AMC
USA(J)MO
USAMO
number theory
modular arithmetic
geometric sequence
arithmetic sequence

What is the series/sequence?  thanks

7, 14, 21, 28, 4, 11, 18, 25, 4

it kind of looks like two separate arithmetic sequences...not sure about that last 4 though.

It stilll goes on, its 7, 14, 21, 28, 4, 11, 18, 25, 4,...
Do you have any ideas on what it might be?  thanks

Look at the numbers in a column of a calendar.

thanks, that seems to make sense.  but how could i write it out?

Why can't it (simply :)) be the columns of a calendar?  Look at this year's calendar, starting with January.

Or... might it be a series of USAMO scores? :D

[quote="xxreddevilzxx"][quote="Hikaru79"]It's simply an arithmetic sequence of the form $\displaystyle t_{n}=(7+(n-1)7) \mod{31}$[/quote]

isnt that simply $t_{n}=7n \mod{31}$?[/quote]
Well, because it's easier to see the general form of an arithmetic sequence if you keep it in the $\displaystyle t_{n} = a + (n-1)d$ form. Simplifying the $\displaystyle 7+(n-1)7$ part is trivial. At least, that's my reasoning =/

[quote="jb05"] Simply put, the previous two posts are both wrong because of the last 4.[/quote]
Oh... oops  :blush: Yeah, see, I was afraid of making mistakes like that. Ionno, I guess I was so excited about finding what looked like a pattern that I didn't notice that the last term doesn't fit.

Back to the drawing board  :)

[quote="jb05"]Mathematicians do so love to say simply! Simply put, the previous two posts are both wrong because of the last 4. All I can determine is that it isn't any modular arithmetic or geometric sequence, because the four indicates that it repeats, which it can't. And it can't be the columns of a calendar unless there is a month with 32 days.[/quote]

If you look at dates at weekly intervals from Jan. 1.

1,8,15,22,29,5,12,19,26,5,12,19,26,2,9,16,23,30,7,14,21,28,
           4,11,18,25,2,9,16,23,30,6,13,20,27,3,10,17,24,1,8,15,22,29,
           5,12,19,26,3,10,17,24,31[b],7,14,21,28,4,11,18,25[/b],4,11,18,25

BTW, this website is useful to patterns: http://www.research.att.com/~njas/sequences/

thanks for all the help.  the answer is the days of the calender.  there's not a really a sequence or series.   again, thanks


hi! I would like to change the header text of the toc and toc of figures from capital letters to small letters, any ideas??.

thanks in advance!!!!!

try
\thispagestyle{fancy}
\lhead{}
\chead{}
\rhead{}
\lfoot{}
\cfoot{}
\rfoot{}

and put what you want in header with the package "fancyhdr"


algorithm
algebra
polynomial
number theory
greatest common divisor

if $ gcd(a,b)\equal{}1$, prove that $ gcd(a\plus{}b, a^2\minus{}ab\plus{}b^2)\equal{}$1 or 3

Thanks.

Are you familiar with how to use the Euclidean algorithm on polynomials?  Just divide with respect to one variable.

In detail, $ a^2 \minus{} ab \plus{} b^2 \equal{} (a \plus{} b)q \plus{} r$.
Then let $ a \equal{} \minus{} b$ we have $ r \equal{} 3b^2$

So the $ gcd \equal{} gcd(a \plus{} b,3b^2) \equal{} gcd(a \plus{} b,3) \equal{} 1$ or $ 3$ because $ gcd(a \plus{} b,b^2) \equal{} 1$

Again this should be moved to High School Basics.

$ gcd(a\plus{}b,a^2\minus{}ab\plus{}b^2)\equal{}gcd(a\plus{}b,(a\plus{}b)^2\minus{}3ab)\equal{}gcd(a\plus{}b,3ab)$. But I have proved [url=http://www.mathlinks.ro/viewtopic.php?t=271863]here[/url] that $ gcd(a\plus{}b,ab)\equal{}1$. So $ gcd(a\plus{}b,3ab)\equal{}1$ or 3. qed

You are looking for the $ gcd$ of 2 homogeneous cyclotomic polynomials at the point $ (a, \minus{} b)$: $ \prod_{d|n} \Phi_d(a,b) \equal{} a^n \minus{} b^n, a \plus{} b \equal{} \Phi_1(a, \minus{} b), a^2 \minus{} ab \plus{} b^2 \equal{} \Phi_3(a, \minus{} b)$.
A standard result about 'normal' cyclotomic polynomials is that the $ gcd$ of $ \Phi_n(a), \Phi_m(a)$ is either 1 or a prime $ p$ (and in the 2nd case, $ \frac {n}{m}$ is a power of $ p$). The proof probably can be generalized to the homogeneous polynomials (with some restrictions such as $ (a,b)\equal{}1$), and it immediately implies the claim.

We can also use the Lifting the Exponent lemma, which is -very- related to cyclotomic polynomials. Let $ a \equal{} x, \minus{} b \equal{} y$. Then $ p|x \minus{} y, p \neq 3 \implies p \nmid \frac {x^3 \minus{} y^3}{x \minus{} y}$. If $ p \equal{} 3$, $ 3\parallel{} \frac {x^3 \minus{} y^3}{x \minus{} y}$. It implies the result.


Let $S_n=\sum_{i=1}^n \frac{1}{\sqrt{i}}$.
Prove: $18 < S_{100} < 19$ by computing only 1 square.
Tony

Though there are some small mistakes in your post (n instead of i in first line), you perfecltly found the point. Not exactly as I did, shorter.
In fact, no computing of square root is required ! Since $\sqrt{101}>10$.
Good !
Tony

thanks... i've found the first line completely unnecessary  :D hope those mistakes can easily be found and understand by other people


For any positive integer n, prove that number of positive integer using only the digits 1,3 and 4 , whose digit-sum is 2n is a perfect square.

I think this may be well-known.  (At least, it seems vaguely familiar to me.)

Define $G(n)$ to be the number of integers whose digits are 1, 3 and 4 and whose digit-sum is $n$.  We wish to show that $G(2n)$ is a perfect square for all $n$.  Let $F_{n}$ denote the $n$th Fibonacci number, so $F_{0}= F_{1}= 1, F_{2}= 2$, and so on.  We compute directly that $G(1) = 1 = F_{0}\cdot F_{1}$, $G(2) = 1 = F_{1}^{2}$, $G(3) = 2 = F_{1}\cdot F_{2}$ and $G(4) = 4 = F_{2}^{2}$.  I claim that this pattern continues.

It is obvious that $G(n+4) = G(n+3)+G(n+1)+G(n)$, enumerating by removing the last digit.  If $n= 2k$ is even, by inductive hypothesis we have $G(n+4) = G(2k+3)+G(2k+1)+G(2k) = F_{k+1}\cdot F_{k+2}+F{k}\cdot F_{k+1}+F_{k}^{2}= F_{k+1}\cdot F_{k+2}+F_{k}\cdot F_{k+2}= F_{k+2}^{2}$, as desired.  If $n = 2k+1$ is odd, $G(n+4) = G(2k+4)+G(2k+2)+G(2k+1) = F_{k+2}^{2}+F_{k+1}^{2}+F_{k}\cdot F_{k+1}= F_{k+2}^{2}+F_{k+1}\cdot F_{k+2}= F_{k+2}\cdot F_{k+3}$, as desired, so we're done.

you are right, i think  :wink: This is my solution:

Define $u_{n}$ to be the number of positive integer numbers whose digits are 1,3,4 and digit-sum is $2n$. We have( by remove last digit ,so) $u_{1}=1,u_{2}=4$ and $u_{n+1}=u_{n-1}+2\sum_{k=0}^{[\frac{n-1}{2}]}u_{n-2k-1}+\sum_{k=0}^{[\frac{n-2}{2}]}u_{n-2k-2}+\sum_{k=0}^{[\frac{n}{2}]}u_{n-2k}$, and problem can soved but not very easy!

Your solution nice more than my solution because :
[quote="JBL"]
Define $G(n)$ to be the number of integers whose digits are 1, 3 and 4 and whose digit-sum is $n$. [/quote]


Let $ a,b,c\geq0$.Prove that:
$ \sqrt{\dfrac{a}{8b\plus{}c}}\plus{}\sqrt{\dfrac{b}{8c\plus{}a}}\plus{}\sqrt{\dfrac{c}{8a\plus{}b}}\geq1$

I don't think it is easy to find a direct proof. :wink:

You can see at $ Easy(3)$ or see at this link:
http://www.mathlinks.ro/viewtopic.php?t=150022

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ \sqrt {\dfrac{a}{8b \plus{} c}} \plus{} \sqrt {\dfrac{b}{8c \plus{} a}} \plus{} \sqrt {\dfrac{c}{8a \plus{} b}}\geq1$[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=194955]See here[/url]
:)


Using the amsmath package, how can you have a matrix inside an equation? \begin{align*}\begin{bmatrix}\end{bmatrix}\end{align*} gives errors.

edit: and [code]\align
\iff \begin{bmatrix} 1 & -5 & 0 & 1\\ -1 & 7 & 0 & 1\\ -3 & 8 & 0 & h \end{bmatrix} &\~ \begin{bmatrix} 1 & -5 & 0 & \\ 0 & 1 & 0 & 1\\ 0 & -7 & 0 & (h+3) \end{bmatrix}\\
&\~ \begin{bmatrix} 1 & -5 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & -\frac{h+3}{7} \end{bmatrix}\\
&\~ \begin{bmatrix} 1 & 0 & 0 & 6\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & -\frac{h+3}{7} \end{bmatrix}
\endalign[/code] takes it from 100 errors to 4, but it still doesn't work...

Assuming that you just mistyped \begin{align*} and \end{align*}, the errors come from using \~ which makes no sense as it is used for accents as in $\~n$.

If you just want ~ then replace \~ by \sim to get
[code]\begin{align*}
\iff \begin{bmatrix} 1 & -5 & 0 & 1\\ -1 & 7 & 0 & 1\\ -3 & 8 & 0 & h \end{bmatrix} &\sim \begin{bmatrix} 1 & -5 & 0 & \\ 0 & 1 & 0 & 1\\ 0 & -7 & 0 & (h+3) \end{bmatrix}\\
&\sim \begin{bmatrix} 1 & -5 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & -\frac{h+3}{7} \end{bmatrix}\\
&\sim \begin{bmatrix} 1 & 0 & 0 & 6\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & -\frac{h+3}{7} \end{bmatrix}
\end{align*}[/code]
\begin{align*}\iff \begin{bmatrix}1 &-5 & 0 & 1\\-1 & 7 & 0 & 1\\-3 & 8 & 0 & h \end{bmatrix}&\sim \begin{bmatrix}1 &-5 & 0 & \\ 0 & 1 & 0 & 1\\ 0 &-7 & 0 & (h+3) \end{bmatrix}\\ &\sim \begin{bmatrix}1 &-5 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 &-\frac{h+3}{7}\end{bmatrix}\\ &\sim \begin{bmatrix}1 & 0 & 0 & 6\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 &-\frac{h+3}{7}\end{bmatrix}\end{align*}


superior algebra
superior algebra unsolved

show that (X) is not maximal in$ %Error. "display" is a bad command.
\mathbb{Z}$[x].

1. If it would be maximal then $ \mathbb Z[X]/(X)$ should be a field, which is not the case.

2. Take for example $ \mathbb Z[X]\supsetneq(2,X)\supsetneq (X)$.


A book with 50 pages numbered 1 through 50 has its pages renumbered in reverse, from 50 to 1. For how many pages do both sets of page numbers share the same ones digit?

The numbers add up to 51 so one is even, the other is odd. Thus unit digit cannot be shared. Thus, 0.


symmetry
inequalities proposed
inequalities

\[ 1 \le a,b,c \le 2
\]àind the maximum
\[ (a \plus{} b \plus{} c)(\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c})
\]

[quote="Lanyes"]
\[ 1 \le a,b,c \le 2
\]
àind the maximum
\[ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]

[hide="hint"]If f(x) is convex and $ a\le x\le b\Rightarrow$ Maximum of f(x) is f(a) or f(b).[/hide]

[quote="Lanyes"]
\[ 1 \le a,b,c \le 2
\]
àind the maximum
\[ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]

use Kantorovich inequality...easy to know
<=(1+2)^2/(4*1*2)*(3)^2=81/8
 :)

I think the maximum is 10
We assume\[ 1 \le a \le b \le c \le 2
\]
\[ \begin{array}{l}
 (2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge 0 \\
  \Rightarrow 8 \minus{} 4(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \minus{} abc \ge 0 \\
  \Rightarrow 8 \minus{} 4(a \plus{} b \plus{} c) \plus{} 2abc(\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}) \minus{} abc \ge 0(1) \\
 X \equal{} a \plus{} b \plus{} c;Y \equal{} \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \\
 (1) \Rightarrow 8 \minus{} 4X \plus{} 2abcY \minus{} abc \ge 0 \\
  \Rightarrow abc(2Y \minus{} 1) \minus{} 4(X \minus{} 2) \ge 0 \\
  \Rightarrow abc(2Y \minus{} 1) \ge 4(X \minus{} 2) \\
 \end{array}
\]
(because X=a+b+c\[ \ge
\]3
\[ \begin{array}{l}
 Then \\
 (1 \minus{} \frac{1}{a})(1 \minus{} \frac{1}{b})(1 \minus{} \frac{1}{c}) \ge 0 \\
  \Leftrightarrow 1 \minus{} Y \plus{} \frac{{X \minus{} 1}}{{abc}} \ge 0 \\
  \Leftrightarrow \frac{{X \minus{} 1}}{{abc}} \ge Y \minus{} 1 \\
  \Rightarrow \frac{{X \minus{} 1}}{{abc}}abc(2Y \minus{} 1) \ge 4(Y \minus{} 1)(X \minus{} 2) \\
  \Leftrightarrow (X \minus{} 1)(2Y \minus{} 1) \ge 4(Y \minus{} 1)(X \minus{} 2) \\
  \Rightarrow XY \le \frac{3}{2}(X \plus{} 2Y) \minus{} \frac{7}{2} \\
 Becau{\rm{s}}e1 \le a \le 2 \\
  \Rightarrow (a \minus{} 1)(a \minus{} 2) \le 0 \Rightarrow a \plus{} \frac{2}{a} \le 3 \\
 b \plus{} \frac{2}{b} \le 3;c \plus{} \frac{2}{c} \le 3 \\
  \Rightarrow a \plus{} b \plus{} c \plus{} 2(\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}) \le 9 \Leftrightarrow X \plus{} 2Y \le 9 \\
  \Rightarrow XY \le \frac{3}{2}.9 \minus{} \frac{7}{2} \equal{} 10 \\
 \end{array}
\]

[quote="Lanyes"]
\[ 1 \le a,b,c \le 2
\]
àind the maximum
\[ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]
Let $ f(a,b,c) \equal{} (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} 3 \plus{} \sum_{sym}\frac {a}{b}$. Then $ \frac {\Delta f(a,b,c)}{\Delta a} \equal{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {b}{a^2} \minus{} \frac {c}{a^2}$. So $ \frac {\Delta^2 f(a,b,c)}{\Delta^2 a} \equal{} \frac {2b}{a^3} \plus{} \frac {2c}{a^3} > 0$. By similar arguments it is convex in $ a$, $ b$ and $ c$. So either $ f(1,b,c) \ge f(a,b,c)$ or $ f(2,b,c) \ge f(a,b,c)$. Because of symmetry we only have to consider the numbers: $ f(1,1,1) \equal{} 9$, $ f(2,1,1) \equal{} 10$, $ f(2,2,1) \equal{} 10$ and $ f(2,2,2) \equal{} 9$. So the max is $ 10$ reached when $ (a,b,c) \equal{} (2,1,1)$ or $ (a,b,c) \equal{} (2,2,1)$ or any permutations.

A generalisation:
Let $ 1 \le a_1,a_2,...,a_n \le 2$
Then the max of
$ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} .. \plus{} \frac {1}{a_n})$
is $ \frac {9n^2}{8}$ for even $ n$ and $ \frac {9n^2 \minus{} 1}{8}$ for odd $ n$.

PS: Ye109: How can we reach $ \frac{81}{8}$ :S

[quote="Mathias_DK"][quote="Lanyes"]
\[ 1 \le a,b,c \le 2
\]
àind the maximum
\[ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]
Let $ f(a,b,c) \equal{} (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} 3 \plus{} \sum_{sym}\frac {a}{b}$. Then $ \frac {\Delta f(a,b,c)}{\Delta a} \equal{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {b}{a^2} \minus{} \frac {c}{a^2}$. So $ \frac {\Delta^2 f(a,b,c)}{\Delta^2 a} \equal{} \frac {2b}{a^3} \plus{} \frac {2c}{a^3} > 0$. By similar arguments it is convex in $ a$, $ b$ and $ c$. So either $ f(1,b,c) \ge f(a,b,c)$ or $ f(2,b,c) \ge f(a,b,c)$. Because of symmetry we only have to consider the numbers: $ f(1,1,1) \equal{} 9$, $ f(2,1,1) \equal{} 10$, $ f(2,2,1) \equal{} 10$ and $ f(2,2,2) \equal{} 9$. So the max is $ 10$ reached when $ (a,b,c) \equal{} (2,1,1)$ or $ (a,b,c) \equal{} (2,2,1)$ or any permutations.

A generalisation:
Let $ 1 \le a_1,a_2,...,a_n \le 2$
Then the max of
$ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} .. \plus{} \frac {1}{a_n})$
is $ \frac {9n^2}{8}$ for even $ n$ and $ \frac {9n^2 \minus{} 1}{8}$ for odd $ n$.

PS: Ye109: How can we reach $ \frac {81}{8}$ :S[/quote]
e...i am wrong the equal conditon can't reached. :(

See [url=http://www.mathlinks.ro/viewtopic.php?t=154015]here[/url] for some generalizations of this.


We are to set $b_n=\frac{1}{2^{n}}\tan \theta _{n}.$


linear algebra
matrix
vector
quadratics
geometry
geometric transformation
reflection

Hello,

I'm sorry for this rather simple question, but I'm quite confused now. :oops:

I was discussing commuting hermitian matrices and the theorem that said that they were simultaneously diagonizable.
We were wondering about the relevance of the matrices being hermitian.

I found a simple example... but one of the two was not diagonizable at all!

So this is my question : let $A$ and $B$ be diagonizable matrices over some field, and let $A B = BA$ .  Can an invertible matrix $Q$ be found such that both $Q^{-1}A Q$ and $Q^{-1}B Q$ are diagonal matrices?

My guess is yes... :|
Please enlighten me  :blush:

Seems, question consists answer in 'let $A B = BA$', naturally.

The talk about the same eigen vectors of two "different" (only by eigen values) matrices. Common reasons: seen, you are free to commutate Any two diagonal matrices $Dd = dD$, if $A$ and $B$ consists the same basis of eigen vectors (columns of Q in $A=Q D Q^{-1}$, $B=Q d Q^{-1}$), it follows (put) $AB = BA$.

PS
There is similar (and famous theorem, used in control system theory) for Any two symmetric matrices and existance of the same transformation for both of them $A=Q D Q^{T}$, $B=Q d Q^{T}$ (leads two quadratic curves to a simple canonical form), where there is a more big freedom to choose Q. Remembers.

[quote="fredbel6"]I'm sorry, are you saying that my "theorem" is true or not? I did not completely get that? :maybe:[/quote]

In what u've not got? In your turn? :) Yes, and in taken prepositions $Q$ is matrix of eigen vectors of diagonalizable (by determination) matrices $A$ or $B$, the same, no difference. It's ivertable.

[quote="Kent Merryfield"]First hint at a proof: suppose $\lambda$ is an eigenvalue of $A$ and let $V_{\lambda}$ be the eigenspace corresponding to that eigenvalue.

Can you prove that $B$ maps $V_{\lambda}$ into $V_{\lambda}?$[/quote]

Well known eigen vectors free taken in the own eigenspace, if the question to me - let everybody put these details to case of equal to eath-other eigen values more preciously with choosen terminology (I less interested in these excercises, it doesn't reflect to result, where were observed only different to eath-other eigen values, why to make a task more complicated then it's? To use eigenspace have to be reasons too).

[quote="Kent Merryfield"]First hint at a proof: suppose $\lambda$ is an eigenvalue of $A$ and let $V_{\lambda}$ be the eigenspace corresponding to that eigenvalue.

Can you prove that $B$ maps $V_{\lambda}$ into $V_{\lambda}?$[/quote]
Yes I think I can! :D
But how can I proceed then?  It would be nice if I could create a basis of eigenvectors for the restriction of $B$ to that space $V_{\lambda}$.

By assumption, I know that $B$ can be diagonalized, but does that mean that that restriction can be diagonalized as well? :|

Extra question : what if I replace "matrices" with linear operators in general vectorspaces, possibly not finitedimensional? What then?

Seems, discussion came out of invertable "matrix" $Q$ :)

Let's go from $A$ and $B$ have the same common basis, therefore it need to build such procedures, to find it observing accurate equations, is the tactical question, seems to be more boring then really interesting. To build need a stimulus (a stimul, in Russian) :)

Not finitedimensional.. mmmm.. linear operators of dynamic systems, they have more often "jordan chains" with absence of "eigen vector" and "eigen value" - convolution operator, for example. Me interesting these generalisations (with 'zeroed' eigen value as a 'main meaning'), seems, this area is free to offers. But it's far out of computational problems.


show that the union of any family of closed disks in R ^2 of positive radii is lebesgue measurable.

Nice problem, isn't it? Well, while using internet is not strictly prohibited, I would prefer you to do it on your own. So, you'll have to wait until May 14 ;)

I'd like to see the solution of this one, tried myself but no luck


This is the first problem from the Tipler book I am having trouble with. It is from the chapter 4, on Newton's laws.

1)Two men pull with equal force on the ends of a rope. A weight w is suspended at the center of the rope.

a)For any given angel theta made by the rope with the horizontal axis, find the tension in the rope.
b)If the mass of each man is m, what force must be exerted by the ground on the man to keep him stationary
c)Evaluate parts a and b for $\theta=5$ degrees, $w=10 lb$, and $mg=160 lb$.
d) Is it possible for the men to hold the weight suspended with the rope horizontal? Is it possible for them to pull in such a way that the rope is horizonal momentarily?

[quote="elchuy"]Hi! :)

Diagram of forces:

[img]http://img149.imageshack.us/img149/193/diag7bg.png[/img]

Like the weigh is static, sum of forces is zero:

$\vec{F_1}+\vec{F_2}+\vec{W}=0$

Forces in $y$ axis:

$F_{1y}+F_{2y}-W=0$

We know that:

$F_1=F_2=T$ (Tension)

Then $2F_y-W=0$, $F_y=T\sin{\theta}$, so $2T\sin{\theta}=W$

$T=\frac{W}{2\sin{\theta}}$

b)Let $\vec{N}$ the force exerted by the ground on the man. For man keep him stationary, sum of forces must be zero:

$\vec{N}+\vec{W}_{man}+\vec{F}_{2r}=0$

$\vec{F}_{2r}=-\vec{F_2}$ (For third Newton law)...

d)
$\lim_{\theta \to 0}T=\lim_{\theta \to 0}\frac{W}{2\sin{\theta}}=\infty$, Therefore is necessary infinite force for to hold the weight suspended with the rope horizontal.

I hope that it helps you :D

:bye:[/quote]Yes, in theory it would take infinite force...

But take a small object, like a ring or pencil, and a piece of string, and try it. It doesn't take infinite force for me...

first, while it may look straight, there will always be deformation, especially because the string is not perfect and has some mass.

second, even if $\theta$ is so small as to be unnoticable, a ring or a pencil would have such a small $W$ that generating a force $T$ corresponding to such a $\theta$ would be really easy.

third, idealizations are nice.  ugly, tedious statics calculations are not much fun. :)

Ok, thank you. I had it all worked out, but when I get wierd answers it bothers me :(

You really helped. Thank you.

[quote="elchuy"]This principle could help us in difficult situations.

If we go in the field and our van remains clogged in the mud, we could remove it we ourself with a rope.

[img]http://img92.imageshack.us/img92/6606/imagen0028ww.png[/img]

Between longer the rope is better.  ;)

:bye:[/quote]
Could you explain this more???


Από όσο ξέρω σε μια συνεπαγωγή "Αν p τότε q" ισχύουν τα εξής:
Αν p=Αληθής και q=Αληθής τότε η συνεπαγωγή είναι αληθής
Αν p=Αληθής και q=Ψευδής τότε η συνεπαγωγή είναι ψευδής
Αν p=Ψευδής και q=Αληθής τότε η συνεπαγωγή είναι αληθής
Αν p=Ψευδής και q=ψευδής τότε η συνεπαγωγή είναι αληθής

Τα 2 πρώτα τα καταλαβαίνω. Δεν καταλαβαίνω τα 2 τελευταία. Γιατί ορίζεται η συνεπαγωγή έτσι ώστε αν η υπόθεση είναι ψευδής τότε η συνεπαγωγή να είναι αληθής. Θα περίμενα να μην μπορούμε να αποφανθούμε για την αλήθεια της συνεπαγωγής λόγω έλλειψης στοιχείων.
Σίγουρα με μια τέτοια πρόταση κανείς δεν είπε ψέματα, αλλά αυτό δε σημαίνει ότι είπε αλήθεια. Θα έλεγα ότι μια συνεπαγωγή με ψευδή υπόθεση....  ας πούμε δεν μου φαίνεται καλά ορισμένη, με την έννοια ότι δε μας δίνει αρκετά στοιχεία για να ελέγξουμε την αλήθεια της.

[b]Υπάρχει κάποιος βαθύτερος λόγος για να επιλέγουμε κάτι τέτοιο (πέρα από το να λέμε απλώς ότι έτσι το ορίζουμε);
[/b]
Συγκεκριμένο παράδειγμα: Αν ο ήλιος χθες ανέτειλε από τη Δύση τότε εγώ αύριο δε θα πάω στη δουλειά μου.

Ο ήλιος όμως χθες δεν ανέτειλε από τη Δύση. Άρα ότι και να κάνω δε λέω ψέματα. Αλλά νομίζω ότι δε λέω ούτε αλήθεια. Βασικά δεν μπορώ να ελέγξω αν είπα αλήθεια λόγω ψευδούς υπόθεσης. Σα να αντιμετωπίζουμε με μη συμμετρικό τρόπο το αληθής και το ψευδής.

Ψευτης στη ζωη αποκαλεισε οταν αθετης μια υποσχεση π.χ λες:

Εαν παω λονδινο θα αγορασω αυτοκινιτο.ΤΩΡΑεχουμε 4 περιπτωσεις.

1) πας λονδινο και αγοραζεις αυτοκινητο. Τοτε αποκαλεισε [b]αληθης[/b]

2)πας λονδινο και δεν αγοραζεις αυτοκινητο.Τοτε αποκαλεισε [b]ψευτης[/b]

3) [b]Δεν πας λονδινο,αλλα αγοραζεις αυτοκινητο[/b] .Πως αποκαλεισε ? [b]ΨΕΥΤΗΣ???[/b].ΤΟΤΕ η αρχικη σου υποσχεση χανει την αξια της ,διοτι ειτε πας λονδινο ειτε οχι αρκει να αγορασεις αυτοκινητο ΓΙΑ να αποκαλεστεις [b]ΨΕΥΤΗΣ[/b]

4) Δεν πας λονδινο ουτε αγοραζεις αυτοκινητο.ΜΠΟΡΕΙΣ ΝΑ ΑΠΟΚΑΛΕΣΘΕΙΣ [b] ΨΕΥΤΗΣ??[/b],ΤΟΤΕ Μπορει να ελθη ο αδελφος σου σπιτι και χωρις λογο ( =υποσχεση) και αιτια και να σε αποκαλεσει ψευτη :

ΓΙΑΤΙ ΔΕΝ ΑΓΟΡΑΣΕΣ ΑΥΤΟΚΙΝΗΤΟ.

ΔΕΝ ΜΠΟΡΩ ΝΑ ΣΚΕΦΤΩ αλλα παραδειγματα με καταστρεπτικες συνεπειες. Αλλα καποιος με δυνατοτερο μυαλο ισως θα μπορεση.

ΒΕβαια τιποτα δεν μας εμποδειζει να αλλαξουμε θεωριτικα βεβαια τον αληθοψευδη πινακα της υποθετικης συνεπειας και να μελετησουμε τα αποτελεσματα.

ΙΣΩΣ Η ΑΛΛΑΓΗ ΝΑ ΕΙΝΑΙ ΚΑΙ ΜΟΝΟ ΖΗΤΗΜΑ ΓΛΩΣΣΑΣ.ΟΠΩΣ ΑΛΛΑΖΟΥΜΕ ΑΠΟ ΤΑ ΕΛΛΗΝΙΚΑ ΣΤΑ ΑΓΓΛΙΚΑ

[quote="τρικλινο"]

3) [b]Δεν πας λονδινο,αλλα αγοραζεις αυτοκινητο[/b] .Πως αποκαλεισε ? [b]ΨΕΥΤΗΣ???[/b].ΤΟΤΕ η αρχικη σου υποσχεση χανει την αξια της ,διοτι ειτε πας λονδινο ειτε οχι αρκει να αγορασεις αυτοκινητο ΓΙΑ να αποκαλεστεις [b]ΨΕΥΤΗΣ[/b]

4) Δεν πας λονδινο ουτε αγοραζεις αυτοκινητο.ΜΠΟΡΕΙΣ ΝΑ ΑΠΟΚΑΛΕΣΘΕΙΣ [b] ΨΕΥΤΗΣ??[/b],ΤΟΤΕ Μπορει να ελθη ο αδελφος σου σπιτι και χωρις λογο ( =υποσχεση) και αιτια και να σε αποκαλεσει ψευτη
[/quote]

Φοβάμαι πως δε με καλύπτει αυτή η απάντηση. Δες το γιατί: Βασικά δε διαφωνώ με αυτό που γράφεις. Κανείς δεν μπορεί να πει ότι είπες ψέματα. Μέχρι εδώ συμφωνούμε.

Αυτό που λέω είναι ότι (αν και δεν είπες ψέματα) κανείς δεν μπορεί να πει ότι είπες αλήθεια. Κανείς δεν μπορεί να βγάλει συμπέρασμα ότι είσαι ειλικρινής για παράδειγμα.
Αυτό που λέω είναι ότι κατά τη δική μου κατανόηση κανείς δεν μπορεί να πει αν είπες αλήθεια ή ψέματα λόγω έλλειψης στοιχείων. Μόνο αν είχες πάει στο Λονδίνο θα μπορούσε.
Ζητάω δηλαδή κάτι παραπάνω από το "έτσι ορίστηκε η αλήθεια της συνεπαγωγής". Θα ήθελα να μάθω αν υπάρχει κάποιος βαθύτερος λόγος που μου διαφεύγει.
Thanks πάντως  :)

Yποτεθεστω οτι αλλαζουμε το:

Aν p=Ψευδησ και q=Ψευδες τοτε η σενεπαγωγη ειναι Αληθης [b]και το κανουμε[/b]:

Av p=Ψευδης και q=Ψευδης τοτε η συνεπαγωγη ειναι [b]Ψευδης[/b].

Τωρα υπαρχει στην λογικη ο κανονας που λεγεται [b]Κανονας Αποσπασεως (M.Ponens)[/b] o οποιος συμβολικα ειναι¨

$ [(p\Longrightarrow q)\wedge p]\Longrightarrow q$  .

ΕΠΙΣΗΣ ξερουμε απο την λογικη οτι¨

[b]Καθε κανονας της λογικης ειναι μια ταυτοτητα και καθε ταυτοτητα ειναι νομος της λογικης[/b]

Με την αλλαλη που εχουμε κανει ,[b]εαν σχηματισουμε τον πινακα αληθειας του παραπανω νομου θα δουμε οτι δεν ειναι πλεον ταυτοτητα και αρα ουτε νομος της λογικης[/b]

ΤΩΡΑ ολοι ξερουμε οτι οποιος πεσει απο το μπαλκονι του οι συνεπειες θα ειναι καταστροφικες.

Ο λογος  [b]που δεν πεφτουμε ειναι γιατι χρησιμοποιουμε τον παραπανω κανονα[/b]

ΞΕΧΩΡΑ απο αυτο [b]χωρις τον παραπανω κανονα ουτε ,σχεδον μια ασκηση στα μαθηματικα δεν θα μπορουσαμε να λυσουμε[/b]

[quote="τρικλινο"]Yποτεθεστω οτι αλλαζουμε το:

Aν p=Ψευδησ και q=Ψευδες τοτε η σενεπαγωγη ειναι Αληθης [b]και το κανουμε[/b]:

Av p=Ψευδης και q=Ψευδης τοτε η συνεπαγωγη ειναι [b]Ψευδης[/b].

[/quote]

Δεν είναι αυτή η απορία μου. Η απορία μου είναι γιατί θα πρέπει να πούμε ότι η συναπαγωγή είναι ψευδής (ή αληθής)... δηλαδή ότι έχει τιμή... και να μην πούμε "δεν μπορούμε να αποφανθούμε για το αν είναι αληθής ή ψευδής λόγω του ότι δεν έχουμε στοιχεία".

Η απάντηση είναι "διότι έτσι επιλέξαμε". Ας προσπαθήσω όμως να πω λίγα περισσότερα:

Πολλές φορές στα μαθηματικά χρησιμοποιούμε έμμεσες αποδείξεις. Για παράδειγμα, για να δείξουμε ότι η $ \sqrt{2}$ είναι άρρητος, υποθέτουμε πως είναι ρητός και καταλήγουμε σε άτοπο. Για να επιτρέπεται να το κάνουμε αυτό πρέπει να έχουμε ένα κανόνα που να λέει ότι για κάθε πρόταση Ρ, είτε η Ρ είναι ορθή, είτε η άρνηση της Ρ είναι ορθή. Με άλλα λόγια είναι επιθυμητό στα μαθηματικά κάθε πρόταση να είναι είτε ψευδής είτε αληθής.

Αυτή είναι η "συνηθισμένη" λογική που χρησιμοποιούμε στα μαθηματικά. Η στάση της πλειοψηφίας των μαθηματικών είναι πως αν απαγορέψουμε αυτό, τότε αρκετά θεωρήματα των μαθηματικών θα πάψουν να ισχύουν και αυτό είναι ανεπιθύμητο. Υπάρχουν όμως και άλλες λογικές όπως για παράδειγμα στην "intuitionistic logic" όπου ο κανόνας του αποκλειόμενου μέσου δεν βρίσκεται στα αξιώματα (και ούτε προκύπτει από αυτά).

Σίγουρα κατάλαβες Δημήτρη ποιο είναι το πρόβλημά μου. Όντως φαίνεται πως θέλουμε να έχουμε τιμή αληθείας για τη συνεπαγωγή ψευδούς υπόθεσης και για αυτό το λόγο ορίστηκε ως αληθής. Απλά είναι κάτι που κάνει το ένστικτό μου να αντιδρά έντονα. Να 2 σενάρια εντελώς αντίθετα μεταξύ τους που δείχνουν τι θέλω να πω:

Λέω στο γιο μου:
«Αν αύριο δεν βρέξει θα σε πάω στο λούνα παρκ».
Τελικά βρέχει και δεν τον πάω. Δεν είπα ψέματα. Εδώ λοιπόν το να θεωρώ αληθή τη συνεπαγωγή με ψευδή υπόθεση είναι κάτι που δε δημιουργεί πρόβλημα.

Κοίτα όμως το παρακάτω παράδειγμα:

«Αν είχα συμμετάσχει σε μαθηματικούς διαγωνισμούς όσο ήμουν μαθητής θα είχα πάρει χρυσό σε ΙΜΟ»
Αλλά δε συμμετείχα ποτέ. Και φαίνεται καθαρά ότι αυτό που λέω είναι κάτι που δεν είναι αλήθεια. Εδώ μου κακοφαίνεται ότι τελικά είπα εκ του ασφαλούς μια πρόταση και λέω η οποία θεωρείται αληθής χάρη στο ότι κρύβομαι πίσω από τον ορισμό της συνεπαγωγής.

Βασικά αυτό που μου θα περίμενα είναι να λέμε «Δεν ορίζεται η συνεπαγωγή με ψευδή υπόθεση και έτσι δεν μπορεί να χαρακτηριστεί ούτε ψευδής ούτε αληθής αφού ουσιαστικά δε λέει τίποτα». Εκεί που λέει κάτι είναι όταν η αλήθεια ή το ψεύδος της υπόθεσης μπορεί δεν μπορεί να ελεγχθεί ακόμα. Εκεί… άντε να το δεχτώ. Αλλά σε περιπτώσεις που γνωρίζουμε εκ των προτέρων ότι η υπόθεση είναι ψευδής (πχ αν χθες δεν έβρεχε ενώ ξέρουμε ότι έβρεξε) τότε δίνουμε δικαίωμα στον καθένα εκ του ασφαλούς να λέει ψέματα και μάλιστα να τα θεωρούμε από μαθηματική άποψη αληθείς προτάσεις.
Πχ
Αν έκανα προπόνηση μικρός θα έτρεχα τα 100 μέτρα σε 9 δευτερόλεπτα.
Αν έκανα προπόνηση μικρός στην άρση βαρών θα σήκωνα 500 κιλά στο αρασέ
Κλπ κλπ

Δεν μου κάθεται όμορφα αυτό… και ξέρεις ότι για μένα τα πάντα πρέπει να εναρμονιστούν τελικά τη διαίσθησή μας, έστω και με modification της διαίσθησής. Αλλά εδώ δε βλέπω τέτοια περιθώρια. Υπάρχει θέμα εδώ δε νομίζεις;

Νομιζω οτι τις καταλαβαινω διαισθητικα:

1) Αν p=Ψευδής και α=Αληθής τότε η συνεπαγωγή είναι αληθής.
2) Αν p=Ψευδής και α=ψευδής τότε η συνεπαγωγή είναι αληθής.

Για την 1, μας λεει καποιος(ο οποιος εχει παντα δικιο) οτι αν ρ τοτε α. Αν δεν γινει το ρ τοτε δεν θα ηταν λαθος να σκεφτουμε οτι το α θα γινει(ισως γινει).

Για την 2, μας λεει καποιος(ο οποιος εχει παντα δικιο) οτι αν ρ τοτε. Αν δεν γινει το ρ τοτε δεν θα ηταν λαθος να σκεφτουμε οτι ουτε το α θα γινει(ισως δεν γινει).

Οποτε πιστευω οτι τις οριζουμε αληθης επειδη δεν μπορει καποιος να μας βγαλει ψευτες αν σκεφτουμε ετσι. Τωρα βεβαια δεν μπορει κανεις στο δικαστηριο να μας βγαλει και σωστους, αλλα ποτε δεν φυλακιζεις καποιον αν δεν εισαι σιγουρος οτι ειναι ενοχος(κυριολεχτικα-εδω, ψευδης).

Διαισθητικα το καταλαβαινω ως "δεν μπορεις να κατηγορησεις καποιον ο οποιος εισαι δεν εισαι 100% σιγουρος οτι ειναι εγκληματιας-ενοχος".

Δες το εξής σενάριο:

Ο Γιώργος λέει: [b]«Αν είχα κάνει προπόνηση μικρός τώρα θα σήκωνα 1000 κιλά».[/b]
Ο Δημήτρης διαφωνεί και λέει: [b]«Αν είχες κάνει προπόνηση μικρός τώρα δε θα σήκωνες 1000 κιλά»[/b].

Επειδή ο Γιώργος δεν έκανε προπόνηση μικρός η φράση του Γιώργου είναι αληθής (ψευδής υπόθεση σε συνεπαγωγή).
Αλλά και ο Δημήτρης, αν και λέει κάτι νοηματικά αντίθετο πάλι είμαστε υποχρεωμένοι να δεχτούμε  ως αληθή την πρότασή του για τον ίδιο λόγο (συνεπαγωγή με ψευδή υπόθεση).

Τελικά ποιος λέει αλήθεια και ποιος ψέματα; Που θα χωρέσουμε τη διαίσθησή μας σε όλα αυτά αφού ο Γιώργος και ο Δημήτρης διαφωνούν και απλά θέλουν να πουν μεταξύ τους το αντίθετο;

Αυτό το μπέρδεμα δημιουργείται καθαρά επειδή θέλουμε πάντα να μπορούμε να χαρακτηρίσουμε μια πρόταση ως αληθή ή ψευδή. Μου φαίνεται πολύ πιο λογικό το να πούμε απλά «οι προτάσεις δεν μπορούν να ελεγχθούν λόγω ψευδούς υπόθεσης στη συνεπαγωγή».

Μπορεί κανείς να βοηθήσει να βγάλω άκρη;

Δημήτρη μπορείς να δώσεις ένα παράδειγμα όσο γίνεται απλό (δηλαδή να μην έχει μαθηματικά προαπαιτούμενα και να το καταλάβω) που να δείχνει ότι πρέπει να μπορούμε πάντα να χαρακτηρίσουμε μια πρόταση ως αληθή ή ψευδή. ΔΗλαδή να φαίνεται τι θα χάναμε αν δεν μπορούσαμε.

[quote="gpapargi"]
Ο Γιώργος λέει: [b]«Αν είχα κάνει προπόνηση μικρός τώρα θα σήκωνα 1000 κιλά».[/b]
Ο Δημήτρης διαφωνεί και λέει: [b]«Αν είχες κάνει προπόνηση μικρός τώρα δε θα σήκωνες 1000 κιλά»[/b].
[/quote]

Το πρόβλημα με τέτοιου είδους προτάσεις της "καθημερινής γλώσσας", είναι ότι μας βάζουν να σκεφτούμε ένα υποθετικό σενάριο. Το πρώτο κομμάτι της πρότασης σου "Αν είχα κάνει προπόνηση μικρός" φαντάζει ως ένα πολύ πιθανό σενάριο. Το δεύτερο κομμάτι "τώρα θα σήκωνα 1000 κιλά" φαντάζει κάτι απραγματοποίητο για τις ανθρώπινες δυνάμεις και ασφαλώς δεν μας αρέσει να δεχτούμε ως ορθή αυτήν την πρόταση. Αλλά και ο θυμόσοφος λαός όταν του λένε τέτοιες προτάσεις συνήθως απαντάει κάτι για την γιαγιά του. (No offence, πλάκα κάνουμε.  :P )

Στην μαθηματική λογική, όταν λέμε "Αν Α τότε Β" το Α είναι είτε ορθό, είτε λάθος και δεν μπορεί να αλλάξει ποτέ η ορθότητά του.

Αν και η πρόταση "Αν 1+1=3 τότε ισχύει η εικασία του Goldbach" είναι ορθή, σίγουρα δεν θα τη γράφαμε σε ένα μαθηματικό κείμενο. (Εκτός ίσως για σκοπούς παραδείγματος.) Ομοίως και δεν θα γράφαμε "Αν 1=1 τότε ισχύει το τελευταίο θεώρημα του Fermat". Ο λόγος είναι πως πάντα όταν λέμε "Αν Α τότε Β" έχουμε στο μυαλό μας μια συνεπαγωγή. Κάπως μπορούμε να αρχίσουμε από την πρόταση Α και να καταλήξουμε στην πρόταση Β, πράγμα που δεν φαίνεται πως μπορούμε να το κάνουμε.  (Ίσως ο Wiles θα μπορούσε να μας δείξει πως από το 1=1 καταλήγουμε στο τελευταίο θεώρημα του Fermat.  :lol: Για την πρώτη πρόταση, βλέπω ένα τρόπο να καταλήξω από το  1+1=3 στην εικασία του Goldbach  :D )

Θέλουμε όποτε η πρόταση "Αν Α τότε Β" είναι ορθή, να είναι και η πρόταση "Αν (όχι Β) τότε (όχι Α)". Νομίζω ότι κανείς δεν θα αμφισβητούσε αυτό. Ομοίως όποτε η πρόταση "Αν (όχι Β) τότε (όχι Α)" είναι ορθή, τότε θέλουμε και η πρόταση "Αν (όχι όχι Α) τότε (όχι όχι Β)" να είναι ορθή. Αυτή όμως η πρόταση είναι η ίδια με την πρόταση  "Αν Α τότε Β". (Υπάρχουν όμως λογικές στις οποίες οι προτάσεις "Α" και "όχι όχι Α" δεν είναι ισοδύναμες.)

Τι συμβαίνει τώρα αν οι Α και Β είναι λανθασμένες; Αν δεκτούμε την "Αν (όχι Β) τότε (όχι Α)" ως ορθή, τότε θα πρέπει να δεκτούμε και την "Αν Α τότε Β" ώς ορθή. (Φυσικά, στην περίπτωση "Αν 1=1 τότε ισχύει το τελευταίο θεώρημα του Fermat" δεν είμασταν απόλυτα ευχαριστημένοι με την ορθότητα της πρότασης, αλλά δεν μπορούμε να μην την δεκτούμε ως ορθή. Άλλωστε θα μπορούσε ο Wiles σε κάποια πρόταση στην απόδειξή του να πει και αφού 1=1, τότε ...)

[quote="Demetres"]
Το δεύτερο κομμάτι "τώρα θα σήκωνα 1000 κιλά" φαντάζει κάτι απραγματοποίητο για τις ανθρώπινες δυνάμεις και ασφαλώς δεν μας αρέσει να δεχτούμε ως ορθή αυτήν την πρόταση. Αλλά και ο θυμόσοφος λαός όταν του λένε τέτοιες προτάσεις συνήθως απαντάει κάτι για την γιαγιά του. (No offence, πλάκα κάνουμε.  :P )
[/quote]

Εννοείται.   :)   Κι εγώ όταν το έγραφα τη γιαγιά είναι στο νου μου αλλά είπα να μην το γράψω για ευνόητους λόγους. Αν και τελικά υπάρχει και το "αν είχα ρόδες θα ήμουν πατίνι"... αλλά μου φαίνεται πως ήμαστε διεστραμμένοι και σκεφτόμαστε αυτό με τη γιαγια  :lol:

Όσο αφορά το διαισθητικό κομμάτι, με ενδιαφέρουν μόνο οι λογικές που δέχονται ότι όχι(όχι(Α))=Α καθώς και την αντιθετοαντιστροφή δηλαδή ότι το
«Αν π1 τότε π2» είναι ισοδύναμο με το «Αν όχι(π2) τότε όχι (π1)». Οπότε μπορούμε να περιορίσουμε την κουβέντα μόνο σε αυτές.

Ας θέσουμε συγκεκριμένες προτάσεις για να το κάνουμε σαφέστερο:
Π1: «Έξω βρέχει»
Π2: «Θα πάρω ομπρέλα»
Οπότε «Αν π1 τότε π2» γίνεται «Αν έξω βρέχει τότε θα πάρω ομπρέλα»
Η ισοδύναμη αντιθετοαντίστροφη είναι «Αν δεν πάρω ομπρέλα τότε έξω δεν βρέχει»

Αν η π1 είναι αληθής και η π2 είναι αληθής τότε η πρόταση «Αν π1 τότε π2» είναι αληθής όπως και η «Αν όχι(π2) τότε όχι (π1)».

Εδώ κάποιος θα μπορούσε να πει (δεν ξέρω αν λες αυτό Δημήτρη) ότι στην αντιθετοαντίστροφη η π2 είναι ψευδής και η π1 επίσης ψευδής. Ωστόσο αυτό που για μένα έχει σημασία είναι ότι μεταξύ του αν και του τότε (δηλαδή πριν το βέλος της συνεπαγωγής) υπάρχει μια αληθής πρόταση... η όχι(π2). Το ίδιο και μετά.

Με άλλα λόγια αν βάλεις κάτι ψευδές ανάμεσα στο «Αν» και στο «τότε» δε βλέπω να βγαίνει κάτι που να έχει νόημα. Και βλέπω ότι γενικά από αυτά που έγραψες ότι κάποιες συνεπαγωγές δεν τις πολυχρησιμοποιούμε ή δε μας αφήνουν ευχαριστημένους. Δε μου κάθεται καλά αυτή η κατάσταση. Μου αφήνει έντονα μια γεύση ότι αυθαιρετούμε.

[quote="gpapargi"]
Ας θέσουμε συγκεκριμένες προτάσεις για να το κάνουμε σαφέστερο:
Π1: «Έξω βρέχει»
Π2: «Θα πάρω ομπρέλα»
Οπότε «Αν π1 τότε π2» γίνεται «Αν έξω βρέχει τότε θα πάρω ομπρέλα»
Η ισοδύναμη αντιθετοαντίστροφη είναι «Αν πάρω ομπρέλα τότε έξω βρέχει»
[/quote]

Δεν έχω παρακολουθήσει ιδιαιτέρως τη συζήτηση, αλλά εδώ μου φαίνεται ότι κάνεις λάθος. Το αντιθετοαντίστροφο της πρότασης: "Αν έξω βρέχει τότε θα πάρω ομπρέλα" είναι "Αν δεν πάρω ομπρέλα, τότε έξω δε βρέχει".

Αυτό που έγραψες παραπάνω είναι το αντίστροφο και όχι το αντιθετοαντίστροφο.

Cheerio,

Durandal 1707

Thanks Pan. Ήταν λαθος από βιασύνη. Το διόρθωσα.

[quote="gpapargi"]
Αν η π1 είναι αληθής και η π2 είναι αληθής τότε η πρόταση «Αν π1 τότε π2» είναι αληθής όπως και η «Αν όχι(π2) τότε όχι (π1)».

Εδώ κάποιος θα μπορούσε να πει (δεν ξέρω αν λες αυτό Δημήτρη) ότι στην αντιθετοαντίστροφη η π2 είναι ψευδής και η π1 επίσης ψευδής. Ωστόσο αυτό που για μένα έχει σημασία είναι ότι μεταξύ του αν και του τότε (δηλαδή πριν το βέλος της συνεπαγωγής) υπάρχει μια αληθής πρόταση... η όχι(π2). Το ίδιο και μετά.

Με άλλα λόγια αν βάλεις κάτι ψευδές ανάμεσα στο «Αν» και στο «τότε» δε βλέπω να βγαίνει κάτι που να έχει νόημα. Και βλέπω ότι γενικά από αυτά που έγραψες ότι κάποιες συνεπαγωγές δεν τις πολυχρησιμοποιούμε ή δε μας αφήνουν ευχαριστημένους. Δε μου κάθεται καλά αυτή η κατάσταση. Μου αφήνει έντονα μια γεύση ότι αυθαιρετούμε.[/quote]

Γιώργο προσοχή. Οι π1 και π2 είπαμε είναι αληθείς άρα οι "όχι π1" και "όχι π2" είναι ψευδείς. Μεταξύ του αν και του τότε υπάρχει μια ψευδής πρόταση.

[quote="gpapargi"]Αυτό δεν ήταν από βιασύνη αλλά λόγω ακριβώς του μπερδέματος που έχω.
Βασικά από ότι καταλαβαίνω λες το εξής:
«Θέλω η αντιθετοαντίστροφη μια πρότασης να έχει πάντα την ίδια τιμή αληθείας ανεξάρτητα από το τι τελικά γίνεται έξω (πχ αν βρέχει ή όχι)."
Ενώ αυτό που εννοώ εγώ βασισμένος στη διαίσθησή μου (και όχι στον ορισμό) είναι ότι η αντιθετοαντίστροφή όντως έχει τον ίδιο βαθμό αληθείας. Ωστόσο οι 2 προτάσεις έχουν διαφορετικά προαπαιτούμενα για να ελεγχθεί η ορθότητά τους.
Έτσι για να είναι δυνατό να ελεγχθεί η ορθότητα της πρότασης «Αν βρέξει τότε θα πάρω ομπρέλα» θα πρέπει έξω να βρέχει.
Αντίθετα για να ελεγχθεί η ορθότητα της ισοδύναμης αντιθετοαντίστροφης «Αν δεν έχω ομπρέλα τότε δεν βρέχει» θα πρέπει να μην έχω ομπρέλα.

Αυτό εννοούσα πριν, ενώ εσύ αν κατάλαβα καλά (με βάση τον ορισμό) θέλεις κάθε φορά οι αντιθετοαντίστροφες να έχουν ίδιο βαθμό αληθείας ανεξάρτητα από το αν γίνεται να ελεγχθεί η ορθότητα νοηματικά (με βάση αυτό που λέμε παραπάνω «διαισθητικά»)
Δηλαδή μέσα μου τις θεωρώ ισοδύναμες ως προς το νόημά τους αλλά όχι ως προς το τι απαιτείται για τον έλεγχο της ορθότητάς του νοήματός τους.
Νομίζω πως σιγά σιγά βγάζω άκρη.[/quote]

Έχουμε λοιπόν δύο προτάσεις οι οποίες είναι «διαισθητικά» ισοδύναμες. Αυτό το «διαισθητικά» θέλουμε να το αφαιρέσουμε και να το κάνουμε "σημασιολογικά" ισοδύναμες. Αυτό ακριβώς επιτυχάνουμε με το να προσθέσουμε τον λογικά κανόνα "Η πρόταση "ψευδης ==> ψευδής" είναι πάντα αληθής".

Όντως… αν έχεις μια συνεπαγωγή «Αν π1 τότε π2» και γνωρίζεις το βαθμό αληθείας των π1 και π2 θα πρέπει να μπορείς να αποφανθείς τελεσίδικα και για την αλήθεια ή όχι της συνεπαγωγής, αφού η συνεπαγωγή εξαρτάται από αυτές και μόνο αυτές.
Και μάλιστα θα πρέπει να μπορεί να αποφανθεί για την αλήθεια ένας υπολογιστής που λειτουργεί μηχανικά και δε διαθέτει καθόλου κατανόηση του τι σημαίνει μια πρόταση σε μια φυσική γλώσσα (πχ νέα ελληνικά)
Αυτό σε συνδυασμό με την ισοδυναμία των αντιθετοαντίστροφων αρκεί για να πούμε ότι η «Αν ψευδής τότε ψευδής» είναι αληθής. Thanks!!!  :)

Ένα ακόμα σχετικό για να κλείσει το θέμα:
Πως προκύπτει η αλήθεια της «Αν ψευδής τότε αληθής»; Εδώ το αντιθετοαντίστροφο είναι πάλι το ίδιο.

[quote="gpapargi"]
Ένα ακόμα σχετικό για να κλείσει το θέμα:
Πως προκύπτει η αλήθεια της «Αν ψευδής τότε αληθής»; Εδώ το αντιθετοαντίστροφο είναι πάλι το ίδιο.[/quote]

Το πιο πειστικό επιχείρημα που μπορώ να σκεφτώ είναι πως θέλουμε να ισχύει πάντα το "οτιδήποτε ==> αληθές".
Στο σχολείο συνηθίσαμε όταν μας δίνουν κάποια δεδομένα για ένα πρόβλημα να τα χρησιμοποιούμε όλα για να αποδείξουμε το ζητούμενο. Αυτό όμως δεν είναι πάντα απαραίτητο. Μπορεί να αποδείξουμε κάτι χρησιμοποιώντας λιγότερα δεδομένα. Στην περίπτωση που προσπαθούμε να αποδείξουμε μια πρόταση P γνωρίζοντας μια πρόταση Q, αν η P είναι αληθής δεν χρειάζεται καν να χρησιμοποιήσουμε το Q. Είναι απλώς ένα περιττό δεδομένο το οποίο είτε ισχύε είτε όχι, εμείς μπορούμε να αποδείξουμε το P.

(Εδώ μπερδεύω λίγο την έννοια της αλήθειας με την έννοια της απόδειξης. Στόχος μας όμως είναι να μπορούμε να αποδείξουμε όλες τις λογικά αληθείς προτάσεις και καμία άλλη.)

[quote="Demetres"]Η απάντηση είναι "διότι έτσι επιλέξαμε". Ας προσπαθήσω όμως να πω λίγα περισσότερα:

Πολλές φορές στα μαθηματικά χρησιμοποιούμε έμμεσες αποδείξεις. Για παράδειγμα, για να δείξουμε ότι η $ \sqrt {2}$ είναι άρρητος, υποθέτουμε πως είναι ρητός και καταλήγουμε σε άτοπο. Για να επιτρέπεται να το κάνουμε αυτό πρέπει να έχουμε ένα κανόνα που να λέει ότι για κάθε πρόταση Ρ, είτε η Ρ είναι ορθή, είτε η άρνηση της Ρ είναι ορθή. Με άλλα λόγια είναι επιθυμητό στα μαθηματικά κάθε πρόταση να είναι είτε ψευδής είτε αληθής.

Αυτή είναι η "συνηθισμένη" λογική που χρησιμοποιούμε στα μαθηματικά. Η στάση της πλειοψηφίας των μαθηματικών είναι πως αν απαγορέψουμε αυτό, τότε αρκετά θεωρήματα των μαθηματικών θα πάψουν να ισχύουν και αυτό είναι ανεπιθύμητο. Υπάρχουν όμως και άλλες λογικές όπως για παράδειγμα στην "intuitionistic logic" όπου ο κανόνας του αποκλειόμενου μέσου δεν βρίσκεται στα αξιώματα (και ούτε προκύπτει από αυτά).[/quote]

Οι ανθρωποι του CERN ΠΟΙΑ λογικη χρησιμοποιουν και γιατι??


group theory
abstract algebra
superior algebra
superior algebra solved

Q and S are two distinct p-sylow subgroups of a finite group G.
If K is <Q,S>  (i.e.generated by two sylow p- subgrops), show that for some k in K,  S is  kQk^(-1). i.e it is conjugate.

[quote="Gockel"]That's trivial if you know the sylow theorems:
K is a finite group, S and Q are p-sylow-subgroups of it and hence conjugated in K.[/quote]

No. the statement is not trivial. they must be conjugate in the set that they generate.

They are still Sylows in the group generated by them, thus it can be applied.

So the idea is to show that they are still sylow groups in the group generated by them and the rest is to show any sylow group in the generated group is conjugate?


i could argue on the basis of planar structures that the compound $ \text{cis 4 \minus{} chloro cyclohexanol}$ and also the $ \text{trans}$forms are achiral but i am not able to do that using the puckered chair structure.anyone please?

And where is the difference? In any structure you draw, you will always have a plane of symmetry passing by carbons 1 and 4 (and "perpendicular" to the ring), or, alternatively, carbons 1 and 4 have two equal groups attached to them.


analytic geometry
linear algebra
matrix
limit
calculus
calculus computations

hi can someone explain this:
Consider the autonomous equation
\[ \frac{dx}{dt}=x, \qquad \frac{dy}{dt}=-2y+x^3  \]

I have shown that the critical point is $(0,0)$ , and also  shown that the system is almost linear. the corresponding linear system is
\[ \left (\begin{array}{ccc} u \\ v \end{array} \right)'= \left(\begin{array}{ccc} 1 & 0 \\ 0 & -2 \end{array}\right) \left(\begin{array}{ccc} u \\ v \end{array}\right)  \]
and also i've shown that the trajectory of this linear system for which $x \rightarrow 0$, $y \rightarrow 0$ as $t \rightarrow \infty$ is given by $x=0$.
------------------------------------------------------------------------------------------
Now the  part of problem that confused me:
Determine the trajectories for the nonlinear system for $x \not= 0$ by integrating the equation $\frac{dy}{dx}$. Show that the trajectory corresponding to $x =0$ for the linear system is unaltered, but that the one corresponding to $y=0$ is $y=\frac{x^3}{5}$.

I confuse about $y=0$, what its mean? it is $y(0)=0$ or $y$ constantly zero? if y is constantly zero then for linear system $y(t)=C$ and $x(t)=C_1e^{t}$ right?? so what is $y= \frac{x^3}{5}$

Yes this is the homework problem, but nevermind I have give my work to my instructor without finishing that part. :(

If you use the linear system and start on a coordinate axis, you will stay on a coordinate axis, either heading in or out. That's what the diagonal matrix does.

The trajectory corresponding to $y=0$ in the full system must be the one in which $x(0)=y(0)=0$ and $\lim_{t\to 0}\frac{y(t)}{x(t)}=0$. (unless there are too many of those; we want the one that's closest to the axis for small $x$)

Thanks Jmerry.
But I still don't know why we must have condition
$x(0)=0$, $y(0)=0$ and $\lim_{t \rightarrow 0} \frac{x(t)}{y(t)} = 0$. Its look like I can't find that in my book.


[url]http://www.foon.co.uk/farcade/?page=balls[/url]

See how far you get. I'm a Sacred Can o' Spam with a whooping 4585

EDIT: Now I'm an Ultimate Banana with a 5080

I'm a Barlow Compound with 7370....
This game is addicting.  :D

From some point on, the problem is to die :D
I stand at $4000000$ points and $2000$ balls at the moment, but it still grows...  :lol:

Edit: $7000000$...
Edit2: $12498850$ and $3142$ balls according to a screenshot... (it stopped growing exponentially cause my computer is to slow)

My first five scores:

355 (Rank: Dead Rat)
375 (Rank: Dead Rat)
60 (Rank : Poo)
225 (Rank: Bum)
785 (Rank: Brick)

Too hard for me.

currently at 30,000 and 800 balls....

I now understand what zetax meant.

2 min later:
1,600,000 and 1800 balls

Now 3 million and 2500 balls.

18,115,410 and 5,856 balls

OK, well, it appears that I died at some point while I was eating lunch and other things.

Score: 126,658,550.

Score 2673535 and like 900-1000 Balls (there are to many balls, i cannot see the number) It's really a problem to die at this point

[quote="ZetaX"]Klebian: how was it possible to die¿[/quote]
Absolutely no clue. I left the computer and when I came back, it was Game "Under"... btw, 126 million is still barlow compound.

Also, one thing I've found is the purple thing shrinks over time.
At one point it was long enough to cover the whole area, but half an hour later there was some space where balls were able to die.

Move to left or right as much as you can
If you keep yourself at the center, the number of balls keeps growing

Well when your paddle (or whatever it's called) spans the whole screen, you can't really move to the left or right.
I made the mistake of trying to make it as big as possible, and wow, that really slows your computer (albeit you might get a higher score).

I'm now a barlow compound with more than 200000 points.

So why did I have that much difficulty in crossing 1000?

Whan you cross the the $2000$, your paddle should already be large enough to to make it hard to die. So this is the bound you have to cross ($\pm1500$ points).
Then the screen looks the following then:

I got to 5000 without doing that. :wink:

1st time:poo
2nd time:poo
3rd time:barlow compound 11950!

this game is awesome

My paddle is nearly as big as the floor
I have $420000$ pts with $500$ balls :)

Edit: $3 000 000$ pts,  $1500$ balls

Now:$14$ $000$ $000$ with $3$ $500$ balls


LaTeX
limit
function
calculus
real analysis
real analysis unsolved

$ f(x),g(x)$ are both continuous on the closed interval $ [a,b]$
let $ h(x)\equal{}max${$ f(x),g(x)$}
show that:
$ h(x)$ is continuous on the closed interval $ [a,b]$

thanks~

Some $ \text{\LaTeX}$ notes: $ \max$ works, and you can use \{ and \} to get curly braces that show up.

Method A: We directly calculate $ \lim_{x\to c} h(x)$ and show it's $ h(c)$
Case 1: $ f(c)>g(c)$. Let $ \epsilon=\frac12{f(c)-g(c)}$, and find $ \delta=\min\{\delta_1,\delta_2\}$ such that $ |f(x)-f(c)|$ and $ |g(x)-g(c)|$ are both $ <\epsilon$ for $ |x-c|<\delta$. Then $ h(x)=f(x)$ for all $ x\in (c-\delta,c+\delta)$, and $ \lim_{x\to c}h(x)=\lim_{x\to c}f(x)=f(c)=h(c)$.
Case 2: $ g(c)>f(c)$. Interchange $ f$ and $ g$ in the preceding argument.
Case 3: $ f(c)=g(c)$. Choose an arbitrary $ \epsilon>0$, and find $ \delta=\min\{\delta_1,\delta_2\}$ such that $ |f(x)-f(c)|$ and $ |g(x)-g(c)|$ are both $ <\epsilon$ for $ |x-c|<\delta$. Then $ f(x)$ and $ g(x)$ are both in $ (h(x)-\epsilon,h(x)+\epsilon)$ for all $ x$ in that interval; since $ h(x)$ is one of those values, it is as well. Repeating for all $ \epsilon$, this is the definition of the limit and $ \lim_{x\to c}h(x)=h(c)$.

Method B: We go through two dimensions.
First, show that $ \max(x,y)$ is a continuous function of $ x$ and $ y$; this is obvious away from the critical line $ x=y$, and true on the line $ x=y$ since $ |\max(t,t)-\max(x,y)|\le \max(|t-x|,|t-y|)$.
Next, $ H(x)=(f(x),g(x))$ is continuous.
Finally, the composition $ h$ is a continuous function.

Hint for another approach: For any $ x$, $ y$ in $ R$, $ \max(x,y) \equal{} (x\plus{}y\plus{}|x\minus{}y|)/2$.

[quote="Null"][quote="WWW"]Hint for another approach: For any $ x$, $ y$ in $ R$, $ \max(x,y) \equal{} (x \plus{} y \plus{} |x \minus{} y|)/2$.[/quote]
how do you prove this?[/quote]

Suppose $ x \leq y$, ...

sorry but i neved had a good course of calculus I, can you full explain it please?

Just look at the formula a little bit:
$ \frac{a\plus{}b}{2}$ is the (arithmetic) mean of $ a$ and $ b$, while $ \frac{|a\minus{}b|}{2}$ is half the distance between $ a$ and $ b$.


probability
geometry
3D geometry
tetrahedron

I think that between does never include the boundaries.

[hide="Junk..."][quote="mr. math"]Three real numbers are randomly chosen between 0 and 10. What is the probability that their sum is less than 10?[/quote]

It doesn't matter: we are concerned with all real numbers, not just integers.

Draw graph bounded by the lines $ x \equal{} 0$, $ y \equal{} 0$, $ x \equal{} 10$, $ y \equal{} 10$, and also $ x \plus{} y < 10$.

We can easily see that the answer is $ \boxed{\frac12}$.[/hide]

We have $ 0\leq x,y,z\leq10$ and we want $ x \plus{} y \plus{} z\leq10$.

Visualize this as a cube in space with side length $ 10$ (satisfies $ 0\leq x,y,z\leq10$), and a tetrahedron with edges $ x \equal{} 0,y \equal{} 0,z \equal{} 0,x \plus{} y \equal{} 10,y \plus{} z \equal{} 10,x \plus{} z \equal{} 10$ (satisfies this and $ s\plus{}y\plus{}z\leq10$).

Now, the probability is easily found to be $ \frac {\frac13\cdot10^3}{10^3} \equal{} \boxed{\frac13}$.


1. Which of the following numbers is smallest?
(A) 2^900 (B) 3^600 (C) 5^400 (D) 7^300 (E) 11^200

Is there a quick solution that I am not seeing? It's easy with a calculator, but I am not sure how to solve this.

[hide="Solution"]
Take the base-2 logarithm of all the answer choices to get:
$ 900, 600\log_2{3}, 400\log_2{5}, 300\log_2{7}, 200\log_2{11}$.

Now we have to approximate. $ 2^1\equal{}2$ and $ 2^2\equal{}4$, so $ \log_23\approx 1.5$ Similarly, $ 2^2\equal{}4$ and $ 2^3\equal{}8$ implies that $ \log_25\approx 2.2$. And $ 2^2\equal{}4$, and $ 2^3\equal{}8$, so $ \log_27\approx 2.8$. Finally, $ 2^3\equal{}8$ and $ 2^4\equal{}16$, so $ \log_211\approx 3.5$

Multiplying, we get the following possible choices:
$ 900, 900, 880, 840, 700$

The large difference between the last choice and the rest indicates that the answer is most likely $ \boxed{\textbf{(E)}\ 11^{200}}$.
[/hide]

[quote="johnfn"]Just take the 100th root of all numbers. That keeps the numbers in proportion, but makes them a reasonable size.[/quote]

See math154's post (above).

@math154: Why is the congruence symbol used in this case? Wouldn't a better choice of notation be the use of $ \min{\{2^{900},3^{600},5^{400},7^{300},11^{200}\}}\equal{}\min{\{2^9,3^6,5^4,7^3,11^2\}}$?

Yeah, probably, but I just suck at using notation.

These solutions stem from the fact that even if you don't know what's greater and what's less,
you can still use basic inequality rules (other than multiplying or dividing by one) to keep
the sign (not found yet) pointing in the same direction.  This way, you can simplify the numbers
until you can actually compute them.  (For example, in this instance, math154 took
the $ 100$th root of all the answers, letting him look at the simplified versions of all of the answers
which he could now compare).


algebra
polynomial
inequalities
ratio
algebra unsolved

Let
\begin{eqnarray*} f(x) & = & a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 \ \ \mbox{and} \\ g(x) & = & c_{n+1} x^{n+1} + c_n x^n + \cdots + c_0 \end{eqnarray*}
be non-zero polynomials with real coefficients such that $g(x) = (x+r)f(x)$ for some real number $r$.  If $a = \max(|a_n|, \ldots, |a_0|)$ and $c = \max(|c_{n+1}|, \ldots, |c_0|)$, prove that $\frac{a}{c} \leq n+1$.

Are you sure it's true?
Notice that $c_{n+1}=a_n$
$c_i=a_ir+a_{i-1}$ for $1 \le i \le n$
$c_0=a_0r$

So then what if you had $a_n=a_{n-1}=...=a_1=1$, $a_0=1000000000$ or some other really big number, and then $r=\frac{1}{n+2}$?  Clearly, $a=1000000000$, and if $n+3<1000000000$, $c=c_0$, and then we have the inequality untrue...

K81o7, you are wrong.
$ c=c_{1} $ and the ratio is smaller than $ 1 $.

like K81o7 said we have:
$ c_0 =ra_0 $
$ c_i =ra_i+a_{i-1} $ for $ 1\le i\le n $
$ c_{n+1}= a_n $.
$ \frac{c}{|c_0|+|c_1|+....+|c_n|}\leq 1/n+1 $.
we will prove that  $ a $ is not greater than $ |c_0|+|c_1|+....+|c_n| $ using induction.
suppose that for $ n-1 $ is true.
if $ a=|a_n|$ then in the last sum we put $ c_{n+1} $ in place of $ c_0 $.
if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true.
now we only have to prove it for $ n=0 $.
$ f(x)=b $ and $ g(x)=bx+br $ so $ a=|b| $ and $ c=|b|x $ where $ x $ is the maximum of $ 1 $ and $ |r| $.we have to prove $ x $ is not greater than 1 which is true.
q.e.d.

[quote=niyinchen]Forgive my poor English.[/quote]
You don't have to write that on post

[quote=Why-][quote=niyinchen][quote=anonymouslonely]we have:
if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true.
[/quote]
When  you add $ |c_n| $ and  $ |a_n| $  but at the same time $ |c_{n-1}| $ changes,thus you,cannot use your induction suppose.That is my qusetion.Have I got anything wrong?Forgive my poor English.[/quote]

I also have your question.But I find a new way to prove it.[/quote]

How can you prove it?


Prove that for all positive values of $k$ less than 1, the equation

\[(x+a)(x+b) = k(x+c)^2\]

has two real roots.

Isn't this a straight forward discriminant check?

Also, I think we need the condition that $a, b, c$ are not all equal.

Anyways, the discriminant works out to be $(a-b)^2 + 4k(a-c)(b-c)$. If $a-c$ and $b-c$ have the same sign, then we are fine. Otherwise, we assume that $a>c>b$. Then by AM-GM
\[
(a-b)^2 = 4\left(\frac{(a-c)+(c-b)}{2} \right)^2 \geq 4(a-c)(c-b) > 4k(a-c)(c-b)
\]
so the discriminant is positive, and the result follows.


Let $ a,b,c\in [1,2]$ and $ ab\plus{}bc\plus{}ca\equal{}6$ . Prove :

$ \frac{a}{\sqrt{b\plus{}c}}\plus{}\frac{b}{\sqrt{c\plus{}a}}\plus{}\frac{c}{\sqrt{a\plus{}b}}\ge \frac{4\sqrt{3}}{3}$

Strictly proven,  we have

$ LHS \ge \frac{\sqrt{3(a\plus{}b\plus{}c)}}{\sqrt{2}}$

The last step is to use

$ a\plus{}b\plus{}c \ge \sqrt{3(ab\plus{}bc\plus{}ca)}$

thanks but I am trying to figure out your inequality because I think it is nice as well. :)


How would explain the first, second, and third law of thermodynamics to a fifth grader (through a poster is what I was thinking, but whatever you suggest I'll appreciate)?

[quote="Aahmadi"]How would explain the first, second, and third law of thermodynamics to a fifth grader[/quote]

You don't: a person of that age is perhaps too young to even understand what a "law" is. Try to show the general experimental facts included in those laws (don't forget the zeroth law) without using fancy words. Say something like "Energy is conserved is all around us" (after giving a proper fealing of what energy is), "All things that happens on Nature increase the disorder ot the world".

(sorry for the double post)
alrighty, thanks for reply, and do u have any ideas for activities that I could have the little kids do, or do you think it'd better to have a pretty poster with basic facts (that are easy to understand, at least with my talking) on it?


calculus
geometry
trigonometry
Gauss
inequalities
number theory
complex numbers

Well, this is quite simple. You just add +1 point to your favorite math category and subtract -1 from your most hated one. This keeps going until only one category survives. You can only vote for surviving categories. If a category reaches 0, it's out of the contest. It would be under the "DEAD" list. Also note that you can only vote once a day, beginning at 12:00 midnight per day on the AoPS clock. Each genre starts with 5 points.
[u]CURRENT STANDINGS:[/u]
Algebra-5
Calculus-5
Combinatorics-5
Geometry-5
Trigonometry-5
Number Theory-5
Statistics-5

Alright, I'll begin!
+Algebra
-Statistics
Now you can begin voting.
[u]Current Standings:[/u]
Algebra-6
Calculus-5
Combinatorics-5
Geometry-5
Trigonometry-5
Number Theory-5
Statistics-4
Please remember to update this list EVERY TIME YOU VOTE! IF YOU VOTE A SECOND TIME IN A DAY, DON'T UPDATE THE LIST, POST CONSECUTIVE VOTES MULTIPLE TIMES IN A ROW, OR VOTE FOR A DEAD CATEGORY, YOUR VOTE WON'T COUNT. REPEATED VIOLATIONS MAY RESULT IN BEING KICKED OUT OF THIS THREAD! So follow the rules, and your vote counts!

+combinatorics, -geometry

Algebra-6
Calculus-5
Combinatorics-6
Geometry-4
Trigonometry-5
Number Theory-5
Statistics-4

arghh, number theory or combanitorics... hmmm....

+number theory, -algebra

Algebra-5
Calculus-5
Combinatorics-6
Geometry-4
Trigonometry-5
Number Theory-6
Statistics-4

+ calculus, - geometry

Algebra-5
Calculus-6
Combinatorics-6
Geometry-3
Trigonometry-5
Number Theory-6
Statistics-4

Hurt Combinatoric, Heal Statistics

Algebra-5
Calculus-6
Combinatorics-5
Geometry-3
Trigonometry-5
Number Theory-6
Statistics-5

+ number theory - statistics

Is algebra abstract algebra or high school algebra?

Algebra-5
Calculus-6
Combinatorics-5
Geometry-3
Trigonometry-5
Number Theory-7
Statistics-4

+geometry
-stats

Algebra-5
Calculus-6
Combinatorics-5
Geometry-4
Trigonometry-5
Number Theory-7
Statistics-3

+Calculus
-Statistics

Algebra-5
Calculus-7
Combinatorics-5
Geometry-4
Trigonometry-5
Number Theory-7
Statistics-2

+ trig, - Geometry.

Just to fill everyone in,
Algebra-5
Calculus-7
Combinatorics-5
Geometry-3
Trigonometry-6
Number Theory-7
Statistics-2

+ Algebra, - Combinatorics.

Just to fill everyone in,
Algebra-6
Calculus-7
Combinatorics-4
Geometry-3
Trigonometry-6
Number Theory-7
Statistics-2

-stat, I have to add 1 um... +calc
Just to fill everyone in,
Algebra-6
Calculus-8
Combinatorics-4
Geometry-3
Trigonometry-6
Number Theory-7
Statistics-1

+combinatorics, -geometry
Algebra-6
Calculus-8
Combinatorics-5
Geometry-2
Trigonometry-6
Number Theory-7
Statistics-1

+trig, -stats

Algebra-6
Calculus-8
Combinatorics-5
Geometry-2
Trigonometry-7
Number Theory-7
[color=red]Statistics-0[/color]

+combinatorics -geometry

Algebra-6
Calculus-8
Combinatorics-6
Geometry-1
Trigonometry-7
Number Theory-7

+Combinatorics, -Algebra

Algebra-5
Calculus-8
Combinatorics-7
Geometry-1
Trigonometry-7
Number Theory-7

Hmmmm, actually I quite like Number Theory.

Algebra 24
Number theory 7

UP WITH THE ALGEBRA!!!!!!!!

Algebra 25
Number theory 5

Algebra:22
Number Theory:5
I don't like algebra at all (no, I don't think complex numbers and inequalities and the like belong in it).

i would so want to be the person killing number theory...

Algebra:23
Number Theory:4

Algebra: 24
Number Theory: 1

Inequalities are totally algebra. And they are totally my favorite subject.

kmmystar: ou can hurt twice, so I added on an extra hurt to NT for you.

HOORAY! NUMBER THEORY IS DEAD!

Algebra-25
Number Theory-[size=200]R.I.P. YOU ARE DEAD![/size]

Congrats! Algebra is now the winner! GO ALG!

Since the contest is over, plz lock this, mods.

KILL THE ALGEBRA!!!

Algebra-22

Heal hunter34

hunter34-1

You know what, please stop that and leave it alone. It's done already. No more additional comments, please.

Mods plz lock this.


calculus
function
logarithms
topology
real analysis
real analysis unsolved

I found this very nice problem in my paper with exercices:
   Let $f$ be a complex function (defined on $C$ with values in $C$) with values on the unit circle and which is $C^{infinity}$. Prove that there exists a function $g$ defined on $C$ and which is $C^{infinity}$ for which $f(z)=e^{ig(z)}$.

"Relevement" is "lifting" :).

$f$ is a continuous function from the space $\mathbb C$, which has all the necessary "nice" properties (like connectedness and path connectedness) to $\mathbb S^1$. Moreover, the homomorphism $f_*$ induced by $f$ on fundamental groups carries the fundamental group of $\mathbb C$ (which is trivial) into the image of the fundamental group of $\mathbb R$ (which is also trivial) through the endomorphism $p_*$ induced by the covering map $p: \mathbb R\to\mathbb S^1,\ p(\theta)=e^{i\theta}$. This means that $f$ has a continuous lifting $g: \mathbb C\to\mathbb R$ such that $p\circ g=f$. Moreover, if we fix $z\in\mathbb C$ and $g(z)$, the function $g$ is unique.

We now have the continuous map $g: \mathbb C\to\mathbb R$ satisfying $f(z)=e^{ig(z)}$. All we need to prove is that $f\in C^\infty\Rightarrow g\in C^\infty$. This, however, follows immediately from the fact that for each fixed $z_0\in\mathbb C$, there is a small neighborhood of $z_0$ where $g$ can be expressed as $g=-i\ln f$, where $\ln$ is the branch of the logarithm satisfying $\ln f(z_0)=ig(z_0)$.


Let a, b, c > 0

1) Show that  $ 2(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \ge 3(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c})$
2) Find the smallest positive real number s for which $ 2(a^s+b^s+c^s)(\frac{1}{a^s}+\frac{1}{b^s}+\frac{1}{c^s}) \ge 3(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c})$ for all a, b, c > 0

This last question was raised in my class. Quite interesting, try to solve it!

Namdung

Let me try to solve the first part first.

\[
\begin{array}{l}
 2(a^2  + b^2  + c^2 )\left( {\frac{1}{{a^2 }} + \frac{1}{{b^2 }} + \frac{1}{{c^2 }}} \right) - 3\left( {\frac{b}{c} + \frac{b}{a} + \frac{a}{b} + \frac{a}{c} + \frac{c}{a} + \frac{c}{b}} \right) \\
  \ge \frac{3}{2}\left( {\frac{{b^2 }}{{c^2 }} + \frac{{b^2 }}{{a^2 }} + \frac{{a^2 }}{{b^2 }} + \frac{{a^2 }}{{c^2 }} + \frac{{c^2 }}{{a^2 }} + \frac{{c^2 }}{{b^2 }}} \right) + 9 - 3\left( {\frac{b}{c} + \frac{b}{a} + \frac{a}{b} + \frac{a}{c} + \frac{c}{a} + \frac{c}{b}} \right) \\
  = \frac{3}{2}\sum\limits_{\scriptstyle \{ i,j\}  \in \{ a,b,c\}  \hfill \atop
  \scriptstyle i \ne j \hfill} {\left( {\frac{i}{j} - 1} \right)^2 }  \\
  \ge 0 \\
 \end{array}
\]

and i think for b the ans is 3/2
am i right?
i used bernouli inequality to get the result, but iam not sure if it is the best constant.

No is not the best constant. I can prove for 4/3 or even 5/4!

Indeed, $\frac{5}{4}$ is a good value for $s$.
I think, the problem can be completed with the following point:
(c) Find the best (maximum) $C$ such that the inequality holds

$(a_1^2+...+a_n^2)(\frac{1}{a_1^2}+...+\frac{1}{a_n^2})+n^2C \ge (1+C)(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})$

for any positive numbers $a_i$.


have they mailed the cmo registration form yet?
and... when are apmo invitations usually sent?

Okay, I just got the APMO invite email today, and I got an email yesterday (?) saying that the CMO reg. form was mailed to my school... I went to the school math office, and they didn't have it yet... I hope that answers your question.  :maybe:

yeah i got both emails today, too

funny how my math teacher hasn't told me about the cmo reg form though

I haven't gotten anything yet, does that mean I didn't make CMO?

Well if you made cmo through cmor then they won't tell you whether or not you made it through email. Instead, you'll know whether you made it if you get a package from cms

well i don't think they sent the invitations to people who made it through cmor already

Yay! I got the APMO invitation!

Now I'm just waiting for the CMO invitation. :)

is this olimpiads online?
and how can i regiter or get this invitation?

wait..
[quote]We are unable to tell you at this time whether or not you will be invited to write the 2009 Canadian Mathematical Olympiad. You will hear directly from the Canadian Mathematical Society if (and only if) you are invited to the CMO.[/quote]

Don't they notify us by email??

I think they mail you a package by courier.

[quote="zepiphanus"]to alpha-beta: did you write comc?[/quote]

no i didn't  =((

ok.. now im getting REALLY worried since I haven't received any email yet...

surely a 40 - 50 will likely get an invitation since last year's cutoff was 42, and this year's problems were harder than last year's... am I correct?? :(

Okay, so my math teacher gives me the CMO letter *finally*, but I notice that there is no book prize... there was one last year... So, is there a book prize this year? Did anyone get one? I was looking forward to the book...  :(  :(

I haven't asked my math head yet about the CMO thing. but I'll do it tomorrow.

I already have two of those CMO books, I don't need another one, although another one wouldn't hurt. I believe I got that book, and form, around the middle of Feb last year while writing Cayley or something. :)

i am :) and at least 5 of my friends from different schools

We're in different ares though. Toronto has some people that I've never heard of.  :|


modular arithmetic
number theory
prime numbers

Hello. I recently looked at CMO #4 and tried to solve it. I was wondering if you guys could give me tips for writing (also I am kind of happy because even though it is sad that this is one of the hardest problems that I have solved, I solved it :) ). I hope that my solution is correct, however I have more faith in the AOPS community than anything else to help me check it over. I will conceal my solution for those who want to try the problem.

[b]Problem[/b]

Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^{n} \minus{} n$.

[hide] Firstly, I looked for a solution in terms of $ p$ with the knowledge that $ p > 1$. Using Euler's Totient Theorem, we get the following.

$ 2^{\varphi(p)} \equiv 1 \pmod{p}$

Since there exist $ p \minus{} 1$ numbers $ k$ such that $ 0 < k < p$ such that $ gcd(p,k) \equal{} 1$ when $ p$ is prime, $ \varphi(p) \equal{} p \minus{} 1$. This gives the following.

$ 2^{p \minus{} 1} \equiv 1 \pmod{p}$

Raising both sides to the power of $ (p \minus{} 1)^{2k \minus{} 1}$, where $ k$ is an integer which satisfies that $ k\ge 1$ gives the following. Note that since $ p > 1$, $ (p \minus{} 1)^{2k \minus{} 1} > 0$.

$ 2^{(p \minus{} 1)^{2k}} \equiv 1^{(p \minus{} 1)^{2k \minus{} 1}} \pmod{p}$

$ 2^{(p \minus{} 1)^{2k}} \equiv 0 \plus{} 1 \pmod{p}$

Now lets examine $ (p \minus{} 1)^{2k} \minus{} 1$ in $ \pmod{p}$.

$ (p \minus{} 1)^{2k} \minus{} 1 \equiv ( \minus{} 1)^{2k} \minus{} 1 \equiv 1 \minus{} 1 \equiv 0 \pmod{p}$

Now we can substitute, yielding that,

$ 2^{(p \minus{} 1)^{2k}} \equiv (p \minus{} 1)^{2k} \minus{} 1 \plus{} 1 \equiv (p \minus{} 1)^{2k} \pmod{p}$

Rearranging yields the following, that

$ 2^{(p \minus{} 1)^{2k}} \minus{} (p \minus{} 1)^{2k} \equiv 0 \pmod{p}$

Therefore for all prime numbers $ p$ and positive integers $ k$, there will exist infinite solutions for $ n$ such that $ p|(2^{n} \minus{} n)$ that take the form $ n \equal{} (p \minus{} 1)^{2k}$. [/hide]

I would say that works, but there is a chance I am hallucinating after tennis.

[hide="alternate solution stuff, with motivation thrown in"] Since this involves powers and a prime, it makes sense that Fermat's Little Theorem will be involved somehow.

From Fermat's Little Theorem, we know $ 2^{p-1}\equiv{1}\pmod{p}$. We suspect that $ p-1$ will be involved in this, since it's the exponent of $ 2$.

Suppose that $ n$ is of the form $ k(p-1)$ for some $ k$.

Then $ 2^{n}\equiv(2^{p-1})^k\equiv{1}^k\equiv{1}\pmod{p}$. We want this to be congruent to $ k(p-1)\pmod{p}$, which is just $ -k\pmod{p}$. So, we want $ k\equiv{-1}\pmod{p}$. Now, we can just let $ k=mp-1$ for positive $ m$, so we are done.[/hide]

[hide="how I would write that"] If $ p=2$, then just take even $ n$.

Now, suppose $ p$ is odd, so $ (2, p)=1$. From Fermat's Little Theorem, we know $ 2^{p-1}\equiv{1}\pmod{p}$.

Let $ n=(mp-1)(p-1)$, for positive $ m$. We have \[ 2^n\equiv{(2^{p-1})^{mp-1}\equiv{1^{mp-1}}\equiv1\equiv{(-1)(-1)}\equiv{(mp-1)(p-1)}\equiv{n}\pmod{p}}\]

as desired.[/hide]

Also, is this from this year's CMO? Assuming you mean the Canadian one, I recall #4 was something involving $ 3^a+7^b$, although I can't check because this laptop won't let me view pdfs for some reason...

Thanks! I took this problem from a CMO in the 80s, and you are right about the Number Theory #4 this year. I like your solution better. :)


Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $ (a,0)$ to $ (3,3)$, divides the entire region into two regions of equal area. What is $ a$?
[asy]size(200);
defaultpen(linewidth(.8pt)+fontsize(8pt));

fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray);

xaxis("$x$",-0.5,4,EndArrow(HookHead,4));
yaxis("$y$",-0.5,4,EndArrow(4));

draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0));
draw((1,0)--(1,2)--(3,2));
draw((2/3,0)--(3,3));

label("$(a,0)$",(2/3,0),S);
label("$(3,3)$",(3,3),NE);[/asy]$ \textbf{(A)}\ \frac12\qquad
\textbf{(B)}\ \frac35\qquad
\textbf{(C)}\ \frac23\qquad
\textbf{(D)}\ \frac34\qquad
\textbf{(E)}\ \frac45$

[hide="Solution"]The area of the shaded region is $ \frac {3(3 \minus{} a)}{2} \minus{} 1$ which we are given equals $ \frac {5}{2}$. Setting $ \frac {3(3 \minus{} a)}{2}$ equal to $ \frac {5}{2}$ gives $ a \equal{} \frac {2}{3}\implies{(C)}$[/hide]

I extended another square, containing vertice $(3,0)$ (so now we have shaded right triangle) and I add a "white" square to other side, which contains vertice (0,2). We simply added a square to both sides, so the areas are still equal. There are now a total of $7$ squares, each color containing $\frac{3}{2}$ of the squares.

So we set an equation for the black squares, the area of the black squares are $\frac{7}{2}$ but it can also be found using the triangle length which would give us our answer, hence $\frac{7}{2}=3\times\frac{N}{2}$, which comes out to $N=\frac{7}{3}$ which is the distance FROM coordinate $(3,0)$ so subtracting we get $3-\frac{7}{3}$ or $\frac{9-7}{3}=\frac{2}{3}$ and we are done.


modular arithmetic
number theory
relatively prime
number theory unsolved

prove that existing infinitely n such that
n+1 /$ 3^n$+$ 2^n$

[quote="kihe_freety5"]prove that existing infinitely n such that
n+1 /${ \{3^n}$+${ \{2^n}$[/quote]I don't understand your notation. :huh:

[quote="Jure the frEEEk"][quote="wangsacl"]Is it $ n \plus{} 1|3^n \plus{} 2^n$?[/quote]
because if it is i can't find a single such $ n$[/quote]

That's not surprising, since there aren't any. I wonder if this is actually what the original problem should have been? It's a nice problem, I enjoyed solving it.

Claim: $ n\plus{}1$ does not divide $ 2^n\plus{}3^n$, for $ n$ a natural number.

Proof: Suppose $ n\plus{}1$ does divide $ 2^n\plus{}3^n$.

We will proceed to show that $ n$ must be even, and is actually divisible by 4, well actually by 8, wait, actually by 16, oh wait, actually .... which will lead to a contradiction by showing that $ n$ must be divisible by an arbitrary power of 2.

First, since $ 2^n\plus{}3^n$ is odd, $ n$ must be even.

Suppose now that a prime $ p$ divides $ 2^n\plus{}3^n$. Obviously $ p$ cannot be 3. Now,

$ 2^n \plus{} 3^n \equiv 0 \pmod{p}$

$ 2^n \equiv \minus{}(3^n) \pmod{p}$

and since $ p$ is relatively prime to 3 we can divide by $ 3^n$: letting $ a \equiv \frac{2}{3} \pmod{p}$, we have

$ \minus{}1 \equiv a^n \pmod{p}$ (*)

but we know that $ n$ is even, so let $ n\equal{}2m$ and note that $ (a^m)^2 \equiv \minus{}1 \pmod{p}$. So the order of $ a^m$ mod $ p$ is precisely 4, which forces $ p \equiv 1 \pmod{4}$ (the order of the multiplicative group mod $ p$ is $ p\minus{}1$, and the order of any element in a group divides the order of the group).

We've shown that under the assumption that $ n$ is even, all prime factors of $ 2^n\plus{}3^n$ are 1 mod 4. But this implies that [i]any[/i] factor of $ 2^n\plus{}3^n$ is 1 mod 4, in particular $ n\plus{}1$ is such, so $ n$ is actually divisible by 4. But now we can repeat the last steps of the argument: Letting $ n\equal{}4k$ in (*) we conclude that $ (a^k)^4 \equiv \minus{}1 \pmod{p}$ and therefore $ (a^k)^8 \equiv 1 \pmod{p}$ and from these two equations it follows that the order of $ a^k$ mod $ p$ is exactly 8, meaning that $ p$ must be 1 mod 8 so $ n$ must be divisible by 8, and continue.... By iterating this procedure we see that $ 2^t | n$ for all $ t$ which is nonsense. So $ n$ doesn't exist, QED.

my point also... only it seems u were faster :) . induction. $ n$ even, than if $ 2^k|n$ than $ 2^{k \plus{} 1}|n$ and therefore no solutions


What does it mean when someone has perfect pitch?  What does relative pitch mean?

You may want to look at this discussion:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1147329178&t=34600[/url]

From what I understand (this may be wrong) it's the ability to identify a random note when heard, among other things.

Relative pitch is a good grasp of intervals, so, if i played a random note, identifying it wouldn't be possible (without perfect pitch) but, playing another note, and telling u what it was, if u had good relative pitch, u could tell me the unknown note based on intervals.

[url]http://en.wikipedia.org/wiki/Relative_pitch[/url]
[url]http://en.wikipedia.org/wiki/Perfect_pitch[/url]

Ha, I've played that piece of which you speak before.  The cadenza's chords aren't that hard compared to those of other concerto cadenzas.  Listen to some recordings and you'll get a general idea of what the chords are supposed to sound like.

[quote="CatalystOfNostalgia"]It is also being able to identify exactly when you are in tune and when you are *just* off, in non-percussive (piano is included in percussive) instruments. My gross lack of perfect pitch is one of the many reasons why I suck at violin. For example my first movement cadenza of Mozart 3 is disastrous as I have no sense of when those chords are in tune (among many, many, many other reasons for which all of my Mozart 3 is disastrous).[/quote]

Actually, I don't think so. Many people with perfect pitch cannot play in tune.

Having a perfect pitch means nothing more than that you have the biological ability to identify pitches. Say, if I generated a 880Hz wave, one with a perfect pitch should be able to tell me "that's A5," provided one has received adequate training in music.
Having a perfect pitch does not make you a musical genius. You may have been born with a perfect body to be a runner, but you still can't be a good runner if you don't run. Surely, it is very convenient to have a perfect pitch when you study music. But if you don't know just what the heck a perfect fifth is, being able to hear a C and a G won't help you at all. Same with playing in tune: it requires training. You might know that you're out of tune, but if you don't have the physical techniques to put yourself back in tune--too bad.


How many ways are there to place $k$ marbles in any of the positions $1,2,\dots,n$ (which are evenly spaced around a circle) such that no two marbles are neighboring each other? (Of course, $k \leq \left \lfloor{\frac{n}{2}}\right \rfloor$, and each position can have at most one marble.)

[quote="paladin8"][hide]
Let O be a marble. We have $k$ of them:

OO...OO.

Then we have $n-k$ spaces, or X's. There needs to be at least one X after each O, but beyond that it doesn't matter. So we have

OXOX...OXOX.

We have $n-2k$ X's remaining, which we can place in any of the $k$ spots between O's. Using our standard argument, this should be

$\binom{n-k-1}{n-2k}$.[/hide][/quote]
I have a few questions:
First of all, are you sure it's $\binom{n-k-1}{n-2k}$ and not $\binom{n-k-1}{n-2k-1}$?
Second, what about accounting for the fact that the marbles are placed in positions $1,2,\dots,n$ on a circle?  :wink:
Third, are you sure this method won't overcount the number of arrangements? It looks to me like it does.  :|

[quote="paladin8"]OK, better idea. Since it's a circle, we want the number of X's between each O to be nondecreasing (that way we won't have any rotated versions of each other and the such). So basically it's the number of partitions of $n-2k$ into $k$ nonnegative integers. Which I'm not sure of how to count at the moment.[/quote]
Well, counting that is equivalent to counting the number of ways to partition $n$ into $k$ [i]nonordered[/i] nonnegative integers. (Hope that simplifies it a bit... but it probably doesn't.  :P ) Unfortunately, this is a lot harder than if they were ordered.  :wink:


A daredevil plans to bungee-jump from a balloon 65.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. (Model his body as a particle and the cord as having negligible mass and obeying Hooke's law.) In a preliminary test,hanging at rest from a 5.00 m cord, he finds that his body weight stretches it by 1.50 m. He will drop from rest at the point where the top end of a longer section of cord is attached to a stationary balloon. What length of cord should he use and what maximal acceleration will he experience?

Okay, well...
firstly: call the guy's mass $m$.  So far, with Hooke's Law, we know that stretching the 5 meter cord 1.5 meteres will equal the guy's weight, so, with a net force of 0, we have $k_0(1.5)=mg$, and $k_0=mg/1.5$.
approximating g as $9.8 m/s/s$, we have $k_0= 6.5m$(with rounding)

Now, let's work with the new bungee cord.  Let's say it's of length $l$.  Then, because you're adding springs in series, the spring constant of this new spring is $k_1=k_0*(5/l)$.
Which is, you know, something.
Actually, not just *something*.  I think it is(or rather, my calculator thinks) $32.5m/l$

So now we know *one* more thing.  The guy stops 10 meters above when he goes splat.

So, working with the full extension of the cord(when it stops him right after the jump).  Let's call the distance it stretches $x$, where $x=55-l$ ( but it's so much easier to write $x$, you see?)
Then you have the energy equation, $1/2k_1x^2=mgh$
or, with the variables we had,

$1/2(32.5m/l)(55-l)^2=539m$

and, well,
cancelling out the $m$'s
and letting my calculator do the dirty work again(i'm sorry!)
I get
$\boxed{l=25.76}$

mm?


Does somebody know where i can find questions from the moroccan mo, Algerian mo etc.

Thanks a lot

you can find the moroccan mo in     http://www.arabmaths.site.voila.fr/
 Meghrabi, will you  participate in moroccan mo in 21 april?


Gauss
linear algebra
matrix
linear algebra theorems

I'm searching for some website, book or article where I can find the comparison of the most important numerical methods for solving systems of linear equations (gaussian LU factorization, Cholesky, and others). What I whant is to see the comparison of computational efficiency of that methods.

The Burden & Faires textbook has reasonable discussions of this. That's probably also true other elementary numerical analysis textbooks, as well. (I'm not a numerical analyst, but I taught the course a couple of times, and that was the text I used.)

If the question is about raw operation count, Gaussian elimination methods (such as LU factorization) tend to run something like $\frac13n^3$ mulitiplications/divisions for a full $n\times n$ system and about the same number of additions/subtractions. Iterative methods (Gauss-Seidel, SOR, etc.) can potentially be significantly faster: $O(n^2)$ computations per iterative step, but the number of iterative steps can be much smaller than $n.$ But all that's subject to the question of whether they converge at all - they can't be used on every system. Special types of sparse matrix - tridiagonal, for example - get their own special counts.

Some of this gets modified when suitability for parellel processing becomes an issue.


AMC
USA(J)MO
USAMO
geometry
trigonometry
inequalities
AMC 12

Ouch, 4 points.  Probably 2 points from problem 1, and 2 points from problem 4.

i'm not a first timer...... but you should add numbers greater than 22 because k18o7 got a 23 as a first timer...

Got a 12. Ahh so much for MOP as a soph. Maybe I'll make it as a junior. At least I now have something to work toward; I thought MOP was out of reach till I took USAMO this year. :lol:

[quote="MysticTerminator"]7 yo since for some reason I failed on alice and bob[/quote]
Hmmm... it was 10 for me since I apparently did not know geometry? Or for some reason was not tipped off by the suggestive angle conditions on #1. That was a pretty crazy year as it was a hard USAMO, yet the top three scores were 42, 41, 40. This year, however, the race is very close.

There's no option for 23 and above...

Yes, I'm a first-timer, a 23.

and dude where's the 42 option.

Yes, I'm a first-timer, a 42.

[quote="probability1.01"]This year, however, the race is very close.[/quote]
No, it's not close at all. tripathy, zhai, lawrence, gong, lars and steinhardt are going to make the team.
btw, i wonder is lars' water bottle will be safe this year...

[quote="MysticTerminator"]7 yo since for some reason I failed on alice and bob[/quote]
Hmm, Eve is what messed me up.

Actually, paladin, I'm quite sure at least a few people put 4013. But that probably would've been their entire solution as well

...Klebian Paladin was referring to #2 not #4...clearly people missed number 4.

Ah, right. Gah, misreading ftl. Yeah, I doubt you'd get any points for saying it's impossible, even if you wrote some random stuff to go with it.

21 as a first timer in 2005 (also my first time taking the AMC 12 and AIME). The sad thing is that I've gotten 15 and 14 since then. I've got one more year...

And sure there is correlation between AMC and USAMO. They both have math problems; it's not like one's a organic chemistry test and the other a European history test or anything. I don't have statistics to back that up, though, and of course bad days and the such can alter either test score significantly.

yeah, i didn't take the exam, but i think i can definiatly get a 42 if i take it :D

This year was my first time taking the USAMO and I got an 11. I'm in 8th grade.

[quote="kstan013"][quote="diophantient"][quote="kstan013"]My first time this year...got a 9...very pleased in spite of the harsh ripping remarks made at the holes in my proofs :D[/quote]
I thought you'd solved 1 and 4 completely?[/quote]Well sorta. 1 was good, 4 was a half handwave, and 6 was just the "if" part, so it was also like a handwave.[/quote]
Hmm, isn't proving the equality case pretty trivial?
I think the major case is for proving it for the rest.
Also, that's right, I forgot about Ono's inequality.

we got comments? Those were just what I thought of me proofs.

Oh i'm sorry i misinterpreted one of your other posts; the remarks you were referring to were made by AoPSers

I got 6 as a first timer, I'm in 8th grade. My proof for #1 was BS, which brought me down from a potential 10ish to a 6. I'm pretty satisfied, however, though I'm pretty sure I could have found the correct solution to #1 had I known the first one I came up with was flawed.


Who got smoked on the AP Stats Exam?
I did.
MC: 28 answered, score of 35.
Free Response: Got 4 and a third right in section 1.  Didn't get a chance to start section 2.  :(  I need better time management skills. That's a score of around 32 ish, I believe. My total score is a 67ish, and you need 68-100 for a 5. So I am pretty worried. Well, that's what I get for cramming the last three-quarters of Statistics into two days!
I think Statistics is easy; its just because everyone else said it easy I didn't study, thinking I would easily get a 5. But really, there's a lot of material, even if it was easy.
I am hoping for the best.

It really isn't as easy as its claimed to be... coming from someone who started studying over Spring Break (middle of April) might be not entirely accurate, but that's ok.

I answered 39 MC questions (in about 30 minutes, they give WAY too much time IMO) and all of the FR (though some of my written explanations could be questioned; I'm just glad there was no designing experiment question). I think I should get a 5.

Well I took a class in school, so it wasn't hard. However, careless mistakes :( like not paying attention to discrete/continuous data, so... I'm 95% confident I got a 5.

Well, of course if you try self studying it in a few weeks you will find it hard ;) .  I self studied it also but I tried staying up with the class at my school (feel behind at inference cause I got lazy though).  On all my practice tests, on the MC, I was always border line 4/5.   I think I did well enough to get a 5 but it could go either way.


If a plane is carpeted with equal regular polygons, and there are $ a$ sides in every one, and there are $ b$ of them which share one common point, find all ordered pairs $ (a,b)$.
[hide="Answer"]
(3,6)
(4,4)
(6,3)
[/hide]

Yes, regular tessellations.
I said it's easy :wink:


Bob and Meena play a two-person game which is won by the first person to accumulate 10 points. At each turn Bob gains a point with probability of $ \frac{1}{3}$ . If he doesn't get a point, then Meena gets a point. Meena is now ahead 9 to 8. What is the probability that Meena will win? Express your answer as a common fraction.

This is 1 minus the probability that Bob will win. $ 1 \minus{} {(\frac {1}{3})}^2 \equal{} \frac {8}{9}$

Why can't it be 2/3? Meena gets the point with probability 2/3.

First, please hide your answers/solutions. There are two ways to do this:
[hide=Sol #1]First, we find the probability that Bob will win. To do that, he has to win 2 times in a row. Each time, he has a $1/3$ chance of winning. So, the probability that Bob wins is $(1/3)^2$ which is equal to $1/9$. Now, the sum of the probabilities are always 1. So, we do $1-1/9=8/9$, the probability that Meena wins.[/hide]
[hide=Sol #2]If you don't like that way, you can use minor casework.
Case #1: Meena wins on the first turn. The chance of that is $2/3$
Case #2: Bob wins this turn, Meena wins the next turn. This probability is $1/3*2/3=2/9$
There are no more cases. Adding these two sums up $6/9+2/9$ gives us a final answer of $8/9$[/hide]


quadrangle $ ABCD$ is written in a circle. The diagonals $ AC\perp BD$ cross in point S.
Prove that a line $ m\perp BC$ such as $ S\in m$ halves $ AD$.

[img]http://img129.imageshack.us/img129/4765/pictnl4.png[/img]

[quote="red_dog"][hide]
Let $ M \equal{} m\cap AD,\ N \equal{} m\cap BC$
We have:
$ \widehat{SAM}\equal{}\widehat{SBC}\equal{}\widehat{CSN}\equal{}\widehat{MSA}\Rightarrow MS \equal{} MA$
$ \widehat{SDM}\equal{}\widehat{SCB}\equal{}\widehat{NSB}\equal{}\widehat{MSD}\Rightarrow MS \equal{} DM$
Thus $ MA \equal{} MD$
[/hide][/quote]

Sorry to ask this, but what does the widehat symbol mean? I think $ \cap$ means something like...not in the set, but I'm not sure of that either.

Does it work to just say that $ \overline{m}$ is the angular bisector of $ \widehat{ASD}$? Angular bisectors do cut the opposite side of a triangle in half, but is $ \overline{m}$ the bisector?

[quote="Kyron"]Angular bisectors do cut the opposite side of a triangle in half[/quote]

No they don't. You are confusing angle bisector with median.

Just consider this picture:

[asy]import cse5; pair A=MP("A",D((0,0))), B=MP("B",D((20,0))), C=MP("C",D(IP(CR(A,5),CR(B,20))),N), D=MP("D",D((4,0))), E=MP("E",D((10,0))); D(A--B--C--cycle,black); D(C--D,red+linewidth(2)); D(C--E,blue+linewidth(2));[/asy]

The blue line is the median, and the red line is the angle bisector.


Hey guys
I am going to count on u guys whenever i need any help...thanx 4 all the support.. :lol:
Do u guys know any good websites other than this one for problems on [b]logic, reasoning and analysis? [/b] Basically brain testers other than Su-doku..
And secondly, has anyone been for the Infosys Students Training Program before?It's supposed to be very good... If so, how is the entrance test?
Thanx

Hello,
  This is not a Logic and Reasoning Problems Site.This is more towards Mathematics.Logic,as a branch of study has quite different foundations.By the way, what you mean seems to be something for competitive examinations here in India like NTSE or CAT.You may refer to any site that is good for CAT.Mathematically,however,the standard of these exams is much lower than IMO.My brother has attended the Infosys Program.I think you wil be intimated about in school.Anyway,the entrance test was fairly simple with basic level Mathematics questions.At Infosys,they were taught Basic Programming Skills and Web Designing.I am not much interested in Computers.So I may not give a very good picture of it.
Sathej

Hey
I should have been clear with punctuation, I meant  any good sites other than this one..hehe
Anyway, have you got any good links? Google doesn't seem to be very helpful on this one..keeps pointing me back here! or to websites like "Build your own Bird Cage" LOL!!  :rotfl:
Links would really hit the spot.
My school is notorious for being like an iceberg- only 1/10th they tell u and hide 9/10th from u...
Anyway...
Yeah, maybe sumthin on the lines of CAT preps..but any level will do yaar! :lol:


I was curious if someone would tell me what a g-mail invite is, what g-mail is and for what level of a student it is designed for.

G-mail, as from the sound of it, is a mail service, recently launched by Google Inc. Its main distinctive features are: 1Gb storage space, no pop-up ads, no fee, ability to search (with Google engine) in your e-mails through the www interface, larger attachement sizes (if not unlimited, I am not sure).

You can make an account only on invitation base, as the system is still in BETA (testing) stage. Some people have invitations available.

There is another thread about G-mail. This is not the thread to be talking about mail services (ie the thread was initially located in Problem Solving -> Getting Started) so I've moved it.

Attachment size limit is 10 MB.

In any case, if you want it check out the thread on Fun and Games


How many prime numbers have squares between 100 and 300?

The perfect squares between 100 and 300 are:

$ 10^2, 11^2, 12^2, 13^2, 14^2, 15^2, 16^2, 17^2$

The prime ones are 11, 13, and 17, so the answer is $ \boxed{3}$.


Let $a,b,c$ be rational numbers. For each $n \in \mathbb Z$ we consider the equation
\[E_{n}: (a+n) x^{2}-2 (b+n) x+c+n = 0 . \]
(a) Prove that if $a,b,c$ are in arithmetic progression, then the equations $E_{n}$ have a common solution.
(b) Prove that if every equation $E_{n}$ has real roots, then $a,b,c$ are in arithmetic progression and each equation has rational solutions.

[hide="(a)"]
Notice that $n=0$ corresponds to the equation
\[ax^{2}-2bx+c=0 \]
We wish to show that the roots of this equation are also roots of all other values of $n$. Notice that $c-b=b-a\iff 2b=a+c$ so that
\[ax^{2}-2bx+c=0 \iff ax^{2}-cx-ax+c=0\iff ax(x-1)-c(x-1)=0 \iff (x-1)(ax-c)=0 \]
Notice that the above factorization could have been "guessed" by considering the unit and $x^{2}$ coefficients respectively. This method used on the general equation yields:
\[(a+n) x^{2}-2 (b+n) x+c+n = (x-1)[(a+n)x-(c+n)] = 0 \]
We verify: $-(a+n)x-(c+n)x=-(a+c+2n)x=-(2b+2n)x=-2(b+n)x$
Thus the common solution 1 since the $x-1$ term always factors out. The other solution is $\frac{c+n}{a+n}$ :P
[/hide]
[hide="(b)"]
It is well known that such an equation with two real roots must have:
\[(b+n)^{2}\geq (a+n)(c+n)\iff b^{2}\geq ac+n(a+c-2b) \]
If $a+c>2b$ then for sufficiently large $n$ we can construct a contradiction.
If $a+c<2b$ then for sufficiently small (large negative) $n$ we can construct a contradiction.
Thus, $a+c=2b$ and we are left with the same equation as before, which has $a,b,c$ in arithmetical progression and each equation has $1, \frac{a+n}{c+n}$ as its solutions which are both rational.
[/hide]


Could anyone post step by step soln of this question :)

A straight line moves in such a way that  the length of the perpendicular on it from the orgin is always p.
Find the locus of the centroid of the triangle which is formed by the line and the axes.

Thanks,
mittal       :roll:

Hello
Well I m sorry as I m solving it quickly,I m afraid I didn't done any mistake.Hope u will understand it.
Consider the equation of line $ xcos\theta\plus{}ysin\theta\equal{}p$ which meet at axis at $ (h,0)$ and $ (0,k)$.
by putting h and k separately in the equation we will get
$ sin\theta\equal{}\frac{p}{k}$
$ cos\theta\equal{}\frac{p}{h}$
the coordinate of centroid =$ (\frac{h}{3},\frac{k}{3})$correspond to $ (X,Y)$
$ {sin\theta}^2\plus{}{cos\theta}^2\equal{}1$
$ \frac{p^2}{X^2}\plus{}\frac{p^2}{Y^2}\equal{}9$
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\equal{}\frac{9}{p^2}$
Thank u.


geometry
geometric transformation
rotation
trigonometry
geometry unsolved

It is not perfect but it's better.
What about the solution?

I think it is an interesting problem.

What come to my mind is:
1. We may drawn some diagonal and some circles.
2. Other think is a rotation probably.
3. I'm sure the problem can be solved using some extension of segments, similaritues and Cosine law
but it require more calculations.

What about you?

I'm sorry but the statement in this problem is not correct.


geometry
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be a triangle with $ O$ and $ H$ its circumcenter and orthocenter, respectively. The bisector of $ \angle BAC$ meets $ BC$ at $ P$. Find the product of the sides of the triangle, given that $ BC\equal{}1$ and $ AHPO$ and $ BHOC$ are concyclic.
(The test was to be solved in two days, each one with 3 problems and 3,5 hours).

Try to do that, please... I wanna see other solutions, because I didn't find any more technics to solve it . The final result is so strange that in the test I had some doubts about my solution, but it was right!

is it $ \angle A\equal{}60, \angle B\equal{}90$? i have a few questions on  the problem: are BHOC and AHPO supposed to be convex quads? also, is it allowed for a couple of points to be the same, for example $ B\equal{}H$?

Yes, campos, you are right. This is exactly the resut. The problem didn't mention conditions about coincidence of points and convexity of quadrilaterals, so it's allowed everything. I doubted myself when I arrived to this result, because it's really strange (we havea degenerated quadrilateral). From where you are it's simple to conclude the answer.
I was the only one from all the students who solved this problem in the olimpiad, but it's more strange than difficult (maybe this is the reason). I was wondering if you could post your complete solution, because I would like to see how you did it.

one should try first the cases when $ H$ coincides with any of the vertices, this is when one of the angles is a right angle... this would lead to the previous solution and also $ \angle A \equal{} 60, \angle B \equal{} 30$ (if one allows $ BCOC$ to be concyclic  :maybe:)...

i'm not used with the notations of oriented angles, so let's make this with bare hands... the values of $ \angle BHC$ are $ A$ or $ 180 \minus{} A$ (if angles B,C are obtuse or not, respectively) and the ones of $ \angle BOC$ are $ 2A$ or $ 360 \minus{} 2A$ (if angle A is acute or not, resp.)... without assuming that $ BHOC$ is convex, we may have $ \angle BHC \equal{} \angle BOC$ or $ \angle BHC \plus{} \angle BOC \equal{} 180$.. checking the possible cases we're only left with $ \angle A \equal{} 60$ and $ \angle A \equal{} 120$...

when $ \angle A \equal{} 60$ and the triangle is acute-angled it's very easy to show that $ \angle AHP \equal{} 90$ (since $ AP$ bissects $ \angle HAO$ and $ AH \equal{} AO$, etc), which is impossible... the case when some of $ B,C$ is obtuse and $ \angle A \equal{} 60$ may be something similar (i haven't tried this)...

when $ \angle A \equal{} 120$ you have that $ AH \equal{} AO$.. also that $ APOH$ is convex, and $ AP$ is an external bisector of $ \angle HAO$, and since $ APOH$ is cyclic we must have that $ PO \equal{} PH$.. since $ A,P$ lie in the same arc $ OH$ and both of them lie in the perpendicular bissector of $ OH$ we must have that $ A \equal{} P$ which is a contradiction...

and that's it, i think (?)... this problem is actually very weird  :o

My solution - Let $ \angle BAC \equal{} \alpha$, so we have $ \angle BOC \equal{}2\alpha$, and form $ BHOC$ be concyclic we have $ \angle BHC\equal{}\angle BOC$, and from there, by chasing angles it's easy to find $ \alpha\equal{} 60^\circ$. Now, $ H$ and $ O$ are isogonal conjugates, and $ AH\equal{}2r \cos \alpha$, where $ r$ is the circunradius, or $ AO$. This implies $ AH\equal{}AO$ and $ AHO$ and $ PHO$ are isosceles. Rememering that $ AHPO$ is concyclic, it's easy to see that $ \angle AHP \equal{}90^\circ$, AN ABSURD when $ H$ is not in side $ BC$, because would implie $ BC$ is parallel to $ HP$, but they meet at $ P$!
So $ B\equal{}H$, and the rest is easy (answer $ \equal{}\dfrac{2}{3}$).
I think when they give you the points of the quadrilateral and that they are concyclic, you should assume its convex hull.

i found the problems from the rioplatense here [url]http://onemperu.wordpress.com/2008/12/13/pruebas-de-la-olimpiada-rioplatense-2008/[/url], and in the problem it is given that $ \angle A$ is acute, so we should not consider the case when $ A\equal{}120$... the rest is the same...

Oh, I forgot this detail.
REMEMBER, PEOPLE: the triangle is acute!
If someone used any different idea, please post it! Thanks in advance.

the triangle can't be acute-angled, because we have that $ \angle B\equal{}90$... the spanish-version says that $ A$ is an acute angle, and that's it...

Here is the official link, wich says that angle $ \angle BAC$ is acute, not necesseraly all the angles of the triangle:
http://www.oma.org.ar/enunciados/omr17.htm
So CORRECTION: $ \angle BAC$ is acute. I hope more solutions...