say it is thursday morning. the man can say "i cannot die today" with his logic. in reality, he does not know if he will be killed either thursday or friday. he doesn't know he cannot be killed until friday morning. if it is friday, then he can't die; if it is any other time, he can die.

Been posted before...

By the man's logic, he is not going to expect death anyways  :D

No, but Captain Sugar is right.  He does not know that he will die on Friday until it is Friday morning.  So then, let's say it's Thursday morning.  What is going through the man's mind?  He could be thinking "I will die today", but he can't be sure, because the judge actually MIGHT put his death on friday.  He can't be completely sure at that instant..

But he knows that he can't die on Friday because if he does, he will know what day he will die before the executer comes... :maybe:

I think that the prisoner is thinking way to much about this. He could just be killed on wednesday or something, and he would argue "no! It's a paradox! You can't kill me!" and then he would just be killed. I bet the prisoner is a lawyer.

[hide="simple solution"]The judge did not say what week he was going to die, so the prisoner could die any week![/hide]

[hide="or maybe"] It was 11:59 on Sunday and the guy thought of it at 12:00 on Monday at the exact time the esecutor came. :D [/hide]

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

gives a serious answer and explains it pretty well.

and daine and laughinghead, there's no trick to the problem. I should have clarified that earlier.

[quote="buzzer11"]A man is convicted of murder, and the judge gives him the death sentance. He says two things. a) the man will die on either Monday, Tuesday, Wednesday, Thursday, or Friday, and b) the man will not know on which day he will die until the exceutioner comes and knocks on his door.

So the man goes to jail on Sunday and thinks to himself: hmm. so if I will not know on which day he will die until the executioner comes, then I cannot die on Friday, because then, because i did not die on Monday through Thursday, I will know I would have to die on Friday.

Following this logic, I cannot die on Thursday.

Following this, I cannot die on Wednesday, or Tuesday, or Monday.

Thus, I cannot die on any day of the week, and thus I cannot be killed!

What is wrong with the man's logic?

btw, the judge is not wrong.

 :)[/quote]

[b]he's actually not supposed to follow the same logic[/b]...he just wants to prove that he shouldn't die! :D  :lol:

He can use that piece of logic only in the case of friday....coz if he uses it for the case of thursday...he knows that either he will be killed on thursday OR friday....but he does not know the exact date...so it shows that the judge is right! He CAN be killed on Monday, Tuesday, Wednesday or Thursday.
Sorry criminal, u lose!  :D  :)

Is that right?? :)

[quote="Padmavathi"][quote="i_like_pie"][quote="buzzer11"]A man is convicted of murder, and the judge gives him the death sentance. He says two things. a) the man will die on either Monday, Tuesday, Wednesday, Thursday, or Friday, and b) the man will not know on which day he will die until the exceutioner comes and knocks on his door.

So the man goes to jail on Sunday and thinks to himself: hmm. so if I will not know on which day he will die until the executioner comes, then I cannot die on Friday, because then, because i did not die on Monday through Thursday, I will know I would have to die on Friday.

Following this logic, I cannot die on Thursday.

Following this, I cannot die on Wednesday, or Tuesday, or Monday.

Thus, I cannot die on any day of the week, and thus I cannot be killed!

What is wrong with the man's logic?

btw, the judge is not wrong.

 :)[/quote]
I read this in some book once. The book's answer is: The prisoner's won't be killed. Explanation: He can't be killed on Friday because as soon as Thursday is over and he is still alive, he knows he will die Friday. He can't die on Thursday either, for the same reason. Continue this process until it comes to today. All other days are ruled out, so he knows he will be killed today, so there is a contradiction.[/quote]

But HOW can u apply the same logic for the rest of the week?[/quote]
He knows, he can't be killed on Friday, so he has to be killed on Thursday or earlier. As soon Wednesday is over, he knows he must die on Thursday, so he can't die Thursday. Now do you get it?

[quote="mustafa"]The answer is that while this prisoner thinks about this, he convinces himself that he can't be killed. When he is killed, he isn't expecting it. Think about it. He could just get killed on Wednesday, and he would be busy thinking "No, you can't kill me! I proved it!"[/quote]
I think you're right. :D

In other words, the prisoner's logic is correct (for those of you who tried to prove otherwise).... Suppose that the execution date is set to Friday. Then, by Thursday night, the prisoner will know that it is on Friday, so that's no surprise. Thus, the prisoner will assume that the execution is on Thursday if it's Wednesday night. However, then there won't be any surprise either, etc.

[hide="a better strategy"] The prisoner should assume that he will executed every day. [/hide]

Assume the execution is on Wednesday. By your logic, the prisoner would only know it couldn't be Friday when he goes to sleep on Thursday night. So the logic doesn't work.

Induction only works one way.

[quote="Phelpedo"][hide="a better strategy"] The prisoner should assume that he will executed every day. [/hide][/quote]

Honestly, the judge is just going to kill the prisoner no matter what, even if the prisoner claims to have expected it. The prisoner should spend his time trying to escape, not proving that he won't die... :P

[quote="i_like_pie"][/quote][quote="Padmavathi"][quote="i_like_pie"][quote="buzzer11"]A man is convicted of murder, and the judge gives him the death sentance. He says two things. a) the man will die on either Monday, Tuesday, Wednesday, Thursday, or Friday, and b) the man will not know on which day he will die until the exceutioner comes and knocks on his door.

So the man goes to jail on Sunday and thinks to himself: hmm. so if I will not know on which day he will die until the executioner comes, then I cannot die on Friday, because then, because i did not die on Monday through Thursday, I will know I would have to die on Friday.

Following this logic, I cannot die on Thursday.

Following this, I cannot die on Wednesday, or Tuesday, or Monday.

Thus, I cannot die on any day of the week, and thus I cannot be killed!

What is wrong with the man's logic?

btw, the judge is not wrong.

 :)[/quote]
I read this in some book once. The book's answer is: The prisoner's won't be killed. Explanation: He can't be killed on Friday because as soon as Thursday is over and he is still alive, he knows he will die Friday. He can't die on Thursday either, for the same reason. Continue this process until it comes to today. All other days are ruled out, so he knows he will be killed today, so there is a contradiction.[/quote]

But HOW can u apply the same logic for the rest of the week?[/quote]
He knows, he can't be killed on Friday, so he has to be killed on Thursday or earlier. As soon Wednesday is over, he knows he must die on Thursday, so he can't die Thursday. Now do you get it?

[quote="mustafa"]The answer is that while this prisoner thinks about this, he convinces himself that he can't be killed. When he is killed, he isn't expecting it. Think about it. He could just get killed on Wednesday, and he would be busy thinking "No, you can't kill me! I proved it!"[/quote][quote="i_like_pie"]
I think you're right. :D[/quote]

Ohhh yeah!! I get it now....WOAH! THANX!!  :lol:  :D
I think that mustafa is right...when the prisoner gets killed, he's  preoccupied thinking he CAN'T get killed....and so it IS unexpected!

The question is, what is wrong with the prisoners logic not if it will be expected or not.He knows he can't die on Friday because then on Thursday night he would know that he would die then. so he can die on Monday or Tuesday.

Also in another solution, they might take away his calender and drug him so he won't know what day it is. :lol:


topology
category theory
abstract algebra
advanced fields
advanced fields theorems

Hi,

I am trying to understand some topology (and I do have an upcoming exam  :roll: )

now my syllabus proves this :

there is an nth homology functor $H_n$ from the category of topological spaces $Top$ to the abelian groups $Ab$

this means that when $f: X\rightarrow Y$ and $g: Y\rightarrow Z$

$H_n(g)\cdot H_{n}(f)=H_{n}(g\cdot f)$

this implies that homeomorphic spaces have isomorphic homology groups.

But my syllabus also states that when to (continuous) $f: X\rightarrow Y$ and $g: X\rightarrow Y$  are homotopic, then
$H_{n}(f)=H_{n}(g)$

I would be interesting in a hint how to prove this.
Interestingly, the syllabus itself says it might help to prove that for the projection $p: X\times [0,1]\rightarrow X$
$H_{n}(p)$ is an isomorphism

Of course, you can find this in the hatcher :D

http://www.math.cornell.edu/~hatcher/AT/AT.pdf , page 110


function
integration
real analysis
real analysis unsolved

If $ f\in L^1(R)$ and $ F(x) \equal{}\int f(x \minus{} y)f(y)dy$
Show that $ F \equal{} 0$ .

:huh: If $ f>0$ everywhere, then $ F>0$ too.

I'm ready to bet 100 against 1 that the OP just forgot to finish the last sentence with the words "implies $ f\equal{}0$". ;)

Let's make it a tiny bit more entertaining:

Show that $ \exists$ nonzero functions $ f$ and $ g$ such that $ \int_{\mathbb{R}}f(x\minus{}y)g(y)\,dy\equal{}0$ for all $ x.$

[quote="hamidreza"]If $ f\in L^1(R)$ and $ F(x) \equal{} \int f(x \minus{} y)f(y)dy$
Show that $ F \equal{} 0$ .[/quote]
sorry,I have typed the problem incorrect  :blush:
please replace with

If $ f\in L^1(R)$ and $ F(x) \equal{} \int f(x \minus{} y)f(y)dy$ and $ F\equal{}0$
Show that $ f \equal{} 0$ almost everywhere.

This is most easily answered through the Fourier transform.

If $ f\in L^1(\mathbb{R}),$ then $ \widehat{f}\in C_0.$

If $ F\equal{}f*f,$ then $ \widehat{F}(\xi)\equal{}\left(\widehat{f}(\xi)\right)^2.$ Hence, $ F\equal{}0$ implies $ \widehat{F}\equal{}0$ implies $ \widehat{f}\equal{}0$ everywhere. Since we can recover $ f$ from $ \widehat{f}$ by Fejer's theorem, that implies that $ f\equal{}0$ (as an element of $ L^1.$)


modular arithmetic
number theory
greatest common divisor
function
induction
totient function
algebra

There's something wrong with #4. Please review the statement.

[hide="1"]$ 32^{801} \plus{} 19 \equiv 2^{801} \plus{} 19 \equiv 2(2^{800} \minus{} 1) \plus{} 21 \equiv 21 \equiv 6 \pmod{15}$, since $ \frac {2^{800} \minus{} 1}{2^4 \minus{} 1} \equal{} 2^0 \plus{} 2^4 \plus{} 2^8 \plus{} \ldots 2^{796}$[/hide]

[hide="2"]$ 41^{82} \equiv ( \minus{} 4)^{82} \equiv 4^{82} \equiv 4((4^3)^{27}) \equiv 4(1^{27}) \equiv 4 \pmod{9}$[/hide]

[hide="3"]$ 1!^{1382!}$ is just $ 1$
$ 3!^{1382!} \equal{} 6^{1382!}$, which ends in $ 6$ like all natural powers of $ 6$.
$ 8!^{1382!}$ ends in a $ 0$, since $ 8!$ ends in a $ 0$ (since $ 2,5|8!$)
$ 2!^{1382!} \equal{} 2^{1382!} \equiv 4^{1382!/2} \equiv ( \minus{} 1)^{1382!/2} \equiv 1 \pmod{5}$, so this ends in either $ 1$ or $ 5 \plus{} 1 \equal{} 6$, but since it is obviously even, it must end in a 6.

It suffices to add the units digits of each term: $ 1 \plus{} 6 \plus{} 0 \plus{} 6 \equal{} 13$, so the units digit is $ 3$[/hide]

[hide="4"]
$ k : \equal{} n \minus{} 1$, so we want to show $ 43 | 6^{3k \plus{} 2} \plus{} 7$

$ 6^{3k \plus{} 2} \plus{} 7 \equiv 36 (6)^{3k} \plus{} 7 \equiv ( \minus{} 7) 6^{3k} \plus{} 7 \equiv 7( \minus{} 6^{3k} \plus{} 1) \pmod{43}$

$ \mathbf{GCD}[7,43] \equal{} 1$, so we must show that $ 43 | 6^{3k} \minus{} 1$

$ 6^3 \equal{} 216 \equal{} 1 \plus{} 5(43)$, so $ 6^{3k} \minus{} 1 \equal{} (6^3)^k \minus{} 1 \equiv 1^k \minus{} 1 \equiv 0 \pmod{43}$, and we are done.
[/hide]

My first three solutions look similar to those from the previous post (naturally), but I have not used cyclicity of exponents or the CRT.

[hide="1"]From Euler's Totient function,

$ 32^{801} \equiv 2 \pmod {15}$

$ 2\plus{}4\equal{}6$

6[/hide]

[hide="2"]$ 41^{82} \equiv 41^4 \pmod {9}$

$ 41^4\equiv 625 \equiv 4 \pmod {9}$

[/hide]

[hide="3"]We find the units digit of each term:

1
6
0
6

The sum of those has a units digit of 3.[/hide]

[hide="4"]Induction is the way:

$ 6^2\plus{}7\equal{}43$

$ 6^{3n\minus{}1} \equiv 36 \pmod {43}$

$ 6^{3(n\plus{}1)\minus{}1}\equiv 6^5 \equiv 36 \pmod {43}$

Yay.[/hide]

I'll post a solution to 5 later.

[hide="#5"]$ 7^3 \equal{} 343 \equal{} 19\cdot 18 \plus{} 1$, hence the given expression is congruent to $ 7^2 \plus{} 7 \plus{} 1 \equal{} 57\equiv 0\pmod{19}$[/hide]

[quote]$ 32^{801} \plus{} 19 \equiv 2^{801} \plus{} 19 \equiv 2(2^{800} \minus{} 1) \plus{} 21 \equiv 21 \equiv 6 \pmod{15}$, since $ \frac {2^{800} \minus{} 1}{2^4 \minus{} 1} \equal{} 2^0 \plus{} 2^4 \plus{} 2^8 \plus{} \ldots 2^{796}$[/quote]

can someone explain please everything in english, what is goin on here, why $ 32$ is replaced by $ 2$. .i am learning number theory from scratch, i have no idea whats goin in those examples.

$ a \equiv b \pmod{n}$ is read as "as is congruent to b in modulo n". In other words, a and b have the same remainder upon division by n, or mathematically, $ n | (a \minus{} b)$. If $ a \equiv 0 \pmod{n}$, then $ n|a$.

This helps us simplify a lot of problems involving remainders. For example, if $ a \equal{} 17$ and $ b \equal{} 32$, then since $ 5 | (32 \minus{} 17)$, we can, [u]in certain very specific situations[/u], use them interchangeably.

Here are some of the situations. Let $ a \equiv b \pmod n$, and let $ c$ and $ k$ be positive integers. Then:

$ a \plus{} c \equiv b \plus{} c \pmod {n}$
[u]Demonstration:[/u] $ a \minus{} b \equal{} kn \implies a \plus{} c \equal{} b \plus{} c \plus{} kn$. $ b \plus{} c \equiv b \plus{} c \plus{} kn \pmod{n}$ because $ b \plus{} c \plus{} kn \minus{} (b \plus{} c) \equal{} kn$ is divisible by $ n$.
[u]Application:[/u] Find all positive integers $ n > 1$ such that $ 17 \equiv \minus{} 1 \pmod {n}$.
Solution: $ 17 \equiv \minus{} 1 \pmod{n} \implies 18 \equiv 0 \pmod{n} \implies n | 18$. Then, $ n \equal{} 2, 3, 6, 9, 18$

$ ac \equiv bc \pmod {n}$
[u]Demonstration:[/u] $ ac \equal{} (b \plus{} kn)c \equal{} bc \plus{} kcn$. Hence, $ ac \minus{} bc \equal{} (bc \plus{} kcn) \minus{} bc \equal{} kcn$, which is divisible by $ n$.

$ a^c \equiv b^c \pmod{n}$
[u]Demonstration:[/u] By the binomial theorem:
$ a^c \equal{} (b \plus{} kn)^c \equal{} \sum_{j \equal{} 0}^{c} \binom{c}{j} b^{c \minus{} j} (kn)^j$ $ \equal{} b^c \plus{} \sum_{j \equal{} 1}^{c} \binom{c}{j} b^{c \minus{} j} (kn)^j \equal{} b^c \plus{} n \left( \sum_{j \equal{} 1}^{c} \binom{c}{j} b^{c \minus{} j} k^j n^{j \minus{} 1} \right)$.

Hence $ a^c \minus{} b^c \equal{} n \left( \sum_{j \equal{} 1}^{c} \binom{c}{j} b^{c \minus{} j} k^j n^{j \minus{} 1} \right)$, which is divisible by $ n$.

In this case, the third one was applied. $ 32 \equal{} 2 \plus{} 2(15)$, so since $ 32 \equiv 2 \pmod{15}$, we can say that $ 32 ^ c \equiv 2^c \pmod {15}$. There's the first step.

the next step was to say that $ 2^{801} \equal{} 2^{801} \minus{} 2 \plus{} 2 \equal{} 2(2^{800} \minus{} 1) \plus{} 2$. This allowed me to use the geometric series formula to show that $ 2^{800} \minus{} 1$ was divisible by $ 2^4 \minus{} 1$, which is conveniently $ 15$, so I could then say $ 2^{801} \equiv 0 \pmod{15}$

thank you very very much The Zuton Force, that helped me alot

[quote="The Zuton Force"]$ a^c \equiv b^c \pmod{n}$
[u]Demonstration:[/u] By the binomial theorem:
$ a^c \equal{} (b \plus{} kn)^c \equal{} \sum_{j \equal{} 0}^{c} \binom{c}{j} b^{c \minus{} j} (kn)^j$ $ \equal{} b^c \plus{} \sum_{j \equal{} 1}^{c} \binom{c}{j} b^{c \minus{} j} (kn)^j \equal{} b^c \plus{} n \left( \sum_{j \equal{} 1}^{c} \binom{c}{j} b^{c \minus{} j} k^j n^{j \minus{} 1} \right)$.[/quote]

Unnecessary.  Either use the fact that $ n | a \minus{} b | a^c \minus{} b^c$ or induct on the second property (that is, if $ a \equiv b \bmod n$ then $ a^2 \equiv ab \equiv b^2 \bmod n$, etc.).


Find all values of $ a$ for which the system

$ \begin{cases} \begin{array}{ll}x^2\plus{}y^2\plus{}2x&\leq 1\\
x\,\,\minus{}\,y\,\plus{}\,\,\,a&\equal{}0

\end{array}\end{cases}$

has only one solution. Find the corresponding solutions.

Nvm.I messed up quite badly. :( I will post another solution tomorrow.

No worries. Better to make mistakes in practice than in competitions.


A simple but nice problem:

For every $n \in \mathbb{N}$ and $\varepsilon > 0$, show that there exists a $c(n,\varepsilon) > 0$ with the following properties. If $a_1 < a_2 < \cdots < a_n$ are positive integers such that $a_n > c(n,\varepsilon)$, then there exists a $k \in \{2,3,\ldots,a_n\}$ for which $a_i \mod{k} < \varepsilon k$ for each $i=1,2,\ldots,n$.

The proof is of course by induction.
We prove the following statement.
For any $n$ and $\epsilon<1$ and reals $x_1, \cdots, x_n$ there exists an integer $C$ and an integer $t<C$.
satisfying $\{a_i t\}< \epsilon$.
Proof is by induction on n.
Base: One of $1,2,\cdots,[ \frac 1 {\epsilon+1}]$ will satisfy hypothesis.
Induction step:
Let $m=[\frac 1{\epsilon}]+1$.
By induction hypothesis there exists a constant $D$ and integer $t<D$ with $\{ta_1\},\cdots,\{ta_{n-1}\}<\frac {\epsilon}m$.
if $\{ta_n\}<\epsilon$ we are done else one of $t,2t,\cdots,mt$ will satisfy the condition.
Moreover we can take $C=Dm$.

Now as the proof is finished take $c(n,\epsilon)=C!$ where $C$ is chosen for our $n, \epsilon$.
Thus we see that by induction there exists an integer $t<C$ satisfying $\{t \frac {a_i} {c(n,\epsilon)}\}<\epsilon$.
Taking $k=\frac {c(n, \epsilon)}t$ which is an integer yields the desired result.
QED


conics
ellipse
geometry
analytic geometry
calculus
calculus computations

Find the acceleration in terms of distance from the origin of a particle moving along an ellipse with its center at the origin (instead of having one focus at the origin) and sweeping out equal area in equal time.

So we probably will have to use the local coordinates $ \bold{u}_r$ and $ \bold{u}_\theta$?

The "equal areas in equal time" condition is conservation of angular momentum, so the acceleration should be purely radial. If you use polar coordinates, you must take into account the fact that the frame changes as you move- this introduces the centrifugal term in your acceleration calculation.

I would actually recommend sticking to Cartesian coordinates in this case. The answer is very simple, and can be explained as the result of a global force field.

oh ok. So from a purely physical view, this makes sense?

Yes. It should be a very familiar model, although you might not have seen it in more than one dimension.


induction
number theory proposed
number theory

Let n=2k+1 and $a_{1},a_{2},...a_{n}$  integers. From these numbers we obtain numbers: $\frac{a_{1}+a_{2}}{2}$, $\frac{a_{2}+a_{3}}{2}$, ...$\frac{a_{n}+a_{1}}{2}$. Frome these we obtain other n numbers, using same method. Prove that if all these numbers are integers, then $a_{1}=a_{2}=...=a_{n}$

By induction $a_{k}=a_{k+2^{m}(mod \ n)}(mod \ 2^{m+1})$. Because n is odd it give $a_{k}=a_{l}(mod \ 2^{m}) \ \forall m$.


[color=cyan]This, I believe, was problem 15 from the 1990 AIME exam.  I may be off by a year or two.

Compute ax:^5: + by:^5:  if:
ax + by = 3
ax:^2: + by:^2: = 7
ax:^3: + by:^3: = 16
ax:^4: + by:^4: = 42

Have fun.

Joel[/color]

Basically I just found that every degree of the equation is equal to the sum of what the equation one degree lower was and the square of the degree of the equation. So... It's in the spoiler font style [hide]42+5^2=67[/hide] Of course if that was #15 on AIME then that couldn't be right otherwise it'd be much harder... I took the lazy way out...

IW.OAV

i dont think thats right, cause then ax :^4: +bx :^4: would equal 32 because 16+4 :^2: =32, not 42.

your thinking only works for the degree of 3.

hope i wasnt impeding. peace

No. It's not quite that easy. Here's a hint for a brute force kind of way to do it:

[hide]

Can you find a (or b) in terms of x and y in three different ways?

[/hide]

Doing the question in a way that makes that hint helpful is going to be really ugly, I think.  Also, finding patterns isn't any good unless you can give a decent shot at justifying why such a pattern should (or even could) exist.

ax^n + by^n = (x+y)(ax^(n-1) + by^(n-1)) - xy (ax^(n-2) + by^(n-2)).

Using this we get
16 = 7(x+y) - 3xy
42 = 16(x+y) - 7xy.

Taking 38 times the first equation minus 14 times the second (or just solve for x,y) we get
20 = 42(x+y) - 16xy

And thus our answer is 20.

[color=cyan]Nicely done.  Can you solve the problem in general?  That is, given:
ax + by = c1
ax:^2: + by:^2: = c2
ax:^3: + by:^3: = c3
ax:^4: + by:^4: = c4
find ax:^5: + by:^5:[/color]

[quote]Taking 38 times the first equation minus 14 times the second [/quote]

Wait, so how did you come up with those two numbers?

Fierytycoon

[color=cyan]Why 38 and 14?  Actually, although Stephen said that an alternative was to solve for x and y, what he really did was solve for the two numbers (x + y) and xy.  If you let (x + y) = a and xy = b, the equations become
16 = 7a - 3b
42 = 16a - 7b.

These are relatively easy to solve and then to plug in the values for a and b, or, find two constants to multiply them by to get the expression
42a - 16b, which (he showed) is equal to the quantity you're looking for.

To find the numbers 38 and 14, you have to solve a seperate system of simultaneous equations:
7m + 16n = 42
3m + 7n = 16

Solving that system will give you two constants, m and n, which you need to multiply the two earlier equations by before adding to get the desired expression.  Do you see why?[/color]

Ok, I get it now. Thanks for the clarification.

Fierytycoon


Let $ A$ be a real $ nxn$ matrix such that $ x^TAx \geq 0$ for every real-$ n$ vector $ x$
Prove that :$ Ker(A)\equal{}Ker(A^T)$
 $ y^T$ is defined transpose of $ y$

$ x^TAx\equal{}1/2(x^TAx\plus{}x^TA^Tx)$ then $ A\plus{}A^T$ is symmetric non negative.$ A\plus{}A^T\equal{}B^2$ where $ B$ is symmetric non negative and $ ker(B)\equal{}ker(A\plus{}A^T)$.
If $ x\in{ker(A)}$ then $ x^TAx\equal{}x^T(A\plus{}A^T)x\equal{}0$ or $ \parallel{}Bx\parallel{}^2\equal{}0$. Therefore $ Bx\equal{}0$ ,$ (A\plus{}A^T)x\equal{}0$ and $ A^Tx\equal{}0$. The opposite inclusion is similar.


for $ 1/a$
Use that:
[url]http://www.mathlinks.ro/viewtopic.php?p=998347#998347[/url]
then multiply each $ x_i$ by $ 1/a$

Thanks and sorry........but I still don't quite understand on your "1/a" and how to transform that to my question.....since the right hand side of your reference is 1 and my listed question is any rational number...Would you please elaborate a bit?

Besides, in the reference question it was to prove only that "at least 1" solution is to be found.but my listed one requires proving infinite possbilities. So does it still apply?

Thanks~~

[quote="HK Angel"]Any advise in solving the following question? Thank you.

The Question:

Prove that any positive rational number, r, can be written as the sum of the reciprocals
of one or more distinct positive integers (for example, 17/12 = 1/1 + 1/4 + 1/6 ). Furthermore, show
there are an infinite number of distinct ways of representing any given r in this way.[/quote]
It is obvicious because the represent in  base b for each $ b\in N$
Each $ a\in Q$ can be reprensent unique in the form :
Enough to consider $ 0 < a < 1$
$ a \equal{} \sum _{i \equal{} 1}^k\frac { c_i}{b^i}$ where $ 0\leq c_i\leq b^i \minus{} 1$

thanks...but ur answer looks difficult to me. where are the reciprocals?? and why 0<1<a suffice??   thankz

ummm...the whole thing still appears quite unclear to me....


$ 1/a\plus{}a/b\plus{}1/{ab}$

Find all integer values of (a,b)

Such that... the above expression is an integer?

Well, the title seems to indicate that the fractions must add up to 1.

Ah.

[hide="Well then..."] We need $ b \plus{} a^2 \plus{} 1 \equal{} ab \implies b(a \minus{} 1) \equal{} a^2 \plus{} 1 \implies b \equal{} a \plus{} 1 \plus{} \frac{2}{a \minus{} 1}$.  It follows that $ a \minus{} 1 \equal{} \minus{}2, \minus{}1, 1, 2 \implies a \equal{} \minus{}1, 0, 2, 3$. [/hide]

[quote="t0rajir0u"]Ah.

[hide="Well then..."] We need $ b \plus{} a^2 \plus{} 1 \equal{} ab \implies b(a \minus{} 1) \equal{} a^2 \plus{} 1 \implies b \equal{} a \plus{} 1 \plus{} \frac {2}{a \minus{} 1}$.  It follows that $ a \minus{} 1 \equal{} \minus{} 2, \minus{} 1, 1, 2 \implies a \equal{} \minus{} 1, 0, 2, 3$. [/hide][/quote]

how did u get  $ b\equal{}a\plus{}1\plus{}\frac{2}{a\minus{}1}$ i got $ b\equal{}\frac{a^2\plus{}1}{a\minus{}1}$ is mine wrong?
how did you find the values of $ a$ from there?

You can simplify that your expression. After you do that, since the expression must be an integer, you have a-1|2, and then it's easy to find the values.

If you can't figure it out, its
$ \frac {a^2}{a \minus{} 1} \plus{} \frac {1}{a \minus{} 1}$ This is just from splitting the fraction.

Little trick, $ \frac {a^2 \minus{} 1 \plus{} 1}{a \minus{} 1} \plus{} \frac {1}{a \minus{} 1} \equal{} \frac {a^2 \minus{} 1}{a \minus{} 1} \plus{} \frac {1}{a \minus{} 1} \plus{} \frac {1}{a \minus{} 1} \equal{} \frac {\left(a \plus{} 1\right)\left(a \minus{} 1\right)}{a \minus{} 1} \plus{} \frac {2}{a \minus{} 1}$ = expression he has.


trigonometry
calculus
derivative
geometry
3D geometry
sphere

Problem: (quite standard)

a boy is sitting on the top of a very very heavy spherical block of ice (no friction at all). Suddendly, he starts sliding on the ice. If the radius of the block is $R$, determine how far is the boy from the ground when he loses the contact with the ice block.

I apologize for my horrible english; if you have doubts about the text, please ask me.

Suppose $h$ is the distance of the boy from the ground (so originally $h = R$). Then we know from conversation of energy that the boy's velocity will be $v = \sqrt{2g(R-h)}$ pointed tangential to the sphere. Consider the $x$-component of this velocity; suppose $\theta$ is the angle between the velocitiy and the horizontal such that the $x$-component will be $v \cos \theta$. Note that $\theta$ is also the angle between $h$ and $R$ (in the hemisphere), so $\cos \theta = \frac{h}{R}$ (alternatively we could have seen this with similar triangles). Then $v_x = \frac{h\sqrt{2g(R-h)}}{R}$. Now, if the boy loses contact, that means that he is not applying a force on the hemisphere, and conversely, there is no normal force on the boy from the hemisphere; thus, the only force will be that of gravity pointed straight down. Thus, at this point, $0 = \frac{F_x}{m} = a_x = \frac{dv_x}{dt} = \frac{dv_x}{dh} \cdot \frac{dh}{dt}$. However, note that $\frac{dh}{dt}$ will never be $0$ after $t = 0$ but before the boy loses contact as he will always be falling, so we have that $0 = \frac{dv_x}{dh}$. Solving this yields $h = \frac{2R}{3}$ (you could take the derivative directly, but I prefer to note that if $\frac{dv_x}{dh} = 0$, then $\frac{d}{dh} v_x^2 = 2v_x\frac{dv_x}{dh} = 0 \Rightarrow 0 = \frac{d}{dh} h^2(R-h) = 2Rh - 3h^2$; the one root $h = 0$ corresponds to the extraneous solution from $\frac{dh}{dt} = 0$, so the root we want is $h = \frac{2R}{3}$).

It's correct; the boy, however, is $5R/3$ far from the ground because the center of your system is the center of the sphere. The problem could also be solved noting that when he loses contact the component of gravity directed towards the center of the sphere is $mv^2r^{-1}$.


linear algebra
matrix
system of equations
superior algebra
circulant matrix

Solve the system of equations in $\mathbb R$

$x^2-yz=a$
$y^2-zx=b$
$z^2-xy=c$

you must find $x,y,z$ in terms of $a,b,c$

Hint: consider $A^{*}$ the comatrice of the permuation matrix $A$

$A=\left[\begin{array}{ccc}x&y&z\\z&x&y\\y&z&x\end{array}\right]$

$A^{*}.A=A.A^{*}=\det(A).I_{3}$

hello, we have
$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(x\plus{}y\plus{}z)^2(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}xz\minus{}yz)^2$
and
$ a^2\minus{}bc\equal{}x(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}xz\minus{}yz)$
so we get
$ x\equal{}\pm\frac{a^2\minus{}bc}{\sqrt{a^3\plus{}b^3\plus{}c^3\minus{}3abc}}$
and analogously
$ y\equal{}\pm\frac{b^2\minus{}ac}{\sqrt{a^3\plus{}b^3\plus{}c^3\minus{}3abc}}$
$ z\equal{}\pm\frac{c^2\minus{}ab}{\sqrt{a^3\plus{}b^3\plus{}c^3\minus{}3abc}}$
Sonnhard.

The interesting case satisfies (*) $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}0$ that implies $ det(A)\equal{}0$.
For instance $ a\equal{}b\equal{}c$; see an application to matrices here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163381
   If (*) is satisfied then the $ 3$ equations are no more algebraically independent !

$\begin{pmatrix}x&y&z\\z&x&y\\y&z&x\end{pmatrix} \begin{pmatrix}a&c&b\\b&a&c\\c&b&a\end{pmatrix}=det(A)\cdot I$
$\begin{pmatrix}a&c&b\\b&a&c\\c&b&a\end{pmatrix} \begin{pmatrix}e&f&g\\g&e&f\\f&g&e\end{pmatrix}=det(A)^2\cdot I$
assuming A is invertible
$\frac{1}{det(A)} \begin{pmatrix}x&y&z\\z&x&y\\y&z&x\end{pmatrix} = \frac{1}{det(A)^2} \begin{pmatrix}e&f&g\\g&e&f\\f&g&e\end{pmatrix}$
$x=\frac{e}{det(A)}=\pm\frac{a^2-bc}{\sqrt{a^3+b^3+c^3-3abc}}$
similarly for y and z.
note that the signs in post #2 cannot be chosen arbitrarily, eg (x,y,z)=(+,+,+) or (-,-,-).


Mr. Smith, Mr. Taylor, and Mr. Elder are teaching 6  different subjects

(Biology, Geography, Math, History, English, and Physics) in the school,

each of them teaching two subjects. We have the following information

about them:

- The Physics and English teachers are neighbors.

- Mr. Smith is the youngest of them.

- Mr. Elder plays poker with the English teacher and the Biology teacher

   every Sunday. The three of them love this game.

- The Biology teacher is older than the Math teacher.

- The Geography teacher, the Math teacher, and Mr. Smith will go to bike

  together next weekend. Mr. Smith already rented the three bikes.

Which teacher teaches which subjects?

If they play poker together or bike together, does that mean they are neighbors?

Don't be an idiot, is everybody you go to school with your neighbor?

[hide="Connecting the jumbo-sized dots"]
It's basically trying to say that Mr. Elder isn't the English or Biology teacher and Mr. Smith isn't the Geography or Math teacher. It also says that Mr. Elder, the English teacher, and the Biology teacher are three different people, same goes for Mr. Smith and the English and Bio teacher.
[/hide]

oh, and I'd say this goes in the MathCounts forum, or at least the Getting Started one, it's simple logic...

Sorry I like being an idiot to annoy arrogants like you.

minor note: nice one, neenal. :rotfl:

What if you are self teaching? I am.


inequalities
rearrangement inequality
inequalities unsolved

For $a,b,c$ positive reals such that $abc=1$ prove that

$\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}+\frac{a^2+b^2}{c}\geq a+b+c +3$

Just use $\frac {a^2}b+\frac {b^2}c+\frac {c^2}a\ge 3$ (AM-GM), and the use rearrangement inequality for $\frac{a^2}c+\frac{b^2}a+\frac{c^2}b\ge a+b+c$.

I'm a new guy in inequalties, can someone say me what is rearrangement?

my proof:
by cauchy in the Engel form we have:
$\frac {a^{2}}{b} + \frac {a^{2}}{c} + \frac {b^{2}}{c} + \frac {b^{2}}{a} + \frac {c^{2}}{b} + \frac {c^{2}}{a} \geq \frac{(2(a+b+c))^{2}}{2(a+b+c)} = 2(a+b+c)
 $

by AM-GM

$a+b+c \geq 3$


Now that I have gotten your attention...

Hello any one out there? :)

Hi - Any one here (In Ohio)   going to  Mathcounts Nationals ?

Me, and also Monkey who I saw today and told to post here.


Let $ ABCD$ be a quadrilateral with $ AB$ extended to $ E$ so that $ \overline{AB} \equal{} \overline{BE}$.  Lines $ AC$ and $ CE$ are drawn to form angle $ ACE$.  For this angle to be a right angle it is necessary that quadrilateral $ ABCD$ have:

$ \textbf{(A)}\ \text{all angles equal}$
$ \textbf{(B)}\ \text{all sides equal}$
$ \textbf{(C)}\ \text{two pairs of equal sides}$
$ \textbf{(D)}\ \text{one pair of equal sides}$
$ \textbf{(E)}\ \text{one pair of equal angles}$

[quote="ocha"][hide]
Let $ AE$ be the diameter of a circle with a center at $ B$. $ \angle ACE$ will be a rightangle if $ C$ is a point on the circle. Therefore $ CB \equal{} AB \equal{} BE$

The only way to ensure this is if all sides of the quadrilateral are equal $ \Rightarrow \boxed{B}$
[/hide][/quote]
But AB = BE is given in the problem? So really the location of point D has no bearing whatsoever on the question.

[hide]I agree with Jtmathlover. Since only the relationship between $\overline{AB}$ and $\overline{BC}$ affect the problem, I think the answer should be $D$.[/hide]


(1)Given $ a,b,c \ge 0$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}6abc \ge ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)\plus{}2abc(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b})$
(2)Give $ a,b,c \ge 0$and $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{ab}{a\plus{}b}\plus{} \frac{bc}{b\plus{}c}\plus{} \frac{ca}{c\plus{}a} \le \frac{a^2\plus{}b^2\plus{}c^2}{6}\plus{}abc$

[quote="babyltv"](1)Given $ a,b,c \ge 0$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \ge ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \plus{} 2abc(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})$
(2)Give $ a,b,c \ge 0$and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \le \frac {a^2 \plus{} b^2 \plus{} c^2}{6} \plus{} abc$[/quote]
The first ineq:
we will prove:
$ \sum (a \minus{} b)^2(\frac {a \plus{} b \minus{} c}{2} \minus{} \frac {abc}{(a \plus{} c)(b \plus{} c)}) \geq 0$
we have $ S_b; S_c \geq 0$ and $ S_a \plus{} S_b \geq 0$. ^^
2) Use SOS and regard this equality:
$ abc \minus{} \frac {4(ab \plus{} bc \plus{} ca) \minus{} 9}{3} \equal{} \sum \frac {(a \minus{} b)^2}{18} (\frac {a \plus{} b \minus{} c}{2})$

2/$ \sum\frac{ab}{a\plus{}b}\equal{}\sum\frac{ab(b\plus{}c)(c\plus{}a)}{\prod(a\plus{}b)}\leq\frac{4abc\sum a}{\frac{8}{9}\sum a\sum ab}\equal{}\frac{9abc}{2\sum ab}$
We must prove that:
  $ \frac{9r}{2q}\leq\frac{9\minus{}2q}{6}\plus{}r\Leftrightarrow r\leq\frac{q}{3},Right$ :)

[quote="karis"]2/$ \sum\frac {ab}{a \plus{} b} \equal{} \sum\frac {ab(b \plus{} c)(c \plus{} a)}{\prod(a \plus{} b)}\leq\frac {4abc\sum a}{\frac {8}{9}\sum a\sum ab} \equal{} \frac {9abc}{2\sum ab}$[/quote]
That's wrong. Take $ c\equal{}0$.

Dear my friend,
(1)<=>(2)
And we need to prove (1)

It's equivalent to :
$ 6a^2b^2c^2 \ge [ \sum ab(a\plus{}b)  ] \prod (a\plus{}b\minus{}c)$
We have :
*$ 9a^2b^2c^2 \ge (\sum a^2)( 2\sum a^2b^2 \minus{}\sum a^4)\equal{} (\sum a^2)(\sum a)\prod (a\plus{}b\minus{}c)$
*$ \frac{(\sum a^2)(\sum a) }{9}\ge  \frac{  \sum ab(a\plus{}b)}{6}$
Hence , we just need to consider 2 case :$ \prod (a\plus{}b\minus{}c) \le 0$ and $ \prod (a\plus{}b\minus{}c) \ge 0$


Consider the proof below:

|x| - |y| <= |x + y|

Proof:

|x| = |x + y -  y| = |x + y| + |-y|

That is |x| <= |x + y| + |y|
    |x| - |y| <= |x + y|

I don’t understand. First he equated x to |x + y -  y|, then manipulated the equation. Then suddenly, he replaced the equation with inequality. Do you think that the equation is wrong? Can you explain why?

[quote="guile"]|x| = |x + y -  y| = |x + y| + |-y|[/quote]

This line is clearly incorrect.

Edit:  [hide="A correct proof"] Begin with the basic inequality $| a+b | \le |a|+|b|$.  Then,

$| (x+y)-y) | \le | x+y |+|-y | \Leftrightarrow |x|-|y| \le |x+y|$. [/hide]

I think the first person just made a mistake on the first equal sign, that should be a less than or equal to sign.

[color=darkblue][b]Lemma.[/b]
1. Let $a,b$ be two real numbers. Then $a\le b\Longleftrightarrow a^{3}\le b^{3}\ .$
2. Let $a,b$ be two positive real numbers. Then $a\le b\Longleftrightarrow a^{2}\le b^{2}\ .$

[b]Remarks.[/b]$|x|\ge 0$ ; $|x|=0\Longleftrightarrow x=0$ ; $|-x|=|x|$ ; $-|x|\le\pm x\le |x|$ ; $|x|^{2}=x^{2}$ ; $|x|\le |y|\Longleftrightarrow x^{2}\le y^{2}$ ;
$|xy|=|x|\cdot |y|$ ; $\left|\frac{x}{y}\right|=\frac{|x|}{|y|}$ ; $|x|=|y|\Longleftrightarrow x=y\ \vee\ x+y=0$ ; $\boxed{\ \parallel x|-|y\parallel \le |x\pm y|\le |x|+|y|\ }\ .$

[b]PROOF.[/b] $\left|\left|x\right|-\left|y\right|\right|\le\left|x\pm y\right|\le |x|+|y|$ $\Longleftrightarrow$ $(|x|-|y|)^{2}\le (x\pm y)^{2}\le (x+y)^{2}$ $\Longrightarrow$
$x^{2}+y^{2}-2|xy|\le x^{2}+y^{2}\pm 2xy\le x^{2}+y^{2}+2|xy|$ $\Longleftrightarrow$ $-|xy|\le \pm xy\le |xy|$, what is truly.[/color]


floor function
number theory
prime numbers
algebra unsolved
algebra

For every sequence $ (x_1, x_2, \ldots, x_n)$ of non-zero natural prime numbers, $ \{1, 2, \ldots, n\}$ arranged in any order, denote by $ f(s)$ the sum of absolute values of the differences between two consecutive members of $ s.$ Find the maximum value of $ f(s)$ where $ s$ runs through the set of all such sequences, i.e. for all sequences $ s$ with the given properties.

I hold in my hands the IMO 1989 LongList brochure, which allows me to correct the statement as being

[color=#0000FF]For each sequence $s= (x_1, x_2, \ldots, x_n)$ of nonzero natural prime numbers, $ \{1, 2, \ldots, n\}$ ordered in an arbitrary way, denote by $ f(s)$ the absolute minimum value of the difference of two consecutive numbers of $ s.$ Find the maximum value of $ f(s)$ (for all sequences $s$ with the given properties). [/color]- proposed by Vietnam.

I doubt the word "prime" has anything to do with this (why then the insistence on the numbers $ \{1, 2, \ldots, n\}$, and then the sequences $s$ are widely uncontrollable).
I think the problem just wants us to compute $\max_{\sigma \in \mathcal{S}_n} \min_{1\leq k \leq n-1} |\sigma(k+1) - \sigma(k)|$ (where clearly $f(\sigma) = \min_{1\leq k \leq n-1} |\sigma(k+1) - \sigma(k)|$).

For $n=2m$ even, the example $\sigma = (m+1,1,m+2,2,\ldots,m-1,2m,m)$ has $f(\sigma) = m$. There cannot exist a $\tau$ with $f(\tau) \geq m+1$, since $m$ must appear in $\tau$, and $|m - \ell| \leq m$ for all $1\leq \ell \leq 2m$.

For $n=2m+1$ odd, the example $\sigma = (m+1,1,m+2,2,\ldots,m-1,2m,m,2m+1)$ has $f(\sigma) = m$. There cannot exist a $\tau$ with $f(\tau) \geq m+1$, since $m+1$ must appear in $\tau$, and $|m - \ell| \leq m$ for all $1\leq \ell \leq 2m$.

Therefore the sought after value is  $\max_{\sigma \in \mathcal{S}_n} \min_{1\leq k \leq n-1} |\sigma(k+1) - \sigma(k)| = \lfloor n/2 \rfloor$.


LaTeX
number theory solved
number theory

can anyone help me with this:
we choose a and b, s.t. ab+2 is a perfect square.
define the recurrence: x1=a,x2=b,x(n)=x(n-1)+x(n-2)-2s, where
s=sqrt(x(n-1)*x(n-2)-2). (sorry i cant use latex  :D )
prove that regardless of a and b, this recurrence is periodic. (note that the terms of x(n) are always integer).
i think we can find a recurrence for x(n), and maybe even a nice formula; but i cant prove that it is periodic. im not exactly sure that it is  :D, but is for most cases ive tried.

Hmm, if a=1,b=7,ab+2=9, x(3)=8-2*sqrt(5) :(
Did you mean ab-2 is square?Or s=sqrt(x(n-1)*x(n-2)+2)? Or maybe both variants will work?

oh sorry
yes i meant ab-2 is a square (for some reason i always make mistakes when posting problems  :blush: )

Then I can prove this even without nice formula for x(n) :)
Let ab=c^2+2, then x(3)=a+b-2c. Now, if c>0,  x(3)<max(a,b): if a>b, we want to prove a>a+b-2c or b<2c. If b>=2c, then a>2c, and ab>4c^2, so c^2+2>4c^2, which is false for c>0. So x(n)<max{x(n-1),x(n-2)} , and sequence m(n)=max{x(n),x(n-1)} is decreasing (strictly speaking, m(n)<=m(n-1) and m(n)<m(n-2)), as long as c>0. So after some iterations we will have c=0, a=1, b=2, and period {2,1,3}
(actually, a and b in this proof denote x(n-2) and x(n-1), not only x(1) and x(2))


function
calculus
integration
real analysis
real analysis unsolved

Let A be a subset of R^n, and let f:A-->R^n and g:A-->R^n be two bounded functions s.t. closure of B={x in A|f(x)=/=g(x)}. I want to show that f is integrable iff g is integrable, and that if f and g are integrable, then their integrals are equal.

Thanks in advance.

What about $B$? Are you perhaps missing a condition, perhaps that $B$ has measure zero?

f integrable implies that the set of discontinuities has measure zero. Never mind, i got it. :)


algebra
polynomial
superior algebra
superior algebra unsolved

I was wondering if there were interesting results for this. I saw it used somewhere, but without proof. Apparently there is very little information about it on the internet. Can someone state the theorem and perhaps make up an example for me :)

Thanks in advance.

[quote="ZetaX"]Tried to use google¿ One of the first results is http://arxiv.org/PS_cache/math/pdf/0612/0612801.pdf .[/quote]

Thanks.

Unfortunately this text is too advanced. I was actually working on a problem of harazi in order to show that the cyclotomic polynomial is irreducible and of degree $\phi(n)$.

[quote]
1) Consider the minimal unitary polynomial with rational coefficients of the primitive root of unity,call it epsilon and this polynomial P.
2) Prove that this polynomial has in fact integer coefficients
3) Using Dirichlet theorem and the fact that Q^p(X)-Q(X^p) has all coefficients divisible by p for any polynomial with integer coefficients and any prime p, deduce that epsilon^k is a root of P for all k relatively prime with n.
4) Deduce an estimation for the degree of P.
5) Conclude.
[/quote]

I got stuck at step 3, although I could prove that $Q^{p}(X)-Q(X^{p})$ has all coefficients divisible by $p$. Maybe someone can help me through step 3.

(Hints are preferred above answers)

You don't need any Dirichlet theorem to conclude in 3).

Some hints:
[hide="Hint 1"]Show that it suffices only to consider those $k$ that are prime [here Harazi probably wanted to use Dirichlet's theorem, but that's not necessary].[/hide]
[hide="Hint 2"]Work woth $\epsilon \mod k$ to get $P(\epsilon^{k}) \equiv 0 \mod k$.[/hide]
[hide="Hint 3"]$x^{n}-1$ has no double roots $\mod k$.[/hide]
[hide="Hint 4"]Assume $P(\epsilon^{k}) \neq 0$ and write $x^{n}-1 = P(x)Q(x)$. Prove that $Q(\epsilon) \equiv 0 \mod k$ and conclude a contradiction with the Hint before.[/hide]
[hide="Complete solution"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=15750 [where you got Harazi's hints from].[/hide]


hi, i was wondering if anyone could share their experiences about NC Governor's School (maybe for math in particular)?


linear algebra
matrix
vector
symmetry
linear algebra unsolved

Given a symmetric $n\times n$ matrix $A$ over field ${\mathbb{F}}_2$ with diagonal $d=(d_1,d_2,\dots,d_n)$.  Prove that there exist a vector $x$ such that $Ax=d$. May it be generalized in any sense? For other fields (I think my proof holds for any field of characteristic 2)?

I guess I never saw the generalization potential of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21394]this[/url].

The system $Ax=d$ is compatible iff there are no column vectors $x$ s.t. $x^TA=0^T,\ x^Td=1$, and this we can prove just like in that thread to which I gave the link (where $d=(1,1,\ldots,1)$). Let's prove that this is indeed the case with our definition of $d$ (and given the symmetry of $A$).

What we must show is that no subset of rows of $A$, an odd number of which cut the diagonal in a $1$, can add up to the null vector. If the rows are, say, $i_1,\ldots,i_k$, then we look at the intersections of these rows with the columns $i_1,\ldots,i_k$, and obtain the following problem: the rows of a symmetric matrix (which here is the submatrix of $A$ obtained from rows and columns $i_1,\ldots,i_k$) in $\mathbb F_2$ which has an odd number of $1$'s on the main diagonal cannot add up to the null vector. This, however, is clear by taking the sum of all the elements in the matrix: we find it to be $1$, so surely, the rows cannot add up to the zero vector (it's actually almost a copy of the proof in that other thread).


geometry
geometric transformation
reflection
geometry proposed

Let $ABC$ be a triangle.  Reflect vertex $B$ over side $AC$ to $B'$, and let the foot of the $B$-altitude be $B''$.  Similarly, reflect vertex $C$ over side $AB$ to $C'$, and let the foot of the $C$ altitude be $C''$.  Prove $B'C'\parallel B''C''$. Enjoy! :)

Masoud Zargar

this is false. if $B'C'$ were parallel to $B''C''$, then $\frac{B''B'}{HB''}= \frac{C''C'}{HC''}$, so $\frac{B''B}{HB''}= \frac{C''C}{HC''}$, so $\frac{B''B-HB''}{HB''}= \frac{C''C-HC''}{HC''}$, so $\frac{BH}{HB''}= \frac{CH}{HC''}$, from which we'd get that $BC$ is parallel to $B''C''$, which isn't true unless $AB=AC$

I think I missed something when working with it. :(


calculus
integration
function
algebra
polynomial
parameterization
real analysis

Let $g: [0,\infty) \to \mathbb{R}$ be continuous such that
\[\int_{0}^{x}g(s) (x^{2}-s^{2})^{k}ds = 0 \]
for all $x>0$ for some fixed integer $k \ge 0$. Show that $g(x) = 0$ for all $x$. [i]How about arbitrary $k >-1$?[/i]

Corrected: $g$ vanishes identically.

(x^2 - s^2)^k is nonnegative on the interval (can be seen from a comparison theorem). Therefore g is zero when k>=0. You can see that g is the 0 function on the interval by trying the mean value theorem since the integrand is continuous.

If k>-1 then the integral for k in (-1,0) does not exist (the integrand is not even defined on the interval) since we must be able to take Stieltjes sums with arbitrary points (specifically, when the intermediate point is x and k is in (-1,0)) and things don't work out. Perhaps though there is a more sophisticated way to deal with such a problem but it seems clear that the usual way of making sure you can take arbitrary Stieltjes sums doesn't work in such a case and so the integral does not exist.

Maybe someone will come by to usurp my opinion.

[quote="§outh§tar"](x^2 - s^2)^k is nonnegative. Therefore g is zero when k>=0. [/quote]I was not completely clear: Show that $g \equiv 0$.

Isn't that what I showed? Maybe I wasn't clear  :P I skipped a few steps in the explanation haha.

No matter what x you pick, using the mean value theorem will ALWAYS show that g has a zero value in the interval [0,x]. Moreover, you can infer from this that g has a zero value in [i]any arbitrary interval[/i] [a,x] (so long as  g is defined on the interval, of course: ie a>=0). This alone should give you a strong hint that g is the zero function since it is a continuous function with zeros in any arbitrary interval. I should have been more explicit.

You may use the first fundamental theorem by differentiating the  integral to show that  the integrand is the zero function on [0,x]. It is not hard to infer from what I said about the polynomial being positive definite that g is identically zero on [0,x), where x is of course an arbitrary non-negative real number.

A different way of reaching the conclusion is using the 'sequence' definition of continuity: see if you can construct a sequence (x(n)) converging to a real number x such that (g(x(n))) converges to g(x) but also such that a subsequence of (g(x(n)) contains only zeros. From this it follows that g(x) is zero. It is not hard to construct such a sequence but you may find it boring although I think its a pretty nice approach. Pick a way you like!

[quote="§outh§tar"]
No matter what x you pick, using the mean value theorem will ALWAYS show that g has a zero value in the interval [0,x]. Moreover, you can infer from this that g has a zero value in [i]any arbitrary interval[/i] [a,x] [/quote]
The first claim here is correct, indeed. But I do not see how the second claim follows from it or from anything else in the post.

Additivity of the interval fedja. Say 0 < a < x.

Then the integral over [0, x] is 0 as is the integral over [0,a].

What does this tell you about the integral over [a,x]?

Now since a was arbitrary, the statement follows.

[quote="§outh§tar"]Additivity of the interval fedja. Say 0 < a < x.

Then the integral over [0, x] is 0 as is the integral over [0,a].

What does this tell you about the integral over [a,x]?
[/quote]
This tells me nothing about $\int_{a}^{x}g(s)(x^{2}-s^{2})^{k}ds$. The integral depends on a parameter, so it doesn't define an additive set function.

I see what you mean. I did not realize that was what fedja was getting at so I did not take a closer look at the integrand. My apologies.  :oops:

The same principle applies though when we convert to a Stieltjes integral.

Define $G: [0,\infty ) \to R$ by $G(x) = \int_{0}^{x}g$ .G is differentiable on $[0,\infty )$.

Then $\int_{0}^{x}{g(s)(x^{2}-s^{2})^{k}ds}= \int_{0}^{x}{(x^{2}-s^{2})^{k}dG(s)}$

When $y > x$ additivity of the interval shows

$\int_{0}^{y}{(y^{2}-s^{2})^{k}dG(s)}= \int_{0}^{x}{(y^{2}-s^{2})^{k}dG(s)}+\int_{x}^{y}{(y^{2}-s^{2})^{k}dG(s)}$

and a comparison theorem gives

$\int_{0}^{x}{(y^{2}-s^{2})^{k}dG(s)}\ge \int_{0}^{x}{(x^{2}-s^{2})^{k}dG(s)}= 0$

Since $\int_{x}^{y}{(y^{2}-s^{2})^{k}dG(s)}\ge 0$ and $\int_{0}^{y}{(y^{2}-s^{2})^{k}dG(s)}= 0$, it follows that: $-\int_{x}^{y}{(y^{2}-s^{2})^{k}dG(s)}\ge 0$

Now. The integrand is continuous on [x,y], the integrator is of bounded variation; the mean value theorem implies that g has a nonpositive value on [x,y].

My above comments should follow now since [x,y] is arbitrary.

[quote="§outh§tar"]... a comparison theorem gives

$\int_{0}^{x}{(y^{2}-s^{2})^{k}dG(s)}\ge \int_{0}^{x}{(x^{2}-s^{2})^{k}dG(s)}= 0$[/quote]Only if $g \ge 0$. The (apparent) difficulty arises if $g$ has infinitely many sign changes.

As I said, it's a quickie. [hide="hint"]Substitute $s^{2}= t$ and write the resulting integral as a function of $y = x^{2}$. Now start differentiating with respect to $y$.[/hide]


Let $I$ be the center of the incircle of a triangle $\triangle ABC$, and $l$ be the perpendicular to the line $AI$ at the point $I$. The line $l$ intersects the side $AB$ in $M$ and the side $AC$ in $N$. Prove that the circle tangent to the line $AB$ at $M$ and to the line $AC$ at $N$ is also tangent to the circumcircle of triangle $\triangle ABC$.

my dear , its one of the basically problems of inversion.

we should only know this fact:

$I$ of triangle $ABC$, changes to $I_a$ of triangle $AB'C'$, after inversion round point $A$.

I invert the shape with center $A$ and radius $AM=AN$.

let name the circle which in tangent to $M,N$, $(C)$.

after inversion, as i said, point $I$ goes to $I'$, which is $I_a$ of $AB'C'$.

then circle $(C)$, changes to circle$(C')$, with center$I'=I_a$.

then circle $(C')$ should be excircle of triangle $AB'C'$, and it means that , it is tangent to segment$B'C'$.

then every thing is ok.

because it means that circle $(C)$ is tangent to the crcum cicle of $ABC$.

how ever if u couldnt figure out it, i can explain each part to u. ;)


function
real analysis
real analysis unsolved

Let $ u: U\rightarrow R$ is locally summable function, and $ V\subset\subset U$. Show by example that if we have $ \|D^hu\|_{L^1(V)}\leq C$ for all $ 0<|h|<\frac{1}{2}dist(V, \partial U)$, it does not necessarily follow that $ u\in W^{1,1}(V)$.

Thanks for help!

I have almost no ideas. I'm undergraduate (my knowledge is poor)student, PDE is hard for me, but i'm studying because my supervisor asking me it.
If you can help me with this problem.


probability
combinatorics unsolved
combinatorics

There is a $n$x$n$ grid of squares, and let us assume that $n$ is also an even integer.  Every second, one of these squares is filled at random, independent of the previous occurrence.  If the squares on the left side of the square are $\frac{2}{3}$ more likely to be filled in than the ones on the right, then what is the probability that the grid is completely filled in in $n^{2}$ seconds?

If $n$ is even, we can say
$P_{L}+P_{R}=1$,
where
$P_{L}$: probability that the square filled in (at one determinate second) will be on the left side,
$P_{R}$: likewise, but on the right side.

I'm not sure which of those 3 interpretations you meant:
- $P_{L}=\frac{2}{3}$, $P_{R}=\frac{1}{3}$;
- $P_{L}=\frac{2}{3}P_{R}$, that is $P_{L}=\frac{2}{5}$, $P_{R}=\frac{3}{5}$;
- $P_{L}=(1+\frac{2}{3})P_{R}$, that is $P_{L}=\frac{5}{8}$, $P_{R}=\frac{3}{8}$.
Anyway, the difficulty of the problem is the same.

In fact, it doesn't matter if it's a $n\times n$ grid, as long as there are $m$ squares on the left side and $m$ squares on the right side ($2m=n^{2}$).

I'll assume all the $m$ squares on the left side have the same probability $P_{L}/m$, and the same for the squares on the right side, but with probability $P_{R}/m$.
Otherwise, I think it would be more difficult to solve the problem in general without more data.

With all these assumptions, I think that the probability you want (each of the $2m=n^{2}$ squares be filled in exactly once, no matter the order) is
$P=(2m)!(\frac{P_{L}}{m})^{m}(\frac{P_{R}}{m})^{m}$,
because there are $(2m)!$ different orderings (in time) that will happen with the same probability, because at the end there are $m$ on the left side and $m$ on the right side.

We can simplify it to
$P=\frac{(2m)!}{(2m)^{(2m)}}(4P_{L}P_{R})^{m}$.

We can see that
$P=P_{0}\sigma^{2m}$,
where
$P_{0}=\frac{(2m)!}{(2m)^{(2m)}}$ is the easiest case $P_{L}=P_{R}$,
and
$\sigma^{2}=4P_{L}P_{R}$ is the variance of the binomial distribution left-right.


I have some trouble understanding/proving/applying the binomial theorem. Can anyone explain in detail? Any help would be greatly appreciated!!! :)

Try the [url=http://www.artofproblemsolving.com/Wiki/index.php/Binomial_Theorem]AoPSWiki[/url] or the [url=http://www.artofproblemsolving.com/Forum/search.php]search function[/url].


Evaluate the sum $ \sum_{k\equal{}0}^{n\minus{}1}\frac{1}{1\plus{}8sin^2\frac{k\pi}{n}}$

It is easy to show, that
 \[ \sum_{k \equal{} 0}^{n \minus{} 1}\frac {1}{1 \plus{} 8sin^2\frac {k\pi}{n}}\equal{}\frac{n}{5}\sum_{k\equal{}0}^{\infty} 0.4^{2k}C_{2k}^k\]

\[ \frac{1}{1\plus{}8sin^2\frac{j\pi}{n}}\equal{}\frac{1}{5\minus{}4cos\frac{2j\pi}{n}}\equal{}\frac{1}{5}\sum_{l\equal{}0}^{\infty}0.4^l(2cos\frac{2j\pi}{n})^l.\]
\[ \sum_{j\equal{}0}^{n\minus{}1}(2cos\frac{2j\pi}{n})^l\equal{}\sum_{j\equal{}0}^{n\minus{}1}0.4^l\sum_{k\equal{}0}^lC_l^kexp{\frac{2\pi ji}{n}(2k\minus{}l)}.\]
Because $ \sum_{j\equal{}0}^{n\minus{}1}exp(\frac{2\pi ijm}{n})\equal{}n\delta_n^m$, were $ \delta_n^m\equal{}0$ if $ n\not |m$ and $ \delta_n^m \equal{}1$ if $ n|m$ we get these formul.


If the medians of a triangle ABC intersect at G, prove that the area of triangle BCG is one third of the area of triangle ABC

Simple - it is well know that where medians intersect, they split each other median into a 2/3 length part and a 1/3 length part. (I don't remember the proof for that, but it is easy cuz I remember learning it in 7th grade school geometry...)

[quote="shobber"]Let $AG$ meets $BC$ at $D$, then $\frac{\triangle{BGC}}{\triangle{BAC}}=\frac{DG}{DA}=\frac{1}{3}$.[/quote]
HEY!! THANX!!!

oh, and it takes a couple seconds to prove the statement I said (about the 2/3rds intersection) using vectors... its in AoPS vol. 2

I never really learned mass points; do you have a link to a topic or someplace that explains them well?

thanx a bunch...
wat r mass points?  :lol:


Let $TABCD$ be a pyramid with top vertex $T$, such that its base $ABCD$ is a square of side length 4. It is given that, among the triangles $TAB$, $TBC$, $TCD$ and $TDA$, one  can find an isosceles triangle and a right-angled triangle. Find all possible values for the volume of the pyramid.

V = 0, V = infinity, V = 8/3, V = 16/[3 sqrt(5)]. Hint: For the hardest scenario, do not solve it but see that the two circles are orthogonal.

Vo Duc Dien


algebra
system of equations
algebra unsolved

Let $ n>1$ be a positive integer an $ a$ a real number. Determine all real solutions $ (x_1,x_2,\dots,x_n)$ to following system of equations:
$ x_1\plus{}ax_2\equal{}0$
$ x_2\plus{}a^2x_3\equal{}0$
…
$ x_k\plus{}a^kx_{k\plus{}1}\equal{}0$
…
$ x_n\plus{}a^nx_1\equal{}0$

From the system of equation we get for $ a \neq 0$   $ \frac {x_2}{x_1} \equal{} \minus{} \frac {1}{a}, \frac {x_3}{x_2} \equal{} \minus{} \frac {1}{a^2}, \dots \; \dots ,\frac {x_1}{x_n} \equal{} \minus{} \frac {1}{a^n}$.Multiplying them and we get for n even $ a^{\dfrac {n(n \plus{} 1)}{2}} \equal{} 1$, so $ a \equal{} 1$ gives
$ \begin{cases}x_i \equal{} k \;\;\; for \;\;\; i \;\;\; odd \\
x_i \equal{} \minus{} k \;\;\; for \;\;\; i \;\;\; even \end{cases}$   are the solutions.
$ a \equal{} \minus{} 1$,  $ 4|n$ gives
$ \begin{cases}x_i \equal{} k \;\;\;for \;\;\;i \equal{} 1,2 \; (mod \; 4) \\
x_i \equal{} \minus{} k \;\;\;for \;\;\;i \equal{} 3,0 \; (mod \; 4) \end{cases}$
For n odd we get $ a^{\dfrac {n(n \plus{} 1)}{2}} \equal{} \minus{} 1$
$ a \equal{} \minus{} 1$, $ n \equal{} 1\;(mod\;4)$  gives
$ \begin{cases}x_i \equal{} k \;\;\;for \;\;\;i \equal{} 1,2 \; (mod \; 4) \\
x_i \equal{} \minus{} k \;\;\;for \;\;\;i \equal{} 3,0 \; (mod \; 4) \end{cases}$
When $ a\equal{}0, all x_i\equal{}0$
The system of equation have no solution for other conditions.


Euler
analytic geometry
geometry proposed
geometry

Let $ABC$ a triangle and $M,N$ two points on sides $AB.AC$ of the triangle $ABC$. Prove that the line $MN$ passes
 through the centroid of the triangle $ABC$ if and only if $\frac{MB}{MA}+\frac{NC}{NA}=1$.

you think,Jeane! We post solutions here. ;)

This is simply the transversal theorem, no?

Let a triangle $ABC$, $D\in BC$, $M\in AB$, $N\in AC$ and $P\in MN\cap AD$. Then:

$\frac{\overline {PD}}{\overline {PA}}\cdot {\overline {BC}=\frac{\overline {MB}}{\overline {MA}}\cdot \overline {DC}+\frac{\overline {NC}}{\overline {NA}}\cdot \overline {DB}}$.

Very good,Levi.How did you prove that? I proved it with a little bit of Menelaus.Can you post a proof? ;)

$\underline {The\ first\ method.}$ Let $U\in MN$, $V\in MN$ such that $BU\parallel CV\parallel AD$.
Thus, $\frac{MB}{MA}=\frac{BU}{AB},\ \frac{NC}{NA}\ =$ $\ \frac{VC}{AP}\Longrightarrow \frac{MB}{MA}\cdot DC+\frac{NC}{NA}\cdot DB=\frac{BU}{AP}\cdot DC+\frac{VC}{AP}\cdot DB=\frac{PD}{AP}\cdot BC.$

$\underline{The\ second\ method.}$ Let d be the line $d$, such that $d\parallel BC,\ A\in d$ and $X\in MN\cap BC,\ Y\in MN\cap d.$
From $\frac{MB}{MA}=\frac{BX}{AY},\ \frac{NC}{NA}=\frac{CX}{AY},\ \frac{MB}{MA}\cdot DC+\frac{NC}{NA}\cdot DB=$ $\frac{1}{AY}(BX\cdot DC+CX\cdot DB)=$
$\frac{1}{AY}\cdot [DC\cdot (XD-BD)+DB\cdot (XD+DC)]=$ $\frac{XD\cdot (BD+DC)}{AY}$
$=\frac{XD}{AY}\cdot BC=\frac{PD}{PA}\cdot BC,$ because $\frac{XD}{AY}=\frac{PD}{AP}.$

$\underline{The\ third\ method.}$ To the transversals $\overline{XMB},\overline{XNP}$ in the triangle $ABD,\ ACD$ respectively, we apply the Menelaus's theorem, where $X\in MN\cap BC:$

$\frac{XB}{XD}\cdot \frac{PD}{PA}\cdot \frac{MA}{MB}=1,\ \frac{XC}{XD}\cdot \frac{PD}{PA}\cdot \frac{NA}{NC}=1$. Thus, $\ \frac{MB}{MA}\cdot DC+\frac{NC}{NA}\cdot DB=$

$\frac{XB}{XD}\cdot \frac{PD}{PA}\cdot DC=\frac{XC}{XD}\cdot \frac{PD}{PA}\cdot DB=\frac{PD}{PA}\cdot \frac{XC.DB+XB.DC}{XD}=\frac{PD}{PA}\cdot XD.$

$\underline{Remarks.}\ P\in [G;\ I;\ O;\ H;\ N(agel);\ E(uler);\ \Gamma(ergogne)]$ a.s.o.

See up!.You have three methods.

[b][u]Caz 1. [/u][/b]If $D$ is the middlepoint of $BC$ $\Longrightarrow$ $\frac{MB}{MA}+\frac{NC}{NA}=2\frac{PD}{PA}\Longleftrightarrow$

$[ 1+\frac{MB}{MA}]+[ 1+\frac{NC}{NA}]=2[ 1+\frac{PD}{PA}] \Longleftrightarrow \frac{AB}{MA}+\frac{AC}{NA}=2\frac{AD}{PA}$ (If $P: =G$).

[b][u]Caz 2.[/u][/b]  If $D$ is such that $m(\widehat {BAD})=m(\widehat {CAD})$ then $(b+c)\cdot \frac{PD}{PA}=b\cdot \frac{MB}{MA}+c\cdot\frac{NC}{NA}$

$\Longleftrightarrow (b+c)\cdot [ 1+\frac{PD}{PA}] =b\cdot [1+\frac{MB}{MA}]+c\cdot [1+\frac{NC}{NA}]\Longleftrightarrow$

$(b+c)\cdot \frac{AD}{PA}=b\cdot \frac{AB}{MA}+c\cdot \frac{AC}{NA}$ (If $P: =I$).

I am sorry, Shobber. I did not want. This is a professional deformation. I am a retired teacher (63 years) and I love all young men.

[quote="shobber"]Let us generize it:
[color=blue]In any triangle ABC, with D being the mid-point of BC. M, N are on AB, AC. AD and MN intersect at O. Prove:
\[\frac{AD}{AO}=\frac{1}{2}(\frac{AB}{AM}+\frac{AC}{AN})\][/color]

If $O=G$, it becomes this problem.[/quote]

Solution in [color=blue][i]barycentric coordinates[/i][/color].
The homogeneous barycentric coordinates of $A, B, C, D, M, N, O$ wrt the reference triangle $\triangle ABC$ are:

$A(1,0,0), \qquad B(0,1,0), \qquad C(0,0,1), \qquad D(0,1,1), \qquad M(1,h,0), \qquad N(1,0,k), \qquad O(h+k,hk,hk)$

Using the distance formula (in barycentrics) we have :

$AM = \frac{{ch}}{{1 + h}}$

$AN = \frac{{bk}}{{1 + k}}$

$AD = \frac{1}{2}\sqrt {2b^2  + 2c^2  - a^2 } $

$AO = \frac{{hk}}{{h + k + 2hk}}\sqrt {2b^2  + 2c^2  - a^2 } $

Thus:

$\frac{{AD}}{{AO}} = \frac{{h + k + 2hk}}{{2hk}}$

$\frac{1}{2}\left( {\frac{{AB}}{{AM}} + \frac{{AC}}{{AN}}} \right) = \frac{{h + k + 2hk}}{{2hk}}$

and we are done. $\blacksquare$


[quote="Interval"]The statement “If x is prime, then it is odd” is false when x equals

(1) 1      (3) 3
(2) 2      (4) 4

Why?[/quote]
[hide]its just knowing your primes. SO it is 2.[/hide]

[quote="Interval"]The statement “If x is prime, then it is odd” is false when x equals

(1) 1      (3) 3
(2) 2      (4) 4

Why?[/quote]
[hide]The definition of prime is a positive integer that has no positive integer divisors other than 1 and itself. The positive integral divisors of the numbers are as follows:
1 = 1
2 = 1,2
3 = 1,3
4 = 1,2,4
Instantly, you can eliminate 1 and 3 because they are odd. Then you can eliminate any even numbers besides 2 because even numbers are always divisible by 2 (by definition). So then the only number that fits that is $\boxed{2}$[/hide]


function
limit
calculus
calculus computations

See rule #2 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184229]this thread[/url] :) Generally speaking, we are not here to do your routine homeworks but you may expect some help if we see clearly what your difficulty is. Just posting the homework problems won't do any good: most likely they'll be just ignored as "not interesting enough to bother about" and even if somebody gives a full solution, it may be quite useless for several reasons (for instance, I can say something like "the limit of a continuous function at a point equals its value, hence..." but what's the use of it if you don't know what a continuous function is yet?)

Now back to the problems: what do [b]you[/b] think the limits are and why? Don't be afraid to be wrong, just say what you think and we'll proceed from there :)

you are right dude, it's my wronge. But I really don't even have the basics of solving these questions. I've missed the first week of classes, and when I came the next week, which is this one, the instructor gave me this homework. I tried to open the book and understand what i 've missed, but it was so hard.

OK, then you should certainly catch your instructor, apologize for missing the first week, and talk to him for an hour or so about the skipped material. For now, let me tell you just three things.

1) If a function is given by a single "reasonable" (I do not want to go into the exact meaning of the word "reasonable" now) formula that makes sense on an interval around the  point, the limit at the point equals the value. So, say, $ \lim_{x\to 3}x^2\minus{}x\equal{}3^2\minus{}3\equal{}6$. Can you do (d) now?

2) If the function is given by two different formulae on two different sides of a point $ a$ then $ \lim_{x\to a\minus{}}$ can be obtained using the formula that describes the function for $ x<a$ and $ \lim_{x\to a\plus{}}$ can be obtained using the one that describes the function for $ x>a$. So, if $ f(x)\equal{}\begin{cases}2x, & x<1\\3, &x\equal{}1\\1\minus{}x, &x>1\end{cases}$, then $ \lim_{x\to 1\minus{}}f(x)\equal{}\lim_{x\to 1\minus{}}2x\equal{}2\cdot 1\equal{}2$. Note that, when looking for a limit, you should never be concerned with the value at the point itself, so that $ 3$ has nothing to do with finding the limit at $ 1$! Can you do (a) and (b) now?

3) In the case described in $ 2$, the (two-sided) limit $ \lim_{x\to a}$ exists if and only if both $ \lim_{x\to a\minus{}}$ and $ \lim_{x\to a\plus{}}$ exist and are equal. So, in my example, I would have $ \lim_{x\to 1\plus{}}f(x)\equal{}\lim_{x\to 1\plus{}}(1\minus{}x)\equal{}1\minus{}1\equal{}0\ne 2$, which means no two-sided limit at $ 1$. Can you do (c) now?

Bear in mind that (1-3) are just "cookbook recipes" and they cannot substitute the understanding of the concept. So, even if you get away with this particular homework, my first advice about talking to your instructor will be still valid :P

thank you dude for this great work. Im rally appreciated


Are you sure this is true?

Taking a=b=c=1/3 seems violate the inequality.

Indeed Fuzzylogic is right.

But the following inequality holds (maybe this is the one Maverick meant):

[i]If a, b, c are positive reals such that a + b + c = 1, then

[tex]\left( ab+bc+ca\right) \left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \leq \frac{a^{2}+b^{2}+c^{2}}{6abc}[/tex].[/i]

Here is my [i]proof[/i]: We have to show

[tex]\left( ab+bc+ca\right) \left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \leq \frac{a^{2}+b^{2}+c^{2}}{6abc}[/tex].

At first, we know that $ab+bc+ca\leq a^{2}+b^{2}+c^{2}$. Now we will prove

[tex]\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\leq \frac{1}{6abc}[/tex].

Since $a+b\geq 2\sqrt{ab}$, $b+c\geq 2\sqrt{bc}$ and $c+a\geq 2\sqrt{ca}$, we have $\left( a+b\right) \left( b+c\right) \left( c+a\right) \geq 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca}=8abc$, and thus

[tex]\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{\left( b+c\right) \left( c+a\right) +\left( c+a\right) \left( a+b\right) +\left( a+b\right) \left( b+c\right) }{\left( a+b\right) \left( b+c\right) \left( c+a\right) }\leq \frac{\left( b+c\right) \left( c+a\right) +\left( c+a\right) \left( a+b\right) +\left( a+b\right) \left( b+c\right) }{8abc}[/tex].

Hence it remains to prove

[tex]\frac{\left( b+c\right) \left( c+a\right) +\left( c+a\right) \left( a+b\right) +\left( a+b\right) \left( b+c\right) }{8abc}\leq \frac{1}{6abc}[/tex],

or, equvialently,

$3\left( \left( b+c\right) \left( c+a\right) +\left( c+a\right) \left( a+b\right) +\left( a+b\right) \left( b+c\right) \right) \leq 4$.

But since a + b + c = 1, this is equivalent to

$3\left( \left( b+c\right) \left( c+a\right) +\left( c+a\right) \left( a+b\right) +\left( a+b\right) \left( b+c\right) \right) \leq 4\left( a+b+c\right) ^{2}$.

Expanding the products, we see that this is equivalent to

$a^{2}+b^{2}+c^{2}\geq bc+ca+ab$,

what is, again, well-known. Hence the inequality is proven.

  Darij

[quote="darij grinberg"]

But the following inequality holds (maybe this is the one Maverick meant):

If $a, b, c$ are positive reals such that $a + b + c = 1$, then

$\displaystyle \left( ab+bc+ca\right) \left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \leq \frac{a^{2}+b^{2}+c^{2}}{6abc}$
[/quote]

If this is what he meant, here is a different proof:
$\displaystyle
(6abc) \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)
\leq  (6abc) \left(\frac{1}{4}\left(\frac{1}{a}+ \frac{1}{b}\right) +  \frac{1}{4}\left(\frac{1}{b}+ \frac{1}{c}\right)+\frac{1}{4} \left(\frac{1}{c}+ \frac{1}{a}\right) \right) $

$= 3 \left( ab+bc+ca\right) \leq (a+b+c)^2 = 1$

and we also have, $ ab+bc+ca \leq a^2+b^2+c^2$.

Yes I am sorry! It was a posting mistake.  Both of the solutions are very nice!


Prove that $p|C^{k}_{p^{n}},\forall k,n>0$, where $p\ge3$ is a prime number.

Yes, it is easy :) Use Lucas's theorem on binomal.


[quote="sri340"]A certain person walks 5 kilometers per hour. If riding a bike 1 kilometer is 8 minutes less than walking the same distance, then the speed at which he rides is how many times the speed at which he walks?[/quote]
[hide]Using $ D\equal{}rt$ we find that it takes him one fifth of an hour to walk 1 kilometer. That is equivalent to 12 minutes. Therefore he bikes 1 kilometer in 4 minutes, which implies a rate of 15 km/h, thus he bikes at a rate 3 times of which he walks.[/hide]


There are 20 points inside a square. They are connected by non-intersecting segments with each other and with the vertices of the square, in such a way that the square is dissected into triangles. How many triangles do we have.
(Hint only if noone answers)

[hide="Not sure"]$ \binom{24}{3}$ ?[/hide]

$ 42.$ :lol:
For $ n$ points the answer is $ 2n\plus{}2.$ :wink:


Is is possible to assign numbers from $1, \ldots, n$ to the edges of a circle $C_n$ in a graph of length $n$ by means of a bijection $f: E \mapsto \{1, \ldots, n\}$ such that the numbers $1, \ldots, \binom{n+1}{2}$ can be represented as sum of edge numbers along a path in $C_n$ or along an edge in $C_n$ itself ?

For a similar question in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22106]another type of a graph.[/url]

[quote="pbornsztein"]Do you know something about it or is it an open question?

Pierre.[/quote]

I don't know whether it is an open problem. At least special cases are known.  ;)


Source: Gauss 2004

1) Let x be the three-digit number with digits ABC and y be the three-digit number with digits CBA. The digits A and C are not 0. If x-y = 495, how many possibilities are there for x?

a) 50
b) 40
c) 24
d) 36
e)32

2) A large block, which has dimensions n by 11 by 10, is made up of a number of unit cubes and one 2 by 1 by 1 block. There are exactly 2362 positions in which the 2 by 1 by 1 block can be placed. What is the value of n?

a) 7
b) 8
c) 9
d) 10
e) 11

3) Water is poured from a full 1.5 bottle into an empty glass until both the glass and the bottle are 3/4 full. What is the volume of the glass?

a) 0.5 L
b) 0.75 L
c) 1.125 L
d) 0.6 L
e) 0.4 L

[color=darkred][size=75][Title changed to "Hard" from "Easy" by moderator rcv.][/size][/color]

Here is the solution of the first problem
[hide] x=ABC=100*A+10*B+C and y=CBA=100*C+10*B+A then x-y=99*(A-C)=495
So A-C=5 or A=5+C <= 9. There are four possibilities to choose A and C and 9 possibilities to choose B.
As the result there are 36 possibilities[/hide]

I wouldn't call those "Easy Questions", especially not in "The Basics".  It will make people feel bad if they can't solve them, and they're certainly not trivial!

I kind of agree with mathmom....

this forum is for middle school( I mean real school ) level section

these problems require some High school knowledge...

These questions are actually 3 of the [b]hardest[/b] questions on the 8th Grade Canadian Gauss 2004 Competition.  I don't have any personal experience with the Gauss competitions, but I presume they are roughly equivalent in style and difficulty to the AMC-8.

[Bumped from Basic to MATHCOUNTS.]

The full competition can be viewed at:
[url=http://www.cemc.uwaterloo.ca/english/contests/past_contest/2004/gauss_grade8.pdf]http://www.cemc.uwaterloo.ca/english/contests/past_contest/2004/gauss_grade8.pdf[/url]

[quote] There are four possibilities to choose A and C and [b]9 possibilities[/b] to choose B. [/quote]

Why 9 and not 10?  (Why can't B  one of  {[b]0,[/b]1,2,..9}?..  (Hope this helps.)  < We really have to have an icon for hope this helps <smile>)

[quote]
Why 9 and not 10? (Why can't B one of {0,1,2,..9}?.. (Hope this helps.) < We really have to have an icon for hope this helps <smile>)[/quote]
Really there are ten possibilies for B so the answer of A) is [hide] 4*10=40[/hide]

on my experiece, it's much harder than AMC-8...

I took many mock AMC-8 and the actual AMC-8 before but this problem is like ten time harder..

[quote="Treething"][quote="ying'n'yang"]Source: Gauss 2004
2) A large block, which has dimensions n by 11 by 10, is made up of a number of unit cubes and one 2 by 1 by 1 block. There are exactly 2362 positions in which the 2 by 1 by 1 block can be placed. What is the value of n?

a) 7
b) 8
c) 9
d) 10
e) 11
[color=darkred][size=75][Title changed to "Hard" from "Easy" by moderator rcv.][/size][/color][/quote]

rcv you font color thief =o =o =o

[hide="solution"]
2(110 + 11n + 10n) + 2(100 + 99 + 10n + 11[n - 1] + 9n + 10[n - 1]) = 2362
110 + 21n + 199 + 19n + 21[n - 1] = 1181
40n + 309 + 21n - 21 = 1181
61n + 288 = 1181
61n = 893

huh?
[/hide][/quote]

Check your equation.


calculus probability AMC USA(J)MO USAMO email geometry

Who's going to Walton next weekend? My school will be bringing at least three people (including me). :)

Yea I hope I'm going...GO SCAVANGER HUNTTTZ (spelled scavenger wrong YEA)

yes.

Hi. :bye:

I'm not scavenger-hunting unless my teammate(s) really want me to.

I understand that His (Non)excellency Calculus shall be there also, to decrease my probability even further of trophy-winning.

I will also be there, possibly with another person from my school.

"His (Non)excellency Calculus"???? What exactly is that referring to?

Perhaps he means that there will be calculus on the written test?

Anyhow, I'm giving a talk!  It will be about geometry, trig, and complex numbers.

Yes, there will supposedly be calculus on the super varsity test, but last year it was minimal (one or two problems).

Miles, will your talk be at the same time as Harrison's? Last year, Phillip's and Harrison's conflicted, I believe. I want to hear them both!

I'm excited about the MathBowl because the one at Princeton earlier this year was so much fun. I hope my team actually gets past the first round this year (though I couldn't have asked for a much better team last year).

Last year, only three trophies were awarded per division, so if this is the same this year, my chance of trophying is not so high.

The final talks schedule is undergoing some revision. I'll try to schedule Miles and Harrison separately, but that might hinder the USAMO practice session.

Can people in denial about not making USAMO go to the practice session? :lol:

If we made USAMO, *must* we go to the practice session? What is idea of the practice session anyway? (Not that I don't think it'll be good, I'll probably come, but if I'm feeling very not-mathy, will Mr. Fulton kill me - I mean, take away my sugar packets - for not coming? Will he give me some sugar packets if I'm good and I come? ;) )

The practice session is for anyone that wants to come. Basically, I'll just provide a room and you guys can work together. If anyone has some particularly good problems, send them to me and I'll be happy to make some copies.

And I don't think there is any sugar left in Cobb County! :-)

I think I'll be going to Walton, but I'm not really sure if I'll find a team :(
Anyone want to come be part of my team?

The only official team competition is the Math Bowl. Well, yeah, I guess the scavenger hunt counts also.

Can we show up unannounced if it turns out we have nothing to do, or do we need to have registered previously? Sounds like something I'd enjoy...

And dude guy there, I'll totally be on your team. As long as it has an awesome name. Something like... Team Slaüghter.

Actually, that might be too awesome.

I think two more people from my school will be going. Stephen, would you mind being listed under Parkview? (I'm not sure how the teams are named...) Another thing, which test do you plan to take? JV, varsity, or supervarsity?

Boru

We would prefer to know that you are coming. We do have to order pizza after all! But you could let me know pretty last minute for a small group.

[quote="gtf"]We do have to order pizza after all![/quote]
And hopefully not run out this time. :wink:

Did they run out last year? (I can't remember)

Yes, but only a few people ( :bye: ) didn't get any.

Cool. Maybe I can talk a few people at my school into going. Doubt it would be more than three or four, if any at all.

If that doesn't work out, Boru, sure. Supervarsity sounds like fun. Wootz.

Hmmmm, is anyone from Northview actually coming? I haven't heard anything from anyone except for Arvind's rambling!

I will be glad to add my rambling to that of Arvind's, though Sitan Chen is not commonly known for rambling. :D

Anyways, I'm pretty sure Northview's taking ppl, though we haven't had a team meeting since JV state, so idk any details about transportation yet.

I need to have a number so that we can order food, etc.

So looks like I won't be going after all. Going to a play at midnight tonight and I have another few hundred pages of The Fountainhead to read... but I'll definitely see everyone at State next week.

[quote="Arvind_sn"]!!!Important!!!z

k, i think northview's bringing abt 10 people, but none of us know when 2 be there or anything. Could some1 give me information about when we have to be there? We need to walk-in register.  Five dollars, rite?[/quote]

I will say what I know...

According to an information sheet that Mr. Fulton sent out less than a month ago, registration is 65 dollars for the first 10 team members and 10 dollars for each additional person.

My teacher instructed our team members to be at Walton by 8:30, so I would imagine that would be a safe time to get there. Maybe the tournament starts at 9:00?

I got a schedule in an email a while ago that said the same thing. So, you are probably correct.

Here are the results that I remember for those who weren't there:

Mathbowl: Northview won, but Woodward was fairly close.
Scavenger Hunt: Northview won, but apparently Athen Academy was very close.

JV: Sitan destroyed everyone.
V: Arvind destoyed everyone. Someone from my school tied for fourth!
SV: Santhosh won by a nice margin (8 points), followed by Miles and I, who tied for second place (and luckily, Mr. Fulton didn't do item analysis). Harrison came in fourth.

All three divisions were won by AoPS frequenters! Coincidence?

Dude this is wut I remember...

JV

1st Sitan Chen (Math Geek), Northview
2nd Yubo Su (yubo56, i think), Northview
3rd Alan Dong, Northview

Varsity

1st Arvind Narayan (Arvind_sn), Northview
2nd Nathan Li, Northview
3rd Umm, some guy from alpharetta, i think

Super Varsity

See Eric's post.

By the way, I'm probably enjoying my "toy" more than I would have enjoyed a trophy. :)

Is that the green thing that twirls backwards when you push it?  Or something like that?  I wanted to get that, but being my azn self, I got the biggest thing on the table instead, the poster of the elements.

I came online after spending an hour or so looking at my pwnzor fridge magnet. And my cube.  :D

[quote="Math Geek"]Is that the green thing that twirls backwards when you push it?[/quote] Yes. :lol:


function
continued fraction
calculus
calculus computations

Is it possible to have an alternating continued fraction representation
$ x \equal{}a_0 \plus{} (a_1 \plus{} 1 \minus{} (a_2 \plus{} (a_3 \plus{} 1 \minus{} ...)^{ \minus{} 1})^{ \minus{} 1})^{ \minus{} 1}$
where $ a_n$ are positive integers

is such an expansion unique (for irrational x)?
does every x have an expansion of that form?

If you want something like that, then yes, each number will have a unique expansion and most other properties of continued fractions will have their analogues. But what do you want to do with your alternating continued fraction expansion that you cannot do with the ordinary one? :?

well if you summed up the terms and treated the series formed by the partial sums as an engle expansion, the function you get should be increasing, which is not true if you did the same with a normal continued fraction.

Or am I just really confused about everything


superior algebra
superior algebra unsolved

Let $K$ be a field and $G$ a group. Suppose that none of $G$'s elements are of finite order. Does $KG$ possess a nontrivial zero divisor?
PS $KG$ is Group Ring. For a definition you can take a look at Hungerford's Algebra, for instance.

As far as I know, this problem has not been solved in general.


Galois Theory
superior algebra
superior algebra unsolved

[quote="zhaobin"]If $\mathbb{F}$ is the finite extension $\mathbb{Q}$ ,then there exist $\alpha \in \mathbb{F}$,such that $\mathbb Q(\alpha)=\mathbb F$[/quote]
Let $[\mathbb{Q}: \mathbb{F}] = n>1$.
Then there exists $\beta_{1}\in \mathbb{F}$ such that $\beta_{1}\not \in \mathbb{Q}$.
Form,
$\mathbb{Q}< \mathbb{Q}(\beta_{1}) \leq \mathbb{F}$
If $\mathbb{Q}(\beta_{1}) \not = \mathbb{F}$ choose $\beta_{2}\in \mathbb{F}$ so that $\beta_{2}\not \in \mathbb{Q}(\beta_{1})$ (otherwise we are done).
And form,
$\mathbb{Q}< \mathbb{Q}(\beta_{1}) < \mathbb{Q}(\beta_{1}, \beta_{2}) \leq \mathbb{F}$.
Continue this until you have exhausted all $\beta_{1},\beta_{2},...,\beta_{k}$ which must happen because $n < \infty$.
So we have that $\mathbb{Q}(\beta_{1},...,\beta_{k}) = \mathbb{F}$. But that doesnt answer your question, sorry  :oops: .
Is the following true? (I seriously doubt it, but a conterexample doesnt enter my mind).
$\mathbb{Q}(\beta_{1},\beta_{2},...,\beta_{m}) = \mathbb{Q}(\beta_{1}+\beta_{2}+...+\beta_{m})$
If that is actually true then choose $\alpha = \beta_{1}+...+\beta_{k}$.

@zhaobin Use the fact that [u]every finite, separable extension is simple[/u].
@sylow_theory Prove your claim, for $m=2$, then go ahead!(note your method of choosing those elements.)

[quote="The Captain"]
@sylow_theory Prove your claim, for $m=2$, then go ahead!(note your method of choosing those elements.)[/quote]
Hi,Can you tell me how to prove $m=2$?

Thank you two.

[quote="The Captain"]
@sylow_theory Prove your claim, for $m=2$, then go ahead!(note your method of choosing those elements.)[/quote]
Can you? I tried to do this. Is there something special about $\mathbb{Q}$? I presume it doesnt work for any arbitrary field.

Is it true?
If $\alpha$ is of degree $n$ over $\mathbb{Q}$ and $\beta$ of degree $m$ then $\mbox{deg}(\alpha+\beta ,\mathbb{Q})= nm$?
That is what I cant show.

[quote="zhaobin"][quote="The Captain"]
@sylow_theory Prove your claim, for $m=2$, then go ahead!(note your method of choosing those elements.)[/quote]
Hi,Can you tell me how to prove $m=2$?
.[/quote]Well, when I said that, I didn't mean you can prove it. I just meant that you should just concentrate on this case. If you know some more advanced theorems, they may help you decide easier which statements are true. That method of choosing elements, helps you omit some trivial cases.

[quote="ma_go"]in fact, you can't show it since it's false :)

you can choose $\alpha^{2}=\beta^{4}=2$ to get a counterexample.
some less trivial counterexample, with extension not one contained in the other can be made by choosing $\alpha = \sqrt{p}+\sqrt{r}, \beta=\sqrt{q}-\sqrt{r}$, where $p,q,r$ are squarefree and distinct. (in this case the degree of the extension is not even the biggest possible, which should be $8$)[/quote]
Hi,but from sylow_theory's solution ,we should just show if $\alpha$ has degree $m$ over $\mathbb Q$,$\beta$ has degree over  $\mathbb Q(\alpha)$,then can we show that $\alpha+\beta$ has degree $mn$ over $\mathbb{Q}$?

Here we have stronger condition.

[quote="ma_go"][quote="ma_go"]choose $\alpha = \sqrt{p}+\sqrt{r}, \beta=\sqrt{q}-\sqrt{r}$, where $p,q,r$ are squarefree and distinct.[/quote]

uhm.. in this case $\beta\notin\mathbb{Q}(\alpha)$, then $\beta$ has degree strictly larger than $1$ over $\mathbb{Q}(\alpha)$. but $\alpha+\beta$ and $\alpha$ both have degree 4 over $\mathbb{Q}$.

but i'm telling you something about primitive element theorem (which is actually what you've stated, in a slight wider generality), that's the tricky point in the proof.
of course, hidden ;)
[hide]i'm sure that what you can show is that in an infinite field there's an element of degree $mn$ of the form $\alpha+t\beta$.[/hide][/quote]
Thank you,we can use the Galois Theory to prove there is finite intermediate field.
Then there exist $t_{1}\neq t_{2}\in \mathbb Q$,such that $\mathbb Q(\alpha+t_{1}\beta)=\mathbb Q(\alpha+t_{2}\beta)$
then:$t_{1}\beta-t_{2}\beta \in \mathbb Q(\alpha+t_{1}\beta)$
then $\alpha,\beta \in \mathbb Q(\alpha+t_{1}\beta)$

[quote="zhaobin"]If $ \mathbb{F}$ is the finite extension $ \mathbb{Q}$ ,then there exist $ \alpha \in \mathbb{F}$,such that $ \mathbb Q(\alpha)=\mathbb F$[/quote]

We can use two theorems.

[b]Theorem:[/b] A field of charachteristic zero is is perfect.

[b]Primitive Element Theorem:[/b] Let $ E$ be a finite seperable extension of a field $ F$. Then there has to exist $ a\in E$ so that $ E=F(\alpha)$.

We have a corollay.

[b]Corollary:[/b] A finite extension of [i]charachteristic zero[/i] is a simple extension.

Now that answers the question cuz $ \mathbb{Q}$ is of charachteristic zero.


Prove combinatorial identities:

$\sum_{m=k}^n {m \choose k} = {n+1 \choose k+1}$

$\sum_{r=1}^{n-1} r(n-r) = {n+1 \choose 3}$

$\frac{1}{n} {n \choose k} = \frac{1}{k} {n-1 \choose k-1}$

[hide="3"]Absorption kills it. ${n \choose k} = \frac{n}{k}{n - 1 \choose k - 1}$
Or $\frac{(n - 1)!}{k(n - k)!(k - 1)!} = \frac{1}{n} \cdot \frac{n!}{k!(n - k)!} = \frac{1}{n}{n \choose k}$[/hide]

[hide="1 - Death by Induction"]
Base case:
$\binom{k}{k} = \binom{k+1}{k+1} = 1$

Induction:
Assume $\binom{k}{k} + \ldots + \binom{n}{k} = \binom{n+1}{k+1}$.
Then $\binom{k}{k} + \ldots + \binom{n}{k} + \binom{n+1}{k} = \binom{n+1}{k+1} + \binom{n+1}{k} = \binom{n+2}{k+1}$.
[/hide]


geometry
3D geometry
prism
geometry theorems

Could anybody give a complete proof of Poncelet Porsim?. Thx   :)

For triangles or for n-agons? Cause I know an elementary proof for triangles, but for the n-agon case...well, it's a bit harder.

oh lepnardo ,i coudnt find the meaning of the word (poncelet porism).could u plz write what the theorem is? :?

Take some circles $C$ and $c$, where $c$ lies inside $C$.

Now, pick a point $P_0$ on $C$ and draw a tangent $PP_1$ to $c$, where $P_1 \in C$. Draw another such tangent $P_1P_2$, with $P_2 \in C$. Continue doing this, and suppose that there is some positive integer $n$ such that $P_0 = P_n$.

Then, for ANY point $Q_0 \in C$, the point $Q_n$ (constructed this way) satisfies $Q_n = Q_0$.

Is this right? I am not so sure about the general case (I know the triangle case...)


Hi!  I was looking at the GIMP website, and it looks pretty neat.  What does it do?  Does it work like Photoshop?  Can you color backgrounds, manipulate photos, and create great graphics wth it?

EDIT: My parents want to know if it's a virus-free program.

Thanks!

Duplicate the layers as many times as needed. Also, you may want to follow some GIMP tutorials, because that helped me when I was a beginner. :) Check out forums like graphicforce.org (I'm IcyFlames there).

I need help making graphical designes on gimp and was wondering if you can use perl to make one along with gimp.

Huh?  I think GIMP is basically a stand-alone software, unless you download the GIMP animation package.

I recently got GIMP also, and used it in conjunction with Apophysis.  See signature.

Please explain the benifits of GIMPAP. (GAP or gimp animation package(w/e))

This is probably the most epicly stupid question ever but here goes...

After I use GIMP, how do I get it onto my AOPS avatar? (yes..it is a stupid question)

Not the stupidest question ever, I've seen much worse.  :D

OK.  So, you start by duplicating your layer.

Then you make a text box and type your username in, and set up all the fonts, colors, etc.  If you want your username to change colors, then you would duplicate your current layer (the one with the text on it) and change the text colors.

[url=http://www.gimp.org/tutorials/Simple_Animations/]Click here for more detailed instructions (with screenshots).[/url]

[quote="westiepaw"]Sorry about the double-posting, but I can't edit my original post now.

I downloaded GIMP and have been using it to try and create animations.  The animations work fine, but I'd like to know how to slow down the frames, because they goby too quickly.  (Like in my avatar.)  Thanks![/quote]
Also, when you save it as an animation, you have an option to choose how many milliseconds between frames.

I'm currently working on an AVATAR for myself...

I have a website for high-quality rendered images, but no one will be interested probably. pm me if you actually are though

I need help figuring out how to make an animation.  Thank you.

Look upwards in the thread.

http://www.gimp.org/tutorials/Simple_Animations/


The absent-minded mathematics professor, asked to read his own gas meter,
was provided with a card similar to the one pictured. The professor
observed the meter and market the card, but forgot to mail it. Days later, in response to a telephone inquiry, he is heard to say, "I did read the meter, but misplaced the card.
Fortunately, I remember that it was the largest possible reading such that
all four pointers on the meter were exactly parallel." What was the
reading on four dials, assuming the professor remembered correctly?

Hmm am I just not seeing something but isn't it just maximise the number of millions and go from there?


I'd just like to get some statistics on this going into the test next Tuesday. :-) Of course, I'd assuming a good majority will be taking it in a week.

No choice... the alternate one falls on spring break.

for me and the rest of the greater houston area the first one fall in spring break so...the alternate it is ....no-life in spring break :D

Well I'm part of "the rest of Greater Houston" and I'm taking the first. :-P So psh...lol jkjk. No actually, I think it's just Fort Bend and a few other districts that aren't taking the first one.

we're taking alternate aime as a whole school from Turkey

My school only offers the first of everything: The 10A and 12A, and the AIME1.  So, I really don't have a choice.

I'm taking the AIMEI, but that means that I miss the graduation tests.  It means I don't get to sleep in next year when they give the tests again :(

Normal here; I don't think the schools have enough money for a gluestick here with all the budget cuts so god knows how we'll pay for the alternates ;)  ;)  ;)  ;) ... and because the AIME1s are usually easier.

Tomorow. so excited. so happy. so mentally prepared. gonna go meditate. actually, gonna go learn to meditate.


calculus
integration
logarithms
trigonometry
algorithm
function
complex numbers

And if Misan thinks it's hard, you know it's one hell of a difficult integral.

We can substitute $ x\mapsto \sin x$ and $ y\mapsto \sin y$ , expand $ \log(1\minus{}z)\equal{}\minus{}\sum_{n\equal{}1}^\infty \frac{z^n}{n}$

and use the formula $ \int_0^\frac{\pi}{2} \sin^{2n\minus{}1} x \,\, dx\equal{}\frac{\sqrt{\pi}\,\, \Gamma(n)}{2\,\Gamma\left(n\plus{}\frac12\right)}$ .

Then showing Misan's integral is equivalent to show that the hypergeometric series

$ \sum_{n\equal{}1}^\infty \left(\frac{\sqrt{\pi}\,\, \Gamma(n)}{2\cdot \Gamma\left(n\plus{}\frac12\right)}\right)^2\cdot \frac{1}{n}\equal{}2\pi K\minus{}\frac72 \zeta(3)$ .

Maybe this is a step in the direction of simplification.

$ \Gamma\left(n \plus{} \frac {1}{2}\right) \equal{} \frac {(2n \minus{} 1)!!}{2^n}\sqrt {\pi}$.

$ \frac {\sqrt {\pi}\,\,\Gamma(n)}{2\cdot\Gamma\left(n \plus{} \frac {1}2\right)} \equal{} \frac {(n \minus{} 1)! \cdot 2^{n \minus{} 1}}{(2n \minus{} 1)!!} \equal{} \frac {(n \minus{} 1)!\cdot 2^{n \minus{} 1}}{\frac {(2n)!}{2^{n}\cdot n!}} \equal{} \frac {(n!)^2}{(2n)!}\cdot \frac {2^{2n \minus{} 1}}{n}$

$ \left(\frac {(n!)^2}{(2n)!}\cdot \frac {2^{2n \minus{} 1}}{n}\right)^2\cdot \frac {1}{n} \equal{} \frac {2^{4n \minus{} 2}}{\binom{2n}{n}^2}\cdot \frac {1}{n^3}$

There exist a Zeilberger's algorithm of summation. Maybe using it we can prove this.

Before proving the identity, we introduce some formulas:

$ (1) \qquad \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} \theta \, d\theta = \frac{2^{2n-1}}{n \binom{2n}{n}}$

$ (2) \qquad \frac{\arcsin x}{\sqrt{1-x^2}} = \sum_{n=1}^{\infty} \frac{(2x)^{2n-1}}{n \binom{2n}{n}}$

$ (3) \log |\tan\theta| = -2 \sum_{n=1}^{\infty} \frac{\cos (4n-2)\theta}{2n-1}$

(1) can be proved by either using the beta function identity or mathematical induction. To prove (2), we use (1).

\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{(2x)^{2n-1}}{n \binom{2n}{n}}
& = & \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{2}} (x \sin \theta)^{2n-1} \, d\theta \\
& = & \int_{0}^{\frac{\pi}{2}} \frac{x \sin \theta}{1 - x^2 \sin^2 \theta} \, d\theta \\
& = & \int_{0}^{x} \frac{1}{1 - x^2 + t^2} \, dt \qquad (t = x \cos\theta)\\
& = & \frac{1}{\sqrt{1-x^2}} \arctan \left( \frac{x}{\sqrt{1-x^2}} \right) \\
& = & \frac{\arcsin x}{\sqrt{1-x^2}}
\end{eqnarray*}

The identity (3) follows from the Fourier expansion of $ \log |\sin \theta|$ and $ \log |\cos \theta|$, which can be obtained by comparing real part of $ \log \sin \theta$ and $ \log \cos\theta$.

Then integrating both sides of (2) and plugging $ x = \sin \theta$ gives

$ \frac{\theta^2}{2} = \sum_{n=1}^{\infty} \frac{4^{n-1}}{n^2 \binom{2n}{n}} \sin^{2n} \theta.$

Multiplying $ 2 \csc \theta$ and integrating from 0 to $ \tfrac{\pi}{2}$ gives

\begin{eqnarray*}\int_{0}^{\frac{\pi}{2}} \frac{\theta^2}{\sin \theta} \, d\theta
& = & \sum_{n=1}^{\infty} \frac{4^{n-1}}{n^2 \binom{2n}{n}} \int_{0}^{\frac{\pi}{2}} 2 \sin^{2n-1} \theta \, d\theta\\
& = & \sum_{n=1}^{\infty} \frac{4^{2n-1}}{n^3 \binom{2n}{n}^2} \\
& = & - \int_{0}^{1} \int_{0}^{1} \frac{\log(1-x^2y^2)}{xy \sqrt{1-x^2}\sqrt{1-y^2}} \, dxdy
\end{eqnarray*}

where the last equal is due to the calculation shown by Doe John and Yuriy Solovyov.

So it remains to calculate this integral. But note that $ \int \csc \theta \, d\theta = \log \tan \frac{\theta}{2} + C$. So we have

\begin{eqnarray*}\int_{0}^{1} \int_{0}^{1} \frac{\log(1-x^2y^2)}{xy \sqrt{1-x^2}\sqrt{1-y^2}} \, dxdy
& = & -\int_{0}^{\frac{\pi}{2}} \frac{\theta^2}{\sin \theta} \, d\theta\\
& = & \int_{0}^{\frac{\pi}{2}} 2\theta \log \tan \frac{\theta}{2} \, d\theta \\
& = & \int_{0}^{\frac{\pi}{4}} 8 \varphi \log \tan \varphi \, d\varphi \\
& = & -16 \sum_{n=1}^{\infty} \frac{1}{2n-1} \int_{0}^{\frac{\pi}{4}} \varphi \cos (4n-2) \varphi \, d\varphi
\end{eqnarray*}

It is easy to evaluate the integrals appearing in the summand by integration by parts, yielding the desired result.


topology
real analysis
real analysis unsolved

Okay, I was reading some analysis book, and it tells me:

[quote="book"]We shall often encounter measures $\mu$ which are carried by a proper subset $F$ of $X$, that is, $\mu(X \setminus F) = 0$. It is not hard to see that in the case where $\mu$ is a Borel measure and $X$ is separable, there exists a unique smallest closed set with this property. [/quote]
Well, despite that, I'm having a hard time seeing it. Can someone sketch the proof for that, please?

Let $\{B_{i}\}_{i\in\mathbb N}$ be a countable base of the topology (every separable metric space has a countable base). Now let $I=\{i: \mu(B_{i}\cap X) = 0\}$. The the set \[A=X\setminus \bigcup_{i\in I}B_{i}\] is the minimum closed set we are looking for. First, it is clear that $\mu(X\setminus A) = \mu(\bigcup_{i\in I}(B_{i}\cap X)) =0$. Second, let $F$ be a closed subset of $X$ s.t. $\mu(X\setminus F) = 0$. Then the complement $C$ of $F$ is open, and $C\cap X$ is a null set. Therefore, $C$ is a union of some $B_{i}$, $i\in I$. That is $A\subset F$. Finally, we prove that $A$ is closed. Pick a set $F$ as in the condition.
Then (since $A\subset F\subset X$) \[A = F\setminus \bigcup_{i\in I}B_{i}\] is closed.

Update: I changed the wording in the last sentence.


Um... my calculator says it is 9. LOL. How do you solve this without a calculator?

One could argue that it is 1, because you said "Find the first digit of 100" and then ended your sentence with an exclamation point.  This is obviously because you were so eager to post the problem and get a solution (I know I would be).

Tee hee hee...

As I tell my sons (and nephews, nieces, their friends etc) , in my days,  when we were not allowed to use any calculators,  we have to remember all logarithms  from 1 to 1000 (Along with memorizing times table to 80  We also had to walk 29Km uphill both ways to school).  :)

We also have to  memorize all the words of wisdom from  old Sanskrit Vedas. :)

So no wonder, this type of problems are easy for me.  :D

And I will tell you how you too  can use this centuries old method for such kind of problems. <smile>. This is straight out from old Sanskrit book of Wisdom and Factorials

[b]In all the following calculations, you take only the fractional part. [/b](Eg if you get  a value
 7.345 just ignore 7 and keep .345  easy enough right?)

Assume the number is N and you want to find first digit of N!
(Remember your logs (to the base 10))
_

[b]Step 1  You multiply  0.565705518 by N[/b] ( and keep the fractional part)
                (For example   for n=100 you have 0.57)
[b]Step 2  You add 0.4  [/b]       (  So now you have 0.97..)

[b]Step 3[/b]  For numbers like 100, 10000  (10^2k)  No action needed.
	  For numbers like 10, 1000,              -  Add 0.5
              In general add    ((N+0.5)*log(N)
(You do see that when N=10^m  log(N) is an integer so things become easier)

[b] Step 4 [/b]- Now all you have to do is to take antilog of the number above (and keep the first digit only)
(IOW if the expression is between  0 and 0.3 the answer is 1, if it is between .3 and .477 the answer is 2 etc..  if you know your logs you dont even have to use a calculator)

     (In our example:            10^(.97) >  9 .. so answer is 9

Hope you find this enchanting
[i]Warning dont misuse this for smaller number, use it only for n>=10)
[/i]
You want to check for 10!

We have   0.65+.4+.5 = 0.55 and 10^.55  is between 3 and 4;  so, answer is 3

HTH
 :)

Thank You.Your knowledge about vedic mathematics is great.This method works fine.But is there any proof  ?

Sriram

It follows from Stirling's approximation $n! \sim n^n e^{-n} \sqrt{2 \pi n}$.