calculus
integration
logarithms
function
algebra
binomial theorem
real analysis

[quote="cimabue"]find an asymptotic formula for $f(x)=\int_2^x\sqrt{\frac{x-t}{t^2-4}}\;dt$ as $x\to+\infty$.[/quote]Write $f(x) = \sqrt{x} \int_h^1 \sqrt{ \frac{1-s}{s^2-h^2}} ds$ with $h = \frac{2}{x}$.

Split the integral into $\int_h^{1/2} + \int_{1/2}^1$.
In the first integral, use the binomial theorem to write ${\int_h^{1/2}\sqrt{ \frac{1-s}{s^2-h^2}} ds = \int_h^{1/2} \frac{1}{\sqrt{s^2-h^2}} ds + \sum_{k \ge 1}}\int_h^{1/2} \frac{c_ks^k}{\sqrt{s^2-h^2}} ds$, where the $c_k$ are binomial coefficients. The first term can be integrated explicitly and behaves like $\ln{\frac{2}{h}} = \ln x$, and all other terms converge to constants and are differentiable in $h$.

The second integral converges to $\int_{1/2}^1 \frac{\sqrt{1-s}}{s}ds = 2 atanh(\frac{1}{\sqrt{2}}) - \sqrt{2}$ and is differentiable in $h$.

Asymptotically therefore, $f(x) = \sqrt{x} (\ln x + const. + O(x^{-1})$. Were you interested in higher order terms?

nice! that's what i was looking for. my guess is that the function itself is related to some special functions but i couldnt get maple to do the work. i was hoping this would come up.


Find all positive integers $n$ such that \[ \frac{3^n - 2^n}{n} \in \mathbb Z \]

Only $1$. Indeed, suppose that $n>1$ and let $p$ be the smallest prime divisor of $n$. Then:
$(\frac{3}{2})^n \equiv 1 \mod{p}$
Let $k=ord(\frac{3}{2})$. We have $k|p-1$ and $k|n$. But $k \leq p-1 < p$ which is impossible because of the minimality of $p$, unless $k=1$. But $k=1$ implies that $\frac{3}{2} \equiv 1 \mod{p}$, so $1 \equiv 0 \mod{p}$, contradiction.

This problem has been discussed here http://www.mathlinks.ro/Forum/viewtopic.php?p=6799#p6799 ;)


geometry
geometric transformation
reflection
circumcircle
trapezoid
geometry unsolved

Let $ ABC$ be an acute triangle.Let $ M$ be a point in the interior of $ \triangle ABC$.
Let $ M_1$ be the reflection of $ M$ wrt $ AB$, $ M_2$ be the reflection of $ M_1$ wrt $ BC$, $ M_3$ be the reflection of $ M_2$ wrt $ AC$.
Let $ K$ be the midpoint of $ MM_3$.
Find the locus of $ K$.

I have some difficulties with this problem.Please help.Thanks.

It is easy to show that M,M2,M1 lie on the circle which have B as the circumcenter
The same as C
Denote E,F are the perpendicular foot of C and B , B’,C’ are the reflection of B,C wrt AB,AC respectively
We have BB’M3M2 is an isosceles trapezoid  =>  BM2=B’M3  =>  KL=LH=LF , where L is the midpoint of BM3 =>  L is the circumcenter of the circumcircle of ⊿FKL
∠HFK=1/2∠KLH=1/2∠MBM2=∠B=∠EFA  => K,F,E are collinear
Thus K moves on the line EF
When M is coincide with B then K=F, when M is coicide with C K=G
The locus of K is segment FG

This is very classical, actualy we get $M_3$ from $M$ by isometry which is composition of 3 reflections thus it is indirect isometry, so by Hjelmslev's theorem $K$ describe (part of) line.


I don't know the correct terminology in english so sorry if I make any mistakes

Give then matrices $ n\times n$ , such that $ A^3\equal{}B^3$ and $ A^2B\equal{}B^2A$ , prove that $ A^2\plus{}B^2$ is non-inversible.

Thanks in advance for any help.

I'm pretty sure this has been posted before, but it's been quite a while.

Hint: what happens when you try multiplying out the product
\[ (A^2\plus{}B^2)(A\minus{}B)\ ?\]

Ah, yes.

triplebig: did you by any chance leave out a condition in the statement of the problem - the condition that $ A\ne B\,?$

I didn't, I checked and I copied it as it is in the book.  The problem is from a WL Putnam Competition exam I think, but am not sure.

As you wrote it, your problem is false in a fairly trivial way. Suppose $ A\equal{}B\equal{}I$ (over a field of characteristic other than 2). Then $ A^3\equal{}B^3\equal{}A^2B\equal{}B^2A\equal{}I$ and $ A^2\plus{}B^2\equal{}2I$ is invertible. You may think you copied it correctly but you didn't. Maybe the condition is verbal: $ A$ and $ B$ are distinct matrices, or some wording to that effect.

Just assume it does have the condition $ A\ne B.$ Can you follow my hint?

I copied it word for word as it is in the list! Perhaps the author was the one that copied it wrong.

Suppose the condition that they are distinct, how would I solve this problem?

I checked with my teacher today and he confirmed that it is indeed missing the condition $ A\neq B$ . I have solved it using your first post. Thank you very much for your help.

The conclusion can be generalized right?

$ XY\equal{}O_n\,;\,Y\neq O_n\,\Rightarrow\,\nexists X^{\minus{}1}$

Right: any two nxn matrices A and B such that $ AB \equal{} O_n$ and $ A,B \neq O_n$ are singular. Proof: Suppose A is non-singular: therefore $ A^{\minus{}1}$ exists and $ A^{\minus{}1}AB \equal{} A^{\minus{}1} O_n <\equal{}> B \equal{} O_n$, contrary to the hypothesis that $ B \neq O_n$. Therefore, A is singular. In a similar fashion, suppose B is non-singular: then $ B^{\minus{}1}$ exists and $ ABB^{\minus{}1} \equal{} O_n B^{\minus{}1} <\equal{}> A \equal{} O_n$. Therefore, B is also singular.


Let $ a,b,c$ are real numbers such that $ a^2\plus{}2b^2\plus{}3c^2\equal{}6$. Find the min of $ A\equal{}a\plus{}b\plus{}c$

We use BCS inequality, we obtain:
$ (a^2\plus{}2b^2\plus{}3c^2)(1\plus{}\frac{1}{2}\plus{}\frac{1}{3}) \geq (a\plus{}b\plus{}c)^2$
Now, we get the result:
$ a\plus{}b\plus{}c \leq \sqrt{11}$


How to do the different arithmetical operations in other number bases? could anyone post links to documents that explain that?

Thanks, but it doesn't show, explain a lot. It is not complete. It discusses only addtion and multiplication. How about division, and subtraction?

another one:

http://mathforum.org/library/drmath/view/55727.html


A penny, nickel, dime, and quarter are simultaneously flipped.  What is the expected value of the amount of the coins which come up heads?  Express your answer in cents, rounded to the nearest tenth of a cent.

Each coin has a value of whatever it's worth (obviously) and a probability of $ \frac12$ of showing that value (otherwise, it shows a $ 0$, so we can discount those possibilities), for a total expected value of $ \frac12(1\plus{}5\plus{}10\plus{}25)\equal{}\frac{41}2\equal{}\boxed{20.5}$.

Perhaps the question could list the value in cents of how much these coins are worth.

A penny is worth 1 cent, a nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents.

P.S.  Can you teach me how to Dougy?  :)

Can't two or three or four coins come up heads?

What can only 1 coin come up heads?

^Yeah;
I did all of the casework of 1,2,3,4 heads but why doesnt that work.

Additionally, if this wasnt casework, there is not (1/2) probability of getting heads, there is (1/2)^4 probability of getting all heads................

If you do the casework you get the same answer.

It follows a binomial distribution for the number of heads. You get 4 heads with a value of 0.41 cents (occurring 1/16), 3 heads with a summated value of 1.23 cents (consider the value of each possible 3 coin configuration and add all the individual configurations), 2 heads with a summated value of 1.23, and 1 head with a summated value of 0.41. After factoring in the probability of each outcome, you get 20.5 cents.
Note - If you add the configuration values together, you do not need to multiply by the number of configurations there are by this solution method.

[hide=Solution]Each coin has a probability $\tfrac{1}{2}$ to come up heads. Thus,
$$\frac{1}{2} \cdot \left(1+5+10+25\right)=\boxed{\frac{41}{2}}$$

oh my god i calculated every single possibility


quadratics
inequalities proposed
algebra
polynomial
three variable inequality
Inequality

[quote="tranthanhnam"]let$a,b,c >0$  and $abc=1$.
 prove that :
$a^2+b^2+c^2 +3 \geq 2(ab+bc+ca)$[/quote]
Too easy. Just use Schur.
$\sum a^2+3\sqrt[3]{a^2b^2c^2}\ge \sum(\sqrt[3]{a^4b^2}+\sqrt[3]{a^4c^2}\geq 2(ab+bc+ca)$

See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=18430
http://www.mathlinks.ro/Forum/viewtopic.php?t=19076
http://www.mathlinks.ro/Forum/viewtopic.php?t=50745

and a more general version in

http://www.mathlinks.ro/Forum/viewtopic.php?t=49049
http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 (beginning with post #2)
http://www.mathlinks.ro/Forum/viewtopic.php?t=40979 (beginning with post #3)

  Darij

Darij Grinberg

I have some questions for you: why dont you post a problem but
 everytime solve them send the problems you had enjoyed ...
Please ...
 - one more question as i remember before the exam in the IMO 2005 you and your teamates were saying womething in your lanquage with a loud voice can you say what were you saying please ...

Davron

[quote="tranthanhnam"]Let $ a,b,c > 0$ and $ abc \equal{} 1,$ prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \geq 2( bc \plus{} ca \plus{} ab).$[/quote]Mr. Qing Song pointed out that this problem is [b]Example 1[/b] in :

Yu-Zhong Weng, Structuring quadratic function to prove inequalities (Chinese), High School Science References (Nanning), 1998, No. 6,  page 14. (ISSN: 1002-6363)

By the way,

the polynomial  $ 2(1 \minus{} bc \minus{} ca \minus{} ab) \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} a^2b^2c^2$ is nonnegetive for real numbers $ a,b,c,$

but cannot be represented as a sum of squares of other polynomials with real coefficients.

See also : http://www.mathlinks.ro/viewtopic.php?t=24093

[quote="Ji Chen"][quote="tranthanhnam"]Let $ a,b,c > 0$ and $ abc \equal{} 1,$ prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \geq 2( bc \plus{} ca \plus{} ab).$[/quote]Mr. Qing Song pointed out that this problem is [b]Example 1[/b] in :

Yu-Zhong Weng, Structuring quadratic function to prove inequalities (Chinese), High School Science References (Nanning), 1998, No. 6,  page 14. (ISSN: 1002-6363)

By the way,

the polynomial  $ 2(1 \minus{} bc \minus{} ca \minus{} ab) \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} a^2b^2c^2$ is nonnegetive for real numbers $ a,b,c,$

but cannot be represented as a sum of squares of other polynomials with real coefficients.

See also : http://www.artofproblemsolving.com/viewtopic.php?t=24093[/quote]
thank Ji Chen.
I come.
Our dialogue ：

http://blog.sina.com.cn/s/blog_4c1131020100yscg.html

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]

The following inequality is also true.
Let $a,b,c >0$ , prove that\[a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]
Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543
Let$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \geq 2(a^2+b^2+c^2).$

[quote=tranthanhnam]let$a,b,c >0$  and $abc=1$.
 prove that :  $a^2+b^2+c^2 +3 \geq 2(ab+bc+ca)$[/quote]
The following inequality is also true.

Let$a,b,c >0$  and $a+b+c+2=abc$. Prove that :
$$a^2+b^2+c^2 +12 \geq 2(ab+bc+ca).$$

[quote=sqing]
Let$a,b,c >0$  and $a+b+c+2=abc$. Prove that :
$$a^2+b^2+c^2 +12 \geq 2(ab+bc+ca).$$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] kills it immediately.

Is this problem true?
Let $a, b, c > 0$ such that $a^2 + b^2 + c^2 = 3$. Prove that:
$$2 + abc \ge a + b + c$$
If this problem wrong, which $0 < k \le 3$ satisfying:
$$k + (3 - k)abc \ge a + b + c$$

[hide=*]Let $a,b,c>0$ and $abc=1.$ For the positive integer $n$ , [url=https://artofproblemsolving.com/community/c6h1598026p9930634]prove or disprove [/url]\[a^{n+1}+b^{n+1}+c^{n+1}+3  \ge 2(a^n+b^n+c^n).\][/hide]
For $a,b,c\geq 0 .$ [url=http://artofproblemsolving.com/community/c6h457354p2569858]Prove that[/url]$$a^3+b^3+c^3+3abc \ge 2\sqrt[3]{abc}(a^2+b^2+c^2).$$
[url=https://artofproblemsolving.com/community/c6h612340p3641832]here
[/url]

[quote=tranthanhnam]let$a,b,c >0$  and $abc=1$.
 prove that :
$a^2+b^2+c^2 +3 \geq 2(ab+bc+ca)$[/quote]

Let $x,y,z>0$, prove

$\sum_{cyc}x^4z^2+3x^2y^2z^2\geq 2\sum_{cyc}x^3y^2z$

Takeya.O

[quote=sqing]Let $a,b,c>0$ and $abc=1.$ For the positive integer $n$ , [url=https://artofproblemsolving.com/community/c6h1598026p9930634]prove or disprove [/url]\[a^{n+1}+b^{n+1}+c^{n+1}+3  \ge 2(a^n+b^n+c^n).\][/quote]
CounterExample
$n=3, a=\frac{3}{2}, b=\frac{2}{3}, c=1$

Let $a,b,c,d >0.$ Prove that$$a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).$$
$$a^2+b^2+c^2 +d^2+abcd+1\geq  a^2+b^2+c^2+d^2+2\sqrt{abcd} \geq ab+ac+ad+bc+bd+cd.$$
[url=https://artofproblemsolving.com/community/c6h1709871p11020577]Turcevichi[/url]

[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ , prove that\[a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).\]
[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]
Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543
Let$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \geq 2(a^2+b^2+c^2).$[/hide][/quote]
Prove that for all positive real numbers $x, y, z,$ the inequality holds
$$x^2 + y^2 + z^2 + 2xyz + 1\geq2 (xy + yz + zx).$$
([b]Training to the Baltic Way 2018 team , Estonian[/b])
Proof:
The net is symmetric with respect to $x, y, z,$ so we can assume that $x> y> z.$
If $y \geq 1, $ then the inequality is equivalent to the equation
$(x-y)^ 2+ (z-1)^2 + 2z (x-1) (y-1) \geq 0.$
If $y <1, $ then the inequality is equivalent to the equality
$(x-y)^2+ (x-1)^2 + 2x (1-y) (1-z) \geq 0.$
Another solution. [b]The difference can be written in the form[/b]
$(x-1)^2+ (y-1)^2+ (z-1)^2 + 2 (x-1) (y-1) (z-1) \geq 0.$
If the expressions $x-1, y-1, z-1$ are exactly $0$ or $2$, then the equation is valid.
So let's assume that there are $1$ or $3$ negative ones.
Let's not narrow the universe $x-1 $ is negative, in this case $(y-1) (z-1) \geq 0. $
Since $x$ is positive, then $-1 <x-1 <0, $ so that $$(x-1)^2+ (y-1)^2+ (z-1)^2 + 2 (x-1) (y-1) (z-1)$$$$> (x-1)^2+(y-1)^2+ (z-1)^2-2 (y-1) (z-1) =(x-1)^2 + ((y-1) - (z-1))^2\geq0.$$

[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ , prove that\[a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).\]
[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url]
Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543
Let$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \geq 2(a^2+b^2+c^2).$[/hide][/quote][hide=Yu-Zhong Weng]Let $a,b,c$ be positive real numbers such that $abc=1 .$ Prove that$$a^2+b^2+c^2+3\geq 2(ab+bc+ca).$$
[b]Proof  A:[/b]
WLOG $(a-1)(b-1)\geq 0,$ so $$a^2+b^2+c^2+3- 2(ab+bc+ca)=(a-b)^2+(c-1)^2+2c(a-1)(b-1)\geq 0.$$
[b]Proof  B:[/b]
$$a^2+b^2+c^2+3- 2(ab+bc+ca)=\frac{1}{a^2(b^2+1)}\left(2b^2c^2(b-1)^2+(bc^2-1)^2+b^2(bc^2+1-c-b^2c)^2\right)\geq 0.$$
https://artofproblemsolving.com/community/c6h1657181p11607206[/hide]

[quote=sqing]
Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][/quote]
Let $a, b, c \ge 0$. For any positive integer n prove that [url=https://artofproblemsolving.com/community/c6h1880582p12793518]the following inequality exist[/url]
$$(2n-1)(a^2+b^2+c^2)+3\sqrt[3]{a^2b^2c^2} \geq 2n(ab+bc+ca).$$
[hide=P]\[2(n-1)(a^2+b^2+c^2-ab-bc-ca)+\left[a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} - 2(ab+bc+ca)\right] \geqslant 0.\][/hide]

[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ , prove that\[a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).\]
[hide=*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=483075]here[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559544]here[/url][/hide]
Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][hide=**][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138]here[/url] [url=http://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]here[/url]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=607569&p=3611330#p3611330
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=612340&p=3642543#p3642543
Let$a,b,c >0$ and $abc=1$.prove that $a^3+b^3+c^3 +3 \geq 2(a^2+b^2+c^2).$[/hide][/quote]
[url]https://artofproblemsolving.com/community/c6h1827390p12225661[/url]
For $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1827390p15189416]Prove that [/url]$$a^2+b^2+c^2+abc+2\ge ab+bc+ca+ a+b+c$$
[hide=Wlog]assume that $(a-1)(b-1) \ge 0$ then
$$a^2+b^2+c^2+abc+2-\left(a+b+c+ab+bc+ca\right)= \left(a-b\right)^2 + \left(c-1\right)^2 + \left(a-1\right)\left(b-1\right)\left(c+1\right) \ge 0$$[/hide]

[quote=tranthanhnam]let$a,b,c >0$  and $abc=1$.  Prove that :
$$a^2+b^2+c^2 +3 \geq 2(ab+bc+ca)$$[/quote]
Let$a,b,c >0$  and $abc=1.$  Prove that
$$a^2+b^2+c^2 +3 \geq a^ 2 +b^ 2 +c^ 2 +\frac{{9}}{{a+b+c}} \geq 2(ab+bc+ca)$$
[quote=sqing]The following inequality is also true.
Let $a,b,c >0$ , prove that\[a^2+b^2+c^2 +2abc+1 \geq 2(ab+bc+ca).\][/quote]
Let $a,b,c$ be three positive real numbers . [url=https://artofproblemsolving.com/community/c1068820h2234630p17094843]Prove that[/url]
$$a^2 +b^ 2+c^ 2+abc+ 4 \geq 2 (ab +bc+ca)$$

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive numbers such that $a_1a_2\cdots  a_n=1.$ Prove that$$a_1^2+a_2^2+\cdots+a_n^2+n\geq2\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right).$$
[url=https://artofproblemsolving.com/community/c6h1064117p11637470]h[/url]

Let $a,b,c,d\geq 0.$ Prove that$$a^2+b^2+c^2+d^2+a+b+c+d+abc+bcd+cda+dab\geq 2(ab+ac+ad+bc+bd+cd)$$
[quote=sqing]For $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1827390p15189416]Prove that [/url]$$a^2+b^2+c^2+abc+2\ge ab+bc+ca+ a+b+c$$[/quote]
[url]https://artofproblemsolving.com/community/c6h518938p12694099[/url]
[url]https://artofproblemsolving.com/community/c6h1827390p12225661[/url]
Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive numbers such that $a_1a_2\cdots  a_n=1.$ Prove that$$a_1+a_2+\cdots+a_n- n\geq\min\{a_1,a_2,\cdots,a_n\}(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}-n)$$
[size=50](p/7303459116)[/size]
[quote=sqing]Let $ {a,b,c}\ge{0}$, prove that\[{a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\geq 2(ab+bc+ca)}.\][/quote]
[url=https://math.stackexchange.com/questions/510104/how-prove-this-abc3-sqrt3abc-ge-2-sqrtab-sqrtbc-sqrtac?rq=1]Solution[/url]


vector
function
integration
Functional Analysis
advanced fields
advanced fields unsolved

Let $ \Omega$ be a bounded, open subset of $ \mathbf R^{n}$ and assume that $ f\colon\mathbf R\to\mathbf R$ is strictly decreasing with $ f(0)=0$. Let $ u \in C^{2}(\Omega) \cap C^{0}(\bar\Omega)$ be a solution of the following problem
\[ \begin{cases}-\Delta u = f(u) & \text{ in }\Omega, \\ u = 0 & \text{ on }\partial\Omega. \end{cases}\]
Show that $ u=0$.

Any $ u$ satisfying the equation $ -\Delta u=f(u)$ cannot have a positive local maximum or a negative local minimum; $ u$ is always increasing in some direction whenever $ u$ is positive since $ \Delta u$ is also positive, and always decreasing in some direction whenever $ u$ is negative since $ \Delta u$ is also negative.

Now, consider the absolute maximum and minimum of $ u$; if the maximum is positive or the minimum is negative, this point must lie in the interior, where it is a local maximum or minimum. This is impossible by the above, and both the maximum and minimum of $ u$ are zero.

Why do you write gradient $ \nabla$ instead of Laplace operator $ \Delta$? As you wrote, the left hand side is a vector function while the right hand side is a scalar function...  :maybe: I guess you mean the following solution (I slightly changed your approach):  :maybe:
[hide="Solution"] Note that $ u$ cannot have positive local maximum and negavite local minimum inside $ \Omega$. Indeed, if $ x_{*}$ is a point of positive maximum, then $ u''(x_{*})$ is negative-semidefinite, yielding that $ -\Delta u(x_{*}) \ge 0$ while $ f(u(x_{*})) < 0$, a contradiction. The situation with negative local minimum is analoguous.

Now, it follows from above reasonings that $ u \le 0$ (because there is no positive abolute maxumum of $ u$ in $ \bar\Omega$). Analoguously, $ u \ge 0$ (because there is no negative absolute minumum of $ u$ in $ \bar\Omega$). Thus $ u=0$. [/hide]

It seems that $ f$ must not be [i]strictly[/i] decreasing, if it were just decreasing, all would be OK. Here're my two solutions not using strict monotonicity of $ f$, and I wonder are there any mistakes?
[hide="Solution 1"] Since $ f(s) \le f(0) = 0$ for $ s\ge0$, one has
	\[ \int_\Omega |\nabla u^{+}|^{2}= \int_\Omega f(u)u^{+}\le 0,\]
	yielding that $ u \le 0$. Now, since $ f(s) \ge f(0) = 0$ for $ s\le0$ and since $ u\le0$, $ -\Delta u \ge 0$, yielding that $ u\ge0$. Thus, $ u=0$.
[/hide]
[hide="Solution 2"]Set $ X : = H_{0}^{1}(\Omega)$ and $ Y : = H^{-1}(\Omega)$. Consider the map $ G \colon X \to Y$ defined by $ G(u) =-\Delta u-f(u)$, then $ G'(u) =-\Delta-f'(u)$. By data, $ f'(0) \le 0$, yielding that $ G'(0) \colon X \to Y$ is a Banach space isomorphism. Since $ G(0)=0$, by Implicit Function Theorem, $ G$ is a diffeomorphism of some neighbourhood $ U_{X}$ of 0 in $ X$ onto some neighbourhood $ U_{Y}$ of 0 in $ Y$. Thus $ u=0$ is the unique solution to our problem.
[/hide]

Mistake fixed.

Also, there's no problem if it's not strictly decreasing; harmonic functions don't have local maxima/minima unless they're constant.


What is the general factorization of $x^{n}-1$ ?

[hide]We know by Bezout's Lemma that since $x = 1 \Rightarrow x^{n}-1 = 0$, $x-1 | x^{n}-1.$ As a result, we have:

$\frac{x^{n}-1}{x-1}= x^{n-1}+x^{n-2}+\cdots+1$

This follows by the formula for the sum of a geometric series. As a result, we can factor this polynomial into:

$(x-1)(x^{n-1}+x^{n-2}+\cdots+1)$[/hide]

[quote="Arvind_sn"]What about $x^{n}+1$?  :)[/quote]
Well it depends on whether x is odd or even I think.

[quote="The QuattoMaster 6000"]That is true, if n is odd, then the factorization is $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}...-x+1)$.  However, if n is [i][even][/i] then there is no [b]real[/b] factorization.[/quote]

no

$4x^{4}+1=4x^{4}+4x^{2}+1-(2x)^{2}=(2x^{2}-2x+1)(2x^{2}+2x+1)$, take $y=x\sqrt{2}$, to get

$y^{4}+1=(y^{2}-\sqrt{2}y+1)(y^{2}+\sqrt{2}y+1)$

@Karth...that is not Bezout's lemma

[quote="Karth"]$x = 1 \Rightarrow x^{n}-1 = 0$, $x-1 | x^{n}-1.$ [/quote]

They call this Bezout's Lemma?   :huh:

http://planetmath.org/encyclopedia/BezoutsLemma.html

[quote="The QuattoMaster 6000"]That is true, if n is odd, then the factorization is $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}...-x+1)$.  However, if n is then there is no [b]real[/b] factorization.[/quote]

Presumably you mean "if $n$ is even."  But what about

$x^{6}+1 = (x^{2}+1)(x^{4}-x^{2}+1)$

Anyway, the general factorization (over the rationals) takes the form

$x^{n}-1 = \prod_{d | n}\Phi_{d}(x)$

http://mathworld.wolfram.com/CyclotomicPolynomial.html

Edit:  And, of course,

$x^{n}+1 = \frac{x^{2n}-1}{x^{n}-1}= \prod_{d | 2n, d \not | n}\Phi_{d}(x)$

Whoops my bad...  :blush: It won't let me edit it though.


inequalities
induction
inequalities proposed

[quote="cezar lupu"]Let $x_1 ,x_2 ,x_3 ,...,x_n >0$ and $n\geq 3$ such that their product is equal with $1$. Prove that:

          $\frac{1}{\sqrt{1+2x_1 }}+\frac{1}{\sqrt{1+2x_2 }}+...+\frac{1}{\sqrt{1+2x_n }} <n-1$ ;)[/quote]

My attempt:

For $n=3$ it is true and well known. Now substitute $x_1=\frac{a_1}{a_2},...$ to get rid of the constraint. [size=75][this map is bijective so we conserve the inequality and still reach all possible values.] [/size]We obtain:
 \[\sum_{cyc}\sqrt{\frac{x_i}{x_i+2x_{i+1}}} > n-1\]
We will now prove the claim by induction. We arrange all $a_i$ in increasing order, starting induction with the lowest 3. Suppose it was true for k. Then it is sufficient to prove that for some $a_i,a_{i+1}$, we can put a new variable $a_{i+1}$ in between them, labeling the old $a_{i+1}$ as $a_{i+2}$, so that the ineq is still true. This is obvious, since we can prove that
\[\sqrt{\frac{a_i}{a_i+2a_{i+2}}} + 1 \ge \sqrt{\frac{a_i}{a_i+2a_{i+1}}} + \sqrt{\frac{a_{i+1}}{a_{i+1}+2a_{i+2}}}\]
and the other terms are invariant. Proof: $1>\sqrt{\frac{a_{i+1}}{a_{i+1}+2a_{i+2}}}$ since $a_{i+2}>0$ and $\sqrt{\frac{a_i}{a_i+2a_{i+2}}}\ge \sqrt{\frac{a_i}{a_i+2a_{i+1}}}$ since $a_{i+1}$ the largest. Which proves the claim. $\Box$

We can remark that for all $x_i$ going to 0, except for $x_1$ going to infinity, we can approximate $n-1$ arbitrarily close.

This inequality can be improved for $n\ge4$ as follows:

Let $x_1 ,x_2 ,x_3 ,...,x_n >0$ and $n\geq 4$ such that their product is equal with $1$. Prove that:
          $\frac{1}{\sqrt{1+x_1 }}+\frac{1}{\sqrt{1+x_2 }}+...+\frac{1}{\sqrt{1+x_n }} <n-1$ ;


In how many ways can 2n indistinguishable balls be placed into n distinguishable urns, if the first r urns may contain at most 2r balls for each r in the set {1,2,3,...,n}?

This is equivalent to the question, "How many paths are there from $ (0, 0)$ to $ (n, 2n)$ composed of steps of the form $ (1, 0)$ or $ (0, 1)$ that never pass above the line $ y \equal{} 2x$?"  This is closely related to a classical problem whose answer is the Catalan numbers.  For those of you who just want to see the answer, it's sequence [url=http://www.research.att.com/~njas/sequences/A001764]A001764[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

This question is simply asking for the number of non-negative integer solutions to the equation:

$ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}2n$

No, it's not:[quote="Ignite168"]the first r urns may contain at most 2r balls for each r in the set {1,2,3,...,n}[/quote] is an additional constraint.

I'm not sure if I'm interpreting this right, but is $ r$ equal to the number of elements in the set containing $ n$, excluding $ n$ itself?

To refine my earlier post, I took into account the restriction.
I think the question is really asking for the number of non-negative integer solutions to the problem:

$ x_1\plus{}x_2\plus{}...x_n\equal{}2n$ where $ 0\le x_1, x_2,..., x_{n\minus{}1}\le 2r$

However, this might be wrong because I'm not sure what $ r$ is exactly.


vector
analytic geometry
graphing lines
slope
LaTeX

I'm not sure if this is correct but is your question similar to something like this?

Vector AB + Vector BC = Vector AC

No, I  think this is what he means.

Let x = (a, b, c) be a vector and let i = (1, 0, 0), j=(0, 1, 0), k=(0, 0, 1).

x = ix + jx + kx

Is this what you are looking for?

I'm sure that's what he means, though that may not know that.

Well, decomposing a vector into components usually means, you are given some other vector and you want to split your vector into two that are parallel and perpendicular respectively to this given vector. Like if you have a side view of a sloping road and a car on it with some gravity going down, you split it into parallel to the slope and perpendicular to the slope.. is that what you mean?

In the following  A,B,C,V are vectors in $ E^3=$ 3-dimensional euclidean space.
Denote $ (A,B,C)= A\cdot (B \times C) .$ It's  easy to show that if
$
\begin{array}{lcl}
  A&=&{ (a_1,a_2,a_3):= a_1  {\bf i}}+ a_2 {\bf j}+a_3{\bf k }\\
  B&=&(b_1,b_2,b_3)\\
  C&=&(c_1,c_2,c_3)\\
  V&=&(x,y,z)\\
\end{array} ,
  $
then
$
(A,B,C)=\begin{array}{|ccc|}
a_1& a_2& a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3
\end{array} .
$
Usiny this equality,  try to show that $ (A,B,C)=0 $ iff the vectors $\{ A,B,C\}$ are
linear dependent in $E^3 $ , i.e. iff these vectors are coplanar.
\par Further suppose that $ (A,B,C)\ne 0 $. This means that $ \{A,B,C\} $ is a basis in $E^3 $ . Therefore, in this case, any vector  $V $ from $ E^3 $ may be written (in an unique way ) as a linear combination of $ A,B,C $. By assuming that
\[
 V=\alpha A+\beta B+ \gamma C ,
\]
then using Cramer's rule, you find that
\[
\alpha=\frac{ (V,B,C)}{(A,B,C)},
\]
\[
 \beta =\frac{(A,V,C)}{(A,B,C)},
\]
\[ \gamma=\frac{(A,B,V)}{(A,B,C)}.
\]
Perhaps, help.
[tex][/tex]

why don't you post a couple of the questions you're having trouble with? may help :)

Hello, there. Well, I don't have much time. So, I'll quickly post brief explanations for the first two problems you've put up.

Firstly, when it says to "break it down into its components" it's just a fancy way of saying allot one component of the vector to the x-axis and the other to the y-axis. I'm sure you remember ordered pairs... cartesian plane. They look something like this: (x,y). Well, it's the same in vectors, except, they just call the x-axis, i, the y-axis, j, and the z-axis, k. With that said, it looks something like this: (i,j,k). With that said, let's get to the questions.

1) This question is basically asking you to draw a right triangle, with the hypotenuse as a 100, and the top angle of the triangle as 60 degrees (b/c 90-30 = 60, as the 100mph vector makes a 30 degree angle with the horizontal). Then, to break it down into it's "components" you just solve the triangle. The base of the triangle will be it's x-component (i-component) and the height of the triange will be its y-component (j-component).

Solving for x-component:

[tex]Sin(\theta)&=&\frac{x}{100}[/tex]

[tex]100Sin(60)&=&x[/tex]

[tex]x&=&50\surd{3}[/tex]

[tex]\Longleftrightarrow\vec{i}&=&50\surd{3}[/tex]

Solving for y-component:

[tex]Cos(\theta)&=&\frac{y}{100}[/tex]

[tex]100Cos(60)&=&y[/tex]

[tex]y&=&50[/tex]

[tex]\Longleftrightarrow\vec{j}&=&50[/tex]

2) In this problem, a force is acting on an object, and the force, if it's pulling somthing [b]down[/b], will be acting straight down in the [tex]y[/tex] direction. So, you draw a right triangle with it's height equaling to 600, and with an angle of elevation as 15 degrees. Then you just solve for the hypotenuse because the hypotenuse takes into consideration both the force downwards and the force required to pull the boat sideways. Solution below:

[tex]Sin(\theta)&=&\frac{Opposite}{Hypotenuse}[/tex]

[tex]Sin(15)&=&\frac{600}{H}[/tex]

[tex]H&=&\frac{600}{Sin(15)}[/tex]

[tex]\Longleftrightarrow&H&=&2318.22&Newtons[/tex]

That's all I have time for now. I hope this helped. I'll see if I can post later. Plus, this was my first time using this latex coding thing. Haha. So, it took me a while.


induction
algebra
polynomial
algebra unsolved

Let,
$ S\equal{}\sqrt{1001^{2}\plus{}1}\plus{}\sqrt{1002^{2}\plus{}1}\plus{}\cdots\plus{}\sqrt{2000^{2}\plus{}1}$

prove that, $ S$ is irrational.

By induction $ \sum_{k\equal{}1}^{n}\sqrt{a_{k}}$ is irrational if $ a_{k}\in N, a_{k}\not\equal{}b_{k}^{2}$.

It is not so obvious. Please write in more details.

These problem (about $ \sum\sqrt{a_{k}}$) posted before.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 .

N. T. TUAN, I think there are some very intelligent people on mathlinks. I saw the ideas on the link. It is far better than before, but I'm just wondering if is there simpler solution to this problem?

P.S.
I like simple and genious problems ... like me :). (For example: If for reals x,y,z>0
xyz(x+y+z)=1 find the smallest possible value of (x+y)(y+z). Russia 1989 - solved in mathlinks)

[quote="borislav_mirchev"][b]N. T. THUAN[/b], I think there are some very intelligent people on mathlinks. I saw the ideas on the link. It is far better than before, but I'm just wondering if is there simpler solution to this problem?
[/quote]
Yes, but i am N.T.TUAN. :P

Suppose contrary that $ S$ is rational.  Consider the following set of polynomials $ \{P_{n}\}$, $ 1\leq{n}\leq1000$.

Let $ P_{1}(x) \equal{} x^{2}\minus{}(1001^{2}\plus{}1)$.  We also recursively define
\[ P_{k\plus{}1}(x) \equal{} P_{k}(x\minus{}\sqrt{(1000\plus{}k)^{2}\plus{}1})\cdot{P_{k}(x\plus{}\sqrt{(1000\plus{}k)^{2}\plus{}1})}\,,\]
for all $ 1\leq{k}\leq999$.  It is easy to see that $ S$ is a zero of $ P_{1000}$.  Since the leading coefficient of $ P_{1000}$ is $ 1$ and all of its coefficients are integers, then the fact that $ S$ is a rational number implies that $ S$ is an integer.

It is clear that $ S > 1001\plus{}1002\plus{}\cdots\plus{}2000$.  Also, $ \sqrt{N^{2}\plus{}1}\equal{} N\sqrt{1\plus{}N^{\minus{}2}}< N\left(1\plus{}N^{\minus{}2}/2\right)$.  Thus,
\[ \sqrt{N^{2}\plus{}1}< N\plus{}1/2N\,,\]
for all $ N \equal{} 1001, 1002,\dots, 2000$.  Consequently,
\[ S < (1001\plus{}\plus{}1002\plus{}\cdots\plus{}2000)\plus{}\frac{(1/1001\plus{}1/1002\plus{}\cdots\plus{}1/2000)}{2}\,.\]
Hence,
\[ S < (1001\plus{}\plus{}1002\plus{}\cdots\plus{}2000)\plus{}\frac{(1/1000\plus{}1/1000\plus{}\cdots\plus{}1/1000)}{2}\,.\]
Thus,
\[ S < (1001\plus{}\plus{}1002\plus{}\cdots\plus{}2000)\plus{}1/2\,.\]
Therefore,
\[ 1001\plus{}1002\plus{}\cdots\plus{}2000 < S < (1001\plus{}\plus{}1002\plus{}\cdots\plus{}2000)\plus{}1/2\,,\]
which is a contradiction!

It's very good! It seems you have lots of experience.


trigonometry
inequalities unsolved
inequalities

let x, y, z be real #, suppose that x^2+y^2+z^2+2xyz=1.
Prove that
x^2+y^2+z^2>=3/4

[hide="hint"]Exist a triangle $ABC$ such that $x=\cos{A},y=\cos{B},z=\cos{C}$[/hide]

[quote="arqady"][quote="N.T.TUAN"]Exist a triangle $ABC$ such that $x=\cos{A},y=\cos{B},z=\cos{C}$[/quote]
[quote="davemiller"]
let x, y, z be real :wink: [/quote][/quote]
If $|x|,|y| \text{or}|z| >1$ then trivial :D

[quote="davemiller"]let x, y, z be real #, suppose that $x^{2}+y^{2}+z^{2}+2xyz=1.$
Prove that
$x^{2}+y^{2}+z^{2}\geq \frac{3}{4}$[/quote]


hello, we have $ \frac{6!}{5!}\minus{}\frac{4!}{3!}\equal{}\frac{5!\cdot6}{5!}\minus{}\frac{3!\cdot4}{3!}\equal{}6\minus{}4\equal{}2$.
Sonnhard.


[quote="Nbach"]An acute triangle $ ABC$. Prove that: $ \sum(\sin2B+\sin2C)^{2}.\sinh \leq 12.\sinh.\sinh.\sinh$[/quote]
$ \sum(\sin2B+\sin2C)^{2}.\sin A\leq 12\sin A\sin B\sin C$


Να  προσδιορίσεται  όλες  τις  πραγματικές  θετικές   τριάδες   x,y  και  z  οι  οποίες  είναι  λύσεις  του  συστήματος

     $x+y+z = 6$

     ${\frac{1}^{x}}+{\frac{1}^{y}}+{ \frac{1}^{z}}= 2-{\frac{4}^{xyz}}$

Αυτά αρέσουν στον αδερφό μου  :D  .
Άσε που άμα την έβαζες στο σχολείο θα σου έλεγαν 2 εξισώσεις με τρεις αγνώστους γίνεται αυτό να λυθεί ? Άλυτο προβλημα για αυτούς . :)  :P
Βάζω λύση μόνο και μόνο επειδή την πρότεινες εσύ Γιώργη . :wink:
Here is my solution
Από Cauchy-Swartz $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$

άρα $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{3}{2}$

και η δεύτερη εξίσωση δίνει
$\frac{3}{2}\leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2-\frac{4}{xyz}$

από όπου $xyz\geq 8$

Με AM-Gm όμως η πρώτη δίνει $6=x+y+z\geq 3\sqrt[3]{xyz}$ άρα
$xyz\leq 8$
Άρα $xyz=8$  kai x=y=z  ara  $x=y=z=2$

Αδερφέ  μπράβο  :o                    καλά  εσύ  έχεις  ελαχιστοποιήσει  τον  χρόνο  .   Αυτήν  την  λύση  είχα  και  εγώ   βρεί

[quote="gdeskos"]καλά  εσύ  έχεις  ελαχιστοποιήσει  τον  χρόνο.[/quote]

Έλα ρε φίλε εντάξει υπερβάλλεις  :)  .
Πάντως ενδιαφέρον έχει η άσκηση στους πραγματικούς . :wink:
Θα βάλλω και εγώ μια άσκηση τώρα αν θες κοίτα την .


Suppose $ A$ and $ B$ are $ n\times n$ matrices satisfying $ A^2 \plus{} A\equal{}O$, $ B^2 \plus{} B\equal{}O$ and $ AB\plus{}BA\equal{}O$. Determine $ \det (A\minus{}B)$, $ \det(A\plus{}B)$.

[hide]
$ (A \plus{} B)^2 \equal{} A^2 \plus{} AB \plus{} BA \plus{} B^2 \equal{} \minus{} A \minus{} B$
$ det(A \plus{} B)^2 \equal{} ( \minus{} 1)^(n).det(A \plus{} B)$
If $ n$ is even : (Write $ det(A \plus{} B) \equal{} x$)
$ x^2 \equal{} x$, then $ x \equal{} 0$ or $ x \equal{} 1$, the example for $ x \equal{} 0$ is $ A \equal{} B \equal{} 0$
and the example for $ x \equal{} 1$ is $ A \equal{} 0$ and $ B \equal{} \minus{} I$
If $ n$ is odd :
$ x^2 \equal{} \minus{} x$,$ x \equal{} 0$ or $ x \equal{} \minus{} 1$, the example for $ x \equal{} \minus{} 1$ is $ A \equal{} 0$ and $ B \equal{} \minus{} I$
Because $ AB\plus{}BA\equal{}0$, then $ BAB\plus{}B^2A\equal{}BAB\minus{}BA\equal{}0$
$ AB^2\plus{}BAB\equal{}0$, then $ \minus{}AB\plus{}BAB\equal{}0$, $ BAB\minus{}BA\minus{}(\minus{}AB\plus{}BAB)\equal{}0$,
$ AB\minus{}BA\equal{}0$, then : $ AB\minus{}BA\plus{}(AB\plus{}BA)\equal{}0$, $ 2AB\equal{}0$, if $ A$ and $ B$ is not a matrix over a field with characteristics $ 2$, we obtain that : $ AB\equal{}0$,$ BA\equal{}0$, then :
$ (A\minus{}B)(A\plus{}B)\equal{}A^2\minus{}B^2\equal{}\minus{}(A\minus{}B)$, $ det(A\minus{}B).det(A\plus{}B)\equal{}(\minus{}1)^n.det(A\minus{}B)$
If $ det(A\plus{}B)\equal{}0$ then $ det(A\minus{}B)$ must be $ 0$ (The example is $ A\equal{}B\equal{}0$
If $ det(A\plus{}B)\equal{}1$ and $ n$ is even then we have many infinitely $ A$ and $ B$ that statisfies the condition such as $ A\equal{}kI$ and $ B\equal{}(1\minus{}k)I$, $ A\minus{}B\equal{}(1\minus{}2k)I$, $ det(A\minus{}B)$ can approach infinity and negative infinity
If $ det(A\plus{}B)\equal{}\minus{}1$ and $ n$ is odd, then we have many infinitely $ A$ and $ B$ that statisfies the condition such as $ A\equal{}\minus{}kI$ and $ B\equal{}(\minus{}1\plus{}k)I$, $ A\minus{}B\equal{}(1\minus{}2k)I$, $ det(A\minus{}B)$ can approach infinity and also can approach negative infinity.
[/hide]

Hi Kruni,
the result about $ det(A\minus{}B)$ is false.
You have shown that if $ charac(K)\not\equal{}2$ then $ AB\equal{}BA\equal{}0$.
$ A,B$ are simultaneously diagonalizable. We may assume that $ A\equal{}diag(a_1,\cdots,a_n),B\equal{}diag(b_1,\cdots,b_n)$ where $ a_ib_i\equal{}0,a_i,b_i\in\{0,\minus{}1\}$. Then $ a_i\equal{}b_i\equal{}0$ or $ a_i\equal{}0,b_i\equal{}\minus{}1$ or $ a_i\equal{}\minus{}1,b_i\equal{}0$. Thus $ a_i\plus{}b_i\equal{}0$ or $ \minus{}1$ and $ a_i\minus{}b_i\equal{}0,1$ or $ \minus{}1$.
Conclusion: $ det(A\plus{}B)\equal{}0$ or $ 1$ with $ n$ even or $ \minus{}1$ with $ n$ odd.
$ det(A\minus{}B)\equal{}0,1,$ or $ \minus{}1$.
If charac(K)=2 then we have only $ AB\equal{}BA$.
$ A,B$ are simultaneously diagonalizable. We may assume that $ A\equal{}diag(a_1,\cdots,a_n),B\equal{}diag(b_1,\cdots,b_n)$ where $ a_i,b_i\in\{0,1\}$.
Conclusion: $ det(A\plus{}B)\equal{}det(A\minus{}B)\in\{0,1\}$.


Yes, or at least use standard forms of functions and operators.

sqroot([i]x[/i]) is the square root of [i]x[/i]
[i]x[/i]^[i]y[/i] is the [i]x[/i] raised to the [i]y[/i]th power
[i]x[/i]*[i]y[/i] is the product of [i]x[/i] and [i]y[/i]

Only use {} for sets, etc.

[quote="hien"]Find all pair of numbers (x,y) that satifies the following equation:
[code](3+2cos(x-y))= \sqrt {3+2*x-x*2}.(1+cos(x-y))+(1-cos(x-y)[/code][/quote]

I think the correct question is:
$3+2cos(x-y)= \sqrt{3+2x-x^2} (1+cos(x-y)) + (1-cos(x-y))$
It's an easy equation... :lol:


function
inequalities
Cauchy Inequality
algebra proposed
algebra

For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\cdots,n, \; j = 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where
\[ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j=1}^{m}a_{ij}^2}. \]

[quote="Fang-jh"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \equal{} 1,2,\cdots,n,j \equal{} 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \equal{} \frac {n\sum_{i \equal{} 1}^{n}(\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} m\sum_{j \equal{} 1}^{m}(\sum_{i \equal{} 1}^{n}a_{ij})^2}{(\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} mn\sum_{i \equal{} 1}^{n}\sum_{i \equal{} j}^{m}a_{ij}^2}$.[/quote]

Dear fang-jh,
I was wondering if there was a typo in the denominator in this term: $ \sum_{i \equal{} 1}^{n}\sum_{i \equal{} j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j\equal{}i$ to $ j\equal{}m$, not from $ i \equal{} j$ to $ i\equal{}m$ ; and the (outer) sum from $ i \equal{} 1$ to $ m$, not $ i \equal{} 1$ to $ n$ (since if $ n \geq i \geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this:
$ \sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij}^2$?

[quote="epitomy01"][quote="Fang-jh"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \equal{} 1,2,\cdots,n,j \equal{} 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \equal{} \frac {n\sum_{i \equal{} 1}^{n}(\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} m\sum_{j \equal{} 1}^{m}(\sum_{i \equal{} 1}^{n}a_{ij})^2}{(\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} mn\sum_{i \equal{} 1}^{n}\sum_{i \equal{} j}^{m}a_{ij}^2}$.[/quote]

Dear fang-jh,
I was wondering if there was a typo in the denominator in this term: $ \sum_{i \equal{} 1}^{n}\sum_{i \equal{} j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j \equal{} i$ to $ j \equal{} m$, not from $ i \equal{} j$ to $ i \equal{} m$ ; and the (outer) sum from $ i \equal{} 1$ to $ m$, not $ i \equal{} 1$ to $ n$ (since if $ n \geq i \geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this:
$ \sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij}^2$?[/quote]
Dear epitomy01,
I check and compare with the Chinese version carefully, find no typo.

[quote="Fang-jh"]For two given positive integers $ m,n > 1$, let $ a_{ij} (i \equal{} 1,2,\cdots,n,j \equal{} 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f \equal{} \frac {n\sum_{i \equal{} 1}^{n}(\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} m\sum_{j \equal{} 1}^{m}(\sum_{i \equal{} 1}^{n}a_{ij})^2}{(\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} mn\sum_{i \equal{} 1}^{n}\sum_{i \equal{} j}^{m}a_{ij}^2}$.[/quote]

A typo exists. It should be

$ f \equal{} \frac {n\sum_{i \equal{} 1}^{n}(\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} m\sum_{j \equal{} 1}^{m}(\sum_{i \equal{} 1}^{n}a_{ij})^2}{(\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij})^2 \plus{} mn\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{m}a_{ij}^2}$

See

shalex is right, please moderator correct it.
thank a lot!

The minimum value is $ \frac{m\plus{}n}{1\plus{}mn}$ while the maximum value is $ 1$.

Correct ?

The maximum value is $ 1,$ the minimum value is $ \frac {m\plus{}n}{mn\plus{}min\{m,n\}}$.

Ahh, I realize it now. In my proof, the equality for the minimum value is $ a_{ij}\equal{}0 \forall i \not\equal{} j$ and $ a_{ii}\equal{}\text{const} \forall i\equal{}1, 2, ..., min(m,n)$.

For sake of convenience, denote
$ \sum_{j \equal{} 1}^{m} a_{ij} \equal{} x_i$
$ \sum_{i \equal{} 1}^{n} a_{ij} \equal{} y_j$

What we have is $ f \equal{} \frac {n \sum ( x_i )^2 \plus{} m \sum ( y_i )^2 }{(\sum \sum a_{ij} )^2 \plus{} mn \sum \sum a_{ij}^2}$

(1) Maximum of $ f$
We shall prove that the maximum is 1.
It suffices to show that
$ n \sum (x_i )^2 \plus{} m \sum ( y_i )^2 \leq (\sum \sum a_{ij} )^2 \plus{} mn \sum \sum a_{ij}^2$

Consider the two following sequence of  $ m^2 \plus{} n^2$ nonnegative real numbers;

$ (x_1 , x_1 , \cdots , x_1 , x_2 , \cdots , x_n , y_1 , y_1 , \cdots , y_1 , y_2 , \cdots , y_m )$
(Where each $ x_i$ appears $ n$ times and each $ y_i$ appears $ m$ times)
$ ( \sum \sum a_{ij} , a_{11}, a_{12} , \cdots , a_{mn} , 0 , 0 , \cdots , 0 )$
(the 0's exist to make the number of terms equal)

It is obvious that the latter sequence majorizes the former sequence (The largest is the largest, and any other term in the latter sequence is smaller in size to an appropriate term in the former sequence)
Since the function $ g(x) \equal{} x^2$ is convex in the nonnegative reals, applying Karamata inequality, we arrive at the result.

(2) Minimum of $ f$
We wish to show that $ f \geq \frac {m \plus{} n}{mn \plus{} min(m,n)}$.
Assume $ m \geq n$, so that $ min(m,n) \equal{} n$
For sake of convenience, denote
$ \sum (x_i )^2 \equal{} A$
$ \sum (y_i )^2 \equal{} B$
$ \sum \sum a_{ij}^2 \equal{} C$
$ (\sum \sum a_{ij})^2 \equal{} D$

What we wish to prove is that
$ \frac {nA \plus{} mB}{mnC \plus{} D} \geq \frac {m \plus{} n}{mn \plus{} n}$ .. (*)

$ (nA \plus{} mB)(mn \plus{} n) \geq (mnC \plus{} D)(m \plus{} n)$

By the Cauchy Inequality, we have $ nA \geq D$, or $ (m \plus{} n)nA \geq (m \plus{} n)D$ .. (1)
Since $ \sum (x_i )^2 \equal{} \sum (\sum a_{ij} )^2 \geq \sum \sum a_{ij}^2$, we have $ A \geq C$
likewise, we have $ B \geq C$
Therefore, we have that
$ n(mn \plus{} n)A \geq n(mn \plus{} n)C$ .. (2)
$ (m^2 n \minus{} n^2 )B \geq (m^2 n \minus{} n^2 )C$ .. (3)
Adding (1), (2), (3) yields (*), the desired result.

My solution for (1) is incorrect - the upper bound solution is nonsensical. I was drowzy when I wrote it =[

Here's a modified version, please edit the post (I don't know why, but I lost my right to do so):

Upper bound - it suffices to show, as mentioned,
$ n \sum x_i^2 \plus{} m \sum y_i^2 \leq (\sum \sum a_{ij} )^2 \plus{} mn \sum \sum a_{ij}^2$
Or, equivalently,
$ n \sum x_i^2 \minus{} (\sum \sum a_{ij})^2 \leq m \sum ( n \sum a_{ij}^2 \minus{} \sum y_i^2 )$
Or, equivalently,
$ n \sum x_i^2 \minus{} (\sum x_i )^2 \leq m \sum (n \sum a_{ij}^2 \minus{} \sum y_i^2 )$ (because $ \sum \sum a_{ij} \equal{} \sum x_i$)
Now, recall the well-known identity $ n \sum a_i^2 \minus{} (\sum a_i )^2 \equal{} \sum (a_i \minus{} a_j )^2$
The inequality becomes
$ \sum (x_i \minus{} x_j )^2 \leq m \sum ( \sum (a_{ik} \minus{} a_{jk})^2 )$
Applying the cauchy inequality in the form
$ (x_i \minus{} x_j )^2 \equal{} (\sum a_{ik} \minus{} \sum a_{jk})^2 \leq m \sum (a_{ik} \minus{} a_{jk})^2$ and summing them up yields the result.

Let $c_i=\sum\limits_{j=1}^ma_{ij},1\leq i\leq n,r_j=\sum\limits_{i=1}^na_{ij},1\leq j\leq m$.
[color=#f00]1 we find the maximum value of f.[/color]
For $1\leq i\leq n,1\leq j\leq m$, let $a_{ij}=1$, then $f=1\Rightarrow f_{\max}\geqslant 1$. We now prove $f\leqslant 1$.
$f\leqslant 1\Leftrightarrow  n\sum\limits_{i=1}^nc_i^2+m\sum\limits_{j=1}^mr_j^2\leqslant \left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2+mn\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2$
$\Leftrightarrow n\sum\limits_{i=1}^n\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left(a_{ij_1}-a_{ij_2}\right)^2\geqslant\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left(r_{j_1}-r_{j_2}\right)^2=\sum\limits_{1\leqslant j_1<j_2\leqslant m}\sum\limits_{i = 1}^{n}\left(a_{ij_1}-a_{ij_2}\right)$
$\Leftrightarrow\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left[\sum\limits_{1\leqslant i_1<i_2\leqslant n}\left(a_{i_1j_1}-a_{i_1j_2}-a_{i_2j_1}+a_{i_2j_2}\right)^2\right]\geqslant 0$ which is obviously true.
[color=#f00]2 we find the minimum value of f.[/color]
Let $m=\min\{m,n\}$. For $1\leq j\leq m$, let $a_{jj}=1$, otherwise $a_{ij}=0$, them $f=\frac{m+n}{m+mn}$. We now prove $f\geqslant\frac{m+n}{m+mn}$.
$f\geqslant\frac{m+n}{m+mn}\Leftrightarrow  (m+mn)\left(n\sum\limits_{i=1}^nc_i^2+m\sum\limits_{j=1}^mr_j^2\right)\leqslant (m+n)\left[\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2+mn\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2\right]$
Cauchy-Schwarz inequality $\Rightarrow n\sum\limits_{i=1}^nc_i^2\geqslant\left(\sum\limits_{i=1}^nc_i\right)^2=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2,m\sum\limits_{j=1}^mr_j^2\geqslant\left(\sum\limits_{j=1}^mr_j\right)^2=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2$
Let $A=n\sum\limits_{i=1}^nc_i^2,B=m\sum\limits_{j=1}^mr_j^2,C=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2,D=\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2.$ We have $A\geqslant C,B\geqslant C,A\geqslant nD,B\geqslant mD$.
$\Leftrightarrow (m+mn)(A+B)\geqslant (m+n)(C+mnD)$
$LHS=(m+mn)A+(m+n)B+(m-1)nB\geqslant (m+mn)nD+(m+n)C+(m-1)nmD=RHS$


I'm sick today, but I'll be there (with a full team!) if I get better. The last two years they've only recognized the top five individuals, so I might not get an award (especially with Miles and Harrison there).

Edit - here is their website with 2006 and 2007 tests: http://www.mathacre.shorturl.com/

Edit again - I disagree with their answer to 2006 written test #2. Their solution discusses three cases: 17 is the first number, the second number, or the third number. The problem I see is that these cases are not independent. 17 could be two of the numbers or all three. The actual number of ways to have a string of three numbers 0-49 containing 17 is $ 50^3\minus{}49^3\equal{}7351$ - the total number of strings minus the number of strings without any 17's. So the probability that one chooses the correct one is $ \frac{1}{7351}$ (which is not an answer choice). Does anyone agree with me, or did I make a mistake?

Are we allowed calculators? (I want to see the look on Billy's face when he sees this mathematically blasphemous post :rotfl: )

No; I distinctly remember having to calculate $ \binom{20}{10}$ by hand last year.

Oh, well, looks like I won't be going after all, from lack of personnel.  :(

[quote="Twiz"]I disagree with their answer to 2006 written test #2.[/quote]
I think you are right, but you should probably post the problem statement if you want people to discuss something with you. :wink:

Arvind :  Why is your post mathematically blasphemous?  And why do you want to see the look on Billy's face in particular?  Anyhow, if you look on the website, it seems that the answer is "no", except on the power round.  That's how we've done Lassiter the past few years, I think.  Hm, well, I guess your post could be construed as blasphemous if you intended to misuse a calculator.

My team has been downgraded to a three person team due to an illness, but fortunately, the team is composed of three of our best people.

[quote]Arvind : Why is your post mathematically blasphemous? And why do you want to see the look on Billy's face in particular? Anyhow, if you look on the website, it seems that the answer is "no", except on the power round. That's how we've done Lassiter the past few years, I think. Hm, well, I guess your post could be construed as blasphemous if you intended to misuse a calculator.[/quote]

Long story...ask me later.

I think the written test was easy, mostly because it had little calculus (yay).

The ciphering was harder, but doable. I noticed that not many people had high scores on it (including me).

And my team didn't get power on varsity, can someone explain it?

For the power:
"Let s(n) be the product of the number's digits in base 4" IE s(3) = 3(base 4) = 3.
Find s(1)+s(2)+..+s(255)

First thing:
255 = 3333(base 4)
So we want the sum of the 1 digit, the 2 digit, the 3 digit, and the 4 digit numbers products in base 4.
1 digit: 1, 2, 3 = 6
2 digit: 1, 2, 3 * (each one of these is multiplied by each one of the numbers above, so 1(*1,2,3)+2(*1,2,3)+3(*1,2,3) or simply 6*6 = 36
3 digit: 1, 2, 3 times each of the above 6*36 = 216
4 digit: 1, 2, 3 times each of the above 6*216 = 1296

Then add them all up and you get 1554. That's how my team did it anyway, didn't realize it until the end that it would work like that, we barely got it in on time.

[quote="gtf"]So what were the results?[/quote]
From what I remember (which isn't very much):
Varsity Individuals:
1. Harrison
2. Me
3. Arvind
Ziggurat got 4th or 5th, but I can't remember which.
I got the highest written test score, but Harrison owned me on ciphering.

Varsity Teams:
1. Northview
2. Alpheretta

Sitan won JV individuals by a lot. Northview won JV team.

I'm surprised you remembered who I was heh, hopefully I'll be placing in a lot more tournaments... the AOPS books are a huge help.
Oh and it was 4th.

Sorry Ziggurat, but what's your name? I didn't catch it at the tournament.

Jeff Powell... it was pronounced "Povveral" I imagine he just read the w as two v's though, heh.
Edit: Anyone know the 3rd place and 4th place teams? The website's not working. I know Etowah got 5th. My coach wanted to see who else placed.

I think GACS and Rockdale placed... :|

I know GACS placed because when they said it a guy came down and they were like "One man team, eh?" and we were all like "Whoa! Oh wait, he's kidding..." But that may have been JV too... :ninja:

[quote="Boy Soprano II"][quote="Arvind_sn"]Well, it's this Saturday. Who's going? I believe it's a new tournament...?[/quote]
No, it's the old Cherokee County tournament, which Creekview hosts this year.

Anyhow, I think I'll be there; I have fond memories (this was the first tournament I ever won).  Plus, they seem to schedule their tournament magically so that every year, I have no schedule conflicts.[/quote]

 :o  :o  :o Miles's first tournament!!! Ahh!!!

Website's mostly back, I know how to do the power question now...I feel so stupid...lol


What exactly is 0? Is it even, odd or neutral.... I think it's even because it's divisble by 2 but other people say it's neutral. Any help would be appreciated.

Thanks

It is neutral.

I really don't remember why but try Wikipedia.

I tried that before I posted and it wasn't very helpful

[quote="cosinator"]I thought it was even.

What's a nuetral number??[/quote]

Well generally it is different from any other number. 0 is a 'special' number and stands out.

0 is even. Just look at it as $0=2(0)$. Pretty simple.

0 is very special  :P
There is a debate of what 0/0 equals (DO NOT START A DEBATE!!!) and also about some other things concerning 0.
0 is neutral in the sense of positives and negatives but I think it would be even because some even numbers are 2, 4, and 6. They all differ by 2 so then 2-2=0 which is even.

[quote="Quevvy"][quote="13375P34K43V312"]A number is defined to be even if it can be expressed in the form $2k,k\in\mathbb{Z}$. You get 0 when you plug in k=0, so 0 fits the definition. So 0 is even.[/quote]
Then would odd numbers be $2k-1,k\in\mathbb{Z}$?[/quote]

That is correct.

Let's start a debate about what 0/0 equals :harhar: :roll:

it's even I think... beecause like every 2 numbers are even and 2 is even so 0 should be.

Even numbers are the integers divisible by $2$.  Zero is such an integer, so it is an even number.  Otherwise the even numbers would not be an ideal.

i agree that 0 is even and definetely not odd.
i don't quite get "neutral" numbers.
it doen't make sense when talking about odd and even

??? :huh:

"Neutral" refers to positive and negative. 0 is neither positive nor negative, but is even.


Do we want to have a stub template for pages which have been created but need expansion (as Wikipedia does)?  There seem to be some pages which have been started and have useful information, but really deserve to be lengthened to do their subject justice.  The stub template would be an easy way to find such pages.

Yes, that's a great idea.  I just made a stub template.  To insert the stub template into a page, just type in {{stub}}.

Also, if anyone has any ideas for other templates that would be useful, post them here or make it and let everyone know.

Wikipedia has a ton of templates, but we aren't wikipedia.  So don't just go and copy every single one of wikipedia's templates unless you think we could benefit from them.

Disambiguation *might* be useful - it's already got a Special: page listing disambiguation pages, so it seems logical to add it.


probability
combinatorics proposed
combinatorics

Since I didn't find a specific place to post a probability problem  :maybe: , I decided to post it here. I hope you could help me:

Suppose that the occurency of rain depends on the conditions of the wheather the day before. If it rains today, then it could rain tomorrow with probability 0,7 and, if it does not rain today, it could raind tomorrow with probability 0,4; Knowing that it rainned today, what's the probability of raining the day after tomorrow?

tanks

[hide]Since we know that it rained today, the probablity of it raining tomorrow is $.7$ and the probability of it no rain is $1-.7 =.3$.  The probability of it raining the day after tomorrow will depend on what it does tomorrow. Here are the cases:
Case 1. Rain, Rain
Case 2. No Rain, Rain

So we have $(.7)(.7) = .49$ for Case 1, and $(.3)(.4) = .12$ for Case 2.
$.49+.12 = \boxed{.61}$[/hide]


anyone else making this decision, i'm having a hell of a time and would appreciate any discussion on it.

[quote="djimenez"]Well, it depends on a lot of variables. I think Duke use to be more famous (and scores higher in the rankins) when we talk about math. Nevertheless, it is not really wise to consider only the ranks as a decision mean. If you are considering grad school, I would say that if you are planning to do something in Algebra, I suppose that Chicago would be your place (they do a lot of things on Algebra, and they have a lot of good people there, or at least that is what a few friends think), but otherwise I would say that probably Duke has more variety (I think). If it is an undergrad, well, they are both great universities, nevertheless I would like better the weather in North Carolina than in Illinois  :D but that is my personal point of view![/quote]

Having 'good people' in algebra at UChicago is like the understatement of the year.  Anyways, I remember recalling that they're quite strong in analysis as well.  They are both excellent schools for Math; Duke probably just tends to be better for non-math related things (such as sports  :D )

Probably Dr. Merryfield has something to add?

i would choose duke, bc i'm from the south :D
but it depends what you like and prefer.
you can take both campus tour, then decide

Well, Duke and Chicago are different as undergraduate institutions (which is what I assume Bubbloy is thinking of; if not, let me know and I'll offer some advice on how to choose a graduate program).  For example, Chicago is the paragon of classical liberal-arts education; if you attend, you'll definitely be well-rounded in a deep way.  Duke has more undergraduates and fewer graduate students than Chicago.  My impression is that Duke has a wider variety of pre-professional programs and the like than Chicago, but I could be mistaken.

hmm it seems like duke does have a wider selection while chicago seem to be more structured. does it matter so much where I attend undergrad, are one of these feeder into grad programs or would i be good at both institutions?

Both schools are academically strong, and both schools send lots of students to good grad schools. You will have an opportunity to learn a great deal at either school.

I think the main differences are in the settings, the campus life, and the general personalities of the students. In those respects, the schools couldn't be more different. You'll have to examine your personality and your study habits to determine which is best for you.

I understand that this is post-factum, and so I first want to find out where you have decided to go.
Now that that's out of the way, allow me to talk about Chicago's math program. If you have been following recent mathematical development, you will notice that the UofC has grown to be one of the strongest math programs in the country and in the world. The biggest development occured 4 years ago when UofC snagged 4 international mathematical superstars from the former Soviet bloc countries and several American institutions. The leader of the group is Vladimir Drinfeld, a Fields medalist and a genious in his field.
As to the question of undergraduate mathematics at the UofC: it is probably one of the strongest programs you can attend, and I guarantee it is stronger than Duke's (I am sorry if I am offending anyone at this moment). Because of the emphasis on theoretical understanding you are guaranteed to come out of the UofC with a solid mathematical understanding on the world. And yes, the University of Chicago is a feeder into the largest graduate programs in the world of academia. Actually, 1 of 7 graduates of the College enter into academia (read as become professors or anything of that sort).
Obviously I am partial as I will be attending the UofC next year, but I do believe that this is a fairly accurate description of the program (also, the 1995 NRC ratings place Chicago in 5th place for mathematics, while DUke is 34rd, ranking below UMCP even - linky: http://www.stat.tamu.edu/~jnewton/nrc_rankings/area31.html )
These are my two cents.


geometry
parallelogram
combinatorics proposed
combinatorics

Here's a cute problem :):

Can a convex polyhedron having all faces parallelograms have exactly $1995$ faces?

I have a conjecture: the polyhedron must have an even numebr of faces...is it right?? ;)

I think so :). I posted it a very long time ago ;).


rotation
geometry
geometric transformation
email

The squares PQRS and LMNO have equal sides of 1 meter and are initially placed so that the side SR touches LM as shown. The square PQRS is rotated about R until Q coincides with N. The square is then rotated about Q until P coincides with O.

It is then rotated about P until S coincides with L and then finally rotated about S until R coincides with M and the square is now back to its original position.

[url=http://img387.imageshack.us/my.php?image=diagramub8.png][img=http://imageshack.us/thumbnmail.png][/url]

Find the exact length of the path traced out by the point P in these rotations.

I was hoping that someone would be able to show me how they get the solution to this question. May you also show clear workings so I understand where the answer comes from

Many Thanks

[quote="thekiterunner"]The squares PQRS and LMNO have equal sides of 1 meter and are initially placed so that the side SR touches LM as shown. The square PQRS is rotated about R until Q coincides with N. The square is then rotated about Q until P coincides with O.

It is then rotated about P until S coincides with L and then finally rotated about S until R coincides with M and the square is now back to its original position.
[img]http://img387.imageshack.us/img387/3041/diagramub8.png[/img]

Find the exact length of the path traced out by the point P in these rotations.

I was hoping that someone would be able to show me how they get the solution to this question. May you also show clear workings so I understand where the answer comes from

Many Thanks[/quote]

[img]9944[/img]

After the first rotation the point $ X$ goes to $ X_{2}$. It is $ \angle QRQ_{2}=180^\circ$, so $ \angle PRP_{2}=180^\circ$ too.
So the point $ P$ draws a semicircle with radius $ RP=\sqrt 2$

At the 2nd rotation (about $ Q_{2}$) we have $ \angle P_{2}Q_{2}P_{3}= 180^\circ$, so the point $ P$ draws a semicircle with radius $ 1$

At the 3rd rotation $ P$ does not move

At the 4th rotation (about $ S$) we have $ \angle P_{3}SP=180^\circ$, so $ P$ draws a semicircle with radius $ 1$

$ \hline$
[b]Length of semicircle[/b] = $ \pi r$

So we have $ \pi\cdot\sqrt 2+\pi\cdot 1+\pi\cdot 1 = \pi(2+\sqrt 2)$

Anyway, this happens to be an Australian Math Competition problem - I think #22 on the 2006 Senior paper.

If this is a question from the Australian Math Competition paper can you please reply with the internet hyper-link as I would like to view the paper and the solution to this question.

Please Reply!!

If you don't mind me asking, did you construct that diagram yourself or did you manage to find it someone on the internet. If so can you please reply with the hyper-link as I would like to view it as well. If you constructed it yourself great job and do you mind if I use it?

Actually, i can send you a soft copy of the 2006 Senior problems and solutions if u give me your email address. I recently typed it out for the Math Society in my school.

[quote="thekiterunner"]If you don't mind me asking, did you construct that diagram yourself or did you manage to find it someone on the internet.[/quote]
I constructed it with a geometric program

[quote="thekiterunner"]If you constructed it yourself great job and do you mind if I use it?[/quote]Hmm, I could get some easy money for the copyright of this picture  :D

Of course you can use it, my friend! Why would I mind?  :)

Once I upload some picture on the internet, I now that everyone can see it (and use it)


find the maximum k=const;
the inequality is right

($ \frac{a}{b\plus{}c}$+$ \frac{b}{a\plus{}c}$+$ \frac{c}{b\plus{}a}$+1)²≥k$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ac}$+$ \frac{25}{4}$ - k

.You don't know this type of ineq ;) ?. Try $ a \equal{} b;c \equal{} 0$ ,we have :$ \frac {11}{4} \ge k$ ;)
To prove :
if $ a,b,c \ge 0$  then :
$ ( \sum \frac {a}{b \plus{} c} \plus{} 1)^2 \ge \frac {11}{4}.\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} \frac {11}{4} \plus{} \frac {25}{4}$
We use Mixing Varibles
[b]P/s[/b] I don't like to use it ^^

why would the earlier solution not work and why do we have to use mixing
I assumed that the inequality has to be true for all a,b,c

[quote="mitdac123"]But we find the maximum value of $ k$ ;) .
If  don't use mixing ;) ,we use pqr ^^ .I can prove it :)[/quote]
could you please post your solution??mitdac123

$ a,b,c \ge 0$ and no two of which are zero .
Prove that :
$ ( \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}1)^2 \ge \frac{11(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)} \plus{}\frac{7}{2}$
[b]My solution[/b]
Assume that :$ a\plus{}b\plus{}c\equal{}1$
<=>$ ( \frac{1\minus{}q\plus{}2r}{q\minus{}r})^2 \ge \frac{11}{4q}\minus{}2$
$ \bigstar$ .$ 4q \le 1$ .
We prove :
$ \frac{(1\minus{}q)^2}{q^2} \ge \frac{11}{4q}\minus{}2$
$ \leftrightarrow  (4 q \minus{} 1) (3 q \minus{} 4) \ge 0$ (Right because $ q \le \frac{1}{4}$)
$ \bigstar$.$ 4q \ge 1$
Use schur ,we have :$ r \ge \frac{4q\minus{}1}{9}\longrightarrow LHS \ge (\frac{q\minus{}7}{5q\plus{}1})^2$
We prove that :
$ (\frac{q\minus{}7}{5q\plus{}1})^2 \ge\frac{11}{4q}\minus{}2$
$ \leftrightarrow (3q\minus{}1)(17q\minus{}11)(4q\minus{}1)  \ge 0$
Which is obvious true because $ \frac{1}{3} \ge q  \ge \frac{1}{4}$
Done ;)


algebra
polynomial
Galois Theory
superior algebra
superior algebra theorems

hi

can anyone give me an example of a four degree irreducible polynomial over Q with a splitting field of degree 24 (so the galois group is isomorphic to S4)

it would help me finding a number with degree 4 over Q without being constructible

as usual, any more general ideas to find galois groups are welcome

fred

theorem:
Let be $f \in K \subset \mathbb{R}$ ($K$ field) a polynomial of degree four and $g$ it's cubic resolvent.
If $f,g$ are both irreducible (over $K[x]$) and $f$ has no real roots, then you have as Galois group of $f(x)=0$:
- $A_4$ if the discriminant of $f$ is a perfect square (in $K$)
- $S_4$ otherwise

for example take $f(x)=x^4+x^2+x+1$ in $\mathbb{Q}$


geometry
rectangle
analytic geometry
algebra unsolved
algebra

Hello everyone

au = area units

The question is, what are the coordinates of the point P.

xcosx = 0,5

How do you calculate x?

[b]The graph ==> y = cosx[/b]

I am sure, it is impossible to solve this transcendental equation.


For $ a>0,\ b>0,\ c>0$, prove the following inequality.

$ \frac{2}{3}(a\plus{}b\plus{}c)\leq \frac{a^3\plus{}b^3}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^3\plus{}c^3}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^3\plus{}a^3}{c^2\plus{}ca\plus{}a^2}\leq 6(a\plus{}b\plus{}c)$

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0$, prove the following inequality.

$ \frac {2}{3}(a \plus{} b \plus{} c)\leq \frac {a^3 \plus{} b^3}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3 \plus{} c^3}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3 \plus{} a^3}{c^2 \plus{} ca \plus{} a^2}\leq 6(a \plus{} b \plus{} c)$[/quote]
The  inequality follows from $ \frac{1}{3}\leq\frac {a^2 \minus{} ab \plus{} b^2}{a^2 \plus{} ab \plus{} b^2}\leq1.$

[quote="kunny"]That's incorrect in part.[/quote]
Where? Explain please. Thank you.

$ \frac{a^2\minus{}ab\plus{}b^2}{a^2\plus{}ab\plus{}b^2}\leq 1$. :lol:

It IS true, since $ a,b,c$ are positive, so $ 2ab>0$, then $ a^2 \minus{} ab \plus{} b^2 < a^2 \plus{} ab \plus{} b^2$

Unless you're saying it is incorrect because the inequality is strict


algebra
polynomial
induction
calculus
integration
abstract algebra
geometry

It is not true, because there are infinetely many solutions $y^{2}-2x^{2}=1$. It mean $P(x)=2x^{2}+1=y^{2}$ for infinetely many x.

Yes. I forgot about monic.
If $P(x)=x^{2n}+a_{2n-1}x^{2n-1}+...+a_{0}$, then
 $4P(x)=Q(x)^{2}+R(x), \ Q(x)=2x^{n}+b_{n-1}x^{n-1}+...+b_{0})^{2}, \ R(x),$ were $deg(R(x))\le n-1.$
Therefore for big x $|R(x)|<2|Q(x)|-1$. Because $4P(x)$ and $Q(x)^{2}$ are perfect square, it mean $R(x)=0$ for infinetely many x, or $R(x)\equiv 0.$
Because all coefficients $Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even.

These [equations of Pell type] are, essentially, the only exceptions. A special case of Siegel's theorem on integral points on algebraic curves ("there are at most finitely many integral points on an affine algebraic curve $C$ with at least three points at infinity, i.e. such that $\# \tilde{C}\setminus C \geq 3$, where $\tilde{C}$ is the projective closure of $C$ [relative to the considered embedding in affine space]") implies: if $P(x) \in \mathbb{Z}[x]$ is an integer polynomial with at least three distinct roots, then the equation $y^{2}= P(x)$ has only finitely many solutions in integers $x,y$. In particular, any irreducible polynomial of degree $\geq 3$ takes at most finitely many perfect square values.

A proof of this is possible via Roth's theorem of diophantine approximation. (And, as well, other results of diophantine approximations, such as the $2$-variable $S$-unit equation, which can be solved effectively). The general version of Siegel's theorem, valid for arbitrary curves, is proven standardly by embedding a curve in its Jacobian variety and then using Roth's theorem and the arithmetic of abelian varieties. But more elegant (and short, granted "the subspace theorem" of W. Schmidt, which is a far-reaching extension of Roth's theorem) is a relatively recent proof, by Corvaja and Zannier, which doesn't at all use the arithmetic of abelian varieties!

But of course, for the special case you have proposed, no algebraic geometry is needed; it suffices to know what is the "2-variable $S$-unit equation" for number fields ("if $K$ is a number field and $S \subset M_{K}$ any finite set of places, then the equation $x+y = 1$ has only finitely many solutions in $S$-units, i.e. with $|x|_{v}, |y|_{v}\leq 1$ for all $v \notin S$"); it is then straightforward to deduce that $y^{2}= P(x)$ has only finitely many solutions, provided that $P(\cdot)$ has at least three complex roots.

It is a much deeper theorem that if $P(x)$ is an irreducible polynomial of degree at least $5$, then $y^{2}= P(x)$ has at most finitely many solutions in [i] rational[/i] numbers $x,y$; this was proven only in 1983 by G. Faltings (other proofs are known since then, but they are also extremely deep).

For the problem that you propose, you want the degree to be even and $>2$ in order to bound a square between consecutive squares.

[quote="Rust"]Because all coefficients $ Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even.[/quote]

Can you show this induction?

[quote="Rust"] then
 $ 4P(x) \equal{} Q(x)^{2} \plus{} R(x), \ Q(x) \equal{} 2x^{n} \plus{} b_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} b_{0})^{2}, \ R(x),$ were $ deg(R(x))\le n \minus{} 1.$
[/quote]

Actually I don't agree with this!

Considering the degree $ x_{2n\minus{}1}$, you get $ 4b_{n\minus{}1}\equal{}4a_{2n\minus{}1}$, so $ b_{n\minus{}1}\equal{}a_{2n\minus{}1}$.

Considering the degree $ x_{2n\minus{}2}$, you get $ 4b_{n\minus{}2}\plus{}b_{n\minus{}1}^2\equal{}4a_{2n\minus{}2}$, so if $ b_{n\minus{}1}\equal{}a_{2n\minus{}1}$ is odd, you cannot choose $ b_{n\minus{}2}$ to be an integer...

Let me certify Rust's correct argument.

Assume $ P(x)\in\mathbb{Z}[x]$ with degree $ 2n$ and leading coefficient $ c^2$ for some $ n,c\in\mathbb{N}$,
and $ P(m)$ is a perfect square for infinitely many $ m\in\mathbb{Z}$.

Then $ p(x) \equal{} x^{2n}P({1\over x})\in\mathbb{Z}[x]$ of degree $ \le 2n$ and $ p(0) \equal{} c^2$.
So we can write $ p(x) \equal{} c^2 \plus{} xt(x)$ for some $ t(x)\in\mathbb{Z}[x]$.
Then using [url=http://en.wikipedia.org/wiki/Binomial_theorem]Newton's generalized binomial theorem[/url] we get
$ p(x) \equal{} \left(\sum_{k\ge 0}\binom{1/2}{k}c^{1 \minus{} 2k}x^kt(x)^k\right)^2$ with $ \sum_{k\ge 0}\binom{1/2}{k}c^{1 \minus{} 2k}x^kt(x)^k\in\mathbb{Q}[ [x] ]$.
So $ p(x)\equiv q(x)^2\pmod{x^{n \plus{} 1}}$ for some $ q(x)\in\mathbb{Q}[x]$ of degree $ \le n$.
Then $ p(x) \equal{} q(x)^2 \plus{} x^{n \plus{} 1}r(x)$ for some $ r(x)\in\mathbb{Q}[x]$,
and $ x^{n \plus{} 1}r(x) \equal{} p(x) \minus{} q(x)^2$ has degree $ \le 2n$, so $ \deg r(x)\le n \minus{} 1$.

And now $ P(x) \equal{} x^{2n}p({1\over x}) \equal{} x^{2n}(q({1\over x})^2 \plus{} {1\over x^{n \plus{} 1}}r({1\over x})) \equal{} Q(x)^2 \plus{} R(x)$
with $ Q(x): \equal{} x^nq({1\over x})\in\mathbb{Q}[x]$ of degree $ \le n$ and $ R(x): \equal{} x^{n \minus{} 1}r({1\over x})\in\mathbb{Q}[x]$ of degree $ \le n \minus{} 1$.

Let $ d\in\mathbb{N}$ be minimal with $ dQ(x)\in\mathbb{Z}[x]$.
Then $ \deg P(x) \equal{} 2n$ gives $ \deg Q(x) \equal{} n > \deg R(x)$,
and therefore $ 2|dQ(x)| \minus{} 1 > |d^2R(x)|$ for sufficiently large $ |x|$.
Then $ d^2P(x) \equal{} |dQ(x)|^2 \plus{} d^2R(x) > |dQ(x)|^2 \minus{} 2|dQ(x)| \plus{} 1 \equal{} (|dQ(x)| \minus{} 1)^2$
and $ d^2P(x) \equal{} |dQ(x)|^2 \plus{} d^2R(x) < |dQ(x)|^2 \plus{} 2|dQ(x)| \minus{} 1 < (|dQ(x)| \plus{} 1)^2$
for sufficiently large $ |x|$.

So if $ P(m)$ is a perfect square for sufficiently large $ |m|$ with $ m\in\mathbb{Z}$,
then so is $ d^2P(m)$, which gives $ d^2P(m) \equal{} (|dQ(m)|)^2$ and $ P(m) \equal{} Q(m)^2$.
But there are infinitely many such $ m$, which gives $ P(x) \equal{} Q(x)^2$.

And now if $ d > 1$ we take $ p\mid d$ prime, and then $ (dQ(x))^2 \equal{} d^2P(x)\equiv 0 \pmod{p}$.
But $ \mathbb{F}_p \equal{} \mathbb{Z}/p\mathbb{Z}$ is a field, so $ \mathbb{F}_{p}[x]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url],
and therefore has no [url=http://en.wikipedia.org/wiki/Zero_divisors]zero divisors[/url].
This gives $ dQ(x)\equiv 0 \pmod{p}$ and $ {d\over p}Q(x)\in\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$.

So $ d \equal{} 1$ and $ Q(x)\in\mathbb{Z}[x]$.

We can generalize this problem:
Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with degree divisible by an integer $d$. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a $d$th power of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^d$.

[quote=Skravin][quote=K.I.M][quote=bilarev]Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$.[/quote]

It would be a nice problem:
Prove that if $P(x)\in Z[x]$ is a perfect square for all $n\in N$ then there exists $Q(x)\in Z[x]$ such that $P(x)=Q(x)^2$
Use this problem to prove the above problem[/quote]

well in that case we need to prove the following statement

for $\exists n \in \mathbb{N}, \exists x \in \mathbb{Z}$, $P(x)=n^2$ then prove that $\forall x \in \mathbb{Z}$, $P(x)=n^2$ for proper $n$s[/quote]

Could you please explain more?
I don't get what you say

[quote=K.I.M]Could you please explain more?
I don't get what you say[/quote]

I used some letters to lessen the statement but it had rather distracted

"For infinitely many(not could be all) positive integers, $P(x)$ becomes the square number. Then prove that $P(x)$ becomes the square number for all $x \in \mathbb{Z^+}$

i dont think its this hard. actually we need to prove that there exists a an positive integer m such that (x+m)^2 < P(x) < (x+m+1)^2 ( for all x>N0) . ofcourse if we assume that p(x) is not Q(x)^2 . and if we assume that there exists infinitely many POSITIVE integer r such that p(r) is square then things will be trivial . if not then there exists
infinitely many negative integers and so we may need to prove there is a m which (x-m)^2 < p(x) < (x-m+1)^2 . which is not hard to prove . its the summary of my solution . but i can write it completely . just inform me

[quote=olorin]
[...]
And now if $ d > 1$ we take $ p\mid d$ prime, and then $ (dQ(x))^2 \equal{} d^2P(x)\equiv 0 \pmod{p}$.
But $ \mathbb{F}_p \equal{} \mathbb{Z}/p\mathbb{Z}$ is a field, so $ \mathbb{F}_{p}[x]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url],
and therefore has no [url=http://en.wikipedia.org/wiki/Zero_divisors]zero divisors[/url].
This gives $ dQ(x)\equiv 0 \pmod{p}$ and $ {d\over p}Q(x)\in\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$.

So $ d \equal{} 1$ and $ Q(x)\in\mathbb{Z}[x]$.[/quote]
Can anybody deal with this part without using field or domain, i.e in more elementary way? Or you have another more simple solution? Thanks very much!


trigonometry
inequalities
inequalities proposed

Prove that in any triangle

$\sum \sin\alpha\sin\beta \leq\left(\sum\cos\alpha \right)^2$

Please, a trigonometric solution.

Thank you very much.

This inequality is equivalant to
    (1-cosA)(1-cosB)(1-cosC)>=cosAcosBcosC. :cool:

Also my solution is based on Treegoner's inequality.

Multiplying the inequality (see the topic: "Only trigonometric solution, please !!")

$\prod\tan\frac{\alpha}{2}\geq\prod\cot\alpha$              $     (\clubsuit)$

by $\prod\sin\alpha$ we obtain

$\prod(1-cos\alpha)\geq\prod\cos\alpha$

I am curious to know if there is a more direct trigonometric solution not based on inequality $ (\clubsuit)$


Hi,

I was wondering what is the best Chinese University (or Japanese) for Mathematics? How do they compare to American Universities within the mathematics that is taught? For example, does the Chinese Universities teach the equivalent of 3rd year mathematics in their first year?

Thanks everyone for your comments/help?

Hm.. Funny. I haven't heard of much Chinese Universities before. I live in the U.S. so I guess that's the problem.

But there are a lot of good colleges in U.K. and the U.S.

Disclaimer: this is purely my personal knowledge.

Beijing University and Qinghua University (both in Beijing) are the two most prestigious tertiary institutes (in pretty much all aspects) in China; some other good math&science schools I've heard of include Zhejiang University, Wuhan University of Science & Technology, and Jiaotong University (in Shanghai - I'm not sure about math, but they seem to be beastly in computer).

It's generally easier for foreign students to get in top Chinese schools than nationals. However, most ranking systems - mostly based on reserach - place Chinese schools in the lower 100 or so, so you should consider carefully if you really want to go there. Also note that typically the best students in these universities vie for the scarce opportunities to transfer to top U.S. schools like MIT.


Hi, I have another chemistry question:

$ \text{In an experiment it was found that the total charge on an oil drop was } 5.93\times 10^{\minus{}18} \text{ C. How many negative charges does the drop contain?}$

I was wondering if anyone could explain this question to me? The answer is 37.

Thank you!

Well the charge on an electron is 1.6 x 10 ^ -19 . total charge is 5.93 x 10^ -18 , so the number of charged particles is total charge  /  charge on each  particle = 37

Okay, I understand!

My textbook didn't tell me the charge of an individual electron.  :mad:  haha


geometry
3D geometry
sphere
trigonometry
integration
probability and stats

We are given a sphere of radius 1 and a tangent plane. We select a point on the surface of the sphere uniformly. How is distributed the distance between this point and the plane? Calculate it's variance.

Let $ \theta$ be the angle between the radius of the sphere to a given point on that sphere and the radius to the point that contacts the plane. This angle is a random variable which is distributed on $ [0,\pi]$ with density proportional to the circumference of the circle of points lying at that angle. That is, the density of the distribution of $ \theta$ is $ C\sin\theta$ for some $ C.$ Since $ \int_0^{\pi}\sin\theta\,d\theta \equal{} 2,$ we have $ C \equal{} \frac12,$

The quantity we are interested in is $ X \equal{} 1 \minus{} \cos\theta.$ That's the distance to the tangent plane.

$ E(X) \equal{} E(1 \minus{} \cos\theta) \equal{} \frac12\int_0^{\pi}(1 \minus{} \cos\theta)\sin\theta\,d\theta \equal{} 1.$

(This result is exactly what we should have expected - on average, the distance in question is the distance to the equatorial circle.)

$ E(X^2) \equal{} \frac12\int_0^{\pi}(1 \minus{} \cos\theta)^2\sin\theta\,d\theta \equal{} \frac43.$

$ \text{Var}(X) \equal{} E(X^2) \minus{} (E(X))^2 \equal{} \frac13,$ so $ \sigma \equal{} \frac1{\sqrt {3}}.$

[hide="Hidden because it is wrong."]This is exactly half of the standard deviation of a uniform $ [0,2]$ distribution.[/hide]

I think it's exactly uniform [0;2] (not "half")
Recently i've met an old but very beautiful problem using this fact:
http://rand.org/pubs/research_memoranda/2006/RM408.pdf

My incorrect comment is still there, but I just hid it.

You're right. The distribution on $ X$ is exactly uniform on $ [0,2].$ This is effectively the same fact as the following: if a spherical loaf of bread is sliced into $ n$ slices of equal thickness, then each slice contains the same amount of crust. This fact was, in essence, known by Archimedes.

Maybe this is quite funny:

There is a possibility to make a sum of $ n$ uniform rv to be uniform.
Kent's post http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1132376#1132376

But there is a possiblity to make a sum of $ n$ uniform rv to be constant! How?
[hide="hint"]With sphere and $ n$ planes [/hide]

I think the question of whether $ n$ is odd or even makes a difference.

I think there are many ways to do this, here is one:

As i remember the sum of all heights in regular $ n$-gon is constant.
So for $ n>2$ we draw regular $ n$-gon, inscribe a circumpherence.
We draw a sphere (this circumpherence should be the equator of the sphere).
We select uniformly a point on the surface of the sphere.
We project this point to the plane of $ n$-gon.
We have $ n$ heights which are uniformly disributed but their sum is constant.

The case of $ n\le 2$ is easy.


conics
ellipse
geometry
hyperbola
parabola
projective geometry
advanced fields

A variable tangent to an ellipse intersects the tangents from vertices A and A' in T and T'. Find put the locus of the intersection point of TA' and T'A.

It's also an ellipse tangent to the 2 tangents through A and A' in the same points (I'm assuming A and A' are on the elipse, right?).

I don't want to get into details right now, but it's true for projective geometry reasons: the map T->T' is a projective transformation from the tangent through A (call it a) and the tangent trough A' (call it a') (this is a well-known theorem in projective geometry). This means that the map AT'->A'T is a projective transformation from the pencil of lines through A to the pencil of lines through A'. It's also well-known that in this case the locus of the intersection pts of 2 analogous lines (i.e. lines s.t. one of them is the image of the other one in the mentioned projective map) is a conic tangent in A and A' to the lines a and a' s.t. a->AA' and AA'->a'.

This shows that our locus is a conic, and since the point of intersection is always inside the ellipse, the conic can't be a hyperbola, a parabola or degenerate (line(s)), so it must be an ellipse tangent in A and A' to a and a' respectively.

I don't know that much on the subject myself, but I got the impression that "De la matematica elementara spre matematica superioara" by Constantin Avadanei, Neculai Avadanei, Constantin Bors and Cristina Ciurea is a very good introduction to projective geometry (everything I know is taken from the chapter in that book :)).

Anyway, I hope I'm not wrong about this..


trigonometry
logarithms
calculus
calculus computations

im lazy.....lol jk ill get to it tomorrow its late


solve the equation:
$ (2\plus{} \sqrt2)^{log_2x} \plus{} x^2(2\minus{}\sqrt2)^{log_2x} \equal{}1\plus{}x^2$

Let $ y\equal{}log_2(x),2\plus{}\sqrt 2\equal{}4^a$, then equation equavalent to $ f(y)\equal{}2^{(2a\minus{}1)y}\plus{}2^{(1\minus{}2a)y}\equal{}2^{\minus{}y}\plus{}2^y$. If $ y\equal{}zln(2)$, then it equavalent to
$ ch((1\minus{}2a)z)\equal{}ch(z)$,  It give unique solution $ z\equal{}0$ or $ x\equal{}1$.