There is a theorem about division algorithm for polynomials in my algebra book:

[i]Given $ f,g\in\mathbb P[x]$ with $ g\neq0$, there exist unique $ q,r\in\mathbb P[x]$ such that $ f = qg + r$, where $ r = 0$ or $ \deg r < \deg g$.[/i]

The author just proves the uniqueness of $ q$ and $ r$. But may we prove the existence of $ q$ and $ r$ like following?

[b][i]Proof.[/i][/b]
Suppose that $ S = \{f - Qg|\deg Q = \deg f - \deg g\text{ for }\deg f\geq\deg g,\text{ otherwise }Q = 0\}$.

Let $ r\in S$ and $ \deg r\leq\deg R$ for any $ R\in S$.

Then there exists $ q$ such that $ r = f - qg$, or $ f = qg + r$.

Claim $ r = 0$ or $ \deg r < \deg g$.

If not, say $ \deg r\geq\deg g$. Let $ r' = r - q'g$, where $ q'\neq0$ and $ \deg q' = \deg r - \deg g$.

Then $ r'\in S$ and $ \deg r'\leq\max\{\deg r,\deg q'g\} = \deg r$.

But $ \deg r\leq\deq r'$, thus $ r = r'$.

It implies $ q' = 0$, which contradicts the assumption that $ q'\neq0$. $ \Box$

That's nonsense that doesn't prove anything. In particular, you haven't used key hypotheses like the coefficient ring being a field.

I'm pretty sure the best way to prove existence is to actually construct it. The division algorithm is a computational tool as well as a theoretical tool.

But we prove the theorem about division algorithm for integers like this:

[b]Theorem.[/b] [i]$ \forall a,b\in\mathbb Z$ with $ a>0$, $ \exists!q,r\in\mathbb Z$ such that $ b=qa+r$, where $ 0\leq r<a$.[/i]

[i][b]Proof for existence.[/b][/i]
Let $ T=\{b-ax\in\mathbb Z_+|x\in\mathbb Z\}$. Then $ T\neq\emptyset$.
Let $ r$ be the least member of $ T$. Then $ r=b-aq$ for some $ q\in\mathbb Z$, i.e. $ b=aq+r$.
It is clear that $ r\geq0$. We claim $ r<a$.
If not, say $ r\geq a$ and take $ r'=r-a$, then $ r'\in T$, contradicting the assumption that $ r$ is the least. $ \Box$

The division algorithm for polynomials is very similar to the division algorithm for integers. Why can't we prove them in the same way?


probability
pigeonhole principle
arithmetic sequence

[quote="guile"]How do you solve these problems?

1. How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?

2. Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?[/quote]

[hide="1"]The second angle has to be 60. So there are 60-1+1=$\boxed{60}$?[/hide]
[hide="2"]
The probability they dont is if they all have different remainders when divided by 5. $n\equiv{0}\mod{5}, n\equiv{1}\mod{5}, n\equiv{2}\mod{5}, n\equiv{3}\mod{5}, and n\equiv{4}\mod{5}$. Probability of this is $120(401/2006)(402/2005)(401/2004)(401/2003)(401/2002)=$ something, and subtract that from 1 and you get the answer? [/hide]

Edit: ..oops, 6 positive integers >.>

[hide="Correct #2"]1[/hide]

[quote="guile"]How do you solve these problems?

1. How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?

2. Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?[/quote]

[hide="1"]
Well like the first person said, 60 is the middle angle. Since the other two can be of any size of angle, we have our first triplet with ${1, 60, 119}$, and the last one with ${59,60,61}$. Thus there are $\boxed{59}$ pairs.
[/hide]

[hide="2"]
There are only 5 different mods of $\mod 5$ and since we are picking 6 numbers there has to be a pair whos difference is a multiple of 5. So $\boxed{1}$[/hide]

[hide="1"]The middle angle must be 60 degree. For any possible smallest angle, we know there is only one possibility for the larger number. There are 59 possibilities for the smallest angle. $\boxed{59}$.[/hide]

[hide="2"]By pigeon hole principle, some two of these (6 integers) must be congruent to same remainder modulo 5 (5 remainders). For two numbers that are congruent modulo n, their difference is divisible by n. $\boxed{1}$.[/hide]

The question says "distinct" positive integers.

[quote="ch1n353ch3s54a1l"]Does 60-60-60 count?[/quote]

Nope, the question states that the angles must be distinct. $60$ was not one of the possible choices on the test so it wasn't meant to trick anyone in that sense  :P

No, the measures of the angles must be distinct.

[hide]1. The anwsers range from (1,60,119) all the way down to (59,60,61).  So the answer is 59.

2.  Consider mod(5), then you will notice by pigeonhole at least two of these will have the same mod(5), so the answer is 1.[/hide]

For problem #1

[hide]

Consider the smallest angle is a positive integer a. The middle sized angle must be a + d, where d is a positive integer, since the angles must be in arithmetic progression. Likewise, the largest angle must be d more than the middle angle: (a + d)+d=a+2d.

Thus, because all these angles are in a triangle, a + (a+d) + (a + 2d) = 180, 3a+3d=180, and a+d=60. Since a and d are positive integers, the number of distinct possible values of a is 59, ranging from 1...59, and 59...1 for d respectively. Note that d must be at least 1, since it is positive, thus there are not 60 possible values for a and d.
[/hide]


inequalities
number theory proposed
number theory

Problem 1.
Solve the equation in $Z^+$
$y^2x^4+zy^2=4x+5z+x^2y^2z^2$

We have that $y^2(x^4-x^2z^2+z)=4x+5z$ and since $x,y,z\in \mathbb{Z_+}$ it means that
$x^4-x^2z^2+z>0 \Rightarrow  x-z>-\frac{z}{x^2(x+z)}>-1$ so $x\geq z$
$y^2=\frac{4x+5z}{x^4-x^2z^2+z}\geq1 \Rightarrow  4x+5z\geq x^4-x^2z^2+z \Rightarrow  4\geq x^2(x-z)$
which together with the previous inequality leads to $x^2(x-z)\in \{0,1,2,3,4\}$
*$x^2(x-z)=0 \Rightarrow  x=z \Rightarrow  y^2=9 \Rightarrow  (x,y,z)=(\alpha,3,\alpha)$ with $\alpha \in \mathbb{Z_+}$ solution.
*$x^2(x-z)\in \{1,2,3\} \Rightarrow  x=1 \Rightarrow  z\leq0$, contradiction to $z \in \mathbb{Z_+}$.
*$x^2(x-z)=4 \Rightarrow  x\in\{1,2\}$, but if $x=1$ we get to the same contradiction as above, so we must have that $x=2 \Rightarrow  z=1 \Rightarrow y=1$ so one more solution is given by $(x,y,z)=(2,1,1)$
It is indeed an easy problem. :)


Dear reader,

Whenever I pressed "Build and view current file (Ctrl+Shift+F5)" in TeXnicCenter in the near past, it was unable to build and view the current file (in PDF, as I have chosen) and

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute the command.

is shown in a new window. May you please help me in this regard?

Thank you

Sincerely,
Daniel Tsai

Have you tried it when Adobe reader is open?

If hunter34's suggestion works then it could be a problem with the version of Adobe Reader that you have; this prblem was resolved in later versions. If that doesn't work, maybe TexnicCenter is looking in the wrong place for Adobe Reader.

In TexnicCenter, Build, Define Output Profiles, LaTeX=>PDF, Viewer tab check that the top box is pointing to the correct place for acrord32.exe and, if not, click on the button next to it to change the location.

Dear hunter34 and stevem,

I followed hunter34's suggestion but TeXnicCenter was still unable to build and view the current file. I also checked as instructed by stevem but TeXnicCenter was still unable to build and view the current file.

Also, hunter34 and stevem, what are your real names?

Sincerely,
Daniel Tsai

Dear hunter34 and stevem,

I apologize, but I typed "\documentclass{document}" rather than "\documentclass{article}" in the first line of the current file. Upon chaning the first line of the current file to "\documentclass{article}", TeXnicCenter was able to build and view the current file. I am a beginner at using LaTeX.

Sincerely,
Daniel Tsai


inequalities
function
calculus
derivative
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
$ \frac{1}{\sqrt{b^2\plus{}bc\plus{}c^2}}\plus{}\frac{1}{\sqrt{c^2\plus{}ca\plus{}a^2}}\plus{}\frac{1}{\sqrt{a^2\plus{}ab\plus{}b^2}} \ge 4\plus{}\frac{2}{\sqrt{3}}.$
Equality holds if and only if $ a\equal{}b\equal{}\frac{1}{2},c\equal{}0$ or any cyclic permutations. :)

@VasC: I created this problem from your problem at Old and new Inequality book
$ \frac{1}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{1}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\plus{}ca\plus{}a^2} \ge \frac{9}{(a\plus{}b\plus{}c)^2}$
This problem can be proved by Cauchy Schwarz Inequality. :)

That problem was discussed here
http://www.mathlinks.ro/viewtopic.php?t=52376

and no nice solution posted .... :(  . Could you post your BCS solution?

2) by Caushy inequality :

$ (\sum a^2 \plus{} ab \plus{} b^2)(\sum \frac {1}{ a^2 \plus{} ab \plus{} b^2}\,\geq\,9$

$ \iff$   $ \sum \frac {1}{a^2 \plus{} ab \plus{} b^2}\,\geq\,\frac {9}{2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} ac \plus{} bc}$

and we have $ a^2 \plus{} b^2 \plus{} c^2\,\leq\,ab \plus{} ac \plus{} bc$   (rearangement)

$ \iff$   $ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} ac \plus{} bc\,\leq\,a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} ac \plus{} bc)$

so :    $ \sum \frac {1}{a^2 \plus{} ab \plus{} b^2}\,\geq\,\frac {9}{(a \plus{} b \plus{} c)^2}$

You're wrong @@ .
$ a^2\plus{}b^2\plus{}c^2 \leq ab\plus{}bc\plus{}ca$ .It's reserve .

[quote="can_hang2007"]@VasC: I created this problem from your problem at Old and new Inequality book
$ \frac {1}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} ca \plus{} a^2} \ge \frac {9}{(a \plus{} b \plus{} c)^2}$
This problem can be proved by Cauchy Schwarz Inequality. :)[/quote]
Let I post my solution for this one :)

This inequality was solved just by using the $ p,q,r$ method (page 135 in my book).
For $ p\equal{}a\plus{}b\plus{}c$, $ q\equal{}ab\plus{}bc\plus{}ca$, and $ r\equal{}abc$, notice that
\[ a^2\plus{}ab\plus{}b^2\equal{}1\minus{}c\minus{}q.\]
The desired inequality reads (in $ p,q,r$)
\[ q(3q\plus{}1)^2\plus{}9r\plus{}1\minus{}4q\geq0\]
which follows immediately from I. Schur inequality.

 The technique you have used with Cauchy Schwarz would be more appreciated with inequalities involving square roots.

[quote="mehdi cherif"]we can also prove de second without using any theoreme :

by using this lemma : $ \frac {a^2 \minus{} ab \plus{} b^2}{a^2 \plus{} ab \plus{} b^2}\geq\,\frac {1}{3}$
 :)
proof : $ \iff$   $ 2(a^2 \plus{} b^2 \minus{} 2ab)\geq\,0$[/quote]
Can you write it clearly?

[quote="mehdi cherif"]i don't have a complet solution for it , but i think that we can solve the probleme with this lemma  :) .[/quote]

If you do not have a solution for it, so why do you think that

[quote="rachid"]for the second inequality ,i find that we have to prove that  a²+b²+c² <= ab+ac+bc which is not true,i think the problem is false. :maybe:[/quote]

I don't know how your proof is but you are so wrong [b]rachid[/b], and if the inequality is not true you just have to give an example

just because can_hang2007 said that we have to use caushy-shwarz inequality,and when i use it i found a wrong result [b]maybe.[/b]

[quote="pvthuan"]This inequality was solved just by using the $ p,q,r$ method (page 135 in my book).
For $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, and $ r \equal{} abc$, notice that
\[ a^2 \plus{} ab \plus{} b^2 \equal{} 1 \minus{} c \minus{} q.
\]
The desired inequality reads (in $ p,q,r$)
\[ q(3q \minus{} 1)^2 \plus{} 9r \plus{} 1 \minus{} 4q\geq0
\]
which follows immediately from I. Schur inequality.

 The technique you have used with Cauchy Schwarz would be more appreciated with inequalities involving square roots.[/quote]
That is the solution of Vasc in Old and New Inequalities.
I wanna do it independent (I think we should use our own solution). And I like it.

[quote="zaizai-hoang"]We have a generalization. (it was posted in VIF)
Let $ a,b,c\ge 0$ and  $ 0 \le k \le \frac {17}{16}$ then:
\[ \frac {1}{\sqrt {a^2 \plus{} kab \plus{} b^2}} \plus{} \frac {1}{\sqrt {b^2 \plus{} kbc \plus{} c^2}} \plus{} \frac {1}{\sqrt {c^2 \plus{} kca \plus{} c^2}} \ge \frac {4 \plus{} \frac {2}{\sqrt {k \plus{} 2}}}{a \plus{} b \plus{} c}
\]
I don't have any nice solution but having a ugly solution for this one (using concave-convex function, I coundn't complete it clearly) :( I have not tried Mixing Variable method but I hope it works well. I think can_hang2007's solution is CYH  :D

My idea is using concave-convex function as I said. Let prove that:
Put $ x \equal{} \frac {a \plus{} b}{2},y \equal{} \frac {a \minus{} b}{2}\to x \plus{} y \equal{} a,x \minus{} y \equal{} b,1 \minus{} 2x \equal{} c$ and:
\[ A \equal{} (x \plus{} y)^2 \plus{} m(x^2 \minus{} y^2) \plus{} (x \minus{} y)^2
\]

\[ B \equal{} (x \minus{} y)^2 \plus{} m(x \minus{} y)(1 \minus{} 2x) \plus{} (1 \minus{} 2x)^2
\]

\[ C \equal{} (x \plus{} y)^2 \plus{} m(x \plus{} y)(1 \minus{} 2x) \plus{} (1 \minus{} 2x)^2
\]
We have
\[ LHS \equal{} \frac {1}{\sqrt {A}} \plus{} \frac {1}{\sqrt {B}} \plus{} \frac {1}{\sqrt {C}} \equal{} f(y)
\]
If we prove that:
\[ f''(y) \equal{} \frac {3xy(m \plus{} 2)(2 \minus{} m)}{\sqrt {A^5}} \plus{} \frac {3}{4}\left(\frac {(2mx \minus{} 2x \plus{} 2y \minus{} m)^2}{\sqrt {B^5}} \plus{} \frac {(2mx \minus{} 2x \minus{} 2y \minus{} m)^2}{\sqrt {C^5}}\right) \minus{} \frac {1}{\sqrt {B^3}} \minus{} \frac {1}{\sqrt {C^3}}\le 0
\]
So we have
\[ f(y)\ge min[f(0),f(x)]
\]
The rest is proving the following inequality
\[ k(x) \equal{} \frac {1}{\sqrt {x^2(m \plus{} 2)}} \plus{} \frac {2}{\sqrt {x^2 \plus{} mt(1 \minus{} 2x) \plus{} (1 \minus{} 2x)^2}} \minus{} 4 \plus{} \frac {2}{\sqrt {m \plus{} 2}}
\]
[/quote]
Please check your solution again, Hoang. I have checked that $ f(y)$ is not even concave or convex.

[quote="Vasc"]Using EV-Theorem, it suffices to prove the original inequality for $ x \equal{} 0$ and for $ y \equal{} z$.
For $ x \plus{} y \plus{} z \equal{} 1$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2p \plus{} 1$, we have to show that
$ \sum\frac 1{\sqrt {2p \minus{} 2x}} \ge 4 \plus{} \frac 2{\sqrt 3}$.
We see that the derivative $ f'(u)$ of $ f(u) \equal{} \frac 1{\sqrt {2p \minus{} 2u}}$ is convex.[/quote]
Thanks, Vasc, my solution is the same. You are really excellent. :)


Determine whether $ \frac{1}{\sqrt{2}} \minus{} \frac{1}{\sqrt{6}}$ is less than or greater than $ \frac{3}{10}$.

hello, it is
$ 47961 <48600 \Leftrightarrow 219 < 90\sqrt {6} \Leftrightarrow \frac {219}{25} <\frac {18}{5}\sqrt {6} \Leftrightarrow 18 < \frac {81}{25} \plus{} 6 \plus{} \frac {18}{5}\sqrt {6} \Leftrightarrow 3\sqrt {2} < \frac {9}{5} \plus{} \sqrt {6} \Leftrightarrow 3\sqrt {2} \minus{} \sqrt {6} < \frac {9}{5} \Leftrightarrow 3\sqrt {2} \minus{} \sqrt {6} <\frac {18}{10} \Leftrightarrow \frac {\sqrt {2}}{2} \minus{} \frac {\sqrt {6}}{6} < \frac {3}{10}$
Sonnhard.

first, u square both sides, which then give you: left:$ (\frac{1}{\sqrt{2}}\minus{}\frac{1}{\sqrt{6}})^2\equal{}\frac{1}{2}\plus{}\frac{1}{6}\minus{}\frac{2}{\sqrt{12}}\equal{}\frac{2\minus{}\sqrt{3}}{3}$ , right:$ \frac{9}{100}$
mutiply both side by 3, so u r actually comparing $ 2\minus{}\sqrt{3}$ with $ 0.27$
u know that $ \sqrt{3}\equal{}1.732$, so $ 2\minus{}\sqrt{3}\equal{}0.268 < 0.27$


[quote="amd2357"]can anybody show me how to evaluate the continued fraction:
1+2/(3+4/(5+6/(7+8/(...))))[/quote]
Are you sure it is possible?  :maybe:

I've checked with a calculator and it seems to converge without a doubt.  Also I found this problem in one of the Arbelos journals where it implied that a solution did indeed exist.

Don't all continued fractions have to converge?  I mean, this one is clearly <1+2/3

THIS fraction converges. Though, indeed, it doesn't imply that we can find its numerical value.

It looks a lot like cosh(1) (the decimal expansion does, anyway)

Problem had been solved by rcv before database crashed.
The key idea was to rewrite that continued fraction in standart way, i.e.
\[a_1+1/(a_2+1/(a_3+1/...)),\]
and then use known representation for $cosh(x)$.

Yes, rcv solved it nicely using that approach.


Let $P$ be a point on the circumcircle of a triangle $ABC$, let $t$ be the tangent to this circumcircle at the point $P$, and let $s$ be the Simson line of the point $P$ with respect to triangle $ABC$. Prove that

$\measuredangle\left(s;\;t\right)=\measuredangle\left(s;\;BC\right)+\measuredangle\left(s;\;CA\right)+\measuredangle\left(s;\;AB\right)$,

where we use directed angles modulo $180^{\circ}$.

[i]Remark.[/i] The [i]Simson line[/i] of the point $P$ with respect to triangle $ABC$ is defined as the line which passes through the orthogonal projections of the point $P$ on the lines $BC$, $CA$, $AB$.

[quote]children resist a return to the burning times,
people be wise to the power of their lies,
be not fooled as those who were fooled before -
children, oh children, be free, be wild...

ps. whether you warn me or do not warn me, i will not believe[/quote]

I said.
 :lol:

I can't do anything you will not believe

[quote="Amir.S"]I said.
 :lol:[/quote]

Guess what, I didn't try to prove you (or the Quran) wrong here. The difference is in the consequences one draws from the assertion.

On the other hand, it wouldn't be inappropriate for a geometry topic to contain a discussion of the problem given ;) .

  darij

[quote="silouan"]Very nice problem Darij . :)
We have (in my figure ). I will write a hint.If anyone want details don't hesitate to ask.
$\angle (s,AB)=A+B$
$\angle (s,AC)=A-C$
$\angle (s,BC)=A$
$\angle (s,t)=2A-2C$

and QED[/quote]

I didn't think on the problem,But your solution can't be correct . az $P$ varies on the circumcircle $\angle (s,AB)$ changes.
Or you take point $P$ a very special point.

maybe this is a solution!
Called $A',B',C'$ are orthor projection of $P$ to lines $BC,CA,AB$
First case $P\in \text{ smallarc }BC$
$2((s,BC)+(c,CA)+(s,AB)) =2((s,OP)+(OP,BC)+(PA',PB)+(PA',PC))(mod \pi) =2((s,OP)+(OP,BC)+(PA',BC)+(BC,PB)+(PA',BC)+(BC,PC))(mod \pi) =2((s,OP)+(OP,BC)+\pi+(BC,PB)+(BC,PC))(mod \pi) =2((s,OP)+(OP,BC))+(OC,OP)+(OB,OP)(mod \pi) =2((s,OP)+(OP,BC))+(OC,BC)+(BC,OP)+(OB,BC)+(BC,OP)(mod \pi) =2((s,t)+(t,OP))+2(OP,BC)+2(BC,OP)(mod \pi) =(s,t)+\pi (mod \pi)=2(s,t)$
similar $P\in \text{ smallarc }CA,P\in \text{ smallarc }AB$  :)

@ Gemath: Your idea is correct, but the way you used it is wrong: Since you are working with directed angles modulo 180°, you can't prove u = v just by showing that 2u = 2v ! For instance, $0^{\circ}\neq 90^{\circ}$, but $2\cdot 0^{\circ}=2\cdot 90^{\circ}$. Actually, your proof can easily be fixed, by applying the central angle theorem in the following form: If A, B, C are three points on a circle with center O, then < BCO = 90° - < CAB.

Here is my solution of the original problem (quite similar to yours; actually all solutions of this problem are somehow equivalent):

Let X, Y, Z be the orthogonal projections of the point P on the sidelines BC, CA, AB of triangle ABC; then, the Simson line s of the point P with respect to triangle ABC is defined as the line passing through these points X, Y, Z.

Since < CXP = 90° and < CYP = 90°, the points X and Y lie on the circle with diameter CP. Consequently, < XYC = < XPC. Hence,

  < (s; CA) = < XYC = < XPC = < (PX; CP) = < (PX; BC) + < (BC; CP).

Since $PX\perp BC$, we have < (PX; BC) = 90°, so that

  < (s; CA) = 90° + < (BC; CP) = 90° + < BCP.

On the other hand, since the line t is the tangent to the circumcircle of triangle ABC at the point P, the tangent-chordal angle theorem yields < (BP; t) = < BCP, so that < (s; CA) = 90° + < BCP becomes

  < (s; CA) = 90° + < (BP; t).

Since < BZP = 90° and < BXP = 90°, the points Z and X lie on the circle with diameter BP, so that < XZB = < XPB. In other words,

  < (s; AB) = < XZB = < XPB = < (PX; BP) = < (PX; BC) + < (BC; BP).
            = 90° + < (BC; BP).

Altogether,

  < (s; BC) + < (s; CA) + < (s; AB)
   = < (s; BC) + (90° + < (BP; t)) + (90° + < (BC; BP))
   = 180° + (< (s; BC) + < (BC; BP) + < (BP; t))
   = 180° + < (s; t) = < (s; t)

(since 180° = 0°, as we are working with directed angles modulo 180°), qed.

  darij

Many sorry I make a computation wrong . :oops:


A floor decoration is a circle with eight rays pointing from the center.  The rays form eight congruent central angles.  One of the rays points due north.  What is the measure of the smallest angle formed between the ray pointing East and the ray pointing Southwest?

[asy]size(3cm,3cm);

draw(unitcircle);

draw((0,0.8)--(0,-0.8),Arrows(HookHead));
draw((-0.8,0)--(0.8,0),Arrows(HookHead));

draw((-0.55,0.55)--(0.55,-0.55),Arrows(HookHead));
draw((-0.55,-0.55)--(0.55,0.55),Arrows(HookHead));[/asy]

Since the $ 8$ rays are congruent, they each form $ \frac{360}{8}=45^\circ$ angles. East to Southwest has $ 3$ of those angles, and hence $ 3\times{45^\circ}=\boxed{135^\circ}$.


probability
USAMTS
articles
induction
expected value

[hide="A start?"]OK, so you win when you get $ 1000$ dollars from your opponent.
To do that, you need to flip the coin $ 1000$ times and have it always come up heads, or you could flip $ 1002$ times for $ 1001$ heads and $ 1$ tails, or $ 1004$ times for $ 1002$ heads and $ 2$ tails, and so on...
Now if you have a sequence of $ 1000 \plus{} 2n$ tosses, of which $ n$ tosses come up tails and the rest come up heads, there are a total of $ \binom{1000\plus{}2n}{n}$ ways to arrange the sequence of tosses. Also, there is a $ \frac{1}{2}$ probability of heads or tails, so each sequence of length $ 1000\plus{}2n$ occurs with a probability of $ \frac{1}{2^{1000\plus{}2n}}$.

Let $ P(A)$ represent the probability of you winning. Then,
\[ P(A)\equal{} \sum_{n\equal{}0}^\infty \binom{1000\plus{}2n}{n} \frac{1}{2^{1000\plus{} 2n}}\]
\[ P(A)\equal{} \frac{1}{2^{1000}} \sum_{n\equal{}0}^\infty \binom{1000\plus{}2n}{n} \frac{1}{2^{2n}}\]
And I have no idea how to simplify that...[/hide]

I am pretty sure that the answer is just
[hide]$ \frac{500}{500\plus{}1000}\equal{}\frac{1}{3}$.[/hide]

I you had the same amount it would have been 1:2.
Since you have twice the amount
it should be
x / x + y
= 500/ 500 + 1000
=1/3

[quote="Ravi B"]The last two posts are missing a rigorous explanation.  Below is a hint.

[hide="Hint"]Remember that the game is fair: The expected winnings is zero.  Let $ p$ be the probability that you win.  Think about the two possible situations at the end.[/hide][/quote]
Ravi bhai , are you sure the player is fair? :lol:

Joking , well yes , intended to do so but didnt type.

[quote="RisingFlame"]Yeah, those solutions make a lot more sense.

I realized that I made a mistake earlier: the upper bound of the summation should be $ 499$ since the opponent can get atmost $ 499$ tails or otherwise he would win. Since the answer is just $ 1/3$, how would you go about simplifying
\[ P(A) \equal{} \frac {1}{2^{1000}}\sum_{n \equal{} 0}^{499}\binom{1000 \plus{} 2n}{n}\frac {1}{2^{2n}} ?
\]
[/quote]
If you would less complicate the solution you woudnt have to resolve it into that.
Its more about logical analysis that using formulas.

[quote="erdogankerem123"]Suppose you and I play a game. You start with 500 dollars and I start with 1000
dollars. We flip a fair coin repeatedly. Each time it comes up heads, I give you a dollar. Each
time it comes up tails, you give me a dollar. We continue playing until one of us has all the
money. What is the probability you will win this game?[/quote]

[hide="Solution."]
-----
We first have to show that the game is fair; that is, we have to show that the amount of money that each player expects to gain after flipping the coin an indefinite amount of times is $ \$0$. Since the coin is fair, the expected value after one flip is $ \$0$, and flipping the coin multiple times will also result in an expected value of $ \$0$.

Now we notice that the game can end in either of two states: me winning your $ \$1000$ with probability $ p$ and you winning my $ \$500$ (equivalently me losing $ \$500$) with probability $ 1\minus{}p$. By the definition of expected value:

\[ E\equal{}0\equal{}1000p\minus{}500(1\minus{}p)\implies p\equal{}\boxed{\frac{1}{3}}\]

See [url=http://www.usamts.org/Forms/ExpectedValue.pdf]this[/url] for the discussion of expected value and the source of the problem.
-----
[/hide]

Thanks everybody for the replies. I just wanted to check the given solution because I saw a thread earlier in which two people agreed that the answer should have been significantly lower, but I couldn't see any flaw in the solution.

Btw, where are the other USAMTS articles like this one? I can't find where the other articles are, as I only followed someone else's link to this one...

um, can someone tell me if there are any more USAMTS guides like this one, except for other subjects? These are excellent guides...

So for a related problem, if a person is 1 block from one place and 4 blocks from another, by moving one block each time before deciding where to go, does he have a $ 4/5$ probability to reach the closer place first?

i think we might be able to think of this recursively

say 1 person has m dollars and the other has n dollars
we can see the process of one persom winning as the process of 1 person getting a favorable flip m (or n) more flips than unfavorable flips

say m=n=1the first flip will decide the winnier, with probability 1/2 for both

edit: ignore the following, i overlooked something
say m=2 and n=1

if the first flip favors the "m" person, he wins. if not, then the second flip decides the winner with probability 1/2
so a total of 1/2 +1/2*1/2 for the "m" person and 1/4 for the other.
we see that if 1 person has 1 dollar and the other has m, the first person has $ (\frac12)^m$ chance of winning
now we move on to both have more than 1 dollar...
say m=n=2
if the first favors m, m has 3/4 chance winning and n has 1/4
if it doesnt, the probabilities are reversed, which (if you do the math) both still haave a total probability of winning of 1/2

is m=3 and n=2
we have a few cases

the first flip favors m:
7/8 for m and 1/8 for b
first favors n:
1/2 for m and n

total 11/16 for m and 5/16 for b

this is where i stop seeing a pattern :D
but this shows that the probability is not simply the dollars over the total

[hide="A Recursive Solution"]
Denote $ p(n)$ as the probability of a person with $ n$ dollars winning the game. Note that $ p(0) \equal{} 0$ and $ p(1500) \equal{} 1$. By the given condition, we have
$ p(n) \equal{} \frac {1}{2}\cdot p(n \minus{} 1) \plus{} \frac {1}{2}\cdot p(n \plus{} 1)$
$ \iff p(n \plus{} 1) \equal{} 2p(n) \minus{} p(n \minus{} 1)$.
Observe that above implies $ p(n) \equal{} n\cdot p(1)$. (It's easy to prove by induction, and guessable after plugging a few values) Therefore $ p(1) \equal{} \frac {1}{1500}$, and $ p(500) \equal{} \frac {500}{1500} \equal{} \frac {1}{3}$.

Surely, the expectation solution is much cleverer and quicker.
[/hide]

Well, If I start with 500 dollars, and you start with 1000, then I have 1 part and you have 2. The odds of me winning a part are the same as you winning a part. You have a 1/2 shot of winning after the first part, and I have a 1/2 shot of leading after the first part. Then I have a 1/2 (of 1/2) shot of winning and you have a 1/2 (of 1/2)...

You:1/2+1/8+1/32....
Me:1/4+1/16+1/64....

As you can see at each part your odds of winning are twice as much as mine, therefore the limit is 2/3 and 1/3.


For a positive integer $ n\geq 5$, let $ a_i,b_i$ ($ i\equal{}1,2,...,n$) be integers satisfying the following two conditions:

(i) The pairs $ (a_i,b_i)$ are distinct for $ i\equal{}1,2,...,n$
(ii) $ |a_1b_2\minus{}a_2b_1|\equal{}|a_2b_3\minus{}a_3b_2|\equal{}...\equal{}|a_nb_1\minus{}a_1b_n|\equal{}1$.

Prove that there exists indices $ i,j$ such that $ 1<|i\minus{}j|<n\minus{}1$ and $ |a_ib_j\minus{}a_jb_i|\equal{}1$.


How do you come up with that ?

anyway very nice

[quote="HTA"]
2/$ x^{2009}-2007x^{2008}+2008=0$[/quote]

[hide="Almost"]
This is a pretty boring problem the way I did it.  :sleeping:

Notice that $ x=-1$ is a trivial solution.
So now we try to show it is the only such solution.

Define $ f(x) = x^{2009}-2007x^{2008}+2008$
With $ f'(x) = 2009x^{2008}-2008\cdot 2007 x^{2007}$

Notice that $ x=0$ and $ x=k=\frac{2008}{2007}{2009}$ are the critical points.
1)For $ x<0$ we have $ f'>0$ so the function is increasing.
2)For $ 0<x<k$ we have $ f'<0$ so the function is decreasing.
3)For $ k<x$ we have $ f'>0$ so the function is increasing.

So if we can show that $ f(k)>0$ then that means there are no other solutions.
[/hide]

Uhh if I'm not mistaken, the second one has more than one solution, 'cause

$ x^{2009}-2007x^{2008}+2008$ factors to

$ (x+1)(x^{2008}-2008x^{2007}+2008x^{2006}+\cdots+2008x^{2}-2008x+2008)$

Sadly the right term has a zero.

[quote="arqady"][quote="HTA"]solve the equations
1/$ x^{7}-x^{2}-2x-1=0$
[/quote]
The easy step: $ x^{7}-x^{2}-2x-1=(x^{2}+x+1)(x^{5}-x^{4}+x^{2}-x-1).$
I think, it is a last step. :D[/quote]
yes , i did exactly like u , but i can't solve $ x^{5}-x^{4}+x^{2}-x-1 = 0$  :blush:


I got the same error just now, searching for "maths". (I didn't check if all the numbers matched up, but it was the same kind of error message anyway).

I am using IE7 on Windows XP, and Mathlinks-style boards.

Okay, I think I got the problem. You guys don't have any friends! :) By that I mean there aren't any people in the board that have selected you as their friend in the Profile panel >> Friends & Foes :P Thus I need to modify the code to consider this situation also ... will do it soon.

Ah, did you modify the code to make friends posts higher up on the search or something of that nature?

Probably it doesn't show your foes' posts...

Problem fixed. Thanks for reporting it.

Chess64, no it doesn't (yet) sort out search results with relevance to friends and foes. It checks whether you have the authorization to view a blog entry or a personal album, since these will be also searchable.


search inequalities geometry perimeter trigonometry function inradius

[quote="Virgil Nicula"][url=http://www.mathlinks.ro/viewtopic.php?search_id=97799964&t=63801][size=134][color=darkred][b]Here[/b][/color][/size][/url] [color=darkblue]is $ \sum\sqrt {\frac {b \plus{} c \minus{} a}{a}}\ge 3$ ([b]acute triangle[/b]) . Prove that  $ \sum\sqrt {\frac {a}{b \plus{} c \minus{} a}}\le 1 \plus{} \frac Rr$ ([b]any triangle[/b]).[/color][/quote]
1. By AM-GM inequality, we have:
$ \sum\sqrt {\frac {b \plus{} c \minus{} a}{a}}\ge 3\sqrt[6]{\frac{(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}{abc}}$
By the well-known inequality $ (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c) \ge abc$ we are done.
2. By AM-GM inequality:
$ \sum \sqrt {\frac {a}{b \plus{} c \minus{} a}}\equal{}\sum \frac{\sqrt{a(b\plus{}c\minus{}a)}}{b\plus{}c\minus{}a} \le \sum \frac{b\plus{}c}{2(b\plus{}c\minus{}a)}\equal{}\sum \frac{b\plus{}c}{4(p\minus{}a)}$
where p is semi-perimeter of the triangle.
$ \sum \frac{b\plus{}c}{4(p\minus{}a)}\equal{}\frac{\sum (b\plus{}c)(p\minus{}b)(p\minus{}c)}{4(p\minus{}a)(p\minus{}b)(p\minus{}c)}\equal{}\frac{abc\plus{}4(p\minus{}a)(p\minus{}b)(p\minus{}c)}{4(p\minus{}a)(p\minus{}b)(p\minus{}c)}$
$ \equal{}1\plus{}\frac{abc}{4(p\minus{}a)(p\minus{}b)(p\minus{}c)}\equal{}1\plus{}\frac{R}{r}$
Equality holds if and only if $ a\equal{}b\equal{}c$ or the triangle is equilateral.

[quote="Virgil Nicula"][quote="NguyenDungTN"]By the well-known inequality $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) \ge abc$ we are done.[/quote]
Sorry, $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) \le abc$[/quote]
Sorry, Virgil Nicula. I have a mistake. I'll try again.

for the second one set $ x \equal{} \sqrt {\frac {a}{b \plus{} c \minus{} a}}$ and etc.
so we have to prove that:

$ x \plus{} y \plus{} z\leq 1 \plus{} \frac {R}{r}$

but we know that in any triangle $ \sum\frac {a}{b \plus{} c \minus{} a} \equal{} \frac {2R \minus{} r}{r}$ or equivalently $ x^2 \plus{} y^2 \plus{} z^2 \equal{} \frac {2R}{r} \minus{} 1$

now we have:

$ x \plus{} y \plus{} z\leq 1 \plus{} \frac {R}{r}$

$ \iff x \plus{} y \plus{} z\leq\frac {x^2 \plus{} y^2 \plus{} z^2}2 \plus{} \frac 32$

but we know that $ x^2 \plus{} y^2 \plus{} z^2\geq \frac {(x \plus{} y \plus{} z)^2}3$ so it's sufficient to show that:

$ x \plus{} y \plus{} z\leq \frac {(x \plus{} y \plus{} z)^2}6 \plus{} \frac 32$

$ \iff \frac {\left(x \plus{} y \plus{} z \minus{} 3\right)^2}{6}\geq 0$

[quote="Virgil Nicula"][quote="Virgil Nicula"] [color=darkred]Prove that in any triangle $ ABC$ exists the inequality  $ \sum\sqrt {\frac {a}{b + c - a}}\le 1 + \frac Rr$ .[/color][/quote]
[color=darkblue][b][u]Proof[/u][/b] (similar to [b]NquyenDungTN's[/b]). I"ll use the well-known relations : $ \{\begin{array}{cc} 1.\ \blacktriangleright & \sqrt {a(b + c - a)}\ \le\ \frac {b + c}{2}\ . \\
 \\
2.\ \blacktriangleright & \frac {p}{p - a}\ = \ \frac {r_a}{r}\ . \\
 \\
3.\ \blacktriangleright & r_a + r_b + r_c\ = \ 4R + r\ .\end{array}$

Therefore,  $ \boxed {\ \sum\sqrt {\frac {a}{b + c - a}}} =$ $ \sum\frac {\sqrt {a(b + c - a)}}{b + c - a}\stackrel {(1)}{\ \le\ }\sum\frac {b + c}{4(p - a)} =$ $ \sum\frac {(p - a) + p}{4(p - a)} =$

$ \frac 34 + \frac 14\cdot\sum\frac {p}{p - a}\ \stackrel {(2)}{\ = \ }\frac 34 + \frac 14\cdot\sum\frac {r_a}{r}\stackrel {(3)}{\ = \ }\frac 34 + \frac {1}{4r}\cdot (4R + r) =$ $ \boxed {\ 1 + \frac Rr\ }$ with equality iff $ b + c = 2a$ .[/color][/quote]
  [color=purple] I have the stronger problem : $ \sum\sqrt{\frac{a}{b+c-a}}\ge \sqrt{1+\frac{4R}{r}}$ :) [/color]

[quote="evarist"]
  [color=purple] I have the stronger problem : $ \sum\sqrt {\frac {a}{b \plus{} c \minus{} a}}\ge \sqrt {1 \plus{} \frac {4R}{r}}$ :) [/color][/quote]
the solution is the same:

let $ x\equal{}\sqrt{\frac{a}{b\plus{}c\minus{}a}}$ and etc.
we have:

$ x^2\plus{}y^2\plus{}z^2\equal{}\frac{2R}{r}\minus{}1\Rightarrow \sqrt{1\plus{}\frac{4R}{r}}\equal{}\sqrt{2(x^2\plus{}y^2\plus{}z^2)\plus{}3}$

thus we have to show that:

$ x\plus{}y\plus{}z\leq \sqrt{2(x^2\plus{}y^2\plus{}z^2)\plus{}3}$

$ \iff (x\minus{}1)^2\plus{}(y\minus{}1)^2\plus{}(z\minus{}1)^2\geq 0$

QED

[quote="Virgil Nicula"][url=http://www.mathlinks.ro/viewtopic.php?search_id=97799964&t=63801][size=134][color=darkred][b]Here[/b][/color][/size][/url] [color=darkblue]is $ \sum\sqrt {\frac {b \plus{} c \minus{} a}{a}}\ge 3$[/quote]
Very nice and hard problem, Mr Nicula. My solution for it (algebraic solution) :)

[quote="Virgil Nicula"][quote="BaBaK Ghalebi"][quote="evarist"]
  [color=purple] I have the stronger problem : $ \sum\sqrt {\frac {a}{b \plus{} c \minus{} a}}\boxed {\ge} \sqrt {1 \plus{} \frac {4R}{r}}$ :) [/color][/quote]
the solution is the same. Let $ x \equal{} \sqrt {\frac {a}{b \plus{} c \minus{} a}}$ and etc.
We have $ x^2 \plus{} y^2 \plus{} z^2 \equal{} \frac {2R}{r} \minus{} 1\Rightarrow \sqrt {1 \plus{} \frac {4R}{r}} \equal{} \sqrt {2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3}$
Thus we have to show that $ x \plus{} y \plus{} z\boxed {\le} \sqrt {2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3}$[/quote]
See $ \boxed {\ge}$ and $ \boxed {\le}$ ![/quote]
then [b]evarist[/b]'s problem is wrong,and the reverse one holds:

$ x \plus{} y \plus{} z\boxed{\geq}\sqrt {2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3}$

$ \iff (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \plus{} (z \minus{} 1)^2\leq 0$

which is true whenever $ x \equal{} y \equal{} z \equal{} 1$ :maybe:

but as I said before the reverse one holds,i.e:

$ \sum\sqrt {\frac {a}{b \plus{} c \minus{} a}}\leq\sqrt {1 \plus{} \frac {4R}{r}}$

[size=117][color=darkred][b]Thank you very much, [u]Can Hang2007[/u] ![/b][/color][/size]

$ 3\cdot\sqrt {\frac {2r}{R}}\le 3\le$ $ \boxed {\ \frac {5R + 2r}{2R}\cdot\sqrt {\frac {R}{2r}}\le\sum\sqrt {\frac {a}{b + c - a}}\le\sqrt {1 + \frac {4R}{r}}\ }\le$ $ 3\cdot\sqrt {\frac {R}{2r}}\le1 + \frac Rr$

[b][u]Remark.[/u][/b]   $ \frac {a}{b + c - a} = \frac {2R}{r}\cdot\frac {(p - b)(p - c)}{bc}$ $ \implies$ $ \boxed {\ \sqrt {\frac {a}{b + c - a}} = \sqrt {\frac {2R}{r}}\cdot\sin\frac A2\ }\ \ (*)$ .

[b][u]Tiberiu  Popoviciu's inequality.[/u][/b] Let $ f\ : \ I\ \rightarrow\ \mathbb R$ be a convexe function and $ \{a,b,c\}\subset I$ . Then

$ 2\cdot[ f(\frac {b + c}{2}) + f(\frac {c + a}{2}) + f(\frac {a + b}{2})]\le f(a) + f(b) + f(c) + 3\cdot f(\frac {a + b + c}{3})$ .

$ 1\blacktriangleright$ Apply the [b]Popoviciu's inequality[/b] to the concave function $ \cos$ on the interval $ (0,\frac {\pi}{2})$ :

$ 2\cdot\sum\cos\frac {B + C}{2}\ge \sum\cos A + 3\cdot\cos\frac {A + B + C}{3}\implies$

$ \sum\sin\frac A2\ge \frac 12\cdot(1 + \frac rR) + \frac 34\implies$ $ \boxed {\ \sum\sin\frac A2\ge \frac {5R + 2r}{4R}\ }\ \ (1)$ .

From the relation $ (*)$ obtain $ \frac {5R + 2r}{2R}\cdot\sqrt {\frac {R}{2r}}\le\sum\sqrt {\frac {a}{b + c - a}}$ .

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

Appling $ (1)$ to the orthic triangle of $ \triangle ABC$ , i.e. $ A: = 180^{\circ} - 2A$ a.s.o.,  obtain $ \sum\cos A\ge \frac {5\cdot\frac R2 + 2r'}{4\cdot\frac R2}$ ,

where $ r'$ is the inradius of the orthic triangle. Thus, $ 4(R + r)\ge 5R + 4r'$ $ \implies$ $ \boxed {\ r - r'\ge\frac R4\ }$ .

Appling $ (2)$ to the orthic triangle of $ \triangle ABC$ obtain $ \sum\sin A\ge \frac {pr}{R^2} + \frac {3\sqrt 3}{4}$ ,

i.e. $ p(R - r)\ge \frac {3\sqrt 3}{4}\cdot R^2$ , what is evidently because $ \{\begin{array}{c} p\ge\frac {3\sqrt 3}{2}\cdot R \\
 \\
R - r\ge \frac R2\end{array}$ .

$ 2\blacktriangleright$ Apply the [b]C.B.S.'s inequality[/b] : $ \sum[\sqrt a\cdot\sqrt {(p - b)(p - c)}]\le\sqrt {\sum a}\cdot\sqrt {\sum (p - b)(p - c)}$ $ \implies$

$ \sum\sqrt {a(p - b)(p - c)}\le \sqrt {2pr(4R + r)}\implies$ $ \sqrt {abc}\cdot\sum\sin\frac A2\le \sqrt {2pr(4R + r)}\implies$

$ \sum\sin\frac A2\le\sqrt {\frac {4R + r}{2R}}\implies$ $ \sqrt {\frac {2R}{r}}\cdot\sum\sin\frac A2\le\sqrt {\frac {4R + r}{r}}\implies$ $ \boxed {\ \sum\sqrt {\frac {a}{b + c - a}}\le\sqrt {1 + \frac {4R}{r}}\ }$ .


linear algebra
matrix
vector
algebra
polynomial
geometry
3D geometry

I dont have a problem here to post but I require a clarification. If there are $ n$ points an euclidean space with distance between any two points =1 or say some arbitrary positive real number, then the dimension of the space must be atleast $ n\minus{}1$.

Is there a proof to this or do we assume it as an axiom ?

take one of these points as the origine $ O$. Then for every other points $ A_i, A_j$,
$ 2(OA_i, OA_j) \equal{} \|OA_j\|^2 \plus{} \|OA_j\|^2 \minus{} \|A_iA_j\|^2$
so $ (OA_i, OA_j) \equal{} 1$ if $ i \neq j$. Then write the associated matrix $ [OA_1, OA_2, \ldots, OA_{n\minus{}1}]$ and you should see that it is invertible, hence the conclusion.

The absolute value of the determinant of the matrix $ M$ of alekk's post is $ \frac1{(n\minus{}1)!}$ times the volume in $ (n\minus{}1)$-dimensional space. If you want to calculate this, look at $ M^TM$; the entries are the dot products of the vectors, so in this case $ M^TM\equal{}\begin{bmatrix}1&\frac12&\frac12&\cdots\\ \frac12&1&\frac12&\cdots\\ \frac12&\frac12&1&\cdots\\ \vdots&\vdots&\vdots&\ddots\end{bmatrix}$

That determinant's not too hard to calculate.

[quote="jmerry"]That determinant's not too hard to calculate.[/quote]
I'll give that a shot. The matrix jmerry wrote down was $ \frac12J\plus{}\frac12I.$

Its determinant is $ 2^{\minus{}n}\det(J\plus{}I).$

The characteristic polynomial of $ J$ if we write that as $ \det(J\minus{}xI),$ is $ (\minus{}1)^n(x^n\minus{}nx^{n\minus{}1})\equal{}(\minus{}1)^nx^{n\minus{}1}(x\minus{}n).$ Evaluating this at $ x\equal{}\minus{}1,$ we get $ (\minus{}1)^n(\minus{}1)^{n\minus{}1}(\minus{}1\minus{}n)\equal{}n\plus{}1.$

That gives us $ \det\left(\frac12J\plus{}\frac12I\right)\equal{}\frac{n\plus{}1}{2^n}$

For instance, with $ n\equal{}2,$ we get that determinant as $ \frac34.$ That checks.

There's an off-by-one discrepancy in the notation. In the original, that's a $ (n\minus{}1)\times (n\minus{}1)$ matrix; Kent's version uses $ n\plus{}1$ points and an $ n\times n$ matrix.

What if we don't require the distance to be exactly $ 1$? Suppose that we want to find $ n$ points in $ \mathbb R^d$ such that the distance between any two of them is in the interval $ [1,L]$, for a fixed $ L>1$. What is the asymptotic lower bound for $ d$ as $ n\to\infty$?

That's a sphere-packing problem: how many spheres of diameter $ 1$ can you fit in a $ d$-dimensional ball of diameter $ L\plus{}1?$
OK, not exactly- but close. The exact scaling constants are hard, but we can start with a simplex packing: each simplex has volume $ \frac1{d!}\sqrt{\frac{d\plus{}1}{2^d}}$ and the ball has volume $ \left(\frac{L\plus{}1}{2}\right)^d\frac{\pi^{n/2}}{\Gamma(\frac d2\plus{}1)}$. For large $ L$, the number of points is approximately the ratio $ (L\plus{}1)^d\frac{d!\sqrt{\pi^d}}{\Gamma(\frac d2\plus{}1)\sqrt{(d\plus{}1)2^d}}$

First conjecture: the maximum is equal to $ (L\plus{}g(L,d))^d\frac{d!\sqrt{\pi^d}}{\Gamma(\frac d2\plus{}1)\sqrt{(d\plus{}1)2^d}} \plus{}d$, where $ 0\le g(L,d)\le 2h(d).$ Here $ h(d)$ is the altitude of the $ d$-dimensional simplex, which is approximately $ \frac1{\sqrt{2}}$ for large $ d$.

You asked for the inverse; $ d$ in terms of the number of points $ n$. That's a complete mess, but the dominant term should be $ \frac{\log n}{\log\log n}$, from the factorials.

On second thought: those are the wrong bounds on $ g(L,d)$. It should be $ \minus{}2h(d)\le g(L,d)\le 2h(d)$, and I don't think either end is sharp. We definitely need negatives on the low end, though- see the case $ L\equal{}1$, where $ g(L,d)$ goes to $ \minus{}1$.


calculus
integration
logarithms
calculus computations

find $ \int \frac{x dx}{x\plus{}\ln (x)}$

According to Mathematica, no such antiderivative exists.

could you express that in terms of the exponential integral?


probability
expected value
combinatorics proposed
combinatorics
random walks
Random walk

There are 94 safes and 94 keys. Each key can open only one safe, and each safe can be opened by only one key. We place randomly one key into each safe. 92 safes are then randomly chosen, and then locked. What is the probability that we can open all the safes with the two keys in the two remaining safes?
(Once a safe is opened, the key inside the safe can be used to open another safe.)

It will be the number of permutations ${1,2,..,n}$ without fix point (take general case)

$D_n=n!(1-\frac{1}{1!}+\frac{1}{2!}-..+(-1)^n\frac{1}{n!})$

plus $2\cdot D_{n-1}$ - number when one fixed point is one from where we have two keys.

sum them and divide by $n!$.

Consider the case when the keys inside the two safes are for each other. In this case there are no fixed points, yet no other safes can be opened.
The only solution I've come up with is using recurrence...

Yes, what is written is fake thing.
Another idea. We need to have one or two cycles not more, such
that we have one key from one of them and one from another.
If we have $n$ number to form full cycle from them there are $\frac{(n-1)!}{2}$ possibilities (hope this estimations is corect).
If there is only one cycle the numberof possibilities is $\frac{(n-1)!}{2}$, and if there two then
$\displaystyle \sum_{i=0}^{n-2}C^i_{n-2}\cdot \frac{i!}{2}\cdot \frac{(n-2-i)!}{2} =\frac{n!}{4}\sum_{i=0}^{n-2}\frac{1}{(n-i)(n-i-1)}=\frac{n!}{4}(1-\frac{1}{n})=\frac{(n-1)!}{4}$.
Total number is $\frac{(n-1)!}{2}+\frac{(n-1)!}{4}=\frac{3(n-1)!}{4}$.

I agree with the general principle, but there are $(n-1)!$ circles, because they are directed (key 1 in safe 2 is a different event from key 2 in safe 1).

Also there's a computation error:

\[ \displaystyle \sum_{i=0}^{n-2} \binom{n-2}i \cdot i!\cdot (n-2-i)! = (n-1) \cdot (n-2)!=(n-1)!  \]

In particular, the probability for success comes out as
\[ \frac{2(n-1)!}{n!}=\frac 2n.  \]

which calls for a much simpler explanation.

I think that the length of cicles does not count, because if one key is for an open safe and the other can open all others safes, we are done.
A GOOD configuration is one in which the permutation has only one or two cycles and the chosen numbers are in diferent cycles. It is easy to see that in this case we can open all safes.
Conversely, if there are more than two cycles, there is a cylcle from which we cannot open any safe. The same thing holds if there are two cycles and both chosen numbers in the same cycle.

The total number of cases is $94!\binom {94} {92}$. The number of GOOD cases is $94 \binom {94} 2$ (when there is only one cycle: 94 ways to define cycle and $\binom {94} 2$ ways to denote the keys) $+ \sum_{i=1}^{94}i \binom i 1\binom {94-i} 1 \binom {94} i$. (\binom {94} i ways to choose the elements of the cycle, i ways to define the cycle, and \binom i 1\binom {94-i} 1 ways to choose the keys ). The calculations are simple now.

Hope I didn't any counting mistakes.

If there are $n$ keys and at the beginning $k$ safes are open at random then the answer is $\frac{k}{n}$.
I have a hypothesis that if one replaces n keys by some bundles of keys (say 3 bundles wich are placed at random and $n-3$ safes are empty) then the answer is again $\frac{k}{n}$.

Any ideas of proof? Or counterexample?

[quote="kunterbunt"]In particular, the probability for success comes out as
\[\frac{2(n-1)!}{n!}=\frac{2}{n}. \]
which calls for a much simpler explanation.[/quote]

You are right, there is a simple explanation, even for the general case of $k$ open safes.

Construct a graph with the safes as nodes and a directed edge for each key. Each node has exactly one in-edge and one out-edge. Merge the nodes corresponding to the $k$ open safes into node $0$.

Each solution maps to exactly one complete traversal of the graph starting from node $0$.

Furthermore exactly $k!$ solutions map to each traversal, because each $0$ in the traversal except for the starting node corresponds to a distinct original node as each node can only be reached once in the original graph, giving exactly $k!$ permutations of the starting nodes for the $k$ $0$'s in the traversal.

Consider the number of such traversals. It is simply $\binom{n-1}{k-1}(n-k)!=\frac{(n-1)!}{(k-1)!}$. So the original number of solutions is $\frac{(n-1)!}{(k-1)!}\times k!=k(n-1)!$, giving probability $\frac{k}{n}$

[quote="roah"]Additional question:
What is the expected proportion of open safes?

Maybe something like $\frac{k(n+1)}{n(k+1)}$, but i'm not sure.[/quote]

Can you post your proof, or any ideas on how to go about it? It would help us a lot...

I don't like my proof. But here it is:

$e(k,n)$ - expected number of open safes

the first open safe may:
with probability $\frac{k}{n}$ give no useful keys
with probability $\frac{n-k}{n}$ give one new useful key

So we have:
$e(k,n)=1+\frac{k}{n}e(k-1,n-1)+\frac{n-k}{n}e(k,n-1)$
Boundary conditions $e(n,n)=n$, $e(0,n)=0$.

Just a guess (based on experiments with small $k$, $n$):
$e(k,n)=\frac{k(n+1)}{k+1}$.
The solution is unique.

By the way, for probability the equation is:
$p(k,n)=\frac{k}{n}p(k-1,n-1)+\frac{n-k}{n}p(k,n-1)$
$p(n,n)=1$, $p(0,n)=0$.
And one may guess $p(k,n)=\frac{k}{n}$

I'd like to reopen this thread because now i know the solution "from the Book" :)

There are $ n$ safes and $ k$ of them are left open. We use this $ k$ keys to open new safes, to get new keys and so on...
When the key will not give a new key?  When it opens the safe in which there is a key from the first $ k$ safes. Let's call such a key "useless".
The probability of opening all the safes is the probability that the key used to open the last safe was "useless". This is exactly the probability that there is a key from the first $ k$ safes in the last safe.

So it is $ \frac{k}{n}$.

No graphs, no recursion, almost two years to find :)

Erm... Sounds great, but what exactly is "the last safe"? It is not a fixed safe as far as I understand, so how can we be sure that the probabilities are equal :? Am I missing something as usual?

The probabilities are exactly the same as if instead of having the keys in safes, we draw $ k$ keys out of a bag containing all the keys, and get to draw a new key out of the bag every time we open a sack.  This in turn is the same as if we randomly ordered all $ n$ keys at the start, and always just took the first remaining key in our ordering when we open a new safe.

In these terms, we win if the safe corresponding to the last key in our ordering is one of the $ k$ safes that are already open.

kevinatcausa solution is awesome!

but mine is a little bit different. i'm indeed considering a random "last safe". Exact definition is: let $ \tau$ be a stopping time - the time when we cannot open more safes (all safes are open or we have no more keys). $ T \equal{} \tau\wedge (n \minus{} 1)$ is also a stopping time. The "last safe" is the safe not opened at $ T$ with minimal possible number. Let $ M_{t}$ be proportion of not opened safes at time $ t$ containing keys from the first $ k$ safes. $ M_{t}$ is a martingale. $ T$ - bounded stopping time. $ E(M_{T}) \equal{} E(M_{0}) \equal{} \frac {k}{n}$ by the optinal stopping time theorem. And $ E(M_{T})$ is the probability we have been searching (the probability of opening all safes is exactly the probability that the last safe contains the key from the first $ k$ safes).

This text corrects all the errors of the intuitive post about "last safe".

There are several nice ways to think of this result in terms of probabilities.
[url=https://dgrozev.wordpress.com/2021/12/23/traps-in-the-probabilistic-method-part-4/]Here[/url] is a brilliant, simple pure combinatorial solution to the problem by my friend Drago.


To show that $ W$ is a subspace of $ V$ , one of the condition states that $ W$ has to be non-empty .

If we want to show that $ W$ is nonempty , do we ALWAYS have to show that $ 0\in W$? Or we can give any example as long as it fulfills the condition ?

Example : $ W\equal{}\{(x,y,z)\in\mathbb{R}^{3}|x\plus{}y\equal{}z\}$

Can we say that $ (1,1,2)$ is an element of $ W$ and hence $ W$ is nonempty ?
Or we must use the example $ (0,0,0)\in W$?

Thanks

This is one of those, "You're thinking too much about the formula and not enough about why it works" questions.  You need to show that a set $ W$ is non-empty, so [i]of course[/i] it would be sufficient to find any point in $ W$ whatsoever.  [i]However[/i], in order to be a subspace, $ W$ must be a vector space, and in order to be a vector space, $ W$ must contain a zero element.  Thus, even if you show that your subspace contains some element other than a zero vector, you still need to show that it contains a zero vector.

In other words kill two stones with one bird and find the zero vector.  :wink:


[quote="AMCDirector"]Recall that the USAMO Index is AMC score (either 10 or 12, either A or B) + 10*AIME score (either AIME I or AIME II).

The first USAMO Selection Index for students who took the AMC 12 (either A-date or B-date, and without considering age or grade level)  is 204.0.
This yields 320 students.

The lowest AIME score among these students was 6 (144.0 + 10*6) so now we look at all students in 10th grade and below who got a 6 or better on the AIME.  There are 468 of them, so we need to use the USAMO Index to discriminate among them.

The 156 highest USAMO indices among them are at Index 202.5 and higher.

This gives 476 students first selected to the USAMO.  To them we add 10 students from U.S. States not already represented.  This gives us 486 invited to take the USAMO.

We still have some late AIME II papers coming in.  We anticipate about another 10 invitees from these late papers, and some U.S. citizens living abroad.  When all is said and done, we will have between 495 and 500 students invited to 2008 USAMO.

Official emails with the official invitation will start going out no later than Friday, April 11, 2008.  A list of USAMO invitees names, schools and state will be posted on our website no later than Friday, April 11.  Paper postal invitations will be mailed shortly thereafter.  Please see the reminder in Section V, page 5 of the AIME Teachers' Manual, no AIME results will be given out by phone (or otherwise) until after April 13.  Please do not call or contact us now , we have much to do yet.

We expect to begin mailing AIME reports this week, either today, April 8 or tomorrow, April 9.  Again, please do not call or email until after April 13.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

LIST OF QUALIFIERS

Shawn Qian 10 Carmel HS Carmel IN
Matthew Croop 12 North Central HS Indianapolis IN
Lauren Stephens 10 North Central HS Indianapolis IN
Richard Ni 9 Park Tudor School Carmel IN
Jeffrey Shen 9 Park Tudor School Carmel IN
Lyndon Ji 8 Sycamore School Carmel IN
Pardha Ponugoti 11 Terre Haute South Vigo HS Terre Haute IN
Sachin Shinde 10 Terre Haute South Vigo HS Terre Haute IN
Mihika Prabhu 9 West Lafayette HS West Lafayatte IN
Yixin Wang 11 West Lafayette HS West Lafayette IN
David Benjamin 11 William Henry Harrison HS West Lafayette IN
Nathan Benjamin 10 William Henry Harrison HS West Lafayette IN
Philip Thomas 11 William Henry Harrison HS West Lafayette IN
Samuel Dittmer 12 Zionsville Community HS Zionsville IN

14 total, but there might be more...they havent graded them all yet.
NOTE: this is just alphabetical order by schools, i just copied and pasted...

Hmm it seems like Indiana with 14 had like top 12-15ish in the US statewise.

lmao even though it was unofficial:

132 + 7*10 = 202

0.5 below cutoff :rotfl:

!! maybe I should've left some blank on the AMC instead of trying to get a perfect XD

Wow I wonder how many 7s and 8s got left out

 :mad:

[code]Park Tudor   IN 0 0 2 2 2 1 0 1 2 1 0 0 0 1 2  14
T.H. S. Vigo IN 0 0 0 0 1 1 1 2 1 1 1 0 2 1 2  13
W Lafayette  IN 0 0 0 0 0 1 1 1 1 0 0 1 2 3 2  12 [/code]

lol go west lafayette!

hehe that last 3 to 2 would be me leaving, mohit graduating, and mihika qualifying with yixin qualifying again XD

I'll predict the next few years:

2009 - 3 (yixin, mihika, anirudh)
2010 - 4 (yixin, mihika, anirudh, wonho)
2011 - 3 (mihika, anirudh, wonho) unless some other genius shows up

hopefully west side won't have to see that string of 0s and 1s again :lol:

uh whos wonho?

anyways, i think carmels only got 3-4  in their history.

and we are probalby going to double our total qualifiers from the last 30 years with next years group of 9th graders...

wait crapp so you guys have a secret army of geniuses too? west side needs to breed more.. jk

wonho = this person in our school who is about as good as me. i think he got a 5 or a 6 this year on AIME but his AMC 10A score was kinda :dry: even though it was just a bit lower than mine. so mine was kinda :dry: too. and sooey,

in maybe like 11th grade this kid named joon might but unprobably could make it. he did okay on 10 but kinda got owned on AIME.

do you think karen/zoe and stuff will make USAMO eventually??

btw MathAndKnowlege is like a beast now. he will own me next year.

lol I'm fine with you or mwpl owning me, but I'd be devastated if mathandknowledge did. that's _not_ going to happen.

oh yeah joon! the tuck-in kid! hehe maybeee.

hmm I don't know many junior high kids.. maybe they will, if they pursue math.

jr high has about as much intellect as a doorknob. no jk here's the list of best ppl in order

1. Me/mathandknowledge
2. the other one
3. joon
4. idk this other girl
5. i guess mustit
6. this other girl

and besides that, we have no-one.  :(

dude mustit is too high up on that list for comfort :D

unless #2, 3, and 4 are really that good?

or mustit is really that good?

its a sad truth and a glaring statement of the intelligence of my peers.

i guess we have one more  qualifier, youkow homma sycamore school 8

hey undefined, since you took the USAMO last year, may I ask where you would get the SIII forms for the USAMO process?

Do they mail you the forms or do you have to find them somewhere.

I asked my teacher, and he said he had the TII (or TIII not sure) forms that he would fill out but i dont know about mine.

thanks.

Yep they mail you the student forms, but I remember my teacher giving me the student forms to fill out beforehand so I already had them completed when I got them again in the mail.

cognos, look at the bright side - michael catlin didn't make it either :P

mwpl, I just got the exeter (*cough*propaganda) DVD in the mail. goddd they make the school look so awesome XD

thats because it is awesome lol

i think you should go, seriously.

yeahhh except the problem is (I think mwpl kinda thinks the same) I don't want to be biased just because it has a million dollar revenue, is a private school, has a beastly math team, etc whatever. I mean will I really benefit from going to a school like exeter?

I STILL WANT TO

but yeah. worries.

i believe the revenue is 1 [b]billion[/b] dollars. its really tempting isn't it. i wanna go too, but i haven't applied yet. hey sooey, did you know that joon (from my school) got admission to exeter?!?!?!?!?!

hehe yeah billion was what I meant

I think any 8 grader who does fairly well on the AIME gets an invitation packet..? didn't you get one?

not yet, but then again, mwpl didn't get one until the beginning of her 9th grade year.


limit
logarithms
real analysis
real analysis unsolved

[b]If $a>0$, find [/b] $\lim_{n\to \infty}\frac{a\sqrt{a}\sqrt[3]{a}\cdots \sqrt[n]{a}}{a+\sqrt{a}+\sqrt[3]{a}+\ldots+\sqrt[n]{a}}$ .

Math analytic Find more different way : he will be find equivalnt of this and after find limit: it give as more good result  :wink:

[quote="N.T.TUAN"]This is easy problem. We need use $C=\lim_{n\to\infty}(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln{n})$ and $\lim_{n\to\infty}\sqrt[n]{a}=1$. That is all :D[/quote]

Have you used the Stolz-Cezaro?
Then what have you done in the case a<e?


Find an integer $n$ such that:
$\sigma(n)=\phi(n)^2$
or show there is no solution.

I suppose you mean $\sigma(n) = \sum_{d \mid n} d$. So let $n = 1$ or $n = 3$ and you're done... :huh:

I think he ment "find [b]all[/b] integers $n$ such that ..." ;)


[img]http://i82.photobucket.com/albums/j274/boyhiphop160/Linh%20tinh/maths.jpg[/img]
Thnz much ^^


The marked price of a book was $ 30\%$ less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?

$ \textbf{(A)}\ 25\% \qquad
\textbf{(B)}\ 30\% \qquad
\textbf{(C)}\ 35\% \qquad
\textbf{(D)}\ 60\% \qquad
\textbf{(E)}\ 65\%$

[hide]Let the retail price be $ r$.  In the marked price, he only paid $ 0.6r$.  Since he only paid half of the marked price, he paid $ (\frac{1}{2})0.6r\equal{}0.3r$.  He paid $ \boxed{30\%}$ of the retail price.[/hide]

[hide="Actually..."]Alice pays $ .5\cdot (1\minus{}.3)\equal{}\boxed{\textbf{(C)}\ 35\%}$[/hide]

[hide=Solution]

Let the original cost be $x$.

This means that the marked price is $x-\frac{30}{100}=\frac{70}{100}x$

Because she got a $50\%$ discount, the percent she got the book at is $\frac{50}{100}\left(\frac{70}{100}x\right)$ which simplifies to $\boxed{35\%}$


group theory
abstract algebra
superior algebra
superior algebra solved

i understand how to find generators of a group such as $Z_{10}$
because gcd(10,1)=1 , gcd(10,3)=1 ect... and i know the order of $Z_{10}$ , but i dont see how to find a generator for $mZ$ + $nZ$ where $m$ and $n$ are integers and $mZ$ + $nZ$  is a subgroup of $Z$ or the generator for the subgroup $12Z$ + $21Z$ ,

A generator for $m\mathbb Z+n\mathbb Z$ is, for instance, $d$, the greatest common divisor of $m$ and $n$.


integration
vector
function
calculus
real analysis
geometry
parallelogram

I have the following query about scalar products:

Everyone knows that in $ l^{2}$, ${ x\cdot y=\sum_{i=1}^{+\infty}x_{i}{\bar y}_{i}}$ defines a scalar product; while in $ L^{2}[0,1]$, ${ (f\cdot g)(x)=\int_{x=0}^{1}f(x){\bar g}(x)\,dx}$ also defines a scalar product. Can we actually prove that they are the only scalar products or are there any other scalar products on the same space exists?

Furthermore, do we have other scalar products on other spaces $ l^{p}$ and $ L^{p}[0,1]$?  :?:

A scalar product $ x\cdot y \equal{}\alpha$, where $ x$, $ y$ are vectors and $ \alpha$ is a scalar, is in general defined through the following properties:

1: $ (x\plus{}y)\cdot z \equal{} x\cdot z\plus{}y\cdot z$ for any vectors $ x,y,z$.
2: $ (px)\cdot y \equal{} x\cdot(py) \equal{} p(x\cdot y)$ for any vectors $ x,y$ and scalar $ p$.
3: $ x\cdot y \equal{} y\cdot x$ for any vectors $ x,y$.
4: $ x\cdot x\geq0$ with equality iff $ x \equal{} 0$.

When the vector space is defined over complex numbers, condition 3 is modified so that $ x\cdot y \equal{} (y\cdot x)^{*}$, where $ ^{*}$ denotes complex conjugate.

Now I ask you, is the following a valid definition of a scalar product on $ \mathbb{R}^{n}$?
\[ \sum_{i\equal{}1}^{n}w_{i}x_{i}y_{i},\]
where the $ w_{i}$ are a set of positive (nonzero!) weights.

What about $ \int_{0}^{1}f(x)g(x)w(x)dx$, where $ w(x)$ is a positive-valued integrable function in (0,1)?

So, how about defining a scalar product on $ L^{p}[a,b]$ space (where $ p\not=2$) and use this scalar product to define a norm $ \lVert\cdot\lVert$ on the space?

a scalar product may be used to define a norm, as long as the vector space over which the norm is defined satisfies certain conditions (I am not an expert on this field, so I am afraid that I cannot answer you very properly - much less tell you exactly which conditions these are).  However, there are certain issuest that I am confident to address:

1) On $ \mathbb{R}^{n}$, being $ n$ finite, we may always define a norm based on a scalar product as the one that I mentioned earlier; it is basically a norm similar to the euclidean distance, except the "distances" along certain directions are "longer" (it is as if we were measuring the distance using the "time" needed to go from one point to another, and advancind along certain directions was "more difficult" than along others).

2) $ L^{p}\left[a,b\right]$ is itself a space where a norm has already been defined, because we already talk about functions whose modulus to the power of $ p$ is integrable, the $ p$-th root of this integral being the norm of the function.  You may always try to define a scalar value for any function in $ L^{p}\left[a,b\right]$ through a formula such as the one that I propose, and call the result a norm, but the integral of $ w(x)f^{2}(x)$ may not converge even if the integral of $ f^{p}(x)$ converges! - I recommend you try to find examples for yourself as an exercise.  Even the definition of a scalar product is then in jeopardy, since even if the integrals of $ f^{p}(x)$ and $ g^{p}(x)$ exist, this does not guarantee the convergence of the integral of $ w(x)f(x)g(x)$!!

3) Using the standard integration, certain non-zero, non-continuous functions may have zero norm; for example, take the set of functions that are square-integrable in $ [0,2]$.  Obviously, a function that takes on value 1 at $ x\equal{}1$ and 0 elsewhere belongs to this set, the result of square-integrating it being 0.  Then, any integral-and-weight-function-based norm that we define will give a 0 norm for this function, which is not identically 0!  So when talking about functions, special care must be taken.  This is also why $ L^{p}$ (and in particular $ L^{2}$) is usually taken as a set, not of functions, but of equivalence classes; the function that I propose above would then be in the same equivalence class as the identically 0 function, and of course the norm and the distances based on the integration of the product would be defined, not between functions, but between equivalence classes, two functions being in the same equivalence class if their difference, when square-integrated, yields 0.

4) One of the main characteristics of the scalar product is that it is "bilinear" (condition 1 on the set that appears on my previous reply).  You may define a norm in $ \mathbb{R}^{n}$ as $ \sqrt[p]{\sum_{i\equal{}1}^{n}x_{i}^{p}}$ for $ p\neq 2$, but good luck finding a scalar product that goes with that norm!  So even if a scalar product may be used to define a norm, it does not necessarily the other way around.

After rereading what I just wrote, I just realized that, more than answering your question, I still gave you some more!  Please let me know if this is useful for you and if I did answer your question or not...

Thanks for your reply. Most of the material could be found in any standrad text of functional analysis, for example Walter Rudin "Real and Complex Analysis" or "Functional Analysis".
A normed space $ (H,\Vert\cdot\Vert)$ has a scalar product if and only if it satisfies the parallelogram rule $ \Vert x\plus{}y\Vert^{2}\plus{}\Vert x\minus{}y\Vert^{2}\equal{} 2(\Vert x\Vert^{2}\plus{}\Vert y\Vert^{2})$ and if it does satisfies, the scalar product is defined by $ {(x,y) \equal{}\frac{1}{4}(\Vert x\plus{}y\Vert^{2}\minus{}\Vert x\minus{}y\Vert^{2}\minus{}i\Vert x\plus{}iy\Vert^{2}\plus{}i\Vert x\minus{}iy\Vert^{2}) }$

I am looking for a full list of examples of norms and scalar products on spaces other than in those standard textbooks.


linear algebra
matrix
linear algebra unsolved

let $ n\ge 2$ and

$ k_1\equal{}(1, n\plus{}1,2n\plus{}1,\dots,n^2\minus{}n\plus{}1)$

$ k_2 \equal{} (2, n \plus{} 2,2n \plus{} 2,\dots,n^2 \minus{} n \plus{} 2)$

$ \dots$

$ k_n \equal{} (n, 2n,\dots,n^2)$.

consider matrix $ K$, where $ k_i$ is the i-th column of $ K$.

for given $ n\ge 2$, compute $ rank(K)$.

The $ k$th column of this matrix can be written as
\[ \begin{bmatrix}1 \\
2 \\
\vdots \\
n\end{bmatrix} \plus{} n(k \minus{} 1)\begin{bmatrix}1 \\
1 \\
\vdots \\
1\end{bmatrix}
\]
Hence every column of the matrix lies is in a particular 2-dimensional subspace. Since the first and second columns are independent by inspection, that means that the rank of the matrix is $ 2.$


logarithms
calculus
integration
derivative
calculus computations

What is the antiderivative of

$ \frac{ln^2x}{x} $

I'm pretty sure you do a u substitution...but how.  Can anyone help, please?

Try letting $u=\ln x,\,du=\frac{dx}x.$

There are two lines of thought that both point towards this.

1. Integration by substitution is the chain rule, read backwards. It's about composition of functions. The "natural" substitution is for the "thing that's inside." In this case, $\ln x$ is inside the square.

2. It's convenient to make substitutions where the derivate is already available to us. In this case we have $\frac{dx}x.$ That's the derivative of $\ln x.$

yep it works... and it's so simple I feel stupid for not seeing it right away.  Thanks for your help!


geometry
geometric transformation
dilation
geometry proposed

Let B and C to fix point and A a variable point on d, where d is a fix line. Find the place where all centroids of triangle ABC lie.

By desarques’s theorem, we can easy to prove that the locus of the centroid $G,$ of the triangle $\bigtriangleup ABC,$ when it’s vertex $A,$ is moved on a fixed line $(d),$ is a line parallel to $(d)$ and passes through a constant point, so be it $S,$ lies on the midperpendicular of the segment $BC.$

Kostas Vittas.

PS. I will post here later, the proof I have in mind about this pretty problem.

$M$ is the midpoint of segment $BC$ so $M$ is fixed. Mind that $AG=\frac23AM$ with $G$ is the centroid of $\Delta ABC$
MN is perpendicular to $(d)$ with $N$ is on $(d)$
Then the locus of $G$ is line $(a)$ which is parallel to $(d)$ and cuts $MN$ at P satisfied $NP=\frac23NM$ (P and N are fixed points)

yeah...it was easy but the other two are harder

Another problem given segment $BC$ and $A$ move on a shape $H$ find the locus of centroid $G$ of triangle $ABC$

I think this is not a hard problem. Let $M$ is midpoint of $BC$.
$(d)$ is an arbitrary line , passes through $M$ cuts $(H)$ at $N$
$P$ is on $(d)$ such that $NP=\frac{2}{3}NM$
Then the locus of $G$ is shape $(H')$ such that
$(H')=f_{\Vec{NP}}((H))$

That's right it is image of $(H)$ through a dilation! but some onther locus I saw on computer is very strange: example locus of $H,I,L$ when $A$ moved on an arbitrary circle ($L$ lemoine point) maybe they have some especial properties!


inequalities
Functional Analysis
advanced fields
advanced fields unsolved

$E$ an infinite dimensional Banach space. Prove that there is a subset $S$ of $E$ that is weakly sequentially closed, but not closed in the weak topology.

Consider the space $l_{1}(\mathbb{C})$. In this space, any sequence is weakly convergent if and only if it is convergent in norm. As an immediate corollary, $\{x;\ ||x|| \geq 1\}$ is weakly sequentially closed but is not weakly closed (the fact that it is not weakly closed is easy to prove).

yes, $L^1$ has the Schur property.
(and the problem was to show that one can find such subsets in every infinite dimensional Banach space)


Suppose that $a$, $b$ and $c$ are positive real numbers. Prove that $\frac{a^{3}}{b^{2}-bc+c^{2}}+\frac{b^{3}}{c^{2}-ca+a^{2}}+\frac{c^{3}}{a^{2}-ab+b^{2}}\ge \frac{3(ab+bc+ca)}{a+b+c}$


Sorry, this problem has some physics but I figured the people on this site are smart enough to do it.

A massless rope is tossed over a wooden dowel of radius $ r$ in order to lift a heavy object of weight $ W$ of the floor.  The coefficient of sliding friction between the rope and the dowel is $ \mu$.  Show that the minimum downward pull on the rope necessary to lift the object is

$ F_{down}$ = $ We^{\pi\mu}$

well this problem ought to be posted in the [url=http://www.mathlinks.ro/index.php?f=332]Advanced Physics[/url] forum
well this has already been solved  [url=http://www.mathlinks.ro/viewtopic.php?search_id=2032861136&t=80502]here[/url] by [b]Djole[/b]


inequalities
geometry
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be an acute triangle with the orthocenter $ H$ and the circumradius $ R$ .

Prove that there is the inequality $ \boxed {\frac {HA^2 \plus{} HB^2 \plus{} HC^2}{HA \plus{} HB \plus{} HC}\ge R}\ .$

Your inequality looks quite attractive in the form you put it. Nice.  :)

Thanks to the identity $ \left|PG\right|^2\equal{} \frac{1}{3} \sum_{cyc} \left|PA\right|^2 \minus{} \frac{1}{9}\sum_{cyc}a^2$ for every point P, we find that  \[ \sum_{cyc} \left|PA\right|^2 \geq \frac{1}{3}\left(a^2\plus{}b^2\plus{}c^2\right)\]

In addition, we can easiley obtain that $ AH \equal{} 2R \cdot \cos{A}$, so therefore:

\[ \frac{AH^2\plus{}BH^2\plus{}CH^2}{AH\plus{}BH\plus{}CH} \geq \frac{\frac{1}{3}\left(a^2\plus{}b^2\plus{}c^2\right)}{2R\left(\cos{A}\plus{}\cos{B}\plus{}\cos{C}\right)}\]

Now, we use the fact that $ \left|OH\right|^2\equal{} 9R^2 \minus{} (a^2\plus{}b^2\plus{}c^2)$, so $ a^2\plus{}b^2\plus{}c^2 \geq 9R^2$
Also, we have the famous inequality $ \cos{A}\plus{}\cos{B}\plus{}\cos{C} \leq \frac{3}{2}$

Combining this, we get \[ \frac{\frac{1}{3}\left(a^2\plus{}b^2\plus{}c^2\right)}{2R\left(\cos{A}\plus{}\cos{B}\plus{}\cos{C}\right)} \geq R\]

Equality occurs when the triangle is equilateral.

Best regards,
Jan

Using the relation $ AH\equal{}2R\cdot cos A$, the problem is quickly rewritten equivalently as \[ \frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{x\plus{}y\plus{}z}\geq \frac{1}{2},\] or $ 2(x^{2}\plus{}y^{2}\plus{}z^{2})\geq x\plus{}y\plus{}z$, where I have denoted $ x: \equal{} cos A$ etc. Using the well known identity $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2xyz\equal{}1$, the above inequality becomes equivalent to \[ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}1\geq x\plus{}y\plus{}z\plus{}2xyz.\] But clearily $ x^{2}\plus{}\frac{1}{4}\geq x$, and the other similar inequalities for $ y$ and $ z$, so by summing we get $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}\frac{3}{4}\geq x\plus{}y\plus{}z$, therefore, it would suffice to prove that $ \frac{1}{4}\geq 2xyz$. But this is clear from the AM-GM inequality:
\[ 1\equal{}x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2xyz\geq 4\sqrt[4]{x^{2}\cdot y^{2}\cdot z^{2}\cdot 2xyz}\] :wink:

Equality occurs for $ x\equal{}y\equal{}z\equal{}\frac{1}{2}$, thus for an equilateral triangle.
Nice problem. :)

[quote="Jan"]Now, we use the fact that $ \left|OH\right|^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2)\ge 0$, so $ a^2 \plus{} b^2 \plus{} c^2\ \boxed {\ \ge\ }\ \ 9R^2$[/quote]
Sorry, here is a bit of mistake  (boxed) ...   Correctly is $ a^2 \plus{} b^2 \plus{} c^2 \le 9R^2$
Nice proof, [b]Andyciup[/b] ! Thanks for it.

My apologies to all, I seem to have made the classical mistake of reversed inequalities, which is sometimes made more quickly than I seem to think. I should have been more precise and have checked my solution.

Thanks for the problem and the correction Virgil Nicula, and for the solution Andrei Ciupan.
Is this also your solution mr. Nicula?

[hide="For Jan Vonk, Andrei Ciupanu & Co , with pleasure !"][quote="Virgil Nicula"]
[color=darkred]Let $ ABC$ be an acute triangle with the orthocenter $ H$ and the circumradius $ R$ .

Prove that there is the inequality $ \boxed {\frac {HA^2 + HB^2 + HC^2}{HA + HB + HC}\ge R}\ .$[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ \{\begin{array}{c} \sum HA^2 + \sum a^2 = 12R^2\ . \\
 \\
\sum a^2 = 2(p^2 - r^2 - 4Rr)\ . \\
 \\
\sum HA = 2R(R + r)\ .\end{array}$ Thus, $ \sum HA^2\le R\cdot\sum HA$ $ \Longleftrightarrow$

 $ 12R^2 - 2(p^2 - r^2 - 4Rr)\ge 2R(R + r)$ $ \Longleftrightarrow$ $ p^2\le 5R^2 + r^2 + 3Rr\ .$

But $ \{\begin{array}{c} p^2\le 4R^2 + 4Rr + 3r^2\ (\ \underline {\mathrm {Gerretsen's\ inequality}}\ ) \\
 \\
4R^2 + 4Rr + 3r^2\le 5R^2 + r^2 + 3Rr\Longleftrightarrow(R - 2r)(R + r)\ge 0\end{array}\|$ $ \implies$

$ p^2\le 5R^2 + r^2 + 3Rr$ $ \Longleftrightarrow$ $ \boxed {\frac {HA^2 + HB^2 + HC^2}{HA + HB + HC}\ge R}\ .$  [/color][/hide]

In: http://www.mathlinks.ro/viewtopic.php?t=24870 , Darij has proved that in any acute triangle $ ABC$, we have $ HA\plus{}HB\plus{}HC\equal{}2\left(R\plus{}r\right)$ and $ HA^2\plus{}HB^2\plus{}HC^2\equal{}12R^2\minus{}\left(a^2\plus{}b^2\plus{}c^2\right)$. So the given inequality is equivalent to $ 10R^2\ge a^2\plus{}b^2\plus{}c^2\plus{}2Rr$, which follows from the fact that $ a^2\plus{}b^2\plus{}c^2\le 9R^2$ and $ 2r\le R$. And we are done!  :D

I do not remember trigonometry well, but in equilateral triangle
sum of squared cosines =3/4, and this is minimum.
Together with the sum of cosines <=3/2 we will get the required inequality?

  M.T.


calculus
derivative
geometry
geometric transformation
rotation
function
linear algebra

Let $f : \mathbb C \to \mathbb C$, $f(z) = f(x+yi) = u(x,y) + i v(x,y)$. If $u$ and $v$ have continuous partial derivatives at $z_0=x_0 + i y_0$, $\displaystyle \frac{\partial u}{\partial x} (x_0,y_0) = \frac{\partial v}{\partial y} (x_0,y_0)$ and $\displaystyle \frac{\partial u}{\partial y} (x_0,y_0) = - \frac{\partial v}{\partial x} (x_0,y_0)$, then $f$ is differentiable at $z_0$.

How do you prove this?

If I remember well my calculus classes, when you extend the definition of derivatives to complex numbers, it has some variations, with geometrical meanings. Hummm...it has something to do with rotation matrix...like, what can I say, to be differenciable, we have to approximate the function by a transformation, such that the matrix associated to it is multiple of something of the form:

$\begin{bmatrix}U&-V\\V&U\end{bmatrix}$

If you try the computations, you'll find a matrix of partial derivatives of u(x,y) and v(x,y). And you'll then set the conditions which will make you conclude what you want...=]

Maybe you're right. But it's hard to understand, especially when you are not too sure about what you're writing  :(

hehee I'm sure about what I am writing =], I was even trying to make the computations here...i wasn't going to answer you something i had no idea about ;-)...the difference is that we can multiply vectors in complex numbers...i.e, the meaning of the division is quite different...when i get home i try to explain better...there are bugs in here uhuhuhuhuh

The mean of a function being differenciable is that we can write it as a linear approximation in "close" points. Mathematically saying:

$f(z+h)=f(z)+f'(z)h+\epsilon(h)$

Okay, but this is for ordinary functions, we actually don't know if it works when we extend it to complex. If you try some computations, you'll see that only this definition is not concise for complex [try to compute this limits in the direction of the real and imaginary "vectors"].

So let's take

$f'(z)=a+bi$

We know that

$f: \mathbb{C}=\mathbb{R}^2\rightarrow \mathbb{C}=\mathbb{R}^2$

which geometrically means that a functions which maps complex into complex is only a plane-plane function. Now let's go to the computations:

$f(z+h)=f(z)+(a+bi)(h_1+h_2i)+\epsilon(h)=f(z)+(ah_1-bh_2,ah_2+bh_1)+\epsilon(h)= f(z)+\begin{bmatrix}a&-b\\b&a\end{bmatrix}\begin{bmatrix}h_1\\h_2\end{bmatrix}+\epsilon(h)$

See? That was the matrix I was talking about. Now, only a little bit more of computations:

$f(z+h)=f(z)+\sqrt{a^2+b^2}\begin{bmatrix}\frac{a}{\sqrt{a^2+b^2}}&\frac{-b}{a^2+b^2}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{bmatrix}\begin{bmatrix}h_1\\h_2\end{bmatrix}+\epsilon(h)$

which means geometrically, that differentiability in complex is associated to an homotety and a rotation in the plane. Now, consider only the functions you want and the matrix associated to it, will be a matrix with partial derivatives of u and v...and take the conditions =]

I hope i could help you and sorry i was late =]
best wishes

Another way to look at this: we have two senses of "differentiable"; the $\mathbb{R}^2$ sense, and the $\mathbb{C}$ sense. A function is (complex) differentiable if and only if it is differentiable as a real function and its derivative has the form of (multiplication by) a complex number. It is a standard theorem that a function is real differentiable if the partial derivatives exist and are continuous, and its derivative is the matrix of partial derivatives. The Cauchy-Riemann equations are the condition that this matrix acts as a complex number.

[quote="cimabue"]my guess is that the original statement is actually wrong. you need the cauchy riemann relations in a neighborhood of $z_0$, pointwise might not be enough.[/quote]

But you also have [quote]  $u$ and $v$ have continuous partial derivatives at $z_0=x_0 + i y_0$ ...[/quote]

In the book I was reading it was said that this is enough.

i misread it, but the original statement was also a bit misleading. so you need to show that $f$ is complex-differentiable.

$f$ is a $C^1$ function, so it is differentiable in the $\mathbb R$-sense.
assume for simplicity $f(0)=0$.
hence  for $z=x+iy$ near $z_0=0\in\mathbb C$, with $a=u_x(0,0)$ and $b=u_y(0,0)$

$u(x,y)=ax+by+o(|z|)$
$v(x,y)=-bx+ay+o(|z|)$

$f(z)=u+iv=a(x+iy)-ib(x+iy)+o(|z|)=(a-ib)(x+iy)+o(|z|)$

let $c=a-ib\in\mathbb C$

then $f(z)=f(0)+cz+o(|z|)$, meaning $\lim\frac{f(z)-f(0)}{z}=c$, so $f$ is complex differentiable at $0$.


analytic geometry
graphing lines
slope
trigonometry

In right triangle $ABC$ with $\angle B=90^\circ ,$ points $M$ and $N$ are on $AB$ and $BC,$ respectively, such that $AM=BC$ and $BM=NC.$ $AN$ and $CM$ intersect at point $P$ within the triangle. Find $m \angle APM.$

Did your friend like tell you an answer?  Could you like hide it so that we can see what the answer is? Thanks

Oh, duh! I can't believe I didn't see this...  :wallbash_red:  :fool: I am retarded.

[hide="Answer"]$45^\circ$[/hide]

[hide="Solution"]Place $A,$ $B,$ and $C$ on a coordinate plane with $AB$ on the y axis and $BC$ on the x axis. Let $BC=a$ and $BM=b$ so that the coordinates of the points are $A=(0, a+b),$ $M=(0,b).$ $B=(0,0),$ $N=(a-b,0),$ $C=(a,0).$

Note that the slope of $AN$ is $\tan ANC=-\frac{a+b}{a-b},$ and the slope of $MC$ is $\tan ACQ=-\frac ba,$ where $Q$ is a point on the x axis with an x coordinate greater than $C.$

Now, $\tan APM=\tan (ANC- ACQ)=\pm 1$, so $m\angle APM=45^\circ.$

[/hide]


A loader has two carts. One of them can carry up to 8 kg, and another can carry up to 9 kg. A finite number of sacks with sand lie in a storehouse. It is known that their total weight is more than 17 kg, while each sack weighs not more than 1 kg. What maximum weight of sand can the loader carry on his two carts, regardless of particular weights of sacks?

[i]Author: M.Ivanov, D.Rostovsky, V.Frank[/i]

Is it $ \frac{9}{10} \cdot 17$ ?

with equal weights of $ \frac{9}{10}\plus{}\epsilon$

i think i have an ugly working solution and it is $ 17\cdot 0.9\equal{}15.3$
first the example mszew gave shows that we can give the loader such weights that he can't guarantee more. now notice that if we give him a weight $ x$ such that $ x\leq 0.85$ then he can obtain $ 15.3$
just fill the carts without using $ x$. if we have no more weights (except $ x$) then we have more that $ 17\minus{}0.85\equal{}16.15$ which is enough. if we simply can't put any more weights(except $ x$) then the sum is atleast $ 15$. if we can put $ x$ we are done, and if we can't then the carts must weigh atleast $ 7.15$ and $ 8.15$ atleast respectively so in total over $ 15.3$
so there is no weight $ \leq 0.85$.
now we count the weights in the sections $ <0.85,0.9], <0.9,0.95>, [0.95,1]$
 if there are at least $ 10$ weights in the first segment then we take them and put them on the bigger cart
their weight is between $ 8.5$ and $ 9$ so they fit in the cart and since we can certainly fill the other carts weight over $ 7$ we are done. so there are at most $ 9$ weights in the first segment. the minimum nuber of weights is $ 18$ so there are at least $ 9$ segments in the interval $ <0.9,1]$. if there are two greater than $ 0.95$ then take those two and six from $ <0.9,0.95>$ and put it on the smaller cart
it fits and weighs atleast $ 6.3$ which is again enough.
so atmost 1 over $ 0.95$ and atleast $ 8$ in $ <0.9,0.95>$
if now there exists atleast one weight smaller than $ 0.9$ put it aside
take the $ 8$ weights from $ <0.9,0.95>$ and put it on the small cart. minimum weight $ 7.2$
now fill the big carts weight arbitrarily over $ 8$.if it's greater than $ 8.1$ we're done otherwise add the weight we put aside (the weight smaller than $ 0.9$) and then we ARE done :)
so now we know there are no weights smaller than or equal to $ 0.9$.
since the loader can surely put at least $ 8$ weights on the small and $ 9$ on the big cart whose total weight is atleast $ (8\plus{}9)\cdot 0.9\equal{}15.3$ we are done... blaaaah :D to long


1.Electrophoretic mobility(without change in primary structure) of a multi-subunit protein sample with interchain disulphide bonds will be affected if treated with:
a)performic acid
b)trypsin
c)NaCl
d)CNBr

2.A polypeptide chain with 60 amino acids residue will have________ number of nucleotides in its DNA?
a)60
b)120
c)180
d)360

Please answer to these above questions with reasons  :!:

think the first answer is CNBr. dunno the reason tho

The solns can be seen in the indian forum .
The answer for 1 is infact a.


function
limit
trigonometry
algebra
polynomial
calculus
calculus computations

Prove that for two functions f(N) and g(N), if lim (n>oo) f(N)/g(N) is a constant, then f(N) = O(g(N)).

I believe that is the definition of the big-O notation.

That's not quite the definition of big-O notation, although many people (incorrectly) say it is.

For two functions $f(N)$ and $g(N)$, to say that $f$ is $O(g)$ as $N \to \infty$ means that there exists a number $C>0$ such that eventually $|f(N)| < C|g(N)|$.

$\lim_{N \to \infty} |f(N)|/|g(N)|$ does not have to exist in order for $f$ to be $O(g)$.

Kent Merryfield wrote a good explanation of big-O notation here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517[/url]

Slammer, did you try to prove this theorem?  How'd it go?

Oh, indeed, DVO is quite right again (and I am quite a fool... again).

From the definitions of $O(g)$ and of the limit, then, it should not be difficult to prove. The limit existing and being finite tells you that for large enough $N$, the expression has to be close to the limit.

The proper notation is $\sim$, as in if $\lim_{x\to\infty} \frac{f(x)}{g(x)}=C$, then $f(x)\sim Cg(x)$.

[quote="Xevarion"]What is $f(x) \asymp g(x)$ then?  :?[/quote]
Some people use this notation to say that $cf(x)\le g(x)\le Cf(x)$ for some $0<c\le C<+\infty$, i.e., to say that $f=O(g)$ and $g=O(f)$ simultaneously. Other people, indeed, use it in the same sense as $\sim$. This all may become pretty confusing but, fortunately, the meaning is usually not hard to guess in each particular case.

if two functions have the same gowth, then the limit is a constant.

[quote="fedja"][quote="slammer"]if two functions have the same growth, then the limit is a constant.[/quote]
You are missing the point: there may be no limit (like for $f(x)=x$ and $g(x)=(2+\sin x)x$ as $x\to+\infty$)[/quote]
if we only consider polynomial.

I guess I missed the memo on where all functions are polynomials.  Seriously, if all we cared about were polynomials, we really wouldn't need big O notation in the first place.


hello

please help solve this

thanx

Determine the number of paths of length n between any two adjacent vertices in K3,3, for all n >= 0.

[hide="h31p"]Adjacent=same side?

3^(n-1) (for even n), 0 (for odd n). Why? You can go to anything on side 1, then anything on side 2, side 1, side 2, ..., then get back to side 2 in the end.

If Adjacent=different side then it is 0 on even and not odd n.[/hide]


probability
search
probability and stats

We are operating in the probability space $ (\Omega,\mathcal{F},\mathbb{P})$. We have $ X,Y\in \mathcal{L}^{1}(\mathcal{F})$, i.e. $ X,Y: (\Omega,\mathcal{F})\rightarrow (\mathbb{R},\mathcal{B})$ are integrable random variables. Now given that $ \mathbb{E}[X|\sigma(Y)]\equal{}Y$ and $ \mathbb{E}[Y|\sigma(X)]\equal{}X$. Prove that $ \mathbb{P}[X\equal{}Y]\equal{}1$.

It is a problem (11.1 I think) from the exercise section of David Williams book Probability with Martingales.

If you use any non obvious result form theory state it.

Notation:
If $ \mathcal{A}$ is a sigma-sub-algebra of $ \mathcal{F}$ then $ Z\equal{}\mathbb{E}[X|\mathcal{A}]$, where $ X$ is a r.v., means exactly that $ Z$ is $ \mathcal{A}$-measurable and that $ \mathbb{E}[X\mathbb{I}_{A}]\equal{}\mathbb{E}[Z\mathbb{I}_{A}]$ for all $ A\in\mathcal{A}$.
$ \sigma(Y)$ denotes the smallest sigma-sub-algebra of $ \mathcal{F}$ such that $ Y$ is measurable under it.
I could go on explaining the notation (and will if absolutely necessary) but I'll leave it at that. The notation is gotten from David Williams book.

It was discussed here before: search for "circular conditional expectation".


linear algebra
matrix
vector
topology
Functional Analysis

Suppose $ A$ is nilpotent. Show that $ I\plus{}A$ is invertible.

So $ \det A \equal{} 0$. Thus $ \det (I\plus{}A) \equal{} 1$. And so $ I\plus{}A$ is invertible?

What theorem do you think you're quoting when you say that $ \det(I\plus{}A)\equal{}1?$

Here's how I'd prove it:

Let $ B\equal{}\sum_{n\equal{}0}^{\infty}(\minus{}1)^nA^n\equal{}I\minus{}A\plus{}A^2\minus{}A^3\plus{}\cdots.$

Note that because $ A$ is nilpotent, this is actually a finite sum and we don't have to worry about convergence.

Then by direct calculation, $ B(I\plus{}A)\equal{}I,$ so $ B\equal{}(I\plus{}A)^{\minus{}1}.$

The proof would work in any ring with identity - it's not limited to just matrices.

I am using the fact that $ \det A \neq 0$ implies that $ A$ is invertible. We know that the determinant of a nilpotent matrix is $ 0$. So if we add the identity matrix we will get a non-zero determinant. This implies that $ I \plus{} A$ is invertible.

It is entirely possible to have a matrix $ A$ such that $ \det A \equal{} 0$ and $ \det(I \plus{} A) \equal{} 0.$ Such a matrix $ A$ would not be nilpotent, but it does reveal that your stated reasoning has a huge hole in it.

[hide="Example"]Let $ A \equal{} \begin{bmatrix} \minus{} 1 & 1 \\
0 & 0\end{bmatrix}$ so that $ I \plus{} A \equal{} \begin{bmatrix}0 & 1 \\
0 & 1\end{bmatrix}.$[/hide]

Also: please don't quote entire messages, and don't quote at all unless there is a specific point. I'm going to go back and delete the quote from your last message - it was completely pointless.

Is $ A$ nilpotent here? If we know that $ A$ is nilpotent, can we say that $ \det(I\plus{}A) \neq 0$?

I can. You can't until you prove it. (characteristic polynomial...)

Of course, I prefer simply constructing the inverse as Kent did.

Also, in a general ring if $ x$ is nilpotent, then for any unit $ u$, $ x\plus{}u$ is a unit. So you can do this completely off of the ring structure and not have to use any facts specific to matrices (although the easiest proof I know of this fact uses the Jacobson radical, which is probably way overkill).

If A is nilpotent then all its eigenvalues are zero. So, -1 is not an eigenvalue of A, which means $ \det(I \plus{} A) \equal{} \det[A \minus{} (\minus{}I)] \neq 0$, and $ I \plus{} A$ is non-singular.

Yes, but that's the weakest solution method.  For example, it fails if $ A$ is an operator on an infinite-dimensional vector space.  The "Neumann series" technique generalizes well and is useful in, for example, functional analysis.

Yes, you are right, I was thinking only in matrices. Three questions:

1) Why does it fail on an infinite dimensional vector space?

2) Why the designation "Neumann Series"?

3) How usefull in functional analysis?

[quote="Carcul"]1) Why does it fail on an infinite dimensional vector space?[/quote]
An operator need not have [b]any[/b] eigenvalues in general, since it need not have any eigenvectors in general.  Invertibility is much more sophisticated in this context.  For nilpotent operators it may be possible to give a separate argument, but the point is that the Neumann series argument becomes much more natural.

[b]Edit:[/b]  To be more precise, a standard way to show that an operator is invertible in functional analysis is to give its inverse explicitly as a diagonal operator on some orthonormal basis, but the eigenvectors of a nilpotent operator $ A$ are just the elements of its nullspace (since $ A$ can have no eigenvectors associated to any eigenvalue except $ 0$), hence in general the eigenvectors of $ I \plus{} A$ don't form a basis and writing down an inverse can't be done in this way.  Instead, one must also consider the nullspaces of $ A^2, A^3, ...$ and then this becomes much harder than the Neumann series approach.

[quote="Carcul"]2) Why the designation "Neumann Series"?[/quote]
Apparently [url=http://en.wikipedia.org/wiki/Neumann_series]von Neumann used them[/url].

[quote="Carcul"]3) How usefull in functional analysis?[/quote]
Very useful!  Neumann series tell us a lot about the topology of the invertible operators and are useful for proving many technical results.


conics
ellipse
geometry
perpendicular bisector

I am not looking for the answer. I want to know the steps to finding the answer on my own. If you find the answers, please EXPLAIN how you find it.

What is an equation of the locus of points equidistant from points A(3, 1) and B(7, 1)?

Two points? Most likely this will end up as a ellipse or a hyperbola. Just plug it into the formula for an ellipse or hyperbola.

[hide="Solution"]The locus of points equidistant from two points is given by the perpendicular bisector of the segment formed by the points.
In this case, we're looking for the equation of the line perpendicular from A and B.
The line going through $(3,1)$ and $(7,1)$ is the horizontal line $y = 1$.
We're looking for the BISECTOR of this segment, so we take the average of 3 and 7 (5), and see that we're looking for a line perpendicular to $y = 1$, and x must be 5.
Thus, the locus of points equidistant from A and B is simply the line
$x = 5$.
[/hide]

DiscreetFourierTransform,

You wrote:

"We're looking for the BISECTOR of this segment, so we take the average of 3 and 7 (5), and see that we're looking for a
line y = 1 perpendicular to , and x must be 5."

I understood your explanation but isn't 7(5) + 3 divided by 2 = 17?


Why not buy it from this site?

http://www.artofproblemsolving.com/Books/AoPS_B_About.php

[quote="devious_"]I would, but the delivery would take a while.[/quote]

The last time I saw the books on Amazon they were being sold used for more than we sell them for new.

We do have a priority delivery option for those willing to pay a few extra dollars for shipping.

[quote="Peter VDD"]What level is the book? are it all questions of at least IMO level or is it rather easy?[/quote]

There are two books and few of the problems are IMO level (though there are a few in Volume 2).  The books are good for helping students train up to Olympiad level problems, but they material covered does no get as deep as the material we cover in our classes.

I see you are in Jordan, so delivery would take a little longer (priority mail is not available outside the US).  If you send an e-mail to us (click on the link on the order page) we can send you international shipping options.  There is an airmail option that takes 5-10 days from the time we receive your order.


geometry
circumcircle
power of a point
radical axis
geometry proposed

This is a problem in Macao Mathematical Olympiad and my solution is very very long. Could  anyone give me some shorter solutions? Thanks!
[size=125]AB is the diameter of a circle. C, D are 2 points on the circle such that B, C are separated by the straight line AD.  Let  M, N, P be the midpoints of AC, DB, CD, and X, Y be the circumcenters of triangles ACP, BDP.  Prove that XY||MN.[/size]

Don't know if this one is shorter:

Let [tex]Q[/tex] be the intersection of [tex]AC[/tex] and [tex]BD[/tex].
We want to show that [tex]PQ[/tex] is the radical axis of circles with diameter [tex]AC[/tex] and [tex]DB[/tex] respectively. Since [tex]ABCD[/tex] is cyclic, we have [tex]QA\cdot QC=QB\cdot QD[/tex]. So [tex]Q[/tex] is on the radical axis.
Now denote by [tex]E,F[/tex] the second intersections of [tex]PQ[/tex] with the circles with diameter [tex]AC[/tex] and [tex]DB[/tex] respectively. Since [tex] \Delta CEA \sim\Delta BDA[/tex] and [tex] \Delta DFB \sim\Delta ACB[/tex], we know: [tex]DF=\frac{DB\cdot AC }{AB}=CE[/tex]. Since [tex] P[/tex] is the midpoint of [tex]CD[/tex], we conclude [tex]PE=PF[/tex] and thus [tex]PE\cdot PC=PF\cdot PD[/tex]


Can I get AoPS Wiki on the google toolbar for search engines?

i have an image of what I'm talking about

[img]
http://www.freewebs.com/a1b9/untitled.PNG
[/img]

Go ask Microsoft and Mozilla if they're feature it as an add on.

In firefox, the second link on http://mycroft.mozdev.org/download.html?name=aopswiki&category=all&country=all&language=all&submitform=Search&sherlock=yes&opensearch=yes works to add the AoPSWiki to your search bar.

[quote="Valentin Vornicu"]You can also use the following addon for Firefox: http://maltekraus.de/Firefox/search-tools/addtosearchbar_1.5.1.xpi to add the search (including the site search!).[/quote]
That link doesn't quite work for me. :maybe:

Use "Save link as ..." (don't click on it) and install from Addons menu.

Actually, what should be done is that the AoPS Wiki people implement the [url=http://www.opensearch.org/Home]OpenSearch[/url] specification.

You just add a little link tag to an OpenSearchDescription file and then, when Firefox visits the page, the search thing will highlight blue and there'll be an option to "Add FOOBAR" to your search engines.

Wikipedia has such a link so you can just copy them, although you'll probably want to remove the reference to [url=http://www.opensearch.org/Specifications/OpenSearch/Extensions/Suggestions/1.0]application/x-suggestions+json[/url] since I doubt you guys have an Google-suggest-like feature enabled. (Incidentally, the default Wikipedia search built into Firefox isn't as good... I recommend getting the one Wikipedia offers and removing the default one, since it does suggests)

I believe IE7 also implements OpenSearch, but I can't confirm this as I have not really touched a Windows machine for a while.


Dei.. Well you didn't ask me not to try it.. I might giving the answer to other probs but its okay. :) I did answer some of your sums but you aren't replying to them.

Here's a slightly non-conventional approach to the sum just to protect the value of my question and rohits question.

Basically what you are doing is like multiplication.

+ x + = +
- x - = +
+ x - = -

So you have a list of numbers with given sign. Which ever order you multiply them, the final answer is the same.

@Rohit: Edhu Eppidi da.

well disguised but its prob 152 da if im not wrong :D  :rotfl: mine is 153 so the answer is wrong :D

Yeah sorry. I think our stupid pardesi is reading hard for the PT exam, he's not even coming online. Please change the title da.

No he's full time on goiit da :D I told him to come here n he said he was taking a break from aops :D

WHAT TITLE SHD I CHANGE?

[quote="Cyber-Shreyas"]Dei.. Well you didn't ask me not to try it.. I might giving the answer to other probs but its okay. :) I did answer some of your sums but you aren't replying to them.

Here's a slightly non-conventional approach to the sum just to protect the value of my question and rohits question.

Basically what you are doing is like multiplication.

+ x + = +
- x - = +
+ x - = -

So you have a list of numbers with given sign. Which ever order you multiply them, the final answer is the same.

[/quote]

yes, in short: it's basically application of associative law of multiplication.
and ofcourse we all know that multiplication is a binary operation in which associative law holds true. :D

Right. So essentially the answer depends on the no. of + and - signs.

@ shre: oops, dint check properly da. i ll reply to all ur attempts at my problems. if i remember correctly ur probability answer (da rec relation ) was wrong. i thot i d give u a lil more time  :D

Crux. If you have even no.of negative signs, the final sign will be positive. Just like in multiplication. I'm still hiding the keyword from here so that people can try mine and rohit's some.

Rag: Why don't you try it da?

And give me your answer for the recursion.... I want to know where I'm wrong/

[quote="Cyber-Shreyas"]Crux. If you have even no.of negative signs, the final sign will be positive. Just like in multiplication. I'm still hiding the keyword from here so that people can try mine and rohit's [color=blue][u][b]sum[/b][/u][/color].

Rag: Why don't you try it da?

And give me your answer for the recursion.... I want to know where I'm wrong/[/quote]

EDIT yours da :D

Also, whoever sees this, cant you find the link between all the three sums :)