For $z \geq y \geq x >0$ such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$prove the ineq

$xy^2z^3 \geq 1$

Cool problem! Write $ y=a, z=abx$ with $a,b$ at least 1 and then find x from the given relation. Replace everything in terms of a,b and you will find a very easy inequality ( after you see of course that the inequality in a,b suffices to be proved for b=1)

I have similiar solution as Harazi.
I think we can prove:  x.y^2.z^3.(x/y)^(1/3)>=1.
I have a question:
              With a>0 satisfy:
                                         x.y^2.z^3.(x/y)^a>=1 ,every 0<x<=y<=z
        x+y=z=1/x+1/y+1/z.
      Is Min a   ?

Another easy way:

The usual techniques give that $xyz \geq 1$ (I suspect it will always be equal to 1 but not proven)
$xyz \geq 1$ implies $x^2y^2z^2 \geq 1$ implies $xy^2z^3 \geq 1$ since $z\geq x$


[tex] \displaystyle{\mbox{for reals a,b,c} \ge 0 \mbox{ Prove that}}[/tex]

[tex]\displaystyle{\large a^5+b^5+c^5 \ge 5abc(b^2-ac)}[/tex]

When you applied AM-GM, what happened to the 1/4 factors?

Yep -- that whole right side gets shrunk by a factor of 2^(8/7), making it smaller than 5.

Hint (in spoiler):  [hide]When a = c = 1, the inequality becomes b5 - 5 b3 + 5 b + 2 :ge: 0.  The left-hand side factors pretty nicely.[/hide]

[hide] The inequality is also homogenous, so perhaps we could let abc = 1 [/hide]

[edit]-Wait, that isn't necessarily a good assumption when a,b,c are reals.

Actually, assuming abc=1 might in fact be a good idea, because we can use it to reduce the inequality to one variable, as follows:

The left side becomes [tex]5(b^2-1/b)[/tex]. Use power-mean on the right side to find

[tex]\displaystyle \sqrt[5]{\frac{a^5+c^5}{2}}\ge\sqrt{ac}=\sqrt{1/b}.[/tex]

So we now have an inequality to prove that includes b alone. This may help.

I think it works a little better to assume that ac = 1.


Does the notion of outer measure change if instead of covering with intervals we cover with euclidian balls in $ R^n$?

Are you talking about the induced outer measure given by

\[ \mu^{*}\left(\textrm{E}\right) \equal{} \inf\left\{\sum\rho\left(\textrm{A}_{k}\right) \: \ \textrm{E}\subset\bigcup\textrm{A}_{k}\right\} ?\]


function
topology
limit
induction
algebra unsolved
algebra

what about the function $f(x)=1$ if $x \geq 0$ and $f(x) = -1$ if $x <0$ ?

Consider the restriction of $f$ to $\mathbb{R}\setminus{0}$.

[quote="PeNtaGoNaL"]For any function $f : \mathbf{R} \to \mathbf{R}$ does there necessarily exist a dense subset of $\mathbf{R}$ for which (the restriction of) f is continuous under the subspace topology?[/quote]

I think the answer is no. You can construct a function which takes all real values in any interval (i.e. $f(I)=\mathbb R$ for any non-degenerate interval $I$). Such a function would clearly not satisfy the condition.

[quote="grobber"]I think the answer is no. You can construct a function which takes all real values in any interval (i.e.  for any non-degenerate interval ). Such a function would clearly not satisfy the condition.[/quote]

I don't think that such function "would clearly not satisfy the condition".

First, there is function which is unbound on every interval $I$, but it satisfies the condition. It could by $f(x)=0$ for x irrational, and $f(x)$ is length of period of decimal representation for every rational $x$. Clearly, $f$ is unbound on every $I$ and continuous on irrationals.

If you had somethinG else on mind, some other argument why such function whouldn't satisfy the condition, then try this: Let $g$ be function such $g(x)=f(x)$ except $g(x)=0$ for all rational x. On every interval $I$ is $g(I)$ is "almost" $R$ (except in countable many points), and any "clear" argument why $f$ wouldn't satisfy the condition whould hold and for $g$ too, but $g$ satisfies the condition. So, I don't think the answer to this problem is such easy.

Indeed, if there is function $f$ with the conditions you asked for, then you could  in every interval $I$ take some x such $f(x)=a$ for some fixed real $a$, and the set of such number $x$ would be dense, and restriction of function $f$ on that set would be just $f(x)=a$, so it would be continuous.

Ok, you have made your point, I am sorry, I wasn;t thinking when I posted that. I am very tired :). I'll try again later, after a nap :).

The answer is yes because the result is a special case of Blumberg’s Theorem. A copy of a paper dealing with this theorem can be procured [url=http://web6.duc.auburn.edu/~brownj4/tatras.pdf]here[/url].

We define a subset $U \subseteq %Error. "reals" is a bad command.
$ to be [i]exhaustible [/i] if it is a countable union of sets that are nowhere dense (i.e. not dense in any interval).

[b]Lemma 1: [/b] The complement of an exhaustible set is dense in $\mathbb{R}$.

Let $S_1, S_2, \ldots$ be nowhere-dense subsets of $\mathbb{R}$; also, as a notational convenience, let $S_0 = \emptyset$.  Suppose that $\mathbb{R}- \cup_{i=1}^\infty S_i$ is not dense; then $\cup_{i=1}^\infty S_i$ contains some open interval and therefore some closed interval $[a_0,b_0]$, where $a_0 < b_0$.  Now we claim there exist real numbers $a_i, b_i\ (i \geq 1)$ such that \[a_0 < a_1 < a_2 < \cdots < b_2 < b_1 < b_0\] and such that, for each $i$, $[a_i,b_i] \cap S_i = \emptyset$.  These will be constructed inductively.  For $i = 0$, we have already chosen $a_0, b_0$.  If $a_i, b_i$ are given, then we know that $S_{i+1}$ is not dense in $(a_i,b_i)$; hence, there is some closed subinterval $[a_{i+1},b_{i+1}]$ whose intersection with $S_{i+1}$ is empty.  The claim follows.  On the other hand, if we let $a = \sup \{a_i\}$, then $a \in [a_i,b_i]$ for all $i$, which forces $a \notin S_i$ for all $i$.  However, $a \in [a_0,b_0] \subseteq \cup_{i=1}^\infty S_i$, a contradiction; hence, the supposition is false, and the lemma follows.

Now let $R$ be a relation between open subsets of $\mathbb{R}$ and real numbers, where we write $R(U,x)$ to mean $(U,x) \in R$ and $\nR(U,x)$ to mean $(U,x) \notin R$.  We say that $R$ is [i]closed[/i] if the set $\{x \in \mathbb{R}\ |\ R(U,x)\}$ is closed for every open set $U$.  We then have the following lemma:

[b]Lemma 2:[/b] If $R$ is a closed relation, let $M$ be the set of real numbers $x$ such that (a) $R(U,x)$ for every open set $U$ containing $x$; (b) there is some open set $V$ with $x$ as a boundary point such that $\nR(V,x)$.  Then $M$ is nowhere dense.

If $x \in M$, let $V$ be as given by (b).  Let $U = \{y \in %Error. "reals" is a bad command.
\ |\ \nR(V,y)\}$; since $R$ is closed, $U$ is open, and also $x \in U$.  In particular, any $y \in U \cap V$ has $V$ as a neighborhood, and $\nR(V,y)$; hence, $y \notin M$.  So every neighborhood $W$ of $x$ contains $U \cap V \cap W$, a nonempty open set whose intersection with $M$ is empty, and the result follows.

Now consider our function $f$.  For real numbers $r_1 < r_2$, define the relation $R_{r_1,r_2}$ as follows: for $U \subseteq %Error. "reals" is a bad command.
$ open and $x \in \mathbb{R}$, $R_{r_1,r_2}(U,x)$ iff there exists a sequence $x_1, x_2, \ldots$ in $U$ converging to $x$, such that $\lim f(x_i)$ exists and lies in the interval $[r_1,r_2]$.  Notice that this is a closed relation.  Indeed, if $\lim y_i = y$ and $R_{r_1,r_2}(U,y_i)$ for each $i$, let $x_{i,1}, x_{i,2}, x_{i,3}, \ldots$ be a sequence in $U$ converging to $y_i$ with $r_1 \leq \lim_j f(x_{i,j}) \leq r_2$ (for each $i$).  Then, for each $i$, we can choose $j_i$ large enough so that $|x_{i,j_i} - y| < 1/i$ and $r_1 - 1/i < f(x_{i,j_i}) < r_2 + 1/i$.  The sequence $(x_{i,j_i})$ then converges to $y$, and it has some subsequence on which $f$ converges, to a limit that is necessarily in $[r_1,r_2]$.  Thus, $R_{r_1,r_2}(U,y)$, and $R_{r_1,r_2}$ is closed.

If $r_1, r_2$ are rational with $r_1 < r_2$, let $T_{r_1,r_2}$ be the set of values $x$ such that $R_{r_1,r_2}(U,x)$ for every open neighborhood $U$ of $x$, but $\nR_{r_1,r_2}(V,x)$ for some open set $V$ with $x$ as a boundary point.  By Lemma 2, this set is nowhere dense.  Hence, the union of $T_{r_1,r_2}$ over all rational $r_1 < r_2$ (which we now denote by $T$) is exhaustible.

Now let us say that $f$ is [i]densely approached[/i] at $x$ if, for every $\epsilon > 0$, there exists a neighborhood of $x$ in which the points $x'$ satisfying $|f(x') - f(x)| < \epsilon$ are dense.  We now have the following lemma:

[b]Lemma 3:[/b] The set of points at which $f$ is densely approached is dense in $\mathbb{R}$.

We will show that the set of points at which $f$ is not densely approached is exhaustible; the result will then follow from Lemma 1.  First consider the isolated points of the graph $\{(x,f(x)) \in \mathbb{R}^2\}$; since these points can be enclosed in non-intersecting open balls in $\mathbb{R}^2$, there are countably many of them.  So it suffices to show that the set of all points $x$ such that (a) $f$ is not densely approached at $x$ and (b) $(x,f(x))$ is not isolated (in the graph of $f$) is exhaustible.  Consider any such $x$.  We have some $\epsilon$ such that every neighborhood $U$ of $x$ contains a nonempty open subset $V$ with $|f(x') - f(x)| \geq \epsilon$ for every $x' \in V$.  Let $U = (x - 1/i, x + 1/i)$ for $i = 1, 2, \ldots$ successively.  Taking the union of the corresponding $V$'s, we obtain an open set $W$ with $x$ as a boundary point, such that $|f(x') - f(x)| \geq \epsilon$ for all $x' \in W$.  Choose rationals $r_1, r_2$ with $f(x) - \epsilon < r_1 < f(x) < r_2 < f(x) + \epsilon$.  Then $\nR_{r_1,r_2}(W,x)$.  On the other hand, $(x, f(x))$ is a limit point of the graph of $f$, and it follows that $R_{r_1,r_2}(U,x)$ for every neighborhood $U$ of $x$.  We conclude that $x \in T_{r_1,r_2} \subset T$.  Since this holds for each choice of $x$, Lemma 3 follows.

We will say that a set $S \subseteq \mathbb{R}$ is $\epsilon$-spaced if it intersects every interval of width $\epsilon$.  We will construct a family of dense subsets \[K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots\] of $\mathbb{R}$, a family of discrete subsets \[D_1 \subseteq D_2 \subseteq D_3 \subseteq \cdots\] of $\mathbb{R}$, and a family of open intervals \[C_{i,j} \quad (i, j = 1, 2, \ldots)\] with the following properties: $D_i \subseteq K_{i+1}$; $D_i$ is $1/i$-spaced; each point of $D_i$ lies in $C_{i,j}$ for some $j$, and conversely each $C_{i,j}$ contains a point of $D_i$; $C_{i,j} \cap C_{i,j'} = \emptyset$ for $j \neq j'$;  for $i > i'$ and any $j, j'$, either $C_{i,j} \subseteq C_{i',j'}$, or $C_{i,j} \cap C_{i',j'} = \emptyset$; $K_i$ does not contain any boundary point of any $C_{i',j}$ for $i' < i$; if $x \in D_i \cap C_{i,j}$ and $x' \in K_{i+1} \cap C_{i,j}$, then $|f(x') - f(x)| < 1/i$.  We will successively construct $K_{i+1}, D_i$, and $C_i$ by induction on $i$.  For the base case, let $K_1$ be the set of points at which $f$ is densely approached (this is dense by Lemma 3), and let $D_0, C_0$ be empty (as a matter of notation).  Now suppose $K_i, D_{i-1}$, and $C_j$ for all $j < i$ are given ($i \geq 1$).  Because $K_i$ is dense and contains $D_{i-1}$, we can choose $D_i \subseteq K_i$ discrete, $1/i$-spaced, and containing $D_{i-1}$.  Write $D_i = \{x_1, x_2, \ldots\}$ (we know $D_i$ is countable because it is discrete).  For each $x_j$, choose $\epsilon_j$ small enough so that no other $x_{j'}$ lies within the interval $[x_j - 2\epsilon_j, x_j + 2\epsilon_j]$ (possible by discreteness); $(x_j - \epsilon_j,x_j + \epsilon_j)$ is either wholly contained in or wholly outside of every $C_{i',j}$ for $i' < i$ (possible because $x_j \in K_i$ is not a boundary point of any such $C_{i',j}$, and the discreteness and disjointness conditions ensure that each bounded neighborhood of $x_j$ contains only finitely many such boundary points); $|f(x') - f(x_j)| < 1/i$ for each $x'$ in some dense subset of $(x_j - \epsilon_j, x_j + \epsilon_j)$ (possible because $x_j \in K_i \subseteq K_1$, so $f$ is densely approached there). Then let $C_{i,j} = (x_j - \epsilon_j, x_j + \epsilon_j)$.  Finally, let $K_{i+1}$ consist of all points $x' \in K_i$ lying in any $C_{i,j}$ such that $|f(x') - f(x_j)| < 1/i$, together with all points in $K_i$ not in the closure of any $C_{i,j}$.  $K_{i+1}$ is then dense, and it is straightforward to check that $D_i, C_i, K_{i+1}$ still meet all our other conditions.  This completes the inductive step and hence the construction.

Now let $D = \cup_{i=1}^\infty D_i$.  Because $D_i$ is $1/i$-spaced, we know that $D$ is dense in $\mathbb{R}$.  Suppose $x \in D$; then $x \in D_i$ for some $i$.  For any $k > i$, we then also have $x \in D_k$, so $x \in C_{k,j}$ for some $j$.  Every point $x' \in D \cap C_{k,j}$ is contained in some $D_l$ (we may assume $l \geq k$), and $D_l \subseteq K_{l+1} \subseteq K_{k+1}$; thus, $x' \in K_{k+1} \cap C_{k,j}$, giving $|f(x') - f(x)| < 1/k$.  Since $k$ was arbitrary, we see that the restriction of $f$ to $D$ is continuous at $x$.  The result is now proven. $\blacksquare$


You've already posted these problems:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197529[/url]

Do not double post.


SORRY NEVER SHOULD HAVE POSTED IN THE FIRST PLACE

You definitely can't just claim that, although the proof is short:  we cannot have either $ a \equal{} 1$ or $ b \equal{} 1$ or else the LHS is greater than $ 1$, so $ a, b \ge 2$ and $ \frac{1}{a} \plus{} \frac{1}{b} \le 1$ with equality if and only if $ a \equal{} b \equal{} 2$.

T0rajirou, do you realize that the question you just answered was part of an ongoing contest? (Don't worry, it's not your fault, I would not have known that if I had not read the question carefully either.)

In any case, I consider this the "fine line" for breaking the rules of the contest mentioned above.

Oh sorry I didn't know we couldn't ask about specific little parts that we are uncertain of, I won't do it again.

:oops:

A mod can delete this thread


USAMTS
summer program
Mathcamp
geometry
ARML
email

Over the next two weeks, your options for prizes will be listed at http://www.usamts.org.

When choosing your prize(s), please confirm that your address is correct in your profile.

Prize options will be listed on the following dates:

Perfect scores: Today
Other gold medalists (96+): April 24
Silver medalists (75-95): April 27
Bronze medalists (60-74): April 29
Honorable mention (40-59): May 2

On or after the date listed, log in to http://www.usamts.org to choose your prize and Tshirt size.

Also, the top 30 men and top 30 women will be awarded $ \$$200 scholarships to Mathcamp (if they have applied to and been accepted to Mathcamp).

Thank you for participating in the USAMTS.

Does "perfect scores" mean perfect scores for the entire year or a perfect score in at least one round?

Entire year, I would think.  As they get to choose ahead of gold winners, who probably did not get 4 24's.

Would a perfect scorer who has selected mind listing out the choices for prizes?

I can't pull the list back up, but I remember:

1) 360 problems for mathematical olympiads (by Andreescu & Andrica)
(I got this one last year and it was very, very good ... I'd highly recommend it to anybody doing Olympaid level math)

2) Challenging problems in geometry
(Also got it last year ... pretty good but I prefer the aforementioned)

3) Count Down
(Good read about the 2001 [?] IMO team ... not much math involvec

4) The elegant universe
(book about string theory & the like)

5) Music of the Primes
(book about Riemann Hypothesis)

6) A book about cryptology
(title escapes me)

7) Mathematica
(get it if you don't already have it)

JB

There's also a book of past ARML problems on the list.

I think I already have most of the stuff, oddly enough (partially due due usamts last year) so I should probably find out if my school or some people i know need any of the books.

[quote="jackb1115"][quote="knexpert"]Do you think that Mathematica will still be around when the silver medalists get to pick?[/quote]

Can't say ... I'd get up early that morning though =)

JB[/quote]

I think so; last year I was a bronze [I think] and I got Mathematica as one of my prizes.

Hehe...no hope for my pitiful "honorable" mention?

Do seniors who are in the top 30 get the $\$$200 scholarship to mathcamp or do the scholarships go to the top 30 in grades 11 and below?

(because seniors can't go to mathcamp)

There is no way to see what you have sent is there?

*ehehheehee hesitantly clicks the submit button* then, phoom! "Thank you for choosing..." replaces the list.

[quote="cosinator"][quote="iostream.h"][quote="cosinator"](because seniors can't go to mathcamp)[/quote]actually seniors CAN go to mathcamp[/quote]

huh?
how?[/quote]they just can... you apply normally, get in, and go.  i know of several people who have done this.

I chose my prizes at 11 PM Mountain Time, 4/26/2007. I took a screenshot of the USAMTS prizes before I submitted them.

[quote="solafidefarms"]I heard about the release of Mathematica 6 today. Will we be getting Mathematica 5 or 6?

[If 5, is the licensing such that I can give it to another student?][/quote]

It will be Mathematica 6.

Supersupersupercool. Means it might actually work on my Linux machine :)... many thanks! Now I'm eagerly looking forward to my prizes coming ;). Thanks again!

Cool! When should we expect to get the licensing email?

oh snap, mathematica 6

you guys are the bomb

Ahh is it possible to change prizes? I might have liked to get Mathematica had I known it was not a version I already had.

Not sure when they'll send the license; we'll send the names to them on Monday.

Unfortunately, no prize swapping (all the Mathematica's are likely spoken for now anyway).

Sorry, but we weren't informed of the upgrade until a couple days ago, either.

I got this e-mail confirming that Mathematica for Students is going to be sent to my address.  However, it also says that it is being billed to my address.  Is this normal?

For me as well, I was notified that it would be billed to my address, but I never supplied credit card/paypal information. I assume that they would only ship it if the billing address was the same as the shipping address and that I will not be charged. Is this correct?


Given two natural numbers $ w$ and $ n,$ the tower of $ n$ $ w's$ is the natural number $ T_n(w)$ defined by

\[ T_n(w) = w^{w^{\cdots^{w}}},\]

with $ n$ $ w's$ on the right side. More precisely, $ T_1(w) = w$ and $ T_{n+1}(w) = w^{T_n(w)}.$ For example, $ T_3(2) = 2^{2^2} = 16,$  $ T_4(2) = 2^{16} = 65536,$ and $ T_2(3) = 3^3 = 27.$ Find the smallest tower of $ 3's$ that exceeds the tower of $ 1989$ $ 2's.$ In other words, find the smallest value of $ n$ such that $ T_n(3) > T_{1989}(2).$ Justify your answer.

Hmm here's a solution.

[hide]
We claim that, for each positive integer $n$, the following holds:
\[4T_n(3)<T_{n+2}(2)<T_{n+1}(3).\]
We proceed by induction. The base case ($n=1$) is equivalent to $12<16<27$, which is true.

Now let $n\geq 2$. By the inductive hypothesis, we have that $4T_{n-1}(3)<T_{n+1}(2)<T_n(3)$. Furthermore,
\[T_{n+2}(2)=2^{T_{n+1}(2)}<2^{T_n(3)}<3^{T_n(3)}=T_{n+1}(3)\]
and
\[T_{n+2}(2)=2^{T_{n+1}(2)}>2^{4T_{n-1}(3)}=16^{T_{n-1}(3)}=\left(\frac{16}3\right)^{T_{n-1}(3)}3^{T_{n-1}(3)}>4T_n(3),\]
and the induction is complete. To finish, note that setting $n=1987$ yields
\[4T_{1987}(3)<T_{1989}(2)<T_{1988}(3),\]
so it is clear that the smallest $n$ is 1988.[/hide]


vector
complex numbers
combinatorics proposed
combinatorics

Let $ABC$ be a triagle.If we paint the points of the plance in red and green,prove that either there exists two red points which are one unit apart or three green points forming a triagle congruent to $ABC$
Sorry if it was posted before!

Isn't it trivial??? :P

Consider the red point $A$ then if we assuming contrary the whole circle $S_1(A)$ is green. Now just inscribe similar triangle to that circle.

You have to find a triangle [b]congruent[/b] to $ABC$ with green points.

Ops... I don't know why I read it as "similar". :blush:

I am going to try something, but I don't know if it's Ok.

Let $a,b,c$ be three complex numbers representing triangle $ABC$. For each complex $z$, at least one of $a+z,b+z,c+z$ is red. Now translate the plane with vector $-a$, and mark all the images of all the red points of the type $a+z$ with $1$. Then, translate the plane with vector $-b$, and mark all the images of all the red points of the form $b+z$ which have not been already marked with $2$. Now repeat this for $c$, marking the respective points with $3$. We have just colored all the plane in three colors, and it is well-known that there must be two points of the same color a distance of one unit apart (i.e. the chromatic number of the plane is at least $4$).

Does it work? :?

More generally, one can prove that :
Given any configuration of four points $\{A,B,C,D \}$ in the plane, and given any two-colouring of the plane, say red and green, with no two red points at distance 1 from each other, then there exists a green configuration which is congruent to $\{A,B,C,D \}$.

Reference :
R. Juhasz, Ramsey type theorems in the plane, Journal of Combinatorial theory, serie A, 27 (1979), p. 152-160.

Pierre.

My students found a really nice elementary solution to this problem.

Which one? The original one or the generalisation?
Could you give this elementary and nice solution?

Pierre.

Pierre, can you give the proof of that book?


limit
logarithms
function
calculus
calculus computations

Prove that, for $ m$ a positive integer and $ n$ a non-negative integer,

$ \lim_{x \to 0} x^{m} (\ln x)^{n}\equal{}0$.

[quote="AndrewTom"]Prove that, for $ m$ a positive integer and $ n$ a non-negative integer,

$ \lim_{x \to 0} x^{m} (\ln x)^{n} = 0$.[/quote]

let  $ t=\ln x$

$ \lim_{x\to 0}x^{m}(\ln x)^{n}= \lim_{t\to -\infty}{\frac{t^n}{e^{-tm}}=0}$,or you can apply L'Hospital law for $ n$ times

Well, strictly speaking the limit does not make sense since $ \ln x$ is defined only for $ x>0$ but the limit isn't for $ x\to0^\plus{}$... Anyway, after the substitution of Deng Qunyi one can use one of the proceedure for the function $ t^ke^{\minus{}mt}$ for
$ t\to\minus{}\infty.$


It does not make sense: if contestents are graded, what exactly are you saying about the pairs of contestants?

[quote="marko avila"]a contest  has  8 judges and the contestants are graded by  yes  or no.  it is known  that  for each  pairs of contestants there are 4 judges  that  grade yes and for  that  grade no . find the maximum numbers of contestants[/quote]
does this mean that for every two contestants 4 grade yes and 4 others grade no, but for another pair there could be the same judges?

is this what you are trying to say?

[quote="funcia"]does this mean that for every two contestants 4 grade yes and 4 others grade no[/quote]

See, this doesn't make any sense: if it means, "both individuals are graded yes by (the same) 4 judges and graded no by (the same) 4 judges," then any system in which 4 judges grade everyone "yes" and the other 4 grade everyone "no" works.  Otherwise, it's not defined how pairs of contestants are graded.


Write every number between 1 and 2500 with four 4s, the operation +, -, *, /, !,  \sqrt , and d(.).

d(n) is the number of divisors of n (I count in these 1 and n).

Another one.
Write every positive integer number using two 2s, the operation +, -, *, /, !, \sqrt , and ln(.).

myth in your problem! I can't figure out how to write 3 !
perhaps it's too obiuos and I can't see it!!
:?

cheers! :D :D

It is not too obvious. I can't give any hint about a way to obtain 3 because it will be solution for all n.

Good luck!


If $ a,b,c>0,$ such that $ a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca\equal{}6,$ prove that $ a^2\plus{}b^2\plus{}c^2\geq3$

[hide]
Case 1
Suppose $ a\plus{}b\plus{}c \le ab\plus{}bc\plus{}ca$

then $ ab\plus{}bc\plus{}ca \ge 3$

From rearrangement $ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ca \ge 3$

Case 2
Suppose $ a\plus{}b\plus{}c \ge ab\plus{}bc\plus{}ca \Longrightarrow a\plus{}b\plus{}c\ge 3$

$ \therefore (a\plus{}b\plus{}c)^2 \plus{} ab\plus{}bc\plus{}ca \ge 3^2 \plus{} 3 \equal{} 12$

$ \therefore 4(a^2 \plus{} b^2 \plus{} c^2) \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab\plus{}bc\plus{}ca) \ge 12$

hence $ a^2 \plus{} b^2 \plus{} c^2 \ge 3$[/hide]

[hide]$ 6\equal{} \sum a \plus{} \sum ab \leq \sum a \plus{} \frac{( \sum a )^2}{3} \leftrightarrow 0 \leq (\sum a \minus{} 3)(\sum a \plus{} 6)  \equal{}> a\plus{}b\plus{}c \geq 3$
so,

$ \sum a^2 \geq \frac{ ( \sum a )^2}{3} \geq 3$[/hide]

[quote="arqady"][quote="ocha"]
Case 2
Suppose $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca \Longrightarrow a \plus{} b \plus{} c\ge 3$

[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.1.$ :wink:[/quote]

$ 6 \equal{} a\plus{}b\plus{}c \plus{} ab\plus{}bc\plus{}ca \equal{} (1 \plus{} 1 \plus{} 0.1) \plus{} (1\cdot 0.1 \plus{} 1\cdot 0.1 \plus{} 1\cdot1) \equal{} 3.3$

$ \therefore 6\equal{}3.3$ :wink:

[quote="ocha"][hide]
Case 1
Suppose $ a \plus{} b \plus{} c \le ab \plus{} bc \plus{} ca$

then $ ab \plus{} bc \plus{} ca \ge 3$

From rearrangement $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca \ge 3$

Case 2
Suppose $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca \Longrightarrow a \plus{} b \plus{} c\ge 3$

$ \therefore (a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca \ge 3^2 \plus{} 3 \equal{} 12$

$ \therefore 4(a^2 \plus{} b^2 \plus{} c^2) \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca) \ge 12$

hence $ a^2 \plus{} b^2 \plus{} c^2 \ge 3$[/hide][/quote]
For case 2, $ ab\plus{}bc\plus{}ca\le 3$, so I think you should justify $ (a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca \ge 3^2 \plus{} 3 \equal{} 12$...


trigonometry
analytic geometry
vector
Pythagorean Theorem
geometry
absolute value

I can do all the problems from the "Are You Ready" PDF for Intermediate Trig, except for this problem:

Evaluate:
$|3+4i|$

The answer is [hide="this"]5[/hide].  I see a potential Pythagorean thoerem, but I don't get the concept.  Please help!

[quote="mysmartmouth"][quote="236factorial"]I don't know why this formula works, but

$|a+bi|=\sqrt{a^{2}+b^{2}}$.

It probably has something to do with the complex plane.[/quote]

I realized that was probably the formula.  However, I'll forget this by the end of the night.  Can anyone explain the derivation, or where this comes from?[/quote]

A complex point thingy is the distance from the origin. So then its just the distance formula, or the pythagorean theorem, or $|a+bi|=\sqrt{a^{2}+b^{2}}$. And since one axis is imaginary and the other is real, you have the formula.

We are working in the complex plane where the real axis is the x-axis and the imaginary axis is the y-axis. So for z = a+bi, the coordinate of z on the complex plane is (a,b). Absolute value of z is $\sqrt{a^{2}+b^{2}}$ or the distance from the origin (0,0) to (a,b).

[quote="236factorial"]
$|a+bi|=\sqrt{a^{2}+b^{2}}$. [/quote]

This is a definition more than anything else, not exactly something that needs to be explained.  You either know it or you don't.

:yoda: Yoda sez that the reason that formula works is simply because using vectors on a number plane creates a right triangle.  Absolute values only mean distance from the center from a number plane regardless of direction.  So 3+4i would be somewhere up and to the right 5 units away from the center.

:jump: has ADD

it's magnitude not absolute value and it's like the equivalent of distance, except in the imaginary plane

i saw an imaginary plane the other day

it was a cloud though

[quote="Treething"]magnitude not absolute value[/quote]

They're defined to be the same thing for a complex number.


Prove that $\sigma(n)\phi(n) < n^2$, but that there is a positive constant $c$ such that $\sigma(n)\phi(n) \ge c n^2$ holds for all positive integers $n$.

If $a,b$ are coprime and we already know $\sigma(a)\varphi(a) < a^{2}$ and $\sigma(b)\varphi(b) < b^{2}$, then multiplying these gives $\sigma(ab)\varphi(ab)=\sigma(a)\varphi(a)\sigma(b)\varphi(b) < a^{2}b^{2}=(ab)^{2}$. Thus we just need to check it for $n=p^{k}$ being prime power of a prime $p$. In that case, $\sigma(p^{k})\varphi(p^{k}) = \frac{p^{k+1}-1}{p-1}\cdot (p-1)p^{k-1}= p^{2k}-p^{k-1}< p^{2k}=(p^{k})^{2}$.

For the other one, set $c=\frac{1}{\zeta(2)}$, $\zeta(2) : = \sum_{k=1}^\infty \frac{1}{k^{2}}<2$. Let $n=p_{1}^{v_{1}}p_{1}^{v_{1}}...p_{s}^{v_{s}}$.
Using multiplicativity and the values at prime powers, we get \[\frac{\sigma(n)\varphi(n)}{n^{2}}=\prod_{i=1}^{s}\frac1{p_{i}^{2v_{i}}}\prod_{i=1}^{s}\frac{p^{v_{i}+1}-1}{p-1}\prod_{i=1}^{s}(p-1)p^{v_{i}-1}= \prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{v_{i}+1}}\right).\]
But now \[\left(\prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{v_{i}+1}}\right) \right)^{-1}\leq \left(\prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{2}}\right)\right)^{-1}= \prod_{i=1}^{s}\left(\frac{1}{1-p_{i}^{-2}}\right) \leq \prod_{p \in \mathbb P}\left(\frac{1}{1-p^{-2}}\right) ,\]
with $\mathbb{P}$ being the set of primes.
But \[\prod_{p \in \mathbb P}\left(\frac{1}{1-p^{-2}}\right) = \prod_{p \in \mathbb P}\sum_{k=0}^\infty (p^{-2})^{k}= \sum_{k=1}^\infty k^{-2}= \zeta (2)\]
by factoring (and some closer look at convergences etc.).

All together, this gives the result.
This also shows that there is no bigger $c$ with that property.

[hide="Solution"]Let $f(n) = \dfrac{\sigma (n)\phi (n)}{n^{2}}$. Notice that $f(n)$ is multiplicative. Take $n=p^{a}$. We have that $f(n) = \dfrac{(\dfrac{p^{a+1}-1}{p-1})(p-1)p^{a-1}}{p^{2a}}$. This simplifies to $\dfrac{p^{a+1}-1}{p^{a+1}}$, which is always less than one. It follows that $f(n)$ is always less than one by the fact that $f(n)$ is multiplicative. (This is excluding $n=1$ which is the one exception to the given identity as $1$ is not less than $1$).
Notice that to maximize the value of $c$, we want to minimize $\dfrac{p^{a+1}-1}{p^{a+1}}$, which clearly occurs at $a=1$. Therefore, the maximum value of $c$ is $\prod_{\text{p is prime}} \dfrac{p^{2}-1}{p^{2}}$. Now refer to [url="http://artofproblemsolving.com/community/u157930h1211489p6047479"]This Post[/url] (credits to MathAwesome123). We have that $c=\dfrac{6}{\pi^{2}}$ satisfies the identity as desired.[/hide]


[quote="DaReaper"]the complement of the supplement of an angle is 20 degrees. What is the angle?

i actually attained answer, just checking to see if its right.

[hide]110?[/hide][/quote]
[hide="algebraic solution"]
Let $ x$ = the angle
$ 180\minus{}x$ would be the supplement
and $ 90\minus{}(180\minus{}x)$ would be the complement of the supplement.
$ 90\minus{}(180\minus{}x)\equal{}20$
$ \minus{}90\plus{}x\equal{}20$
$ x\equal{}110$
Yes.
[/hide]


probability
real analysis
function
integration
probability and stats

Let $ X$ and $ Y$ be two random variables defined on a probability space $ (\Omega, \mathcal{F}, P)$. Suppose that both $ X$ and $ Y$ have continuous distribution.
Question: Are there any probability measure, say $ Q$, on $ (\Omega, \mathcal{F})$ such that under $ Q$, $ X$ and $ Y$
i) still have continuous distribution, and
ii) have the same distribution.

This might be possible if both random variables are independant
I don't see any other solution nor counter-example for it
May be it can be seen that is not possible to do so using Fourier transform for $ X$ and $ Y$ because
you must then have under $ Q$ $ \mathcal{F}_X(\lambda)\equal{}\mathcal{F}_Y(\lambda)$

what if $ X \in [1;2]$ and $ Y \in [3,4]$ almost surely ?
what do you exactly mean here by continuous distribution? do you mean that $ X,Y$ have continuous densities over the Lebesgue measure: there exists continous function $ \phi_X, \phi_Y$ such that for any test function $ f$: $ E_P[f(X)] \equal{} \int_{\mathbb{R}} f(u) \phi_X(u) du$, $ E_P[f(Y)] \equal{} \int_{\mathbb{R}} f(u) \phi_Y(u) du$. Or maybe you mean that there is no $ x_0$ such that $ P(X\equal{}x_0)>0$ ?

[quote="alekk"] there exists continous function $ \phi_X, \phi_Y$ such that for any test function $ f$: $ E_P[f(X)] \equal{} \int_{\mathbb{R}} f(u) \phi_X(u) du$, $ E_P[f(Y)] \equal{} \int_{\mathbb{R}} f(u) \phi_Y(u) du$...
  there is no $ x_0$ such that $ P(X \equal{} x_0) > 0$ ?[/quote]
It is my custom to call the former the absolute continuous distribution and the later the continuous one.
Anyway, I knew that the answer is negative in general (a simple counter-example is alekk's; a more complicate one is TheBridge Two's). What exactly I would like to know is with which condition, we can change the probability measure to make some continuous distribution (under the old P.M.) random variables to become e.g a normal distribution RVs. Furthermore, suppose that we have a continuous time process say $ X(t)$ on a probability space, then the question is with which conditions, $ X(t)$ becomes a Wiener process under another P.M. , etc.
I hope every member consider my question as an open one and freely give his/her opinions.

Regarding your comments on Alekk's answer personnaly I'm ok for absolute continuous conditions denomination
but for  the second one I call them diffuse law.
(there exist diffuse laws that are not absolutly continuous with repsect to lebesgue measure)

Returning to the topic,

For the process view of the problem, you can take a look at the Dambis Dubions-Schwartz theorem where everything is seen through a time change procedure. ( Karatzas and Shreve's book on brownian motion for a reference)

You can also take a look at the so called Skorohod Problem where your are given a brownian motion and a distribution $ d$ and you are looking for a stopping time $ \tau$ such that $ B_{\tau}$ has the distribution $ d$. A refrence is "Obloj - The Skorokhod embedding problem and its offspring".

But the question you should adress first is the framework of your problem which is a little "fuzzy" at the moment. Espescially on the question of joint law of (X,Y) under P.

see ya

Thank for your kind comment. I will have a look at your reference book and paper.


At a state contest, 21 Mathletes stay in the same hotel. Each Mathlete gets his/her own room and the room numbers are 1 through 21. When all the Mathletes have arrived, except the ones staying in rooms 12 and 13, what is the median room number of the other 19 Mathletes?

The median of a 19-element set is the 10th element, which is 10.

The middle room, or the median, is (1+19)/2=10

Alternatively, you can count 9 rooms from 1 and 19 to get room 10.


conics hyperbola geometry incenter symmetry Asymptote

Given two externally tangent circles, what is the locus of the centers of all circles externally tangent to both of them?

Yakko

Call the circles $C_1, C_2$ and their radii $R_1, R_2$ respectively. Let their centers be $O_1, O_2$, and the point of tangency $Q$. Suppose $P$ is a point such that there's a circle with center $P$ tangent to $C_1, C_2$ and this circle has radius $r$. Let $X, Y$ be the points where $O_1P$ and $O_2P$ intersect $C_1, C_2$ respectively. Then we have

\[O_1X=O_1Q=R_1 \\
O_2Y=O_2Q=R_2 \\
PX=PY=r \]

In other words, $Q, X, Y$ are the points where the incircle of triangle $O_1O_2P$ touches its sides. Now take the line perpendicular to $O_1O_2$ through $Q$; the incenter $I$ of the triangle must be on this line. Now take any point $I$ on the line, and draw the circle with center $I$ touching $Q$. It is clear by symmetry on $O_1I, O_2I$ that if the points where the circle intersects $C_1, C_2$ are $A, B$ respectively, then $O_1A, O_2B$ are tangent to the incircle. Then the intersection of the lines $O_1A, O_2B$ is a possible point $P$. Well, that gives a method for constructing the points, but the locus is indeed an ugly hyperbola.

Call the centers of the given circles A and B. Call the center of a circle that is tangent to both C. AC-BC is equal to the difference between the radii of A and B. The difference between the radii of the given circles is constant. The hyperbola is the locus of points such that the difference between the distances from the foci is constant. The foci of the hyperbola are A and B.

Edited, based on JBL's suggestion

Bictor, I think you mean "foci," not "loci."

So, if the locus is an hyperbola, it only gives us one half of the hyperbola.  How would you get the other half?

Thanks for correcting me again. I think some of my wires have crossed.

The other half of the hyperboloid is the locus of centers of circles that contain the two given circles, such that the given circles are internally tangent to them.

Good.  So we've got both branches of the hyperbola and its foci -- what are its asymptotes?

The asymptotes are the perpindicular bisectors of the segments externally tangent to both circles.  To figure it out, think of the problem in terms of a limit to infinity: as the radius of the third circle nears infinity, the arc between the two circles become more and more like a segment, and the center lies on the perpindicular bisector of that segment/arc.

Here's an interesting problem based on the one on this topic:

Given 3 mutually tangent circles, 2 of which have a fixed radius, find the loci of points that are the centers of circles that are tangent to all 3 other circles.

*Shudders at the thought of a formula for it*

Okay, one final one: the hyperbola crosses the line joining the centers of the two circles at two points.  One of them is the point of tangency of the two circles; what is the significance of the other?

[hide="Towbomb"]If you're talking about circles externally tangent to all 3, it would just be a portion of the hyperbola. On each side, the limit of the third circle is the tangent to the fixed circles. The portion of the hyperbola is bounded by the center of the circle that is tangent to this line and the fixed circles.[/hide]

[hide="JBL"]It would be the center of the smallest circle that contains both of the given circles. Extend the centerline to the given circles. This double diameter is the diameter of the smallest circle, centered at the intersection of the hyperbola and centerline.[/hide]


How can I devide a circle in 2 shapes with equal area by drowing an arc with the compass. Is it possile?

Fastest way for calculation (?):
draw a diameter $AB$, then draw a circle of diameter $AO$ ( $O$ - the centre of initial circle); the perpendicular bisector of $AO$ intersects the smaller circle at 2 points, let $C$ be any of them; draw an arc centered at $A$, radius $AC$, it will intersect $AB$ at $D$. The circle $(D, DA)$ is the required 'arc'.

Best regards,
sunken rock

draw radii of a circle and through its centre draw three 60degree angles which results in semicircle and hence circle divides in two equal areas........


Let ABC be a triangle whose circumcircle is (O) and such that AC<AB .The tangent lines to (O) at B,C intersect at T.The line through A and perpendicular to AT meet BC at S.Choose the points $ B_1$ and $ C_1$ on ST such that $ TB_1\equal{}TC_1\equal{}TB$ and that $ C_1$ lies between S and T.Prove that ABC and $ A_1B_1C_1$ are similar.

To mod: Pls lock this topic because it's in Vietnam magazine last month: [b]Mathematics and Youth[/b]

why???do you think I post it to ask and send it to mathematic and youth magazine?you are wrong
this exercise is in month 6 magazine,and it is out of date to send answer to mathematic and youth magazine.I post it  because I want some people who don't know it to solve it.

You are wrong. The date to send answer is [b]two months[/b]! May be you don't ask and send to this magazine but someone reads the solution andx they will cheat.

[color=red]EDIT: Deadline for submitting solutions to Mathematics and Youth expired on 8/31/2009.[/color]


How to prove $ \aleph^\aleph\equal{}2^\aleph$?
Thanks

For any infinite cardinal $ X$, $ 2^X\equal{}2^{X\cdot X}\equal{}(2^X)^X\ge X^X\ge 2^X$.

[quote="jmerry"]For any infinite cardinal $ X$, $ 2^X \equal{} 2^{X\cdot X} \equal{} (2^X)^X\ge X^X\ge 2^X$.[/quote]
sorry, can you tell how to prove that $ X \equal{} X\cdot X$, if cardinal $ X$ is infinite?


Let [tex]f(x)[/tex] be convex on [tex](-\infty,0][/tex], and concave on [tex][0,b], x_i\leq b.[/tex] Prove that: [tex] \sum^n_{i=1}f(x_i)\leq f(s_1)+(n-1)f(\frac{s_2}{n-1})[/tex]
where [tex]s_1=\sum_{x_i\leq0}x_i [/tex] and [tex]s_2=\sum_{x_i>0}x_i. [/tex]

Conseder
          x1<=...<=xk<=0
and
         0<y1<=...<=yn-k.
We have
     f(x1)+...+f(xk)<=(k-1)f(0)+f(s1)  (Majorization Inequality)
and
    (k-1)f(0)+f(y1)+...+f(yn-k)<=(n-1)f(s2/(n-1))  (Jensen's Inequality)
Summing these inequality, we obtain the given inequality.
I hope it is correct.


A graph plotter gives the graph of x^3 + 3xy + y^3 =1 as the line x + y = 1? Is this a mistake?

[hide]$ x^3 \plus{} 3xy \plus{} y^3 \equal{} 1$
$ (x)^3 \plus{} (y)^3 \plus{} (\minus{}1)^3 \minus{} 3(x)(y)(\minus{}1) \equal{} (x\plus{}y\minus{}1)(x^2\plus{}y^2\plus{}1\minus{}xy\plus{}x\plus{}y) \equal{} 0$.

So the graph is $ x\plus{}y \equal{} 1$ and $ x^2 \plus{} y^2 \plus{} 1 \plus{} x \plus{} y \equal{} xy$. The second equation is equivalent to $ (x\plus{}1)^2 \plus{} (y\plus{}1)^2 \equal{} (x\plus{}1)(y\plus{}1)$, which by AM-GM has only the solution $ (\minus{}1,\minus{}1)$. So you should get $ x\plus{}y \equal{} 1$ and the point $ (\minus{}1,\minus{}1)$. Almost not a mistake.[/hide]


Hi!
I am a freshman, and would like to know if you can take the AMC 10A, and the AMC 10B.
Are the tests going to be different?
Will my result for the AMC 10B count??

thanks guys,
jukjukenator

Yes, you can take both the "A" date AMC 10 and the "B" date AMC 10 in the same year, as long as you are eligible by age/grade for the AMC 10. The tests are different. They are meant to be of approximately equal difficulty (although this is hard to make exact) and they tend to cover the same topics each year.

Good luck. Read the rest of this forum to see more answers to frequently asked questions.

you can, of course, but instead i would do AMC 10A and AMC 12B, that way you can get a taste of 10 and 12.


Let ABC be a triangle. Then prove that:

$ sin^2(A) \plus{} sin^2(B) \plus{} sin^2(C)\equal{}2 \leftrightarrow$ ABC is right angled

[quote="Kouichi Nakagawa"][hide="Hint"]
\[ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C \equal{} 2(1 \plus{} \cos A \cos B \cos C)
\]
[/hide][/quote]

Cool  :lol:

There are a lot of other methods but this one is the shortest one.

Thanks


Can it be proved that if a determinant is zero, ( consider a $3 \times 3$ determinant!) , then one row/column is a linear combination of the other two? I hope you understand what I mean by 'linear combination of rows etc' - it means that one element of one column in a row is a linear combination of the elements of the same column in other rows and vice-versa ! Note: -This question arises in the context of Vectors 'decomposition'! If The result is true, I'd appreciate if someone can post or guide me to a proof!

That's true. Do you know what the effect of elementary row/column operations is?

Yes, it is all the interest of constructing the determinant, and it works for upper dimensions.

If $A = Mat(u,B)$, where $B$ is the canonical basis of $\mathbb{R}^{n}$, then $det(u(e_{1}),\dots,u(e_{n}),B)=det(u)det((e_{1},\dots,e_{n}),B)$ where $det(u)=det(A)$ is independent of $B$, and $(e_{i})$ denotes the $i$-th vector of the canonical basis. Now since $det(A)=0$ you have $det(u(e_{1}),\dots,u(e_{n}),B)=0$.
But in general $det((x_{1},\dots,x_{n}))=0 \Rightarrow (x_{1},\dots,x_{n})$ not free (proof: decompose the vector in $B$ and write its determinant), and reciprocally (use the fact that the determinant is an alternated multilinear form). Now you're done.

I forgot: to prove the same result with the rows, use $det(A)=det(A^{t})$ (can be obtained by the definition of the determinant itself, involving the sum over the permutations ...).

Arne: That's good that the result is true. I do know the basic row/column transformations etc. But I wasn't aware that the converse result was true as well!

julien_santint :  I'm so sorry I cant understand 90 % of your proof. I'm not even aware of the terms you are using! Can you please give a very simple, basic level proof of the result possibly for the simpe case of a $3 \times 3$ determinant! Thanks a lot!!!

Try this: Each elementary row operation in a row-elimination algorithm can be represented by left multiplication of the given matrix by an "elmenetary matrix" $E.$ There are three types of elementary matrices. Each type of row operation has a predictable effect on the determinant, which we can summarize as $\det(EA)=\det(E)\det(A).$

In particular, $\det(E)=-1$ if $E$ is a row interchange, $\det(E)=r$ if $E$ is the operation of multiplying a particular row by a nonzero constant $r,$ and $\det(E)=1$ for a "type 3" row operation (adding a multiple of one row to another row.) All of these values are nonzero.

Starting with a given matrix $A$ we can apply a sequence of elementary row steps to get it into  row-echelon form. That is, $E_kE_{k-1}\cdots E_1A=R$ where $R$ is in row-echelon form. $\det(A)=0$ iff $\det(R)=0.$ That only happens if $R$ has at least one row of zeros across the bottom.

One of the rows (columns) can be writting as a linear combination of the other rows (columns)

iff

the rows (columns) are linearly dependent

iff

the matrix is row-equivalent to a matrix with a row of zeros

iff

the determinant is zero.

Thanks!!! I got a simple proof for the $3 \times 3$ determiannt myself! Maybe its fundamentally the same as yours!! I am not aware of the theory you used but have understood the proof! Thanks!

This looks like a particular case of a more general theorem:

"the maximum number of linearly independent rows equals the maximum order of a $\neq 0$ minor"

I think an approach similar to Kent's would be useful, but I have troubles with some minors when I add a multiple of a row to another (they change).


I thought that somebody here might be able to help me; I have a Win XP computer since a few days and I'm constantly being attacked by viruses, worms and dialers. When I surf on the web, a nasty "Security Warning" ("Sicherheitswarnung") window regularly pops up, asking me whether I want to install some strange "IE Plugin" by "DOWNLOADS UNLIMITED". The Plugin is said to be certified by "Thawte Code Signing CA". However, by the kind the window behaves (and the web address of the Plugin: http://www2.skoobidoo.com ), it is clear that the software is a virus. I am clicking on "No", but the window is regularly popping up again. Which configurations should I modify in order to block such windows (generally: all programs certified by "Thawte Code Signing CA", cuz all such programs seem to be faked) automatically? And: how can I close ports and which ports should I close?

I really have no clue on Win XP or on any antivirus program either. (I used Win 98 before, and the chance of being infected by a virus on surfing in Win 98 is really low, unless one opens some idiot spams and virus mails. But now I went online and got a virus in the first minute.)

Thanks!
  Darij

(PS. Sorry, some more newbie questions on security are still to come, but now I go offline to get rid of some spyware on my disk...)

Hmm... I never did anything particular, and I don't have it...

All I can recommend: a good anti-virus (like Norton), and visiting less underground sites :D

(j/k m8 ;))

Darij, please install all the latest upgrades of WinXP (including SP2). They will close all the major security leaks. Also put up your winXP firewall. That helps also.

And finally get a resident antivirus (since you are running now winXP, I assume you have also upgraded your computer) like Norton Antivirus (usually such an antivirus comes when buying a full system).

Thanks for the infos, Valentin. I have not upgraded my computer, I have bought a new one. It actually has a Norton Antivirus, but the virus definitions are from August 2003 and the update won't install. I have also tried the WinXP upgrades but it either takes a very looooong time to download them or the download just hangs. We have ordered a CD from Microsoft now. What belongs the firewall, I was idiot enough to deactivate it when I began surfing (actually, I still have many of my old Win98 surfing habits...) and now I've activated it again, but it doesn't look very active now :?  ...  I have used Ad-Aware to kill some spyware programs and a dialer, but actually I don't see any sign of antivirus when I'm surfing with my new computer; the download windows keep popping up as before.

Ah yes, but fortunately I still have my old Win98 computer, from which I am typing this message...

  darij

You ought to install Spybot: Search and Destroy. It has the function to disable installations of certain BHOs, so those dialogs may disappear.

Me, personally I just use Firefox.

Thanks, Osiris. I have also been said now that the windows firewall is not a real firewall, so people have recommended me a better one. Thanks to all of you.

  Darij

It is, if you know how to use it. In fact the whole firewall business is generally made so that only experienced users on the internet can use it to its full potential ... I guess this will be the case, as internet becomes more and more complicated each day  :maybe:

With regards to firewalls, one ought to install something along the lines of Tiny Personal Firewall, Kerio Personal Firewall or Sygate Personal Firewall. Lots of people use ZoneAlarm, but personally I dislike it because it uses way too much system resource.

Windows Firewall is not a true firewall in the sense that it only blocks unwanted inbound connections. Once spyware gets onto your computer, its outbound activity is not restricted at all. It can then 'phone home' with information from your computer. This is a far greater threat than, say, a simple port scan or a DoS attack, which Windows Firewall does protect your computer against.

You're welcome.  :lol:

[quote="Valentin Vornicu"]Normally a vanilla winXP does not pose problems when installed ok. :) There must be something you've done.[/quote]

Sure..  and there are people who believe in my proof that,  all odd numbers are prime ...if you  can't follow that  proof there is something wrong in your math education ... :lol:

Okay ..  I meant, that if you have vanilla winXP installed "correctly" and  pick and an odd number it will be a priime.. if it is not there must be something you have done which is wrong... :D

Eum... Gyan, I don't really get the relevance for the prime numbers... since the way I understood it you would be saying that almost every good system is a vanilla XP? :D

Anyway Darij, just don't give up. An anti-virus like norton as said; stuff like Ad-Aware SE, Spybot: Search & Destroy, Pest-Ad, ... are all very useful too. As for the Norton updating, you definitely need new virus updates then! If all methods fail, I'd recommend uninstalling and re-installing norton, if possible in safe mode... since the reason it doesn't work is probably that the virus (or any other program) is blocking it :)

Firewalls usually just do fine with the normal settings, but as soon as a malicious program is in your comp, it's easy for it to disallow proper working of the firewall :)

What I am trying to say, is that unlike other people, I have been running my current Windows XP installation for about 2 years now, and yet no black screen, no virus, no interrupted connections. I do not use a firewall, but just upgrade to the latest Norton Anti-virus and Windows XP upgrades.

Also there are no apparent delays in Windows loading, although I have almost 150 programs installed (and around 50 uninstalled) ... :)

Seriously? :) Then winXP does have some advantages... I got my winME for half a year now, I gotta defrag regularly etc just to keep a decent speed, and the thing blocks from time to time.

But yeah for some reason spammers all around the world got my email, I get like 0.5 infected message per day (on average), I can hardly believe there ain't one sometimes Norton can't deal with.
Same for the web, even though I have my antivirus on and updated, I find multiple spyware every time I check... yes, sometimes even after I only visited this site... :?

I got no clue where they keep coming from :)

Thanks again to everybody. I have decided to postpone the work on the new computer until this weekend, since I need much time for mathematics now and I have to reply to a dozen of emails which I should have replied to a month ago...

Yeah, Peter, I get around 30 spam mails every day. In fact, my first idiotic mistake was posting in a newsgroup where email addresses are openly displayed, without properly munging my email address. Now I have noted that although I have set the option "Always show my e-mail address" to No in my MathLinks profile, my address still appears in those subfores where I am a moderator, namely in the "Geometry" subfores. I really hate this, since it only helps spammers. Why not allow the members to mung the email addresses, e. g. writing << darij_grinberg@ebw.de with 'web' instead of 'ebw' >> ? I think this would be helpful.

  Darij

perhaps the email is displayed to you... I can't see it anywhere

but you're right about hiding it, Orl's got a contaminated mail from my, probably the bot found the match on this site... :?

[quote="Peter VDD"]perhaps the email is displayed to you... I can't see it anywhere[/quote]

Indeed, sorry. I see the emails of all posters in the Geometry fores since I am a moderator of these fores. But others don't see them (I've checked this using my "isotomion" account). You should see the email addresses at the Pre-Olympiad forum.

  Darij


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a torsion-free group having a cyclic subgroup of finite index. Prove that G is cyclic.

This is probably not the simplest approach, but it is how I first thought of it.

You can show your group is finitely generated, and thus it suffices to show it's abelian.  You can show the centralizers of all the generators are finite index, and thus the center of your group is finite index.  By Schur's Lemma, the commutator is finite, thus trivial.

Your problem can be generalized as follows:
If $ G$ is a torsion-free group, and $ H\le Z(G)$ is a finite-index subgroup, such that $ H$ is free abelian, then $ G$ is free abelian.

For a proof, see Lyndon and Schupp, "Combinatorial Group Theory", Proposition III.7.13.

Steve


inequalities
induction
number theory solved
number theory

Prove that sigma(n!) \leq 1/2*(n+1)! for all natural numbers n and determine the case when equality holds. (Here sigma(k) denotes the sum of all positive divisors of k.)

First, it is easy to check that equality occurs for n = 1,2,3,4,5 and strict inequality occurs for n = 6.
Now, suppose that n  \geq  6, and that strict inequality occurs.
Let n+1 = (p_1) x_1 *...(p_k) x_k  be the prime decomposition of n+1.
Let p be a prime, and v_p(a) be the exponent of p in the prime decomposition of a. Then :
- If p does not divides n+1, we have v_p( (n+1)! )  = v_p( n! )
- If p = p_i for some i in {1,...,k}, then v_p( (n+1)! ) = v_p( n! ) + x_i.
Let a_i = v_(p_i)(n!).

Lemma :
For each i, we have x_i  \leq  a_i + 1.

Proof :
If x_i > 1, in {1,2,...,n} there are p_i, (p_i) 2 ,..., (p_i) x_i - 1 so that :
a_i + 1  \geq  1 + 1 + 2 + ... + (x_i - 1) = 1 + x_i(x_i - 1)/2 , and this remains valid for x_i = 1.
It is easy to verify that in that case a_i + 1  \geq  x_i, as desired.

Now, we return to the original problem :

From the well-known expression of sigma, it follows that
sigma( (n+1)! ) = sigma(n!)*(product{1 \leq i  \leq k} ((p_i) x_i + a_i + 1  - 1)/( (p_i) a_i + 1  - 1) ) )
<(product{1 \leq i  \leq k} ((p_i) x_i + a_i + 1  - 1)/( (p_i) a_i + 1  - 1) ) )*((n+1)!)/2.
Thus it suffices to prove that
product{1 \leq i  \leq k} ((p_i) x_i + a_i + 1  - 1)/( (p_i) a_i + 1  - 1) )  \leq  n + 2 = 1 + product{1 \leq i  \leq k} (p_i) x_i .
that is (the products are over i = 1 to k) :
product( (p_i) x_i ((p_i) a_i + 1  - 1) + (p_i) x_i  - 1)  \leq  product( (p_i) a_i + 1 - 1)*(1 + product((p_i) x_i ).

Let X_i = (p_i) x_i  and Y_i = (p_i) a_i+1  - 1.
We have X_i, Y_i  \geq  1 and from the lemma Y_i \geq X_i - 1 for each i.
We want to prove that :
product(X_i*Y_i + X_i - 1)  \leq  product(Y_i)*(1 + product(X_i)).

By induction on k :
For k = 1, the desired inequality reduces to X_1Y_1 + X_1 - 1  \leq  Y_1(1+X_1) which holds since Y_1  \geq  X_1 - 1.
Now suppose that it holds for some given k-1. Thus :
product(X_iY_i + X_i - 1)  \leq  (X_nY_n + X_n - 1)*[Y_1*...Y_(k-1))]*[1+ X_1*...*X_(k-1))
= (Y_1*...*Y_k)(X_k + (X_k - 1)/Y_k)*(1 + X_1*...X_(k-1))
 \leq  (Y_1*...*Y_k)*X_k*(1 + X_1*...*X_(k-1))
 \leq Y_1*...Y_k*(1 + X_1*...*X_k), hence result.

It follows by induction that for each n > 0 : sigma(n!) \leq (n+1)!/2 and that equality occurs if and only if n is in {1,2,3,4,5}.

Pierre.

Note that :

-From above, we have sigma(n!)/n! \leq (n+1)/2.

- In another hand, we have sigma(n!)/n!  \geq  1 + 1/2 + 1/3 + ... 1/n (since sigma(m)/m is the sum of the reciprocals of the positive divisors of m), thus :
sigma(n!)/n! > c*ln(n) for some positive constant c (from which it follows that sigma(n!)/n! - > +oo as n - > +oo).

Does anybody know anything more precise about the range of sigma(n!)/n! ?

Pierre.

First of all, 1!+2!+3!=9<12=1/2(4!).  I believe that equality should hold only for n=1, 2.

Whether this solution allows for easier analysis is unclear.  But it is certainly much shorter.

Observe that sum(n*n!)=sum((n+1)!-n!)=(n+1)!-1, by telescoping.  Therefore, n*n!+(n-1)*(n-1)!+...+2*2!+2*1!=(n+1)!.  However, since the coefficient for each k! term on the left hand side >=2, 1/2(n+1)! = LHS >= sum(n!).

Equality occurs only when the coefficient on every LHS term is 2, i.e. for n=1, n=2.

I think you misunderstood the question, because in the question what is meant by sigma n! is not 1!+.... +n!.

What does it mean? Second fault during one day?

Ok, it's time to go to holidays....

Pierre.

Before going, and in order to save face, I will drop the atomic bomb :

The result follows easily from an inequality due to Duncan [1] :
For each n > 0, we have
sigma(n) < n*( 7f(n) + 10)/6 where f(n) is the number of prime divisors of n.
Since f(n!) =  \pi (n), it follows that
sigma(n!) < (n!/2)*( 7 \pi (n) + 10)/3.

Thus it suffices to prove that ( 7 \pi (n) + 10)/3  \leq  n+1, which can be made by usual not-too-sharp evaluations of  \pi (n).
It leaves a finite but not too large number of n to check directly to get the desired inequality (but I have no time to do it now).

Reference :
R.L. Duncan, 'Some estimates for sigma(n)', Amer. Math. Monthly, 1967, p.713-715.

Pierre.

I am sure that there is a mistake in the following lines, but I will dare to write, since if it is true, it's quite interesting.
  Struggling to solve this problem, I have found the following:
  If m>1 has n prime factors, then f(m)<2m \sqrt 2n-1, where f(x) is the sum of divisors of x. And if this is true, then it's stronger than Duncan's result.
  The proof:
   Suppose m=(p_1)^(k_1)*...*(p_n)^k_n, with p_1<...<p_n primes and k_1,...,k_n>=1. Then f(m)/m=(1+1/p_1+...+1/(p_1)^k_1)*...(1+1/p_n+...+1/(p_n)^k_n)< p_1/(p_1-1)*p_2/(p_2-1)....p_n/(p_n-1)=
=(1+1/(p_1-1))...(1+1/(p_n-1)). But p_1>=2, p_2>=3 and generally p_i>=2i-1 for any i>=2. Thus, (1+1/(p_1-1))*...*(1+1/(p_n-1))<=2(1+1/2)(1+1/4)*...*(1+1/(2n-2))=2*3/2*5/4*...*(2n-1)/(2n-2). By an easy induction we establish that 3/2*5/4*...*(2n-1)/(2n-2)< \sqrt 2n-1, which proves the claim.
   Hoping there is no mistake, I will continue. Since n! has at most n prime factors, it follows that f(n!)<2n! \sqrt 2n-1, which is clearly smaller than (n+1)!/2  except for a few number of cases.
   And now I apologise if there is any mistake and you lost your time with this.

Then I guess this can be called Harazi's inequality, right? Of course not.


trigonometry
function
inequalities
inequalities unsolved

[b] Prove that
\[ (*)\; \; \; \; \; \begin{array}{|c|} \hline \\ \sin\left( \cos(x)\right) < \cos\left( \sin(x) \right) \\ \\ \hline \end{array} \; ,\; \; \; \forall x \in {\mathbb R}  \]
( Source=  [color=green]  Olimpíada Internacional De las Matematicas, Belgica 1976   [/color]  )
[u] Find another pair (f,g) of functions ${\mathbb R} \to {\mathbb R}$ satisfying (*) [/b] [/u]

[quote="flip2004"][b]
[u] Find another pair (f,g) of functions ${\mathbb R} \to {\mathbb R}$ satisfying (*) [/b] [/u][/quote]
There are no $f$ or $g$ in the formula you gave so what do you mean?
$f,g\in \mathbb{R}^\mathbb{R}$ satisfying $f(g(x))<g(f(x))$ for all real numbers? Well, take $f(x)\equiv 0,g(x)\equiv 1$.

[quote="puuhikki"][quote="flip2004"][b]
[u] Find another pair (f,g) of functions ${\mathbb R} \to {\mathbb R}$ satisfying (*) [/b] [/u][/quote]
There are no $f$ or $g$ in the formula you gave so what do you mean? ... Well, take $f(x)\equiv 0,g(x)\equiv 1$.
[/quote]
[b]How we can describe the set of all pairs $(f,g)$ of functions $f,g: {\mathbb R}\to {\mathbb R}$ having this property ? [/b]

I don't think there is a simple description - for example, if $f(x)$ is bounded above and $g(x)$ is bounded below, then $f(x), h(x) = g(x) + c$ would satisfy the inequality trivially for sufficiently large $c$, although this is clearly not the only such family of solutions.


Let a,b,c distinct integers. Can the polynomial (x-a)(x-b)(x-c)-1 be factored into the product of two polynomials with integer coefficients? Justify.

[hide]The polynomial $ (x \minus{} a)(x \minus{} b)(x \minus{} c) \minus{} 1$ is a cubic, so if it can be expressed as a product of two polynomials, then one term must be linear, having the form $ (x \minus{} d)$, since the leading coefficient is $ 1$. But since $ d$ is a root, then $ (d \minus{} a)(d \minus{} b)(d \minus{} c) \minus{} 1 \equal{} 0$ and $ (d \minus{} a)(d \minus{} b)(d \minus{} c) \equal{} 1$, which is impossible since $ d \minus{} a$, $ d \minus{} b$, and $ d \minus{} c$ are distinct integers.

Thus, it's not possible.[/hide]


Is the sequence a mathematical or a trick sequence?

[hide="pattern recognition"]58, involves sums of digits and a particular power :D[/hide]

[hide]  I also recognized that pattern and found that the first digit was 5 ( we need to get the "number - the sum of digits= 45) but how did u find out 8 was the last digit???.[/hide]

palytoxin has the correct answer, but what do you mean 'a particular power?' lol

It seems like it involves a power of 2...

This is what I see:
The sum of the digits in the nth term is equal to the sum of the digits in the nth power of 2.

The next term is the 8th term, so the sum of the digits in the term should be equal as teh sum of the digits in $2^8$ or 13.
There are many possibilities with this condition. 49, 58, 67, 76, etc.

I'm trying to find another condition. If the number 47 is changed to 38, then the answer should be 49. (Each one is the next such number greater than the previous number)

[quote="Barnacle"]find the next number in the sequence:

11 13 17 25 32 37 47 ...[/quote]
two patterns i found:

first pattern:
$11, 13, 17, 25, 32, 37, 47, 58, 71, 79, 95, 109, 119, 130, 134, 142,\ldots={a_n}$
where each $a_n$ is obtained by adding $a_{n-1}$ and the sum of the digits of $a_{n-1}$

second pattern:
$11, 13, 17, 25, 32, 37, 47, 58, 71, 79, 86, 100, 119, 139, 161, 187,\dots={a_n}$
where each $a_n$ is obtained by adding $a_{n-1}$ and the sum of the digits of $2^{n-1}$

wow lol never thought of the 2nd one.  it's hilarious that there's 2 answers; after having given so many terms, you'd think only one algorithm for getting the terms persists.

you like that eh?  ;)   i like it too because that kills off your intuition  :play_ball:

How in the world did you find that 2nd one? :huuh:

yeah, definitely a 'w007' for creativity.  as soon as I find one, I usually stop assuming that the problem maker ensured only one solution.


For $ 0<a,b,c<1$, let $ u: \equal{}abc$ and $ v: \equal{}(1\minus{}a)(1\minus{}b)(1\minus{}c)$. Prove
\[ \frac1u\plus{}\frac1v\ge\frac {12}{1\minus{}u\minus{}v}\]

Are there nicer proofs than with mixing variables?

[quote="spanferkel"]For $ 0 < a,b,c < 1$, let $ u: \equal{} abc$ and $ v: \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$. Prove
\[ \frac1u \plus{} \frac1v\ge\frac {12}{1 \minus{} u \minus{} v}\]
Are there nicer proofs than with mixing variables?[/quote]
I think AM-GM Inequality works here. Indeed, we have
$ 1 \minus{} u \minus{} v \equal{} 1 \minus{} abc \minus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \equal{} a(1 \minus{} b) \plus{} b(1 \minus{} c) \plus{} c(1 \minus{} a) \ge 3\sqrt [3]{abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}.$
On the other hand, we also have
$ \frac {1}{u} \plus{} \frac {1}{v} \ge \frac {2}{\sqrt {uv}} \equal{} \frac {2}{\sqrt {abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}.$
Multiplying these inequalities, we get
$ (1 \minus{} u \minus{} v)\left(\frac {1}{u} \plus{} \frac {1}{v}\right) \ge \frac {6}{\sqrt [6]{abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}.$
Furthermore, one has
$ abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \equal{} [a(1 \minus{} a)][b(1 \minus{} b)][c(1 \minus{} c)] \le \frac {1}{4^3} \equal{} \frac {1}{2^6}.$
Combining this with the above inequality, the conclusion follows.

[quote="can_hang2007"]
I think AM-GM Inequality works here. Indeed, we have
$ 1 \minus{} u \minus{} v \equal{} 1 \minus{} abc \minus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \equal{} a(1 \minus{} b) \plus{} b(1 \minus{} c) \plus{} c(1 \minus{} a) \ge 3\sqrt [3]{abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}.$
[/quote]

Very nice!  :thumbup:


Find the sum:

$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+...+\frac{1}{99 \cdot 100}$

I've seen many variations on this problem, but I don't remember how to solve it :(

Could someone please provide an explanation :blush:

Also, how about the general case?

$\frac{1}{k(k+1)}= \frac{A}{k}+\frac{B}{k+1}$ for some $A$ and $B$.

If we add those two fractions together, we see that $A(k+1)+Bk$ must equal $1$.

If we let $k=0$ we notice that B cancels and $A=1$, if we let $k=-1$ we notice that A cancels and $B=-1$ which shows us that:

$\frac{1}{k(k+1)}= \frac{1}{k}-\frac{1}{k+1}$

How can you just "let" k equal something? Is it because the statement is supposed to be true for all k?

By the "general" case, I meant something like

$\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 4}+\frac{1}{3 \cdot 5}+blah...$

or

$\frac{1}{1 \cdot 4}+\frac{1}{2 \cdot 5}+blah...$

Yes, exactly, because it is true for all $k$, if we choose $k$ strategically so as to eliminate $A$ or $B$ we can find what they are equal to. For example, if we had let $k=2$ then we would have noticed that $3A+2B=1$, but that wouldn't have been too useful.

As for your second question, I don't fully understand what you mean. To solve those you would just find the partial fraction decomposition of $\frac{1}{k(k+2)}$ and $\frac{1}{k(k+3)}$ respectively.

[quote="max_tm"]As for your second question, I don't fully understand what you mean. To solve those you would just find the partial fraction decomposition of $\frac{1}{k(k+2)}$ and $\frac{1}{k(k+3)}$ respectively.[/quote]

Let's see if I understand: in that case I would do

$\frac{1}{k(k+2)}=\frac{A}{k}+\frac{B}{k+2}$
$1=A(k+2)+Bk$

So I would let k=0 and find $A=\frac{1}{2}$, and then let k=-2 and find $B=1$... then plug those in and telescope the series?

Heh, doesn't seem to be working out ahaha... hmm... I'll leave this off to someone who can correct my mistake, maybe I'm making a wrong assumption somewhere.

EDIT: Yeah so the mistake was in arithmetic as skimnc pointed out below. But kunny destroyed the problem with a generalization so... I guess this is a moot point  :D.

mathnerd314.. if $k=-2 \implies-2B=1 \iff B = \boxed{\frac{-1}{2}}$

[quote="mathnerd314"]How would you derive $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$? I know how to prove it, but how would I derive the formula using partial fraction decomposition?[/quote]

What max_tm did works great and is basically a shortcut.  When you break a fraction into partial fractions your are "matching" coefficients of like terms by means of a system of equations.  I'll work this out the long way to make it more clear:

$\frac{1}{k(k+1)}= \frac{A}{k}+\frac{B}{k+1}$
 Now observe that in the original fraction the numerator has no $k$ term, in other words the coefficient of $k$ is $0$.  So we have:
$A(k+1)+Bk$ which gives the system:
$A(1) = 1 \qquad$                   (This equation represents the constant term of the original numerator.)
$Ak+Bk = (0)k \quad$          (This equation represents the coefficient of $k$ in the original numerator.)
It follows that $A=1$ and $B=-1$ so that  $\frac{1}{k(k+1)}= \frac{1}{k}-\frac{1}{k+1}$

Hello, everyone, have you seen my post? :)

Another thing is that
$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$
This can be proved with induction.


Prove that the sequence $< \frac{(n+1)^n n^{2-n}}{7n^2+1}>_{n=0,1,\ldots}$ is strictly increasing.

It is the product of two strictly increasing sequences : $\left(\frac{n+1}{n} \right)^n$ and $\frac{n^2}{7 n^2 +1 }$ .


logarithms
limit
ratio
calculus
calculus computations

Hello,
since it has been taking me forever to solve this
Un=((n/2) power (n/2))/n!
I decided to ask you for some help, please.

$u_{n}=\frac{\left(\frac{n}{2}\right)^{\frac{n}{2}}}{n!}$

Stirling's approximation (even a weak version of Stirling, such as $\ln(n!)\approx n\ln n-n$) settles this pretty quickly. The limit is zero.

Here's another idea: compute $\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}$ as in the ratio test. Yes, I know that this is a sequence, not a series, but go ahead and think about the series $\sum u_{n}.$

That ratio simplifies to

$\lim_{n\to\infty}\frac{\sqrt{n+1}}{2(n+1)}\cdot \left(\frac{n+1}n\right)^{\frac{n}{2}}$

Within that, $\left(\frac{n+1}n\right)^{\frac{n}{2}}$ tends to $\sqrt{e}$ but the whole ratio tends to zero.

We conclude that not only does $u_{n}$ tend to zero, it tends to zero so rapidly that the power series $\sum u_{n}x^{n}$ has an infinite radius of convergence.

[quote="Kent Merryfield"]$u_{n}=\frac{\left(\frac{n}{2}\right)^{\frac{n}{2}}}{n!}$[/quote]
Shame on me :blush:
I am not good at using WORD yet, but YES, that is how it looks.


What did you get for XMAS?
As for me, we don't celebrate Xmas.

My gift will come in a few more days in the mail...  :(

It is magic cards (MTG, not a deck of playing cards). The online store ran out of stock, so I am still waiting.

I got;

Intro to Geometry
A few other books.
A few games.
A video recorder.
Some clothes.

I got a go set and a really good book which teaches you how to  kill problems using complex numbers.

what book?
do you play go?
Play me at gokgs.com
pm me if you're online.

I got Mario party 8 for my wii :( . And I also got some other stuff including a ncaa football encyclopedia.

Hm... I got one of those Flip Video thingys...

I got 2 shirts, sandals, and cash (like 35 dollars in colombian money) :sleeping: .

A nice vacation and free rating from Ttocs


The sum of the coefficients of the cocient in this division:
(2x^(n-1)+3x^(n-2)+4x^(n-3)+,...+nx+n+1)/(2x-1)  is 28. Find the remainder

Thanks
leon-sotelo@hotmail.com

If we let $ P(x) \equal{} 2x^{n \minus{} 1} \plus{} 3x^{n \minus{} 2} \plus{} \cdots \plus{} nx\plus{}n\plus{}1$ (I think that's what you mean?), then we have

$ P(x) \equal{} (2x \minus{} 1)Q(x) \plus{} r$ ($ \deg R(x) \equal{} 0$ since the divisor has degree $ 1$).

We are also given that $ Q(1) \equal{} 28$ (why?), so $ P(1) \equal{} Q(1) \plus{} r \equal{} 28 \plus{} r$, and because $ P(1) \equal{} 2 \plus{} 3 \plus{} 4 \plus{} \cdots \plus{} n\plus{}n\plus{}1 \equal{} \frac {n(n \plus{} 1)}2\plus{}n$,

$ r \equal{} \frac {n(n \plus{} 1)}2 \plus{}n \minus{} 28 \equal{} \boxed{\frac {n^2 \plus{} 3n \minus{} 56}2}$.

Edit: Forgive my stupidity, I was thinking something else. :blush:

The independent term of P (x) is n +1

León-Sotelo

one start would be to assign a random integer value to n (like 10) and find the quotient and remainder that results.
there is a neat pattern that results from dividing this out.

Wait does that mean my solution is wrong???

[hide="pretty sketchy solution"]
i'm not that good at showing work when dividing polynomials, but i'll just tell you that the quotient (not cocient) is
$ x^{n\minus{}2}\plus{}2x^{n\minus{}3}\plus{}3x^{n\minus{}4} \ldots \plus{} (n\minus{}2)x\plus{}(n\minus{}1)$ with a remainder of $ \boxed{2n}$
this is mainly found out by experimentation with the division and finding a pattern
[/hide]

Oh wait--both of us are correct.

It is not difficult to show that $ n\equal{}8$ is the only possibility, yielding $ r\equal{}2(8)\equal{}\frac{(8)^2\plus{}3(8)\minus{}56}2\equal{}\boxed{16}$.


Let $ x,y$ satisfying $ x\ge y \ge1$$ xy\ge 2$
Find maximum value of $ P\equal{}\frac{1}{x}\plus{}\frac{1}{y}$
It is from our school contest

[quote="Math pro"][quote="Allnames"][quote="Math pro"]It is quite easy!Here is my solution for my test:
$ x\geq y \geq 1$ so $ (x \minus{} 1)(y \minus{} 1)\geq 0$
$ < \minus{} > xy \minus{} x \minus{} y \plus{} 1\geq 0$
$ < \minus{} > 1 \plus{} \frac {1}{xy} \geq \frac {1}{x} \plus{} \frac {1}{y}$
we have $ xy \geq 2 \minus{} > \frac {1}{xy} \le \frac {1}{2}$
$ \minus{} > P_(max) \equal{} \frac {1}{2} < \minus{} > x \equal{} 2,y \equal{} 1$ :wink:[/quote]
Very nice,any other solution?[/quote]
post your solution,all names.I think your solution is nice,too! :oops:[/quote]
My solutiuon
Let$ xy\equal{}a$ then $ p\equal{}\frac{y}{a}\plus{}\frac{1}{y}$But $ f'(y)\le 0$ then $ f(y)\le f(1)$
With $ a\ge 2$ we have $ f(y)\le \frac{3}{2}$
your proof is nicer ,thanks

[quote="Math pro"]It is quite easy!Here is my solution for my test:
$ x\geq y \geq 1$ so $ (x \minus{} 1)(y \minus{} 1)\geq 0$
$ < \minus{} > xy \minus{} x \minus{} y \plus{} 1\geq 0$
$ < \minus{} > 1 \plus{} \frac {1}{xy} \geq \frac {1}{x} \plus{} \frac {1}{y}$
we have $ xy \geq 2 \minus{} > \frac {1}{xy} \le \frac {1}{2}$
$ \minus{} > P_{max} \equal{} \frac {1}{2} < \minus{} > x \equal{} 2,y \equal{} 1$ :wink:[/quote]
i think $ P_{max} \equal{}1\frac{1}{2}$ ; you have a mistake in

[quote]we have $ xy \geq 2 \minus{} > \frac {1}{xy} \le \frac {1}{2}$
$ \minus{} > P_{max} \equal{} \frac {1}{2} < \minus{} > x \equal{} 2,y \equal{} 1$ :wink:[/quote]

Thanks for nice sollution  :)


find all integer numbers x,y,z >0 such as: x^3+y^3=z^3 .

It's very nasty, if I remember correctly. It includes lots of casework in the ring $\mathbb Z[\frac{1+\sqrt {-3}}2]$. You can find the solution in Hardy and Wright's "Introduction to the Theory of Numbers", for example.


[i][u]Problem :  [/u]

   Let $ x , y , z$ be positive integers such that :

  $ x \ \plus{} \ y \ \plus{} \ z \ \equal{} \ 3$ and $ x^{2} \ \plus{} \ y^{2} \ \plus{} \ z^{2} \ \equal{} \ 7$

[u]   Prove that :[/u]

 $ 1 \ \plus{} \ \frac{6}{xyz} \ \geq \ \frac{1}{3} \left( \frac{x}{z} \ \plus{} \ \frac{y}{x} \ \plus{} \frac{z}{y} \right)$[/i]

$ 1 \ + \ \frac{6}{xyz} \ \geq \ \frac{1}{3} \left( \frac{x}{z} \ + \ \frac{y}{x} \ + \frac{z}{y} \right)$
$ \Leftrightarrow 3xyz+18 \geq x^2y+y^2z+z^2x$
we have
$ xy+yz+zx=1$
$ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=18$
it's easy to prove:
$ x^3+y^3+z^3\geq x^2y+y^2z+z^2x$
then:
$ 18=x^3+y^3+z^3-3xyz\geq x^2y+y^2z+z^2x-3xyz$
$ \Leftrightarrow 18+3xyz\geqq^2y+y^2z+z^2x$


[quote="bertram"]Prove that each natural number $ n$ has either $ 0$ or $ \varphi(\varphi(n))$ primitive roots.[/quote]
 If $ n$ has no primitive root the proof is complete. Otherwise, if $ a$ is a primitive root for $ n$ then so is $ a^k$ for $ 1\leq k\leq n$ where $ \gcd(k,n)\equal{}1$ in total there are $ \phi (\phi (n))$ such possibilities.

[quote="t0rajir0u"]In fact, we can explicitly describe $ n$ with primitive roots; that could also be written as a series of exercises, for example

0) $ n \equal{} 1, 2, 4$,
1) $ n$ is an odd prime,
2) $ n$ is a power of an odd prime,
3) $ n$ is twice a power of an odd prime,
4) No other $ n$.[/quote]
Its a theorem by Gauss  :omighty: . I think.

[quote="t0rajir0u"]In fact, we can explicitly describe $ n$ with primitive roots; that could also be written as a series of exercises, for example

0) $ n \equal{} 1, 2, 4$,
1) $ n$ is an odd prime,
2) $ n$ is a power of an odd prime,
3) $ n$ is twice a power of an odd prime,
4) No other $ n$.[/quote]
This is necessary and sufficient, correct?

[quote="diophantient"]
This is necessary and sufficient, correct?[/quote]
:yup: that's what [b]t0rajir0u[/b] said.


geometry
parallelogram
geometric transformation
rotation
homothety
vector

On the sides of the triangle $ABC$, three similar isosceles triangles $ABP$ $(AP=PB)$, $AQC$ $(AQ=QC)$, and $BRC\ (BR=RC)$ are constructed. The first two are constructed externally to the triangle $ABC$, but the third is placed in the same half-plane determined by the line $BC$ as the triangle $ABC$. Prove that $APRQ$ is a parallelogram.

[hide="My solution"]It's easy to get $\widehat{ACB}=\widehat{RCQ}$
so ABC and QRC are similar triangles. DO the same we get ABC, PBR are similar too.
RB=RC $\rightarrow \Delta PBR=\Delta QRC \rightarrow$ parallelogram[/hide]

[quote="Jan"]I think you also need to prove that $RP \parallel QA$, but that's trivial anyway. Short and easy solution hikaru123!

My solution used rotation and homothecy:[list]
[*]First we perform a rotation with center $A$ and angle $\measuredangle PAB$, combined with a homothecy with factor $\frac{AB}{AP}$.

[*]Then we perform a rotation with center $C$ and angle $\measuredangle BAP$, combined with a homothecy with factor $\frac{AP}{AB}$.[/list]
Because $\measuredangle BAP =-\measuredangle PAB$, we obtain that all points are translated over the same vector when we apply these transformations to them.

We see that $P \rightarrow R$ and $A \rightarrow Q$ so it is indeed a parallellogram.[/quote]
No. Because we have $\Delta PBR=\Delta QRC$
So $PR=QC=QA$ and $QR=PB=PA$ That's enough to prove a parallelogram.

Indeed, now we can see you used the fact that the two pairs of opposite sides have equal lengths, you didn't mention it at first. Anyway, the sketch of your proof was clear.


Let A, B diferent positives integers. Find the minimun value of a-b (a>b) iff
(A+B)/2,   (AB)^1/2   and 2AB/(A+B) are integers too.

Wait your solutions...

some computer educated guessing gives me the minimum value of a-b=40 occuring for a=45, b=5.

Note that: $H=\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}$ is the Harmonic mean of a and b;

$G=\sqrt{ab}$ is the Geometric mean of a and b;

and $A=\frac{a+b}{2}$ is the Arithmetic mean of a and b.

Also, we have for all a,b>0, H<G<A and the equality $G^{2}=AH$ holds as well.

For a=45, and b=5 we have H=2, G=15, and A=25.

Thanks for this maddening problem,

                   -Ben


exist  or not exist function continuous and satisty withall x in real number such that :
$ f(f(x))\equal{}e^{\minus{}x}$

There is no real-valued $ C^1$ function satisfying this.

Note that $ f'(f(x))f'(x)\equal{}\minus{}e^{\minus{}x}.$ From this, we see that $ f'(x)$ can never be zero.  I was assuming that $ f\in C^1(\mathbb{R}).$ That would mean that either $ f'(x)>0$ for all $ x$ or $ f'(x)<0$ for all $ x.$ But either way, we would have $ f'(f(x))f'(x)>0,$ contracting that it equals $ \minus{}e^{\minus{}x}.$

I don't yet see how to get this if we weaken the hypothesis to what is stated, that $ f\in C(\mathbb{R}).$

On the other hand, if we allow complex-valued functions on $ \mathbb{R},$ then we would have the example $ f(x)\equal{}e^{ix}.$

Another  observation: No monotone function exists either since $ e^{ \minus{} x}$ is decreasing.

That observation does it, since any solution would have to be injective.

This can be  generalized I think : There exists no continuous function $ f: I\rightarrow I$ so that $ f^{2n}$ is decreasing.
This would work for a Darboux function instead of a continuous one as well.


Let $ABCD$ be a quadrilateral incribed in $(O,R)$ . $K$ is a point inside the quadrilateral $ABCD$ . Let $M, N , P , Q$ respectively be the projections of $K$ on $AB , BC, CD ,DA$ respectively .
  a) Prove that : $S_{MNPQ} \leq \frac{1}{2}S_{ABCD}$
  b) Put $OK =d$ . Prove that : $p_{MNPQ} \geq \sqrt{8S_{ABCD}(1-\frac{d^2}{R^2})}$ , where $p_{MNPQ}$ is the perimeter of quadri. $MNPQ$

The first one is easy . I used the formula $S =\frac{1}{2}AB.AC.\sin(AB,AC)$ to solve it .
  The second is harder...

Jarod , can you write down your solution for a) ? Ihaven't solved it yet .
 The expression in b) reminds me of Euler's formula .
  In brief , I haven't solved this problem .  :(  :(

Here is something I manage with b)
 Let $I$ be the feet of the perpendicular from $K$ to $AC$ .
 Recall Euler's formula : $S'=\frac{1}{4}.S.|1-\frac{d^2}{R^2}|$ , where $S'$ is the area of the pedal triangle .
 So $S_{ABCD}.(1-\frac{d^2}{R^2}) = 4(S_{PQI}+S_{MIN})$

  Please help me to finish this one !


If the largest possible circle is drawn in a 27 x 27 square array  of dots, how many of the dots would the circle pass through?
(The circle can pass through dots on the edge of the array.)

[hide]It's acutally a 26x26 square if there are 27x27 dots.

So we have: x^2 + y^2 = 13^2 if the middle dot is the center (14th dot in both horizontally and vertically)

We think of 5-12-13 triangles

x = -5, 5, -12, 12 and y = -12, 12, 5, -5

So we have:

2C1 * 2 * 2 = [b]8[/b]

Then we have to add in the edge ones. There are 4. [b]12[/b] is the answer. [/hide]


$ \triangle ABC$ is a right isosceles triangle. Choose points $ K$ and $ M$ from the hypotenuse $ AB$, such that $ K \in AM$ and $ \angle KCM \equal{} 45$º. Prove that $ (AK)^2 \plus{} (MB)^2 \equal{} (KM)^2$

Thanks for any help.

http://www.mathlinks.ro/Forum/viewtopic.php?t=27981 (scroll down!) for a solution.
http://www.mathlinks.ro/Forum/viewtopic.php?t=371 for the converse.

  darij

Thanks! Now that I know that I got almost to the end of it, I'm wondering why I didn't see the last stretch...

The idea is how can we draw the subject points?!

Take a quarter of a circle centered C, radius CA, draw a radius CN inside the angle ACB; the angle bisectors of angles ACN and NCB cut AB at K and M. See tt KN=AK and MN=MB and angle KNC= angle KAC = angle MBC=angle MNC=45 and we are done (proof without words?)

Best regards,
sunken rock

A blind man sees better the details


inequalities
logarithms
inequalities proposed

$ a,b,c,$ are positive reals such that $ abc\equal{}1$ Prove that
$ {a^2\plus{}b^2\plus{}c^2\plus{}9(ab\plus{}bc\plus{}ac)}\ge{10(a\plus{}b\plus{}c)}$

$ f(a) \equal{} a^2 \plus{} \frac {9}{a} \minus{} 10a \plus{} 17\ln a\ (a > 0),$

$ f'(a) \equal{} \frac {(a \minus{} 1)(2a^2 \minus{} 8a \plus{} 9)}{a^2}\Longrightarrow f(a)\geq f(1) \equal{} 0.$

$ \therefore \sum_{cyc} \left(a^2 \plus{} \frac {9}{a} \minus{} 10a \plus{} 17\ln a\right)\geq 0.$

[quote="Litlle 1000t"]$ a,b,c,$ are positive reals such that $ abc \equal{} 1$ Prove that
$ {a^2 \plus{} b^2 \plus{} c^2 \plus{} 9(ab \plus{} bc \plus{} ac)}\ge{10(a \plus{} b \plus{} c)}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197


Let $ n\geq 3$ and $ a_1,a_2,...,a_n$ be positive real numbers. Define $ S \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n, P \equal{} a_1a_2...a_n$. Prove that

$ \frac {1}{S \minus{} a_1} \plus{} \frac {1}{S \minus{} a_2} \plus{} ... \plus{} \frac {1}{S \minus{} a_n}\leq \sqrt[n \minus{} 1]{\frac {S}{P}}$

$ S \minus{} a_1 \ge (n \minus{} 1) \sqrt [n \minus{} 1] {\frac {P} {a_1}} \Rightarrow \frac {1} {S \minus{} a_1} \le \frac {1} {n \minus{} 1} \cdot \sqrt [n \minus{} 1] {\frac {a_1} {P}}$

Now, it is enough to prove that: $ \sqrt [n \minus{} 1] {a_1} \plus{} \sqrt [n \minus{} 1] {a_2} \plus{} ... \plus{} \sqrt [n \minus{} 1] {a_n} \le (n \minus{} 1) \cdot \sqrt [n \minus{} 1] {S}$

$ \Leftrightarrow \sqrt [n \minus{} 1] {\frac {a_1} {S}} \plus{} \sqrt [n \minus{} 1] {\frac {a_2} {S}} \plus{} ... \plus{} \sqrt [n \minus{} 1] {\frac {a_n} {S}} \le n \minus{} 1$

$ \Leftrightarrow \sum \sqrt [n \minus{} 1]{\frac {a_1} {S} \cdot \frac {1} {n^{n \minus{} 2}}} \le (n \minus{} 1) \cdot \sqrt [n \minus{} 1] {\frac {1} {n^{n \minus{} 2}}}$

But $ \sum \sqrt [n \minus{} 1]{\frac {a_1} {S} \cdot \frac {1} {n^{n \minus{} 2}}} \le \sum \frac {1} {n \minus{} 1} \cdot (\frac {a_1} {S} \plus{} \frac {n \minus{} 2} {n}) \equal{} \frac {1} {n \minus{} 1} \plus{} \frac {n \minus{} 2} {n \minus{} 1} \equal{} 1$

It is enough to prove that $ 1 \le (n \minus{} 1) \cdot \sqrt [n \minus{} 1] {\frac {1} {n^{n \minus{} 2}}} \Leftrightarrow n^{n \minus{} 2} \le (n \minus{} 1)^{n \minus{} 1} \Leftrightarrow (\frac {n} {n \minus{} 1})^{n \minus{} 1} \le n$

The last inequality results easily from the relation:

$ (1 \plus{} \frac {1} {n \minus{} 1})^{n \minus{} 1} \equal{} \dbinom{n \minus{} 1}{0} \plus{} \dbinom {n \minus{} 1} {1} \cdot \frac {1} {n \minus{} 1} \plus{} \dbinom{n \minus{} 1}{2}\cdot\frac {1}{(n \minus{} 1)^2} \plus{} ... \plus{} \dbinom{n \minus{} 1}{n \minus{} 1}\cdot$

$ \cdot\frac {1}{(n \minus{} 1)^{n \minus{} 1}}\le1 \plus{} 1 \plus{} 1\plus{}... \plus{} 1 \equal{} n$


can anyone explain to me why if p is prime and p in not equal to 3 when divided by 3 the remainder is either 0, 1, or 2.

thanks!!!

Because if the remainder was three, then it would be divisible by 3 again.

(I'm not sure what your question is - that was the answer to what I think  your question is)

I think your problem is to express it clearly.

As I understand the question, you're saying that $ p$ is prime, and $ p\neq{3}$, and you're asking why $ p\mod{3} \equal{} 0, 1, \text{or} 2$?
First of all, if $ p$ is prime, and $ p > 3$ then $ 3$ does not divide $ p$ and thus the remainder will never be $ 0$.
Now, to answer your question about why $ p\mod{3} \equal{} 1 \text{or} 2$: $ p$ is greater than $ 3$. That means that, by the Division Algorithm, $ p/3$ can be expressed as $ 3q \plus{} r$ for some integers $ q$ and $ r$, which represent quotient and remainder respectively. $ r$ will NEVER be greater than or equal to the coefficient of $ p$, because if it were, you'd be able to do this:
$ p/3 \equal{} 3q \plus{} 4$ (suppose that $ r \equal{} 4$ for this case)
$ p/3 \equal{} 3q \plus{} 3 \plus{} 1$
$ p/3 \equal{} 3(q \plus{} 1) \plus{} 1$
So, the largest $ r$ can be if $ r$ is an integer is $ 2$ (in this case). This means that you'll have either $ 1$ or $ 2$ as $ p\mod{3}$

(Apologies for the bad LaTeX).

Dear Dubito...
if the remainder be 3 then it would b divisible by 3 again.
and if the remainder be 4 then if 4 divided by 3 then the remainder is 1.
and continue like this.

Thanks grn_trtle that answers my question.

@happyme

"if the remainder was three, then it would be divisible by 3 again" I know that was one of my problems but the book said that if p is prime and p is not equal to 3 then when divided by 3 the remainder is either 0,1, or 2.


[quote="fredfox"]Fine all (x,y,z) satisfy:  2^x+3^y=5^z
                                 x,y,z  are   integers[/quote]

We see that $x, y, z$ must be non-negative.

If $x=0$ then we have $5^z-1=3^y$ but this equation doesn't have any solution since $5^z-1$ is divisible by $4$ and $3^y$ is not.

If $x=1$ then we have $2+3^y=5^z$. After checking $\mod 4$ and $\mod 3$ we see that $y=2m+1$ and $z=2n+1$. So we have: $2+3^{2m+1}=5^{2n+1}$
$3^{2m+1}-3=5^{2n+1}-5$
$3(3^m+1)(3^m-1)=5(5^n+1)(5^n-1)$
Equality only holds when both sides are $0$ so $m=n=0$. Thus we have solution $(x,y,z)=(1,1,1)$

If $x\geq2$ when after checking $\mod 4$ we see that $y=2m$. After cheking $\mod 5$, we see that $x=2k$ and after $\mod 3$ that $z=2n$
Then we have:
$2^{2k}=(5^n-3^m)(5^n+3^m)$
We see that one of the multipliers is divisible by $4$ and other by $2$. Say that $5^n+3^m$ is divisible by $4$ and $5^n-3^m$ by $2$. So if you devide $5^n-3^m$ by $2$ you get odd number which must be power of two. So the only posibility is $1$. Thus $5^n-3^m=2$. We have solved this similarly before so the only sulution is $m=n=1$ and then $k=2$. We have one more solution $(x,y,z)=(4,2,2)$.

So the solutions are $(x,y,z)=(1,1,1), (4,2,2)$

----------

I am not very sure about this [quote="eLeM"]$3(3^m+1)(3^m-1)=5(5^n+1)(5^n-1)$ Equality only holds when both sides are $0$ so $m=n=0$. Thus we have solution $(x,y,z)=(1,1,1)$[/quote], but I doubt it is correct. I was trying to prove this but failed so maybe someone has a nice proof? Or maybe nice proof to whole problem because I don't realy like my solution :)


geometry
3D geometry
inequalities
floor function
calculus
integration
number theory unsolved

Prove:
$\{\sqrt[3]{n}\}>\frac{1}{3\sqrt[3]{n^2}}$

where $n$ is a positive integer not equal to a cube of any integer  ;)

If you want the proof for inequality here is it :

Since n is a positive integer it is obvioust that $n \geq 1$
Thus we have $3n \geq 3>1$.

I don't see how that solves the problem...
Note that $\{x\}$ (the fractional part of $x$) is $x-\lfloor x \rfloor$ with $\lfloor x \rfloor$ (the integral part of $x$) being the greatest integer $\leq x$.


abstract algebra
calculus
integration
linear algebra
superior algebra
superior algebra unsolved

According to my linear algebra book, a free module V over a commutative ring R with identity with n generators is the same as a free module V over R with rank n, where 'rank' is "the smallest integer k such that k elements span V". I can see that if V is a free R-module with n generators then V has rank n but I can't see how the converse is true. According to my book a finitely generated module may not have a finite basis (any examples?) so this statement seems to be in contradiction. Unless I misunderstand the definition of rank?

I know very little algebra and what I know is from my linear algebra book, so try to explain in laymans terms.

the remark applies to free modules over integral domains. you can see this by tensoring with the field of fractions.


geometry
geometric transformation
combinatorics proposed
combinatorics

You have an amount of sugar whose weight is unknown, you have a scales and you have a given weight whose weight is 1 g What is the fastest way (or: what is a fastest way) to obtain a 1 kg portion of sugar ?

[b]Note by Darij:[/b] Since my translation is very poor, let me make some little explanations: The scales has two pans, and it allows you see whether the two masses on the pans have equal weight or not, in the latter case showing which of the masses is heavier. Moreover, you can continuously fill up the pan with the lower weight with sugar until the scales is in balance; this latter action (continuously filling up until balance) will be called \textit{balancing}. The fastest way is defined as the way with the minimal number of balancings.

Can you put 2 heaps at a time in the pan?

Like can you weight 9g from 4g and 5g without mixing up the 4g and 5g?

makes a difference here :)


Here's a question that's been on my mind for a while. It's a bit... contrived (at least, I think it is) and I'm hitting a wall. While I will continue working on the problem, here's a chance for others to have a go at it:

Let $p_{1}, p_{2},p_{3},...$ be the primes listed in increasing order, and let $x_{0}$ be a real number between $0$ and $1$. For a positive integer $k$, define:

$x_{k}=0$ if $x_{k-1}=0$ otherwise $x_{k}=${$p_{k}/x_{k-1}$} if $x_{k-1}\ne0$ where {$x$} denotes the fractional part of $x$. Find with proof, all $x_{0}$ such that the sequence $x_{0},x_{1},...$ eventually becomes zero.

Isn't it this one?
http://www.kalva.demon.co.uk/usa/usoln/usol971.html

Pierre.

Pierre

You're right! Thanks for the info.


a,b,c is nonnegative and a+b+c=1, prove that
        a/(a+b)^1/2+b/(b+c)^1/2+c/(c+a)^1/2<=5/4

 :blush: sorry I can't use Latex.
(3/4,1/4,0) and sys.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=138851  :wink: