere. Similarly, if you have a cylinder, and you quotient out by one of the circles on the end, then you get a cone.

LkNsngth, thanks for the help. An example I have seen reguarding quotient topology is as follows:

Let $ S^n$ be the unit $ n$-sphere of points $ x\in R^n$ such that $ |x|\equal{}1$. Let $ RP^n$ be the set of all one-dimensional linear subspaces of $ R^{n\plus{}1}$. Let $ q: S^n\rightarrow RP^n$ send a point $ x\in S^n$ to its corresponding subspace in $ R^{n\plus{}1}$. (Therefore each element of $ RP^n$ is the image under $ q$ of two antipodal points in $ S^n$) This induces a quotient topology on $ RP^n$ such that $ q$ is an identification map.

In this example what is the intepretation of the quotient topology?

So the example you've given is the construction of "real projective space". What the quotient topology on this means is that you're taking opposite points on the sphere and sort of gluing them together. To better imagine this, first consider the case $ RP^1$. Here we have a circle with opposite points identified. You can imagine taking half the circle, twisting it and gluing it onto itself giving a circle again, hence $ RP^1\equal{}S^1$. Now, consider $ RP^2$, this is the sphere with opposite points identified. Thus we again imagine sort of gluing the bottom half of the sphere into the top. However, now on the equatorial boundary circle we also have to glue together boundary points. Projecting this down into the equatorial plane allows us to think of $ RP^2$ as a disc in the plane with antipodal points on the boundary identified. This construction is true in general: if we take $ D^n$ the solid $ n$-ball and glue antipodal points on the boundary together we get $ RP^n$. In terms of cells this means $ RP^n$ can be thought of as $ RP^{n\minus{}1}$ with a $ n$-ball glued down onto it.

Thanks so much for the help. Topology seems very intuitive when explained well.

I find topology tends to fall into two parts: (1) intuitive understanding so that you can think about things pictorially, and (2) rigorous basis so you can translate your intuitive arguments.

Sadly, it's hard to develop models which satsify (1) while giving an elegant (2).

I highly approve of teaching both :).


[quote="moldlee"]Using 6 given colors, you color each side of a cube to a

different color, then you write the six numbers on it so that

the numbers 6 and 1; 2 and 5 ; 3 and 4 are facing each other.

How many different cubes can you make?

(Two cubes are considered to be the same if you can rotate one cube into

the position of the other.)[/quote]
[hide="hint"]All that matters is the placement of the colors. Any cube can be rotated into any other one.[/hide]
[hide="answer if you can reuse colors"]There are $6^{6}$ possibilities[/hide]
[hide="answer_if_you_cannot"]There are 6!=720 ways[/hide]

[size=75][color=darkred]Don't forget you can't use spaces and apostrophes in hide tags! Use an underscore, "_" , instead (for spaces).
-nebula42[/color][/size]
[color=blue][size=75]kyyuanmathcount: Well, I think spaces work in hide tags, or at least for me[/size][/color]

[hide="I am assuming that you are not allowed to re-use the colors"]
Since there are 6 sides, there are $6!=\boxed{720\text{ ways to make different cubes.}}$[/hide]


What's the method for solving this problem using algebra?

A merchant has coffee worth $ 20 a pound that they wish to mix with 40lbs of coffee worth$80 a pound to get a mixture that can be sold for $ 60 a pound.
How many pounds of the$20 dollar coffee should be used?

[quote="learningfish"]A merchant has coffee worth $ \$20$ a pound that they wish to mix with 40 lbs of coffee worth $ \$80$ a pound to get a mixture that can be sold for $ \$60$ a pound. How many pounds of the $ \$20$ dollar coffee should be used?[/quote]
[hide="Solution"]Let $ x$ be the pounds of coffee needed.

We can write the equation $ \frac {20x \plus{} 80\cdot40}{x \plus{} 40} \equal{} 60$, where the numerator of the fraction is the two types of coffee mixed, the denominator is the total number of pounds used, and the RHS is the resulting cost.

This equation can be expanded into $ 20x \plus{} 3200 \equal{} 60x \plus{} 2400$, and then simplified to $ 40x \equal{} 800$. Therefore, $ x \equal{} \boxed{20}$.[/hide]


For some reason it shows " heta" instead of "\theta" in the code view window.
[asy]pair A=origin, B=(1,2), C=(-2,3);
D(B--MA("\theta",B,A,C,0.3)--C);[/asy]
All other Greek letters behave hardly any better: \ disappears and you get some strange symbols instead of the first letter (try \beta, for instance).

Maybe it gets parsed as a special character?

Hm so "\t" is becoming a tab, which becomes a space.  This is interestingly exploitable... :|

Fixed. Thanks ccy for the input :police:


inequalities
trigonometry
inequalities proposed

Assume that $s=a+b+c>3$. Put $x=3a/s,\,x=3b/s,\,z=3c/s$. Then $x+y+z=3$ and $x<a,\,y<b,\,z<c$, hence $2(x^2+y^2+z^2)+3xyz<9$. Make it homogeneous: $2(x^2+y^2+z^2)+9\frac{xyz}{x+y+z}<(x+y+z)^2$, after simplifying: $x^3+y^3+z^3+3xyz<xy(x+y)+yz(y+z)+zx(z+x)$, a contradiction with Schur inequality.

Right solution!
This problem seems to be much easier than it's brother here
http://www.mathlinks.ro/Forum/viewtopic.php?p=186512&highlight=#186512
I hope you find interest in that one

Fedor Petrov ,Your solution is resemble my one:
My idea is a method from a book which my friend read and he tell me it.I'm interested in it.
Supposing that :
[tex]a+b+c>3[/tex]
Then there exist [tex]M,x,y,z>0[/tex] satisfying:
[tex]M>3,x+y+z=1[/tex] and [tex]a=Mx,b=My,c=Mz[/tex]
Therefore:
[tex]2(a^2+b^2+c^2)+3abc=M^2(2(x^2+y^2+z^2)+3Mxyz)>9(2(x^2+y^2+z^2)+9xyz)[/tex]
(Since [tex]M>3[/tex])
But since [tex]x+y+z=1[/tex],we can prove that:
[tex]2(x^2+y^2+z^2)+9xyz\geq\ 1[/tex]
Hence:
[tex]2(a^2+b^2+c^2)+3abc>9.[/tex]
Contradiction!
And it follows.

Here it is I think the easiest solution. The main idea is to use the inequality
  a^2+b^2+c^2>= :sqrt: 3abc(a+b+c) whisch can be proved very easy.With this inequality we get
   2(a^2+b^2+c^2)>=2 :sqrt: 3abc(a+b+c) or 9-3abc>=2 :sqrt: 3abc(a+b+c) or
      9/ :sqrt: 3abc- :sqrt: 3abc>=2 :sqrt: (a+b+c).So we have to prove that
             9/ :sqrt: 3abc- :sqrt: 3abc<=6.By notation :sqrt: 3abc=t we have to prove that

              t^2+6t-9>=0,which is trivial.

   I hope I didn't make a mistake. :D

[quote="cezar lupu"]
              t^2+6t-9>=0,which is trivial.
 [/quote]
i dont' see this trivial.. and i think it is wrong, can u elaborate?

Hint for my solution
$2(a^2+b^2+c^2)\frac{a+b+c}{3}+3abc\ge 9(\frac{a+b+c}{3})^3$

[quote="keira_khtn"]Similarly,we have following inequality:
For a,b,c be positive numbers  such that:[tex]a^2+b^2+c^2+abc=4[/tex]
Prove that:[tex]a+b+c\leq 3[/tex][/quote]
but this one is much simpler. Just substitute $a=2\cos x$ and so on and finished.

[quote="minhkhoa"]Let $a,b,c$ be positive numbers such that
$2(a^2+b^2+c^2)+3abc=9$
Prove that
$a+b+c\le 3$[/quote]
this one can aslo be sloved by $siuhochung$'s substitution?
I think it can. :)


Prove that

$ a^4\plus{}b^4\plus{}c^4\leq\frac{8}{3}R(R\minus{}r)(4R\plus{}r)^2$

[quote="Ligouras"]Prove that

$ a^4 \plus{} b^4 \plus{} c^4\leq\frac {8}{3}R(R \minus{} r)(4R \plus{} r)^2$[/quote]

The following inequality is equivalent to

$ \frac {8}{3}R(R \minus{} r)(4R \plus{} r)^2 \minus{} 2(s^4 \minus{} 8Rrs^2 \minus{} 6s^2r^2 \plus{} 16R^2r^2 \plus{} 8Rr^3 \plus{} r^4)\geq 0$

$ \Longleftrightarrow 6(s^2\plus{}4R^2\minus{}3r^2\minus{}4Rr)(4R^2\plus{}4Rr\plus{}3r^2\minus{}s^2)\plus{}8(R\minus{}2r)(2R\plus{}r)(2R\minus{}3r)(R\plus{}r)\geq 0$

by Gerretsen $ s^2\plus{}4R^2\minus{}3r^2\minus{}4Rr\ge16Rr\minus{}5r^2\plus{}4R^2\minus{}3r^2\geq 0$

$ 4R^2\plus{}4Rr\plus{}3r^2\minus{}s^2\geq 0$


What is the slope of a line perpendicular to the line through $ (4,3)$ and $ (7,\minus{}5)$? Express your answer as a common fraction.

Using the formula for slope, the slope is:

\[ \frac{3\minus{}(\minus{}5)}{4\minus{}7}\equal{}\frac{8}{\minus{}3}\equal{}\frac{\minus{}8}{3}\].

Thus the slope of the line perpendicular to that line is $ \boxed{\frac{3}{8}}$.


calculus
function
derivative
real analysis
real analysis unsolved

My calculus professor made a statement today in class:
If the limit of f(x) as it approaches infinity exists, then the limit of f'(x) as it approaches infinity also exists.

He said that it was intuitively false, as if it were true, it would be well known and he would know it.  However, he was unable to think of a counter-example of a function that would disprove this theory.

I've been looking at this for a while, but haven't gotten very far - I ask you all - can you think of a function that limit exists at infinity but the derivative does not?

Try a function which oscillates increasingly rapidly, but shrinks in amplitude.

[hide="Answer"]$\frac1x\sin(x^{2})$[/hide]

Great solution, jmerry.

Another question: Does this work?

[hide]f = 1/[x]

since f is defined for all real, whereas f'(x) = -[x]' / [x]^2 isn't defined when x is an integer?[/hide]

There really should be a condition on that conjecture: for a differentiable function $f$ such that $f(x)$ tends to a limit as $x\to\infty$ ...

That would rule out your example, and it would also rule out even sillier things, such as a nowhere differentiable function or even a nowhere continuous function. (Example: let $f(x)=\frac1x$ for $x$ rational and zero for $x$ irrational.)

However, you could start with your step function example and "round off the corners" of the steps to make a differentiable function - the "risers" are now steep but not vertical. Depending on the details of how you do this, you could either get an $f'$ that oscillates between zero and some fixed negative value, or we could have $f'$ oscillate between zero and increasingly negative numbers.


analytic geometry
geometry proposed
geometry

Given a circle with equation $x^{2}+y^{2}=r^{2}$ ,consider in general, inscribed  polygons having sides and diagonals of RATIONAL lenghts.

Prove /disprove/find :

a)We can always pick n>4 points on the circle's perimetry such that they are vertices of a rational-sided polygon (ie. "n-tagon")?

b)For what values of n inscribed rational sided polygons  have also rational lenghts diagonales?

c)If n vertices on circle $x^{2}+y^{2}=1$ are picked such that they are rational coordinates for what values of n distances among all the points are also rational lenghts?

[quote="Eros88"]Given a circle with equation $x^{2}+y^{2}=r^{2}$ ,consider in general, inscribed  polygons having sides and diagonals of RATIONAL lenghts.

Prove /disprove/find :

a)We can always pick n>4 points on the circle's perimetry such that they are vertices of a rational-sided polygon (ie. "n-tagon")?[/quote]

Does this help you?
[url=http://www.problem-solving.be/pen/viewtopic.php?t=210]PEN problem F 12[/url].
In fact, this problem gives you a circle and infinitely many points on it with all pairwise distances between them being rational. The circle is not your circle but the circle with center $\left(\frac12,0\right)$ and radius $\frac12$, but you can translate the center of this circle to the origin and stretch the radius by the factor $2r$ so that you get your circle with equation $x^{2}+y^{2}=r^{2}$, and the distances still remain rational.

[quote="Eros88"]b)For what values of n inscribed rational sided polygons  have also rational lenghts diagonales?[/quote]

ALL such polygons?

  Darij


Eleven percent of what number is seventy-seven?

First, What i do is get the number to 1% by doing 77/11 =7
so 7=1% now just multiply 7 by 100 because then it makes it 100%
7*100=700
The answer is 700 :welcome:  :welcomeani:

$ \frac{100}{11}(77)\equal{}\fbox{700}$, using the procedure for inverse percents.


function
trigonometry
calculus
calculus computations

Find a bijection from (0, infinity) onto (0,1).

Those are open intervals.
I am wondering if there is a straight-forward method to solve this kind of problems, I have tried many functions, but I couldn't have onto (0,1)

for example:

f(x) = ln x
g(x) = x + 1
h(x) 1/x
Then, h(g(f(x))) = 1/ (ln x + 1), which is not onto (0,1)

Any hint and help would be great, thank you.

I think $f(x)=\frac{1}{x+1}$ also works.


I forgot how to make multiple columns, but \newpage is the command to begin a new page.

The simplest way to use 2 columns is to use \twocolumn[put text across the columns here]. But this will put a page break before the columns start and after the \onecolumn command (which does precisely what it says).

You can also use \documentclass[twocolumn]{article} if you want 2 columns in the whole document.

For more control and more columns (up to 10) use the multicol package


I think math is fun, and you can find beauty in the unlikeliest of places.

I wonder why you put your location as what you did.  And I realize that's a good way to get words you want to appear on this forum, because many of the words I want to post get censored.

Yes I also think that you meant what we think of your sig..but I'm not sure.

[quote="Amir.S"]What do you think?[/quote]
I don't think that I think. :ninja:

[quote="fredbel6"]Can I be honest?

That it's good you changed your signature but[hide] it's "congratulate"[/hide][/quote]

Why is it important for you?

Do you want it cause of spelling mistake or somthing else? :?:

Not actually. But I can't speak for fredbel...

[quote="Amir.S"][quote="fredbel6"]Can I be honest?

That it's good you changed your signature but[hide] it's "congratulate"[/hide][/quote]

Why is it important for you?

Do you want it cause of spelling mistake or something else? :?:[/quote]

Giving a hint about possible spelling or grammar corrections is in my opinion good for the person who made an error.  The power of correction is amazing :"I congratulate the Iranians and especially the scientists on their recent success in Nuclear Tec." could be better. :maybe:

More seriously :!:  : I think things like "Down with USA and Israel" are a bit too direct, even though I'm not sure what exactly you meant with it  :huh: .  There are people of all around the world here...
However, I cannot speak for the Mathlinks administrators.

Please, Amir S., what is this topic about?  What do we think .... of what??

I think that Amir.S wanted us to write what we think right now while reading the topic and that's my answer to both Amir.S and fredbel6


ratio
number theory solved
number theory

Let $p,q,n$ be three positive integers with $p+q<n$. Let $(x_{0},x_{1},\cdots ,x_{n})$ be an $(n+1)$-tuple of integers satisfying the following conditions :

[b](a)[/b] $x_{0}=x_{n}=0$, and
[b](b)[/b] For each $i$ with $1\leq i\leq n$, either $x_{i}-x_{i-1}=p$ or $x_{i}-x_{i-1}=-q$.

Show that there exist indices $i<j$ with $(i,j)\neq (0,n)$, such that $x_{i}=x_{j}$.

(Sorry, this may be really confusing, but I'm new at this.)

[hide]

Let $a_k$ and $b_k$ be positive integers such that $x_k = a_k p - b_k q$.

So $x_i = x_j$ for $i<j$ implies that $p(a_i-a_j) = q(b_i-b_j)$.

We claim that we can find $a_i, a_j, b_i, b_j$ such that $a_i-a_j=q$ and $b_i-b_j=p$.

Consider a string of $n$ letters that are either $p$ or $q$. We have a total of $a_n$ $p's$ and $b_n$ $q's$. The k-th letter corresponds to $x_k$.

Call a sequence of $p+q$ elements of this string a stretch.
Consider the first stretch. If there are $q$ $p's$ and $p$ $q's$, then we are done. Otherwise, suppose without loss of generality that there are more than $p$ $q's$ and less than $q$ $p's$.
When we move a stretch by one to the right, such that each index is replaced by its successor, we call this a right-shift. By continually right-shifting, we either add or subtract from either the number of $p's$ or the number of $q's$ by one. But we must still make sure that not all stretches have more than $p$ $q's$. This is impossible because then the ratio $\frac{a_n}{b_n}$ would be greater than $\frac{q}{p}$, a contradiction.

So there exists a stretch with $p$ $q's$ and $q$ $p's$ - suppose it has indices $i$ and $j$ for the beginning and end, respectively. Then $x_{i-1} = x_j$ (since each element of the stretch adds to the previous x). Furthermore, since $p+q<n$, we know that a stretch cannot contain $x_0$ and $x_n$ simultaneously.

[/hide]


I was just wondering a few things about the new Google experiment.
1)  How many of you guys have heard about it?
2)  How many of you guys have it already?
3)  If you don't have it, would you want it? (having seen the demo video of course)
here
[youtube]p6pgxLaDdQw[/youtube]

4)  Lastly, what do you think of the concept?  Will it really be the next "wave?"

In my opinion, it's been pretty useful, but many of the functions I don't find on say, gmail I don't use often, like being able to see pictures.  However, the ability to reply to a previous part of the wave and how one can easily drop another person into the conversation are things I have liked.

(if you guys want it, you can send me a pm, i have plenty of invites. (it is in beta stage and I would need your email...))

I have it.
It kinda failed since all of my RL friends don't want to make a gmail :(.
jjfun sent me the invite...
so like I have 8 left.
If zapi runs out...... meewhee's here :)

I know multiple people that have invites, b ut for me, it just seems like two much.
Too many little extra things to do.  And so many places for people to do bad things.
Just too open for me I guess.

[quote="bluecarneal"]And so many places for people to do bad things.
Just too open for me I guess.[/quote]

like real-time spelling errors!  :lol:

Well, it's still in beta, so I hope they get some kinks out of the system.

I have it, but I'm too lazy to actually learn how to use it; too complicated for me  :P

I'm fine with just Gmail.

I second xp, for me it seems pretty useless since most of my communication doesn't need to be in real time.

I'm sure it'll be helpful if I'm snowed in and need to work on a group project. So far though, I haven't been snowed in, and I can just walk to my friend's dorm (and we can play Rock Band when we're done). I'm sure project groups that span across states and countries will find more use for it than a project group contained within a single campus.

Still, I think it's great because it has the [i]ability[/i] to communicate in real-time. I'm sure it has lots of potential.

It's potentially helpful with [url=http://www.artofproblemsolving.com/Forum/index.php?f=581]NIMO[/url], but otherwise, I have no use for it. Gmail is good enough for me.

They lied; it was 7 minutes and 52 seconds.  It sounds pretty cool, especially the language part.  But google translate isn't completely accurate.. =/  I would want it because of fast publishing and just because I guess.   :P

Personally, gmail is enough for me.
I have google wave but I've only used it once.
I don't think it'll ever become really popular because, truthfully, not that many people know about it since you need invites anyways. And as for the ability to communicate in real-time, you can do that with Gmail, too.

[quote="AceOfDiamonds"]real-time as in you can see what they're typing before they hit enter... gmail is only real time like that in vid chats[/quote]
Oh. Well, whenever that happens for me, it lags up my compy. So yeah, I still prefer gmail.

One sec. What is google wave??What's an invite ...like to google wave..?

Yeah, it's still in beta mode.  It's sort of similar to when gmail was in beta and the only way you could get it was via invite.

Google wave is nice, but not extremely useful i guess.

And some reason I don't have a sudoku button under the "other options" menu :(.  In fact, I have nothing under the ... menu.

[quote="abcak"]Google wave is nice, but not extremely useful i guess.
[/quote]
That brings up a good point. What is the purpose of google wave?

It's something that's meant to be used for collaborative projects. Forcing it to be a social networking engine just doesn't work.


Let $\ A,B\in M_{2n}(\mathbb{R})$ two invertible matrices such that the equality
                            $\ A^m+A^{-m}+(A^*)^m=B^m+B^{-m}+(B^*)^m$
is true for $\ m\in\{1,2,3\}$. Prove that $\ A=B$.

[quote="Thor"]
                           ... $A^*$...[/quote]

What does this mean ?

Probably the adjugate / classical adjoint, i.e. $A A^\ast = \left( \det A \right) I$.

As the matrices have real entries, it would have been pretty redundant to write $A^\ast$ instead of $A^T$.


trigonometry
number theory proposed
number theory

Given triangle $ABC$ such that $\angle B=60^\circ$ and the side length $b$ is integer, $a,c$ be prime numbers.
Prove that the triangle is equilateral triangle.

[hide]
Using cosine law we get

$b^2=a^2+c^2-ac$ .

If $a=c$ then $b^2=a^2$  :arrow: $a=b=c$

If $a\neq c$ or $(a,c)=1$ , we have $b^2=(a-c)^2+ac$ :arrow: $(b-a+c)(b+a-c)=ac$

Hence there are two possible
(i)$b-a+c=a$ , $b+a-c=c$
(ii)$b-a+c=c$, $b+a-c=a$

But both of them yields $a=b=c$ .

Hence $ABC$ is equilateral triangle .
 :)
[/hide]

[quote="kunny"]shyong,are you sure that there are two possibilities? :?[/quote]

then I guess there are two more which are

(i)$b-a+c=1$ ; $b-c+a=ac$
(ii)$b-a+c=ac$ ; $b-c+a=1$

and we can work out both and obtain $a=b=c=1$ which is also true .

EDIT: I think these two cases shouldnt be considered since $a,c$ are prime which is $>1$ :blush:


Find all triples $ (x,k,n)$ of positive integers such that

$ 3^k\minus{}1\equal{}x^n$

Apart from the obvious $ n\equal{}1$ familly of solutions we may assume $ n>1$
assume first that $ n$ is even ($ n\equal{}2m$) than $ 3^k\equal{}(x^m)^2\plus{}1^2$ which is a contradiction because $ (x^m)^2\plus{}1^2$ can have no prime factors with reminder $ 3$ when devided by $ 4$. therefore $ n$ is odd

we have $ 3^k\equal{}(x\plus{}1)(x^{n\minus{}1}\minus{}x^{n\minus{}2}\plus{}x^{n\minus{}3}\minus{}....\minus{}x\plus{}1)$ neither $ x\plus{}1$ nor $ \frac{x^n\plus{}1}{x\plus{}1}$ can equal $ 1$ (because that would imply x=1 which gives no solutions) therefore $ x\plus{}1$ and $ \frac{x^n\plus{}1}{x\plus{}1}$ are both powers of $ 3$

now because of the last conclusion and following lemma, $ gcd(x\plus{}1,\frac{x^n\plus{}1}{x\plus{}1})$ devides $ n$, we can conclude that $ 3$ devides $ n$ ($ n\equal{}3a$).

therefore $ 3^k\equal{}(x^a\plus{}1)(x^{2a}\minus{}x^a\plus{}1)$. obviously $ x^a\equal{}2,5,8\mod9$ and consequently $ x^{2a}\minus{}x^a\plus{}1\equal{}3\mod9$. and since $ x^{2a}\minus{}x^a\plus{}1$ is a power of $ 3$ we can conclude that $ x^{2a}\minus{}x^a\plus{}1\equal{}3$ meaning $ x^a\equal{}2$ which yields $ x\equal{}2,n\equal{}3,k\equal{}2$ as the only non-trivial solution.


Hi all this is breaking my head

If:

$ \overline{a_1a_2a_3a_4a_5a_6a_7\ldots} \equal{} 2009^{2008^{2007^{2006^{2005^{2004^{2003^{2002^{2001^{2000}}}}}}}}}$

Find:

$ a_{2009^{2009}}$

Exactly the requested digit of position $ 2009 ^ {2009}$ counting from left to right.

Greetings  :D

Could you  explain the notation? Usually overline denotes the repeating decimal value but your number is finite.

hi Someone an idea please.

Greetings  :D

[quote="TALENTO"]hi Someone an idea please.

Greetings  :D[/quote]

Is this a real problem found in a book, contest, ... so supposed to have a solution, or just an invented one ?


limit
probability
conditional probability
probability and stats

suppose that $ X$ is a symmetric random walk on $ \mathbb{Z}$, starting from $ 0$. For $ A,B > 0$, one defines the stopping time
$ T_{A,B}(\omega) \equal{} \inf( n \geq 0 : X_{\omega} \in \{A,B\})$
What is the law of $ T_{a,b}$? and the joint distribution of $ (T_{A,B}, X_{T_{A,B}})$ ?

And if now $ X$ is a brownian motion ?

I don't understand why you have two positive numbers $ A$ and $ B$ here. Would not the random walk always stop at the smaller one?

:blush:  I meant $ T_{A,B} \equal{} \inf\{ t: W_t \in \{A, \minus{}B \} \}$

Having done this before, I know that $ E(T_{A,B})\equal{}AB$ and $ P(X_{T_{A,B}}\equal{}A)\equal{}\frac{B}{A\plus{}B}.$

But that isn't the full information you asked for.

I will give the proof of the results Kent Merryfield mentioned, so maybe somebody will solve the full question:
First $ T_{A,B} < \infty$ because $ \limsup W_t = - \liminf W_t = + \infty$. $ W^{T_{A,B}}$ is a bounded martingale (hence uniformy integrable martingal), hence the optional stopping theorem applies:
$ E(W^{T_{A,B}}) = A P(X{}_{T{}_{A,B}}=A) + B P(X{}_{T{}_{A,B}}=B) = E(W_0) = 0$
From here one gets the first result: $ P(X{}_{T{}_{A,B}}=A) = \frac{B}{A+B} = 1 - P(X{}_{T{}_{A,B}}=B)$

The second result follows from the fact that $ W_t^2 - t$ is a martingale. Hence $ E[(W_t^{T_{A,B}})^2]=E[\min(T_{A,B},t)]$ for all $ t$. To conclude remark that $ E[T_{A,B}] = \lim E[\min(T_{A,B},t)]$ (monotone convergence theorem), and that $ E[W_t^{T_{A,B}}^2]$ tends towards $ E[(W{}_{T{}_{A,B}})^2] = A^2 P(X{}_{T{}_{A,B}}=A) + B^2 P(X{}_{T{}_{A,B}}=B) = AB$ (dominated convergence)

The problem for the random walk is not easy, the solution can be found in the book P. Billingsley "Convergence of probability measures", in the section devoted to the Donsker Theorem in $ C[0,1]$. It is also shown how the result for Brownian motion follows from the result for random walk.

There's a more elementary approach to what I said earlier, based on conditional expectations.

We're doing a random walk until we hit either $ A$ or $ \minus{}B,$ whichever comes first.

Let $ x(k)$ be the conditional probability that we will finish at $ A,$ given that our current location is $ k.$

Let $ y(k)$ be the conditional expectation of the remaining time, given that the current location is $ k.$

$ x(\minus{}B)\equal{}0,x(A)\equal{}1,$ and $ y(\minus{}B)\equal{}y(A)\equal{}0.$

Consider what happens on one step.

$ x(k)\equal{}\frac12x(k\plus{}1)\plus{}\frac12x(k\minus{}1).$

$ y(k)\equal{}1\plus{}\frac12y(k\plus{}1)\plus{}\frac12y(k\minus{}1).$

Clearing the fractions, we get:

$ x(k\plus{}1)\minus{}2x(k)\plus{}x(k\minus{}1)\equal{}0$

$ y(k\plus{}1)\minus{}2y(k)\plus{}y(k\minus{}1)\equal{}\minus{}2.$

From these difference equations, we get that

$ x(k)\equal{}\alpha\plus{}\beta k$ and $ y(k)\equal{}\gamma\plus{}\delta k\minus{}k^2.$

Putting in the boundary conditions gives us that

$ x(k)\equal{}\frac{k\plus{}B}{A\plus{}B}.$ In particular, $ x(0)\equal{}\frac{B}{A\plus{}B}.$

$ y(k)\equal{}AB\plus{}(A\minus{}B)k\minus{}k^2.$ In particular, $ y(0)\equal{}AB.$


So Here's the prob.
$ \ 39A_{12}\plus{}488_12$
I got $ \ 907_12$
The book said i was wrong.

Add up the two and you get 7 in the hundreds place (144 place) 17 in the 10s (12 place) and 18 in the 1s place.

Carrying over the 18, you get 18 in the 10s place and 6 in the ones. then carry over the new 18 to get 6 in the ones, 6 in the tens, and 8 in the hundreds.

866.

ahh i see where i went wrong i thougt 18-12=7.


Do the roots of the polynomial $ P(x)\equal{}1^3z\plus{}2^3z^2\plus{}3^3z^3\plus{}\cdots \plus{}n^3z^n$ lie inside, on, or outside the unit circle $ |z|\equal{}1$?

Just a clarification

Does your problem possibly imply that some can lie inside, some can lie outsie, some can lie on

Cause obviouslz z=0 lies inside the unit circle

From the wording of the problem, I am assuming that all of them lie inside the unit circle (since $ z\equal{}0$ is a root).  My problem is proving that all of them lie inside the unit circle.

This problem was discussed in the following link for a long time ago! :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=12915


You seem to have been getting a quick answer. Take it easy! :)

Except there are errors:

[hide="See it to believe it"]For this

[quote="Altheman"]$y_{n+1}=2\cdot y_{n}$[/quote]

to hold, it must be

[quote="Altheman"]$y_{n}=\frac{x_{n}}{\underbrace{n^{2}}_{\textrm{not}\,\,(n+1)^{2}}}\underbrace{-}_{\textrm{not\,\,+}}1$[/quote]

therefore $y_{1}={2\over 1^{2}}-1=1$, hence $y_{n}=2^{n-1}$, and the final answer should be $x_{n}=n^{2}(2^{n-1}+1)$.[/hide]


$ \{a,b,c,d\}\subset \mathbb{R}^ \plus{}$
Show that
$ \sum \frac {ab}{cd}\ge 4$
and $ 4\left[\sum (ab)^2\right]\ge 1$ when $ \sum ab\equal{}1$

All sums are cyclic.

AM-GM kills the first one... the second one is obviously false - set $ a\equal{}b\equal{}c\equal{}d\to 0$ :huh:

Sorry, I think I remembered that one wrong.

And 0 isn't positive, but it doesn't work for very small numbers....

bertram means we can take $ a, b, c, d$ arbitrarily close to $ 0$.  Depending on whether the required sum is cyclic or symmetric, we take either $ a, b, c, d < \frac{1}{4}$ or $ a, b, c, d < \frac{1}{2 \sqrt{6} }$.

[quote="Temperal"]$ \{a,b,c,d\}\subset \mathbb{R}^ \plus{}$
Show that
$ \sum \frac {ab}{cd}\ge 4$
and $ 4\left[\sum (ab)^2\right]\ge 1$

EDIT: I remembered the second wrong, I think, I'll try to find it again...[/quote]
I think $ 4\left[\sum (ab)^2\right]\ge 1$ when $ ab\plus{}bc\plus{}cd\plus{}da\equal{}1$ (By Cauchy-Schwarz)

Maybe that's what I meant, so I'll edit the first post with it...


Which gas has properties that are most similar to those of an ideal gas?

$ \textbf{(1)} \ \text{N}_{2} \qquad \textbf{(2)} \ \text{O}_{2} \qquad \textbf{(3)} \ \text{He} \qquad \textbf{(4)} \ \text{Xe}$

Thanks!

Keep in mind the assumptions we make for the ideal gas law: PV = nRT...

duh, helium because an ideal gas is:

monoatomic (He, Xe)
the size of the particles is minimum (He)
.
.
.

Sorry... my grammar sucks :stink:

The answer is helium.

The two assumptions of the molecules of an ideal gas are:

1. They take up no volume.
2. They exert no attractive forces on each other.

Helium gas is the smallest of the four (only one energy level), so it is the closest to satisfying condition 1.

Also, all four gases are nonpolar, so they have London dispersion forces. Helium has the weakest of these forces because it has the fewest electrons, therefore it is the closest to satisfying condition 2.


induction
number theory proposed
number theory

Let $ x_1,x_2,...,x_n$ and $ y_1,y_2,...,y_n$ be two sets of pairwise different natural numbers.
$ x_1^{x_1}\plus{}x_2^{x_2}\plus{}...\plus{}x_n^{x_n}\equal{}y_1^{y_1}\plus{}y_2^{y_2}\plus{}...\plus{}y_n^{y_n}$

Prove that the two sets are permutations of each other.

[quote="doicu"]do you have a solution?[/quote]

Yes, I do. But why don't you try it first?  :)


:( I have two problems:
1.A football match, may be either won,drawn or lost by the home team,so there are three ways of forecasting the result of any one match,one correct and two incorrect.Find
the probability of forecasting at least three correct results.
(what confuses me most is how many matches are held,it's unknown,or i assume there are some rules that i don't know.I suppose the number of the matches are n,then replace n by 6,then i get the exact answer given,0.3196 to 4 decimal places.So why
the number of the matches held is 6?)

2.In a card game for four players, a pack of fifty-two cards is dealt round so that each player receives thirteen cards.A hand that contains no card greater than nine is called a yarborough.How many hands are necessary for the probability of at least one hand to be a yarborough is greater than 1/2?

能否翻译成中文？（中国人用中国字嘛！）

能否翻译成中文？（中国人用中国字嘛！）

[quote="gsh91best"]能否翻译成中文？（中国人用中国字嘛！）[/quote]

她不像是中国人，可能只是因为她们学校的中国人数学功课好，就上来这里问功课。

 I have two problems:
1.A football match, may be either won,drawn or lost by the home team,so there are three ways of forecasting the result of any one match,one correct and two incorrect.Find
the probability of forecasting at least three correct results.

球赛有胜平负三种结果，求预测至少三场结果正确的概率。
你看中文多简洁，还把原题的朦胧不清也译出来了。

2.In a card game for four players, a pack of fifty-two cards is dealt round so that each player receives thirteen cards.A hand that contains no card greater than nine is called a yarborough.How many hands are necessary for the probability of at least one hand to be a yarborough is greater than 1/2?

4人打牌每人得13张。谁手上的牌都不大于9就称他有一手“押宝啦”牌。需要多少手牌才能使其中必有一手“押宝啦”牌的概率大于1/2？
也不知道一手牌指的是啥。

谁说我不是中国人啦。。。
我问这些题目只是觉得题目有问题，也或者是答案错了。
很早问的。
因为原题是英文题目，我翻译不好。我也不知道一手牌是什么意思。
所以拿来问你们咯。
真是。
 :rotfl: 你怎么会那样想的，哈哈。搞笑。


If

 $ a,b,c$ distinct

then

$ \sum\frac {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}{(a^2 \minus{} b^2)^2}\geq\frac {3}{2}$

Try $ a\equal{}1\ ,\ b\equal{}\sqrt{2}\ ,\ c\equal{}\sqrt{3}$ .

[quote="alex2008"]Try $ a \equal{} 1\ ,\ b \equal{} \sqrt {2}\ ,\ c \equal{} \sqrt {3}$ .[/quote]

Sorry my friend  :(

the next version is correct

$ \sum\frac {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}{(a^2 \minus{} b^2)^2}\geq\frac {3}{2}$

Here it is

$ 2(a^2\plus{}b^2) \ge (a\minus{}b)^2$

$ 2(a^2\plus{}b^2) \ge (a\plus{}b)^2$

We are done!

[quote="great math"]Here it is

$ 2(a^2 \plus{} b^2) \ge (a \minus{} b)^2$

$ 2(a^2 \plus{} b^2) \ge (a \plus{} b)^2$

We are done![/quote]

dear G.M., can you post your full proof? Thank you


Let $f: \mathbb{R}\to\mathbb{R}$ a $C^{2} [a,b]$ function. Prove that there is $\xi\in (a,b)$ such that

                                              \[ f(b)=2f(\frac{a+b}{2})-f(a)+\frac{(b-a)^{2}}{4}\cdot f''(\xi).  \]

let $F(x)=f(x+\frac{b-a}2)-f(x)$
so use largrange formula:
we get
$F(\frac{b+a}2)-F(a)=(\frac{b-a}2)F'(\xi_1))=(\frac{b-a}2)(f'(\xi_1+\frac{b-a}2)-f'(\xi_1))$
then apply the largrange twice,it is easy to get you result :)

Let $h=\frac{b-a}{2}$, put $g(x)=f(a+2x)-2f(a+x)+f(a)-Ax^2$, $x \in [0,h]$.
Then $g \in C^2[0,h]$. Since $g(0)=g(h)=0$, by Roll Th., there exists $c\in (0,h)$ such that
\[ g'(c)=2f'(a+2c)-2'f(a+c)-2Ax=0,  \]
i.e.,
\[ f'(a+2c)-f'(a+c)=Ac.  \]
Since $f' \in C^1[a,b]$, apply mean value theorem to $f'$ in $[a+c,a+2c]$, the conclusion follows.

Note that the condition can be changed to $f''$ exists on $[a,b]$.


calculus
derivative
logarithms
integration
algebra
partial fractions
calculus computations

I can't seem to figure these two problems out. I haven't tried the second really hard, but I can't think of a way.

1. (dy/dx) = (1 + ln x) y ;; if y = 1 when x = 1, then y = ?

I don't know the derivative of ln x, because there isn't one I think.

EDIT: Oops. I meant ANTIderivative. :P

2. If f is the antiderivative of (x^2)/(1 + x^5) such that f(1) = 0, then f(4) = ?

Any help would be much appreciated. Much thanks.

[quote="blahblahblah"]$\frac {d}{dx}\ln x=\frac {1}{x}$

$\int \ln xdx=x\ln x - x + C$[/quote]

^o^ Much thanks man. Now I gotta figure out the second one or wait for someone to get how to do it.   :)

[quote="adidasty"]a problem similar to #2 was posted in this forum earlier...check it out[/quote]

Was it from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30320]here[/url] by any chance?

No, Tarento, it's not similar to that. I don't see any trick substitutions that work. The answer can be expressed in closed form, but it's going to be truly painful. I'll give  you the first step (leading to partial fractions as the next step):

$\frac{x^2}{x^5+1}=\frac{x^2}{(x+1)\left(x^2-\left(\frac{1+\sqrt{5}}2\right)x +1\right)\left(x^2-\left(\frac{1-\sqrt{5}}2\right)x +1\right)}$

Yes, it is hideous.  I refrained from posting in hopes that there would be a trick substitution that someone else would see.

that happens to be one partial fractions problem that im NOT going to do. lol

OMG. Is there a way to use the calculator for it? I didn't mention a computer can be used since I didn't expect it to be that long.  :?

[quote="Gandalf"][hide="problem 2"] Yes, the problem is truly hideous. You can define f(x) as the integral from 1 to x of (x^2)/(1 + x^5) since the initial condition is f(1) = 0, and in this case the integral simply disappears. Then f(4) = the integral from 1 to 4 of (x^2)/(1 + x^5). I don't know of any simple way to evaluate it - I would guess you would need to manipulate it so that you could apply partial fractions, but it will be messy. If you're interested in just the numerical answer (even the fully simplified anti-derivate expression is quite long), Mathematica gives 0.37566. [/hide][/quote]

How would you punch it in the calculator?

Thank you everyone for the huge help!

EDIT: Nevermind! I get it now. Thank you everyone!

Any computeralgebra software can solve that...

Here's the closed form if of any use. Beware, manual simplification may still be needed :)

${\displaystyle \frac {1}{5}} \,\mathrm{ln}(5) - {
\frac {1+\sqrt{5}}{20}} \,\mathrm{ln}(30 + 4\,\sqrt{5})
\mbox{} - {\displaystyle \frac {2}{\sqrt{5}}} \,{ \frac {
\mathrm{arctan}({ \frac {15 + \sqrt{5}}{\sqrt{10 + 2
\,\sqrt{5}}}} )}{\sqrt{10 + 2\,\sqrt{5}}}}  -
{\frac {1-\sqrt{5}}{20}} \,\mathrm{ln}(30 - 4\,\sqrt{5}) + {\frac {2}{\sqrt{5}}} \,
{\displaystyle \frac {\mathrm{arctan}({\frac {15 -
\sqrt{5}}{\sqrt{10 - 2\,\sqrt{5}}}} )}{\sqrt{10 - 2\,
\sqrt{5}}}}$

[quote="Tarento"]

^o^ Much thanks man. Now I gotta figure out the second one or wait for someone to get how to do it.   :)[/quote]

Just use integration by parts:

$\int \ln(x) dx$
$u=\ln(x)$
$dv=dx$
$du=\left(\frac{1}{x}\right)dx$
$v=x$
$\int \ln(x) dx = x\ln(x)-\int \frac{x}{x} dx= x\ln(x) - x +C$.


geometry
incenter
ratio
trigonometry
circumcircle
angle bisector
power of a point

Anyone helps me to work out this problem???

Let $A', B', C'$ be midpoints of $BC, CA, AB$ of triangle $ABC$ and let $D, E, F$ be the feet of altitudes, $I, I_a, I_b, I_c$ are incenter, excenters. Suppose that the incircle touches $AB, BC, CA$ at $Z, X$ and $Y$. The excircle center $I_a$ touches $AB, BC, CA$ at $Z_a, X_a$ and $Y_a$, etc. $I I_a$ cuts $BC$ at $P$. $I I_b$ cuts $CA$ at $Q$ and $I I_c$ cuts AB at R. Show that
(1) $A'$ is midpoint of $XX_a$.
(2) Inversion center $A'$ and radius $A'X$ maps the 9-point circle to a straight line through $P$ at right angles to the diameter $A'A''$ and tangent to the incircle.
(3) Deduce from (2) that the $9$-point circle touches the incircle and excircles.
(4) What are touching points in (3).

Denote a = BC, b = CA, c = AB the triangle sides, $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ its internal angles and s the semiperimeter.

[color=blue][b](1)[/b][/color] Since $BA' = CA'$ and $BX = s - b = CX_a$, it trivially follows that $XA' = CA' - BX = CA' - CX_a = X_aA'$, i.e., the midpoint $A'$ of $BC$ is also the midpoint of $XX_a$.

[color=blue][b](2)[/b][/color] P is the foot of the angle bisector of the angle $\angle A$, hence, it divides BC in the ratio $\frac{BP}{CP} = \frac{AB}{CA} = \frac c b$ and $BP = \frac{ca}{b + c}$, $CP = \frac{ab}{b + c}$. WLOG, assume that b > c. Then the points $D, X, P, A', X_a, C$ follow on the line BC in this order. Calculating the power of the point P to the 9-point circle (N) passing through the points D, A':

$PD \cdot PA' = (BP - BD)(BA' - BP) = \left(\frac{ca}{b + c} - c \cos \beta\right) \left(\frac a 2 - \frac{ca}{b + c}\right) =$

Substituting the cosine theorem formula $\cos \beta = \frac{c^2 + a^2 - b^2}{2ca}$:

$= \frac{2ca^2 - (c^2 + a^2 - b^2)(b + c)}{2a(b + c)} \cdot \frac{a(b + c) - 2ca}{2(b + c)} =$

$= \frac{b^3 - c^3 - a^2(b - c) + bc (b - c)}{2a(b + c)} \cdot \frac{a(b - c)}{2(b + c)} =$

$= \frac{(b - c)^2(b^2 + 2bc + c^2 - a^2)}{4(b + c)^2} = \frac{(b - c)^2[(b + c)^2 - a^2]}{4(b + c)^2} =$

$= \frac{(b - c)^2(b + c + a)(b + c - a)}{4(b + c)^2} = \frac{(b - c)^2s(s - a)}{(b + c)^2}$

Calculating the power of the point P to the circle (A') with radius $A'X = A'X_a$ passing through the points $X, X_a$:

$PX \cdot PX_a = (BP - BX)(CP - CX_a) = \left(\frac{ca}{b + c} - (s - b)\right) \left(\frac{ab}{b + c} - (s - b)\right) =$

$= \frac{ca - (s - b)(b + c)}{b + c} \cdot \frac{ab - (s - b)(b + c)}{b + c} =$

$= \frac{[2ca - (c + a - b)(b + c)][2ab - (c + a - b)(b + c)]}{4(b + c)^2} =$

$= \frac{(b + c)(b -c) - a(b - c)][(b + c)(b - c) + a(b - c)]}{4(b + c)^2} =$

$= \frac{(b - c)^2(b + c - a)(b + c + a)}{4(b + c)^2} = \frac{(b - c)^2(s - a)s}{(b + c)^2}$

Since the powers of the point P to the circles (N) and (A') are the same, this point lies on their radical axis. The 9-point circle (N) passes through the inversion center A', hence, in an inversion in the circle (A') with positive power $A'X^2$, it is carried into a straight line, the radical axis of the circles (A'), (N), perpendicular to their center line A'N.  Let (O) be the circumcircle of the triangle $\triangle ABC$. The triangle $\triangle A'B'C'$ is the medial angle of the triangle $\triangle ABC$ and the 9-point circle (N) is its circumcircle. Consequently,

$\angle NA'B = \angle NA'C' + \angle C'A'B = \angle OAC + \angle ACB = 90^\circ - \beta + \gamma$

Since the radical axis of the circles (A'), (N) is perpendicular to to their center line $A'N$, it forms the angle $\angle NA'B + 90^\circ = 180^\circ - \beta + \gamma = 2 \gamma + \alpha$ with the line BC tangent to the incircle (I) and to the excircle $(I_a)$. But since the angle $\angle BPI \equiv \angle BPA = \angle PCA + \angle PAC = \gamma + \frac \alpha 2$, the radical axis of the circles (A'), (N) is the other tangent to the incircle (I) from the point P. For a similar reason, this radical axis is also the other tangent to the excircle $(I_a)$ from the point P. In other words, the radical axis of the circles (A'), (N) is the other common internal tangent (other than BC) of the circles $(I), (I_a)$,

[color=blue][b](3)[/b][/color] Since $A'X \perp IX$ are perpendicular radii of the circles (A'), (I) at their intersection point X, these 2 circles are perpendicular to each other. Likewise, since $A'X_a \perp I_aX_a$ are perpendicular radii of the circles $(A'), (I_a)$ at their intersection point $X_a$, these 2 circles are also perpendicular to each other. But inversion with a positive power carries the circles perpendicular to the inversion circle into themselves. Hence, the incircle (I) and the excircle $(I_a)$ are their own images in the inversion in the circle (A'). Since inversion preserves angles between curves, in particular, angles equal to zero (i.e., tangencies), the 9-point circle is tangent to both the incircle (I) and the excircle $(I_a)$.

[color=blue][b](4)[/b][/color] Let $T', T_a'$ be the tangency points of the radical axis of the circles (A'), (N) with the incircle (I) and the excircle $(I_a)$. The tangency point T of the circles (N), (I) is the other intersection of the line $A'T'$ with these circles (other than A' for the 9-point circle (N) and other then T' for the incircle (I)). Likewise, the tangency point $T_a$ of the circles $(N), (I_a)$ is the other intersection of the line $A'T_a'$ with these circles (other than A' for the 9-point circle (N) and other then $T'_a$ for the excircle $(I_a)$).


Let a matrix $ A$ belong to $ \text{M}_n\left(\mathbb{C}\right)$ with $ n\geq 2$. We also know that $ \det\left(A\right) \equal{} d\neq 0$.

Show that:
[b]i)[/b] $ \det\left(A^*\right) \equal{} d^{n \minus{} 1}$.
[b]ii)[/b] $ \left(A^*\right)^* \equal{} d^{n \minus{} 2}A$.
[b]iii)[/b] Find all $ A$ such that $ \left(A^*\right)^* \equal{} A$.
[b]iv)[/b] Find all $ A$ such that $ \left(\left(A^*\right)^*\right)^* \equal{} A$.

[i]Hint.[/i] We know that $ A^{ \minus{} 1} \equal{} \frac {1}{\det\left(A\right)}\cdot A^*$.

it's simple actually  just apply  A^(-1)=1/det(A) A*

cheers! :D

actually I belive he wants some proof for A-1 = A* / detA.
I belive it's much easier to prove that A.A*= (det A) . In

you can find the demonstration in the 11th algebra manual


group theory
abstract algebra
superior algebra
superior algebra unsolved

p is a prime.
Prove or disprove: For any group G of order $ p^2$, any subgroup (of G) of order p is normal.

Every group of order $ p^2$ is abelian, so...

Let $ H$ be a subgroup of order $ p$. Since $ (G : H) \equal{} p$ and $ (N[H] : H) \equal{} (G : H)$ mod p, we cannot have $ N[H] \equal{} H$. Hence $ N[H] \equal{} G$, so $ H$ is normal...


since they are the absolute values, you can assume that they are positive. Therefore, you are able to factor them out. This results in $(y^{2})(y^{2}+2y-3)$

Factoring further results in
$(y^{2})(y+3)(y-1)$

Now substituting $(x-1)$ for $y$ you get

$(x-1)(x-1)(x+2)(x-2)$. Of these 4 values, an even number must be negative and positive. All four could be negative. The only value of real concern here would be $x+2$. To be less than 0, x must be [i][b]less than -2[/b][/i]. The other case to consider is if all 4 are positive. The only one you need to worry about here is x-2. To be greater than 0, x must be [i][b]greater than two[/b][/i] in this situation. Therefore we have our two cases, $x>2 or x<-2.$

[quote="myc"]solve for x:
$|(x-1)^{4}|+2|(x-1)^{3}|-3|(x-1)^{2}|>0$

letting y=x-1, i have
$|y^{4}|+2|y^{3}|-3|y^{2}|>0$  but what do I do with all the absolute values?   :o

thanks :) :P[/quote]

Your inequality is the same as $|y|^{4}+2|y|^{3}-3|y|^{2}>0$.  Factor our a $|y|^{2}$ to get $|y|^{2}(|y|^{2}+2|y|-3)>0$.  The $|y|^{2}$ doesn't affect the sign of the left side, so discard it.  We have $|y|^{2}+2|y|-3>0$, so $(|y|+3)(|y|-1)>0$.  Therefore, $|y|<-3$ or $|y|>1$.  But $|y|$ is positive, so we have $|y|>1$ as the only possibility.

The answer is then $x<0$ or $x>2$.


projective geometry
geometry proposed
geometry

If $A,C, E$ are three points on one line, and $B,D,F$ are three points on another line, and if the two lines $AB$ and $CD$ are parallel te $DE$ and $FA$, respectively, then prove that $EF$ is parallel to $BC$.

Pappus theorem for the collinearity line at infinity.


Let $ P(x) \equal{} x^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} \cdots \plus{} a_0$ be a polynomial such that $ {a_0,a_1, \cdots , a_{n\minus{}1}} \in \{0, 1\}$. Show that if $ P(2)$ is prime, then $ P(x)$ is irreducible in $ Z[x]$.

See here http://www.mathlinks.ro//viewtopic.php?t=32826 .


find all $ a \in \mathbb {R}$ for which there is exists a nonconstant function $ f: (0.1] \to \mathbb {R}$ satisfying:
$ a \plus{} f(x\plus{}y\minus{}xy) \plus{} f(x)f(y) \leq f(x)\plus{}f(y)$ for all $ x,y \in \mathbb {R}$

Well for any $ a \leq 0$ you can choose, for example, $ f(x) \equal{} x$.  Not sure about $ a>0$.


linear algebra
matrix
vector
algorithm
function
algebra
domain

I was wondering if anybody knows what is the configuration that a matrix has to have so that when we do FFT on it, we get just real values in the spectrum map and no imaginary part. (For instance, for one dimension vector, we need a configuration of C0,C1,C2,C3,C2,C1, for instance - what is the configuration for 2D matrix). Thanks

Paulo

First, a hopeless attempt at linguistic purity: to me the FFT is an algorithm, or a family of algorithms. The object being computed by these algorithms is the "discrete Fourier transform" or "Fourier transform on $\mathbb{Z}_N$". Your question is about the latter.

The right way to look at a vector for which we are computing the DFT is to view it as a function whose domain is the additive group $\mathbb{Z}_N.$ The criterion for such a function $f$ to have a real Fourier transform is that $f(n)=\overline{f(-n)}$ In the event that $f$ is itself real-valued that simplifies to $f(n)=f(-n),$ which is to say that $f$ is even. The example you wrote out was simply that: a generic even function on $\mathbb{Z}_6.$

Your question is not specific to the discrete case; the answer is the same in any Euclidean group. By Euclidean group, I mean any of the additive groups $\mathbb{R},\mathbb{Z},\mathbb{T}\,(=\mathbb{R}/\mathbb{Z}),\mathbb{Z}_N,$ or any direct prodoct of these. In fact, the answer will be the same in any locally compact abelian group.

To illustrate the answer to your question, I'll try it not in $\mathbb{Z}_N\times\mathbb{Z}_M,$ which is what you asked, but in $\mathbb{R}^2.$ The results are the same, anyway.

$\hat{f}(s,t)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y) e^{-2\pi i(sx+ty)}dx\,dy$

So, $\overline{\hat{f}(s,t)}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \overline{f(x,y)} e^{2\pi i(sx+ty)}dx\,dy$

$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \overline{f(-x,-y)} e^{-2\pi i(sx+ty)}dx\,dy$

So $\hat{f}(s,t)=\overline{\hat{f}(s,t)}$ iff $f(x,y)=\overline{f(-x,-y)}.$

In the case that $f$ is real, then $\hat{f}$ is real iff $f(x,y)=f(-x,-y).$ That is, the only thing we require of the function is that it be symmetric about the origin.

The same applies in the discrete case, or for that matter in any locally compact abelian group.


prove that : not exist 6 positive integers  such that : the sum of any 4 of 6 numbers is divisible by the sum of the other two numbers.


a circle of radius 9 cm is inscribed in an equilateral triangle and is tangent at three points. three smaller circles are inscribed so that they are each tangent to two sides of the triangle and to the larger circle. the process is continued with circles of ever decreasing diameters. what is the number of centimeters in the sum of the circumferences of all such circles? i dont know where to start.
thanks

[color=cyan]A place to start -- find the radius of the first circle, and then of the smaller circles, and then of the next set of smaller circles, and hopefully a pattern will emerge.  You can help it out by drawing tangents to the big circle that are parallel to the sides of the triangle, which will put the new circles in smaller triangles, which might help you out.[/color]

Keep in mind that when you draw the tangents, you'll be left with a bunch of new [i]equilateral[/i] triangles. You should get a geometric series to sum.