[quote="Mateescu Constantin"]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac {x}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} xy \plus{} y^2}\ge \frac {3}{x \plus{} y \plus{} z}$ .[/quote]
The following inequality is also true.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy\plus{}xz\plus{}yz\neq0.$ Prove that:
\[ \frac {x}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} xy \plus{} y^2}\ge\sqrt{ \frac {3}{xy\plus{}xz\plus{}yz}}\]

[quote="arqady"][quote="Mateescu Constantin"]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac {x}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} xy \plus{} y^2}\ge \frac {3}{x \plus{} y \plus{} z}$ .[/quote]
The following inequality is also true.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy \plus{} xz \plus{} yz\neq0.$ Prove that:
\[ \frac {x}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} xy \plus{} y^2}\ge\sqrt {\frac {3}{xy \plus{} xz \plus{} yz}}\]
[/quote]

from 2#'s ineq we get

$ \sum_{cyc}\frac{x}{y^{2}\plus{}yz\plus{}z^{2}}\ge\frac{\sum_{cyc}x}{\sum_{cyc}xy} \ge \frac{\sqrt{3(xy\plus{}yz\plus{}zx)}}{xy\plus{}yz\plus{}zx}\equal{}\sqrt {\frac {3}{xy \plus{} xz \plus{} yz}}$

 :lol:

kuing, what about the following inequality?
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy + xz + yz\neq0.$ Prove that:
\[ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge{\frac {2}{\sqrt{xy + xz + yz+\sqrt[3]{x^2y^2z^2}}}}\]
 :wink:

[quote="arqady"]kuing, what about the following inequality?
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy + xz + yz\neq0.$ Prove that:
\[ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge{\frac {2}{\sqrt {xy + xz + yz + \sqrt [3]{x^2y^2z^2}}}}\]
:wink:[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-98-m-rozenberg.html

In fact, this problem also true:
Let $ x,y,z$ be real numbers such that $ \left(x^{2}\plus{}y^{2}\right)\left(y^{2}\plus{}z^{2}\right)\left(z^{2}\plus{}x^{2}\right)>0,$ $ xyz\ge0.$ Prove that
$ \sum\frac{x}{y^{2}\plus{}yz\plus{}z^{2}}\le\frac{x^{3}\plus{}y^{3}\plus{}z^{3}}{x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2}}.$

[quote="wya"]In fact, this problem also true:
Let $ x,y,z$ be real numbers such that $ \left(x^{2} \plus{} y^{2}\right)\left(y^{2} \plus{} z^{2}\right)\left(z^{2} \plus{} x^{2}\right) > 0,$ $ xyz\ge0.$ Prove that
$ \sum\frac {x}{y^{2} \plus{} yz \plus{} z^{2}}\le\frac {x^{3} \plus{} y^{3} \plus{} z^{3}}{x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2}}.$[/quote]
Again another nice result, wya. I found that this inequality is true according to the hypothesis and the Iran 1996 Inequality.

[quote="wya"]In fact, this problem also true:
Let $ x,y,z$ be real numbers such that $ \left(x^{2} \plus{} y^{2}\right)\left(y^{2} \plus{} z^{2}\right)\left(z^{2} \plus{} x^{2}\right) > 0,$ $ xyz\ge0.$ Prove that
$ \sum\frac {x}{y^{2} \plus{} yz \plus{} z^{2}}\le\frac {x^{3} \plus{} y^{3} \plus{} z^{3}}{x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2}}.$[/quote]
\[ \Longleftrightarrow \sum{\left(x^3\minus{}\frac{x(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)}{y^2\plus{}yz\plus{}z^2}\right)}\geq 0\]
\[ \Longleftrightarrow \frac{x^2\minus{}yz}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{y^2\minus{}zx}{z^2\plus{}zx\plus{}x^2}\plus{}\frac{z^2\minus{}xy}{x^2\plus{}xy\plus{}y^2}\geq 0\]
which is true by Chebyshev.

[quote="Tourish"][quote="wya"]In fact, this problem also true:
Let $ x,y,z$ be real numbers such that $ \left(x^{2} \plus{} y^{2}\right)\left(y^{2} \plus{} z^{2}\right)\left(z^{2} \plus{} x^{2}\right) > 0,$ $ xyz\ge0.$ Prove that
$ \sum\frac {x}{y^{2} \plus{} yz \plus{} z^{2}}\le\frac {x^{3} \plus{} y^{3} \plus{} z^{3}}{x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2}}.$[/quote]
\[ \Longleftrightarrow \sum{\left(x^3 \minus{} \frac {x(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)}{y^2 \plus{} yz \plus{} z^2}\right)}\geq 0\]

\[ \Longleftrightarrow \frac {x^2 \minus{} yz}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y^2 \minus{} zx}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z^2 \minus{} xy}{x^2 \plus{} xy \plus{} y^2}\geq 0\]
which is true by Chebyshev.[/quote]
Very nice, Tourish. Thanks.

[quote="Mateescu Constantin"]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge \frac {3}{x + y + z}$ .[/quote]

$ LHS = \frac {x^2}{xy^2 + xyz + z^2x} + \frac {y^2}{yz^2 + xyz + x^2y} + \frac {z^2}{zx^2 + xyz + y^2z}\ge$

$ \ge\frac {(x + y + z)^2}{(x + y + z)(xy + yz + zx)} = \frac {x + y + z}{xy + yz + zx}\ge \frac {3}{x + y + z}$ it is true because

$ (x + y + z)^2\ge3(xy + yz + zx)$
[quote]
The following inequality is also true.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy + xz + yz\neq0.$ Prove that:
\[ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge\sqrt {\frac {3}{xy + xz + yz}}\]
[/quote]

Use $ LHS\ge \frac {x + y + z}{xy + yz + zx}\ge\sqrt {\frac {3}{xy + xz + yz}}$ it is true because

$ x + y + z \ge \sqrt {3(xy + yz + zx)}$

[quote]
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy + xz + yz\neq0.$ Prove that:
\[ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge{\frac {2}{\sqrt {xy + xz + yz + \sqrt [3]{x^2y^2z^2}}}}\]
[/quote]

Use $ LHS\ge \frac {x + y + z}{xy + yz + zx}\ge{\frac {2}{\sqrt {xy + xz + yz + \sqrt [3]{x^2y^2z^2}}}}$ it is true becase
\[ \left( {x + y + z} \right)^2 \left( {xy + yz + zx + \sqrt [3]{{x^2 y^2 z^2 }}} \right) \ge 4\left( {xy + yz + zx} \right)^2 ,\]
.it was proved by [b]kuing[/b] :)

[quote="kuing"]
\[ \left( {x \plus{} y \plus{} z} \right)^2 \left( {xy \plus{} yz \plus{} zx \plus{} \sqrt [3]{{x^2 y^2 z^2 }}} \right) \ge 4\left( {xy \plus{} yz \plus{} zx} \right)^2 ,\]
$ \iff f(v^2) \equal{} \minus{} 36v^4 \plus{} 27u^2v^2 \plus{} 9u^2w^2 \ge 0$

just need to prove that when $ y \equal{} z$.

setting $ x \equal{} a^3,y \equal{} z \equal{} b^3,a\ge 0,b > 0$, ineq becomes

$ \frac {a^2 (a \minus{} b)^2 \left(2 a^5 \plus{} 5 a^4 b \plus{} 8 a^3 b^2 \plus{} 4 a^2 b^3 \plus{} 4 a b^4 \plus{} 4 b^5\right)}{b^6 (a \plus{} b) \left(2 a^2 \minus{} a b \plus{} b^2\right) \left(2 a^3 \plus{} b^3\right)^2} \ge 0,$

true... :blush:[/quote]
It can be write into: $ 1 \plus{} \frac {\sqrt[3]{x^2y^2z^2}}{xy \plus{} yz \plus{} zx} \ge \frac {4(xy \plus{} yz \plus{} zx)}{(x \plus{} y \plus{} z)^2}$
Using AM-GM, we'll prove the stronger:
$ 3 \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)} \ge \frac {12(xy \plus{} yz \plus{} zx)}{(x \plus{} y \plus{} z)^2}$
Obviously trues by AM-GM again  :)

[quote=Mathias_DK][quote="Mateescu Constantin"]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac {x}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} zx \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} xy \plus{} y^2}\ge \frac {3}{x \plus{} y \plus{} z}$ .[/quote]
$ \left ( \sum_{cyc} x(y^2\plus{}yz\plus{}z^2) \right ) \left ( \sum_{cyc} \frac{x}{y^2\plus{}yz\plus{}z^2} \right ) \ge \left ( \sum_{cyc} x \right )^2$
And $ \left ( \sum_{cyc} x(y^2\plus{}yz\plus{}z^2) \right ) \equal{} \left ( \sum_{cyc} x \right ) \left ( \sum_{cyc} xy \right )$
So:
$ \sum_{cyc} \frac{x}{y^2\plus{}yz\plus{}z^2} \ge \frac{\sum_{cyc} x}{\sum_{cyc} xy}$
And $ 3\sum_{cyc} xy \le \left ( \sum_{cyc} x \right )^2$ hence:
$ \frac{\sum_{cyc} x}{\sum_{cyc} xy} \ge \frac{3}{\sum_{cyc} x}$ and we are done :)[/quote]
Let $a,b,c>0 .$  [url=http://www.artofproblemsolving.com/community/c6h1268102p6615011]Prove that [/url]$$ \frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}\ge \frac{a+b+c}{ab+bc+ca}\ge\frac{3}{a+b+c}$$

[quote=Mateescu Constantin]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac{x}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{y}{z^2\plus{}zx\plus{}x^2}\plus{}\frac{z}{x^2\plus{}xy\plus{}y^2}\ge \frac{3}{x\plus{}y\plus{}z}$ .[/quote]

WLOG let $xyz=1$ then T2(make $\frac{x}{y^2\plus{}yz\plus{}z^2}$ into $\frac{x^2}{xy^2\plus{}xyz\plus{}xz^2}$)

$\frac{(x+y+z)^2}{(xy+yz+zx)(x+y+z)}  \geq \frac{3}{x+y+z}$

too simple game

btw by coincidence I also proved sqing's one

wait I didn't have to convert $xyz=1$ :read:

[quote=Mateescu Constantin]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac{x}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{y}{z^2\plus{}zx\plus{}x^2}\plus{}\frac{z}{x^2\plus{}xy\plus{}y^2}\ge \frac{3}{x\plus{}y\plus{}z}$ .[/quote]

Bad proof:

<=>

\[\frac{1}{2}\sum{(x-y)^2(2y^4+5xy^3+xz^3+6x^2y^2+4x^3y)}+\frac{3}{2}\sum{(x^2-yz)^2xz}\]

[quote=xzlbq][quote=Mateescu Constantin]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac{x}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{y}{z^2\plus{}zx\plus{}x^2}\plus{}\frac{z}{x^2\plus{}xy\plus{}y^2}\ge \frac{3}{x\plus{}y\plus{}z}$ .[/quote]

stronger

\[{\frac {x}{4\,{y}^{2}+yz+4\,{z}^{2}}}+{\frac {y}{4\,{z}^{2}+zx+4\,{x}^
{2}}}+{\frac {z}{4\,{x}^{2}+xy+4\,{y}^{2}}}\geq  \left( x+y+z \right) ^{-1
}\][/quote]

again T2

$\frac{(x+y+z)^2}{4(xy+yz+zx)(x+y+z)-9xyz} \geq \frac{1}{x+y+z}$

equality case is $x=y=z$ so we can apply our "lovely" formula

$x+y+z=3$
$xy+yz+zx=3-3t^2$
$xyz \leq (1+2t)(1-t)^2$

$\frac{27}{36(1-t^2)-9(1+2t)(1-t)^2} \geq 1$

and the denominator decreases in $t \in [0,1)$

proved!

[quote=xzlbq][quote=Mateescu Constantin]Let $ x,y,z$ be positive real numbers . Prove that :

$ \frac{x}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{y}{z^2\plus{}zx\plus{}x^2}\plus{}\frac{z}{x^2\plus{}xy\plus{}y^2}\ge \frac{3}{x\plus{}y\plus{}z}$ .[/quote]

stronger

\[{\frac {x}{4\,{y}^{2}+yz+4\,{z}^{2}}}+{\frac {y}{4\,{z}^{2}+zx+4\,{x}^
{2}}}+{\frac {z}{4\,{x}^{2}+xy+4\,{y}^{2}}}\geq  \left( x+y+z \right) ^{-1
}\][/quote]
Cauchy and schur inequality

then the general form:

prove that:$\sum_{cyc}\frac{x}{ky^2+yz+kz^2} \geq \frac{9}{(2k+1)(x+y+z)}$ for all positive real number $k$

I don't know when it's 0 or even negative... but would be apparent in positive

but the ways I used would be rather boring in this case so waiting for some exciting ones

[quote=luofangxiang]k>4 wrong[/quote]

proof?

[quote=luofangxiang][quote=kuing][quote="arqady"]kuing, what about the following inequality?
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ xy + xz + yz\neq0.$ Prove that:
\[ \frac {x}{y^2 + yz + z^2} + \frac {y}{z^2 + zx + x^2} + \frac {z}{x^2 + xy + y^2}\ge{\frac {2}{\sqrt {xy + xz + yz + \sqrt [3]{x^2y^2z^2}}}}\]
:wink:[/quote]

from 2#'s ineq $ \sum_{cyc}\frac {x}{y^{2} + yz + z^{2}}\ge\frac {\sum_{cyc}x}{\sum_{cyc}xy}$, just need to prove that
\[ \left( {x + y + z} \right)^2 \left( {xy + yz + zx + \sqrt [3]{{x^2 y^2 z^2 }}} \right) \ge 4\left( {xy + yz + zx} \right)^2 ,\]
$ \iff f(v^2) = - 36v^4 + 27u^2v^2 + 9u^2w^2 \ge 0$

just need to prove that when $ y = z$.

setting $ x = a^3,y = z = b^3,a\ge 0,b > 0$, ineq becomes

$ \frac {a^2 (a-b)^2 \left(2 a^5 + 5 a^4 b + 8 a^3 b^2 + 4 a^2 b^3 + 4 a b^4 + 4 b^5\right)}{b^6 (a + b) \left(2 a^2 - a b + b^2\right) \left(2 a^3 + b^3\right)^2} \ge 0,$

true... :blush:[/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy767XdyBRh9b&690[/img][/quote]

$x,y,z\geq 0,xy+yz+zx=1,$
\[x^2+y^2+z^2+\frac{9xyz}{x+y+z}\geq 2\]

[quote=xzlbq]
$x,y,z\geq 0,xy+yz+zx=1,$
\[x^2+y^2+z^2+\frac{9xyz}{x+y+z}\geq 2\][/quote]

It's Shcur Inequality

$x,y,z$ are positive reeal numbers so we use Titu's lemma:
$\frac { { x }^{ 2 } }{ x({ y }^{ 2 }+yz+{ z }^{ 2 }) } +\frac { { y }^{ 2 } }{ y(z^{ 2 }+zx+{ x }^{ 2 }) } +\frac { { z }^{ 2 } }{ z({ x }^{ 2 }+xy+{ y }^{ 2 }) } \ge \frac { { (x+y+z) }^{ 2 } }{ xy(x+y)+yz(y+z)+zx(x+z)+3xyz } \ge \frac { 3 }{ x+y+z } \Longleftrightarrow $

${ (x+y+z) }^{ 3 }\ge 3[xy(x+y)+yz(y+z)+zx(z+x)+3xyz]\Longleftrightarrow$

${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\ge 3xyz$

which is trivial.

[quote=xzlbq][quote=xzlbq]

$x,y,z\geq 0,xy+yz+zx=1,$
\[x^2+y^2+z^2+\frac{9xyz}{x+y+z}\geq 2\][/quote]

$x,y,z\geq 0,xy+yz+zx=1,$
\[x^2+y^2+z^2+\frac{9xyz}{x+y+z}\sqrt{\frac{\sqrt{3}}{x+y+z}}\geq 2\][/quote]

[img]http://s4.sinaimg.cn/middle/006ptkjAzy769vbJ3sTb3&690[/img]

[quote=xzlbq][quote=luofangxiang]
[img]http://s4.sinaimg.cn/middle/006ptkjAzy769vbJ3sTb3&690[/img][/quote]

\[\frac{x^2+y^2+z^2}{xy+yz+zx}+\frac{8xyz}{(x+y)*(y+z)*(z+x)}\geq2+4\frac{(x-y)^2(y-z)^2(z-x)^2}{(x+y)^2(y+z)^2(z+x)^2}\]

$\in B$ type,stronger[/quote]

[img]http://s8.sinaimg.cn/middle/006ptkjAzy77b0XxDHV27&690[/img]


It takes Franz 45 minutes to walk between her home and school. One morning she walked half way to school and remembered that she had left her calculator at home. She ran home and then ran all the way to school. It took 5 minutes at home to get her calculator. She runs twice as fast as she walks. How many minutes more than usual did it take her to get to school? Express your answer as a mixed number.

Franz walks for 45/2=22.5 minutes.  Then, she takes 22.5/2=11.25 minutes to run home.  She takes 5 minutes to get her calculator, and then runs to school in 45/2=22.5 minutes.  In total, it takes her 22.5+11.25+5+22.5=61.25 minutes to get to school.  Then, it takes her 61.25-45=16.25=16 1/4 minutes longer than usual for Franz to get to school.


Prove that in any set of 30 natural numbers there will exist any  of these numbers that is divisible by 30 or some of them such that their difference is divisible by 30 .

If 2 numbers have same remainder mod 30, then their difference is 0 mod 30.
If not, then they have remainders: 0,1,2,...,28,29 so 1 number is 0 mod 30.


Let a,b,c be any real number and a,b,c>0,then prove that
$ 3(a+ \sqrt {ab}\sqrt [3]{abc})\leq (8+\frac {2\sqrt {ab}}{a+b}} ){\sqrt[3]{a\frac{a+b}{2}\frac{a+b+c}{3}}$

A counterexample : $ a\equal{}b\equal{}c\equal{}3$.

I think you should rewrite
$ 3(a + \sqrt {ab}+\sqrt [3]{abc})\leq (8 + \frac {a+b}{2\sqrt {ab}} ){\sqrt [3]{a\frac {a + b}{2}\frac {a + b + c}{3}}}$

[quote="bunhiacovski"]I think you should rewrite
$ 3(a + \sqrt {ab} + \sqrt [3]{abc})\leq (8 + \frac {a + b}{2\sqrt {ab}} ){\sqrt [3]{a\frac {a + b}{2}\frac {a + b + c}{3}}}$[/quote]
If so, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1820

If $ x,y,z$ be positive real numbers, then

$ 7(x + \sqrt {xy}+\sqrt [3]{xyz})\leq 3(5 + \frac {4\sqrt {xy}}{x+ y}}){\sqrt [3]{x\cdot \frac {x + y}{2}\cdot \frac {x + y + z}{3}}.$

[quote="yunxiu"]If $ x,y,z$ be positive real numbers, then

$ 7(x + \sqrt {xy} + \sqrt [3]{xyz})\leq 3(5 + \frac {4\sqrt {xy}}{x + y}}){\sqrt [3]{x\cdot \frac {x + y}{2}\cdot \frac {x + y + z}{3}}.$[/quote]
sorry,I think you are wrong  :)

used the inequality know

$ \sqrt [3]{x.\frac {x + y}{2}.\frac {x + y + z}{3}}\ge \frac {x + \sqrt {xy} + \sqrt [3]{xyz}}{3}$

the inequality $ \Leftrightarrow (\sqrt {x} - \sqrt {y})^2\le 0$ (wrong)

[quote="bunhiacovski"][quote="yunxiu"]If $ x,y,z$ be positive real numbers, then

$ 7(x + \sqrt {xy} + \sqrt [3]{xyz})\leq 3(5 + \frac {4\sqrt {xy}}{x + y}}){\sqrt [3]{x\cdot \frac {x + y}{2}\cdot \frac {x + y + z}{3}}.$[/quote]
sorry,I think you are wrong  :)

[/quote]
What your counterexample, bunhiacovski? Thank you!

sorry，I'm wrong.It should be:
[quote="xiaohui"]Let a,b,c be any real number and a,b,c>0,then prove that
$ 3(a + \sqrt {ab}+\sqrt [3]{abc})\leq (8 + \frac {2\sqrt {ab}}{a + b}} ){\sqrt [3]{a\frac {a + b}{2}\frac {a + b + c}{3}}$[/quote]


NOTE:  What is the Factor Theorem in simple terms?

Use the factor theorem to prove that x + c is a factor of
x^n + c^n if n is > or = to 1 is an odd integer.

The simplest form of the Factor Theorem states that if $f(a) = 0$ then $x-a$ is a factor of $f(x)$.

Obviously, you are defining the posted theorems using textbook jargon.  I can simply read this stuff in my math book and go nowhere.  Do you know why?

I need someone to explain it to me in simple terms.

Can you help me with this question or not?

If we have any expression of the form, say, $f(x) = (x-a)(x-b)$ then clearly

$f(a) = (a-a)(a-b) = 0 \cdot (a-b) = 0$
$f(b) = (b-a)(b-b) = (b-a) \cdot 0 = 0$

In other words, [b]If $(x-a)$ is a factor of $f(x)$ then $f(a) = 0$.[/b]

The Factor Theorem merely states that the reverse is also true for [b]polynomials[/b] $f(x)$.  For example, given

$f(x) = x^{2}+4x-5$

We observe that

$f(1) = 1+4-5 = 0$

By the Factor Theorem, we can conclude that $x-1$ must divide the polynomial.  And indeed,

$f(x) = (x-1)(x+5)$

If we want to figure out whether $x+c$ divides $f(x) = x^{n}+c^{n}$ then we need to check what $f(-c)$ is (since $x+c = x-(-c)$.  Well,

$f(-c) = (-c)^{n}+c^{n}$

For even $n$ this becomes $2c^{n}$, but for odd $n$ this becomes

$f(-c) =-c^{n}+c^{n}= 0$

So indeed $x+c$ divides $x^{n}+c^{n}$.

All well and good, but you asked why this is true.  The answer is not too simple, but I will try to explain it in simple terms.

We need to perform [b]polynomial division.[/b]  This is like regular division, only everything - the divisor, the dividend, the quotient, the remainder - is a polynomial.

Recall how normal division happens.  For example, if I want to divide $41$ by $5$, one way of expressing this division is as

$41 = 5 \cdot 8+1$

Where

- $41$ is the [b]dividend.[/b]
- $5$ is the [b]divisor.[/b]
- $8$ is the [b]quotient.[/b]
- $1$ is the [b]remainder.[/b]

Notice that a property of the remainder is that [b]it must be less than the divisor[/b], or else we could divide into it again.

We're going to keep this pattern, but we're going to apply it to polynomials.  In the above example of $f(x) = x^{2}+4x-5$, that is our dividend and $x-1$ is our divisor.  The division pattern takes the form

$x^{2}+4x-5 = (x-1) Q(x)+R(x)$

Where

- $x^{2}+4x-5$ is the [b]dividend.[/b]
- $x-1$ is the [b]divisor.[/b]
- $Q(x)$ is the [b]quotient.[/b]  (We don't know it yet.)
- $R(x)$ is the [b]remainder.[/b]  (We don't know that yet either.)

Now, one of the properties of the remainder polynomial $R(x)$ is that it [b]must have degree less than that of the divisor[/b], or it is $0$. That's because otherwise we could divide into it again, just like with numbers.

In this case, the divisor is of degree $1$, so $R(x)$ must be of degree zero - in other words, it is a constant $R$.

However, if in the division expression we let $x = 1$, we obtain

$1-4+5 = 0 = (1-1) \cdot Q(x)+R = R$

So $R = 0$.  In other words,

$x^{2}-4x+5 = (x-1) Q(x)+0$

So $x-1$ evenly divides $x^{2}-4x+5$ as we wanted.

That is a good explanation that[b] anyone[/b] should understand.

He said it clearly the first time too.


construction
geometry
IMO Shortlist
IMO Longlist

Given a point $P$ lying on a line $g,$ and given a circle $K.$ Construct a circle passing through the point $P$ and touching the circle $K$ and the line $g.$

We draw a line $d$ through the center of the circle (we can construct it as the intersection of two chords if it's not given) and perpendicular to the given line $g$. $d$ cuts the circle at a two points. Through one of them we construct a line $g'\|g$. $g'$ is parallel to $g$ and tangent to $K$. We may now construct the images of two points on $g'$ under the inversion of center $P$ which preserves $K$. Call the two images $X,Y$. The circle $(PXY)$ is what we are looking for.

Trace a generic circle C tangent in P to g. If R and S are the common points to K and C let T=RS^g. Let T1 and T2 be the point where tangents from T to K touch K.
Let b1 and b2 the angle bisectors of <PTT1 and <PTT2 respectively. The intersections of b1 and b2 with the perpendicular to g in P give us the centers of the two circles wich are solution of the problem.


vector
abstract algebra
algebra
polynomial
linear algebra
linear algebra theorems

suppose T is a linear operator defined on a n dimensional vector space. if T is nilpotent show that T^n=0

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83234[/url]
my answer in this topic (your own topic) establishes that if the kernel stops rising, it never goes up after that

that should do it

Nilpotent means all the eigenvalues are 0. That means the characteristic polynomial is $x^n$. The minimal polynomial always divides the characteristic polynomial (Cayley-Hamilton theorem), so surely $T^n$ must be 0.


function
limit
integration
calculus
calculus computations

find  :D

$ \prod_{n\equal{}1}^\infty \; \dfrac {\left(1\plus{}\dfrac{20}{n}\right)} {\left(1\plus{}\dfrac{7}{n}\right)\cdot \left(1\plus{}\dfrac{13}{n}\right)}$

[quote="misan"]find  :D

$ \prod_{n \equal{} 1}^\infty \; \dfrac {\left(1 \plus{} \dfrac{20}{n}\right)} {\left(1 \plus{} \dfrac{7}{n}\right)\cdot \left(1 \plus{} \dfrac{13}{n}\right)}$[/quote]
just note that :$ 7/n\plus{}13/n\equal{}20/n$ and use gamma function property . :wink:

[quote="Yuriy Solovyov"]$ \prod_{n \equal{} 1}^{\infty}{\frac {(n \plus{} 0)(n \plus{} 20)}{(n \plus{} 7)(n \plus{} 13)}} \equal{} \frac {\Gamma (8)\Gamma (14)}{\Gamma (1)\Gamma (21)}$[/quote]
Is this a property? How's this generally?

If $ \alpha_1 \plus{} \alpha_2 \plus{} ... \plus{} \alpha_k \equal{} \beta_1 \plus{} \beta_2 \plus{} ... \plus{} \beta_k$ and any of $ \beta_i$ is not a negative integer then :

$ \prod_{n \equal{} 1}^{\infty}{\frac {(n \plus{} \alpha_1)(n \plus{} \alpha_2)...(n \plus{} \alpha_k)}{(n \plus{} \beta_1)(n \plus{} \beta_2)...(n \plus{} \beta_k)}} \equal{} \frac {\Gamma (1 \plus{} \beta_1) \Gamma (1 \plus{} \beta_2)...\Gamma (1 \plus{} \beta_k)}{\Gamma (1 \plus{} \alpha_1) \Gamma (1 \plus{} \alpha_2) ... \Gamma (1 \plus{} \alpha_k)}$

This is easy to prove using formula : $ \Gamma (x) \equal{} \lim_{m\to\infty}{\frac {m^{x} m!}{x(x \plus{} 1)(x \plus{} 2)...(x \plus{} m)}}$

[quote="Yuriy Solovyov"]
This is easy to prove using formula : $ \Gamma (x) \equal{} \lim_{m\to\infty}{\frac {m^{x} m!}{x(x \plus{} 1)(x \plus{} 2)...(x \plus{} m)}}$[/quote]
Ahh, that's pretty nice! And how'd you prove this last result?

We know that $ \Gamma (z) \equal{} \int_{0}^{\infty}{e^{ \minus{} t}t^{z \minus{} 1}dt}$.
And $ e^{ \minus{} t} \equal{} \lim_{n\to\infty}{\left(1 \minus{} \frac {t}{n}\right)^{n}}$.

Now try to calculate integral : $ P_{n}(z) \equal{} \int_{0}^{n}{\left(1 \minus{} \frac {t}{n}\right)^{n}t^{z \minus{} 1}dt}$

Put $ t \equal{} nx$ we can write : $ P_{n}(z) \equal{} n^{z}\int_{0}^{1}{(1 \minus{} x)^{n}x^{z \minus{} 1}dx}$
Now use partial integration : $ \int_{0}^{1}{(1 \minus{} x)^{n}x^{z \minus{} 1}dx} \equal{} \frac {n}{z}\int_{0}^{1}{(1 \minus{} x)^{n \minus{} 1}x^{z \minus{} 1}dx} \equal{} \frac {n}{z(z \plus{} 1)}\int_{0}^{1}{(1 \minus{} x)^{n \minus{} 1}dx^{z \plus{} 1}}$
And again : $ \int_{0}^{1}{(1 \minus{} x)^{n}x^{z \minus{} 1}dx} \equal{} \frac {n(n \minus{} 1)}{z(z \plus{} 1)}\int_{0}^{1}{(1 \minus{} x)^{n \minus{} 2}x^{z \plus{} 1}dx}$

In general we get : $ P_{n}(z) \equal{} \int_{0}^{n}{\left(1 \minus{} \frac {t}{n}\right)^{n}t^{z \minus{} 1}dt} \equal{} \frac {1\cdot 2\cdot ... \cdot n}{z(z \plus{} 1)...(z \plus{} n)}n^z$

And when $ n\to\infty$ : $ P_{n}(z)\to \Gamma (z)$  :D

This formula was proved by Euler.

Nice product [b]misan[/b] and nice solution [b]Yuriy Solovyov[/b]

Actually i dont agree that 13 is the number that brings unluckyness! :D  its my favourite number and a beautiful one!


Every space point is colored with exactly one of the colors red, green, or white. The sets A, B, C consist of the lengths of those segements in space with both endpoints red, green, anhd white, respectively. Show that at least one of these sets contains all nonnegative real numbers.

I think i have seen this problem somewhere and I think this is true for the plane


inequalities
function
triangle inequality
calculus
calculus computations

Let $ (M,d)$ and $ (N,\rho)$ be metric spaces, and define a function $ d_{2}$ on $ M\times N$ by $ d_{2}((a,x),(b,y))\equal{}\sqrt{d(a,b)^{2}\plus{}\rho(x,y)^{2}}$.  Show that this satisfies the triangle inequality.  That is, $ d_{2}((a,x),(c,z))\leq d_{2}((a,x),(b,y))\plus{}d_{2}((b,y),(c,z))$.

Since you don't know anything about metric spaces, there isn't much you can do to estimate $ \sqrt{d(a,c)^2 \plus{} d(x,z)^2}$ except to use the triangle inequality:
$ \sqrt{d(a,c)^2 \plus{} d(x,z)^2}\le \sqrt{(d(a,b)\plus{}d(b,c))^2 \plus{} (d(x,y)\plus{}d(y,z))^2}$. Now the problem reduces to an inequality of the form
$ \sqrt{(A_1\plus{}B_1)^2 \plus{} (A_2\plus{}B_2)^2}\le  \sqrt{A_1^2\plus{}B_1^2}\plus{}\sqrt{A_2^2\plus{}B_2^2}$
which you are probably familiar with.

Thanks.  My problem was how to get the initial expression in terms of the final one, and your inequality did the trick (I was actually not familiar with that inequality, which is probably why I couldn't figure this out).


trigonometry
trig identities
Law of Cosines

:( i need help with the following problems using the inscribed quad.

1. in terms of x, write a simplified expression for cos B
2. in terms of x, write a simplified expression for cos D
3. find x

Try using the law of cosines:

$x^2=6^2+9^2-2\cdot6\cdot9\cos B$

Similarly

$x^2=2^2+7^2-2\cdot2\cdot7\cos D.$

Oh, and one more thing: $\cos B=-\cos D,$ since the angles are supplementary.

Where do these expressions lead you?

[hide]Use the law of cosines.

1. $\cos B=\frac{6^2+9^2-x^2}{2*6*9}=\frac{117-x^2}{108}$

2. $\cos D=\frac{2^2+7^2-x^2}{2*2*7}=\frac{53-x^2}{28}$

3. Since $\angle B$ and $\angle D$ intercept arcs which combine to make the whole circle, $2\angle A+2\angle B=360^\circ \Rightarrow$ $\angle B$ and $\angle D$ are supplementary. Then their cosines are opposites.

$\frac{117-x^2}{108}=\frac{x^2-53}{28}\Rightarrow \frac{117-x^2}{27}=\frac{x^2-53}{7}$

$\Rightarrow 819-7x^2=27x^2-1431\Rightarrow 34x^2=2250\Rightarrow x^2=\frac{1125}{17}$

$\Rightarrow x=\boxed{\frac{15\sqrt{85}}{17}}$[/hide]


calculus
integration
trigonometry
calculus computations

Show by finding the integral in two different ways, that for constant C and D

$\int sinxcosx\,dx = \frac{1}{2}sin^{2}x+c=\frac{-1}{4}cos2x+D$

b)How may the two answers be reconciled?

What does part B mean?

It means:  the answers look different, but they are supposed to be equal.  How so?

And the answer is that (edited:  :oops: )

$\cos 2x = 1-2 \sin^{2}x$

Of course, t0rajir0u meant to say

$\cos 2x=1-2\sin^{2}x$

Or maybe

$\sin^{2}x=\frac{1-\cos 2x}{2}$

ok thanks for your helps.

p.s Kent Merryfield are you a phd professor?


How many different three-digit security codes are possible using the digits 1-5, if the second digit cannot be the same as the first, and the third digit cannot be the same as the second?

There are $ 5*4*4 \equal{} \boxed{80}$ security codes.

There are $ 5$ possibilities for the first digit, $ 4$ for the second digit, and $ 4$ for the last digit. Thus, we have $ 5\times{4}\times{4}\equal{}\boxed{80}$.

Gaussintraining, although your input is helpful, try to restrain from reposting answers to questions that have already been resolved. :) it can get kind of annoying for some people, as well as the moderators.

When $LaTeX$ first posted, there was something incorrect with his solution, so I posted the correct answer. If you'll look at his post, he edited after I had posted. Thanks anyway though  :)

[hide="answer"]there are:

5 possiblities =1st lock #
4possibilites =2nd lock #
4possibilites =3rd lock #

if you're wondering why the 3rd isn't 3 possibilities is because the problem said that the lock # cannot be the sane as the 2nd # but it CAN be the same as the 1st #

so the equation is 5*4*4 =80[/hide]

[b][i][u]remember to rate my post ! thanks[/u][/i][/b]

but please be fair  thanks [=

 :bomb:   :bomb:   :bomb:   :bomb:

[hide]5x4x4=80 Answer 80[/hide]
Problem and solution above.

What do you mean? I'm not the only one doing it, Half of these problems more than 1 person gave a solution or answer. :)

[quote=bluecarneal]DO NOT re-answer questions that have already been answered.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=503&t=221525]READ THIS before posting in this forum.[/url][/quote]

"-- Do not make a post saying, 'Hey! That's spamming! Don't do it, __________!'"

Sorry about the necro but I found this extremely hilarious


probability
geometry
USAMTS
LaTeX
number theory
greatest common divisor
absolute value

First and most, I just realized that I have a trouble with my schedule on March 7th and early March so that I may not be able to hold the contest during that date. I'm going to be on my local math meet and I'd like to get prepare for it.

Yet, I do not want to end my COP 3rd without beginning it so I'll post the Round 1 Problems which I collected sometimes ago.

[b]Answers are due to me 2/25/05[/b]

Rules are simple for this one. There are 5 questions. You're given about a week to do this. If you get ALL right, you'll be going to next round. If you get at least ONE wrong, you're out.

So, I suggest you use your time wisely and checking your answer carefully.

Most of all, good luck to you all!

-----------------------------------------------------
1. Six distinct points are plotted in a circle. Connecting some or all points, polygons are formed. This number is denoted as $m$. If there was one more point, the number would've increased by $n$ and $\frac {m}{m+n} = \frac {a}{b}$ with $(a,b) = 1$, find $a+b$.

2. A number is called "intelligent" if and only if it is:

\[\text {i. 3-digit positive real number}\]
\[\text {ii. In form of abc with}\] \[a<b<c\]
\[\text {iii. a+b=c}\]

Denote $S$ as the sum of all intelligent numbers. Let $D$ equals the sum of all positive divisors of $S$. Find the remainder when $D$ is divided by $S$.

3. Two points are randomly and simulataneously chosen from 5 x 4 grid of 20 lattice points. The probability that the two points have rational distance is $\frac {a}{b}$ where $(a,b) = 1$. If $\frac {a+b}{b-a} = \frac {p}{q}$ in lowest terms, find $p+q$.

4. Twelve points are equally spaced on the circumference of a circle of radius 1. What $a+b$ if the distance between each of adjacent points is in form of $\frac {a \pi}{b}$ where $a,b$ may or may not equal to 1.

5. The area between two concentric circles are called annulus. Suppose that annulus has area of 103. Find the absolute value of differences between the two concentric circles' radii.
----------------------------------------------------

Note: These problems are made harder than what I expected. You may use calculator, mathworld, and any tools you need exception of other people.

Another Note: Even you didn't sign-up, you may still send the answers in and you don't have to wait until the deadline. You can always send in early.

Yes. Can we use calculators or books or websites?

Books, calculators, website such as (mathworld.com, etc..) are perfectly accepted.

But OTHER PEOPLE or ONLINE TUTOR is not accepted. Those who did USAMTS should know what this means.

For answers, yes, you send me the answers using private message.

Please try not to split the problems. I have quite number of contestants and it gets hard for me to keep track of qualifiers if there are separate posts for each person.

And make sure label the answers with Round 1 or something like that.

Thanks and Regards,

Silverfalcon

What do you mean by decimals?

Do you mean its positive divisors? Those divisors would be integers.

If you mean the answer, answer is an integer.

I mean, can a non-integer number such as 3.58 be considered intelligent?
If not, you should word condition i as 'three-digit positive integer' as opposed to 'three-digit positive real'.

Yes.

3-digit integer is better description.

[quote="sky9073"]<AEC = 60 in number 4 doesn't seem right.... or is it just me?[/quote]

Yeah, I got that too.  I pmed Silverfalcon, but he wanted to see if more people would realize this before changing the problem.

[quote="Silverfalcon"]

4. Twelve points are equally spaced on the circumference of a circle of radius 1. What $a+b$ if the distance between each of adjacent points is in form of $\frac {a \pi}{b}$ where $a,b$ may or may not equal to 1.[/quote]

Do you want the linear straight line distance?  Or the distance going along the circle (like an arc)?

I'm guessing since there's a pi, he wants it as an arc.. but i dont wanna take any risks

for number 5.. do you want it to be 103(pi) or just 103?

Do you want us to send you the final answers only? I still do not not how to use latex.

For #4, the $a$ and $b$ are in lowest terms.

So, please simplify them before adding those two.

You don't need to use Latex to answer any of these questions. I think you can just use your number pad in computer. :)

Even after the extended date, I have not received anymore answers. :(

Please send your answers to COP Round 1.

I'll be appreciated.

Remember, you only need 5 answers. Just check your work and send them.

I might send the remind private message again.

[quote="Silverfalcon"]
4. Twelve points are equally spaced on the circumference of a circle of radius 1. What $a+b$ if the distance between each of adjacent points is in form of $\frac {a \pi}{b}$ where $a,b$ may or may not equal to 1.[/quote]

Do you mean that the GCD of a and b may or may not be equal to 1?  If that is the case, wouldn't there be an infinite number of possible answers?

Ok.

$a,b$ are simplified.

Sorry for the confusing wording.

Can someone explain this competition to me, or tell me where I can find more information?

Jimmy, here, read this thread: :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24488

The Round 1 for the COP 3rd has not ended yet. You can still solve the problems and send them to me! :)

In the first question, does $(a,b)$ mean $a\choose{b}$?

No, it just means that $a$ and $b$ are relatively prime.

Meaning, you just have to simplify $\frac{a}{b}$

[quote="Jimmy"]When are the problems due, and do I PM them to you?[/quote]

It's due this Friday and you pm your answers to me.

Yes, $(a,b) = 1$ states $a$ and $b$ are relatively prime integers.

Do you want just answers, or do we have to show our work?

Just 3 more questions, in #5, do you want integer answers?  (I don't think the problem has a unique solution otherwise).

In #1 the six points are not [b]on[/b] the circle right?  And is the maximum number of polygons formed when the points are connected?

Thanks.

Points are on the circles. That's how you make the polygons.

Yes, the number is maximum.

For the integer part, um, the answer is integer. BIG HINT:

[size=150]Answer is from 000 to 999[/size] :D  :D

Please note that the answer is due Friday.

Sorry for sending the first one before it was finished Silverfalcon. That was my bad--pressed Submit instead of Preview by accident lol.

white_horse_king88: Don't worry heh. You'll do fine.


factorial
floor function
number theory
prime factorization

How many consecutive zeros are at the end of 32 factorial?

$\lfloor{32/5}\rfloor+\lfloor{32/5^{2}}\rfloor = 7$.

There are $7$ zeros at the end.

We get a zero at the end of a number if it has a factor of both $2$ and $5$.  I hope you can see that there are more $2$s than $5$s in a factorial, so we are going to count the powers of $5$.

There are $\lfloor \frac{32}{5}\rfloor = 6$ multiples of five from $1$ to $32$, and since

$32! = 1 \cdot ... \cdot 5 \cdot ... \cdot 10 \cdot ...$

It follows that each multiple of five contributes $6$ factors of $5$ to the prime factorization of $32!$.

However, there are also $\lfloor \frac{32}{25}\rfloor = 1$ multiple of $25$.  This will contribute an extra factor of $5$, so we add it in.

[quote="physics"]32 factorial has 7 consecutive zeros.[/quote]

welcome physics,

please do not just post the answers...we are more intrested in the solution

thanks,
jli

[quote="now a ranger"]25 has two zeros

and 5 10 15 20 and 30 each with 1. so that's 7
Let's say 2006 factorial. It would be 500 consecutive zeros because some numbers ending in 5 have more 0s than others.[/quote]
Actually it has 499 consecutive zeros...

no, it is 500...

unless I have been really stupid

and i saw that on a math practice anyway

yeah i think bpms forgot one of them.

I've seen a question asking the number of consecutive zeros twice this year, once in practice, once on a real test.

[quote="Klebian"]I'm getting 500... 401 + 80 + 16  + 3[/quote]
Oh, my mistake I did 400+80+16+3...


algorithm
number theory
Euclidean algorithm

"A 3-letter word has been encrypted using the RSA scheme with public keys n = 4717 and e = 2961. The encrypted message is 4110, 3555, 1925. Find primes p and q with n = p x q (done, p = 53 and q = 89). Use the Euclidean algorithm to find d such that d x e = 1 mod (p - 1)(q - 1) (done, d = 17). [b]Decrypt the message by calculating the standard representative of x^d mod n for x = 4110, 3555 and 1925 and converting your answers to letters using A = 1, B = 2....Z = 26[/b]"

I've done the first couple of parts of this question, but am stuck on the last part. There is a similar worked example on Wikipedia which I'm trying to follow but the numbers are HUGE. The formula I should be using is:

m = c^d mod n

Where m is the message I want, c is the encryption (i.e. 4110, 3555 and 1925). But d is 17 so c^d = 4110^17 which is huge. Help!

You don't need to calculate $ x^d$ to calculate $ x^d \bmod n$.  Instead, you can calculate $ x^2 \bmod n$, multiply that by $ x$ to calculate $ x^3 \bmod n$, and so forth.

An even faster method is available called [url=http://en.wikipedia.org/wiki/Exponentiating_by_squaring]exponentiation by squaring[/url], which is how PowerMod is usually implemented in most programming languages.

OK I just did it but I think I used a different method altogether.

I rewrote $ 4110^{17} mod 4717$ as $ (4110^2)^8 * 4110 mod 4717$ and repeated those iterations until I got down to a number in the range of 1 to 26.

Christ that took me AGES to get to grips with.


abstract algebra
induction
topology
superior algebra
superior algebra unsolved

Let $ I$ be a nilpotent ideal in a commutative ring $ R$, let $ M$ and $ N$ be $ R$-modules and let $ \varphi: M \to N$ be an $ R$-module homomorphism. Show that if the induced map $ \bar{\varphi} : M/IM \to N/IN$ is surjective, then $ \varphi$ is surjective.

Let $ K\equal{}N/Im(\phi)$.  We have the following exact sequence:
$ M\rightarrow N\rightarrow K\rightarrow 0$.
Tensoring with $ R/I$ gives the exact sequence
$ M/IM\rightarrow N/IN\rightarrow K/IK\rightarrow 0$.
By hypothesis, the first part is surjective, so $ K\equal{}IK$, which since $ I$ is nilpotent, implies $ K\equal{}0$, that is,
$ N \equal{} Im(\phi)$.

Steve

Sorry, this is an exercise in the chapter of Dummit and Foote introducing module homomorphisms and quotients, so the tensor product hasn't been introduced (in fact, neither have exact sequences; I happen to know what those are but I don't know anything about the tensor product). Could you translate that into a more elementary argument?

If you want to tough it out, you can probably translate it yourself.  If M → N → K → 0 is exact, then I think you can prove directly that M/IM → N/IN → K/IK → 0 is exact.  Once you have done that, you've eliminated all use of the tensor product. Most of the "exact sequence" stuff is just symbolic shorthand for a couple of equalities, K is just a name for N/φ(M), etc. so there is no trouble translating that when you turn in a solution or otherwise discuss the work with other people still learning that chapter.

Really the key part is that N = φ(M) + IN and I is nilpotent.  N = φ(M) + I(φ(M) + IN) = φ(M) + IIN, etc.

The exact sequences just keep track of everything nicely, and the tensor product is just a convenient way to remember that if you have an exact sequence, then you can put "bars" on all the modules to denote M/IM style quotient modules and still stay exact on the right.

It is not true that $ I(N/\varphi(M)) \equal{} IN/\varphi(M)$, because we do not necessarily have $ \varphi(M) \subset IN$.

This should read $ I(N/\varphi(M)) \equal{} IN/(\varphi(M) \cap IN)$, but then we cannot conclude $ K/IK\equal{}N/IN$.

More down-to-earth: For every $ n\in\mathbb{N}$, prove (by induction) that the induced map $ \bar{\varphi}: M\slash I^nM\to N\slash I^nN$ is surjective.

  darij

[quote="JustSee12"]One more thing: I'm having a little difficulty with the symbol $ K/IK$. Doesn't that just mean $ K/IK \equal{} (N/\varphi(M)) / I(N/\varphi(M)) \equal{} (N/\varphi(M)) / (IN/\varphi(M)) \cong N/IN$?[/quote]

LydianRain already pointed out the problem, but I prefer to write it as K/IK = (N/φ(M)) / I(N/φ(M)) = (N/φ(M)) / ((IN+φ(M))/φ(M)) ≅ N/(φ(M)+IN).

You know that N = φ(M)+IN, and then you can rewrite the second N in the equality using the equality to get N =  φ(M)+I( φ(M)+IN ) = φ(M)+IIN, and then do it again, and again, etc.

For what it is worth, I record a straightforward generalization (which I "invented" in order to prove a theorem in Etingof's representation theory notes, but which turned out to be totally unrelated):

Let $ I$ be a nilpotent left ideal in a (not necessarily commutative) ring $ R$. Let $ M$ and $ N$ be two left $ R$-modules, and let $ \varphi: M \to N$ be a left $ R$-module homomorphism. Show that if the induced map $ \bar{\varphi} : M/IM \to N/IN$ is surjective, then $ \varphi$ is surjective.

  darij

[quote="Hanno"]It seems sufficient that $ M,N$ are $ I$-adically complete and Hausdorff instead of assuming that $ I$ is nilpotent.[/quote]

Yes, it is even enough to require that $ N$ is Hausdorff with respect to the $ I$-adic topology (i. e., we have $ \bigcap_{n\in\mathbb{N}}I^nN\equal{}0$), and that every Cauchy sequence in $ M$ converges to a (not necessarily unique) limit.

This is mildly useful in proving the (formal-series) Weierstrass preparation theorem, but it still doesn't make the proof trivial.

  darij


algorithm
combinatorial geometry
calculus
calculus computations

So I'm new to this problem: does there exist a formula for volume of a convex hull determined by say m given points in n-dimensional space? and yes, I googled it to no avail (yet I found many algorithms for such a computation). Or is there even a 2- or 3-d version of this formula?

Thanks,

   --Ben

Split the hull into simplexes, calculate the volumes of those, and sum them up. In two dimension the formula is known as Pick's theorem (or Pick's formula).

Not Pick's Theorem - that's about counting lattice points in the interior, and benorin didn't want to go in there. For the correct two-dimensional formula, see perfect_radio's first post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87063]this topic.[/url]

[quote="benorin"]So I'm new to this problem: does there exist a formula for volume of a convex hull determined by say m given points in n-dimensional space? and yes, I googled it to no avail (yet I found many algorithms for such a computation). Or is there even a 2- or 3-d version of this formula?
[/quote]

I think there is no reasonable formula of such type. In 2- or 3-dimensional case you can write useless formula by using Heron formula (or its analog for 3-dimensional situation) and conglomeration of maximums and minimums. But this formula will be extremely complicated and similar to formulas of n-th prime number.


is there a name for a figure that is made by rotating the half of an ellipse around its axis?

EDIT:

yes, rotating a semiellipse

...so i guess there is no name for that figure?

Why would you want to know that? :?:
I think its a subglobose or something like that.


[quote="hydro"]are you sure? try $F_n=5$, it becomes $(5*13-8)^2\neq 1$[/quote]

sorry , there is a typo , it should be $F^2_{n+1}$

 :oops:  :blush: Now it is ok  ;)

[quote="shyong"]For fibonanci sequence , $F_n=F_{n-1}+F_{n-2}$ , $F_0=F_1=1$ , prove that

$(F_nF_{n+2}-F^2_{n+1})^2 =1$

 :)[/quote]
Written in a slightly different form, this is Cassini's Identity:

$F_n F_{n+2} - F^2_{n+1} = (-1)^{n+1}$.

Induction is the standard way to prove an identity like this, as was done by hydro.  I have a slightly different way:
[hide]
Note the following series of transformations:

$F_n F_{n+2} - F_{n+1}^2 = F_n (F_{n+1} + F_n) - F_{n+1}^2 \\ \\ = F_n F_{n+1} + F_n^2 - F_{n+1}^2 \\ \\ = F_n^2 + F_{n+1} (F_n - F_{n+1}) \\ \\ = F_n^2 - F_{n+1} F_{n-1}$

What we've shown is this:

If $I_n = F_n F_{n+2} - F^2_{n+1}$, then $I_n = - I_{n-1}$.

This recurrence relation has the solution $I_n = (-1)^{n+1} I_1$.  We further note that $I_1 = 1$, which means that $I_n = (-1)^{n+1}$, as desired.[/hide]


[b]The $ n$ members of a committee are numbered 1 through $ n$. One of the members is designated as the "Grand Pooh-Bah." The $ n$ members sit in a row of $ n$ chairs, but no member with a number greater than the Grand Pooh-Bah may sit in the seat to the immediate right of the Grand Pooh-Bah. Suppose that the Grand Pooh-Bah is member number $ p$, where 1<=p<=n. Find a formula, in terms of $ n$ and $ p$, for the number of ways for the committee to sit.[/b]

[hide] There are $ n \minus{} p$ members who could be sitting to the right of Grand Pooh-Bah with a higher number than $ p$
There are $ n \minus{} 1$ different chairs that Grand Pooh-Bah could be sitting in with someone on his right (since he could be sitting on the end, in which case there won't be).
For each of these there are $ (n \minus{} 2)!$ permutations.
The total number of committees without restriction is $ n!$ so the total number of ways allowed is: $ n! \minus{} (n \minus{} 1)(n \minus{} p)(n \minus{} 2)! \equal{} (n \minus{} 1)!(n \minus{} (n \minus{} p)) \equal{} p(n \minus{} 1)!$[/hide]


LaTeX
function
calculus
derivative
calculus computations

Given that $ a_{0} \equal{} 1, a_{1} \equal{} 1$ and $ n(n \minus{} 1)a_{n} \equal{} (n \minus{} 1)(n \minus{} 2)a_{n \minus{} 1}\minus{}(n \minus{} 3)a_{n\minus{}2}$  find  $ \sum_{n \equal{} 0}^{\infty}{a_{n}}$.

[hide="Hint"]Find $ a_{2}, a_{3}$ and deduce a formula for $ a_{n}$. Then, prove it by induction. After this, find $ \sum_{n \equal{} 0}^{\infty}{a_{n}}$. The result is $ e$.[/hide]

Thanks, but the hint leaves little or nothing to do.

Inductive step with some LaTeX (my computer's back-slash key (to the left of the space-bar doesn't seem to work, or else I'd have used frac):

$ n(n\minus{}1) a(n) \equal{} (n\minus{}1)(n\minus{}2)/(n\minus{}1)! \minus{} (n\minus{}3)/(n\minus{}2)!$

$ \equal{} 1/(n\minus{}3)! \minus{} (n\minus{}3)/(n\minus{}2)!$

$ \equal{} 1/n!$

[quote="AndrewTom"]Inductive step with some LaTeX (my computer's back-slash key (to the left of the space-bar doesn't seem to work, or else I'd have used frac):

$ n(n \minus{} 1) a(n) \equal{} (n \minus{} 1)(n \minus{} 2)/(n \minus{} 1)! \minus{} (n \minus{} 3)/(n \minus{} 2)!$

$ \equal{} 1/(n \minus{} 3)! \minus{} (n \minus{} 3)/(n \minus{} 2)!$

$ \equal{} 1/n!$[/quote]Exactly! Could you now find:
 $ \sum_{n \equal{} 0}^{\infty}{\frac {1}{n!}}$ ?  :)

[quote="enb081"] ...Exactly! Could you now find:
 $ \sum_{n \equal{} 0}^{\infty}{\frac {1}{n!}}$ ?  :)[/quote]

Hi :) !!!

$ \sum_{n \geq 0} \frac{x^n}{n!} \equal{} e^x$ because $ x \to e^x$ is analytic function than:

$ \sum_{n \geq 0} \frac{1}{n!} \equal{} e$

and thanks

I have been taught that $ e^{x}$ is the function whose derivative is itself so that, using the Maclaurin expansion and substituting $ x \equal{} 1$, we have that your series is $ e$.


calculus
integration
trigonometry
limit
real analysis
calculus computations

Think of this as a spinoff inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=269818]this problem[/url] and the fact that I originally misread that other problem and tried to solve something else:

For $ n\in\mathbb{N},$ let $ u_n\equal{}\int_0^{\infty}\frac{\sin^nx}{x}\,dx.$

We have $ u_n\equal{}\infty$ for all even integers $ n.$ But the interesting question is about odd integers:

Does $ \lim_{k\to\infty}u_{2k\plus{}1}$ exist?

[quote="Kent Merryfield"]One comment: it isn't really the Riemman-Lebesgue Lemma. That's $ \sin^{2n}x,$ not $ \sin(2nx),$ and it doesn't oscillate any more rapidly as $ n\to\infty.$ It gets smaller in other ways.[/quote]

Oh my god! you're right. I made a ridiculous mistake :blush:

It goes to 0 because $ \sin^{2n} x \searrow 0$ (except for $ x \equal{} \tfrac{\pi}{2}$).


integration
calculus
real analysis
real analysis unsolved

Is true:
If $ f \in L^1([0,n],\lambda)$ for $ n \in \mathbb{N}$ and \[ x^{\minus{}1}\int_0^xf(t)dt\geq f(x) \Rightarrow\] $ f$ is monoton decreasing?

False. Start with a decreasing piece and then, once you accumulate some noticeable integral, do whatever you want using very small values only.


[quote]$\displaystyle \log_2{(\log_3{(\log_4{x})})} = \log_3{(\log_4{(\log_2{y})})} = \log_4{(\log_2{(\log_3{z})})} = 0$[/quote]
[hide]
$\displaystyle \log_3(\log_4x)=1 \Rightarrow \log_4x=3 \Rightarrow x=64$
$\displaystyle \log_4(\log_2y)=1 \Rightarrow \log_2y=4 \Rightarrow y=16$
$\displaystyle \log_2(\log_3z)=1 \Rightarrow \log_3z=2 \Rightarrow z=9$
$x+y+z=64+16+9=89$
The answer is $\displaystyle C$.

[/hide]

A shortcut is noticing that you can simplify it by changing 0 to 1 and removing the first layer of each expression.

[quote="hiitsme"]Isn't that what I did?[/quote]Yes, but normally you would hide your solution so as not to give it away.


[quote="kunny"]4 points $ A,\ B,\ C,\ D$ are on a circle $ O$ in this order.Suppose that the diagonals $ AC$ of a quadrangle $ ABCD$ are perpendicular to that of $ BD$ , denote the intersection point by $ P$, and suppose that the line which passes through the point $ P$ and perpendicular to the side $ AB$ intersects with the sides $ AB,\ CD$ at $ Q,\ R$ respectively. If $ AC \equal{} 8,\ BP \equal{} 3,\ DP \equal{} 4,\ AP < BP$, then find the values of $ PQ^2,\ PR^2.$[/quote]
 [b][color=blue]my solution :
PQ^2 =  36 / 13
PR^2 = 144 / 13
If right i will post complet solution
almonkez :read: [/color][/b]

sorry for this mistak
PR^2 = 13  isnt it?
almonkez :P

[quote="kunny"]4 points $ A,\ B,\ C,\ D$ are on a circle $ O$ in this order.Suppose that the diagonals $ AC$ of a quadrangle $ ABCD$ are perpendicular to that of $ BD$ , denote the intersection point by $ P$, and suppose that the line which passes through the point $ P$ and perpendicular to the side $ AB$ intersects with the sides $ AB,\ CD$ at $ Q,\ R$ respectively. If $ AC \equal{} 8,\ BP \equal{} 3,\ DP \equal{} 4,\ AP < BP$, then find the values of $ PQ^2,\ PR^2.$[/quote]
 :play_ball:
[url=http://www.servimg.com/image_preview.php?i=263&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk12.jpg[/img][/url]


1) If A is an m x n matrix, show that if b is in the column space of A then there is a unique x in the row space of A for which Ax = b.

I would like to prove this [i]without[/i] using concepts about basis. That is, I would like to solve this just by using the properties of row reduced matrices with respect to inhomogenous linear equations.

2) Given an affine hyperspace in R(m), how can we construct a matrix A and find a vector b so that the solution space of the system Ax = b is the affine hyperspace?

A counterexample to (1):
(over $\mathbb{C}$)
$A=\begin{bmatrix}1&i\\1&i\end{bmatrix}$. If $x=\begin{bmatrix}a\\ia\end{bmatrix}$ is in the transpose of the row space, $Ax=0$.

In other words, this is about specific properties of the field, and you need a different statement.

Hmm.. you are quite right jmerry. That's weird

Here is the proof given in my book:

Let {v(1),...,v(r)} be a basis for the row space of A. Then Av(1),...,Av(r) are r linearly independent vectors in the column space of A. Therefore they span the column space. Hence every b in the column space of A is of the form b = Ax for some x in the row space of A. There can be only one such x because the intersection of the rowspace and the null space is the 0 vector.

I'm not quite sure about the last statement, if that's a theorem or if that's some unmentioned caveat. I don't see it stated anywhere. Maybe that's something that holds true in R(n)? If it's just an assumption then the proof is easy but I don't see it stated anywhere.   :maybe:

One last question: If V and W are infinite dimensional vector spaces, is it possible for the space of linear operators on V to be isomorphic to the space of linear operators on W?

My book asks a question about this but neglects to specify that V and W are finite dimensional so I'm not sure.

That statement is true in $\mathbb{R}^{n}$. The null space is the orthogonal complement of the transpose of the row space (with the standard inner product), and no vector is orthogonal to itself. If a sum of squares can be zero, this fails- and that happens in most fields.

I have no idea what you're trying to ask in that last question.


Is there a function with 2 actual points (not infinity) that’s not affected by a horizontal Graph Shift?
And
Is there a function with 2 actual points (not infinity) that’s not affected by a vertical Graph Shift?

Unless i'm missunderstanding you, then any periodic function works for number 1 (ie, sin x, cos x, ...). and there are none for number two because that would imply two points each with the same x coodinate, thus making it not a function


hello, we have
$ \frac{17\plus{}\sqrt{33}}{2}\minus{}\sqrt{\frac{17\plus{}\sqrt{33}}{2}}\equal{}\frac{17\plus{}\sqrt{33}}{2}\minus{}\sqrt{\left(\frac{\sqrt{33}\plus{}1}{2}\right)^2}\equal{}\frac{17\plus{}\sqrt{33}}{2}\minus{}\frac{\sqrt{33}\plus{}1}{2}\equal{}8$.
The right hande side can be prove analogously.
Sonnhard.


In the box we have 5 white, 3 black and 4 red balls.

I take from box 5 balls.

How is probability that 3 of taken balls are white?

[quote="pluristiq"]how is this conditional?[/quote]

Maeby it is not, maeby I am wrong.

I thoug that this is conditional, because you have new situation in box after taken 1st, 2nd, and 3rd ball from box.

if you remove without replacement, the probability is $ \frac{\binom{5}{3}\binom{7}{2}}{\binom{12}{5}}$

[quote="pluristiq"]if you remove without replacement, the probability is $ \frac {\binom{5}{3}\binom{7}{2}}{\binom{12}{5}}$[/quote]

Exuse me, but $ {\binom{5}{3}}$ is combination? So 5!/(3!*2!) ? Or it is 1,66666 ? Or something different?

[quote="pluristiq"]yes, combination.[/quote]

I am confused.

At first, after evaluation your equation I get very low numer, something like 0,00005 - but ok, more interestingly:

I am not understand, how there can be just one answer = one number if there can be few different scenarios, like

1. first was taken White, second was White, third was not white and for fourth was P equal 4/6 , so at all for scenario WWnW is P = 5/12 * 4/11 * 8/10 * 7/9 - what is different from combination upper

2. first was White, second was NotW, third was White and for fourth was P equal 4/6, so for scenario WnWW is P  = 5/12 * 4/11 * 3/10 * 7/9  what is different from combination and 1.

3. ...

ect.

So, there is in my opinion set of results/probabilities for different scenarios.

Whats wrong?

well, the probability is computed by adding the probabilities of the mutually disjoint scenarios that can occur...


limit
integration
HMMT
logarithms
calculus
real analysis
calculus computations

([b]7[/b]) Evaluate the limit $ \lim_{n\rightarrow\infty}
n^{\minus{}\frac{1}{2}\left(1\plus{}\frac{1}{n}\right)}
\left(1^1\cdot2^2\cdot\cdots\cdot n^n\right)^{\frac{1}{n^2}}$.

$ L \equal{} \lim_{n\rightarrow\infty} n^{ \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right)} \left(1^1\cdot2^2\cdot\cdots\cdot n^n\right)^{\frac {1}{n^2}}$

Taking ln of both sides, and using continuity and properties of ln(x)

$ ln(L) \equal{} \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right) \cdot ln(n) \plus{} \frac {1}{n^2} \sum_{k \equal{} 1}^n k \cdot ln(k)$

Now,

$ \int_0^{n \minus{} 1} x \cdot ln(x) dx < \sum_{k \equal{} 1}^n k \cdot ln(k) < \int_1^{n} x \cdot ln(x) dx$

and

$ \int x \cdot ln(x) dx \equal{} \frac {x^2}{4}\cdot(2 ln(x) \minus{} 1)$

so,

$ \frac {(n \minus{} 1)^2}{4} \cdot\left(2 ln(n \minus{} 1) \minus{} 1 \right) < \sum_{k \equal{} 1}^n k \cdot ln(k) < \frac {n^2}{4} \cdot (2ln(n) \minus{} 1) \plus{} \frac {1}{4} .$

This implies that

$ \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right) \cdot ln(n) \plus{} \frac {(n \minus{} 1)^2}{4n^2} \cdot\left(2 ln(n \minus{} 1) \minus{} 1 \right)\le ln(L)\le\lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right) \cdot ln(n) \plus{} \frac {1}{4} \cdot (2ln(n) \minus{} 1) \plus{} \frac {1}{4n^2}$

Expanding,

$ \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\cdot ln(n) \minus{} \frac {1}{2n}ln(n) \plus{} \frac {1}{2}ln(n \minus{} 1) \minus{} \frac {1}{n}ln(n \minus{} 1) \plus{} \frac {1}{2n^2}ln(n \minus{} 1) \minus{} \frac {1}{4} \plus{} \frac {1}{2n} \minus{} \frac {1}{4n^2}\le ln(L) \le \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}ln(n) \minus{} \frac {1}{2n}ln(n) \plus{} \frac {1}{2}ln(n) \minus{} \frac {1}{4} \plus{} \frac {1}{4n^2}$

It is easy to see that both of the limits are $ \minus{} \frac {1}{4}$, so

$ ln(L) \equal{} \minus{} \frac {1}{4}$
$ L \equal{} e^{\minus{}\frac {1}{4}}$

[quote="CaptainCanada"]$ ln(L) \equal{} \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right) \cdot ln(n) \plus{} \frac {1}{n^2} \sum_{k \equal{} 1}^n k \cdot ln(k)$[/quote]

Write $ \ln n$ instead.  It's much nicer-looking.

In any case, at this point we should write $ \minus{} \frac {1}{2} \left( 1 \plus{} \frac {1}{n} \right) \ln n \equal{} \minus{} \frac {1}{n^2} \sum_{k \equal{} 1}^{n} k \ln n$ so that we can write

$ \ln L \equal{} \lim_{n \to \infty} \frac {1}{n^2} \sum_{k \equal{} 1}^{n} k \ln \left( \frac {k}{n} \right) \equal{} \lim_{n \to \infty} \sum_{k \equal{} 1}^{n} \frac {k}{n} \ln \left( \frac {k}{n} \right) \frac {1}{n}$

which is the Riemann sum of $ \int_{0}^{1} x \ln x \equal{} \minus{} \frac {1}{4}$.  ;)

Edit:  Something in the way of motivation.  First of all, summations like this look an awful lot like Riemann sums.  Second of all, $ \lim_{n \to \infty} n^{ \frac{1}{n} } \equal{} 1$, so there was no reason to include that term in the original limit unless it helped make a rewrite easier.  :)

[quote="Extremal"][quote="worthawholebean"]([b]7[/b]) Evaluate the limit $ \lim_{n\rightarrow\infty} n^{ \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right)} \left(1^1\cdot2^2\cdot\cdots\cdot n^n\right)^{\frac {1}{n^2}}$.[/quote]
what is HMMT 2008 ? :maybe:[/quote]

GIYF

[quote="Extremal"][quote="worthawholebean"]([b]7[/b]) Evaluate the limit $ \lim_{n\rightarrow\infty} n^{ \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right)} \left(1^1\cdot2^2\cdot\cdots\cdot n^n\right)^{\frac {1}{n^2}}$.[/quote]
what is HMMT 2008 ? :maybe:[/quote]

See http://web.mit.edu/hmmt/www/ .

[quote="t0rajir0u"][quote="CaptainCanada"]$ ln(L) \equal{} \lim_{n\rightarrow\infty} \minus{} \frac {1}{2}\left(1 \plus{} \frac {1}{n}\right) \cdot ln(n) \plus{} \frac {1}{n^2} \sum_{k \equal{} 1}^n k \cdot ln(k)$[/quote]

Write $ \ln n$ instead.  It's much nicer-looking.

In any case, at this point we should write $ \minus{} \frac {1}{2} \left( 1 \plus{} \frac {1}{n} \right) \ln n \equal{} \minus{} \frac {1}{n^2} \sum_{k \equal{} 1}^{n} k \ln n$ so that we can write

$ \ln L \equal{} \lim_{n \to \infty} \frac {1}{n^2} \sum_{k \equal{} 1}^{n} k \ln \left( \frac {k}{n} \right) \equal{} \lim_{n \to \infty} \sum_{k \equal{} 1}^{n} \frac {k}{n} \ln \left( \frac {k}{n} \right) \frac {1}{n}$

which is the Riemann sum of $ \int_{0}^{1} x \ln x \equal{} \minus{} \frac {1}{4}$.  ;)

Edit:  Something in the way of motivation.  First of all, summations like this look an awful lot like Riemann sums.  Second of all, $ \lim_{n \to \infty} n^{\frac {1}{n} } \equal{} 1$, so there was no reason to include that term in the original limit unless it helped make a rewrite easier.  :)[/quote]

That takes care of the nasty indefinite integral CaptainCanada didn't account for. What justifies the initial equality though?

$ \minus{} \frac{1}{2} \left( 1 \plus{} \frac{1}{n} \right) \equal{} \minus{} \frac{n \plus{} 1}{2n} \equal{} \minus{} \frac{ \frac{n(n\plus{}1)}{2} }{n^2}$.  :)

[quote="§outh§tar"]

That takes care of the nasty indefinite integral CaptainCanada didn't account for. What justifies the initial equality though?[/quote]

There is nothing particularly nasty about $ \int_0^{n \minus{} 1} x \cdot ln(x) dx$, because $ x \cdot ln(x)$ goes to 0 with x.

[quote="t0rajir0u"]$ \minus{} \frac {1}{2} \left( 1 \plus{} \frac {1}{n} \right) \equal{} \minus{} \frac {n \plus{} 1}{2n} \equal{} \minus{} \frac {\frac {n(n \plus{} 1)}{2} }{n^2}$.  :)[/quote]

Ah sorry. I didn't realize that in your original statement there were different variables k and n. My mixup.  :blush:


An even number of persons are seated around a table.After a break,they are again seated around the same table,not necessarily in the same places.Prove that at least two persons have the same number of persons between them as before break.   :lol:

let $ n$ be the number of persons,the problem also holds when $ n\equal{}6k\plus{}3$.

anyway,first of all label the seats with numbers $ 1,2,\ldots ,n$,now assume that the person which sat in the chair number $ i$ before the break,sits in the chair $ f(i)$ after the break ($ i\equal{}1,2,\ldots, n$).
now for the sake of contradiction,assume that for every $ 1\leq i,j\leq n$ we have:

$ f(i)\minus{}f(j)\not\equiv i\minus{}j\pmod n$

and

$ f(i)\minus{}f(j)\not\equiv j\minus{}i\pmod n$

hence the sets $ \{f(i)\plus{}i : 1\leq i\leq n\}$ and $ \{f(i)\minus{}i : 1\leq i\leq n\}$ must be two complete residue systems modulo $ n$.but $ n$ is even,contradiction.


What did everyone get on the 2007 USA Biology Olympiad Open Exam?

I got 30/50!  :D

My teacher said the highest score in the nation is 43/50. One person at my school got 39/50.

37/50, i want to see the answers and past semifinal exams dunno how to find them

If you took it at your school, the teacher administering your test should have your score. If you took it at a different location, I believe the results will be mailed to you. (not entirely sure)

My school was cancelled yesterday and today (March 2); unfortunately, I haven't obtained the results yet. What was the cutoff for semifinals?

Curious. I though last year's test was easier (cutoff=26).

38!  I was so surprised I passed, considering in the last minute I filled C C C C C

39/50 :)

Last year, I barely made the cutoff with a 26/50, so I was really  surprised to see that I had improved by so much.

33/50

I am surprised I made it because I haven't touched a biology book since last May during the week of AP tests.

Random question... which is easier... USABO Test or the AP Bio Test?

Dang...I missed the cutoff by a couple points...if I hadn't had the flu while I was taking the test... I couldn't even pay attention in AP Biology class afterwards. That's okay; I would have been squashed by semis, though.
But they cut question #14, right? So it's out of 49?

Did that happen, really? What did question #14 ask?

That's what it said on the official answer list. It was the one with the pedigree, and you had to choose which genetic inheritance modes were possible. Supposedly, none of the given answers were correct.

someone from my school got 43.
i got 38.

but finalists will be announced april 9. which means they should be at least done correcting the multiple choice part of the exam.


$a,b,c$ are the sides of triangle $ABC$, and : $2 ( ab^{2}+bc^{2}+ca^{2}) = a^{2}b+b^{2}c+c^{2}a+3abc$
Prove that $ABC$ is equivalent.

it is equivalent
$4(ab^{2}+bc^{2}+ca^{2})=2(a^{2}b+b^{2}c+c^{2}a)+6abc \Leftrightarrow (a+b-c)(b-c)^{2}+(b+c-a)(c-a)^{2}+(c+a-b)(a-b)^{2}=0$
Because $a,b,c$ are three side of triangle $\Rightarrow a=b=c$ :D


well... can any one help me to solve this problem?

given n numbers, N1, N2...Nn. u have to find out least postive integer m so that for it the numbers have all different residues.

1 <= Ni <= 99999.

I hope i am going to get help!

 :) Mahbub

Heh, seems like everybody's been asking the same question today... :) I'll try to find out from a friend of mine who's solved this problem.

One thing comes to mind:

m > n

Because from m you have m-1 different residues. Therefore, m must be greater than the number of all input numbers - n.

I hope that helped.

brute force is the only way i can think of for now thats pretty easy to code... start at n+1 and just increment...

another thing is maybe take pairwise differences? and m cannot be a divisor of any of the differences... so maybe make a list of all divisors of the differences and sort it and take the smallest m that isn't there.

OK, I know a person who solved this problem using brute force (trying all numbers from n to 1000000) and it passed. But I'm almost sure there exists a better solution...

well, where did he solved this problem? it was given at UVA contest. and the brute force got TLE.

but that flagging method might work.

At the UVA contest you're talking about, the one that was few days ago. He wrote it in C++ and probably used a lot of code optimization... but it worked.


Log2x=2log2 to the second +log2 to the fifth

In x= In 5 + In 8[/b]

Seems a little too easy for intermediate, and also looks like homework.

[hide="Oh well."]
$\log 2x = 2 \log 2^2 + \log 2^5 = \log 2^4 + \log 2^5 = \log 2^9 = \log (2 \times 2^8)$
$x = 2^8$

$\ln x = \ln 5 + \ln 8 = \ln 40$
$x = 40$
[/hide]


linear algebra
matrix
superior algebra
superior algebra solved

[quote="Kuba"]Find the inverse of $
\left( {\begin{array}{*{20}c}
   a & b & b & b  \\
   b & a & b & b  \\
   b & b & a & b  \\
   b & b & b & a  \\
\end{array}} \right)
$ if you know that it exists.[/quote]

Let A=
$
\left( {\begin{array}{*{20}c}
   a & b & b & b  \\
   b & a & b & b  \\
   b & b & a & b  \\
   b & b & b & a  \\
\end{array}} \right)$

B=
$
\left( {\begin{array}{*{20}c}
   x_{1}   \\
   x_{2}   \\
   x_{3}   \\
   x_{4}   \\
\end{array}} \right)$

C=
$
\left( {\begin{array}{*{20}c}
   b_{1}   \\
   b_{2}   \\
   b_{3}   \\
   b_{4}   \\
\end{array}} \right)$

Then we have $AB=C$

and you compute AB and then find $x_{k}$ in terms of $b_{k}$ and for example you find: $x_{1}=a_{1}b_{1}+\cdots+a_{4}b_{4}$ and then coeficients of $b_{k}$ are on the first row of $A^{-1}$ and the same for the others.

I hope my explanation is correct and I am sure there is another smarter way to find the inverse!

For b=/=0

A= (a-b)I_4 + bJ where

J=
(1  1  1  1)
(1  1  1  1)
(1  1  1  1)
(1  1  1  1)

its eigenvalues are 0 with multiplicity 3,
and 4 with multiplicity  1.

I_4 is the identity matrix

detA = b^4det((a-b)/bI_4 + J)

A is invertible

iff  -(a-b)/b is not an eigenvalue of J

iff a=/=b and a-b+4b =/=0

A^(-1) = 1/(a-b).I_4 - b/(a-b).1/(a-b+4b).J

For the n dimension

A^(-1) = 1/(a-b)I_n  - b/(a-b).1/(a-b+nb).J


geometry
geometric transformation
combinatorics unsolved
combinatorics

There are $ k$ rooks on a $ 10 \times 10$ chessboard. We mark all the squares that at least one rook can capture (we consider the square where the rook stands as captured by the rook). What is the maximum value of $ k$ so that the following holds for some arrangement of $ k$ rooks: after removing any rook from the chessboard, there is at least one marked square not captured by any of the remaining rooks.

[quote="April"]There are $ k$ rooks on a $ 10 \times 10$ chessboard. We mark all the squares that at least one rook can capture (we consider the square where the rook stands as captured by the rook). What is the maximum value of $ k$ so that the following holds for some arrangement of $ k$ rooks: after removing any rook from the chessboard, there is at least one square not captured by any of the remaining rooks.[/quote]

$ k\equal{}81$.  It seems fairly trivial unless I'm missing something.  If the chessboard consists of the $ 100$ squares $ (i,j)$ where $ 1\le i,j\le 10$ then put the rooks on the $ 81$ squares $ (i,j)$ where $ 2\le i,j\le 10$.  The square $ (1,1)$ is then not captured by any rook and certainly will not be after we remove a rook.  On the other hand if $ k\equal{}82$ then after removing one there must by one square--we can assume without loss of generality that it is $ (1,1)$ not captured by any of the remaining rooks.  So the remaining $ 81$ rooks must cover the $ 81$ squares $ 2\le i,j\le 10$.  Removing one of those $ 81$ rooks will still leave any square except $ (1,1)$ captured by one of the other $ 80$--and the square $ (1,1)$ is captured by the $ 82nd$ rook--all squares are captured in this case and the arrangment is not possible with $ k\equal{}82$.  $ k\equal{}81$ is the best we can do.

[quote="NoYoo-hooForMe"][quote="April"]There are $ k$ rooks on a $ 10 \times 10$ chessboard. We mark all the squares that at least one rook can capture (we consider the square where the rook stands as captured by the rook). What is the maximum value of $ k$ so that the following holds for some arrangement of $ k$ rooks: after removing any rook from the chessboard, there is at least one square not captured by any of the remaining rooks.[/quote]

$ k \equal{} 81$.  It seems fairly trivial unless I'm missing something.  If the chessboard consists of the $ 100$ squares $ (i,j)$ where $ 1\le i,j\le 10$ then put the rooks on the $ 81$ squares $ (i,j)$ where $ 2\le i,j\le 10$.  The square $ (1,1)$ is then not captured by any rook and certainly will not be after we remove a rook.  On the other hand if $ k \equal{} 82$ then after removing one there must by one square--we can assume without loss of generality that it is $ (1,1)$ not captured by any of the remaining rooks.  So the remaining $ 81$ rooks must cover the $ 81$ squares $ 2\le i,j\le 10$.  Removing one of those $ 81$ rooks will still leave any square except $ (1,1)$ captured by one of the other $ 80$--and the square $ (1,1)$ is captured by the $ 82nd$ rook--all squares are captured in this case and the arrangment is not possible with $ k \equal{} 82$.  $ k \equal{} 81$ is the best we can do.[/quote]You are missing a big thing. Each square must be attacked by at least one rook at the initial arrangement. :D

[hide="Please check if I am missing something or not"]We claim that the answer is 10. Assume that there are more than 10 rooks. If each row has at least one rook, then there exist a row with exactly two rooks. If we remove one of these two rooks, all squares are still attacked by the rooks because there is a rook in each row. So there must be a row with no rook, similarly there must be a column with no rook. But the intersection of the empty row with the empty column is not attacked by any rook, contradiction. Hence the maximum value cannot exceed 10. To construct an example, simply put the rooks on a diagonal of the chessboard.[/hide]

[quote="Johan Gunardi"]You are missing a big thing. Each square must be attacked by at least one rook at the initial arrangement. :D [/quote]

The problem statement didn't really state that--at least not clearly.  I suspected this might be an implied condition though.  Perhaps the translation from the original Russian was imperfect.

[hide="proposed solution by Johan plus my comments"][quote="Johan Gunardi"]We claim that the answer is 10. Assume that there are more than 10 rooks. If each row has at least one rook, then there exist a row with [b]exactly two rooks[/b]. If we remove one of these two rooks, all squares are still attacked by the rooks because there is a rook in each row. So there must be a row with no rook, similarly there must be a column with no rook. But the intersection of the empty row with the empty column is not attacked by any rook, contradiction. Hence the maximum value cannot exceed 10. To construct an example, simply put the rooks on a diagonal of the chessboard.[/quote]

Yes I think this works.  Minor correction:  where you say [b]exactly two rooks[/b] it should say [b]at least two rooks[/b].  But your solution succeeds with only this minor fix.
[/hide]

[quote="hsiljak"]Let me and my poor Russian knowledge clear some doubts:

...

[quote]after removing any rook from the chessboard, there is at least one [b]marked [/b]square not captured by any of the remaining rooks.[/quote][/quote]

Ahhh...this is yet a different problem...distinct both from the one Johan solved and the one I solved.

[hide="my solution for this interpretation of the problem"]I claim that in this case $ k \equal{} 16$.  We can establish $ k\ge 16$ by the specific example of putting $ 16$ rooks on $ (1,i)$ and $ (i,1)$ for $ 2\le i\le 9$.  If we remove the rook on $ (1,i)$ then square $ (10,i)$ is no longer attacked--similarly if we remove the rook on $ (i,1)$.  Note that there is one square $ (10,10)$ not "marked" at the beginning--but this is okay under your understanding of the problem statement.

Now if $ k \equal{} 17$ then each of the $ 17$ rooks must have one row or one column in which it is the only rook.  We can assume without loss of generality that there must be $ 9$ rows each with one rook, and the other $ 8$ rooks must be in the other row.  If we remove one of those $ 8$ rooks we are still left with all $ 100$ squares under attack, and hence we have a contradiction where $ k \equal{} 17$.

So $ k \equal{} 16$ is the maximum in this case.[/hide]

[quote="hsiljak"]Let me and my poor Russian knowledge clear some doubts:

[quote]В некоторых клетках доски 10×10 поставили k ладей, и затем отметили все клетки, которые бьет хотя бы одна ладья (считается, что ладья бьет клетку, на которой стоит). При каком наибольшем k может оказаться, что после удаления с доски любой ладьи хотя бы одна [b]отмеченная[/b] клетка окажется не под боем?[/quote]

The bold word is missing in the translation, but it could have been deduced from the rest of the text (if it wasn't [i]отмеченная[/i], why  have we [i]отметили[/i] the fields in the first place  :lol: )
So, the corrected statement should say

[quote]after removing any rook from the chessboard, there is at least one [b]marked [/b]square not captured by any of the remaining rooks.[/quote][/quote]
Thank you very much, [b]hsiljak[/b]. I don't know Russian, and I translated this problem from a Vietnamese version of this contest (provided by Tran Nam Dung). What a pity, in this version, the word 'marked squares' was missed. I am really sorry for this mistake, I will try to do better next time.
[quote]Trên các ô của bàn cờ 10 x 10 người ta xếp k con xe. Sau đó người ta đánh dấu tất cả các ô mà có ít nhất một con xe ăn được (ta coi rằng ô mà con xe đứng cũng là ô bị nó ăn). Với k lớn nhất bằng bao nhiêu có thể xảy ra trường hợp: sau khi bỏ đi bất kỳ một con xe nào tìm được ít nhất một ô không bị các quân xe ăn?[/quote]

The answer is $k=16$.For example,put $8$ squares on $(1,i)$,where $i<9$ and on $10,i$,where $1<i<10$.Now,if $k>16$,we will have at least $9$ marked squares that are attacked once horizontaly and zero verticaly,so from this all other rooks must be in the row that has $0$ rooks and since we have only one such,we can't have $17$ rooks.


limit
real analysis
real analysis unsolved

Find the limit $ \lim_{n \rightarrow \infty} n^2 \left ( \left (1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}\minus{}\left (1\plus{}\frac{1}{n}\right)^{n} \right)$.

Hint: Take the log of $ (1\plus{}1/x)^x$, expand the log term in a power series out to about the $ O(1/x^3)$ term, exponentiate, subtract, etc. My first pass through gave me $ e/2$ as the limit.


What is the sum of the next three terms in the geometric sequence $ 1,\minus{}2,4,\ldots$?

hello, by the definition of a geometric sequence we get
$ a_4\equal{}4(\minus{}2)\equal{}\minus{}8$
$ a_5\equal{}\minus{}8(\minus{}2)\equal{}16$
$ a_6\equal{}16(\minus{}2)\equal{}\minus{}32$
so the searched sum is $ \minus{}8\plus{}16\minus{}32\equal{}\minus{}24$.
Sonnhard.


Given$ a,b,c,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ {\frac {\sqrt [10]{a^{11}}}{b \plus{} c \plus{} d}}\plus{}\frac{\sqrt[10]{b^{11}}}{c\plus{}a\plus{}d}\plus{}\frac{\sqrt[10]{c^{11}}}{d\plus{}a\plus{}b}\plus{}\frac{\sqrt[10]{d^{11}}}{a\plus{}b\plus{}c} \ge \frac {4}{3}$
It easy but nice.I hope you will like it. :)

[quote="quykhtn-qa1"]Given$ a,b,c,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ {\frac {\sqrt [10]{a^{11}}}{b \plus{} c \plus{} d}} \plus{} \frac {\sqrt [10]{b^{11}}}{c \plus{} a \plus{} d} \plus{} \frac {\sqrt [10]{c^{11}}}{d \plus{} a \plus{} b} \plus{} \frac {\sqrt [10]{d^{11}}}{a \plus{} b \plus{} c} \ge \frac {4}{3}$
It easy but nice.I hope you will like it. :)[/quote]
arqady, you don't like this inequality? :(

Please, please write deDicated and not deLicated.^^
It's dedicated, deDicated.

I wonder if I may not reply this topic because it is dedicated to someone.
In fact my solution is not completed. =_=

well, by CBS,
$ \big( \frac{a^{\frac{11}{10}}}{b\plus{}c\plus{}d} \plus{} \frac{b^{\frac{11}{10}}}{c\plus{}d\plus{}a} \plus{} \frac{c^{\frac{11}{10}}}{d\plus{}a\plus{}b} \plus{} \frac{d^{\frac{11}{10}}}{a \plus{} b\plus{}c} \big) \big( (b\plus{}c\plus{}d)a^3 \plus{} (c\plus{}d\plus{}a)b^3 \plus{} (d \plus{} a \plus{} b)c^3 \plus{} (a \plus{} b\plus{} c)d^3 \big)$
$ \ge (a^{21/10} \plus{} b^{21/10} \plus{} c^{21/10} \plus{} d^{21/10} )^2$

and here Jensen :
$ f(x) \equal{} x^{21/20}$ has concavity, so $ f(a^2) \plus{} f(b^2)\plus{}f(c^2) \plus{} f(d^2) \ge 4 f(\frac{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 }{4}) \equal{} 4$
$ \therefore a^{21/10} \plus{} b^{21/10} \plus{} c^{21/10} \plus{} d^{21/10} \ge 4$

And how can I show:
$ (b\plus{}c\plus{}d)a^3 \plus{} (c\plus{}d\plus{}a)b^3 \plus{} (d \plus{} a \plus{} b)c^3 \plus{} (a \plus{} b\plus{} c)d^3 \le 12$
I can't make a proof. It seems that it is true but I don't know.

My solution is little bit boring and too long (even it is not completed). Please post here a poof. :)

[quote="khj920214"]I wonder if I may not reply this topic because it is dedicated to someone.
In fact my solution is not completed. =_=

well, by CBS,
$ \big( \frac {a^{\frac {11}{10}}}{b + c + d} + \frac {b^{\frac {11}{10}}}{c + d + a} + \frac {c^{\frac {11}{10}}}{d + a + b} + \frac {d^{\frac {11}{10}}}{a + b + c} \big) \big( (b + c + d)a^3 + (c + d + a)b^3 + (d + a + b)c^3 + (a + b + c)d^3 \big)$
$ \ge (a^{41/20} + b^{41/20} + c^{41/20} + d^{41/20} )^2$

and here Jensen :
$ f(x) = x^{41/40}$ has concavity, so $ f(a^2) + f(b^2) + f(c^2) + f(d^2) \ge 4 f(\frac {a^2 + b^2 + c^2 + d^2 }{4}) = 4$
$ \therefore a^{41/20} + b^{41/20} + c^{41/20} + d^{41/20} \ge 4$

And how can I show:
$ (b + c + d)a^3 + (c + d + a)b^3 + (d + a + b)c^3 + (a + b + c)d^3 \le 12$
I can't make a proof. It seems that it is true but I don't know.

My solution is little bit boring and too long (even it is not completed). Please post here a poof. :)[/quote]
Nice proof :lol:
$ (b + c + d)a^3 + (c + d + a)b^3 + (d + a + b)c^3 + (a + b + c)d^3 \le 12$
$ \leftrightarrow \sum{(b + c + d)a^3} \le 3(\sum{a^2})^2$
$ \leftrightarrow \sum{ (a - b)^4} \ge 0$
Which is obvious true :)

[quote="quykhtn-qa1"]
Nice proof :lol:
$ (b + c + d)a^3 + (c + d + a)b^3 + (d + a + b)c^3 + (a + b + c)d^3 \le 12$
$ \leftrightarrow \sum{(b + c + d)a^3} \le 3(\sum{a^2})^2$
$ \leftrightarrow \sum{ (a - b)^4} \ge 0$
Which is obvious true :)[/quote]

Nice problem and nice solution , too, quykhtn-ga1 . :P  :roll:

[quote="arqady"]Thank you, quykhtn-qa1 for your gift. I am sorry, but I am very busy recently.[/quote]
I hope you have an other solutions :)


inequalities
limit
calculus
calculus computations

Let $a\geq b>0,\ x\geq 0.$ and $n$ be positive integers.Prove the following inequalities.

(1) $0\leq e^x-(1+x)\leq \frac{x^2e^x}{2}.$

(2) $a^n-b^n\leq n(a-b)a^{n-1}.$

(3) $e^x-\left(1+\frac{x}{n}\right)^n\leq \frac{x^2e^x}{2n}.$

[quote="kunny"]
(3) $e^x-\left(1+\frac{x}{n}\right)^n\leq \frac{x^2e^x}{2n}.$[/quote]

I'm a bit confused. Let's look at $e^x-\left(1+\frac{x}{n}\right)^n\leq \frac{x^2e^x}{2n}$. Let us take limits of both sides as $n\rightarrow\infty$. So we get \[ \lim_{n\rightarrow\infty}{\left[e^x-\left(1+\frac{x}{n}\right)^n\right]}\leq\lim_{n\rightarrow\infty}{\left[\frac{x^2e^x}{2n}\right]}  \]
On the right side, we can take $e^x$ out as a constant, since it is unaffected by $n$. The remaining limit is $-e^x$ (this is the definition of $e$ !), so $e^x-e^x=0$. We are left with \[ \lim_{n\rightarrow\infty}{\left[\frac{x^2e^x}{2n}\right]}\geq0  \] Obviously since the numerator is constant and the denominator is growing without bound, this approaches zero regardless of what $x^2e^x$ is. So the two sides are always equal; in what situation would there be an inequality?

Thanks :)

Both sides of $e^x-\left(1+\frac{x}{n}\right)^n\leq \frac{x^2e^x}{2n}$

do indeed tend to zero as $n\to\infty.$

But that's not the question; the question is to show that this inequality holds for each finite $n.$ Think of it as a question about contolling how fast the left hand side tends to zero.


algebra
system of equations
algebra proposed

Find all real solutions to the following system of equations:

$ \frac{4x^2}{1\plus{}4x^2}\equal{}y \quad \quad \frac{4y^2}{1\plus{}4y^2}\equal{}z \quad \quad \frac{4z^2}{1\plus{}4z^2}\equal{}x$

1996 Canada, #2:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77671[/url]


inequalities
ratio
topology
analytic geometry
geometry
integration
real analysis

In $\mathbb{R}^{n}$, how can one prove that if a subset has diameter $d$, its volume is less or equal to the volume of a ball of diameter $d$ ?

(I need this to compute the Hausdorf measure of a ball in $\mathbb{R}^{n}$. Is there any other way to do this ? )

Suppose $E$ is a set with diameter $d$. Then $\overline{E}$ also has diameter $d$ and its volume is at least as big as that of $E$ and since it is compact there is a segment of length $d$ whose endpoints are in $\overline{E}$. Let that segment be a diameter of a ball of diameter $d$. If $\overline{E}$ has a point outside of the ball, consider the triangle formed by the two endpoints of the diameter and the point outside...

I hope that's legitimate....

[quote="alekk"]doesn't work. Try with an equilateral triangle (or maybe I have not understand what you mean ..)[/quote]
Oh you are right, I'm an idiot.  :blush:

This is taxing my memory a bit, but you can probably do it with Steiner symmetrization.  Start with your set $E$, and choose any plane $P$.  For every line $L$ perpendicular to $P$, look at the line segment of length $|L\cap E|$ that is bisected by $P$.  The union of all these line segments is called (I think) is called the Steiner symmetrization of $E$ with respect to the plane $P$.  This procedure preserves measure and does not increase diameter (check this).  The idea is that if you do this repeatedly for lots of different planes through the origin, then the isodiametric ratio approaches that of the ball.  (Because intuitively, the symmetrization procedure is making $E$ more ball-like.)

IF this is the right proof, then it's not very easy in the sense that it would probably be hard to come up with the idea of Steiner symmetrization on your own.

Do you really need this to prove that simple fact about Hausdorff measure?  I would have thought it was easier than that.

In any case, I'm sure you can find all of this stuff in an accessible book such as Evans and Gariepy or something like that.

[url=http://eom.springer.de/S/s091730.htm]Steiner symmetrization[/url] is really cool. But you don't have to use it a lot of times: having symmetrized $E$ with respect to all coordinate planes $\{x_{i}=0\}$, you'll get a centrally symmetric body $\widetilde E$. Then just observe that $\widetilde E\subset B(0,d/2)$.

In $\mathbb R^{2}$ there is an amazingly simple proof (I read it in Littlewood's "Miscellany"). WLOG $E$ is convex and its boundary is a polar curve $r=r(\theta)$, $0\le\theta\le\pi$, where $r(0)=r(\pi)=0$. The area of $E$ is
\[\frac12\int_{0}^{\pi}r(\theta)^{2}d\theta=\frac12\int_{0}^{\pi/2}(r(\theta)^{2}+r(\theta+\pi/2)^{2})d\theta\le \frac12\int_{0}^{\pi/2}d^{2}d\theta=\frac{\pi d^{2}}{4}\]
where the middle inequality follows from a picture. I don't know if this idea works in higher dimensions. :noo:

Regarding the original post: how do you define "the" Hausdorff measure: by coverings with balls or general sets?

oh, Thank you very much for these answer, I have never heard about this Steiner symetrization. Seems awesome.

PS: for the definition of Hausdorf with arbitrary set, is there a simple approach to compute the Hausdorf measure of a ball. Indeed, it suffices to compute the $n$-Hausdorf measure of any body with positive Lebegue measure since the $n$-Hausdorf measure is proportional to the Lebesgue measure, but I cannot find a simple example where the $n$-Hausdorf measure is easily computable ..

the fact that the symmetrization process decreases the diameter is quite easy, but I have some questions:
1/ is it true that if $E$ is measurable, $\widetilde E$ is also measurable ?
2/ it is quite intuitive that volume is preserved, but what about a proof ? I think it easier in the case $E$ is convex, and I only need that for my problem, but I wonder how to prove it in the general case ?

For measurability, try using the fact that symmetrization of an open set is open, and measurable sets can be approximated with open supersets. On a second thought: symmetrized body $E^{*}$ is of the form (say) $\{|x_{1}|\le f(x_{2},\dots,x_{n})\}$, where $f$ is a measurable function, equal to $\int \chi_{E}(x)dx_{2}\dots dx_{n}$.

For volume-preserving: Volume of $E$ is an integral $1\,dx$ over $E$. Use Fubini.

(I will post more later; in a hurry right now.)

a compact subset of $R^{n}$ of diameter $d$ is included in a ball of radius $\frac{\sqrt{3}}{2}d$ (take the middle of a diameter as center). What is the best constant $\alpha$ such that a set of diameter $d$ is included in a ball of diameter $\alpha d$ ?


induction
algebra
polynomial
calculus
integration
inequalities
Rational Root Theorem

Let $p$ be a prime such that $p\geq 5$. Prove that for any integer $x$, $x^{p-1}+x^{p-2}+\cdots +x^{2}+x+2$ is not a perfect square.
            I look forward to seeing your solutions. This problem proved to be the hardest one from the contest.

hmm, really nice problem, havent solved it yet, but induction on x should work

for $x=1$ it is equal to $p+1$. Obviously, if $s^{2}= p+1$ then $p = s^{2}-1 = (s-1)(s+1)$, contradiction.

then we assume that $P(x) = x^{p-1}+x^{p-2}+...+x^{2}+x+2$ is not a perfect square, we have to prove that $S(x) = (x+1)^{p-1}+(x+1)^{p-2}+...+(x+1)^{2}+x+1+2$ is not a perfect square. ok, so we have
$S(x) = (P(x)-2)+D(x)+p+1$

D(x) is some weird sum with loads of Newton symbols :)) It should have some regular form such that we can imply that S(x) cant be a perfect square, maybe again contradiction analogically to $x=1$ case or some modulo work?

What do you think guys?

edit:

$S(x) = \frac{(x+1)^{p}-1}{x}+1$, so let $s^{2}= \frac{(x+1)^{p}-1}{x}+1$ then $x(s-1)(s+1) = (x+1)^{p}-1$. for any $s > 0$ polynomial $(x+1)^{p}-1$ is irreducible over $x(s-1)(s+1)$, obviously...

looks weird and not complete, im definately not satisfied with it. can i have a look at the author's solution?

I don't think that we can use induction.
because $x$ is not positive integers.....

gollywog// I think you must prove $P(x-1)$ or $P(-x)$ is not square.

oh jeez this cause pretty much of s**t with even and odd powers... wont even bother continuing this way of thinking.

any ideas?

I think that induction is not the case here too. However, I consider that this problem deserves much more atention :P  :(  Maybe someone will post a complete solution in the end or at least will try to come up with some ideas. :)

This is an idea which I think works:

By way of contradiction assume that there is $a\in N$ and $k\in N$ such that

$a^{p-1}+a^{p-2}+\cdots+2=k^{2}$.

Then we consider the polynomial

$P(x)=x^{p-1}+x^{p-2}+\cdots+2-k^{2}$. Observe that the preceding condition says that $P(a)=0$ in other words that our polynomial $P$ is factorable. Now, we concentrate to prove that this is false by using Eisenstein criteria.

Observe that $P(x)=\frac{x^{p}-1}{x-1}+1-k^{2}$, so we prove that

$P(x+1)=\frac{(x+1)^{p}-1}{x}+1-k^{2}$ is not factorable by using Eisenstein Criteria with $p$ for the prime number.
To show this we need to prove that $p^{2}$ does not divide $k^{2}-2$, which I think it holds. This would meas $k^{2}=2$  modulo $p^{2}$, which I think does not hold.
What do you think?

Your assumptions is false.I don't think that holds generally.For example $100\equiv 2(mod 49)$

We all tried to "kill" the problem with rather complicated methods at the contest. I really want to see if someone finds the solution, so please allow yourself a little time more with this problem  :) (for the participants at the contest: please do not post yet the solution :P )

No harazi, this is the solution, but it was amazing that no one saw it at the contest. We all tried complicated facts and disregarded this easy approach. Even now i feel strange and bad :(
    At least this topic is closed :D

[quote="harazi"]I don't get why it is considered so difficult. Isn't it a pretty standard result that all prime divisors of $1+x+...+x^{p-1}$ are either $p$ or equal to $1$ mod $p$? From here, the problem is obvious: if $a^{2}=2+x+...+x^{p-1}$, any prime divisor of $(a-1)(a+1)$ is either $p$ or equal to $1$ mod $p$. This is clearly impossible for $p>3$. Am I missing something?  :maybe:[/quote]

Am I wrong?
Let $q$ is prime factor of $(a-1)(a+1)$.
why can't $q \equiv-1 \mod p$ and $q^{2}\mid (a-1)(a+1)$ ???

If I'm not wrong, then harazi just show
"There is no integer $a$, such that $a^{2}\equiv 1, 2 \mod p$, when $p>3$" but this is wrong....

We know that all primes $q$ dividing $(a+1)(a-1)$ are $\equiv 1 \mod p$ (the case $q=p$ is impossible by very similar reasoning). Thus $a-1 \equiv 1 \equiv a+1 \mod p$, impossible.
Especially, $q\equiv-1 \mod p$ will not occure.

Read Harazis post again, he did never say what you state there.

Sorry, I can't understand that all primes $q$ dividing$(a-1)(a+1)$ are  $q \equiv 1\mod p$:blush:
Who anyone prove it for me?

I think I'm in a maze -_-;;
Please, point out my mistake. :blush:
8≡1(mod 7)
3² = 1 + 8 ≡ 2 (mod 7)
3² - 1 = (3-1)(3+1) ≡ 1 (mod 7)
but 2 is not 1(mod 7)

[quote="harazi"]Isn't it a pretty standard result that all prime divisors of $1+x+...+x^{p-1}$ are either $p$ or equal to $1$ mod $p$?[/quote]
Don't forget this ;)

Then, $1+x+\cdots+x^{p-1}\equiv 0 or 1 (mod p)$ isn't true?

What has this to do with the problem¿

Lemma [without proof, well known]: all prime divisors of $1+x+x^{2}+...+x^{p-1}$ are $\equiv 1 \mod p$ or equal to $p$ ($p$ a prime).

Assuming that $2+x+x^{2}+...+x^{p-1}=a^{2}$, we get $(a+1)(a-1)=1+x+x^{2}+...+x^{p-1}$ has only prime divisors $\equiv 1 \mod p$ or equal to $p$.
Thus the same holds for $a+1$ and $a-1$ alone each, thus they are $\equiv 0$ or $\equiv 1 \mod p$ (depending on if a factor $p$ occures).
But that's not possible for both if $p>3$ (they have difference $2$).

:blush:  I think I am crazy -_-;
I saw "don't forget" => "can forget"
sorry.  :blush:  :blush:

[quote="harazi"]I don't get why it is considered so difficult. Isn't it a pretty standard result that all prime divisors of $1+x+...+x^{p-1}$ are either $p$ or equal to $1$ mod $p$? From here, the problem is obvious: if $a^{2}=2+x+...+x^{p-1}$, any prime divisor of $(a-1)(a+1)$ is either $p$ or equal to $1$ mod $p$. This is clearly impossible for $p>3$. Am I missing something?  :maybe:[/quote]

o well, thats exactly what i made some time ago, but due to end of the term i didnt have time to post :)

the beggining of this way of thinking is in my first post...

i began with induction just becouse polynomials often work fine with this tool.

[quote="harazi"]Ok, let's make the problem more difficult: can we find all solutions to $\frac{x^{n}-1}{x-1}=y^{2}-1$ for $n$ odd number? Or even? I don't know...[/quote]
for odd n we have finite solution.
see: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=711405#711405[/url]

[quote="ali666"][quote="harazi"]Ok, let's make the problem more difficult: can we find all solutions to $\frac{x^{n}-1}{x-1}=y^{2}-1$ for $n$ odd number? Or even? I don't know...[/quote]
for odd n we have finite solution.
see: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=711405#711405[/url][/quote]

Sorry, I do not see how the problems are equivalent

Can someone solve Harazi's problem?

(sorry for the big revive)

The problems are not equivalent, but the solutions given also answer harazi's problem.

Edit/repost: I was asked to expand this a bit. Here it goes:

[b]If $n$ is odd[/b], then the problem is of type $x^{2k}+... = y^2$. As mentioned by Rust in the link we can now write this as $(x^k+...)^2 + r(x) = y^2$ with $r$ of degree lesser than $k$ and all coefficients rational (the proof of that is mostly a little bit of algebraic manipulations). The polynomials may not be integral, but after multiplication with a sufficiently large perfect square they are. Thus we get a problem of type \[f(x)^2 + s(x)  = g(y)^2,\] where $f,g,s$ have integer coefficients and $\deg(s)<\deg(f)$ (actually, $g(y) = a y$ for some integer $a$, but we won't need that).

Now we use that there is no perfect square between $n^2$ and $(n+1)^2$. We want $f(x)^2+s(x)$ to be a square. By assuming that $|x|$ is large enough we get that $s(x) \neq 0$ and $|s(x)| \leq 2 |f(x)|$, the latter by $\deg(s)<\deg(f)$. Lets assume that $f(x), s(x) \geq 0$ (all other cases are similiar and/or easily reduced to this one). Now we get \[f(x)^2 < f(x)^2+s(x) < f(x)^2+2f(x)+1 = (f(x)+1)^2\]
and conclude that  $f(x)^2+s(x) $ lies between consecutive perfect squares and can thus not be a perfect square $g(y)^2$ itself.

To make this a bit more explicit: as shown, the only solutions can be those where $s(x)=0$ or $|s(x)| > |2 f(x)|$; the former can be easily checked by the rational root theorem and the latter e.g. by some easy inequalities involving the coefficients (left to the reader, I am too tired and lazy now).

[b]If $n$ is even[/b], we get a more difficult problem.  There are several theorems implying there are only finitely many solutions. Most prominently, this is a Thue equation (http://en.wikipedia.org/wiki/Thue_equation). We could also throw Falting's theorem (http://en.wikipedia.org/wiki/Faltings%27s_theorem) at it to even get the finiteness of rational solutions if $n \geq 6$ (for $n=4$ we get an elliptic curve about which the theorem says nothing: they have genus $0$).
Note that Thue's theorem is in principle effective, i.e. we get again a bound on $x$ and would only need to check the smaller ones. But this bound will be much worse than what we got in the odd case.


function
logarithms
integration
calculus
calculus computations

Can somebody explain me what is a function described parametrically?I came across it studying a book about differential equations-at the homogeneus equations,in this book,they give the solution parametrically,but i don't understand
what they say here.
So,the unknown function is x(t),the variable is t,and the equation is
x'=g(x/t),t in an interval I,where g:J->R is a continuous function.
And they make the substitution x/t=y,and obtain:
dy/(g(y)-y)=dt/t,they integrate,and obtain t=function of y ,and x=function of y ,where c is a constant,for every y in J,and say that this is the parametrical description of the function.I think that there's something wrong here,because they doesn't make difference between y as a function and y as a variable-first y is a function,but then they say "for every y in J",which means that y is a variable.
Can somebody explain me why they do that,and what is that parametrical description of the function,because i don't understand.(well,this is a college course,and i'm only highschool,but i want to understand that.)Tnahks.

They just the use of the method of solving adifferential equation by separating variables. The aim of such method is to have an expression of the form $dxf(x)=dtg(t)$ which you can integrate and then express x with t to get x(t).
In your particular case they have the differential equation: $\frac{dx}{dt}=g(\frac{x}{t})$. Then they make the substitution $y(t)=\frac{x(t)}{t}$. To express the left side with y(t) (without x(t)) you calculate the derivate $\frac{dy}{dt}$:

$\frac{dy(t)}{dt}=\frac{t\frac{dx(t)}{dt}-x(t)}{t^2}$ from which you get:

$\frac{dx(t)}{dt}=\frac{dy(t)}{dt}t+y(t)$.

The differential equation becomes:

$\frac{dy(t)}{dt}t+y(t)=g(y(t))$.

Then we separarte the variables:

$\frac{dt}{t}=\frac{dy}{g(y(t))-y(t)}$

Now you can integrate the equation and get:

$\ln t+c=\int{\frac{dy}{g(y(t))-y(t)}}$.

For a given function g you have to calculate this integral and solve the equation in y you get to become $y=\phi(t)$ with some function $\phi$. $x(t)$ is then: $\phi(t)t$.

Misha


inequalities induction calculus function logarithms

$m$ and $n$ are two natural numbers such that $m>n$. Prove that:

$(1+\frac{1}{m})^{m}>(1+\frac{1}{n})^{n}$.

A hint if needed.
[hide]$m \geq n+1$[/hide]

[quote="perfect_radio"][hide="Classical Proof."]
By $AM-GM$ we have that \[\sqrt[n+1]{1 \cdot \underbrace{\left( 1+\frac1{n}\right) \cdot \ldots \cdot \left( 1+\frac1{n}\right)}_{n \, \textrm{terms}}}< \frac{1+n+\frac{n}{n}}{n+1}= 1+\frac1{n+1},\] so \[\left( 1+\frac1{n}\right)^{n}< \left( 1+\frac1{n+1}\right)^{n+1}, \, \forall n \geq 1 .\] (Note that we have a strict inequality since the terms are not all equal.)
Now, a simple induction solves the problem.
[/hide][/quote]
 :huh: Now how come I have never thought of that. My solution is crazy now I have seen your one.

you could always be lame and use calculus to show that it is a strictly increasing function using calculus

Well, you could also claim that as $m$ and $n$ grow larger and larger, their respective functions approach the value of $e$, but never truly get to $e$.  :)

yes, that's exactly what I was about to post...
here is my similar problem: prove if m is an integer, the greatest value of m^{1/m} occurs at m=3

Actually, I think people first showed that the sequence \[a_{n}=\left(1+\frac{1}{n}\right)^{n}\] does actually converge, and then assigned its limit the notation $e$.

[quote="me@home"]yes, that's exactly what I was about to post...
here is my similar problem: prove if m is an integer, the greatest value of $m^{1/m}$ occurs at $m=3$[/quote]

If $m$ is real it occurs at $m=e$. Thus you can guess the answer since $3$ is closer to $e$ than $2$. :D :D

Can you show that if $m$ is real, it occurs at $m=e$?  :)

yes, that's basic calculus (btw the way you prove the answer is 3 for integers is by induction, because for each n>3 you can show that f(n+1)>f(n) using basic algebra and the fact that (1+1/n)^n converges on e)

We could also show that \[\frac{d}{dx}\left(1+\frac{1}{x}\right)^{x}= \frac{-\left(1+\frac{1}{x}\right)^{x}\ln\left(1+\frac{1}{x}\right)}{x^{2}},\] (if I didn't make any computational error) which is clearly positive for all $x\geq 1$. Therefore, $\left(1+\frac{1}{x}\right)^{x}$ is increasing when $x\geq 1$ and because $m>n\geq 1$, the desired result follows.

actually, I this problem was posed in a number theory/proof technique class at a camp I went to, which is kind of hilarius