topology
complex analysis
complex analysis unsolved

Can anyone help me solve this problem ?

Let $\mathbf{X^{*}}$ be the dual space of the Banach space. A sequence $\{f_{n}\}$ in $\mathbf{X^{*}}$ is said to converge weakly to $f \in\mathbf{X^{*}}$ if $f_{n}(x)\to f(x)$ as $n\to\infty$ , for every $x\in\mathbf{X}$ .
Suppose $\mathbf{X}$ is separable , and $\{f_{n}\}$ is a sequence in $\mathbf{X^{*}}$ such that $\{||f_{n}||\}$ is bounded .

Prove that $\{f_{n}\}$ contains a weakly convergent subsequence.

Let the sequence $(x_1,...,x_n,...)$ be a countable dense subset of $X$($X$-separable). Now since all our functionals are bounded by constant $C$ the sequence $f_1(x_1),...,f_n(x_1),...$  is bounded, which means that we can extract a convergent sequence $f_1^{(1)},...,f_n^{(1)},...$ such that $f_1^{(1)}(x_1),...,f_n^{(1)}(x_1),...$ is convergent. Now let's in the same way extract the subsequence $f_1^{(2)},...,f_n^{(2)},...$ of the above sequence such that $f_1^{(2)}(x_2),...,f_n^{(2)}(x_2),...$ is convergent and so on...
$f_1^{(1)},...,f_n^{(1)},...$
$f_1^{(2)},...,f_n^{(2)},...$
$......................................$.
Now, just take the diagonal $f_1^{(1)},...,f_n^{(n)},...$ and get the sequence such that $f_1^{(1)}(x_n),...,f_n^{(1)}(x_n),...$ converges for all $n$, which means that $f_1^{(1)}(x),...,f_n^{(1)}(x),...$ is convegent for all $x \in X$-we used here the denseness of $x_1,x_2,...,x_n,...$ in $X$

[quote="Chun"]Thanks  :)  . But
[quote="eugene"]Now since all our functionals are bounded by constant $C$ the sequence $f_1(x_1),...,f_n(x_1),...$  is bounded, which means that we can extract a convergent sequence $f_1^{(1)},...,f_n^{(1)},...$ such that $f_1^{(1)}(x_1),...,f_n^{(1)}(x_1),...$ is convergent.[/quote]
I think it is the most difficult part in your proof. Could you tell me how to solve this ?  :blush:
 The rest of your proof is the diagonal process.[/quote]
Or, it is easy: you know that in every compact subspace(of the metric space) all the infinite sequences have accumulation points, so if our functional take the values in $\mathbb{R}$ and $|f_1(x_1)|<C,...,|f_n(x_1)|<C,...$, so all the values $f_1(x_1),...,f_n(x_1),...$ are in $[-C,C]$ and since every bounded closed set is compact(Heine-Borel theorem) this set has accumulation point and we can extract such a subsequence.


Find a number of the form 200..07 (with n 0’s, n > 2) which is divisible by 2007.

[hide]
Actually, let there be $ n\minus{}1$ 0's.  Then the number is equal to $ 2\cdot 10^n \plus{} 7$.  So we want to find $ n$ such that $ 2\cdot 10^n \equiv \minus{}7\mod{2007}$.  We can reduce this to $ 10^n \equiv 1000\mod{2007}$.  Taking $ m \equal{} n\minus{}3$, we find that $ 10^m \equiv 1\mod{2007}$.  Obviously $ m \equal{} 0$ is a trivial solution.  Since $ \phi(2007) \equal{} 2\cdot 3 \cdot 222 \equal{} 1332$, Euler's theorem gives us $ m \equal{} 1332$ as a solution.  We can actually reduce this to $ m \equal{} 222$ with multiplicative orders and a good computer algebra system, so the number $ 200\cdots 007$ with 224 0's satisfies the given condition.  So do similar numbers of 0's for 446, 668, etc.
[/hide]

We know that $ 9|(10^k\minus{}1)$ always, so it sufficies to consider $ \text{ord}_{223} (10)|\phi (223)\equal{}222$.

From there we have to test out all factors of $ 222\equal{}2*3*37$.

What does $ \text{ord}_{223} (10)|\phi$ mean?

$ \phi(223)$ is the number of positive integers $ 1 \le a < 223$ such that $ \gcd(223, a) \equal{} 1$.  By Euler's Totient Theorem,

$ 10^{ \phi(223) } \equiv 1 \bmod 223$

However, this is not the smallest exponent such that this is true.  $ \text{ord}_{223}(10)$ is defined as the smallest positive integer such that

$ 10^{ \text{ord}_{223}(10) } \equiv 1 \bmod 223$

From here, it's not hard to prove that $ \text{ord}_{223}(10)$ must divide $ \phi(223)$.

Let $ k$ be the minimal positive integer ($ a>1$) such that $ m|(a^k\minus{}1)$. If $ m|(a^n\minus{}1)$, then $ k|n$.

We can apply the division algorithm so we have $ n\equal{}qk\plus{}r$. Then $ a^n\equiv a^r\equiv 1\mod m$. But then $ 0\le r <k$. If $ r>0$, then $ k$ is not minimal, so $ r\equal{}0$, and we are done.


vector
analytic geometry
geometry
parallelogram
number theory unsolved
number theory

The $Farey$ sequence is defined by all the irreducible fractions $\frac{a}{b}$, where $0\leq\frac{a}{b}\leq1$ and $1\leq b\leq n$.For example some of the first terms of some first sequences are:
$\frac{0}{1},\frac{1}{1}$   ;
$\frac{0}{1},\frac{1}{2},\frac{1}{1}$   ;
$\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1}$   ;
(note : the numbers in this sequence are written in increasing form)
Prove that if $\frac{a_{1}}{b_{1}}$ and $\frac{a_{2}}{b_{2}}$ are two adjacent fractions of a $Farey$ sequence then: $a_{2}b_{1}-a_{1}b_{2}=1$.

I know it must be a famous problem, but I tried my best to solve it....but i couldn't :(

Consider the vectors $\left< a_{1}, b_{1}\right>, \left< a_{2}, b_{2}\right>$.  If they are two adjacent fractions of a Farey sequence then there does not exist a lattice point within the triangle that they form (otherwise the coordinates of that point form a fraction that is between them in the Farey sequence).

By Pick's Theorem the area of that triangle is therefore $\frac{1}{2}$, so the area of the parallelogram that they form (by vector addition) is $1$.

This area can also be calculated using the cross product; ergo,

$a_{2}b_{1}-a_{1}b_{2}= 1$

(It's not hard to verify that this is the positive value.)  QED.

Can anyone recommend any good texts/resources about lattice points? I have seen in used in a few places and would like to learn more about it, however I am not sure where to look.

I think  http://www.cut-the-knot.com   would help, It's cute :)


could we have a southwest u.s. community? right now the closest thing to me is texas, and i'm not from texas. i'm from oklahoma.

Where in Oklahoma? I haven't lived there in more than 30 years, but I grew up in Bartlesville.

i live in duncan. it's a half-hour drive from lawton.

[geography lesson]

Coffeyville, Kansas, is about 40 miles from Bartlesville, Oklahoma. Bartlesville is 50 miles from Tulsa and Coffeyville is 70 miles from Tulsa. Also in the general neighborhood: Caney, KS, Independence, KS, Nowata, OK.

[/geography lesson]

my cousins who live in coffeyville are the relatives who are nearest to us, but we've never visited them. that's kind of odd.


When does equalities occurs for [url=http://www.artofproblemsolving.com/Wiki/index.php/H%C3%B6lder%27s_inequality]this[/url] form of Hölder's Inequality?

lemma: if $ a$ and $ b$ are positive real numbers and  $ p,q>1$, $ 1/p\plus{}1/q\equal{}1$ then $ ab\equal{}a^p/p\plus{}b^q/q$  if and only if $ a^p\equal{}b^q$

so we have

If $ p,q>1$, $ 1/p\plus{}1/q\equal{}1$ , $ \parallel{}fg\parallel{}_1\equal{} \parallel{}f\parallel{}_p\parallel{}g\parallel{}_q$. if and only if

$ |f|^p\equal{}c|g|^q$ almost everywhere.


geometry
rotation
integration
calculus
calculus computations

Consider the curve  C  with parametric equations  x = 5 + 4t - 1/3(t^3) and y = 2 - 2t^2  .

a) Determine the values of  t  where the curve is concave upward or downward.

b) Find the area of the surface obtained when the portion of  C  with  0  <=  t  <= 1  is rotated about the x-axis.


[quote="sonny"][hide]$5+\frac{1}{a+\frac{1}{b}}=5.16$

$5+\frac{1}{\frac{ab+1}{b}}=5.16$

$5+\frac{b}{ab+1}=5.16$

$\frac{b}{ab+1}=\frac{16}{100}$

Now reduce the fraction so that the denominator is odd

$\frac{b}{ab+1}=\frac{4}{25}$

from here b=4 and a=6 therefore a+b=10[/hide][/quote]

Correct, but I'd do

[hide]$a+\frac1b=\frac{100}{16}=\frac{25}{4}$, which gives the same answer.[/hide]


$ f\in C[\minus{}1,1], f^{(n)}(0)\equal{}0$ for any positive integer n.
And $ |f^{(n)}(x)|\leq n!C^n$ holds for any x and n>0, where $ C>1$ is a constant.
Show that $ f(x)\equiv f(0)$

I think the original problem is different from this one I posted,
it is $ |f^{(n)}(x)|\leq n!C^n$, which in the original one is $ C\equal{}1$


Firefox often freezes or just stops responding when it's loading a page from this site. It would go like halfway and then you just can't see the rest of it. And...hmm...hopefully it's not a problem on my side. It's never happened before...so...yea

On which page specifically did that happen?

I use Firefox also and when I just came out of the classroom it would say "(Not Responding)", i.e. it froze.  I don't see any problems getting into the classroom or while I'm in it so I can live with it.

Other than that, I've noticed a few times over the past few days where it would not load for a long time (a couple minutes).  I would try other sites while waiting for it to load and they would load in a matter of seconds.  I specifically remember this happening when I was browsing the AIME PS forum but I think it's happened in a few other forums.  It only happens every once in a while so it's not too big of a problem.

I am the most "active" browser (that is the pages feed to my computer add up to the largest size / ip in average / day) and I use Firefox in 95% of these cases, however I haven't yet encountered what you are experiencing. Do you have the lastest Firefox  / Java versions?

My Firefox should be the latest but I don't know about Java.

Yes, my browser is up-to-date.

It happens mostly when I'm loading a forum or the forum index.

First of all, let me just mention that although I use Firefox allmost all the time, I do not recommend it for browsing. I use it more for debugging, and other advanced features, that normal users have no use for. It takes a [b]HUGE[/b] amount of memory from your computer, so if you don't have a really strong machine, I don't recommend it at all. Instead I recommend you use IE 6.0 and/or Opera 7.5 for browsing :)

Secondly, I would need more data to actually try to understand what's going on and fix it if you really want to stick with Ff :P. So if you please post your configuration (OS version, but also CPU, RAM) that might help. Also please post the PGT (page generation times located at below each page). I assume you are using the AoPS skin right?

Windows XP Prof. SP2
AMD Athlon XP 2600+
512 mb RAM

[quote][ Time: 0.3576s ][ Queries: 11 (0.0169s) ][ GZIP enabled - Debug on ][/quote]
That time there is actually on the slow side. I'm usually at less than .3 (if it loads at all) but it shouldn't matter too much.

Yes, I'm using the AoPS skin.

Also, while you're at it...would you possibly know why my username/password does't seem to save? I've tried clearing cookies and using different browsers (IE 6, Firefox 1, Netscape 7.2).

Ok, what I need you to try is first to see if settings Options >> Board Layout >> Categories Packing >> Yes, helps solve the problem (will change the view of the forum). If that doesn't do it, try using the same settings on the ML skin. Tell me if it's still happening.

As for the password / username, I really can't say. They are working perfect on all those browsers (+Opera 7.54) on my computer :?, so it's not server-side. I must be a setting you have on the those browsers, that prevents password from being cached. Try looking at the cookie / password settings . (btw it's known that Ff has some issues with cookies, so it's best you try clearing out cookies the hard way on IE 6.0 and see if it works there first).

Problem seems to be fixed on IE now. I guess it could be just a Ff thing. Or maybe my ISP? (Richard had said roadrunner wasn't working perfectly with AoPS some time ago)

The cookie settings look fine, but I'll keep testing it. It's never been a problem before...so I'm thinking maybe I downloaded a bad update or something...

Ah yes, roadrunner strickes back. Clone Attack II :D


inequalities proposed
inequalities
algebra

Looks like Lagrange multipliers will do the trick if we find $ \nabla S$ and $ \nabla g$, which we can call $ a\plus{}b\plus{}c\plus{}d$.

[quote=chien than]Given $ a;b;c;d>0$ satisfying $ a+b+c+d=10.$ Find min of
$ S=\frac{a^3+36}{a}+2\cdot \frac{b^4+12}{b^2}+3\cdot\frac{c^5+144}{c^3}+4\cdot\frac{d^6+28}{d^4}$[/quote]
Any one?

[hide=Nice.][quote=dragonheart6][quote=chien than]Given $ a;b;c;d>0$ satisfying $ a\plus{}b\plus{}c\plus{}d\equal{}10$
Find min of
$ S\equal{}\frac{a^3\plus{}36}{a}\plus{}2.\frac{b^4\plus{}12}{b^2}\plus{}3.\frac{c^5\plus{}144}{c^3}\plus{}4.\frac{d^6\plus{}28}{d^4}$[/quote]

We have
$$\frac{a^3 + 36}{a} \ge 21 + 2(a - 3)$$
since $\mathrm{LHS} - \mathrm{RHS} = \frac{(a + 4)(a - 3)^2}{a}$.
(Note: Alternatively, we may use the convexity of $\frac{a^3 + 36}{a}$.)

Similarly, we have
$$2\cdot \frac{b^4 + 12}{b^2} \ge 14 + 2(b - 2)$$
and
$$3\cdot\frac{c^5 + 144}{c^3} \ge 43 + 2(c - 3)$$
and
$$4\cdot \frac{d^6 + 28}{d^4} \ge 23 + 2(d - 2).$$

Thus, we have $S \ge 101$.

Also, when $a = 3, b = 2, c = 3, d = 2$, we have $S = 101$.

We are done.[/quote]

[/hide]
Given $ a,b,c,d>0$ and $ a+b+c+d=10.$ Then
$$\frac{a^3+36}{a}+2\cdot \frac{b^4+12}{b^2}+3\cdot\frac{c^5+144}{c^3}+4\cdot\frac{d^6+28}{d^4}\geq 101$$
[url=https://artofproblemsolving.com/community/c6h365309p2008863]here[/url]


\[ \begin{gathered}
  z_1  \equal{} a \plus{} ib \hfill \\
  z_2  \equal{} \frac{a}
{{\sqrt 2 }}(1 \minus{} i) \plus{} \frac{b}
{{\sqrt 2 }}(1 \plus{} i) \hfill \\
  PT: \arg z_2  \minus{} \arg z_1  \equal{}  \minus{} \frac{\pi }
{4} \hfill \\
\end{gathered}
\]


inequalities
function
geometry
circumcircle
inequalities proposed

Knowing that in any triangle $a+b+c \geq 3\sqrt{3} R$ (this was already posted - even recently - look for topic in "inequalities unsolved) it is enough to prove:

$\sum \frac{a^2}{b+c-a} \geq a+b+c$

Consider casual substitution: $x=b+c-a$, $y=c+a-b$, $z=b+a-c$, so we have:

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

as inequality is homogenous we can add constraint $x+y+z=1$ so we're about to prove:

$\frac{1}{3} \sum \frac{(1-x)^2}{x} \geq \frac{4}{3}$

But this follows from Jensen as function $f(x)=\frac{(1-x)^2}{x}$ is convex

By the way,Megus you are totally wrong here. It is $a+b+c\leq 3\sqrt{3} R$ ;)

Aaaaa, by the way Megus it is too complicated and it isn't worth to do that substitutions since the inequality you supposed it is stronger follows imediately from Cauchy-Schwarz inequality. ;)

You're right Cezar - sorry for this nonsenses above  :blush:  It was like I just remember I've seen this $3\sqrt{3} R$ I opened that topic and looked just at first post which firmly stands $\geq$ so I didn't read further ... thanks for pointing out

There is even a stronger one,but kind of ugly. ;)

[quote="cezar lupu"]Come on,guys.The inequality that Lagrangia posted is too easy.It follows immediately from Chebyshef, $a^2+b^2+c^2\geq 18Rr$ and from $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$.

 The last inequality can be easy deduced from this inequality $xy+yz+zx\geq\sqrt{3xyz(x+y+z)}$ by putting $x=p-a,y=p-b,z=p-c$.

 By the way I have another two solutions for this one. ;)[/quote]

  Oh Cezar , I infer nothing from two your ineqs  :P

Cezar can you post the other two sols you found???? :)
It looks interesting.... :D

[quote="jarod"]Hold on ! This problem is still unsolved  :)[/quote]
why?!

Why is this problem still unsolved? Jarod, I think there is a misunderstanding here. :mad:
 Can't you do all the calculations? I solved almost 85% of the problem. ;)

Socrates I will post those two solutions very soon. ;)

Megus, i think that $(a+b+c) \leq 3\sqrt{3}R$  ;)

[quote="Valentin Vornicu"]That is true e.lopez. It is called Mitrinovici's inequality ;)[/quote]

i know.  :)

[quote="cezar lupu"]e.lopes have you seen post number 5? :P[/quote]

sorry  :D

[quote="cezar lupu"]Come on,guys.The inequality that Lagrangia posted is too easy.It follows immediately from Chebyshef, $a^2+b^2+c^2\geq 18Rr$ and from $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$.

 The last inequality can be easy deduced from this inequality $xy+yz+zx\geq\sqrt{3xyz(x+y+z)}$ by putting $x=p-a,y=p-b,z=p-c$.

 By the way I have another two solutions for this one. ;)[/quote]
can you detail it? and i don't think $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$ is right...

Yes, you are right. Actually, it is $\sum\frac{1}{b+c-a}\geq\frac{\sqrt{3}}{2r}.$ ;)

The solution i gave can be found in my book of inequalities. It is problem GD.56 ;)

[quote="Lagrangia"]Prove that in any triangle $ABC$ the following inequality holds:
$\frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R$.
[/quote]
The above inequality can be refined as follows:
\[\sum\frac{a^{2}}{b+c-a}\geq\frac{R}{2r}(a+b+c)\geq\frac{1}{4}\sum\frac{(b+c)^{2}}{b+c-a}\geq 3\sqrt{3}R, \]
where $a,b,c$ are the sides of triangle $ABC$, $R$ is radius of the circumcircle and $r$ is radius of the incircle.

The second inequality see: http://www.mathlinks.ro/Forum/viewtopic.php?t=63848

[quote="Lagrangia"]Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \]

[i]Laurentiu Panaitopol[/i][/quote]
Let $a+b-c=z$, $a+c-b=y$, $b+c-a=x$, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $\sum_{cyc}\frac{a^2}{b+c-a}\geq3\sqrt{3}R\Leftrightarrow\sum_{cyc}\frac{a^2}{b+c-a}\geq\frac{3\sqrt{3}abc}{4S}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(y+z)^2}{4x}\geq\frac{3\sqrt3(x+y)(x+z)(y+z)}{8\sqrt{xyz(x+y+z)}}\Leftrightarrow$
$\Leftrightarrow2\sum_{cyc}xy(x+y)^2\geq3(x+y)(x+z)(y+z)\sqrt{\frac{3xyz}{x+y+z}}\Leftrightarrow$
$\Leftrightarrow2(27u^2v^2-15uw^3)\geq3(9uv^2-w^3)\sqrt{\frac{w^3}{u}}\Leftrightarrow f(v^2)\geq0$, where $f$ is a linear function.
Hence, by $uvw$ it remains to check only one case: $y=z=1$, which gives
$2(2x^3+2x+2+4x^2+2)\geq3\cdot2(x+1)^2\sqrt{\frac{3x}{x+2}}$,
which is $(x-1)^2(4x^5+32x^4+89x^3+118x^2+85x+32)\geq0$, which is obvious and very very not strong.
Indeed, even the [b]Ji Chen[/b]'s following stronger inequality is very easy.
[quote="Ji Chen"]
The above inequality can be refined as follows:
\[\sum\frac{a^{2}}{b+c-a}\geq\frac{R}{2r}(a+b+c)\][/quote]
It's equivalent to $\sum_{cyc}(x^3y+x^3z+2x^2y^2-4x^2yz)\geq0$. :wink:

My proof for Ji Chen's following inequality:
\sum\frac{a^2}{b+c-a}\geq\frac{R}(2r}(a+b+c)                                          (1)
Using the usual notations ,(1) is equivalent to:
\a^2r_a+b^2r_b+c^2r_c\geq\2s^2R                                                             (2)
Applying  Chebyshef's inequality we get:
\a^2r_a+b^2r_b+c^2r_c\geq\frac{1}{3}(a^2+b^2+c^2)(r_a+r_b+r_c)     (3)
It suffices to show that:
\(a^2+b^2+c^2)(r_a+r_b+r_c\geq\6s^2R                                                     (4)
Are well-known relationships a^2+b^2+c^2=2(s^2-r^2-4Rr and r_a+r_b+r_c=r+4R,so
(4) is rewritten as:
\(s^2-r^2-4Rr)(r+4R)\geq\ 3s^2R                                                                   (5)
After calculations we obtain:
\ s^2\ geq\ frac{r(r+4R)^2}{R+r}                                                                   (6)
which is well-known.

[quote=Lagrangia]Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \]

[i]Laurentiu Panaitopol[/i][/quote]
[url=https://artofproblemsolving.com/community/c6h1710538p11027842]Bosnia and Herzegovina Team Selection Test 1999
[/url]

[quote=Lagrangia]Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \]

[i]Laurentiu Panaitopol[/i][/quote]

this hold:

In triangle,prove that

\[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b
-c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot
 \left( C \right)  \right) \]

[quote=Lagrangia]Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \]

[i]Laurentiu Panaitopol[/i][/quote]

hard is this:
\[ \sum{\frac{a^{2}}{b+c-a}\cos^5{\frac{B-C}{2}}}\geq 3\sqrt{3}R. \]

[quote=xzlbq]
In triangle,prove that
\[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b
-c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot
 \left( C \right)  \right) \][/quote]
Note that
\[\sum \frac{a^2}{b+c-a} = \frac{2p(R-r)}{r},\]
and
\[\sum \cot A = \frac{p^2-4Rr-r^2}{2pr}.\]
Therefore we need to prove
\[12R^2r+Rp^2+3Rr^2-4p^2r \geqslant 0,\]
equivalent to
\[(4Rr+p^2+3r^2)(R-2r)+2r(4R^2+4Rr-p^2+3r^2) \geqslant 0.\]
Which is [url='https://nguyenhuyenag.wordpress.com/2017/01/05/mot-so-bat-dang-thuc-hinh-hoc/']true[/url].

[quote=xzlbq]

this hold:

In triangle,prove that

\[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b
-c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot
 \left( C \right)  \right) \][/quote]
In triangle,Let
$Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{
\it h_c}-r}} \right) $
then prove that

\[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b
-c}}\geq Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot
 \left( C \right)  \right) \]

[quote=xzlbq]
In triangle,Let
$Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{
\it h_c}-r}} \right) $
then prove that

\[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b
-c}}\geq Q \][/quote]
The Ravi's substitution gives
$$(x+y+z)\sum_{sym}(3x^4y^2-x^4yz+3x^3y^3-2x^3y^2z-3x^2y^2z^2)\geq0,$$ which is Muirhead.

[quote=xzlbq]Right side is:

In triangle,Let
$Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{
\it h_c}-r}} \right) $
then prove that

\[Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot
 \left( C \right)  \right) \]

<=>

\[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2}
 \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b
 \right) }{a+b}}\geq 0\]

How to do?[/quote]

\[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2}
 \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b
 \right) }{a+b}}\geq 0\]

<=>

\[{\frac {{a}^{2}}{b+c}}+{\frac {{b}^{2}}{c+a}}+{\frac {{c}^{2}}{a+b}}\geq \frac{3}
{4}\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}}{s}}\]
But prove this stronger:

\[{\frac {{a}^{2} \left( a+b+c \right) }{ \left( b+c \right)  \left( {b}
^{2}+{c}^{2}+{a}^{2} \right) }}\geq \frac{3}{2}\,{\frac {{a}^{2} \left( a \left( b
+c \right) - \left( b-c \right) ^{2} \right) }{{\it \sum} \left( {a}^{
2} \left( a \left( b+c \right) - \left( b-c \right) ^{2} \right)
 \right) }}\]


Find all functions $f: \mathbb R \to \mathbb R$ such that for all reals $x$ and $y$,
\[f(x^2\minus{}y^2)\equal{}xf(x)\minus{}yf(y).\]

[quote="mkn19"]Find all real valued functions defined on the reals such that for every real x,y: $ f(x^2 \minus{} y^2) \equal{} xf(x) \minus{} yf(y)$[/quote]
Let $ P(x,y)$ be the assertion $ f(x^2\minus{}y^2)\equal{}xf(x)\minus{}yf(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}xf(x)$ and so $ P(x,y)$ becomes $ f(x^2\minus{}y^2)\equal{}f(x^2)\minus{}f(y^2)$
$ P(0,x)$ $ \implies$ $ f(\minus{}x^2)\equal{}\minus{}xf(x)\equal{}\minus{}f(x^2)$
These two lines imply that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$

Then $ P(x\plus{}1,0)$ $ \implies$ $ f(x^2\plus{}2x\plus{}1)\equal{}(x\plus{}1)f(x\plus{}1)$ so $ f(x^2)\plus{}2f(x)\plus{}f(1)\equal{}xf(x)\plus{}f(x)\plus{}xf(1)\plus{}f(1)$ and, since $ f(x^2)\equal{}xf(x)$, we get : $ f(x)\equal{}xf(1)$
And it is immediate to check back that this indeed is a solution.

Hence the general solution : $ \boxed{f(x)\equal{}ax}$

[quote="erfan_Ashorion"]we know that f(x+y)=f(x)+f(y) by the proof up;)!!so we can use coshi function and see that f(x)=cx!!sorry for my bad english!![/quote]

no, it's not true. if for a function $f$ we have $f(x+y)=f(x)+f(y)$ and it's [b]continuous[/b] then we can apply cauchy's theorem to state that $f(x)=ax$ for all real variables $x$.

Also from f(x)=0 and xf(x)=f(x^2) we can divide by x^2 to and have g(y)=f(y)/y, but since g(x)=g(x^2) g is a constant say a and thus g(x)=a=f(x)/x and therefore f(x)=ax

Sorry to revive this topic, but can someone please explain why we can conclude from $ g(x) = g(x^2)$ that $ g $ is a constant?

[quote="souheibsolver"]These two lines imply that $ f(x+y)=f(x)+f(y) \forall x,y $

here you've proved it just when one of $ x $ and $ y $ is negative and the other is positive
you haven't done it when they're from the same signal[/quote]
Read more carefully. The second line of "these two lines" prove that $f(x)$ is odd.So ....

Why does this problem look so familiar? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=337857&sid=fb2d18932f71cab5bf436781078d7a5d#p337857

[quote=pco][quote=john10]Sorry to revive . @ Pco can you explain more how did you get cauchy's equation . :blush:
These two lines imply that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$[/quote]

Let $Q(x,y)$ be the assertion $f(x^2-y^2)=f(x^2)-f(y^2)$

Let $x\ge 0$ and $y\ge 0$ : $Q(\sqrt{x+y},\sqrt y)$ $\implies$ $f(x+y)=f(x)+f(y)$
Let $x\ge 0$ and $y\le 0$ : $Q(\sqrt x,\sqrt{-y})$ $\implies$ $f(x+y)=f(x)-f(-y)=f(x)+f(y)$
Let $x\le 0$ and $y\le 0$ : $Q(\sqrt{-x-y},\sqrt {-y})$ $\implies$ $f(x+y)=f(x)+f(y)$
Let $x\le 0$ and $y\ge 0$ : $Q(\sqrt y,\sqrt{-x})$ $\implies$ $f(x+y)=f(y)-f(-x)=f(x)+f(y)$

Hence the claim[/quote]
@Pco , in the seconde case , although y is negative you substitute it by $ \sqrt{-y} $ ( wich is positive) .Similary in the third and last case ? Can you explain please

[quote=yassino]@Pco , in the seconde case , although y is negative you substitute it by $ \sqrt{-y} $ ( wich is positive) .Similary in the third and last case ? Can you explain please[/quote]
$Q(x,y)$ is true $\forall x,y\in\mathbb R$, positive or negative.

So the only caution is to be sure that the square roots are defined ...
And they are in each subcase.

It did not say that $f$ has to be injective.

[quote=Verit]Sorry to revive yet again... but what about $f(x) = 0 \, \forall x \in \mathbb{R}?$[/quote]

That works and [b]pco[/b] addressed that function in his solution.

$f(x)=ax$, let $a=0$ and we have $f(x)=0$.

[quote=erfan_Ashorion]we know that f(x+y)=f(x)+f(y) by the proof up;)!!so we can use coshi function and see that f(x)=cx!!sorry for my bad english!![/quote]

As said thousands of times : Cauchy does not allow to directly conclude linearity from additivity.
Some conditions more are needed (mononotne, continuous, locally lowerbounded or upperbounded, ...)

In this example, $f(-x^2)=-f(x^2)$ shows that function is upperbounded over $\mathbb R^-$ and it would have been sufficient to conclude linearity. I prefered the elementary two lines proof I gave.

Why do we follow all the time that f(x+y)=f(x)+f(y) and f(0)=0??

[quote=pco]In this example, $f(-x^2)=-f(x^2)$ shows that function is upperbounded over $\mathbb R^-$ and it would have been sufficient to conclude linearity. I prefered the elementary two lines proof I gave.[/quote]
[b][u]Wizard_32[/u][/b] just alerted me upon the fact that this phrase is wrong.
And indeed it is.
The key in my first post in this thread is that from $f(x)$ additive and $f(x^2)=xf(x)$ we easily get $f(x)=xf(1)$ (see my first post).

And forget the apparently-simpler-but-unfortunately-wrong phrase quoted.

Is this solution correct?
Basic observations-
$P(0,0)$ gives $f(0)=0$
$P(x,y)$ and $P(y,x)$ give $f$ is odd
$P(x,0)$ gives $f(x^2)=xf(x)$
So the equation transforms to
$f(x^2-y^2)=f(x^2)-f(y^2)$
As $f$ is odd, $-f(y^2)=f(-y^2)$
So the equation again transforms to
$f(x^2-y^2)=f(x^2)+f(-y^2)$
As $f$ is bounded, $f(x)=ax$

I am doubtful with the end


Prove that every rational number can be represented in this form :

$\frac{a^2+b^3}{c^5+d^7},(a,b,c,d) \in N^4$

 :D

You cann't.
I want to know how did you create negative rational numbers? :rotfl:


I have a problem solving this problem, can anyone help assist me on it

Can you guys help me solve this problem
For certain years from 1950 to 1990, the average annual per capita consumption C of cigaretts by Americans (18 and older) can be modeled by C=3248.89+108.64t-2.98t^2 for 0≤t≤40, where t is the year, with t=0 corresponding to 1950.

a) In 1960, the U.S population (18 and over) was 116,530,000. Of those, about 48,500,000 were smokers. What was the average annual cigarette smokers. What was the average annual consumption per smoker in 1960? What was the average daily cigarette consumption per smoker?

Since t=0 corresponds to 1950, t=10 corresponds to 1960. The equation gives the number of cigarettes consumed in a particular year. Plugging in t=10 gives the number of cigarettes consumed in 1960.

Average annual consumption per smoker is the total number of cigarettes consumed that year divided by the total number of smokers.

Average daily consumption per smoker is the total number of cigarettes per day divided by the total number of smokers. To find total number of cigarettes consumed each day, divide the total number of cigarettes consumed that year by the number of days in a year.


geometry
geometric transformation
probability
algebra
polynomial
conditional probability
probability and stats

two players A and B play a game: they lance a coin. Each lance, if the coin is prone, B gives A one unit, and if the coin is face, A gives B one unit. The possiblity of being prone of this coin is p, and all these lances are independent. Initially A has i units and B has N-i units. So which is the possibility that A win? or that A gains his mount of money?

This is the original version of this problem, in french. I hope my translation would be exact.

"Deux joueurs A et B misent sur les résultats successifs du jet répété d’une pièce. A chaque jet, A reçoit une unité de la part de B si Pile est sorti tandis qu’il paie une unité à B dans le cas contraire. Ils poursuivent le jeu tant qu’aucun des deux n’est ruiné. On suppose que les jets sont indépendants et que le côté Pile de la pièce apparaît avec une probabilité p. Soient encore i et N − i les fortunes initiales de A et B respectivement. Quelle est la probabilité que A gagne ? Quelle est la probabilité qu’il y ait un gagnant ?"

Thanks!

The standard English: "prone" is "tails", "face" is "heads", and "lance" is "toss".

This is known as the "gambler's ruin", and the case of a fair coin has appeared before.

The simplest way I know:

Define $ x_j$ to be the conditional probability that $ A$ will win, given that the current state is that $ A$ has $ j$ units of money. Clearly $ x_0\equal{}1$ and $ x_N\equal{}0.$

For $ 1\le j\le N\minus{}1,$ we have the following expression: $ x_j\equal{}px_{j\plus{}1}\plus{}(1\minus{}p)x_{j\minus{}1}$

Write this as $ px_{j\plus{}1}\minus{}x_j\plus{}(1\minus{}p)x_{j\minus{}1}\equal{}0.$

This recursion is a linear constant coefficient second order linear difference equation. Its characteristic polynomial is $ p\lambda^2\minus{}\lambda\plus{}(1\minus{}p)\equal{}0,$ which has roots $ \lambda\equal{}1$ and $ \lambda\equal{}\frac{1\minus{}p}{p}.$

The case of $ p\equal{}\frac12,$ which is to say that of a fair coin, is special as then the characteristic equation has a double root. For now I will assume that $ p\ne\frac12.$ We have then that

$ x_j\equal{}C_1\plus{}C_2\left(\frac{1\minus{}p}{p}\right)^j.$

Putting in the boundary conditions to determine $ C_1$ and $ C_2,$ we obtain

$ x_j\equal{}\frac{(1\minus{}p)^N\minus{}p^{N\minus{}j}(1\minus{}p)^j}{(1\minus{}p)^N\minus{}p^N}$

Setting $ j\equal{}i$ then gives us the answer to the original question.

(The case $ p\equal{}1\minus{}p$ gives us a different, and simpler, algebraic form.)

A closely related question which can be solved by closely related methods: what is the expected length of the game?

[Note: looking at that, It's possible that I have $ p$ and $ 1\minus{}p$ reversed. If I have made that mistake, it should be clear how to fix it.]

thanks very much. I have lost my english since i learnt french!
Kent, your answer is clear and easy to know. The second question is, which is the possibility that there is at least one winner.

This can be answered without even looking at the details of the problem. Since we have a finite Markov process in which every state can be reached from every non-terminal state with positive probability, the probability that the game will terminate is $ 1$. A more interesting question is
[i]What is the expected duration of the game?[/i]

I thought that's what I was asking. "Length," the word I used, and "duration," the word you used, should be synonyms.

Yes, Kent you're perfectly right :blush: I should confess that I didn't bother to read your post to the very end and reacted just to yevon's last post. I'll try to be more attentive :).

i cant solve your problem: what is the expected duration. can you give me a hint. sorry i am not good at math

See my long post. Now let $ y_j$ be the expected remaining duration of the game given that $ A$ currently holds stake $ j.$ We have $ y_0\equal{}y_N\equal{}0.$ Looking at what happens in one game give us

$ y_j\equal{}py_{j\plus{}1}\plus{}(1\minus{}p)y_{j\minus{}1}\plus{}1$

The rest of the problem consists of solving that second order linear constant coefficient nonhomogeneous difference equation.

ohhhhhhhh, ok, iam so stupid.
thank you very much.


geometry
circumcircle
Online Math Open
power of a point
radical axis
geometry proposed

Cyclic quadrangle $ ABCD$.$ O$ is the circumcenter.Diameters $ AC$ , $ BD$ intersect each other at $ E$ and circumcircles of $ \triangle OAD$ and $ \triangle OBC$ intersect each other at $ F$.prove that $ OF$ $ \perp$ $ EF$

I think I got a solution:

[hide]
EDIT: Have not proven that $ O$, $ O_1$ and $ P$ are collinear.
Let $ O_1$ be the circumcenter of $ \triangle OAD$ and $ O_2$ be the circumcenter of $ \triangle OBC$. The radical axis of these two circles is the line through the points $ O$ and $ F$. Since this is the radical axis, $ O_1 O_2$ is perpendicular to $ OF$.

Let the intersection of $ O_1 O_2$ with $ OF$ be $ M$. Since both $ \triangle OO_1 F$ and $ \triangle OO_2 F$ are isosceles, $ M$ is the midpoint of $ OF$. Now draw the line through $ E$ and $ F$ and let the intersections with the circumcircles of triangle $ OAD$ and triangle $ OBC$ be $ P$ and $ N$ respectively.Since $ \angle OPF \equal{} \frac {1}{2}\widehat {OF} \equal{} \angle OO_1 M$ and by similar argument $ \angle ONF \equal{} \angle OO_2 M$. So $ \triangle OO_1 O_2$ and $ \triangle OPN$ are similar and $ \angle O_1 MO \equal{} \angle OFE \equal{} 90^{o}$.[/hide]

[quote="kaitos"] So $ \triangle OO_1 O_2$ and $ \triangle OPN$ are similar and $ \angle O_1 MO \equal{} \angle OFE \equal{} 90^{o}$.[/hide][/quote]

Are u sure about this part?

$ \triangle OO_1 O_2$ and $ \triangle OPN$ are similar, because $ \angle OO_1 M \equal{} \angle OPN$ and $ \angle OO_2 M\equal{}\angle ONP$ (I use the fact that $ O_1 O_2$ is perpendicular to the radical axis and that $ M$ is the midpoint of $ OF$). So $ O_1 O_2$ is parallel to $ PN$ and $ EF$ as $ P,E,F,N$ all lie on the same line. And since $ \angle OMO_1 \equal{}90^{o}$ ($ M$ is the midpoint of $ OF$ and $ OO_1 F$ is isosceles), we have $ \angle OFE \equal{}90^{o}$.

[quote="kaitos"]$ \triangle OO_1 O_2$ and $ \triangle OPN$ are similar, because $ \angle OO_1 M \equal{} \angle OPN$ and $ \angle OO_2 M \equal{} \angle ONP$ (I use the fact that $ O_1 O_2$ is perpendicular to the radical axis and that $ M$ is the midpoint of $ OF$). So $ O_1 O_2$ is parallel to $ PN$ and $ EF$ as $ P,E,F,N$ all lie on the same line. And since $ \angle OMO_1 \equal{} 90^{o}$ ($ M$ is the midpoint of $ OF$ and $ OO_1 F$ is isosceles), we have $ \angle OFE \equal{} 90^{o}$.[/quote]

I think this is wrong

[quote="kaitos"]well I can't see the error, but can you elaborate why it is wrong? (the fact that $ O_1 O_2$ is perpendicular to the radical axis is a lemma).[/quote]

I know that is a lemma but we have to prove $ O$,$ P$,$ O_1$ are collinear then your proof will be complete.


Hello, let $\phi: \mathbb{R}^m\rightarrow \mathbb{R}^n$ be a linear application. What are the conditions for $\phi$ to be open ?
I think it's "surjectiv" but I didn't manage to show it properly, could you help me ?

Thanks

Hnit : May a strict sub-linear space be open ?

This is true in the more general setting of banach spaces:

A continuous linear map $T: B_1\to B_1$, where $B_i$ are Banach spaces is open iff it is onto, but the interesting implication isn't the one for which $\mbox{t\mu t\mu}$ gave a hint :).


One of my friends was wondering about how brain damage effects intelligence. I started wondering too, about how much force to the head it would take to decrease your intelligence. Would preventing the daily few taps to the head I get or stop accidently running into walls improve my intelligence over the long term? Would getting hit with a textbook over the head once ruin someones chances for the IMO? Or is intelligence pretty fixed and nothing but a good high speed crash can decrease it? How much protection does your skull provide?

Come on, not even any opinions or stories???

[quote="ccy"]My 6th grade PE teacher said that every time you hit your head in a wall, you lose 3 neurons...[/quote]
Just 3?
u know sometimes i think a mild hit probly increases yor temporary IQ by increasing the blood circulation.

[quote="rgep"]There are numerous resources at [url=http://dmoz.org/Health/Conditions_and_Diseases/Neurological_Disorders/Trauma_and_Injuries/Brain_Injury/]dmoz.org/Health/Conditions_and_Diseases/Neurological_Disorders/Trauma_and_Injuries/Brain_Injury/[/url][/quote]
Thank you! Does anyone want to discuss those materials?

It seems as if you don't pass out, nothing major happened.

Comments on this? I know getting hit on the head means something is getting hurt, but probably just the skull.

[quote="PenguinIntegral"] Would getting hit with a textbook over the head once ruin someones chances for the IMO? Or is intelligence pretty fixed and nothing but a good high speed crash can decrease it? How much protection does your skull provide?[/quote]

Most people, including IMO medalists, have had a concussion or something close to it at some point in their first two decades of life (and probably multiple such events as infants), not to mention holding their breath for extended periods underwater, riding roller coaster, etc.   To answer your question more directly, I certainly know of people who did well at the IMO after suffering serious head injuries some years earlier.   If you have a serious car crash shortly beforehand, yes, you may have headaches, dizziness or other symptoms that make it hard to concentrate.   The earlier the age of the injury, the more plastic and resilient the brain (or so goes the current theory; more plasticity is discovered each year in adults).

At the professional level, there have been mathematicians who, during their productive years, changed or ended their careers due to brain or brain-affecting illness.

[quote="fleeting_guest"]
At the professional level, there have been mathematicians who, during their productive years, changed or ended their careers due to brain or brain-affecting illness.[/quote]

That is sad...

I would hate nothing more than my entire lifes work to be torn away from me because of such a simple matter as age.

Many of the IMO people I knew were quite clumsy & hit their heads often. They seemed none the worse for it. It is better to worry about diet/ exercise/ actually using the brain, than the occasional bump on the head, as these have a much greater effect on intellegence. Also, the stress/anxiety of worrying about hitting your head may have a worse effect than the actual injury.

[quote="ccy"]My 6th grade PE teacher said that every time you hit your head in a wall, you lose 3 neurons...[/quote]
i should have lost all my neurons :)


If Anna flips 8 coins, what is the probability that she gets more heads than tails?

hello
[hide="solution"]
Bernoulli trials
p=success to get a head
required probability
=${ \frac {1}{2}[1 - \binom{8}{4}\frac {1}{2^8}}$]
[/hide]
Thank u.

This is how I did it: I am not sure if this correct or I was just lucky...

So obviously number of heads = at least 5 and at most 8. so 8 choose 5 + .... + 8 choose 8 (which equals 93)

there are 2^8 = 256 total possibilites, so probabiliy = 93/256

this is correct but my way of doing it is diff from the solution given so is  my way mathematically correct?

Here's how I did the problem.

[hide="Solution"]Either there will be more heads than tails, or vice versa, or there will be the same number of heads and tails.

The number of possibilities with more heads than tails is the same as the number of possibilities with more tails than heads (it's symmetry!)

Now, the total number of possibilities is $2^{8} = 256$. The number of possibilities with the same number of heads and tails is the number of ways to choose 4 heads (or to choose 4 tails) from among the 8 coins, which is $\dbinom{8}{4} = 70$.

So, that means that there are $256 - 70 = 186$ possibilities that there will be more heads than tails or vice versa.

Taking half of that, we get the number of possibilities that there will be more heads than tails: $\frac{186}{2} = 93$.

Since the total number of possibilities was $256$, and since each possibility is equally likely to happen, the probability is $\boxed{\frac{93}{256}}$.[/hide]

Lol imagine being dumb and bashing this with binomial probability, totally couldent be me  :(

I think Alan Liang The... is correct. I did the same way. The way you can add up 8 choose 5 and so on, is to use the formula of n choose x = n choose n - x. But I think this is correct. Then we add them up, getting 56 + 28 + 9 = 93. And just because your solution differs from theirs, doesn't mean it is incorrect.


Find all functions $f: \mathbb R \to \mathbb R$ which satisfy
\[ f(y\minus{}f(x))\equal{}f(x^{2010}\minus{}y)\minus{}2009yf(x)\]
for all reals $x$ and $y$.

[quote="Algadin"]Find all $ f: R\to R$ which satisfy:
$ f(y \minus{} f(x)) \equal{} f(x^{2010} \minus{} y) \minus{} 2009yf(x)$[/quote]

Let $ P(x,y)$ be the assertion $ f(y \minus{} f(x)) \equal{} f(x^{2010} \minus{} y) \minus{} 2009yf(x)$

$ P(x,\frac{f(x)\plus{}x^{2010}}2)$ $ \implies$ $ f(x)(f(x)\plus{}x^{2010})\equal{}0$ and so $ \forall x$ : either $ f(x)\equal{}0$, either $ f(x)\equal{}\minus{}x^{2010}$

As a consequence, $ f(0)\equal{}0$.
Suppose now $ \exists u\ne 0$ such that $ f(u)\equal{}\minus{}u^{2010}$ : $ P(u,u^{2010})$ $ \implies$ $ f(2u^{2010}) \equal{}  2009u^{4020}$ and, since $ RHS\ne 0$, we get $ f(2u^{2010}) \equal{}\minus{}(2u^{2010}) ^{2010}$ and so :

$ 2009u^{4020}\equal{}\minus{}(2u^{2010}) ^{2010}$ which is impossible since $ LHS>0$ while $ RHS <0$

Hence the unique solution : $ \boxed{f(x)\equal{}0}$ $ \forall x$ (which, indeed, is a solution).


two circles of radius 10 are tangent to each other. A tangent is drawn from the center of one circle to the other circle. Find the area of the shaded region. This is one of the best geometry questions I have ever done :!:
If you are stuck then no worries :) ,
[hide="hint"] Right-angled triangle[/hide]

[hide] Let the center of the left circle be $A$, and the center of the right circle be $B$. Let the point of tangency between the line and circle $B$ be called $C$. Let the point of tangency between the 2 circles be $D$. Let the tangent line and circle $A$ intersect at $E$. We can find the area of $\triangle ABC$, then subtract sectors $AED$ and $BCD$. The triangle $ABC$ is a right triangle with a leg $BC=10$ and a hypotenuse $AB=20$. The area of the triangle is $\frac{1}{2}\cdot10\cdot10\sqrt3=\boxed{50\sqrt3}$.
The triangle is a $30-60-90$ triangle, with $\angle CBD=60^\circ$ and $\angle EAD=30^\circ$. Therefore, the area of sector $AED$ is $\frac{30}{360}\cdot10^{2}\pi=\frac{25}{3}\pi$. And the area of sector $BCD$ is $\frac{60}{360}\cdot10^{2}\pi=\frac{50}{3}\pi$. $\frac{25}{3}\pi+\frac{50}{3}\pi=\frac{75}{3}\pi=\boxed{25\pi}$.
The area of the shaded region is $\boxed{\boxed{50\sqrt3-25\pi}}$. :D [/hide]

[quote="236factorial"][hide]Draw a line connecting the two centers (length=20). we see a 30-60-90 triangle (a tangent to a circle is perpendicular to the radius drawn to the point of tangency) and several sectors. The shaded area is

[Area of rt. triangle]-[area of sectors]=${50\sqrt3-(\frac{30}{360}\cdot100\pi+\frac{60}{360}\cdot100\pi})=50\sqrt3-25\pi$. [/hide][/quote]
Nice solution :) , it's just like mine.


linear algebra
matrix
inequalities
linear algebra unsolved

let A be a square matrix over a field

define B=adj(A), the adjungated matrix : the element Bij is the determinant when you remove row j and column i
in A and correct with (-1)^(i+j)

now one can easily  proof that A*B=det(A)*I

however, what if A is singular
i want to know the determinant of B
is it always zero
the formula above doesn't help anymore

fred

That formula does help- since $AB=0$ and the only zero-divisors in the rank of matrices are singular, $B$ must be singular. In fact, $B$ has rank 1 if $A$ has rank $n-1$ and $B=0$ if $A$ has rank less than $n-1$. These should be pretty easy to show.

of course! silly of me

however your statement of rank n-1 bugs me :blush:

Of course, $n$ is the size of the matrix here.

For $n\times n$ matrices, $\min(\mathrm{rank}(A),\mathrm{rank}(B))\ge\mathrm{rank}(AB) \ge\mathrm{rank}(A)+\mathrm{rank}(B)-n$
If $AB=0$, this implies that $\mathrm{rank}(A)+\mathrm{rank}(B)\le n$
When $A$ has rank $n-1$, this means that either $B=0$ or $\mathrm{rank}(B)=1$.

hmm i kinda feared that equation would be the trick

but i refrained from it because i don't even know how to prove that

i'm not asking to prove it but could you give me its name? ( the first inequality is obvious)

i'm interested because you can dedue nullity(AB)<=nullity(A)+nullity(B) from it, am i correct? :?

and how about this problem :
prove adj(AB)=adj(B)adj(A) for all A,B in $M_n(K)$
when A,B are not singular then easily to show
but when they are singular, i think it's very difficult.Who can do it?

you can suppose your field is infinite, otherwise take the closure of $K$ for example.
If $B$ is invertible:
Then consider $M(x)=Adj((A-xI)B)-Adj(A-xId)Adj(B)$. This is a matrix whose coefficient are in $K[X]$, each coefficient of degree at most $n$, with infinitely many roots. So $M(0)=0$.
This is why $Adj(AB)=Adj(A)Adj(B)$ if $B$ is invertible. You can conclude in the same way for the general case.


geometry
perimeter
trigonometry
inequalities
combinatorics proposed
combinatorics

Given a convex polygon $ G$, show that there are three vertices of $ G$ which form a triangle so that it's perimeter is not less than 70% of the polygon's perimeter.

I got an easy bound of $ \frac 2 {\pi}$  :(

Pierre.

In fact, Schneider proved that
Let $ C$ is a convex set in the plane and $ n \geq 3$.
Let $ Q_n$ (resp. $ q_n$) denotes an $ n$-gon of minimum (resp. maximum) perimeter circumscribed about $ C$ (resp. inscribed in $ C$). Then :

$ Per(Q_n) \leq Per(C) \frac n {\pi} \tan( \frac {\pi}n)$ and $ Per(q_n) \geq Per(C) \frac n {\pi} \sin(\frac {\pi}n)$.

Both inequalities are tight if $ C$ is an ellipse.

It gives a better bound of $ \frac {3 \sqrt{3}}{2 \pi} \equal{} 0.8269 \cdots$.

Anyway, does anyone have a proof for the proposed problem?

Reference :
R. Schneider, Zwei extremalaufgaben für konvexe bereiche, Acta Math. Acad. Sc. Hung, 1971, 22, p.379-383.

Pierre.


A water tank is in a shape of an inverted right cone. The radius of its base is 16 and it height is 96. What is the height, in feet of the water in the tank if the amount of water is 25% of the tank's capacity? Express your answer in simplest radical form.

[hide]The volume of a cone is $\frac{r^{2}h\pi}3,$ so this particular cone has volume $8192\pi.$ 25% of it's volume would be $2048\pi.$

The triangle formed by the radius, the height, and a segment going from the top to the other endpoint of the radius is similar to the other one formed by the cone of water.

We already know the volume of the smaller cone, so we can find $h,$ its height.

$2048\pi=\frac{\left(\frac{16h}{96}\right)^{2}h\pi}3$

$2048=\frac{h^{3}}{108}$

$h=\sqrt[3]{221184}$

$h=\boxed{48\sqrt[3]{2}}$[/hide]

i used basically the same method.........
[hide]
the volume of the small cone would be 2048(pi)
since the big cone's height and radius share the same
ratio with the small cone's h and r, you get
(r/x)^2(h/x)(pi)/3=2048(pi) where x is the ratio
simplying you get
r^2(h)/x^3=6144
solving for x yields x=cube rt of 4
so the h of the small cone is 96/x
simplified to 48(cube rt of 2)
[/hide]

[hide]
Because the total area of the water is 1/4 of the whole cone, each dimension is $\sqrt[3]{\frac{1}{4}}= \frac{1}{\sqrt[3]{4}}= \frac{\sqrt[3]{2}}{2}$ of the big cone. Multiply this by 96 to get $48\sqrt[3]{2}$.[/hide]

I have no idea how to make cube roots.

[quote="Ubemaya"]I have no idea how to make cube roots.[/quote]
$\sqrt[b]{a}$

[code]$\sqrt[b]{a}$[/code]


function
Gauss
limit
probability
set theory

What are all the degrees of infinity?

When I found out that there are more irrational numbers than rational the reason was the degree of infinity.  So I was just wondering how many degrees there are and what are they.

Why, there are infinitely many, of course ;).

The basic problem of infinity is this:  there are infinitely many natural numbers and there are infinitely many real numbers and yet I can show that it impossible to make a bijection (that is, a relationship that pairs each member of one set to exactly one member of the other, and vice versa) between the two sets.  We don't usually think of this as interesting in finite sets, but it becomes much more interesting and problematic in infinite sets.

[tex]\aleph_0[/tex], pronounced aleph-naught, is the "smallest" size of infinity; it is equal in size to the natural numbers, integers and rational numbers.  The size of the real numbers is C, for the continuum.  [tex]\aleph_1[/tex], pronounced aleph 1, is the size of the set of all subsets of the integers, also known as the power set of the integers.

There are (infinitely :)) many degrees of infinity. The smallest is [tex]\aleph_0[/tex], which is the number of integers. The number of real numbers is [tex]2^{\aleph_0}[/tex], which is equal to [tex]\aleph_1[/tex] iff the Continuum Hypothesis is true. (Actually, that's what the Continuum Hypothesis says.) The number of functions [tex]f:\mathbb{R}\to\mathbb{R}[/tex] is [tex]2^{2^{\aleph_0}}[/tex], etc.

Edit: Joel beat me to it.

Yes, basically, the infinity of counting numbers is smaller than the infinity of real numbers or irrational numbers.  Aleph null is a fancy way of saying the cardinality (i.e. number of elements in the set) of the set of counting numbers.  However, the set of counting numbers is not the only set with cardinality equal to aleph null. Consider:

2, 4, 6, 8... which has a one to one corespondance w/
1, 2, 3, 4...

Therefore the set 2,4,5,6,8... has cardinality equal to aleph null.  Another way of putting it is that the infinity of 2,4,6,8... is the same as the infinity of 1, 2, 3,4... which is contrary to what has been said before -- there is less of 2,4,6,8... than there is of 1,2,3,4... because you skip a number?  I have confused myself and probably should not have gotten into this.

Maybe someone can clarify?

(unrelated:) The strange thing is, JBL and ComplexZeta not only answered the post at the same time, but both made the same joke!  Also, did you notice there was a RealZeta running around AoPS?  Maybe this should be the next match for your chess tournament!

I don't know the answer to JS1527's question, but here's what it seems to me like.

Since we can associate infinitely many real values to every integer (e.g., for an integer n, all values n :le: r < n+1), the infinity of real values is infinitier than the infinity of integers.

Here's a question that I ran into recently while trying to solve a problem. Let [tex] A_0 = \aleph_0[/tex] and [tex]A_n = 2^{A_{n-1}}[/tex]. What is the cardinality of [tex]A = A_0 \cup A_1 \cup A_2 \cup \ldots[/tex]. I ended up having to do some research to answer this question. It has an interesting and surprising answer.

(By the way, the original question was, if A is an uncountable set is there a set B s.t. |A|=2<sup>|B|</sup>?)

The test as to whether two sets are the same size or not is whether or not one can set up a bijection between them.  There is a very simple bijection between the set of counting numbers 1, 2, 3, ... and the set of even numbers 2, 4, 6, ....  Thus, those two sets have the same cardinality.  There is a classical proof that the real numbers between 0 and 1 are [i]not[/i] the same cardinality as the natural numbers, known as Cantor's Diagonal Proof:

Every real number has a decimal expansion.  Assume that there are only countably many (that is, as many as the counting numbers) reals between 0 and 1.  In that case, we can put them in a list:

0.1546492...
0.9153471...
0.1439754...
0.7465984...
0.4851487...
0.6341890...

and so on.  Now, construct a real number as follows: take your list and select the first digit of the first number, the second digit of the second number, and so on:

0.[b]1[/b]546492...
0.9[b]1[/b]53471...
0.14[b]3[/b]9754...
0.746[b]5[/b]984...
0.4851[b]4[/b]87...
0.63418[b]9[/b]0...
and so on.  Put those digits together, in order:

0.113549...

Now, take a new number which is different in every single place (all infinitely many of them :)) that the number we just made -- say by adding 1 to each digit and changing 9's to 0's:

0.224650...

Now, our list contained all real numbers between 0 and 1, right?  But think about this new number -- it [i]can't[/i] be on our list, since then it would differ from itself in some digit.  Thus, our original assumption, that we could put the reals on such a list, must be false.

Now, with respect to what probability said -- it raises the question, are there as many rationals as reals, as many rationals as integers, or somewhere in between?

Gauss: how is your (second) question different from the continuum hypothesis?
That first question ... icky.  I'm inclined to say that it is something like [tex]\aleph_0^{\aleph_0}[/tex], but I don't even have a good idea what that would mean.

Hm. Is the answer [tex]\aleph_{\aleph_0}[/tex]?

If we take the Continuum Hypothesis to be false, then the answer to the original question should immediately be ``no'' since we can take [tex]|A|=\aleph_1<2^{\aleph_0}[/tex] unless I'm being silly (or uninformed).

[quote="JBL"]

Now, our list contained all real numbers between 0 and 1, right?  But think about this new number -- it [i]can't[/i] be on our list, since then it would differ from itself in some digit.  Thus, our original assumption, that we could put the reals on such a list, must be false.

[/quote]

I don't follow the digit part.

If the new number is the same as one of the old ones, it must have the same first digit, second digit, etc.

Imagine that our number were the 1,000,007th number on the list.  (It has to be somewhere on the list, after all.)  But we know that our number has a different 1,000,007th digit than the 1,000,007th number on the list -- thus, they aren't the same number.  (Actually, I have to admit -- you need to tinker a little bit to avoid the .9999... problem.  But that very minor.)

aleph-aleph-naught?  That sounds better than mine, but I still don't know what it would mean.

ComplexZeta wrote:Hm. Is the answer [tex]\aleph_{\aleph_0}[/tex]?

What is your definition of [tex]\aleph_{\aleph_0}[/tex]?

ComplexZeta wrote:If we take the Continuum Hypothesis to be false, then the answer to the original question should immediately be ``no'' since we can take [tex]|A|=\aleph_1<2^{\aleph_0}[/tex] unless I'm being silly (or uninformed).

Yes, you're right. The question isn't equivalent to the Continuum Hypothesis (as I had originally thought) though, because of sets like the one I described above. Even if the Continuum Hypothesis is assumed to be true the answer is still [hide]undecidable in ZFC[/hide].

[hide]If you'd like to learn more about this, cardinals that can't be expressed as 2<sup>|B|</sup> for any cardinal |B| are called Inaccessible Cardinals. But in order to tell if they exist or not in ZFC you need a new (independent) axiom called the Axiom of Inaccessible Cardinals.[/hide]

http://ii.best.vwh.net/math/cardinals/

[quote="Alison"]I don't know too much set theory, but I believe that the subscripts in alephs should be ordinals, not cardinals.  So the aleph you would be looking for is not [tex]\aleph_{\aleph_0}[/tex], but [tex]\aleph_\omega[/tex].  However, I don't know how you intend to take limits there, except by some process equivalent to unioning them.

[quote="ComplexZeta"]Ok. So let's assume CH. Therefore [tex]2^{\aleph_n}=\aleph_{n+1}[/tex]. (I didn't mess this up yet, did I?) Then I want to have

[tex]\aleph_{\aleph_0}=\lim_{n\to\infty} \aleph_n.[/tex]

That probably didn't make any sense. I need to learn set theory properly.[/quote][/quote]

Well, if you assume the Generalized Continuum hypothesis, it's 2^(aleph_1) or the cardinality of the set of all subsets of the reals, and also the cardinality of the set of functions from the reals to the reals.

If not, well, who knows!

[quote="Yasha"]I have always been plagued by the question: What's aleph_2?[/quote]

Regarding Gauss's question, It seems to me that since A(n) is infinitesimal compared to A(n+1), then the cardinality of A(n) U A(n+1) should just be A(n+1), so the cardinality of the union of all the A's would be equal to A(omega).  However, wouldn't this be the "highest aleph" that satisfies 2^x=x and *does not exist*?  Or would the next number in the sequence just be A(omega+1)?

How would one prove that there are [tex]\aleph_0[/tex] integers as well as [tex]\aleph_0[/tex] rational numbers?

Also, is it true that there are no "infinities" between [tex]\aleph_0[/tex] and [tex]2^{\aleph_0}[/tex]?

Thanks.

[quote="JSRosen3"]How would one prove that there are [tex]\aleph_0[/tex] integers as well as [tex]\aleph_0[/tex] rational numbers?[/quote]

You can easily count the integers: 0, 1, -1, 2, -2, 3, -3, 4, -4, ... You can use this list to make a one-to-one correspondence between the naturals and the integers.

For the rationals you need to make a grid:

1/1 1/2 1/3 1/4 1/5...
2/1 2/2 2/3 2/4 2/5...
3/1 3/2 3/3 3/4 3/5...
4/1 4/2 4/3 4/4 4/5...
5/1 5/2 5/3 5/4 5/5...
... ... ... ... ... ... ... ... ...

Then you can count them like this: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1,... It's OK that you're repeating some, because at least you've counted them all. As shown above, you can show that there is a one-to-one correspondence between the positive rationals and all the rationals.

[quote="JSRosen3"]Also, is it true that there are no "infinities" between [tex]\aleph_0[/tex] and [tex]2^{\aleph_0}[/tex]?[/quote]

It is if you assume the Continuum Hypothesis, which I'm pretty sure was proven to be an axiom. Otherwise, it isn't.

[tex]\omega \neq \omega +1[/tex].
Instead, [tex]\omega = 1 + \omega.[/tex]

[quote="probability1.01]
Err... how do you prove something is an axiom?[/quote]

I think what he is referring to is Paul Cohen's proof that the Continuum Hypothesis is independent of the axioms of Zermelo-Fraenkel set theory (even if you add the axiom of choice) -- so if you want to work (in ZF) with anything that concerns the Continuum hypothesis, you need either to add it as an axiom or add its negative as an axiom.

[quote="ComplexZeta"]But [tex]\omega+1=\omega[/tex] or [tex]\aleph_0+1=\aleph_0[/tex] (I don't know which is correct).[/quote]

[tex]\aleph_0+1=\aleph_0[/tex], but [tex]\omega+1 \neq \omega[/tex]. There's this confusing distinction between cardinal and ordinal numbers. To read more about this try here: http://www.ii.com/math/ch/#math

But Sly Si is sort of on the right track. Assuming that the Continuum Hypothesis is true, the cardinality of each of the sets An is [tex]\aleph_n[/tex]. (If the CH is not assumed, then, well.. lets not even go there right now).

As Allison pointed out, the cardinalities of sets are ordered by [i]ordinals[/i], not by other cardinals. So I believe the cardinality of the set A would be [tex]\aleph_\omega[/tex] - the smallest cardinal larger than the cardinality of all of the An's.

Surprisingly though, the power set of A cannot have the same cardinality as A itself. It can be shown that the power set of [i]any[/i] set (even large and strange ones like this one) is always larger than the original set. So 2|A| > |A|. Assuming the Generalized Continuum Hypothesis is true, I believe the cardinality of 2|A| would be [tex]\aleph_{\omega + 1}[/tex].

However, its undecidable in regular set theory if A is expressable as the power set of some other set. This is all very new (and confusing to me too), so I may not be completely right about of this. I'm still learning about these strange things myself.

[quote="Alison"]

I think what he is referring to is Paul Cohen's proof that the Continuum Hypothesis is independent of the axioms of Zermelo-Fraenkel set theory (which includes the axiom of choice) -- so if you want to work (in ZF) with anything that concerns the Continuum hypothesis, you need either to add it as an axiom or add its negative as an axiom.[/quote]

Zermelo-Frankel set theory (denoted ZF) does not include the axiom of choice. When the axiom of choice is added to ZF you denote it by ZFC. Surprisingly, ZF + the generalized continuum hypothesis (GCH) implies the axiom of choice.

[quote="probability1.01"][quote="Yasha"]

[quote="JSRosen3"]Also, is it true that there are no "infinities" between [tex]\aleph_0[/tex] and [tex]2^{\aleph_0}[/tex]?[/quote]

It is if you assume the Continuum Hypothesis, which I'm pretty sure was proven to be an axiom. Otherwise, it isn't.[/quote]

Err... how do you prove something is an axiom?[/quote]

If you wanted to prove statement p is an axiom, you would have to prove p is independent from all other axioms. Being independent from the other axioms means that you can't prove p from those axioms, (if you could p would be a theorem), and you can't disprove p from those axioms (if you could assuming p would cause a contradiction).

So in order to prove p is an axiom, you'd have to prove that you can neither prove nor disprove p from your original set of axioms. (How you would actually go about doing this I wouldn't know).

The way that you show a statement is independent of a set of axioms is to add the statement to the set of axioms, and then show that a model of this system is consistent if the original set of axioms is consistent.

Then you add the negation of the statement to the axioms and show that a model of [i]this[/i] system is also consistent if the original set of axioms is consistent.

Hence, a statement is independent of a set of axioms if the consistency of the original set of axioms is unchanged when either the statement or the negation of the statement is added.

To be precise, showing that p is independent of your given axioms doesn't really prove that p is an axiom.  Sure, if you needed to use p for some reason, one way to do it would be to assume p as an axiom.  But it might be better to add one or more other, more "natural" axioms that would imply p as a consequence as well as yielding other, hopefully also useful, results.  I believe that is what some people have done recently in constructing interesting models of set theory in which the continuum hypothesis does not hold.

Thanks gauss, I meant to say that the Continuum Hypothesis is even independent of ZF + Axiom of Choice (this is true, I hope?)  Will edit.

Yes, this is correct Allison. What's interesting though, is that the axiom of choice (AC) plays a very crucial role in how we interpret the generalized continuum hypothesis (GCH). Because without AC we have no way of comparing the relative sizes of infinte cardinals. So rather than them forming a "line", they could branch out into "trees", where certain cardinalities are neither larger nor smaller than other cardinalities.


geometry
3D geometry
octahedron
pyramid
number theory
least common multiple

could you guys provide solutions to 10,15, and 29?

http://www.yashar.biz/mathcounts/mathcounts_files/2001/2001%20State%20Sprint.pdf

Thanks in advance.

[hide]10. just in case you don't know... an octahedron is a shape with 8 faces made of triangles... so it looks like
two pyramids stuck together. The four edges you color is just the square bottom where the two pyramids intersect.[/hide]

[hide]15. There are 12 pentagons and 20 hexagons. Since the seams are where two shapes have one edge being the same.. you just find the number of edges and divide it by 2.
12*5+20*6=180.
180/2=90[/hide]

[hide]29.After Janene runs the 16 minutes, she is 16/9 miles ahead.
The LCM of 8 1/4 and 9 is 99.
Divide the numbers and ....
So for every 99 minutes, Janene runs 11 miles, Emily runs 12.
In every 99 minutes, Emily catches up by 1 mile... OMG!!!!
Janene is 16/9 miles ahead so the time would be:
(16/9)(99)=176 minutes[/hide]


Have you ever heard about the Euler line of quadrangle?? :huh:

I have searched but I haven't found anythings.
Can you show me the link having this or tell me about Euler line of quadrangle,please?

I don't know what it is :blush: , so I don't know what link will help, but maybe you should enter "quadrilateral" instead of "quadrangle" in your search.

Oh, thank you very very much.
Your suggestion worked. I have found it!

[quote="Kyoshiro"][url=http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf]Here[/url][/quote]

I think this link is die!  :(

[hide="@@@Kyoshiro"]Can you translate carefully[/hide]

You can try on [url=http://209.85.175.104/search?q=cache:qCSjtWryIaUJ:forumgeom.fau.edu/FG2006volume6/FG200634.pdf+Euler+line+in+quadrilateral&hl=en&ct=clnk&cd=4]this link[/url]

1st of all, I don't understand what it is.
2nd of all, I think I should just wait until I am older and more capable of understanding it.
3rd of all, I have no clue why I set up this post in this format.


[hide]First, we have that the equation is $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=65$. The LCM of all the denominators is 12, so   I multiplied both sides of the equation. I simplified, and I got $6x+4x+3x=780$. Adding the x's, we get $13x=780$. Using the Division Property of Equality, we get $x=60.$ Therefore, there were sixty guests.[/hide]

Sorry if I'm getting this in late... Happy Cinco de Mayo everybody!!! :lol:


Will someone please provide a solution to this? I read the actual solution provided and did not understand.

For all positive integers x, let

$f(x)=\begin{cases}1 & \text{if x=1}\\ \frac x{10}& \text{if x is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}$

and define a sequence as follows: $x_{1}=x$ and $x_{n+1}=f(x_{n})$ for all positive integers $n$. Let $d(x)$ be the smallest n such that $x_{n}=1$. (For example, $d(100)=3$ and $d(87)=7$.) Let $m$ be the number of positive integers $x$ such that $d(x)=20$. Find the sum of the distinct prime factors of $m$.

Ugh, this is an awful problem;; iirc, I think you're supposed to work backwards from 1 to possible starting positions...it's really tedious and annoying, one of the worst #15's I've ever seen.

uhhh...not really...

it is the number of sequences of A,B such that there are 19 symbols, and not more than 9 As in a row

and you count the complement...10,11,...,19 As in a row...

10A, B, etc
B, 10A, B, etc
etc, B, 10 A

so thats 2*2^{19-10-1}+2^{19-10-2}

do that for 10<-11,12,13,14,15,16,17, i.e.

$S=\sum_{k=10}^{17}2^{19-k}+2^{17-k}$

add in 18, 19 cases, which are just 2,1

the total is 2^{19}-(S+3)=...oO this does not work out...w/e

Swell. So I'm getting $109$ using a recursion for $f(x)$... where the total number was $283284$, but I don't think that's right, esp since that's bigger than $2^{19}$, which our number should be just less than...

[b]Edit:[/b] So K81o7 got [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=378142]511[/url]
Ok, that make's sense now!

Could you give your solution please?

The answer is $511$

This thread is a repost of a problem which already exists on the forum.  The Zuton Force gave a link to one of its previous appearences.  There is a full solution there.


search function Harvard college \/closed

I was looking for an old post with the site search function just now, and it appears that site search has been badly corrupted by the forum merger. I tried a search term, "Harvard" that should have got me quite a few different threads, but it only got me two Fun and Games threads, and NO threads from College Forum. I tried some other searches and got no-match results.

Is the database still reliably searchable, or are most threads now invisible to search?

It appears that search now mostly serves up posts formerly hosted by MathLinks, and misses most (all?) posts formerly hosted by AoPS. That is a major drag.

[quote="rrusczyk"]We're working on it![/quote]

Thank you. I knew something was seriously amiss when a search on "Rusczyk" turned up no hits. By the way, I've been wondering about the ethnic provenance of your family name. I had a schoolteacher in eighth grade who was remarkably reliable in guessing students' ethnic heritage from their family names, which is what got me started in this interest. Your name is plainly eastern European, from the spelling pattern, but I can't figure out if the "y" in the name is a better clue to origin or the "cz."

My real-world family name is English, but not easily guessed as such. That's the smallest part of my ancestry, actually.

Best wishes in ironing out all the technical difficulties.

We hope to get the Search fixed over the weekend.

Rusczyk used to be Ruszczyk  I've only met one non-family person in the US who knew this immediately - a Polish immigrant, like my dad's grandfather, who dropped the extra Z to make the name 'easier'.

It's probably unrelated, but since they are both "Search" issues, I thought I would mention this weirdness.

The prominent "View posts since last visit" link at the top of the Forums page seems to behave differently than it used to.

Previously, it appeared to use a time-stamp to show what had been posted since your last visit.

Now, it seems that if you have viewed a thread, it is filtered out of the responses.  So, for example, I posted something about 30 minutes ago.  My own recent posting does not appear in my "View posts since last visit", although posts prior to mine and posts subsequent to mine appear.

After viewing rrusczyk's response in this thread, this thread disappeared from my "View posts since last visit".  After tokenadult responded, this thread reappeared.

[quote="Valentin Vornicu"]It's the unread flags mod that's doing that. What basically happens is that if you have not yet seen a thread it marks it. If you log out, it will still remain marked, so when you come back you can still see the threads that you haven't yet seen. (usually logging out would mean that you have seen all the topics which is not true in general).[/quote]

Is there a way to give users the option to set this unread flagging system to the way it was before for AoPS? Better yet, would it be possible to add a "mark all topics read" button? I usually don't read every thread posted here, so I would like those to be removed from the unread threads list.

Thanks.

[quote="Fierytycoon"]Is there a way to give users the option to set this unread flagging system to the way it was before for AoPS? Better yet, would it be possible to add a "mark all topics read" button? I usually don't read every thread posted here, so I would like those to be removed from the unread threads list.

Thanks.[/quote]There already is a "mark all topics read" on all the pages :)

[quote="Valentin Vornicu"]There already is a "mark all topics read" on all the pages :)[/quote]

Sorry, but I can't seem to find it, hehe. Where on the page is it (I am using the AoPS skin by the way)?

[quote="rrusczyk"]You should see it now right next to the Font size thing at the top of the page (I just added it).[/quote]

I see it. Thanks.

[quote="Valentin Vornicu"]It's the unread flags mod that's doing that. What basically happens is that if you have not yet seen a thread it marks it. If you log out, it will still remain marked, so when you come back you can still see the threads that you haven't yet seen. (usually logging out would mean that you have seen all the topics which is not true in general).[/quote]

I think I understand the change.  But I think there is a use for both modes.  The Mathlinks entry-point has a scrolling window containing recent posts.  The AoPS entry-point has no such window.

You have done away with the old method by which "View posts since last visit" worked.  If this was intentional, and if it is permanent, then I would suggest the link should be renamed to "View unread posts".

It would also be helpful to have a new method to "View all [or recent] posts, sorted by date".  This method would not hide previously read posts.

My mode of browsing has been that I would bring up the "View posts since last visit".  Typically, I would browse the many posts of interest to me.  If I was logged on for a while, I would refresh the page.  Typically, I don't immediately respond to a posting.  The questions and issues roll around in my head for a while before I fire off a response.  But now, when I am ready to respond, it is tedious to find the posts that I have recently read.  Thanks in advance for considering this suggestion.

[quote="rrusczyk"]We hope to get the Search fixed over the weekend.[/quote]

It looks like search is already back in working order on Saturday afternoon. Thank you for fixing it. This is getting to be quite a cool database for math learners.

[quote="tokenadult"]
It looks like search is already back in working order on Saturday afternoon. Thank you for fixing it. This is getting to be quite a cool database for math learners.[/quote]

Thanks for letting me know you see it operating correctly.  I think we've fixed all the in-post links (which is particularly good for you, as you're one of our most effective post cross-linkers!).  Please let me know if I mised any.


geometry
geometric transformation
reflection
rotation
modular arithmetic

Lines $l_1$ and $l_2$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l$, the transformation $R(l)$ produces another line as follows: $l$ is reflected in $l_1$, and the resulting line is reflected in $l_2$. Let $R^{(1)}(l)=R(l)$ and $R^{(n)}(l)=R\left(R^{(n-1)}(l)\right)$. Given that $l$ is the line $y=\frac{19}{92}x$, find the smallest positive integer $m$ for which $R^{(m)}(l)=l$.

[hide="Solution"]Let the angle any line makes with the positive x-axis be $\theta$. After two reflections, the angle the line makes with the positive x-axis will be $\theta-\frac{\pi}{35}-\frac{\pi}{27}=\theta-\frac{62\pi}{945}$.

Repeating this process $m$ times gives $\frac{62m\pi}{945}$. We need the smallest $m$ such that $\frac{62m\pi}{945}$ is a multiple of $\pi$, so our answer is $m=\boxed{945}$.[/hide]

A reflection of $l$ over $l_1$ isn't necessarily $\frac{\pi}{35}$...it changes by twice the angle between $l$ and $l_1$...

I don't see why getting a multiple of $\pi$ produces line $l$ again.  Could someone please explain this more clearly?

Thanks

Because a line is the same backwards and forwards. It can be rotated any mutiple of $\pi$ and return to the original.

Let $a$ and $b$ be the angles $l_1$ and $l_2$ make with the x axis, and $x$ be the angle of $l$.

To reflect a line, we subtract the difference between the lines from the line reflecting over.  So reflecting $l$ over $l_1$ is a line with angle $a - (x - a) = 2a - x$.  Similarly, using this formula for reflecting this line over $l_2$, and we have$R(l) = 2b - 2a + x$, in terms of angles.  This simplifies to $x +\frac{8 \pi}{945}$, and we have $R^n(l) = x + \frac{8n \pi}{945}$.

In this notation of angles, a line is defined by an angle $- \frac \pi 2 \le \theta < \frac \pi 2$, so it has a period of $\pi$.  So we want $\frac{8n\pi}{945} \equiv 0 \pmod \pi$.  For this to be true, we have to cancel out the denominator, thus $n \equiv 0 \pmod{945}$ and the smallest positive solution is $\boxed{n=945}$

I know people don't like it when old threads get bumped but I actually have a question. If you look at the two solutions by 4everwise and deej21, both get the same answer of 945, but one of them is wrong as to how to reflect the line $ y\equal{}\frac{19}{92}x$ about lines L1 and L2. My solution was the same as deej21's. I also found that reflecting the line twice increases the angle of that line by $ \frac{8\pi}{945}$ and I did it the same way. But 4everwise is forever wise so I want to be sure.


[quote="wly3298456"]how to you find the remiander when an integer like a^b is divided by 17?[/quote]
You mean remainder(not remiander), right? :D
We can use Fermat's Little Theorem. It says that,
For a prime number, p,
\[a^{p-1}=1(mod\ p)\]
if (a,p)=1.
So, $a^{b+16}=a^b(mod\ 17)$ if a is not a multiple of 17.

$a^{42}=a^{16*2+10}=a^{10}(mod\ 17)$
This is the simplest you can get. The remainder of $a^{42}$ is same as the remainder of $a^{10}$.

[quote="wly3298456"]yeah..I meant remainder...lol

so if the number is a^42, when it's divided by 17, what would be the remainder?[/quote]

[hide]
since $a^{16} \equiv 1\mod{17}$, then $a^{32} \equiv 1\mod{17}$
so you are left with $a^{10} \equiv x\mod{17}$ to find.

shouldnt it depend on what a is?
[/hide]

[quote="wly3298456"]say if the number is like 1577^10, so how can you find the remiander when it's divided by 17?[/quote]

[hide]
First of all $1577 \equiv 13 \equiv -4 \mod{17}$.

So $1577^{10} \equiv (-4)^{10} \equiv 16^{5} \equiv (-1)^5 \equiv -1 \equiv 16 \mod{17}$.[/hide]

are there any good websites that teach Elementary Number Theory?


I can't view the output file in PDF. Does anybody know what went wrong?

Is this coming out of TeXnicCenter?  Namely, did a little dialog box come up that said "Cannot open the output file [somethingorother]"?  If so, then Adobe Acrobat Reader is either acting up or being slow - maybe it's asking you to check for updates or something.  See if this is the problem.


calculus
integration
calculus computations

can you solve :D

$ \int_0^{2\cosh^{\minus{}1}(1/\pi)}\;\;\dfrac{\textbf dx}{1\plus{}\pi^4\sinh^2\,x}$

[hide]
I was thinking of using tanh half angle substitution, and turning sinh^2 into cosh. But I realised the upperbound is complex? Is there a typo
[/hide]

javaman, you are right, the integral was set up incorrectly.
my apologies...  :( here is the [i]corrected[/i] integral...

$ \int_0^{2\cosh^{ \minus{} 1}\pi}\;\;\dfrac{\textbf dx}{1 \plus{} \dfrac{\sinh^2\,x}{\pi^4}}$


algebra
polynomial
number theory proposed
number theory

Find all polynomials $ P(x)\in\mathbb{Z}[x]$,such that $ P(q)$ is prime number,for all $ q\in\mathbb{P}$.

Let $ P(x)\equal{}a_nx^n\plus{}...\plus{}a_1x\plus{}a_0$, and $ p\equal{}P(q)\not\equal{}q$. Then for any k $ p|P(kp\plus{}q)$
Exist infinetely many k, suth that $ kp\plus{}q$ is prime, therefore for suth k we get $ P(kp\plus{}q)\equal{}p$. It give $ P(x)\equiv p$.
If $ P(q)\equal{}q \ \forall q\in P$, then $ P(x)\equiv x$.

[quote="Rust"]Let $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \plus{} a_0$, and $ p \equal{} P(q)\not \equal{} q$. Then for any k $ p|P(kp \plus{} q)$
Exist infinetely many k, suth that $ kp \plus{} q$ is prime, therefore for suth k we get $ P(kp \plus{} q) \equal{} p$. It give $ P(x)\equiv p$.
If $ P(q) \equal{} q \ \forall q\in P$, then $ P(x)\equiv x$.[/quote]

What about $ P(x)\equal{}2$ or $ P(x)\equal{}c, c \in \mathbb{P}$?

See http://www.mathlinks.ro/viewtopic.php?t=91236 for a generalisation.

@SimonM: ne, he didn't miss that case. Read the second line again.

[quote="Rust"]Let $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \plus{} a_0$, and $ p \equal{} P(q)\not \equal{} q$. Then for any k $ p|P(kp \plus{} q)$
Exist infinetely many k, suth that $ kp \plus{} q$ is prime, therefore for suth k we get $ P(kp \plus{} q) \equal{} p$. It give $ P(x)\equiv p$.
If $ P(q) \equal{} q \ \forall q\in P$, then $ P(x)\equiv x$.[/quote]
My solution is the same :) Thank you


Find all pairs $(a,b)$ of positive integers that satisfy the equation \[a^{a^{a}}= b^{b}.\]

The first we has a lemma:
Consider the equation:
$ a^m\equal{}b^n$
Call $ d\equal{}\gcd(m,n)$ then exist $ c\in N$ such that $ a\equal{}c^{\frac{n}{d}},b\equal{}c^{\frac{m}{d}}$
[u]Proof[/u]
We has for each $ p\in P$ then
$ m{v_{p}a}\equal{}nv_{p}b$
Imply that $ \frac{n}{d}|v_{p}a$ $ \frac{m}{d}|v_{p}b$
This lemma was be prove.
Now
Let $ d\equal{}\gcd(a^a,b)$ then $ a^a\equal{}dm,b\equal{}dn$ where $ \gcd(m,n)\equal{}1$
Exist $ c\in N$ such that :
$ a\equal{}c^n,b\equal{}c^m$
It mean that
$ (c^n)^{c^n}\equal{}dm \Rightarrow c^{nc^n}\equal{}dm$
And $ dn\equal{}c^m$
So we have:
$ nc^{nc^n}\equal{}mc^m$
[u]Case 1[/u]$ nc^n\geq m$
$ nc^{nc^n\minus{}m}\equal{}m$
Imply that $ n\equal{}1$ so
$ c^{c\minus{}m}\equal{}m\Rightarrow c\equal{}m\equal{}1$
So $ a\equal{}b\equal{}1$
[u]Case 2[/u]$ m>nc^n$ so $ m\equal{}1\Rightarrow c<1$ tradition!
So the equation has only solution :$ (a,b)\equal{}(1,1)$

Oh , it is the question number 4 of the second round of Iran 2014 !!!

:coool:  :huh:  :|  :no:  :roll:  :wacko:  :noo:


What's the coefficient of the $ m^3n^5$ term in the expansion of $ (m\plus{}n)^8$?

The answer is $ \binom{8}{3}\equal{}56$.

[quote="isabella2296"]In case you don't know how to do: $ \binom{8}{3}$, it's like this.

$ 8 \times 7 \times 6 \implies 336$

$ 336 \div 3! \implies 336 \div 6 \implies \boxed{56}$

If you had, say, $ \binom{5}{2}$ it would be like this:

$ 5 \times 4 \implies 20$

$ 20 \div 2! \implies 20 \div 2 \implies \boxed{10}$

(By the way, the "!" is called factorial. $ 2!$ is $ 2 \cdot 1$. $ 4!$ is $ 4 \cdot 3 \cdot 2 \cdot 1$.)[/quote]

lol izzy

everytime i c 1 of ur posts here, it has a long explanantion  :rotfl:  :rotfl:  :rotfl:


【Help】How can i use "SmallCap" in \section ?

Some method listed follow will not suffice my need.

[size=150]i want "SmallCap" style with "Bold Face" in "\section" in "\documentclass{artilce}" mode, other than \documentclass{amsart}"[/size]

some other [b]suggestions[/b]:

【1】. \section{\normalfont\textsc{Introduction}}, while it is not the Bold Face.
【2】. Using "\documentclass{amsart}". but this make a lot changes in my artilce.

[size=150]So i want "SmallCap" style with "Bold Face" in "\section" in "\documentclass{artilce}" [/size]

Could any one help me?

[b]thanks[/b].

[quote="DPatrick"]Try \section{\scshape Introduction}[/quote]

Thank you for helping me. but it still has the same effect as \section{}, and does not show SmallCal Style.

Could you give me other suggestions? thanks.

use the package [b]titlesec[/b] and then use then:

\titleformat*{\section}{\scshape}

if you want more info on that package, visit:

http://www.ctan.org/tex-archive/macros/latex/contrib/titlesec/titlesec.pdf

ledurt


[quote="dgreenb801"]Can't we just say $ (2009/2008)^{2008} \le e \equal{} 2.71828...$ QED?

By the way, if the left side is 2016/2008 instead of 2009/2008, the inequality reveres.[/quote]

 :huh:  :huh:

Okay, e is the limit of $ (\frac {x \plus{} 1}{x})^x$ as x approaches infinity. This is a strictly increasing function. If you use a calculator to evaluate $ (2009/2008)^{2008}$ you will see that it is indeed very close to but a little less than e.


algebra
polynomial
geometry
3D geometry
analytic geometry
advanced fields
advanced fields unsolved

I have what I think is a proof of this, and I was hoping for some constructive criticism.

The problem is:
Prove the "homogeneous Nullstellensatz," which says if $ \mathfrak{a} \subseteq S: \equal{}k[x_0, \ldots, x_n]$ is a homogeneous ideal, and if $ f \in S$ is a homogeneous polynomial with $ \mbox{deg } f >0$, such that $ f(P)\equal{}0$ for all $ P \in Z(\mathfrak{a})$ in $ \mathbb{P}^n$, then $ f^q \in \mathfrak{a}$ for some $ q>0$.

My attempt at a proof:
The homogeneous ideal $ \mathfrak{a} \subseteq S$ defines a cone over $ Z(\mathfrak{a})$ in $ \mathbb{A}^{n\plus{}1}$ via the map $ \theta: \mathbb{A}^{n\plus{}1}\minus{}\{(0, \ldots, 0)\} \rightarrow \mathbb{P}^n$ sending the point with affine coordinates to the point with homogeneous coordinates.  Since $ f$ vanishes at (the images under $ \theta$ of) all points of this cone (including $ (0, \ldots, 0)$ as $ \mbox{deg } f>0$), we have $ f^q \in \mathfrak{a}$ by the affine Nullstellensatz.

My problem with this is that it seems like I can't really apply the ordinary Nullstellensatz to the affine cone, since I'm not sure if it's an ideal.  Any help or advice would be very appreciated.

Thanks.

I think we should carefully distinguish between the homogeneous ideal $ \mathfrak{a}$ and the ideal $ \mathfrak{a}'$ (without any grading) defining the cone. After that apply the affine Nullstellensatz to $ \mathfrak{a}'$ to get your result.

So what you're saying is that $ \mathfrak{a}'$ is the same ideal as $ \mathfrak{a}$, but without the grading so that its zero set lives in $ \mathbb{A}^{n\plus{}1}$?

Wait, just to clear up the last bit of confusion: is it right that the cone itself is actually just $ Z(\mathfrak{a}')$?  These arguments reducing projective questions to affine questions kind of confuse me.  What is the general procedure?  There are a lot of problems in Hartshorne's AG that say, "reduce to the affine case and use blah blah blah."  It would probably help to be comfortable with this process.

That's right: The general process is to first forget about the grading, apply some theorem, then use it in the graded world - maybe after applying the assumption that all the objects are homogeneous or graded.

There is another approach if the property you need to prove is closed, i.e. where the locus where the property is fulfilled is a closed set. Then it might help to restrict to some affine subset, prove it there and take the topological closure again.


function
integration
geometry
quadratics
area of a triangle
Heron\u0027s formula
algebra

Hey uh im doing some community service at my church and i have to write down some formulas. Does any one know some really cool math formulas you could post?

If you could that would be great   :lol:

The Pancake Function: $ \displaystyle\int^1_0 \frac{x^4(1\minus{}x)^4}{1\plus{}x^4} dx \equal{} \frac{22}{7}\minus{}\pi$ or

Euler's formula: $ e^{i \pi} \plus{} 1 \equal{}0$,

are the two that come to mind.

EDIT: Sorry--just seen that this is classroom maths, so I'm not sure if they will make sense! :maybe:

The pancake function is actually with $ 1 \plus{} x^2$ in the denominator, not $ 1 \plus{} x^4$.

Well, these are kind of simple, but useful:

$ \frac{n(n \plus{} 1)}{2}$

for sum of numbers $ 1 \sim  n$

and

$ n^2$

for sum of numbers $ 1 \sim n\text{th odd number}$

Oh, and also

$ \frac{n(n \minus{} 3)}{2}$

for the number of diagonals in a $ n$th-sided polygon.

Some simple ones

Heron's formula
The area of a triangle given sides lengths a, b, and c, where s=(a+b+c)/2.  The area of the triangle is $ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$

Quadratic formula
$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$
is the solution to $ ax^2\plus{}bx\plus{}c$

The cubic formula would also be a nice one (but don't memorize it if you don't want to waste 329847298570855489213532750 yours of your life!)

Also, the quartic formula, but don't memorize that either unless you have 324897120573986573079281047321632749821362319748923167592013 hours of free time!

yeah,

thanks. I really apriciate it.

p.s. i already knew and agree that the number formula is pretty basic.

My church will really apriciate that.

TYVM   :D

Here are some links for the cubic and quartic formulas:

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

http://en.wikipedia.org/wiki/Cubic_equation

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

http://en.wikipedia.org/wiki/Quartic_equation

Hope these will help!  :)

Yeah, those links really help.

My curch loved the equations guys.

TYVM :lol:

Here are two formula sheets that contain a complete list of [url=http://www.geometry-formulas.net]geometry formulas[/url] and [url=http://www.algebra-formulas.com]algebra formulas[/url].
Hope this is helpful.


geometry
3D geometry
sphere
geometry unsolved

Three mutually perpendicular rays $O_x,O_y,O_z$ and three points $A,B,C$ on $O_x,O_y,O_z$, respectively. A variable sphere є through $A, B,C$ meets $O_x,O_y,O_z$ again at $A', B',C'$, respectively. Let $M$ and $M'$ be the centroids of triangles $ABC$ and $A'B'C'$. Find the locus of the midpoint of $MM'$.

[quote="M4RI0"]Three mutually perpendicular rays $ O_x,O_y,O_z$ and three points $ A,B,C$ on $ O_x,O_y,O_z$, respectively. A variable sphere є through $ A, B,C$ meets $ O_x,O_y,O_z$ again at $ A', B',C'$, respectively. Let $ M$ and $ M'$ be the centroids of triangles $ ABC$ and $ A'B'C'$. Find the locus of the midpoint of $ MM'$.[/quote]
Suppose that $ A ";B ";C "$ be midpoint of $ AA';BB';CC'$
We have:
$ 3\vec{OM} \equal{} \vec{OA'} \plus{} \vec{OB} \plus{} \vec{OC}$
$ 3\vec{OM'} \equal{} \vec{OA} \plus{} \vec{OB'} \plus{} \vec{OC'}$
So $ 3\vec{OS} \equal{} \frac {3}{2}(\vec{OM} \plus{} \vec{OM'})$
$ \equal{} \frac {1}{2}(\vec{OA} \plus{} \vec{OA'}) \plus{} \frac {1}{2}(\vec{OB} \plus{} \vec{OB'}) \plus{} \frac {1}{2}(\vec{OC} \plus{} \vec{OC'})$
=>$ 3\vec{OS} \equal{} \vec{OA "} \plus{} \vec{OB "} \plus{} \vec{OC "}$
[hide]I can't continue :blush:[/hide]

[b]Hint:[/b] $ \mbox{Solve first plane analogous: }$
$ \mbox{Let } Ox \perp Oy \mbox{ and } A\in Ox, B\in Oy. \mbox{ A variable circle } (c) \mbox{ through A and B meets } Ox \mbox{ and } Oy \mbox{ again at } A^{\prime} ,B^{\prime} \mbox { respectively.} \\
\mbox{Let } M \mbox{ and } M^{\prime } \mbox{ midpoints of } [AA^{\prime}] \mbox{ and } [BB^{\prime}], \mbox{ respectively.} \mbox{ Find the locus of the points S, midpoint of } [MM^{\prime}].$

$ \mbox{If } A^{\prime \prime} \mbox{ midpoint of } [AA^{\prime}] \mbox{ and } B^{\prime \prime} \mbox{ midpoint of } [BB^{\prime}], \mbox{ then }$ $ A^{\prime \prime} MB^{\prime \prime} M^{\prime} \mbox{ - parallogram} \mbox{ and the point S is midpoint of } [A^{\prime \prime}B^{\prime \prime}]; \dots ?$
--------------------------------------------------------------------------------------------------------------
[b]1).[/b]$ \mbox{ Let }B_1\in Oy; AB_1\parallel A^{\prime}B^{\prime} \Rightarrow \angle{BAO}\equiv \angle{A^{\prime}B^{\prime}O\equiv \angle {AB_1O}.}$
[b]2).[/b]$ \mbox{ Let }M_1 \mbox{ the midpoint of } [AB_1] \mbox{ and } S_1 \mbox{ the midpoint of } [MM_1].$
$ \mbox{If A, B fixe and } A^{\prime}\mbox{ - variable}\Rightarrow \mbox{the points } M, B_1, M_1, S_1 \mbox{ are fixed and, we: } \\
M^{\prime}\in OM_1; S_1S\parallel OM_1\Rightarrow \mbox{ locus of point S is ray(half - straight line)}\parallel OM_1.$


probability
calculus
integration
vector
linear algebra
matrix
expected value

Does anybody know how to do any of these problems involving random variables?

1. You have 4 light bulbs whose lifetimes are independent normal random variables with mean 100 (hours)
and standard deviation 5 (hours). Use the central limit theorem to approximate the probability that
your 4 light bulbs together live for at least 402 hours. The answer should be written as an explicit
integral involving the normal density; not a numerical answer.

2. Suppose $X_{1},X_{2},X_{3}$ and $X_{4}$ are independent with common mean 1 and common variance 2. Compute
$Cov(X_{1}+X_{2},X_{2}+X_{3})$.

3. Suppose X is a standard normal random variable. What is the density of $\frac{1}{X^{2}}$?

4. A pair of fair dice are cast independently from one another. If you are told that the total sum of the
dots is 10, then find the probability that the first die was 6.

The 4th question is just standard discrete probability.  There are 3 equally likely outcomes, (4, 6), (5, 5) and (6, 4), so the answer is $\frac{1}{3}$.

Let's look at #2 in complete generality. Consider random vectors $X=[X_{1},X_{2},\dots,X_{n}]^{T}$ and $Y=[Y_{1},Y_{2},\dots,Y_{n}]^{T}.$ (The transpose sign means I'm considering them to be column vectors.)

We can define the covariance matrix of $X$ and $Y$ to be $\text{Cov}(X,Y)=E(XY^{T})-E(X)E(Y)^{T}.$ The $(i,j)$ entry of this matrix is the covariance of $X_{i}$ with $Y_{j}.$ The most important special case of this is the covariance of $X$ with itself, $\text{Cov}(X)=E(XX^{T})-E(X)E(X)^{T}.$ In this case, the diagonal elements of the matrix are the variances of the $X_{i},$ and if the $X_{i}$ are independent, that implies that the covariance matrix is diagonal.

Here's the most useful theorem. Let $a=[a_{1},a_{2},\dots,a_{n}]^{T}$ and $b=[b_{1},b_{2}\dots,b_{n}]^{n}$ be constant vectors. Let $U=a^{T}X$ and $V=b^{T}Y.$ $U$ and $V$ are scalar random variables formed as linear combinations of the components of $X$ and $Y.$ The covariance of these two vectors is:

$\text{Cov}(U,V)=E(UV)-E(U)E(V)=a^{T}\text{Cov}(X,Y)b.$

As a very important special case of this theorem,

$\text{Var}(U)=\text{Cov}(U,U)=a^{T}C(X)a.$

From the setup I've given the proof of this theorem is almost mindlessly simple - we need only the linearity of the expected value (it's an integral, after all). Most of the work has been done in setting up the notation.

Now, for your particular problem: you have a 3-vector $X$ and it is given that $C(X)=2I.$ (Twice the identity matrix.) We have $U=[1,1,0]X$ and $V=[0,1,1]X.$ Then by the above theorem,

$\text{Cov}(U,V)=[1,1,0](2I)[0,1,1]^{T},$ which is twice the dot product of those two vectors, or $2.$

Similar calculations show that $\text{Var}(U)=\text{Var}(V)=4.$ Hence, the correlation coefficient of $U$ and $V$ is $\frac{2}{\sqrt{4}\sqrt{4}}=\frac12.$

#1 is a lousy problem; you don't need the central limit theorem when adding independent normal variables (since the sum is itself exactly normal), the central limit theorem is a terrible approximation for a sum of four things anyway, and the lifetime of a lightbulb will not be described by anything resembling a normal distribution (it will be nonnegative, and significantly skewed toward small values).

#3: Densities transform like integrals; the new density is the old density times a scaling factor, and that scaling factor is exactly what you need to keep the integral constant.


Suppose that a,b,m and n are natural numbers. Prove/explain why GCD(m,n)= GCD(m,am+n)= GCD(am+bn,n).

Well I'm not sure about this and I'm also not sure if you can get an advantage from this http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity

And by the way don't post the same topic in multiple forums

its not true, let m=2 n=7 a=7 and b=3 then GCD(m,n)=1 and GCD(am+bn,n)=GCD(35,7)=7=/=1


geometry
geometric transformation
reflection

Line m is given by the equation y=(square root of 3)x, and line n by x = (square root of 3)y . A set of points has the property that for any point a in the set, the reflections of a over lines m and n are also in the set. If the set contains the point (6,5), what is the least number of points the set could contain?

[quote="cdymdcool"] the reflections of [point] a over lines m and n are also in the set. [/quote]
I don't have it either.
If you can tell me how the point (6,5) is reflected over a line,
AND is it reflected over one of the lines and then in a
consecutive fashion over the other line (from this quote)?

Other than that, I see the equivalent of the lines:

y = [sqrt(3)]x for line m, and y = (1/3)[sqrt(3)]x for line n.

Well if the line contains one point, it must contain the point reflected over line m and line n. And then You reflect those two points and continue in this way until the reflection merely brings back a point to another you already had.
Sorry if it doesn't make sense, but drawing it out will help


Prove that the equation $ x^3\plus{}y^3\equal{}7z^3$ has infinitely many solutions in integers $ x,y,z$, where $ \gcd (x,y,z)\equal{}1$

This is an elliptic curve, and the above solution gives the doubling formula (based on the group law).


Positive numbers $a,b,c$ satisfy the equality $ab+bc+ca=3$. Prove that $a^3+b^3+c^3+6abc\ge 9$.

We have already discussed it...

http://www.mathlinks.ro/Forum/viewtopic.php?t=63308
http://www.mathlinks.ro/Forum/viewtopic.php?t=68089
http://groups.google.de/group/sci.math/browse_thread/thread/ad5f7fdf3cc117bb/c9e40d4dc3dce798

  darij

it's easy.SO here is my proof:
(ab+bc+ca)^3=<(a^3+b^3+c^3)^2*3
Then we have:a^3+b^3+c^3>=9
and 3abc>=0(a.b.c is positive numbers)
the we have this inequality

[quote="ghjk"]it's easy.SO here is my proof:
(ab+bc+ca)^3=<(a^3+b^3+c^3)^2*3
Then we have:a^3+b^3+c^3>=9
[/quote]
It is easy,but I think you are wrong :rotfl: