vector
induction
linear algebra
linear algebra unsolved

This is not so bad, nice & easy, I got this question in a practice test today.

Let $U$ and $W$ be subspaces of a vector space $V$ such that $U \cap W = \{0\}$.
Prove that there exists a subspace $W'$ of $V$ satisfying \[ W \subset W',\ \textrm{ and }\ U \oplus W' = V.  \]

[quote="Peter VDD"]... I don't really see the point in this question... or maybe you start from another point than we did?

But let {e1..ek} be the vectors spanning U, {f1..fm} the vectors spanning W. They thus form an n+k-dimensional subspace of V. Now we add the minimum number p of vectors {g1..gp} so that all vectors together span V. Thus W' = span(fi,gi)?[/quote]
$V$ may not be finite dimensional

so uhm arne, infinite dimension seems a bit too hard for a test? or is it?

We do cover infinite dimensional vector spaces, but the problem was actually for a finite dimensional vector space.

Peter, I did NOT say that it is hard, right?

[quote="Arne"]Peter, I did NOT say that it is hard, right?[/quote] Well, I assumed so since it was in problems forum, not in computations and tutorials... (last time you told me that was a distinction you made between trivial and nontrivial)

I indeed assumed V was finite, however I suspect the theorem to be true for infinite dimensions as well. Anyone has a more abstract proof for the general case? My proof holds completely, except for the part "the minimum number p of ..." where this would fail for infinity. Actually, there's just a theorem saying you can complete the basis of a subspace to a basis of the larger base, but I don't know wether it's true for infinte dimension aswell. (I guess it's also true, but counterexamples can come out of nowhere here)

Anyway if it's true, there's no real difference between the finite and the infinite case here.

"the minimum number p of..."

That's induction, and to extend to the infinite case it needs to become transfinite induction. Index everything by the ordinals (any initial segment of the ordinals is a well-ordered set) and the proof works in the infinite case as well.

The basis extension theorem: if $A$ is a linearly independent set in $V$, then there is a basis $B$ of $V$ containing $U$.
Proof: Let the elements of $A$ be indexed by ordinals as $a_1,a_2,\dots$. Extend this indexing to contain all elements of $V$. Now, let $B=\{a_i: a_i\not\in\text{span }\{a_j: j<i\}\}$. Clearly, $B\supset A$. We want to show that $B$ is linearly independent and that $B$ spans $V$.
Linear independence is easy: if $\sum_{j=1}^nc_ja_{i(j)}=0$ with each $c_j$ nonzero, there is some $m$ such that $i(m)>i(j)$ for each other $j$. We then have $a_{i(m)}=\sum_{j\neq m}\frac{c_j}{c_m}a_{i(j)}$, which is impossible if $a_{i(m)}\in B$ by our choice of $B$.
For the span, we use (transfinite) induction. The induction hypothesis: For each $i$, $\text{span }\{a_j: j< i\}=\text{span }\{a_j: j< i,a_j\in B\}$.
Suppose this hypothesis is true for $i<k$. Let $v=\sum_{r=1}^nc_ra_{j(r)}$, where $j(r)<k$ for each $r$ be some vector in the first span. Let $m$ be the maximum of the $j(r)$, with index $s$. If there is some $k'$ with $m<k'<k$, $v\in\text{span }\{a_j: j< k',a_j\in B\subset\text{span }\{a_j: j< k,a_j\in B\}$ by the induction hypothesis. If not we split into two cases. If $a_m\in B$, $v-c_sa_m\in \text{span }\{a_j: j< m\}=\text{span }\{a_j: j< m,a_j\in B\}$ by the induction hypothesis, and therefore $v\in\text{span }\{a_j: j< k,a_j\in B\}$. If $a_m\not\in B$, we write $a_m$ as a linear combination of the $a_j$ with $j<m$; combining with the other terms, $v\in\text{span }\{a_j: j< m\} = \text{span }\{a_j: j< m,a_j\in B\} \subset\text{span } \{a_j: j< k,a_j\in B \}.$
These cover all cases, and we have proved the induction hypothesis at $k$
This completes the induction. Taking the union of all these sets, the span of $B$ is $V$ and we are done.


geometry
circumcircle
incenter
geometry proposed

Let A and B lie on the circle (O) and C be a point inside the circle.Suppose that (I) is a circle tangent to segments AC,BC and (O).Let (I) touch AC and (O) at P and Q.Show that the circumcircle of APQ passes through the incenter of ABC

[quote="jayme"]Dear Mathlinkers,
see for example
http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf   p. 227
Sincerely
Jean-Louis[/quote]
My problem said that the incenter of triangle [b]ABC[/b]. Can you post the solution which uses this lemma: "Let I be the incenter of triangle ABC, and let its incircle touch sides BC,CA,AB at M,N,P.CI meet NP at T then T,I,M,B,P are concyclic and T lies on the line connecting the midpoints of AB and BC"? :D

Dear "Love" and Mathlinkers,
my reference works for your problem. You have to adapt...
For your lemma, this result is known as the "Lascases's theorem" that you will found the enonciation on my site:
http://perso.orange.fr/jl.ayme   vol. 1  Another unlikely concurrence.
Sincerely
Jean-Louis

Oh, thanks you very much, jayme! I am finding the applications of this lemma :D
For my problem, I know the solution which uses your reference, but I wish someone to post other solution which uses my lemma :)

Does anybody have the solution for this that uses livetolove212's lemma? I am looking for it as well.

I think this lemma is true if $C$ doesnt lie on circle too.for a proof you can see http://perso.orange.fr/jl.ayme vol. 2 Un remarquable résultat de Vladimir Protassov.


geometry circumcircle algebra polynomial calculus derivative inequalities

[b]5015:[/b] In triunghiul ABC se duc inaltimile AP si CQ, mediana BM si diametrele MK si ML ale cercurilor circumscrise triunghiurilor AQM si respectiv CPM.
Demonstrati ca punctul de intersectie al dreptelor KL si BM se gaseste pe cercul circumscris triunghiului BPQ. ([i]I.I. Bogdanov[/i])
[b]5016:[/b] Un polinom de grad $ n > 1$ are $ n$ radacini distincte: $ x_1, x_2, \dots , x_n$; iar derivata polinomului are radacinile $ y_1, y_2, \dots ,y_{n \minus{} 1}.$
Demonstrati ca are loc inegalitatea: $ \frac {x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2}{n} > \frac {y_1^2 \plus{} y_2^2 \plus{} \dots \plus{} y_{n \minus{} 1}^2}{n \minus{} 1}.$ ([i]M.V. Muraskin[/i]).

Can anyone translate the first problem???

Thanks
Dimitris

[b]5015[/b]: Given triangle $ ABC$, altitudes $ AP$ and $ CQ$, median $ BM$ and diameters $ MK$ and $ ML$ of the circumcircles of $ \Delta AQM$, respectively $ \Delta CPM$, prove the meeting point of lines $ KL$ and $ BM$ lies on the circumcircle of $ \Delta BPQ$. ([i]I.I. Bogdanov[/i])

[b]5016[/b]: A degree $ n > 1$ polynomial has $ n$ distinct real roots $ x_1, x_2, \dots , x_n$; while the derivative of the polynomial has roots $ y_1, y_2, \dots ,y_{n \minus{} 1}$.
Prove the inequality
\[ \frac {x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2}{n} > \frac {y_1^2 \plus{} y_2^2 \plus{} \dots \plus{} y_{n \minus{} 1}^2}{n \minus{} 1}.\]
([i]M.V. Muraskin[/i])


thats cool , i didnt think about it  :)
what i did is after we get

$z^{10}+\bar{z}^{10}=z+\bar{z}$ , i change it into

$2\cos 10\theta=2\cos\theta$ and use the sum to product rule and conclude that

$\sin 9\theta=0$ , $\sin 11\theta=0$

Admittedly, my first idea was the same as yours :) but then I worked backwards from $(z^{11}-1)(z^{9}-1)=0$ and "thought of" multiplying with $z^{10}$  :D


linear algebra
matrix
function
algorithm
probability
graph theory
combinatorics open

Some time ago I have tried to prove Hall's matching/marriage theorem on my own. As a result, I obtained a rather tedious linear-algebraic proof which I don't deem particularly interesting on its own, but it contains an idea that I would love to see applied at some more serious questions.

The proof can be found in a note called "An algebraic approach to Hall's matching theorem" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url]. [url=http://www.cip.ifi.lmu.de/~grinberg/Hall.pdf]The note itself[/url] is a pain to read and is just there to make sure that the proof has no flaws. If you are interested in the proof, try out [url=http://www.cip.ifi.lmu.de/~grinberg/HallAbridged.pdf]the shortened version of the note[/url], which is still 7 pages long (the first 2 of them being prerequisites, however). Given the fact that the Hall theorem can be shown on half a page (though I don't find the proofs quite natural, honestly), this is not quite a sensation.

Anyway, as I said, what matters is the idea, which is basically to fill an adjacency matrix or another kind of characteristic matrix not with $ 0$'s and $ 1$'s but with $ 0$'s and free variables. In the case of the Hall theorem, we have a bipartite graph with $ n$ blue vertices $ B_1$, $ B_2$, ..., $ B_n$ and $ n$ green vertices $ G_1$, $ G_2$, ..., $ G_n$. (Have you guessed that B and G stand for blue and green?) We take an arbitrary field $ K$, for instance $ K \equal{} \mathbb{Q}$, and denote by $ L$ the field of all rational functions of $ n^2$ indeterminates $ X_{1,1}$, $ X_{1,2}$, ..., $ X_{n,n}$ (one indeterminate $ X_{i,j}$ for each pair $ \left(i,j\right)\in\left\{1,2,...,n\right\}^2$) over $ K$. Then, we consider the matrix $ S\in\mathrm{M}_n\left(L\right)$ which has, in its $ i$-th row and $ j$-th column, the entry $ \left\{\begin{array}{c} X_{i,j},\text{ if }G_{j}\text{ is connected with }B_{i} ; \\
0,\text{ if }G_{j}\text{ is not connected with }B_{i} \end{array}\right.$. This matrix "represents" the bipartite graph and has the nice property that it is invertible if and only if the bipartite graph has a perfect matching. My proof of Hall's theorem exploits the properties of this matrix (alas, in a rather ugly way).

Note that $ S$ is not an adjacency matrix, but we could easily define a similar variation on the concept of adjacency matrices, or incidence matrices. It is less clear to me, however, under which circumstances such variations can be of use. Therefore two questions:

- Is the idea/trick of filling characteristic matrices with free variables new?
- Can anyone find a less simple application of the trick?

  Darij

[quote="Ravi B"]The idea of free variables for matching has been used before.  For example, Tutte (1947) showed an analogous if-and-only-if result for non-bipartite graphs, in which the matrix is a certain skew-symmetric matrix.

The idea became popular in computer science about 20 years ago, because it led to randomized parallel algorithms for matching.  The trick is to assign random values to the free variables.  With high probability, if the graph has a matching, then the numerical determinant will be non-zero.  See Mulmuley, Vazirani, and Vazirani (1987), "Matching is as easy as matrix inversion", [i]Combinatorica[/i].[/quote]
hello friend I am working on similar concept so can you tell about the key points which I must take care of?Waiting for reply thanks in advance:)


Let be $ n\in N$ and $ z\in C\minus{}R$ such that $ z^n$ and $ (z\plus{}1)^n$ are reals. Find the values of $ z$.

Use trigonometric representation of complex number and a half angle formula for sine and cosine.

Obviosly n even, and \[ z\equal{}\frac{\minus{}exp(\frac{2\pi i k}{n})}{2\cos\frac{2\pi k}{n}}, |k|<\frac n4 .\]

[quote="Adriana N."] Let $ n\in\mathbb{N}\ ,\ n\ge 3$ . Find $ z\in\mathbb{C}\backslash\mathbb{R}$ so that the numbers $ z^n$ and $ z^{n \plus{} 1}$ are real. [/quote]

[u][b]Proof.[/b][/u] $ z^n\in\mathbb{R}\iff z^n \equal{} \overline{z}^n\iff\left(\frac {z}{\overline{z}}\right)^n \equal{} 1$ . Therefore , $ \frac {z}{\overline{z}} \equal{} \omega_k$ , where $ \omega_i \equal{} \cos\frac {2k\pi}{n} \plus{} \text{i}\sin\frac {2k\pi}{n}\ ,\ k \equal{} \overline{0,n \minus{} 1}$ .

Similarly, since $ (1 \plus{} z)^n\in\mathbb{R}$ we get: $ \frac {1 \plus{} z}{1 \plus{} \overline{z}} \equal{} \omega_p\ ,\ p \equal{} \overline{0,n \minus{} 1}$ . If $ k \equal{} p$ , then $ \frac {1 \plus{} z}{1 \plus{} \overline{z}} \equal{} \frac {z}{\overline{z}}\iff z \equal{} \overline{z}\iff z\in\mathbb{R}$

which is clearly a contradiction . For this reason $ k\ne p$ . But $ \overline{z} \equal{} \frac {z}{\omega_k}$ and plugging it in the relation $ \frac {1 \plus{} z}{1 \plus{} \overline{z}} \equal{} \omega_p$ we finally obtain:

$ \boxed{z \equal{} \frac {\omega_k\omega_p \minus{} \omega_k}{\omega_k \minus{} \omega_p}}$ , for all $ k,p\in\{0,1,2,\ldots ,n \minus{} 1\}\ ,\ k\ne p$ .


trigonometry
logarithms
search
real analysis
function
calculus
integration

Can anyone help me to show that $ \sum_{k\equal{}1}^{\infty}\frac{\sin{kx}}{k}\equal{}\frac{\pi\minus{}x}{2}$?

It is not true. Substitute $ x\equal{}0$ and astonish.

The equality holds for $ 0 < x < 2\pi$:
\[ \minus{} \log \left(1 \minus{} e^{ix}\right) \equal{} \sum_{k \equal{} 1}^\infty \frac {\left(e^{ix}\right)^k}{k} \equal{} \sum_{k \equal{} 1}^\infty \frac {\cos kx}{k} \plus{} i\sum_{k \equal{} 1}^\infty\frac {\sin kx}{k} \\
\Rightarrow \sum_{k \equal{} 1}^\infty \frac {\sin kx}{k} \equal{} \Im \left [ \minus{} \log \left(1 \minus{} e^{ix}\right)\right ] \equal{} \frac {\pi \minus{} x}{2}
\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=755036626&t=176509]This post[/url] of mine contains a slightly expanded version of what Joe Blow just said.

I was searching for the proof that the series does converge where we want it to, but there were too many possibilities and I didn't locate it. I must have at some time posted the proof to the following theorem (which contains the Alternating Series Test as a special case):

[b]Theorem:[/b] Suppose that $ b_k$ is a non-increasing sequence of positive numbers with limit zero. ("Eventually non-increasing" will suffice as a hypothesis, with obvious slight modifications of the statement). Then the series
\[ \sum_{k\equal{}0}^{\infty}b_kz^k\]
converges for all complex numbers $ |z|\equal{} 1,x\ne 1.$  Furthermore, the convergence is uniform  $ |z|\equal{}1,\ |z\minus{}1|\ge\delta$ for any $ \delta>0.$ In fact,
\[ \left|\sum_{k\equal{}n\plus{}1}^{\infty}b_kz^k\right|\le\frac{2b_{n\plus{}1}}{|z\minus{}1|}.\]

The proof involves summation by parts, also known as the Abel summation formula. As for my comment about the Alternating Series Test - what do you see if you let $ z\equal{}\minus{}1?$

Rudin contains a generalization:  $ z^k$ can be replaced by a sequence $ a_k$ such that $ \sum_{k\equal{}1}^{n} a_k$ is bounded.

In that case, you would replace the $ \frac {2}{|z \minus{} 1|}$ in my estimate of the tail-end sum by an $ M$ such that $ \left|\sum_{k \equal{} 1}^na_k\right|\le M\,\forall\,n.$

The key words I was searching for just before I wrote my previous reply were "summation by parts."

[quote="PeterP"]It is not true. Substitute $ x \equal{} 0$ and astonish.[/quote]
The series converges for all $ x.$ The function it converges to is periodic of period $ 2\pi$ but not continuous. In particular, it has a jump discontinuity at $ 0.$ The value it converges to at zero is the midpoint of the jump - halfway between the right-hand and left-hand limits.

Since the partial sums are continuous and the limit function is discontinuous, the convergence cannot be uniform. However (see my claim above), the convergence is uniform on the interval $ [\delta,2\pi\minus{}\delta]$ for any $ \delta>0.$

Square the function, and integrate the square over $ [0,2\pi].$ Do this integral twice: once as $ \left(\frac{\pi\minus{}x}{2}\right)^2$ and once as the series (taking advantage of orthogonality). You'll have one of many computations of $ \zeta(2).$

[quote="t0rajir0u"]Rudin contains a generalization:  $ z^k$ can be replaced by a sequence $ a_k$ such that $ \sum_{k \equal{} 1}^{n} a_k$ is bounded.[/quote]

Isn't that just Abel-Dirichlet criterion for convergence?


geometry
rectangle
trigonometry
analytic geometry
graphing lines
slope
perimeter

I found a book that is about geometry, but I'm not sure if it actually is what you learn in the geometry course, so could someone outline the standard geometry curriculum?

It would really depend on what you look at. Different states might have different curricula, and it depends on whether your book is high school level geometry. The method I learned is proving everything from 24 postulates, but the curriculum is hard to define.

In New Jersey, we begin with some postulates, basic 2-column proofs, then move onto triangles, rectangles, circles, polygons, common formulas and problems, basic-basic-basic trigonometry. Essentially, it's a watered down version of Introduction to Geometry (AoPS.)

I take Geometry this year and, according to my text book, the curriculum (summarized) looks like the following:
 - Defining points, lines, planes, and angles
 - Starting logic and reasoning, leading into two-column proofs
 - Perpendicular and parallel lines
 - Congruent Triangles (not similar yet) - This is where we are right now in the year
 - Intro to Quadrilaterals
 - Proportion and Similarity
 - Intro to right triangles, leading into the idea of very basic trig (like what isabella2296 said)
 - Intro to circles
 - Polygons and their areas
 - Surface area and volume
 - Intro to Coordinate Geometry
 - Coordinate Transformations and such

Basically, you define the things that make up geometry, discover how to prove things (very basic compared to high school competitions, I would say), and take a look at different shapes, including right triangles, which is where trig comes in. My curriculum finishes with coordinate geo.

I, like PowerofPi, am taking Geometry this year. However, it's 8th grade geometry, and is very basic. This is what we have in our textbooks.

-Defining angles, lines, line segments, etc.
-Inductive and Deductive Reasoning, Conditional and Biconditional statements
-two column proofs, flowchart proofs, algebraic proofs, geometric proofs and paragraph proofs
-Parallel Lines and Transversals, Perpendicular Lines, Slope
-Congruent Triangles
-Properties of Triangles i.e. Angle Bisectors of Triangle are concurrent(This is where we are)
-Intro to polygons and quadrilaterals
-Similarity
-Basic Trigonometry and Right Triangles
-Perimeter, Circumference, Area
-3-dimensional figures(spatial reasoning)
-Properties of Circles
-Transformational Geometry
-Coordinate Geometry

This is my basic Geometry course. We look into shapes, and get an intro into 3-dimensional figures. We do some basic trigonometry and later do some coordinate geometry.


linear algebra
matrix
algebra
polynomial
college

here is a problem i made up(probably someone thought about this before):
if A is a square matrix with n rown and n columns and A^2+I=2A, then Tr(A)=n and det(A)=1.where I is the unit matrix(elements on the main diagonal set to 1 and all other elements set to 0).

$A^{2}+I = 2A \Rightarrow A^{2}-2A+I = O \Rightarrow \lambda^{2}-2\lambda+1= 0 \Rightarrow (\lambda-1) = 0 \Rightarrow \lambda = 1$
$(\forall \lambda \in \text{eigenbalue of A})$
by Frobenius Theorem

Thus,
$\text{Tr}(A) = \sum_{i=1}^{n}\lambda_{i}= \sum_{i=1}^{n}1 = n$
$\det(A) = \prod_{i=1}^{n}\lambda_{i}= \prod_{i=1}^{n}1 = 1$

Tr=trace and is the sum of the elements on the main diagonal.
 concerning the demonstration, to wich frobenius theorem is the referrence?or is it to the perron-frobenius theorem?

Frobenius theorem
About $A\in M(n, K)$, we assume $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}$ of solution of the n unit that put repetition of the characteristic equation.
A solution of a characteristic equation of $f(A)\in M(n, K)$ puts repetition for polynomial $\forall f(x)$ then, and it is $f(\lambda_{1}), f(\lambda_{2}), \cdots, f(\lambda_{n})$.
In specific,
$\text{Tr}(f(A))=\sum_{i=1}^{n}f(\lambda_{i})$
$\det (f(A))=\prod_{i=1}^{n}f(\lambda_{i})$

By a problem about Frobenius theorem, there are the following things.
[url=http://math.berkeley.edu/~desouza/Prelims/Fall82/Fall82.html]Berkeley Preliminary Exams - Fall 1982 Problem 6[/url]


superior algebra
superior algebra unsolved

Is there an easy condition for an element of $\hat{\mathbb{Z}}$ or $\mathbb{Z}_{p}$ to come from $\mathbb{Z}$?

[quote="spix"]What to you mean by coming from $Z$? :huh:[/quote]
probably talking about inverse limits.

One has the canonical embedding of $\mathbb{Z}$ into these rings.


LaTeX
limit
real analysis
real analysis theorems

Suppose that  limit (n goes to infinity) a(n) = 0 and b(n)  be a non negative sequence with limit (n goes to infinity) ( b(1) + b(2) + ... + b(n-1) + b(n) ) = infinity.

Prove that

limit (n goes to infinity) ( a(1)b(1) + a(2)b(2) + ... a(n)b(n) ) / ( b(1) + b(2) + ... + b(n-1) + b(n) ) = 0.

More general case:

Suppose that  limit (n goes to infinity) a(n) = A and b(n)  be a non negative sequence with limit (n goes to infinity) ( b(1) + b(2) + ... + b(n-1) + b(n) ) = infinity.

Prove that

limit (n goes to infinity) ( a(1)b(1) + a(2)b(2) + ... a(n)b(n) ) / ( b(1) + b(2) + ... + b(n-1) + b(n) ) = A.

Split the fraction as follows: $ \frac{\sum_{k\equal{}1}^n a_kb_k}{\sum_{k\equal{}1}^n b_k}\equal{}\frac{\sum_{k\equal{}1}^m a_kb_k}{\sum_{k\equal{}1}^n b_k}\plus{}\frac{\sum_{k\equal{}m\plus{}1}^n a_kb_k}{\sum_{k\equal{}1}^n b_k}$. As $ n\to\infty$, the first term goes to zero. The second does not exceed $ \max_{m\plus{}1\le k\le n}|a_k|$, which is small when $ m$ is large.

Use Latex!  Reading that gobbeldygood is painful.

Also, the second follows immediately from the first: if $ \lim a_n = A$, set $ c_n = a_n - A$.  Then $ c_n \to 0$ and we have \begin{eqnarray*} \frac {\sum_{i = 0}^n a_i b_i}{\sum_{i = 0}^n b_i} & = & \frac {\sum_{i = 0}^n (A + c_i) b_i}{\sum_{i = 0}^n b_i} \\
& = & A + \frac {\sum_{i = 0}^n c_i b_i}{\sum_{i = 0}^n b_i} \\
& = & A. \end{eqnarray*}

I think the case $ b_i \equiv 1$ is (part of?) what gets called the Cesaro-Stolz theorem around here.


[quote="me@home"]So you mean, $a,b,c \in \mathbb{Z^{+}}$ find $_{cyc}\sqrt{a^{2}+b^{2}},_{cyc}\sqrt{a^{2}+b^{2}+c^{2}}\in \mathbb{Z^{+}}$
Probably move this to number theory, this could get math-heavy...[/quote]

Ok, I'll make a new thread in Number Theory. :) Thanks

[quote="Altheman"][quote="cincodemayo5590"]Let $a,b,c \in \mathbb{Z^{+}}$ be edge lengths of a rectangular parallelepiped. Prove (or disprove) that there exist no $(a,b,c)$ such that all face and main diagonals are integers.

Source: PreCalc class. The teacher told us that he was not sure whether a "perfect box" was possible, but knew that 6 of the 7 lengths (edges, face diagonals, and main diagonal) could be made integers -- this was for extra credit :D  .  I'm not planning to use anyone's work other than for my own curiosity (dont worry :P).[/quote]

tsk tsk sam[/quote]

 :rotfl: hehe.. I was just wondering if anyone knew how to prove/disprove it. C'mon alex...  :D

the 7 side case is an unsolved problem (so says wolfram), i found a solution for the 6 of 7 measures being integers case, note that i found the solution with relatively little case work


Let $n$ be a positive integer. Prove that \[\frac{1}{3}+\cdots+\frac{1}{2n+1}\] is not an integer.

Suppose $ H(n)=\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}$ is natural number.

There exist a natural number $ k$ which satisfy $ 3^{k}\leq n < 3^{k+1}$
Let $ P(n)= 3^{k-1}\times$ (all positive numbers which satisfy both non-multiple of 3 and smaller than $ n$)

Then $ H(n) \times P(n)$ is not natural number. ($ H(n)P(n)=m+\frac{1}{3}$ where $ m \in N_{0}$ because of $ 3^{k}$)

Contradiction!

(sorry for bad English)

[quote="nicetry007"]Let $ 3^{k}\;\leq \; 2n+1 \;<\; 3^{k+1}\;$.

It is to be noted that every positive odd integer $ m \;\neq\; 3^{k}$ and $ 3\;\leq \;m \;\leq\;2n+1$ has a power of $ 3$ strictly smaller than $ k\;$.

Suppose not. Let $ m \; = \; 3^{l}\cdot s$ where $ l \;\geq \;k$ and $ s$ is odd and $ s \;\geq\; 1\;$.

We have $ m \; = \; 3^{l}\cdot s \;\leq \;2n+1\;< \;3^{k+1}\;\Rightarrow s \;< \;3^{k+1-l}\;\leq \; 3\;$ ( as $ l \;\geq\; k$) $ \Rightarrow s = 1$ (as $ s$ is odd ) $ \Rightarrow m \; = \; 3^{l}\;<\; 3^{k+1}\;\Rightarrow \; l \;\leq \; k$.

But $ l\;\geq \;k$. Hence, $ l \; = \; k \;\Rightarrow \; m \; = \; 3^{k}$, which is a contradiction as $ m \;\neq\; 3^{k}$.

Thus, we can conclude that every odd positive integer $ m \; \neq \; 3^{k}$ and $ 3\;\leq \;m \;\leq \; 2n+1$ has a power of $ 3$ strictly smaller than $ k$.

The denominator of the sum of fractions is $ lcm(3\;,\;5\;,\;\cdots\;,\;2n+1)$ which has $ 3^{k}$ as a factor. Thus, when we take the sum of all the unit fractions, the numerator of every fraction other than $ \frac{1}{3^{k}}$ gets multiplied by a power of $ 3$ and the numerator of $ \frac{1}{3^{k}}$ gets multiplied by an odd number which is not divisible by $ 3$. Hence, the numerator of the sum of all fractions is a number that is not divisible by $ 3$.

Therefore, the sum can never be an integer as the numerator is not divisible by $ 3$ and the denominator is a multiple of $ 3$.[/quote]

Does this work?
By Bertrand's postulate, there exists some $ p$ with $ n \plus{} 1 < p < 2n \plus{} 2$. Also, $ p$ is clearly odd, so we have that $ \frac{1}{p}$ is among our sum. However, it is obvious that $ \frac{1}{p}$ is the only term in our fraction with a denominator divisible by $ p$. As such, if we put all our terms over a common denominator, our number will be in the form $ \frac{pa \plus{} b}{pc}$, where $ p \not | b$. But this fraction clearly cannot be an integer, since $ p \not | pa \plus{} b$, so we are done.

The method user by a Euler- Is can be generalizaded
to show that the sum of the recrocals of the two or more First terms of the arithmetic progression (1,p,2p-1,3p-1,...), where p is a prime, can never be an integer.


algebra
polynomial
number theory proposed
number theory

is there infinite subset $A$such that for each natural number sufficienty large the number of expressing this number in the form $\sum_{a_i,a_j\in A}a_i+a_j$ is constant.

Yes, the idea is known. Use generating functions. We take $ f(x)= x^{a_1}+x^{a_2}+... $ where $ a_i$ are the elements of a. In this case there must exist a constant k and a polynomial P such that $ f^2(x)+f(x^2)=P(x)+\frac{k}{1-x} $. Now we take x to tend to -1 and we get that infinity is bounded.


The following proof was confusing me so any help would be appreciated!

Let $V$ be a vector space and let $W_{1}$ and  $W_{2}$ be subspaces of  $V$.  Prove that $W_{1}\cap W_{2}$ is a subspace of  $V$.  (Do not forget to show that $W_{1}\cap W_{2}$ is nonempty.)

To enter and exit math mode, use the dollar sign rather than [math][/math]. I have edited your post; click on the "edit" button yourself to see what it looks like.

As for the proof: A nonempty subset of a vector space is a subspace if and only if it is closed under addition and closed under scalar multiplication. An immediate corollary is that zero is a member of any subspace. So that's what you know about $W_{1}$ and $W_{2}$ and that's what you want to show about $W_{1}\cap W_{2}.$

Just write out explicitly what "closed under addition and scalar multiplication" means.

Thanks!  So would it be right if I were to say something like:

Let a and b be vectors in the intersection.  This means that a is in W1 and a is in W2, and b is in W1 and b is in W2.  So this means that a + b has to be in W1, as W1 must be closed.  The same goes  for W2.  Therefore, a + b is in W1 and a + b is in W2.  So a + b must be in the intersection.

Then for scalar multiplication.

Let a be a vector in the intersection.  This means that a is in W1 and a is in W2.  Therefore, a times a scalar x will be in W1 as well as W2 by the closure property of scalar multiplication.  So a*x must be in the intersection.

[quote="Kent Merryfield"]That is correct. And the intersection is nonempty because ____[/quote]

Okay, so I'm guessing the intersection is nonempty because it contains both a+b as well as ax?

Even easier. Every subspace (every vector space actually) has to have this...


calculus
integration
calculus computations

I noticed a pattern that for odd integers $n$,

$\int_0^1 x^n e^{-x^2} \ dx = \prod_{k=3}^{\frac{n-1}{2}} k -\left[ \frac{5}{2} \cdot \prod_{k=3}^{\frac{n-1}{2}} k +\sum_{a=4}^{\frac{n-1}{2}} \left(\frac{1}{2} \cdot \prod_{k=a}^{\frac{n-1}{2}} k \right) +\frac{1}{2}\right]e^{-1}, n\ge7$

Let $m = \frac{n-1}{2}$

The RHS of the above equation is equivalent to $-\frac{1}{4}\left(2\sum_{a=4}^{m}\left(\frac{a}{a!}\right)m! -m!(2e-5)+2\right)e^{-1}$

I can't simplify this any further right now, and I'm sure there is a more elegant solution, so I'll keep trying.

[quote="kunny"]Evaluate

\[\int_0^1 x^{2005}e^{-x^2}dx\][/quote]

My solution is as follow:
Let $I_n=\int_0^1 x^{2n+1}e^{-x^2}dx\/$.
It is not difficult to get $I_0=\frac{1}{2}\left(1-\frac{1}{e}\right)\/$.
I fount that $I_n= \frac{1}{2} \left(2nI_{n-1} -\frac{1}{e} \right)\/$ for $n=1, 2,3,\dots$ and hence (if my calculation was correct)
$I_n=\frac{1}{2}\left(n!-\frac{n!\left(1+\sum_{r=1}^{n-1}\frac{1}{r}\right)+1}{e}\right)\/$ for $n=1, 2,3,\dots$

[quote="kunny"][quote="pipi"][quote="kunny"]Evaluate

\[\int_0^1 x^{2005}e^{-x^2}dx\][/quote]

My solution is as follow:
Let $I_n=\int_0^1 x^{2n+1}e^{-x^2}dx\/$.
It is not difficult to get $I_0=\frac{1}{2}\left(1-\frac{1}{e}\right)\/$.
I fount that $I_n= \frac{1}{2} \left(2nI_{n-1} -\frac{1}{e} \right)\/$ for $n=1, 2,3,\dots$ and hence (if my calculation was correct)
$I_n=\frac{1}{2}\left(n!-\frac{n!\left(1+\sum_{r=1}^{n-1}\frac{1}{r}\right)+1}{e}\right)\/$ :?  for $n=1, 2,3,\dots$[/quote][/quote]

$\int_0^1 x^{2n+1}e^{-x^2}dx= \frac{n!}{2} \biggl(1 -\frac{1}{e} \sum_{r=0}^{n}\frac{1}{r!}\biggr)$ for $n=0,1, 2,3,\dots$


inequalities
trigonometry
inequalities proposed

Prove the following inequality:

If $ \sum_{i=1}^{n}{a_i}=\pi$ then $ \frac{\cot{\frac{\pi}{n}}}{n}\geq \frac{\sum_{i=1}^{n}{\sin{a_i}\cos{ai}}}{(\sum_{i=1}^{n}{\sen{a_i}})^2}$

Sorry!!!!!
Didn't realized I missed the sinus!!!!!

$ \frac{\cot{\frac{\pi}{n}}}{n}\geq \frac{\sum_{i\equal{}1}^{n}{\sin{a_i}\cos{ai}}}{(\sum_{i\equal{}1}^{n}{\sin{a_i}})^2}$


induction
number theory unsolved
number theory

Prove that any nonnegative even integer can be uniquely represented as $ (x\plus{}y)^2\plus{}3x\plus{}y$ where $ x,y$ are nonnegative integers

[quote="jbmorgan"]Prove that any nonnegative even integer can be uniquely represented as $ (x \plus{} y)^2 \plus{} 3x \plus{} y$ where $ x,y$ are nonnegative integers[/quote]
[hide="Solution"]
We first prove that any even integer can be represented in the form and then we will prove uniqueness.  If the even integer is in the form of $ n^2 \minus{} n$, then it can be represented as $ n^2 \minus{} n \equal{} (n \minus{} 1 \plus{} 0)^2 \plus{} 3\cdot 0 \plus{} n \minus{} 1$.  Otherwise, we proceed with induction starting at $ n^2 \minus{} n$ and going to $ n^2 \plus{} n$.  The base case was just proven.  Now, if $ 2k$ can be represented as $ (x \plus{} y)^2 \plus{} 3x \plus{} y$, then $ (x \plus{} 1 \plus{} y \minus{} 1)^2 \plus{} 3\cdot (x \plus{} 1) \plus{} (y \minus{} 1) \equal{} 2k \plus{} 2$.  Notice that $ y$ starts out as $ n \minus{} 1$ and it can go to $ 0$, meaning that this process repeats $ n$ times, so can we go from $ n^2 \minus{} n$ to $ n^2 \minus{} n \plus{} 2(n \minus{} 1) \equal{} n^2 \plus{} n \minus{} 2$.  We can replace $ n$ with $ n \plus{} 1$ and proceed in the same manner from $ n^2 \plus{} n$ to $ n^2 \plus{} 3n \plus{} 2$ and keep going, so induction is complete.

We will now prove uniqueness.  Assume that $ (a \plus{} b)^2 \plus{} 3a \plus{} b \equal{} (c \plus{} d)^2 \plus{} 3c \plus{} d$, where $ a\ne c$ or $ b\ne d$.  If $ a \equal{} c$, then we have that $ a^2 \plus{} 2ab \plus{} b^2 \plus{} 3a \plus{} b \equal{} a^2 \plus{} 2ad \plus{} d^2 \plus{} 3a \plus{} d$, so $ 2ab \plus{} b^2 \plus{} b \equal{} 2ad \plus{} d^2 \plus{} d$, so $ (b \minus{} d)(2a \plus{} b \plus{} d \plus{} 1) \equal{} 0$.  Yet, this means that $ d \plus{} 2a \plus{} b \plus{} 1 \equal{} 0$, which contradicts that $ a,b,d$ are positive.  Then, $ a\ne c$ (and analagously, $ b\ne d$), so we may say that $ c > a$.  It follows that $ (a \plus{} b \minus{} c \minus{} d)(a \plus{} b \plus{} c \plus{} d \plus{} 1) \equal{} 2(c \minus{} a)$.  Since $ c \minus{} a > 0$, we have that $ a \plus{} b \minus{} c \minus{} d > 0$.  If $ a \plus{} b \minus{} c \minus{} d \equal{} 1$, then we would see that $ a \plus{} b \plus{} c \plus{} d \plus{} 1 \equal{} 2c \minus{} 2a\implies 3a \plus{} b \plus{} 1 \equal{} c$ and $ a \plus{} b \minus{} d \minus{} 1 \equal{} c$.  This means that $ 3a \plus{} b \plus{} 1 \equal{} a \plus{} b \minus{} d \minus{} 1$, so $ 0 > \minus{} d \minus{} 1 \equal{} 2a\ge 0$, which is a contradiction.  It follows that $ a \plus{} b \minus{} c \minus{} d\ge 2$, so $ a \plus{} b \plus{} c \plus{} d \plus{} 1 \equal{} \frac {2(c \minus{} a)}{a \plus{} b \minus{} c \minus{} d}\le c \minus{} a$, meaning that $ 0 < 2a \plus{} b \plus{} d \plus{} 1 \equal{} 0$, which is a contradiction, so an even integer cannot be represented in the form in two different ways, establishing uniqueness. [/hide]

[hide]
Call $ (x\plus{}y)^2\plus{}3x\plus{}y$ $ 2L$. Then $ L\equal{}{ x\plus{}y\plus{}1\choose 2}\plus{}x$, so $ { x\plus{}y\plus{}1\choose 2}<L\leq {x\plus{}y\plus{}2\choose 2}$
Thus given $ L$, $ x\plus{}y$ is determined. The results follows.
[/hide]


inequalities
function
inequalities proposed

Let $x_{i}\in[a,b], i=1,2,...,n$ are real numbers. Prove the following inequality:
$(\sum_{i=1}^{n}{x_{i}})(\sum_{i=1}^{n}{\frac{1}{x_{i}}})^{ab}\le{(\frac{a+b}{1+ab}\cdot{n})^{1+ab}}.$
In my solution I've used that $a,b$ are natural. I think it can't be proved without that, am I right?

Here is a proof for positive $a,b,x_1,...,x_n$:

With weighted AM-GM inequality we get:

$LS^{\frac{1}{ab+1}}\leq \frac{1}{ab+1}(\sum_1^n x_i)+\frac{ab}{ab+1}(\sum_1^n\frac{1}{x_i})$

where LS is the left side of your inequality.

Since the $x_1,...,x_n$ can be chosen independent from each other we are looking for the maximum of

$f(x):=\frac{1}{ab+1}x+\frac{ab}{ab+1}\frac{1}{x}$.

The function $f$ is convex on $[a,b]$ and attains its maximum at a range of this interval. But we have: $f(a)=f(b)=\frac{a+b}{ab+1}$. Now it is clear that the maximum of $LS^{\frac{1}{1+ab}}$ is $n\cdot\max_{[a,b]} f=n\frac{a+b}{ab+1}$ which implies your inequality.

What was your proof for natural $a,b$ ?

Misha

Ok,Misha, I've solved it with the same method. In the beginning I  used AM-GM but how do you use it for example for ab not natural? But now I understand that my question was stupid
because in this case of course we can use Yung(maybe wrong name) inequality for which we don't need anything on a,b.

[quote="Armo"]Ok,Misha, I've solved it with the same method. In the beginning I  used AM-GM but how do you use it for example for ab not natural? [/quote]

I used the weighted AM-GM: $\prod_1^nx_i^{\alpha_i}\leq\sum_1^n\alpha_ix_i$ for all positive reals $x_1,...,x_n$ and $\alpha_1,...\alpha_n$ with $\sum_1^n\alpha_i=1$.

I hope it is clear now.

Misha

Ok, I understand, this was exactly what I meant but I didn't know that it is called weighted AM-GM. Thank you.
Best regards,
Armo.


Joe and Bob and 7 other kids are going to the movie. There are 9 seats, how many different combination of seating are there? BUT Joe hates Bob so they cant sit next to each other... How many Combination there will be?

Thanks!

[hide="Hint"]You can count the total possible combinations and subtract those with the 2 people sitting next to each other, which can be found by grouping the 2 people together as 1 unit.[/hide]

[quote="ch1n353ch3s54a1l"][quote="kidwithshirt"][quote="ch1n353ch3s54a1l"][hide]There are $9!$ combinations total. But there are $16$ arrangements where Joe and Bob sit next to each other. In each arrangement, there are $7!$ arrangements for the other people. $9!-16\times7!=\boxed{282240}$? [/hide][/quote]

Sorry i need some clear explaination here. Let try to infer wat you just said

9!(total combo for 9 seats)-16(9C2, where how many combo bob and joe can be in 9 seats)-7!(Where hmm... Arrangement without bob and joe?)[/quote]

Um I don't think it's 9C2. It might be, but the way i did it is that there are 16 options for Joe and Bob to sit next to each other: (1,2), (2,1), (2,3), (3,2)...etc. and for each one of those for the other 7 seats there are 7! arrangements, so the 16 arrangements (1,2), (2,1), (2,3), (3,2)...etc. each are multiplied by 7! for a total of 80640 arrangements for Joe and Bob to sit next to each other, and that is subtracted from total, 9!.[/quote]

Pascals pascals... quick wheres my pascals

[quote="kidwithshirt"][/quote][quote="ch1n353ch3s54a1l"][quote="kidwithshirt"][quote="ch1n353ch3s54a1l"][hide]There are $9!$ combinations total. But there are $16$ arrangements where Joe and Bob sit next to each other. In each arrangement, there are $7!$ arrangements for the other people. $9!-16\times7!=\boxed{282240}$? [/hide][/quote]

Sorry i need some clear explaination here. Let try to infer wat you just said

9!(total combo for 9 seats)-16(9C2, where how many combo bob and joe can be in 9 seats)-7!(Where hmm... Arrangement without bob and joe?)[/quote]

Um I don't think it's 9C2. It might be, but the way i did it is that there are 16 options for Joe and Bob to sit next to each other: (1,2), (2,1), (2,3), (3,2)...etc. and for each one of those for the other 7 seats there are 7! arrangements, so the 16 arrangements (1,2), (2,1), (2,3), (3,2)...etc. each are multiplied by 7! for a total of 80640 arrangements for Joe and Bob to sit next to each other, and that is subtracted from total, 9!.[/quote][quote="kidwithshirt"]

Pascals pascals... quick wheres my pascals[/quote]

k anyway the 16 is definitely not a nCp thing... i guess the 16 is right because u have 8 seating u can give to bob and joe and they can have 2 diff. combo on the seatings


Mine sure is. :)

Edit: Hey look, 100% for yes! (I just voted) :P

Mine is, but only so I remember to check for stuff to do. Otherwise it would probably be Warpig or Slashdot.

[quote="sponge008"][quote="kezyxes"]I have a epiphany profile to automiatcaly open 3 googles, one slashdot, one apos, and one gmail on start.[/quote]
THREE googles? Why in the world?[/quote]

Multitasking  :lol:

Nope, google.com

Thinking of changing it into wikipedia or gmail though.

my home page is google, i might change it though because i find myself constantly wandering back to this site

Having the google search bar is essentially the same as having a google page up, for all my intentions at least.   ;)

No homepage? That's just weird. Google is mine, but I'm gonna download the toolbar and change it to AoPS :)

[quote="kezyxes"]I have a epiphany profile to automiatcaly open 3 googles, one slashdot, one apos, and one gmail on start.[/quote]

How do you make multiple windows come up automatically?

[quote="ashpotter"]I have yahoo. I only come to AoPs for classes and sometimes for transcript reading. Most people want search engins for their home page  :ninja:[/quote]

Or you can just use the search bar instead.

[quote="jmadsen"]Mine is google but I have the toolbar so I should change it.

2 people don't have home pages? How do you not have a home page??[/quote]
about:blank

That's my homepage, it just loads a blank page, therefore starts up a lot faster.

Mine is Blank Page. But if I open the Explorer, I'll go to the Mathlinks first.

I just changed mine to AoPS, but I have Google as a tab.


probability
logarithms
probability and stats

in a prison, there are $ 2n$ ($ n$ is very large) prisoners numbered from $ 1$ to $ 2n$. In a room there are $ 2n$ boxes, each one containing one number from $ 1$ to $ 2n$. The next day, each prisoner will enter the room and will be allowed to open $ n$ boxes, and then leave. If each prisonner find his number, they all get realased, if not they are all killed. Today, they can discuss to elaborate a good strategy ..

can you find a strategy that gives them a probability $ P_n \approx \ln(2)$ to get released ? Can you do better (?) ?

Does each prisoner know the the results of the previous prisoners' searches? Or is that information not communicated?

There is an unavoidable $ \frac12$ probability of losing on the first search. $ \ln 2$ is not going to happen.

This looks like the probability decays exponentially. Even if we assume the numbers in the boxes are a permutation, it's hard to beat $ 2^{\minus{}2n}$ by much. I can do $ \frac1{\binom{2n}{n}}$ pretty easily, and I suspect that's optimal.

I meant $ 1 \minus{} \ln(2)$  :D  very surprising isn't it .. (and of course we assume here that the numbers in the boxes are a permutation of $ [1;2n]$)

Very wrong it is. An upper bound on the probability: $ \frac{n}{2n}\cdot\frac{n}{2n\minus{}1}\cdots \frac{n}{n\plus{}1}\approx\frac{e^n}{4^n}$.

A proof: We give the prisoners an extra piece of information: after each success, the location of that winning box is revealed. If the first $ k$ prisoners succeeded, the next knows $ k$ locations that don't win, and knows nothing about the contents of the other $ 2n\minus{}k$ boxes. The obvious best strategy is to pick $ n$ of the boxes that have a chance, for a $ \frac{n}{2n\minus{}k}$ probability of winning if $ k<n$, and certainty if $ k\ge n$. These are independent, and the probability of an overall win is as above.
In the original, with less information, the probability of winning can't be any higher. I'm pretty sure $ \frac1{\binom{2n}{n}}$ is the actual optimum.

I vaguely remember something like that where the strategy was just to go over the cycle in the permutation (i.e., open the box with your own number, then the box with the number in the first box, etc.) and hope that there are no cycles longer than $ n$ in the random permutation but that required to know that the permutation in the boxes is random. This problem is completely different as far as I understand: the content of the boxes is chosen by a sheriff who doesn't like his prisoners much and you have to invent a randomized strategy...

ok, in fact the sheriff is not that clever  :D  he chooses a permutation at random, and the solution I had is the same as the one you said. Is this optimal ?

Well, that strategy does apply to the version I was thinking of- I neglected the possibility of using the numbers to modify strategy on the fly.

In fact, it's not the numbers that make a strategy better than random possible, but the fact that the $ 2n$ tries by prisoners are not independent events (but have the same permutation).
Vaguely speaking, they can try to "either loose all or win all", as all of them loosing isn't worse than one loosing.

And yes, that result is optimal in asymptotics. This is a result by Curtin und Warschauer,


geometry
trigonometry
analytic geometry
trig identities
Law of Cosines
power of a point
AMC

1. Three coplanar circles with respective radius-lengths of 1, 2, and x are externally tangent to each other and internally tangent to a circle of radius 3.  What is x?

2. Quadrilateral ABCD is inscribed in a circle.  If AB = 8, AD = 5, BC = 3 and angle BAD = 60 degrees, compute the lengths of CD, AC and BD.

3. In triangle ABC, AB=AC=17.  Point D is on BC such that CD = BD + 8.   If AD = 16, what is BC?

4. The lengths of the legs of a right triangle are 6:rt3: and 8:rt3:.  An equilateral triangle is constructed on each of the legs of the triangle.  A line is drawn connecting the centers of the two triangles.  Compute all possible values for the area of a square having that segment as one side.

JBL wrote:2. Quadrilateral ABCD is inscribed in a circle. If AB = 8, AD = 5, BC = 3 and angle BAD = 60 degrees, compute the lengths of CD, AC and BD.

Spoiler: [hide]Apply law of cosines to triangle ADB to get BD = 7. Now the angle C is 120 since ABCD is cyclic, so we apply law of cosines in triangle CDB to get 40 = x^2+3a. Solving, CD=5. Now apply Ptolemy's Theorem to the quadrilateral. We have 5*8 + 3*5 = 7 * AC, so solving, AC = 55/7. In conclusion, DB = 7, CD =5, AC = 55/7[/hide]

JBL wrote:3. In triangle ABC, AB=AC=17.  Point D is on BC such that CD = BD + 8.   If AD = 16, what is BC?

[hide]Draw Altitude AM.

AM = h

MD = y

BM = CM = x

we have CD = BD + 8 so,

x + y = x - y + 8

y = 4

and we also have by pythagorean,

x :^2:  + h :^2:  = 289

h :^2:  + 16 = 256

h :^2:  = 240

x :^2:  = 289 - 240

x = 7

2x = BC = 14.[/hide]

JBL wrote:4. The lengths of the legs of a right triangle are 6:rt3: and 8:rt3:.  An equilateral triangle is constructed on each of the legs of the triangle.  A line is drawn connecting the centers of the two triangles.  Compute all possible values for the area of a square having that segment as one side.

[hide]I got

4(25 + 12 :rt3: )

and

73/4[/hide]... im not sure on these...

I agree with your first number. Im not so sure about the second.

Chinaboy, how did you get those? I got 4 solutions that look very different.. although my method was a bit unclean so I may have made many errors. I'll post them in a little bit (let me double check first).

there are four solutions, i think chinaboy has one of them. i did the problem by putting a coordinate system on the problem, and calculating the coordinates of the centers. (The center of the equilateral triangle is one third of the way up its altitude.) Afterwards, i used distance formula.

umm... im not very clear on this myself.

You can have the the equilateral triangles with one pointing outward, and one inward... like inward intersects with the right triangle...

for the outwards one, i used law of cosine...

the last one...

i now get [hide]4(25 - 12 :rt3: )[/hide] ( conjugate of the first one...)

Uh oh, I now count only 3 solutions:

[hide]100 :pm: 48 :rt3: , 100.

6 :rt3: leg has median M, 8 :rt3: leg has median N. Draw perpendicular bisectors to the legs. They meet at P (on hypotenuse). MP = 4 :rt3: , NP = 3 :rt3: .

In the triangles, the distance from base to center is s :rt3: / 6, where s is the side length.

4 cases for positions of triangles: out/out, out/in, in/out, in/in. The out/in and in/out yield the same value, so I got 3 solutions...? You can find the values using Pythagoras.[/hide]

Yah, my answers are the same as wyra's now.

[quote="Chinaboy"][quote="JBL"]3. In triangle ABC, AB=AC=17.  Point D is on BC such that CD = BD + 8.   If AD = 16, what is BC?
[/quote]BD = CD = x
[/quote]

how can cd=bd when it says in the question that cd=bd+8

yeah, i was pouring over the aops books to see if i was just really bad at geometry or something.

Everybody so far is correct.  Number 1 giving you trouble?

Hehe.  Think carefully about the drawing.  Hint follows:

[hide]Get rid of the circles.  Find segment lengths instead.[/hide]

well... all i can figure out is that the centers of the two known circles are on the diameter of the largest one.

Good.  You're half way there.  Now take the hint from the previous problem to get yourself a triangle with a cevian (line drawn from one vertex to the opposite side).  From there, it shouldn't be bad.

i got [hide]5/6[/hide], is that right?

it might be very dumb, because i did it with laws of cosines, and from your sentence, it looks as if u use Ceva... but anyways, ill post it...

We have the points

O: center of circle with radius 3

A: center of circle with radius 1

B: center of circle with radius x

C: center of circle with radius 2

AC = 3

AO = 2

CO = 1

BC = x+2

BO = 3 - x

AB = x+1

we have from laws of cosine...(TriangleOBC),   :theta:  is angleOCB

1 + x^2 + 4x + 4 - 2(2+x)Cos :theta: = x^2 - 6x + 9

we solve for Cos :theta:

cos :theta:  = (10x - 4)/2(2+2x)

using it on TriangleABC, we have...

x^2 + 4x + 4 + 4 - 2(10x - 4) = x^2 + 2x + 1

2x + 7 = 20x - 8

18x = 15

x = 5/6

What is the easy way to do this?

now just go edit that and turn it into a solution :)

i dunno, but could we use power of a point and similar triangles. blah, ive been eating too many oatmeal raisin cookies, that may be part of it

naw, power of a point isn't gonna cut it.  Chinaboy has it, he just needs to write his solution.  Think triangles, not circles.

Added in later edit:
Oh, yes, so he did post it.  Excellent.  Chinaboy, what you did, which can be described as "the law of cosines twice on the center angle" can also be summed up in a nice way as Stewarts Theorem.  If you have triangle ABC and D inbetween B and C, so that AB = c, AC = b, BC = a, CD = m, BD = n and AD = d, then
b2m + c2n = d2a + man, also written man + dad = bmb + cnc, which can be remembered by the mnemonic device "a man and your dad put a bomb in the sink."

i did post it... and it might not be the best way, look up.

I don't believe Chinaboy's answer is correct. There's a mistake in the line where he solves for Cos :theta:. And then again in his law of cosines for Triangle ABC.

This problem is very similar to one that I posted a long time ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6744

The easiest way to solve this is using an advanced technique of geometry called iversion.

well, i didn't simplify cos :theta: , if that's what you mean... but it works out... (and JBL said it was correct)

can u tell me where i made my mistake?

[quote="Chinaboy"]
we have from laws of cosine...(TriangleOBC),   :theta:  is angleOCB

1 + x^2 + 4x + 4 - 2(2+x)Cos :theta: = x^2 - 6x + 9

we solve for Cos :theta:

cos :theta:  = (10x - 4)/[b]2(2+2x)[/b]

using it on TriangleABC, we have...

x^2 + 4x + 4 + [b]4[/b] - [b]2[/b](10x - 4) = x^2 + 2x + 1

2x + 7 = 20x - 8

18x = 15
x = 5/6
[/quote]

1. I believe the denominator in your expression for cos :theta: should be 2(2+x).

2. Then when you were applying the law of cosines to Triangle ABC; side AC = 3, so that 4 should be a 9 and the 2 should be a 3.

Despite those calculation errors your answer still ends up being surprisingly close, but it's not quite right.

can someone solve it by inversion? im curious to see how it works out.

oops, lol, i had it right on my paper...

oh well, what is inversion anyways?


What is the 39th number in the row of Pascal's triangle that has 41 numbers?

The nth row has n+1 numbers, so the row with 41 numbers is row 40.

The 39th number of this row, is the third to last. Using combinations,

$ \binom{40}{38} \equal{} \binom{40}{2} \equal{} \frac{40 \times 39}{2} \equal{} 780$


Four points are arranged on a plane such that a maximum number of
lines can be determined by them. What is the maximum number of
lines?

Maximum number of lines is if no two points are collinear, which means all lines are distinct. Therefore, there are 4C2 = 6 lines


How many of the following numbers are greater than 3 and less
than 4? \\
$ \sqrt[3]{3} \hspace{1cm} \sqrt{7} \hspace{1cm} \frac{8}{3} \hspace{1cm} \frac{25}{7} \hspace{1cm} \sqrt{20}  \hspace{1cm} \frac{3\pi}{2}$

By counting, we see that $ \frac{25}{7}$ is the only number that is both greater than 3 and less than 4. So our answer is $ 1$.

$ \sqrt[3]{3}$

less than 1.5. 1.5 *1.5 *1.5 > 3

$ \frac{8}{3}$

less then 9/3 so less then 3.

$ \frac{25}{7}$

more than 3 less than 28 / 4

$ \sqrt{20}$

greater than sqrt(16) so greater than 4.

$ \frac{3\pi}{2}$

greater than 9 / 2 so greater than 4

Only 25 / 7 fits the conditions.


induction
real analysis
real analysis unsolved

Hi,

 The question below seems really obvious, but how could I show it in a more rigorous way?

  Let S={ a1, a2, a3.....an) be a nonempty, finite set of real numbers.
  a.)Show that S is bounded; b.) Show that least upper bound(lub) S and greatest lower bound(glb) S are elements of S.

 Also, another question: Let S be a bounded set of real numbers and suppose that lubS=glbS. What can you conclude about S?

I think S is a set with only one element, how do you prove it though?

Thanks a million.

You said "Let $S=\{a_1,a_2,\dots,a_n\}.$"

Those dots in the middle are a hint. [b]Any[/b] construction or definition that uses those dots in the middle, [b]any[/b] argument that employs such dots in the middle, or which uses language like "continue likewise until ..." is a construction, definition, or argument by mathematical induction. Very often that underlying induction can be left as informal language, but if you wanted excruciating rigor, you would need to make the induction explict.

That's a hint that perhaps you should try proving this theorem by induction on $n,$ the number of elements in the set. What if the set has two elements? Then there are only two cases: $a_1<a_2,$ or $a_1>a_2.$ Now, can you extent from two elements to $n$ elements with an inductive proof?


[b]The Seven Circles Theorem (see [2]-1974).[/b] Let $A_0, A_1, A_2, A_3, A_4, A_5$  be six consecutive points around the circumference of a circle $({\mathcal O })$.  Suppose circles can be drawn internally tangent to circle $({\mathcal O})$  at these six points so that they are also externally tangent to each other in  pairs (that is, the circle at $A_i$ is tangent to the circle at $A_{i-1}$  and the circle at $A_{i+1}$ , where subscripts are reduced modulo 6  ). [b]Then segments $A_0A_3, A_1A_4$,  and $A_2A_5$ concur.[/b]

 [b]  References: [/b]
[1] H. Martyn Cundy, [i]The Seven-circles Theorem[/i],  Math. Gazette. 62(1978)200–203. MR 80c:51019.
[2] Cecil John Alvin Evelyn, G. B. Money-Coutts and J. A. Tyrrell,[i]  The Seven Circles Theorem and Other New Theorems[/i],  Stacey International. London: 1974.
[3] Cecil John Alvin Evelyn, G. B. Money-Coutts and J. A. Tyrrell,  [i]Le Theorem des Sept Cercles,[/i]  Collection Formation des Maitres en Mathematique, No. 21 Editions CEDIC, Paris: 1975.
[4] Stanley Rabinowitz, [i]The Seven Circles Theorem,[/i]   Pi Mu Epsilon Journal, 8(1987)441--449.
[5] J. A. Tyrrell and M. T. Powell, [i]A Theorem in Circle Geometry[/i],  Bull. London Math. Soc. 3(1971)70--74.

As beautiful as it is, this theorem has been already discussed in this section: http://www.mathlinks.ro/Forum/viewtopic.php?t=5190 .

  darij

I suppose that this proof is well known:
$t_{A_iA_{i+1}} = \frac{A_iA_{i+1}}{R} \sqrt{(R+r_i)(R+r_{i+1})}$
Where: $A_iA_{i+1}$ = chord between touch points at $(O, R)$
$t_{A_iA_{i+1}}$ = external tangent to two circles $(O_i, r_1)$ and $(O_{i+1},r_{i+1})$= $2\sqrt{r_ir_{i+1}}$
Then: $A_0A_1.A_2A_3.A_5A_6$ = $A_1A_2.A_3A_4.A_6A_0$ = $\frac{8R^3 \sqrt{r_0r_1r_2r_3r_4r_5}}{\sqrt{(R+r_0)(R+r_1)(R+r_2)(R+r_3)(R+r_4)(R+r_5)}}$
Therefore: $A_0A_3,A_1A_4,A_2A_5$ concur by Ceva in the circle $(O)$.


geometry
algebra
system of equations
geometry open

[quote="pontios"]what do the blue dots mean???  :huh:[/quote]
They mean that the two angles are equal.

Then what are the ohter symbols on the angle $C$?

I thought that the other symbols (the small line segments) were to indicate that $\angle ACD = \angle BCD$

I don't know this problem solution.
And...
m(DBE)=m(EBC)

m(ACD)=m(DCB)
?????????????????(very hard questions)

Say we have $\triangle ABC$ where $D\in AC$ so $\angle ABD=\angle DBC$. We can derive that:

$BD^{2}=BA\cdot BC\cdot \left(1-\frac{AC^{2}}{BA^{2}\cdot BC^{2}}\right)$

The problem is just a system of equations then.

ihsan_einstein,

from one of your posts in this thread,

"m(DBC) = m(EBC)"....  Wrong.  You can see from the diagram

that angle EBC < angle DBC.

"m(ACD) = m(DCB)"....  This is redundant information because we first see the hash marks indicated on the diagram at the top of the thread indicating that those two angles have the same measure.  So as part of "And..." it is not new information.

in my opinion this given information is not enough.
Because we assume that $\angle EAB=\alpha$, $\angle ABE=\beta$ and $\angle ACD=\theta$.if we use sinus theorem in $\triangle ABE$ we obtain that $\frac{14\sqrt{6}}{Sin(\alpha)}=\frac{35}{Sin(\pi-\alpha-\beta)}$.From $Sin(\pi-\alpha-\beta)=Sin(\alpha+\beta)$ we can write $\frac{14\sqrt{6}}{Sin(\alpha)}=\frac{35}{Sin(\alpha+\beta)}$.Thus

$\frac{Sin(\alpha)\cdot Cos(\beta)+Sin(\beta)\cdot Cos(\alpha)}{Sin(\alpha)}=\frac{35}{14\sqrt{6}}$.Therefore we find that this equation.
$Cos(\beta)+Sin(\beta)\cdot Cot(\alpha)=\frac{5}{2\sqrt{6}}$
we assume $Sin(\beta)=x$ $Cot(\alpha)=y$.Then $Cos(\beta)=\sqrt{1-x^{2}}$ and
$\boxed{\sqrt{1-x^{2}}+x\cdot y=\frac{5}{2\sqrt{6}}\ \ \ \ \ \ \ \ \ \circle d{%Error. "bold" is a bad command.
{(1)}}}$

Now we will use sinus theorem in  $\triangle ACD$.
$\frac{30}{Sin(\alpha)}=\frac{AC}{Sin(\pi-\alpha-\theta)}$.From    $\theta=\frac{\pi-\alpha-2\beta}{2}$
$\frac{30}{Sin(\alpha)}=\frac{AC}{Sin(\frac{\pi}{2}-(\frac{\alpha}{2}-\beta))}=\frac{AC}{Cos(\frac{\alpha}{2}-\beta)}$

$\boxed{AC=\frac{30\cdot Cos(\frac{\alpha}{2}-\beta)}{Sin(\alpha)}=\frac{30\cdot Cos(\frac{\alpha}{2})\cdot Cos(\beta)+30\cdot Sin(\frac{\alpha}{2})\cdot Sin(\beta)}{Sin(\alpha)}\ \ \ \ \ %Error. "bold" is a bad command.
{(2)}}$

we know that if
$Cot(\alpha)=y ,\ Cos(\alpha)=\frac{y}{\sqrt{1+y^{2}}}$.
Then from
$2\cdot Cos^{2}x-1=Cos \ 2x$
$Cos(\frac{\alpha}{2})=\sqrt{\frac{y+\sqrt{1+y^{2}}}{2\sqrt{1+y^{2}}}}$ $\text{and}$ $Sin(\frac{\alpha}{2})=\sqrt{\frac{\sqrt{1+y^{2}}-y}{2\sqrt{1+y^{2}}}}$
if we write these informations in $%Error. "bold" is a bad command.
{(2)}$
we obtain this equation.
$\boxed{AC=\frac{\sqrt{\frac{y+\sqrt{1+y^{2}}}{2\sqrt{1+y^{2}}}}\cdot \sqrt{1-x^{2}}+\sqrt{\frac{\sqrt{1+y^{2}}-y}{2\sqrt{1+y^{2}}}}\cdot x}{\frac{1}{\sqrt{1-y^{2}}}}\ \ \ \ \ \ \ \ %Error. "bold" is a bad command.
{(3)}}$
$\boxed{\sqrt{1-x^{2}}+x\cdot y=\frac{5}{2\sqrt{6}}\ \ \ \ \ \ \ \ \ \circle d{%Error. "bold" is a bad command.
{(1)}}}$

Now if we examine $%Error. "bold" is a bad command.
{(2)}$ and $%Error. "bold" is a bad command.
{(3)}$, for to find $AC$ we can understand that we required to know one of $x, y$. For example, we assume that we know $x$.Then from $%Error. "bold" is a bad command.
{(1)}$ equality we will find $y$ after that we write them in $%Error. "bold" is a bad command.
{(3)}$ equality and also we will be find $AC$.
If we know neither of them, we have got three unknowns($x,y,AC$) and two equations$(%Error. "bold" is a bad command.
{(1)},\bold{(3)}$ equalities$%Error. "bold" is a bad command.
{)}$.So we cannot find $AC$

[size=117][u]Therefore [b]ihsan_einstein[/b] must check given informations[/u][/size]

this problem is correct. And prapered by mustafa yağcı

ihsan_einstein,

"this problem is correct...."
--------------------------

Respond to my prior post in this thread.

Namely, . . . . "m(DBC) = m(EBC)"

This means angle EBC < angle DBC, which is

IMPOSSIBLE according to the diagram supplied by you.

[quote="ihsan_einstein"]this problem is correct. And prepared by mustafa yağcı[/quote]
ok.Now look my post.Where is false in my post. Do u read my post?

ok osman but \cos(a-b)=cosa*cosb+sina.sinb\       .....

you are right.İ wrote false.But it isn't important and doesn't change result.


Prove that the electron of a hydrogen atom doesn't interfere with its nucleus.(using [url=http://en.wikipedia.org/wiki/Uncertainty_principle]Heisenberg's relation[/url] )

[hide="hint"]
if the elctron lies within the nuclei then the $ \Delta x$ of the position of electron $ \equal{}2r_{n}$ where $ r_{n}$ is the radii of nuclei...now use heisenberg's relation[/hide]

Does "interfere" have some special meaning here?  The electron exerts a force on the nucleus, surely.  I suppose that doesn't count as "interference"?

In this case, "interfere with" means, "spiral into" or "occupy the same space as."


For n real numbers $a_{1},\, a_{2},\, \ldots\, , a_{n},$ let $d$ denote the difference between the greatest and smallest of them and $S = \sum_{i<j}\left |a_i-a_j \right|.$ Prove that \[(n-1)d\le S\le\frac{n^{2}}{4}d\] and find when each equality holds.

Not totally sure, but this might be a solution.

[hide]

Let $s_{1}, \dots, s_{n}$ be a sequence so that $a_{s_{1}}\le \dots \le a_{s_{n}}$. Considerer sum $S=\sum_{i<j}|a_{i}-a_{j}|$ and arbitrary $s_{i}, i \le n-1$. Because every pair of $a_{i}, a_{j}$ are counted exactly once $a_{s_{i}}-a_{s_{i}+1}$ is counted $i(n-i)$ times. This implies that we can permute the indices of sequence $(a),$ as we wish.

WLOG we can thus assume $a_{1}\ge a_{2}\ge \dots \ge a_{n}$ and write $S=\sum_{i < j}(a_{i}-a_{j})$.

Now define a sequence $b_{1}= a_{1}-a_{2}, b_{2}= a_{2}-a_{3}, \dots, b_{k}= a_{k}-a_{k+1}, k < n$.

In the sum $b_{1}$ appears $n-1$ times, $b_{2}$ appears $2(n-2)$ times and so on, in general $b_{k}$ appears $k(n-k)$ times.

The sum becomes $S = \sum_{i=1}^{n-1}i(n-i) b_{i}$.

Now let's see what happens if we modify $b_{j}, 1 \le j \le n-1$ by some length m.

$S' = j(n-j)(b_{j}+m)+\sum_{i \neq j, 1 \le i \le n-1}i(n-i) b_{i}$
$= S-j(n-j)b_{j}+j(n-j)(b_{j}+m) = S+j(n-j) m$.

On the other hand $d' = b_{j}+m+\sum_{i \neq j, 1 \le i \le n-1}b_{i}= d+m$.

Examining the difference $\epsilon' = \frac{n^{2}}{4}d'-S' = \frac{n^{2}}{4}d+\frac{n^{2}}{4}m-S-j(n-j) m = \epsilon+m(\frac{n^{2}}{4}-j(n-j))$.

Assuming $n \ge 2$ we have $m(\frac{n^{2}}{4}-j(n-j)) \ge 0$ when $m > 0$, strengthening the inequality and similarly weakening the inequality when $m < 0$.

Obviously when we weaken the inequality to its limit $b_{j}= 0$ for every j and $S = 0 = d$.

To construct every possible situation we can start from $b_{1}, \dots, b_{n}= 0$ and start adding some m to them.
To keep the equality we must have $m(\frac{n^{2}}{4}-j(n-j)) = 0$ Resulting $j=\frac{n}{2}$.

That is, only $b_{n/2}$, if it exists,  may be greater than zero for the equality.

Now considering the other inequality, $(n-1) d \le S$:

We have $\epsilon' = S'-(n-1)d' = S+j(n-j)m-(n-1)d-(n-1)m = \epsilon+m(j(n-j)-n+1)$.
Like above, we have $m(j(n-j)-n+1)$ \ge 0 when $m > 0$, strengthening the inequality and weakening when $m < 0$.

Again, at its limit $(n-1)d = 0 = S$.

Here for the equality we want $m(j(n-j)-n+1) = 0$, which implies $j=1$ or $j=n-1$.

That is, only $b_{1}$ and $b_{n-1}$ may be greater than zero for the equality.

[/hide]

[b]1)[/b] WLOG $ a_{1}\leq a_{2}\leq ... a_{n}$

$ \left |a_{i}\minus{}a_{1}\right|\plus{}\left |a_{n}\minus{}a_{i}\right| \equal{}a_{n}\minus{}a_{1}\equal{}d$

$ S \equal{}\sum_{i < j\not\in\left\{1,n\right\}}\left |ai\minus{}aj\right|.\plus{}\sum (|a_{i}\minus{}a_{1}|\plus{}|a_{n}\minus{}a_{i}|)\plus{}a_{n}\minus{}a_{1}\equal{}$ $ (n\minus{}1)d\plus{}\sum_{i < j\not\in\left\{1,n\right\}}\left |ai\minus{}aj\right|\geq (n\minus{}1)d$
and the equality holds when $ a_{2}\equal{}a_{3}\equal{}...\equal{}a_{n\minus{}1}$

[b]2)[/b] If $ i\leq\frac{n}{2}$ and $ a_{i\minus{}1}\equal{}a_{1}$:
 $ \sum |a_{i}\minus{}a_{j}|\leq\sum |a_{1}\minus{}a_{j}|$

If $ i\geq\frac{n\plus{}2}{2}$ and $ a_{i\plus{}1}\equal{}a_{n}$:
 $ \sum |a_{i}\minus{}a_{j}|\geq\sum |a_{n}\minus{}a_{j}|$

with this smoothing you have that the equality holds when $ a_{1}\equal{}a_{2}\equal{}...\equal{}a_{\frac{n}{2}}$ y $ a_{\frac{n\plus{}2}{2}}\equal{}...\equal{}a_{n}$, and evidently doesn't hold for $ n$ odd.

Suppose WLOG that $ a_{1}\leq ...\leq a_{n}$. Then

$ S\equal{}(n\minus{}1)(a_{n}\minus{}a_{1})\plus{}\sum_{i\equal{}2}^{[n/2]}(n\plus{}1\minus{}2i)(a_{n\plus{}1\minus{}i}\minus{}a_{i})$.

Hence,

1) $ (n\minus{}1)d\leq (n\minus{}1)d\plus{}\sum_{i\equal{}2}^{[n/2]}(n\plus{}1\minus{}2i)(a_{n\plus{}1\minus{}i}\minus{}a_{i})\equal{}S$, and

2) $ S\leq (n\minus{}1)d\plus{}\sum_{i\equal{}2}^{[n/2]}(n\plus{}1\minus{}2i)d\leq\frac{n^{4}}{4}d$.

Sorry for reviving , but doesn't the fact that $S$ is convex in each of $a_i$'s give an easier solution?

It does only for searching maximum. With order $a_1\le a_2\le...\le a_n$ it suffices to prove this inequality for all $k=0,1,...,n$ where $a_1=...=a_k\wedge a_{k+1}=...=a_n$


For $a,b,c$ arbitrary reals such that $a^2b+b^2c+c^2a=0$ prove that

$a^2+b^2+c^2 \geq \root_3 \of{4}(ab+bc+ca)$

Let's consider the numbers $a,b,c \neq 0$ because it's trivial otherwise. If so the condition can be seen as a second degree equation in $a$ or $b$ or $c$ this means that:

$a^{4}-4c^{3}a \geq 0$ and the other two simmilar ineqs. Then we have for example:

$a^{4}-4c^{3}a \geq 0 \Leftrightarrow a(a^{3} - (\sqrt[3]{4}c)^{3}) \geq 0 \Leftrightarrow a^{2} \geq \sqrt[3]{4}ac$ and using the other two simmilar inequalities we get what we needed! :D :D

cheers!


induction
symmetry
group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose G  is agroup of order n and A1,A2,...,Ap are elements in G s.t.
 (A1)(A2)=(A2)(A3)=...=(A(p-1))(Ap)=(Ap)(A1) where p is a prime
 show 2p divides n.p isodd prime

Here's what I have: $n$ is divisible by $2$, $\forall i,j, A_{i}^{2}=A_{j}^{2}$, and all the $A_{i}$ have same order in $G$. I still cannot prove that $p$ divides $n$. Can anyone finish from here ?

====
Partial proof:
====

The fact that all $A_{i}'s$ have the same order is trivial (they are all conjugated).

Now $A_{2}^{-1}A_{1}A_{2}=A_{3}$, and $A_{4}=A_{3}^{-1}A_{2}A_{3}=A_{2}A_{1}^{-1}A_{3}$. By induction, since $p$ is odd, $A_{p}=A^{-1}_{2}A_{1}A_{p-1}$, and $A_{2}=A^{-1}_{2}A_{1}A_{1}$, giving $A_{2}^{2}=A_{1}^{2}$. By symmetry, $\forall i,j, A_{i}^{2}=A_{j}^{2}$.

From here, it's clear that all $A_{i}'s$ must have even order (otherwise they are all equal). Whence $2$ divides $n$.

How to finish ?

More generally, $A_{i}A_{i+1}=A_{j}A_{j+1}$ (where it is understood that $A_{p+1}=A_{1}$ and $A_{0}=A_{p}$), and $A_{i-1}A_{i}A_{i+1}=A_{i-1}A_{j}A_{j+1}$, whence $A_{i}A_{i+1}^{2}=A_{i}A_{j+1}^{2}=A_{i-1}A_{j}$, and $A_{i}A_{j+1}=A_{i-1}A_{j}$.

This tells you that $\forall i,j, A_{i+1}A_{j+1}=A_{i}A_{j}$.

Now, if someone can prove that $A_{i}^{2}=1$, then we will be done since the set of all $A_{i}A_{j}$ (where $i$ can equal $j$) will be easily proved to be a subgroup of $G$ with order $p$.

Well, some other person on sci.math seems to assert the result is false, and gave me some alleged couterexample, but I don't have the courage to check it. It's my very last post on sci.math, you can look it up easily (Author: Julien Santini).

I didn't check the example given in sci.math, because James did not give a3.

But there is a group of order 12 with elements a, b, c, s.th. 1,a,b,c are p.w. distinct, ab = bc = ca and s.th. a^2, b^2 and c^2 are all not equal to 1: The semidirect product of Z/3 and Z/4. Take a, b, c as suitable elements of order 4.

LiMa

There must be some hypothesis missing then !


Why is $ o((x\minus{}x_{0})^{n}) \equal{} O((x\minus{}x_{0})^{n\plus{}1})$,
and why is $ o((x\minus{}x_{0})^{n})$ weaker then $ O((x\minus{}x_{0})^{n\plus{}1})$,as $ x\rightarrow x_{0}$?

Hi.
question about $ O$ and $ o$ was posted here many times.
$ o(f(x))$ in point $ x_{0}$ mean that :
$ \frac{o(f(x))}{f(x)}\equal{}\alpha (x)$
where $ \alpha (x)\to 0$ when 4$ x\to x_{0}$
and
$ O(f(x))$ in point $ x_{0}$ mean that
$ |\frac{O(f(x))}{f(x)}|<const$
your property is true by definition.
$ O(x^{n\plus{}1})$ in point zero mean that
in point zero we have
$ \frac{O(x^{n\plus{}1})}{x^{n}}\equal{}O(x)$ in point zero which also $ \to 0$ when $ x\to 0$ (bu definition $ O(X)$ is $ \alpha (x)$)
or $ \frac{o(x^{n})}{x^{n}}\equal{}\alpha(x)$ which also is $ O(x)$ and e.t.c.


calculus
integration
logarithms
algebra
functional equation
calculus computations

solve it:
$ \int^{xy}_1 f(t)\,dt \equal{} y\int^x_1 f(t)\,dt \plus{} x\int^y_1 f(t)\,dt$
and $ f(1) \equal{} 3$,then $ f(x) \equal{} ?$

[color=green][Use {}. Click on your work to see the edit.][/color]

Differentiating with respect to $ x$ gives

$ \frac {\partial}{\partial x}\int_1^{xy}f(t)dt \equal{} yf(x) \plus{} \int_1^yf(t)dt$

Plugging in $ x \equal{} 1$:
$ 0 \equal{} yf(1) \plus{} \int_1^yf(t)dt$
$ \Rightarrow \int_1^yf(t)dt \equal{} \minus{} 3y$
$ \Rightarrow \frac {d}{dx}\int_1^yf(t)dt \equal{}\frac {d}{dx}( \minus{} 3y)$

$ \Rightarrow \boxed{f(y) \equal{} \minus{} 3}$ for all $ y$.

[quote="BlueVelvet"]Differentiating with respect to $ x$ gives

$ \frac {\partial}{\partial x}\int_1^{xy}f(t)dt \equal{} yf(x) \plus{} \int_1^yf(t)dt$

Plugging in $ x \equal{} 1$:
$ 0 \equal{} yf(1) \plus{} \int_1^yf(t)dt$
$ \Rightarrow \int_1^yf(t)dt \equal{} \minus{} 3y$
$ \Rightarrow \frac {d}{dx}\int_1^yf(t)dt \equal{} \frac {d}{dx}( \minus{} 3y)$

$ \Rightarrow \boxed{f(y) \equal{} \minus{} 3}$ for all $ y$.[/quote]
i think it's wrong because we have $ f(1)\equal{}3$ and  i solved that $ f'(x)\not\equal{}0$

$ \boxed{f(y) \equal{} \minus{} 3}$ for all $ y$ is wrong indeed, because for $ y\equal{}1$ we should have $ f(y)\equal{}3\neq \minus{}3$  :rotfl:

I wonder where I made a mistake though.

PS: it's unnecessary to quote my whole message. Who else would you be replying to?

The error is here:
[quote="BlueVelvet"]Plugging in $ x \equal{} 1$:
$ 0 \equal{} yf(1) \plus{} \int_1^yf(t)dt$[/quote]
$ \frac {\partial}{\partial x}\int_1^{xy}f(t)dt \equal{} yf(xy)$, so plugging in $ x \equal{} 1$ would give
\[ yf(y) \equal{} 3y \plus{} \int _1^yf(t)dt \\
\Rightarrow yf'(y) \equal{} 3 \\
\Rightarrow \boxed{f(y) \equal{} 3 \plus{} 3\log y}
\]
solves the original integral-functional equation if we restrict $ x,y > 0$.


(αφιερωμένη στον r_boris :))

Θα λέμε ότι ένας αριθμός $ x\in [0,1]$ είναι [i]σατανικός[/i] όταν η δεκαδική του αναπαράσταση $ x \equal{} 0.x_1x_2\ldots x_n\ldots$ περιέχει την ακολουθία $ 666$.

Να δείξετε ότι σχεδόν κάθε αριθμός στο $ [0,1]$ είναι σατανικός! :diablo:

Δηλαδή, αν επιλέξετε τυχαία έναν αριθμό στο $ [0,1]$, αυτός θα είναι σατανικός με πιθανότητα $ 1$! :diablo:

Cheerio,

Durandal 1707

PS: Ούτε που θέλω να ξέρω τι θα γινόταν αν αυτό ήταν γνωστό στους ιεροεξεταστές η στη Λουκά ;)

ευχαριστώ για την αφιέρωση Παναγιώτη αν και μου φαίνεται ότι οι λόγοι σου είναι σατανικοἰ :diablo: ΧΑ-ΧΑ

Θα βρούμε την πιθανότητα για ν ψηφία και στη συνέχεια θα βρούμε το όριο όταν το ν τείνει το άπειρο.

Σκέψη 1η:
Γράφω τον αριθμό έτσι ώστε το πλήθος των δεκαδικών ψηφίων του να είναι πολλαπλάσιο του 3. Αυτό μπορεί να γίνει αν προσθέσω στο τέλος των δεκαδικών ψηφίων του 0, 1 ή 2 μηδενικά. Άρα φαντάζομαι πια έναν αριθμό της μορφής 0,xxx.xxx.xxx.xxx κλπ.

Σκέψη 2η:
Έστω p1 η πιθανότητα να υπάρχουν στον αριθμό 3 συνεχόμενα 6ρια ξεκινώντας από οποιοδήποτε σημείο του. Και έστω p2 η πιθανότητα να υπάρχουν 3 συνεχόμενα 6ρια ανάμεσα σε 2 διαδοχικές τελείες (δηλαδή σε καθορισμένες μόνο θέσεις).
Ισχύει ότι p1>p2 αφού οι ευνοϊκές περιπτώσεις για την p1 είναι περισσότερες. (Αυτό ισχύει γιατί κάθε αριθμός που έχει 3 συνεχόμενα εξάρια ανάμεσα σε 2 διαδοχικές τελείες προσμετράτε και στην p1 και στην p2. Αντίθετα αν υπάρχουν 3 συνεχόμενα εξάρια που εκτείνονται εκτός των ορίων που θέτουν οι τελείες προσμετρώντε στην p1 αλλά όχι στην p2 πχ ο αριθμός 0,126.660.

Σκέψη 3:
Θα δείξω ότι όταν το πλήθος των ψηφίων τείνει στο άπειρο η p2 τείνει στο 1. Άρα και η p1 θα τείνει στο 1. Δηλαδή θα εγκλωβίσω την p1 από την p2 που υπολογίζεται ακριβώς πολύ πιο εύκολα.

Η πιθανότητα να έχω 3 συνεχόμενα 6ρια ανάμεσα σε 2 τελείες (μια ή περισσότερες φορές) είναι ίση με 1 μείον την πιθανότητα να μην εμφανιστούν ποτέ τα 3 συνεχόμενα 6ρια στις συγκεκριμένες θέσεις. Αλλά η τελευταία πιθανότητα είναι $ \frac{999}{1000}\frac{999}{1000}\frac{999}{1000}$... = $ (\frac{999}{1000})^k$ όπου κ το πλήθος των τριάδων στις οποίες έχουμε χωρίσει τον αριθμό. Η τελευταία πιθανότητα τείνει στο 0 άρα η p2 για άπειρο πλήθος ψηφίων τείνει στο 1 (αν ο αρχικός αριθμός δεν έχει πλήθος ψηφίων που είναι πολλαπλάσιο του 3 τότε το κλάσμα δεν είναι υψωμένο στην κ αλλά στην κ-1 αφού τα τελευταία ψηφία είναι fake αλλά το όριο δεν αλλάζει). Άρα και η p1 τείνει στο 1.

Well done George :)

Πάει και ανάποδα:

1. Προφανώς αρκεί να δείξουμε ότι οι αριθμοί που έχουν το 666 σε μία από τις θέσεις που περιγράφει ο Γιώργος είναι σχεδόν παντού.

2.1 Το σύνολο των αριθμών που έχουν το 666 στην πρώτη τριάδα έχει μέτρο $ 1/1000$.
2.2 Μετά, έχουμε 999 διαστήματα που έχουν το 666 στη δεύτερη τριάδα, αλλά όχι στην πρώτη. Αυτά έχουν μέτρο $ \frac {1}{1000^2}$.
2.3 Τρίτη τριάδα, κ.ο.κ.

Καταλήγουμε στο ότι το σύνολο των σατανικών αριθμών έχει συνολικό μέτρο $ I$ για το οποίο ισχύει:
$ 1\geq I \geq \frac {1}{1000} \plus{} \frac {999}{1000^2} \plus{} \cdots \equal{} \frac {0.001}{1 \minus{} 0.999} \equal{} 1$

Εξίσου εύκολα, να δείξουμε τώρα και ότι όσοι απαρνιούνται το Σατανά είναι εξίσου πολλοί?  :yinyang:

Δηλαδή, αν αφαιρέσουμε τους σατανικούς αριθμούς από το $ [0,1]$, θα μας μείνει ένα σύνολο με υπεραριθμήσιμο πλήθος στοιχειών που τίθεται σε 1-1 αντιστοιχία με το αρχικό διάστημα $ [0,1]$! ;)

Cheerio,

Durandal 1707

PS: Και πάλι, αν το έχετε δει, αφήστε για όσους δεν το έχουν δεί :)

[quote="Durandal"]Well done George :)
Εξίσου εύκολα, να δείξουμε τώρα και ότι όσοι απαρνιούνται το Σατανά είναι εξίσου πολλοί?  :yinyang:

Δηλαδή, αν αφαιρέσουμε τους σατανικούς αριθμούς από το $ [0,1]$, θα μας μείνει ένα σύνολο με υπεραριθμήσιμο πλήθος στοιχειών που τίθεται σε 1-1 αντιστοιχία με το αρχικό διάστημα $ [0,1]$! ;)
[/quote]

Το είδα καθυστερημένο ότι έπεσε και δεύτερο ερώτημα  :lol:

Λοιπόν... αυτό το "εξίσου πολλοί" είναι κάπως παραπλανητικό. Για να το δείξουμε νομίζω πως πρέπει να φέρουμε σε 1-1 αντιστοιχία τους σατανικούς με τους μη σατανικούς. Αυτό γίνεται αλλά η πιθανότητα να πέσουμε σε μη σατανικό είναι 0 αφού η πιθανότητα να πέσουμε σε σατανικό είναι 1.

Παναγιώτη ξενικινάω και με διορθώνεις όπου χρειαστεί :)

Κάθε διαφορετικός σατανικός μπορεί να αντιστοιχηθεί σε ένα διαφορετικό μη σατανικό αλλάζοντας τα 6αρια πχ σε 5άρια.
Κάθε διαφορετικός μη σατανικός μπορεί να αντιστοιχηθεί σε διαφορετικό σατανικό εισάγοντας τρία 6άρια αμέσως μετά την υποδιαστολή.

Δεν ξέρω αν έκανα κάποιο λάθος. Ωστόσο αξίζει να σημειωθεί ότι αν και εξίσου πολλοί οι σατανικοί με τους μη σατανικούς… οι πρώτοι είναι πολύ πιθανότεροι από τους δεύτερους. Μου φαίνεται πολύ δύσκολο η πεπερασμένη διαίσθησή μας να χωρέσει την έννοια του απείρου.

Αυτό που με παίδεψε πολύ στο αρχικό ερώτημα ήταν όταν πήγα να βρω την ακριβώς την πιθανότητα για ν ψηφία. Εκεί έμπλεξε το πράγμα πολύ. Το πήγα μέχρι κάποιο σημείο αλλά δυσκόλευε η αναλυτική  έκφραση μέσα από τύπο. Έχεις υπόψη σου αν λύνεται; Μη μπλέξουμε τζάμπα.

[quote]Λοιπόν... αυτό το "εξίσου πολλοί" είναι κάπως παραπλανητικό. Για να το δείξουμε νομίζω πως πρέπει να φέρουμε σε 1-1 αντιστοιχία τους σατανικούς με τους μη σατανικούς. Αυτό γίνεται αλλά η πιθανότητα να πέσουμε σε μη σατανικό είναι 0 αφού η πιθανότητα να πέσουμε σε σατανικό είναι 1.[/quote]

Ίσα-ίσα, το "εξίσου πολλοί" είναι right to the point ;) Αυτό που ήθελα να δείξω με αυτήν την άσκηση είναι ακριβώς το ότι το "μέγεθος" ενός συνόλου δεν έχει καθόλου μα καθόλου να κάνει με τον αριθμό των στοιχείων του!

[quote]Κάθε διαφορετικός σατανικός μπορεί να αντιστοιχηθεί σε ένα διαφορετικό μη σατανικό αλλάζοντας τα 6αρια πχ σε 5άρια.
Κάθε διαφορετικός μη σατανικός μπορεί να αντιστοιχηθεί σε διαφορετικό σατανικό εισάγοντας τρία 6άρια αμέσως μετά την υποδιαστολή.[/quote]

Ok.

Για να ξεκαθαρίσει το τοπίο, έστω $ S$ το σύνολο των σατανικών αριθμών και $ G \equal{} I\backslash S$ το σύνολο των μη σατανικών αριθμών. Για να δείξεις ότι τα $ S,G$ έχουν τον ίδιο αριθμό στοιχείων, πρέπει σε κάθε σατανικό αριθμό να αντιστοιχήσεις έναν μη σατανικό και αντιστρόφως.

(Formally, χρειαζόμαστε δύο συναρτήσεις $ S\to G$ και $ G\to S$ που να είναι 1-1. Να σημειωθεί ότι, αν και διαισθητικά προφανές, εδώ κρύβεται ένα πολύ χρήσιμο και cute θεώρημα της θεωρίας συνόλων, το Schroeder-Bernstein)

Αυτό πράγματι το κάνεις και είμαστε ok :) Στο hint όμως δίνω την extra πληροφορία ότι το $ G$ έχει τον ίδιο αριθμό στοιχείων με το αρχικό διάστημα $ [0,1]$. Χρειάζεται να πούμε κάτι παραπάνω για να καταλήξουμε στο συμπέρασμα αυτό?

[quote]Ωστόσο αξίζει να σημειωθεί ότι αν και εξίσου πολλοί οι σατανικοί με τους μη σατανικούς… οι πρώτοι είναι πολύ πιθανότεροι από τους δεύτερους.[/quote]

Ακριβώς! Isn't math cool? :)

Cheerio,

Durandal 1707

ΟΚ τώρα κατάλαβα αυτό που σε ένοιαζε ήταν το μέρος των συνόλων και όχι το να μετρήσεις πόσες είναι οι ευνοϊκές πιθανότητες στανικού αριθμού για ν ψηφία. Αυτό είναι ένα άλλο θέμα.

[quote="Durandal"]
Αυτό πράγματι το κάνεις και είμαστε ok :) Στο hint όμως δίνω την extra πληροφορία ότι το $ G$ έχει τον ίδιο αριθμό στοιχείων με το αρχικό διάστημα $ [0,1]$. Χρειάζεται να πούμε κάτι παραπάνω για να καταλήξουμε στο συμπέρασμα αυτό?
[/quote]

Νόμιζα πως θέλαμε απλά να δείξουμε ότι οι σατανικοί με τους μη σατανικούς είναι το ίδιο πολύ, οπότε το πήγα έτσι. Αν θέλουμε να δείξουμε ότι η μη σατανικοί είναι το ίδιο πολύ με το [0,1] πρέπει όντως να το τεκμηριώσουμε.

Λοιπόν… οι σατανικοί είναι το ίδιο πολύ με τους σατανικούς, άρα ή και οι 2 είναι αριθμήσιμοι ή και οι 2 μη αριθμήσιμοι και ίσοι με το [0,1]. Αλλά το άθροισμά τους (το [0,1] δηλαδή) είναι μη αριθμήσιμο και ίσο μ ε το [0,1]. Άρα και οι 2 είναι μη αριθμήσιμοι γιατί η ένωση 2 αριθμήσιμων συνόλων είναι αριθμήσιμο σύνολο (αντιστοιχείς τα στοιχεία τους πρώτου συνόλου στους μονούς και του άλλου στους ζυγούς).

[quote="Durandal"]
[quote]Ωστόσο αξίζει να σημειωθεί ότι αν και εξίσου πολλοί οι σατανικοί με τους μη σατανικούς… οι πρώτοι είναι πολύ πιθανότεροι από τους δεύτερους.[/quote]

Ακριβώς! Isn't math cool? :)[/quote]

Όντως είναι πολύ "math cool"  :)  Αυτό που καταλαβαίνω ότι οι φυσικές γλώσσες (πχ ελληνικά/αγγλικά) εμπεριέχουν ασάφειες. Τι θα καταλάβει ένας μη μαθηματικός αν τους πεις ότι οι σατανικοί είναι το ίδιο πολύ με τους μη σατανικούς αλλά απείρως πιθανότεροι... αφού ξέρει πως πιθανότητα είναι ευνοϊκές/συνολικές και οι ευνοϊκές είναι ίδιες (μη αριθμήσιμες); Εκεί πήγαινε το "παραπλανητικό" που είπα στην αρχή. Δηλαδή δεν μπορείς να στηριχτείς στην καθημερινή ερμηνεία που δίνουμε στις λέξεις. Ακόμα και τώρα που το βλέπω με τα μάτια μου... δεν μπορώ να το πιστέψω. Είναι από τα παραδείγματα που προσφέρονται για να δείξεις ότι τα πράγματα δεν είναι πάντα όπως φαίνονται. Καμιά μέρα να κάνουμε μια τέτοια λίστα από τέτοια παραδείγματα βγαλμένη από τα μαθηματικά και τη φυσική.

Λοιπόν, επειδή διάβασα χθες πολύ βιαστικά την απάντησή σου, υπάρχει ένα προβληματάκι.

Έχεις δείξει μία χαρά ότι υπάρχει 1-1 συνάρτηση των μη σατανικών στους σατανικούς με την απλή πρόσθεση του 666.

Όταν πας όμως να απεικονίσεις τους σατανικούς στους μη σατανικούς, οι σατανικοί αριθμοί 0.5666, 0.6666, 0.6665 αντιστοιχίζονται και οι τρεις στον αριθμό 0.5555, οπότε το σχήμα σου τριπλομετράει μερικές περιπτώσεις. Έτσι, η απεικόνισή σου από τους σατανικούς αριθμούς στους μη σατανικούς δεν είναι ούτε 1-1 ούτε επί. Χρειάζεται κάτι ακόμη...

[quote="gpapargi"]ΟΚ τώρα κατάλαβα αυτό που σε ένοιαζε ήταν το μέρος των συνόλων και όχι το να μετρήσεις πόσες είναι οι ευνοϊκές πιθανότητες στανικού αριθμού για ν ψηφία. Αυτό είναι ένα άλλο θέμα.[/quote]

Ναι, τώρα θυμήθηκα και την ερώτησή σου, για το αν βγαίνει η πιθανότητα "ακριβώς" για έναν n-ψήφιο αριθμό να (μην) είναι σατανικός. Αυτό είναι άσκηση σε δοκιμές Bernoulli, αλλά χωρίς ιδιαίτερα κομψή απάντηση αν θυμάμαι καλά... αν έχεις όρεξη να ασχοληθείς, fire away ;)

Cheerio,

Durandal 1707

[quote="Durandal"]Λοιπόν, επειδή διάβασα χθες πολύ βιαστικά την απάντησή σου, υπάρχει ένα προβληματάκι.

Έχεις δείξει μία χαρά ότι υπάρχει 1-1 συνάρτηση των μη σατανικών στους σατανικούς με την απλή πρόσθεση του 666.

Όταν πας όμως να απεικονίσεις τους σατανικούς στους μη σατανικούς, οι σατανικοί αριθμοί 0.5666, 0.6666, 0.6665 αντιστοιχίζονται και οι τρεις στον αριθμό 0.5555, οπότε το σχήμα σου τριπλομετράει μερικές περιπτώσεις. Έτσι, η απεικόνισή σου από τους σατανικούς αριθμούς στους μη σατανικούς δεν είναι ούτε 1-1 ούτε επί. Χρειάζεται κάτι ακόμη...

[/quote]

Έχεις δίκιο... μου διέφυγε. Να κάτι που μου ήρθε τώρα. Αλλάζεις όλα τα 6ρια ανεξαρτήτου θέσεως σε 5ρια όπως πριν αλλά θέλεις και κάτι άλλο έτσι ώστε οι διαφορετικοί σατανικοί να έχουν διαφορετική μη σατανική εικόνα. Το καλύτερο είναι να κωδικοποίησεις τις θέσεις των αρχικών 6αριων και να τις βάλεις στην αρχή του αριθμού. Αφού οι αρχικοί μη σατανικοί ήταν διαφορετικοί θα έχουν και διαφορετικοί εικόνα.

Λοιπόν... αφού αλλάξεις τα 6αρια σε 5ρια δεν υπάρχουν πια 6ρια καθόλου. Τα 6ρια τα έχουμε τώρα για διαφορετική χρήση... για διαχωριστές πεδίου. Σε κάθε αριθμό αφού του αλλάξεις τα 6ρια σε 5ρια  θα βάλεις μετά την υποδιαστολή το 6....6 που σηματοδοτεί την αρχή και το τέλος του κωδικοποιημένου μέρους. Ανάμεσα στα 2 ακραία 6ρια θα βάζεις για κάθε αρχικό 6αρι τη θέση του αρχικού 6ριου που έγινε 5άρι και μετά αμέσως ένα 6αρι για να μπορούμε να ξεχωρίσουμε τις θέσεις των διαφορετικών ψηφίων μεταξύ τους. Αλλά επειδή υπάρχει περίπτωση να έχουμε αλλαγή στη θέση πχ 66 και μαζί με το διαχωριστικό 6αρι να πάμε σε σατανικό αριθμό... οι θέσεις θα είναι γραμμένες στο δυαδικό σύστημα που έχει 0 και 1 και δεν θα έχουμε 6αρια. Δες αν και τώρα μου ξέφυγε κάτι

[quote="gpapargi"]Έχεις δίκιο... μου διέφυγε. Να κάτι που μου ήρθε τώρα. Αλλάζεις όλα τα 6ρια ανεξαρτήτου θέσεως σε 5ρια όπως πριν αλλά θέλεις και κάτι άλλο έτσι ώστε οι διαφορετικοί σατανικοί να έχουν διαφορετική μη σατανική εικόνα. Το καλύτερο είναι να κωδικοποίησεις τις θέσεις των αρχικών 6αριων και να τις βάλεις στην αρχή του αριθμού. Αφού οι αρχικοί μη σατανικοί ήταν διαφορετικοί θα έχουν και διαφορετικοί εικόνα.

Λοιπόν... αφού αλλάξεις τα 6αρια σε 5ρια δεν υπάρχουν πια 6ρια καθόλου. Τα 6ρια τα έχουμε τώρα για διαφορετική χρήση... για διαχωριστές πεδίου. Σε κάθε αριθμό αφού του αλλάξεις τα 6ρια σε 5ρια  θα βάλεις μετά την υποδιαστολή το 6....6 που σηματοδοτεί την αρχή και το τέλος του κωδικοποιημένου μέρους. Ανάμεσα στα 2 ακραία 6ρια θα βάζεις για κάθε αρχικό 6αρι τη θέση του αρχικού 6ριου που έγινε 5άρι και μετά αμέσως ένα 6αρι για να μπορούμε να ξεχωρίσουμε τις θέσεις των διαφορετικών ψηφίων μεταξύ τους. Αλλά επειδή υπάρχει περίπτωση να έχουμε αλλαγή στη θέση πχ 66 και μαζί με το διαχωριστικό 6αρι να πάμε σε σατανικό αριθμό... οι θέσεις θα είναι γραμμένες στο δυαδικό σύστημα που έχει 0 και 1 και δεν θα έχουμε 6αρια. Δες αν και τώρα μου ξέφυγε κάτι[/quote]

Το κωδικοποιημένο μέρος δεν θα έχει απαραίτητα τέλος: αποκτάς προβλήματα a la Hilbert's hotel σε αριθμούς που είναι απείρως σατανικοί (δηλ. έχουν άπειρες τριάδες 666) ή που είναι πολυσήμαντα σατανικοί (όπως ο 0.6666 όπου υπάρχει πιθανότητα στις άκρες του κωδικοποιημένου μέρους να βγει κάτι σατανικό και να σου χαλάσει το σχήμα).

Cheerio,

Durandal 1707

[quote="gpapargi"]Φοβάμαι πως δεν κατάλαβα.
Ο 0,6666 θα γίνει 0,61610611610065555. Που σημαίνει ότι τα 6αρια είναι στις θέσεις 1 (1 στο δυαδικό),2 (10 στο δυαδικό) ,3 (11 στο δυαδικό) και 4 (100 στο δυαδικό) Πως θα προκύψει κάτι σατανικό; Τα γράφω στο δυαδικό σύστημα ώστε να μην χρειαστεί να χρησιμοποιήσω το 6 που θα μου προέκυπτε κάτι σατανικό με τους delimeters.
Το κωδικοποιημένο μέρος δεν έχει απαραίτητα τέλος όπως ούτε και ο αριθμός. Αν πχ έχει άπειρα δεκαδικά ψηφία ούτε να το γράψω μπορώ, ούτε να αλλάξω τα 6ρια σε 5ρια μπορώ αφού έχω να κάνω άπειρα πράγματα. Το μόνο που έχω να κάνω έτσι κι αλλιώς είναι να περιγράψω έναν αλγόριθμο αντιστοίχησης δηλαδή τι αντιστοιχεί με τι. Τα άλλα έτσι κι αλλιώς δε γίνονται. Έτσι κάπως έτσι τα καταλαβαίνω τα πράγματα.[/quote]

Ααα, μάλιστα, τώρα κατάλαβα τι έχεις στο μυαλό σου!!! Sorry, όλα καλά :)

Cheerio,

Durandal 1707

Η δική μου λύση για το ότι οι μη σατανικοί αριθμοί είναι όσοι και οι αριθμοί στο $ I\equal{}[0,1]$.

Έστω $ G$ το σύνολο των μη σατανικών αριθμών και $ G'\subseteq G \subseteq I$ το υποσύνολο αυτών που δεν έχουν καθόλου 6ρια (base-10). Προφανώς $ |G'|\leq |G|\leq |I|$.

Όμως, αν γράψεις έναν αριθμό $ x\in I$ σε base-9 αναπαράσταση (χαζομαρούλες του στυλ $ 0.88888\ldots_9 \equal{} 1$ δε μας απασχολούν) και θέσεις όπου 6 το 9, έχεις μία 1-1 απεικόνιση $ I\hookrightarrow G'$. Άρα $ |I|\leq|G'|\leq |G|\leq |I|$ και... τέλος :)

Cheerio,

Durandal 1707


Do you support zero tolerance? why or why not?

Edit: Zero tolerance is mega punishment for anything

Would you care to enlighten me on what zero tolerance is. I have heard of it, but I'm not sure what it means and why it is a big deal.

It would be quite helpful if you explained what you are talking about.  Zero tolerance [i]what[/i]?  If you ask a question like this, it is helpful to put in (1) what you are talking about, (2) a little bit of background (perhaps why you are interested or where you heard about it), and maybe (but not necessarily) (3) an argument or two in one direction or the other.

Instead of making a new post, edit your original one.

My friend likes to add "your mom" and "in bed" alot to what everybody says.

Do we really need to be talking about boxers and girl underwear?

Zero Tolerance: I think it is necesary to inflict pain - there is no other way of impressing a memory that works for the general public. In our school people cuss, fight, shout, and do all sorts of obscene things that nobody cares about. I like Rep's school code because it does offer some way of dealing w/ situation.

I un-support zero-tollerance. It is useless and stupid for there are people running around hardware stores and MIT labs with explosive materials freely while there are kids with little crackers and stuff getting in trouble. That's just weird.

On the other hand I have never been the victim of zero-tolerance, although I have been screwed because of the heightened school computer security and the no-calculator policy, but oh well.

Btw that teacher would probably mysteriously dissapear here in my school for we have a Jewish extremist club here (a la Freedom of Assembly+Jewish community) but again, oh well.

Our teacher's don't care at all.. half of them curse in the middle of class =) it just makes it all the more interesting.. lol

[quote="hawaiianglaciers"]my school's pretty loose on these things. Teachers just overlook it when people break the dress codes, cussing, or a small bit of violence. My friend hit me on the head lightly and no one cared. According to the school rules, that would result in a behavior report. A guy cusses everyday during class and the teacher just ignores it.

Weapons are another matter though. I think that the swiss army knife rules are pretty dumb. If you forgot, then you forgot. Just because you accidently brought it to school and didn't know you did until a teacher saw doesn't mean that you should get expelled. THe puishment should fit the crime and circumstances.[/quote]

haha, right, glaciers this guy (anonymous) got suspended for three days for saying another guy was hot ( i really hope he was joking) that's not exactly loose

At my school, there is more than enough cursing in the hallways. There is, however, none in the classroom. Cursing in the classroom results in a detention, or, if it's a third offense or more, it may result in an after school, but that depends on the teacher and student.

Violence is not permitted at my school either. Fights often result in a Saturday school. Kicking in the hallways, or in class for that matter, isn't a big issue with teachers, unless it's hurtful. Or if it's an often occurence between students. Thre is a ZERO tolerance for weapons. Weapons of any sort, pocket knives, knives, guns, are not tolerated. A student was recently given a 4 week suspension for having a pocket knife. Another student received a 3 week suspension for a lighter. Those kind of things aren't tolerated for the safety of the students.

[quote="churchilljrhigh"][quote="pinkbubblez90"][quote="churchilljrhigh"]I really wish people had some sense of decency, especially math students. Yesterday I was at a track meet and this crazy guy at my school had his boxers down alot and so his butt was showing.  A month ago this fellow 12th grader who was a pretty good math student also had his boxers down alot, and his butt was showing. It was with some junior high kids and so the junior high teacher says pull up your pants and he refues her.

Then there are these messed up girls that wear whatever you call those underwear (I'm not a girl so I don't know). And their skirts aren't always high enough and long enough.[/quote]

i agree...omg why do guys sag it's so wrong[/quote]

huh? what is omg and what is sag?[/quote]

Because I am one stone cold sober conservative. And why do you think that is ok?

omg : oh my god
sag: when guys wear their pants way too low ( like around their knees) and walk by pulling their pants up with every step. It really shows the boxers.

I agree with the girls wearing too short skirts. Our school (pinkbubblez and mine) has a rule that says that shorts and skirts may not be too short, but people think it's fine if shorts barely cover what is necessary to cover, if you get what i mean. THe shirts are too short and tight. The skirts/shorts are too short and low. that results in TOO MUCH EXPOSURE.  it's sick, what those people can wear to school.

schools are so messed up. YOu can get into huge trouble chewing gum, but people can walk around wearing things that britney spears would wear in music videos. and it doesn't help that the counseler has a belly button ring, tattoos, and breaks the dress code EVERY DAY.

Yeah, my school's dress code is pretty simple, no stomach can be shown, and all skirts/shorts/whatever have to be at the highest, two inches above the knee, and there's no sagging allowed at my school, except most people manage to avoid those rules anyways, but oh well... and i think the punishment definetely has to consider the circumstances, for example, i got suspended, because me and my friends were knocking each others binders out of our hands....imean, who gets suspended for that?!?!

hmm.  Theres really not too much of a problem in the public school where I go now.  At my old school there was though and it was a private school.  We wore uniforms but people still managed to break the dress code.  There were several weapons in school while I was there but they didn't get introuble for them (whats with that?!?!?!).  The only way u could get in trouble at my old school is talking in the hall!  for that u got a detention but thats pretty weird.

Metal detectors? My school is pretty relaxed about this kind of stuff, if they see a knife, the'll take it away and suspend you, but no one searches the students. Also our school cop always carries a gun with him.

A weapon means suspended, right?
Even if you didn't know about right?
So what if I hated somebody and slipped a lighter in their pocket? (Not saying I would...)

And...
[quote="JBL"][quote="Rep123max"]Our school cop has no power at all, they dont even trust her with a gun.[/quote]

Which, if you think about it, makes a whole lot of sense.  It would almost certainly be easier to acquire a gun already in the school by grabbing it from a security guard then it would be to sneak one in yourself.  Even in NYC, where the school security guards are part of the NYPD, they don't carry weapons.  (New York also hasn't had a shooting in school in years.  Maybe only 1 on school grounds in the past decade or so.)[/quote]
So, the Secret Service shouldn't carry guns because somebody might snatch it away from them?

Just wondering, are you reading Reader's Digest, DarkShadow?

[quote="Dark_Shadow"]So, the Secret Service shouldn't carry guns because somebody might snatch it away from them?[/quote]

I think we can both agree that the secret service is in a rather different position than a security guard.  I don't happen to know whether secret service carries guns, although I would be somewhat surprised if they didn't.  However, the specific job of many people protecting one or two individuals does not make a good analogy with the more general job of patrolling a school.  The threats are entirely different, the training is entirely different, and the situations likely to be encountered are totally different.  I cannot think of one single good reason to bring a gun into a school.  It creates an enormous potential for destruction and mayhem while creating a very small potential to possibly prevent something bad from happening.  Furthermore, it is worthwhile considering the advantages of having a school guard system which is not purely adversarial (guards vs. students), and arming guards could only create greater tensions between students and guards.

By the way, since when is a lighter a weapon?

Yes JBL the secret service carries guns.  they are also trained in martial arts though and many can kill a person in like 10 seconds without guns so with them it's not a maqjor problem.

I was wondering the same thing...A lighter is a weapon?  I suppose it can be but I never thought of it as one.

Whoa!!!<gapes in awe>. I'm never getting myself involved with the secret service. Darn - it seems kind of amazing that they go through all that training for the purpose of defending the president. What I'm trying to say is, what good does martial arts do if a guy is firing at the president. In that case, wouldn't all the training secret agents go through result in a waste when they end up throwing themselves in front of the president only to get killed by a gun?

Well, in my personal opinion, zero tolerance is a very delicated topic. Nevertheless, I think that basically every violent crime should have zero tolerance. I also consider that some of those crimes should have a way to more agresive punishment, and I mean death. Some of such crimes, in my opinion, should be Rape and Kidnaping (and white traffic). Also I think that there should be a more simplified way to apply the death penalty without having that many problems. I mean, most of the time it is more expensive to execute an inmate than to mantain him alife the rest of his life because of all the legal problems. I vote for the eighteen cents penalty!

Best,

Wow... this thread was definitely 1 year old. And my perspective on zero tolerance has changed much since then...

Zero tolerance=impossibility here. In fact, some students are not only allowed, but [i]required[/i] to have box cutters, exacto knives, etc. Way fun. Dress code? It's loose but supposed to be respected. And cussing? *laughs* Try reading Hunter S. Thompson and Allen Ginsberg and NOT cussing. It's grand.

[quote="mcalderbank"]Er, there is no dress code in my school, no cops, no metal detectors, but the board is thinking about extra security.

Also, there is no punishment for swearing, in fact, most teachers are granted "poetic license" when lecturing.

However, my school is very broad. I go to a public school (dress code unenforceable). There are really delinquent students, very smart students, and average students, with extremes for both ends of the spectrum. Hence, I gaurantee that the policy is different for those in less advanced classes. Often, the more advanced you are, the more leniency is granted. For example: You are not allowed to leave school campus and go into town (for lunch, or smoke, or whatever) but many do.[/quote]

that's pretty much how it is at my high school. my health teacher decided that she could use as much "poetic liscense" as she wanted while lectureing us, which got quite annoying. as for leniency, same here. more advanced, the more wiggle room you get where as if ur kinda a nobody or on the opposite end you get very little wiggle room.

Our middle school is 19th in the WORLD in the AMC 10 Results, yet we have so much violence that we have 8 armed police officers as well as 12 adult monitors at all times.  Talk about contrast!

Either Virginia [as a state] or Charlottesville [as the city] recently passed a law that if any of your underwear is showing, it is a 50 dollar fine.

As far as the rest of this thread, I think the important thing is that whatever rules are in force should be enforced, and fairly among all students.

I say if you are going to have a dress code, or whatever, it should be enforced, and if you aren't, then you should be open and admit that you don't have the energy to enforce it.

Our school the dress code is, if you let your hands rest by your legs and your shorts aren't longer than your hands, then you are supposed to be sent home and allowed to change, and if the pants of guy's sag, and boxers can be seen, then you are supposed to be sent home and allowed to change.  The dress code is find, but I've never actually seen it enforced.

If somebody curses in class at my school, the teacher gets pretty mad and there is maybe a "see me after class" but no real punishment unless it's repetitive. Depends on the teacher of course. The funniest was my English leacher last year, who was awesome, talking about a character in To Kill a Mockingbird, and called him straight out a "turd." He actually said, "The one word to describe this character is "turd."" That was pretty darn funny.

Oh yeah, and my English teacher typed up this example book report and took a quote from the book that had the word "sh**" in it without realizing it.

Our dress code is pretty funny actually.  If a teacher sees something wrong with it, like the shirt shows the belly, or the shorts are too short, then you get sent to the office to wear other clothes.  The catch is that the pants are sweat pants, and post the pants and shirts say "RJHS Dress Code" (RJHS is our school), so then everyone can see who breaks it, and wo didn't.

Hey, it's like the Scarlet Letter! That's actually really funny.