pigeonhole principle
geometry
rectangle
AMC
USA(J)MO
USAMO
inequalities

Given a 13 x 13 grid of points, 53 of which are coloured red. Prove that there exists a rectangle such that all of its vertices are red grid points and its sides are parallel to the sides of the grid.

(Is this clear enough ? I can imagine my English is not entirely corrrect :).)

I'll post some more pigeonholes later ...
To be continued ...

I don't know how to do it, but the question is clear, your english is perfect.

This is really not so trivial :mrgreen:
I'm joking (the others are not always trivial), but this is certainly not easy ...

This is an olympiad style problem.

Maybe this could be problem 1 or problem 4 on an USAMO ...
It's that level of difficulty I think. It's really a very nice one.

Yes...I'm working on a proof...quite interesting realli

the way the problem is designed creates a lot of possibilities of how stuff could come out...

No solution not yet ? A (small) hint maybe :

[hide]I didn't attach my signature for a while, but now I suddenly started to use it again ...  [/hide]

I'll clarify a bit :

[hide]It comes down to prove a certain inequality from which the conclusion can be derived by the pigeonhole principle. To prove this inequality you can use Cauchy's inequality for example. But the main problem is : how do you get the inequality ? [/hide]

I think I've got it  :mrgreen:

If ai denotes the number of red points on line i.

:Sigma: ai = 53.

Now there are :Sigma: C(ai,2) ways of choosing 2 points on the lines. There are C(13,2) ways of choosing 2 colunms (C = binomial coefficient)

It took me [size=150][b]a lot of time[/b][/size] to see it but if :Sigma: C(ai,2) > C(13,2) then by PHP there will be too few combinaisons of columns and we will be able to find the rectangle.

:Sigma: C(ai,2) = (:Sigma: ai - 53)/2, C(13,2) = 78.
What we have to prove is :Sigma: ai > 209

Here is where Cauchy appears (your first hint was a bit unfair  :wink: ) !

13 * Sum ai > (Sum ai) = 53
Now 53/13 > 209 so it's proved  :D

[quote="belenos"]It took me [size=150][b]a lot of time[/b][/size] to see it but if :Sigma: C(ai,2) > C(13,2) then by PHP there will be too few combinaisons of columns and we will be able to find the rectangle.[/quote]

That's the main idea indeed. Once you think of it, it appears to be obvious but this problem is offputting if you don't realize this.

This is really a fantastic problem, isn't it ?

[quote="Arne"]This is really a fantastic problem, isn't it ?[/quote]

Indeed  :!:

PHP problems are always nice (and very often very tricky).

I saw an example somewhere of a situation with only 52 points, in which you couldn't find any rectangle.

Beautiful problem...it took me about an hour and a half to see it...but when I finally did it was nice. Where did you get it?

I believe Arne writes some problems...plus he's an IMO'er...so he's got quite a bit of resources...

It was given to me at IMO training last year.
I believe our pigeonhole instructor took it from a math magazine.

So ... I didn't write this one myself :mrgreen:


(J.-L.Ayme, France) Points $ P$, $ Q$ lie on the circumcircle $ \omega$ of triangle $ ABC$. The perpendicular bisector $ l$ to $ PQ$ intersects $ BC$, $ CA$, $ AB$ in points $ A'$, $ B'$, $ C'$. Let $ A"$, $ B"$, $ C"$ be the second common points of $ l$ with the circles $ A'PQ$, $ B'PQ$, $ C'PQ$. Prove that $ AA"$, $ BB"$, $ CC"$ concur.

Why is it necessary for $ P$ and $ Q$ to lie on $ \omega$ ?  :huh:

  dg

Dear Darij and Mathlinkers,
I found this situation by solving another problem. Perhaps you could see any developpment with your remark.
Sincerely
Jean-Louis

I've found a solution for this problem by using Inversion. But i think there is another solution.

[quote="jayme"]Dear Mathlinkers,
this problem is in relation with the Seimiya's line.
Sincerely
Jean-Louis[/quote]

Can you give the solution Jayme? The problem is very interesting by itself and the point where the lines concurr makes it even more.

Dear MLs

In the proposed form this problem is a very particular case of Blaikie theorem

.........................................
Why is it necessary for P and Q to lie on \omega ? :huh:

dg
......................

Again,  Darij the Great  is precise   in his , so appealing to me , vagueness.

M.T.

[quote="jayme"]Dear Mathlinkers,
for a synthetic proof of my result you can see
http://perso.orange.fr/jl.ayme  vol. 3 La droite de Seimiya p. 5.
Sincerely
Jean-Louis[/quote]

I will wait until someone else posts a solution since I don;t understand french.

Dear Armpist (M.) and Mathlinkers,
I am always ready for new and nove approach of a problem.
Dear Armpist, how do you imagine the proof by using the Blaikie theorem and perhaps can you recall this result?
Thanks in advance.
Sincerely
Jean-Louis

Dear J-L

This is the link to the problem proposed by DG (Darij the Great)
some time ago:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=38178

M.T.

Dear Armpist (M.) and Mathlinkers,
thank you for the links I have lost.
But how do you use this resultat for the problem in question?
Sincerely
Jean-Louis

We may crack this problem by Desargues's involution theorem, although it was quite possible that the projective solution would  not be recognized in the contest.  :oops:


abstract algebra
group theory
function
superior algebra
superior algebra unsolved

Let [b]Tors [/b]be the category whose objects are torsion abelian groups; if $ A$ and $ B$ are torsion abelian groups, we define $ Mor_{\text{Tors}}(A, B)$ to be the set of all (group) homomorphisms $ \phi : A \rightarrow B$. Prove that direct products exist in [b]Tors[/b]; that is, show that given any indexed family $ \{A_i\}_{i \in I}$ where $ A_i$ is a torsion abelian group, there exists a torsion abelian group which serves as a direct product for this family in [b]Tors[/b].

Let $ \{A_i\}_{\{i \in I\}}$ be a family of torsion abelian groups. Let $ A$ be their ordinary product in the category of (abelian) groups; it comes equipped with projection homomorphisms $ p_i : A \to A_i$ for each $ i \in I$. Now $ A$ may fail to be a torsion group (example: consider $ \prod_{n \in \mathbb{N}} \mathbb{Z} / n\mathbb{Z}$), so let $ T \leq A$ be its torsion subgroup (that is, the subgroup of all elements of finite order). Let $ \iota : T \to A$ be the natural inclusion homomorphism, and define $ q_i : T \to A_i$ simply as the composition $ T \stackrel{\iota}{\to} A \stackrel{p_i}{\to} A_i$. That's it: $ T$, along with the homomorphisms $ q_i$, is the desired product in the category [b]Tors[/b].

[b]Proof[/b]: Let $ B$ be any torsion abelian group and let $ f_i : B \to A_i$ be homomorphisms. Since $ A$ is a direct product, there is a unique homomorphism $ f : B \to A$ which satisfies

(*) $ p_i \circ f \equal{} f_i$.

Since $ B$ is torsion, the image of $ f$ must in fact be contained in $ T$, so we obtain a unique homomorphism $ B \to T$ which satisfies the requirement of $ T$ as a universal object. More formally, we may write $ f \equal{} \iota \circ \tilde{f}$ where $ \tilde{f} : B \to T$ (note that $ \tilde{f}$ is essentially the same function as $ f$, only considered as having range $ T$ instead of $ B$). From (*) we obtain

$ p_i \circ \iota \circ \tilde{f} \equal{} f_i$, so

$ q_i \circ \tilde{f} \equal{} f_i$

which is just what we need for $ T$, along with the $ q_i$, to satisfy the definition of a product. Note that $ \tilde{f}$ is unique since $ f$ is, by the definition of a product applied to $ A$. QED

It's interesting to note that this construction doesn't work if we drop "abelian" from the statement of the problem, due to the fact that $ T$ cannot be defined as the torsion subgroup of $ A$ (this doesn't make any sense for an arbitrary group). I suspect that products do not, in fact, exist in the category of torsion groups in general (although the failure of the above method, of course, is not a proof of that claim).


monovariant
number theory
greatest common divisor
invariant
least common multiple

Several positive integers are written on a blackboard. One can erase any two distinct integers and write their greatest
common divisor and least common multiple instead. Prove that eventually the numbers
will stop changing.

P.S. it puzzles me and i cant understand the solution given.

If I'm not mistaken....

[hide]Lemma: If we have a list of positive integers $ a_1, \ldots, a_n$ written on a blackboard, and at each stage, for any $ a_i$ and $ a_j$, we replace $ a_i$ with $ \min(a_i, a_j)$ and $ a_j$ with $ \max(a_i, a_j)$, then the numbers on the board will eventually remain the same.

Proof: Suppose we define our sequence of pairs $ (i, j)$ with $ i < j$ to operate on to be $ s_1, s_2, \ldots$, and call this operation operation S. We say that we apply operation S to a term in this sequence $ s_n$, we are applying operation $ S$ on $ a_i$ and $ a_j$, where $ s_n \equal{} (a_i, a_j)$.

Observe that if we perform operation S on a pair with $ i < j$ and $ a_i < a_j$, the numbers on the board will stay fixed. Observe that if at some point in the sequence, applying operation S to $ s_{n\plus{}1}, s_{n\plus{}2}, \ldots$ does nothing, we will be done (since our board will remain the same after these operations.)

Define the number of inversions of this sequence to be the number of pairs $ a_i$, $ a_j$ such that $ i < j$ and $ a_i > a_j$. We claim that if an application of operation S to some term $ s_n$ swaps two terms of a sequence, the number of inversions decreases. Indeed, if $ i < j$ and $ a_i > a_j$, then $ a_i$ and $ a_j$ will have their positions swapped. I could go through this all more rigorously, but I'm lazy, and really, it's easy to verify that the number of inversions does decrease.

If there were indeed infinitely many terms in $ \{s_n\}$ that changed the board, then we would have an infintiely decreasing sequence of integers, a contradiction. Thus, our board must eventually remain constant, as desired.

Let our integers be $ a_1, a_2, \ldots, a_n$. We now prime factorize each of the integers. Suppose that for some prime $ p$ dividing some integer on the blackboard, we take another blackboard and write down in order $ v_p(a_1), v_p(a_2), \ldots, v_p(a_n)$ (where $ v_p(n)$ denotes the greatest exponent of $ p$ dividing $ n$). Note that when we take $ a_i$ and $ a_j$, and replace them with their greatest common factor and least common multiple, respectively, we are replacing $ v_p(a_i)$ and $ v_p(a_j)$ with $ \min(v_p(a_i), v_p(a_j))$ and $ \max(v_p(a_i), v_p(a_j))$, respectively. Applying our lemma, we see that at some point, the exponents of $ p$ in our sequence will stay fixed when this operation is performed on them. Since we can use this argument on every prime factor, it follows that all the numbers on the blackboard must eventually stay fixed as well. [/hide]

[quote="Zhero"]Lemma: If we have a list of positive integers $ a_1, \ldots, a_n$ written on a blackboard, and at each stage, for any $ a_i$ and $ a_j$, we replace $ a_i$ with $ \min(a_i, a_j)$ and $ a_j$ with $ \max(a_i, a_j)$, then the numbers on the board will eventually remain the same. [/quote]  This result is not actually the one you used.  (Your lemma as stated is just silly: replacing $ \{a_i, a_j\}$ with their min and max doesn't ever change the set of numbers ;).)  Just the inversions monovariant argument is fine :)

Good point. :blush:

I think I meant to say the sequence of numbers on the board will stay the same, i.e., taking order into account.

Well the gcd of two distinct numbers is less than both unless one is a multiple of the other. But if one is a multiple of the other the two numbers will just stay the same, so we can assume that is not the case. So we keep picking two numbers, neither of which is a multiple of the other, and the smaller of the two keeps decreasing. As long as there exists two numbers, neither of which is a multiple of the other, we will keep decreasing some of the numbers until they can't decrease anymore (1 is the smallest natural), and then move on to the rest, and this process will countinue until for any two numbers, one is a multiple of the other, and then they all stay the same.
Is this rigorous?

Perhaps some of those numbers could be made more explicit?

Another invariant argument, because $ \text{gcd}(a,b) \le a,b \le \text{lcm}(a,b)$ and $ \text{gcd}(a,b)\text{lcm}(a,b) \equal{} ab$, we (because, for example $ \text{lcm}(a,b) \minus{} \text{max}(a,b) \equal{} \frac{\text{max}(a,b)}{\text{gcd}(a,b)}[\text{min}(a,b) \minus{} \text{gcd}(a,b)]$, etc) have $ \text{gcd}(a,b) \plus{} \text{lcm}(a,b) \ge a \plus{} b$, equality iff equality held in the earlier condition. So $ S \equal{} \sum a_i$, which is a positive integer, will strictly increase, but it is bounded above [say, by $ S \le n\text{lcm}(a_1, \ldots, a_n)$].

[quote="Zhero"]Proof: Suppose we define our sequence of pairs $ (i, j)$ with $ i < j$ to operate on to be $ s_1, s_2, \ldots$, and call this operation operation S. We say that we apply operation S to a term in this sequence $ s_n$, we are applying operation $ S$ on $ a_i$ and $ a_j$, where $ s_n \equal{} (a_i, a_j)$.

Observe that if we perform operation S on a pair with $ i < j$ and $ a_i < a_j$, the numbers on the board will stay fixed. Observe that if at some point in the sequence, applying operation S to $ s_{n \plus{} 1}, s_{n \plus{} 2}, \ldots$ does nothing, we will be done (since our board will remain the same after these operations.)

Define the number of inversions of this sequence to be the number of pairs $ a_i$, $ a_j$ such that $ i < j$ and $ a_i > a_j$. We claim that if an application of operation S to some term $ s_n$ swaps two terms of a sequence, the number of inversions decreases. Indeed, if $ i < j$ and $ a_i > a_j$, then $ a_i$ and $ a_j$ will have their positions swapped. I could go through this all more rigorously, but I'm lazy, and really, it's easy to verify that the number of inversions does decrease.

If there were indeed infinitely many terms in $ \{s_n\}$ that changed the board, then we would have an infintiely decreasing sequence of integers, a contradiction. Thus, our board must eventually remain constant, as desired.

Let our integers be $ a_1, a_2, \ldots, a_n$. We now prime factorize each of the integers. Suppose that for some prime $ p$ dividing some integer on the blackboard, we take another blackboard and write down in order $ v_p(a_1), v_p(a_2), \ldots, v_p(a_n)$ (where $ v_p(n)$ denotes the greatest exponent of $ p$ dividing $ n$). Note that when we take $ a_i$ and $ a_j$, and replace them with their greatest common factor and least common multiple, respectively, we are replacing $ v_p(a_i)$ and $ v_p(a_j)$ with $ \min(v_p(a_i), v_p(a_j))$ and $ \max(v_p(a_i), v_p(a_j))$, respectively. Applying our lemma, we see that at some point, the exponents of $ p$ in our sequence will stay fixed when this operation is performed on them. Since we can use this argument on every prime factor, it follows that all the numbers on the blackboard must eventually stay fixed as well. [/hide][/quote]

i really dont understand your proof but i did understand what the book had. What is an inversion exactly?i dint understand para 3,5. someone please elaborate :(

suppose $(a,b)$ is transformed to $(gcd(a,b),lcm(a,b))=(c,d)$. $d \ge max(a,b) $ it can be easily shown that $c+d \ge a+b$ .So we can say that the sum is non decreasing as well as bounded.As the sum is bounded at some point d will stop changing and as ab=cd so c will stop changing. Hence the numbers will stop changing.


abstract algebra
group theory
superior algebra
superior algebra theorems

$\phi$ is a homomorphism from a group $G$ [b][u]onto[/u][/b] $Z_{6}\oplus Z_{2}$, and $\text{Ker}\phi$ has order 5. I am trying to show that $G$ must have normal subgroups of orders 5, 10, 15, 20, 30 and 60.

This is what I've done (not very much :( ):
Using the First Isomorphism Thm, $G/\text{Ker}\phi \approx \phi(g)$, so $|G/\text{Ker}\phi|=|\phi(g)|$, and thus $|G|/|\text{Ker}| = |\phi(g)|$. So $|G|=60$.

I don't know where to go from here.

I have a theorem that states that:
Every normal subgroup of a group G is the kernal of a homomorphism of G. In particular, a normal subgroup N is the kernel of the mapping g->gN from G to G/N.

But I don't think that's going to help me.

Any help would be appreciated. Thanks

What are the quotients of $Z_{6}\oplus Z_{2}$? All of those are also homomorphic images of $G$ by composition.
If you can find groups of order $12,6,4,3,2,1$ which are homomorphic images of $Z_{6}\oplus Z_{2}$, you get what you want by the converse of the theorem you quoted and basic counting.

Such groups exist. Construct them.

I found normal subgroups of $Z_{6}\oplus Z_{2}$ with orders $12, 6, 4, 3, 2 ,1$.

Now using some properties of subgroups under homomorphisms:
1) If $|\text{Ker}\phi| = n$, then $\phi$ is an $n\text{-to-1}$ mapping from $G$ onto $\phi(G)$.
2) If $\overline{K}$ is a normal subgroup of $\overline{G}$, then $\phi^{-1}(\overline{K}) = \{k \in G | \phi(k) \in \overline{K}\}$ is a normal subgroup of $G$

So using those, and the fact that $\phi$ is onto, I have that for each element of each normal subgroup of $Z_{6}\oplus Z_{2}$, there are $5$ elements ($\text{Ker}\phi = 5$) in the corresponding normal subgroup of $G$. So the orders of the normal subgroups of G are $5\cdot\{12, 6, 4, 3, 2, 1\}$.

Is that correct?


When I go ahead and begin an unordered list with something like \begin{itemize} and then have \item first sentence and \item second sentence...  The bulleted list begins quite a ways down from the previous line as shown here in the picture below and also the second bullet (though not shown) is a little bit farther below than the first that I probably need.  I am looking to reduce a bit of the white space.

[img]http://www.orangesynergy.com/pics/itemize.jpg[/img]

This document is going to have a lot of bullets and therefore I'd like to reduce this space so that I don't have too many extra pages.  Any ideas?  Thanks.

Design:

\vspace{-xpt}
\begin{itemize}
...

Set the value you desire to reduce the space.

Sweet that works.  For some reason I couldn't get the pt to work, but using something like \vspace{-0.6cm} worked great.  Thanks! :)


function
induction
algebra unsolved
algebra

Find all functions $f: \mathbb N \to \mathbb N$ such that $ f^{19}(n)+97f(n)=98n+232$. Notice in this problem that $f^{19}(n)$ means the composition of $f$ with itself $19$ times.


Prove that existence of a unique sequence $\{u_{n}\}_{n\geq 0}\subset\mathbb{N}$ such that \[u_{n}^{2}=\sum_{r=0}^{n}\binom{n+r}{r}u_{n-r}\forall n\geq 0.\]

Why unique?
There are infinetely many real sequenses.
Let $u_{n,k}, k=0,1,...$ mean sequense defined by
$u_{n,k}=0, k<n$, $u_{n,n}=1$, \[u_{n,k+1}=\frac{1+\sqrt{1+4\sum_{r=1}^{k+1}C_{k+1+r}^{r}u_{n,k+1-r}}}{2},k\ge n.\]
For example $u_{\infty ,k}\equiv 0$. and $u_{0,k}=2^{k}$.

[quote="Rust"]Why unique?
There are infinetely many real sequenses.
Let $u_{n,k}, k=0,1,...$ mean sequense defined by
$u_{n,k}=0, k<n$, $u_{n,n}=1$,
\[u_{n,k+1}=\frac{1+\sqrt{1+4\sum_{r=1}^{k+1}C_{k+1+r}^{r}u_{n,k+1-r}}}{2},k\ge n. \]
For example $u_{\infty ,k}\equiv 0$. and $u_{0,k}=2^{k}$.[/quote]
Please read carefuly problem.

Because I give all real solutions. Only one $u_{o,k}=2^{k}\in N \ \forall k\in Z_{+}$.

Because $u_{n}>0$ therefore
[quote="Rust"]
\[u_{n}=\frac{1+\sqrt{1+4\sum_{r=1}^{n}C_{n+r}^{r}u_{n-r}}}{2}. \]
[/quote]
Ok, thanks. This is combinatoric problem.


Find the value of $ \tan20^{o} \cdot\tan40^{o} \cdot\tan60^{o} \cdot \tan80^{o}$


function
vector
calculus
calculus computations

[url=http://img245.imageshack.us/my.php?image=flux.jpg][img]http://img245.imageshack.us/img245/5735/flux.th.jpg[/img][/url]

according to my observation the level surface will be a paraboloids opening on the z-axis. Is this correct?

From the attachment, I'm guessing you're meaning level surfaces of $ f$, and those are spherical, centered in the origin, since $ f$ is a function only of  $ \rho \equal{} \sqrt {x^2\plus{}y^2\plus{}z^2}$.

Is the question in the problem related do the gradient of $ f$ ?

yes I am sorry.. I meant level surfaces of $ f$.. I thought it was spherical but wasn't sure... how do I find F in this case? F here is grad$ f$... and f is a function of g(rho)...

Again from the attachment, seems the question is for the points $ \rho \equal{} {1\over 3} \text{ and } \rho\equal{} 1$, if one don't make it graphically. F will be ever radial, and its intensity is given by $ f'(\rho)$.

maybe the question wasn't complete... it asks for F at (1,2,-2), so therefore rho when plugging in here is 3  and g'(3) is 1/3 so F is 1/3r?

The gradient is a vector:

 \[ \vec{F}(3) \equal{} {{\vec{u}_\rho}\over 3}\text{, where } \vec{u}_\rho\]
is the unitary spherical radial vector.


parameterization
parametric equation
calculus
calculus computations

Surface 1: x^3+3x^2y^2+y^3+4xy-z^2=0
Surface 2: x^2+y^2+z^2=11

Find parametric equation for the line tangent to the curve of intersection of the surfaces at the given point.
The given point is (1, 1, 3).

These two surfaces are written as $ f(x) \equal{} a$ and $ g(x) \equal{} b.$ That is, they're written as level surfaces for functions.

Here's the geometric question for you to visualize: in what directions do $ \nabla f$ and $ \nabla g$ point at the point in question, and how are either or both of them related to the direction of the tangent line to the curve of intersection?

(One note: I didn't check, but we'd better have $ \nabla f$ and $ \nabla g$ not be scalar multiples of each other, including that neither should be zero. If that condition is not met, then the notion of "curve of intersection" may be in trouble.)


If $a,\,b,\,c$ are positive real numbers, prove that: \[\frac{bc}{a^{2}+2bc}+\frac{ca}{b^{2}+2ca}+\frac{ab}{c^{2}+2ab}\le 1\]

[quote="April"]If $a,\,b,\,c$ are positive real numbers, prove that:
\[\frac{bc}{a^{2}+2bc}+\frac{ca}{b^{2}+2ca}+\frac{ab}{c^{2}+2ab}\le 1 \]
[/quote]
And $\frac{bc}{2a^{2}+bc}+\frac{ca}{2b^{2}+ca}+\frac{ab}{2c^{2}+ab}\geq 1.$ :wink:

[quote="April"]If $a,\,b,\,c$ are positive real numbers, prove that:\[\frac{bc}{a^{2}+2bc}+\frac{ca}{b^{2}+2ca}+\frac{ab}{c^{2}+2ab}\le 1 \][/quote]
[quote="arqady"]And $\frac{bc}{2a^{2}+bc}+\frac{ca}{2b^{2}+ca}+\frac{ab}{2c^{2}+ab}\geq 1$[/quote]
[quote="white_horse_king88"]That means $\sum_{cyc}\frac{a^{2}+2bc}{bc}\geq 1$[/quote] :huh: i feel left out...
if anyone has any solutions to these, can you post them?
sry im noob :|

[hide="hint"]
\[a^{2}+2bc\leq a^{2}+b^{2}+c^{2}\]
(of course, you'll have to figure out how to manipulate the inequality before using this; doing it right away is BAD!)
[/hide]

[hide="solution for the second"]
One-liner!
\[\left( \sum_\text{cyc}2a^{2}bc+b^{2}c^{2}\right) \left( \sum_\text{cyc}\frac{bc}{2a^{2}+bc}\right)\geq (bc+ca+ab)^{2}=\sum_\text{cyc}2a^{2}bc+b^{2}c^{2}\]
[/hide]

for the first inequality is equivalent to 1/(2+A)+1/(2+B)+1/(2+C) <=1 when
ABC=1 ,wich is equivalent after devlopment AB+BC+CA>=3 wich is true by IAG ( i take A=a²/bc...) :lol:

[quote="hikaru123"]Applying AM-GM inequality we have $a^{2}+2bc\le a^{2}+b^{2}+c^{2}$
So $\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ac}+\frac{c^{2}}{c^{2}+2ab}\ge 1$
$\leftrightarrow 3-\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ac}+\frac{c^{2}}{c^{2}+2ab}\le 2$
$\leftrightarrow \frac{bc}{a^{2}+2bc}+\frac{ac}{b^{2}+2ac}+\frac{ab}{c^{2}+2ab}\le 1$
Equality occurs if and only if a=b=c[/quote]
You forgot some brackets.

[quote="hikaru123"]Applying AM-GM inequality we have $a^{2}+2bc\le a^{2}+b^{2}+c^{2}$
So $\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ac}+\frac{c^{2}}{c^{2}+2ab}\ge 1$
$\leftrightarrow 3-(\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ac}+\frac{c^{2}}{c^{2}+2ab})\le 2$
$\leftrightarrow \frac{bc}{a^{2}+2bc}+\frac{ac}{b^{2}+2ac}+\frac{ab}{c^{2}+2ab}\le 1$
Equality occurs if and only if a=b=c[/quote]
I see. but i cannot edit my post


Does AoPS really help you do better in MathCounts? Like a whole lot better?

If you take learning the math seriously (including possibly enrolling in the Problem Series, whose quality is unknown to me but probably very good) and don't waste your time reading/posting spam, it should.

[quote="tinytim"]Yes, It helps you if you don't spend your entire time spamming.[/quote]

You don't mean me, right???


As part of my assignment I am required to look up applications of parametric curves. However, a google search reveals nothing useful (cars and fonts - but not in the form of equations). Please suggest some applications (with equations), links, or relevant search terms. No "famous curves" (e.g. cycloids) alone please...thanks!


calculus
derivative
function
analytic geometry
graphing lines
slope
real analysis

Solve the following differential system:
dx/x=dy/y=dz/z;x>0;y>0;z>0;

Of course it is veery simple,but please explain me what kind of diffrential system is this one;i mean a differential system contains derivatives such as x',y',z',etc.
And i even don't know what are the unknown functions in the above system,because it doesn't say.
I,m having serious difficulties in understanding this stuff with dx,dy,dz in equations.
Can you please help me?
The relation x'=dx/dt is only a symbolic notation for x' or  dx/dt is a real fraction?
What mathematical mechanisms let us write a diff. equation such:
dx/dt=2x/t (for example) in the form
dx/2x=dt/t,and then "integrate"?
Thanks a lot.

PS:and there is some other strange stuff i,ve seen in a book:
you have the sistem:
dx/p(x,y,z)=dy/q(x,y,z)=dz/r(x,y,z)=(adx+bdy+cdz)/(ap(x,y,z)+bq(x,y,z)+cr(x,y,z)),
where we choose a,b,c such that
ap(x,y,z)+bq(x,y,z)+cr(x,y,z)=0,AND THEY ARISE THE CONCLUSION THAT adx+bdy+cdz=0.
Is there in mathematics such a theorem wich says that if we have a fraction a/b,if b=0 then a=0?
It is really strange.I knew that the fraction a/0 has no sense in maths,but i  see here that is we have a/0 than a=0?
Very,very,very strange!
Please someone help me!

Nobody knows the answers to my questions?

as far as I know. for your question about $\frac{dy}{dx}$ , it is the leibniz symbol for derivative. but in some sense many enginering say that it is a fraction of two arbitary small number called "infintesimal".
But In mathematics science, the notation $\frac{dy}{dx}$ involve more than a fraction. note that althought the $-$ in this notation has the behaviour same as the fraction operator, but it is not operator, it just a sign that occurs in the leibniz notation.

Then, for some reason mathematician, need to define the meaning $dy$ and $dx$ separately.
you can thinks as follow. in a cartesian system draw "new"  coordinate system $dy$ paralell to $y$ coordinate and $dx$ parallel to $x$. with a point $P(x_0,y_0)$ as the new origin. then if there is a curve $y=f(x)$ so that it has a tangent line in P. then the slope in P relative to coordinate $dy$-$dx$  and $x$-$y$ is the same. thus we have
$dy=f'(x_0)dx$ since the slope in P relative to coordinate $x$-$y$ is $f'(x_0)$
this is call the defenition of differential(its different from derivative). if we divide both side by $dx$ we have excatly leibniz notation. this defenition state the relation between the change of $y$ by $dy$related to  the change of $x$ by $dx$


Should b) be $ (1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{b}) \geq 9$?
If it is then:
$ (1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{b}) \equal{} (\frac {a \plus{} 1}{a})(\frac {b \plus{} 1}{b})
\equal{} \frac {ab \plus{} a \plus{} b \plus{} 1}{ab} \equal{} \frac {ab \plus{} 2}{ab} \equal{} 1 \plus{} \frac {2}{ab}$

and since $ ab$ has a maximum value of $ \frac {1}{4}$ then $ \frac {2}{ab}$ has a minimum value of $ 8$ and $ 1 \plus{} \frac {2}{ab} \geq 9$

[quote="PUTALE"]need help setting up the proof:

1. suppose that a and b are positive numbers whose sum is 1.
a. find the maximum value of the product ab.
b. use your answer to part a) to prove that (a + (1/a))(1+(1/b)) >= 9

I just need help with the 1B since I have already found out that the max of ab is 1/4.

thanks a lot.[/quote]
a. By AM-GM inquality, we have: $ ab \le \frac{(a\plus{}b)^2}{4}$ ( or inequality becomes $ (a\minus{}b)^2 \ge 0$)
b. By AM-GM:
$ 1\plus{}\frac{1}{a}\equal{}1\plus{}\frac{1}{2a}\plus{}\frac{1}{2a} \ge 3\sqrt[3]{\frac{1}{4a^2}}$
So:
$ (1\plus{}\frac{1}{a})(1\plus{}\frac{1}{b}) \ge 9\sqrt[3]{\frac{1}{16a^2b^2}} \ge 9(because ab \le \frac{1}{4})$


algebra
function
domain
calculus
integration
abstract algebra
Ring Theory

Hi everybody,

I hope you can help me with this:

I have to prove that if R is a semilocal domain, Pic(R) is trivial.

Thx

1. What is a semilocal domain?
2. What does Pic(anything) mean?

Sorry, but I don't know any of these terms.

semilocal means that there are only finitely many maximal ideals. the picard group of an integral domain $ R$ is the quotient group {fractional invertible ideals} / {principle ideals}. you can even define the picard group of a scheme, but this is not important here.

thus $ Pic(R) \equal{} 0$ means: every $ R$-submodule $ \mathfrak{a}$ of $ Q(R)$ such that there is some other $ \mathfrak{b}$ with $ \mathfrak{a}*\mathfrak{b} \equal{} R$ (then $ \mathfrak{a}$ and $ \mathfrak{b}$ are automatically finitely-generated) can be generated by a single element. I've got a solution when $ R$ is local:

choose a preimage $ x$ of $ 1$ with respect to the canonical epimorphism $ \mathfrak{a} \otimes_R \mathfrak{b} \to R$. a calculation shows that $ 1 \to x$ defines an inverse homomorphism $ R \to a \otimes_R \mathfrak{b}$ (which is not true for arbitrary homomorphisms onto $ R$!). thus $ \mathfrak{a} \otimes_R \mathfrak{b} \cong R$. if $ \mathfrak{m}$ is the maximal ideal of $ R$, we have $ R/\mathfrak{m} \cong \mathfrak{a}/\mathfrak{m}\mathfrak{a} \otimes_{R/\mathfrak{m}} \mathfrak{b}/\mathfrak{m} \mathfrak{b}$, thus $ \dim_{R/\mathfrak{m}} \mathfrak{a}/\mathfrak{m} \mathfrak{a} \equal{} 1$. nakayama's lemma shows that $ \mathfrak{a}$ is principle, as desired.

but what do we do in the case of a semilocal ring? the above argument shows that $ \mathfrak{a}_\mathfrak{m}$ is a principle $ R_m$-module. so we may choose some generator $ x_\mathfrak{m} / 1$. if $ x_\mathfrak{m}$ is not contained in any other maximal ideal, we're done: the product $ x$ of the finitely many $ x_\mathfrak{m}$ differs from $ x_\mathfrak{m}$ in $ \mathfrak{a}_\mathfrak{m}$ only up to a unit, thus $ x$ generates $ \mathfrak{a}$ locally and thus, globally. but I don't see how the property of the $ x_\mathfrak{m}$ can be achieved.

Maybe you can use the hint in exercise 2.42, page 26  from the course notes

http://websites.math.leidenuniv.nl/ant2008/

Bibliography: http://www.regulations.leiden.edu/content_docs/students/plagiarism.pdf

in the hint, why may we choose $ x_m \in I \backslash \mathfrak{m} I$? this follows from nakayama if $ R$ were local.

[b]General remark:[/b] The notion of the Picard group $ Pic(R)$ applies to general commutative unitary rings $ R$, not only to integral domains. A $ R$-module $ M$ of finite presentation is called projective of rank 1 iff the natural homomorphism
\[ M\otimes_RHom(M,R)\rightarrow R
\]
is an isomorphism. Sometimes they are called line bundles. This is equivalent to that $ M$ is projective of rank 1 (the latter means that the localization in every prime in free of rank 1). This definition coincides with fractional ideals in case $ R$ is an integral domain. $ R$-modules of rank 1 set up a group where multiplication is the tensor product; the inverse of $ M$ is the dual $ Hom(M,R)$.
As -oo- has already pointed out the $ Pic(R)$ is defined as quotient of the group of the projective $ R$-modules of rank 1 modulo free $ R$-modules of rank 1.

That $ Pic(R)$ is trivial for a semilocal ring $ R$ stems from the following general

[b]Lemma:[/b] Let $ R$ be a semilocal ring and $ M$ a finite projective $ R$-module such that there is some $ n > 0$ such that $ M_\mathfrak{m}$ is a free $ R_\mathfrak{m}$-module of rank $ n$ for every maximal ideal $ \mathfrak{m}$: $ M_\mathfrak{m}\simeq R_\mathfrak{m}^s$. Then $ M$ is free of rank $ s$.

[hide="Proof:"] If $ \mathfrak{m}_1,\ldots, \mathfrak{m}_r$ denote the maximal ideals of $ R$, then let $ J \equal{} \bigcap_{i \equal{} 1}^r\mathfrak{m}_i$ the so called Jacobson ideal.

[u]Step 1:[/u]  $ J \equal{} 0$. The isomorphism $ R\simeq \oplus_i R/\mathfrak{m}_i$ (Chinese remainder theorem) implies $ R_{\mathfrak{m}_i}\simeq (R/\mathfrak{m}_i)_{\mathfrak{m}_i} \simeq R/\mathfrak{m}_i$, hence $ R\simeq \oplus_i R_{\mathfrak{m}_i}$. All these are $ R$-module isomorphisms. Tensoring the last isomorphism with $ M$ yields
\[ M\simeq \oplus_i M_{\mathfrak{m}_i} \simeq \oplus_i R_{\mathfrak{m}_i}^s\simeq R^s.
\]
[u]Step 2:[/u] In the general case Step 1 gives that $ M/JM$ is a free $ R/J$-module of rank $ s$. Taking elements $ m_1,\ldots,m_s\in M$ such that they are a basis modulo $ JM$ we have a surjection $ R^s\rightarrow M$, by Nakayama. Since $ M$ is projective, hence flat, the short exact sequence
\[ 0\rightarrow K\rightarrow R^s\rightarrow M\rightarrow 0
\]
induces a short exact sequence
\[ 0\rightarrow K/JK\rightarrow R^s/JR^s\rightarrow M/JM\rightarrow 0
\]
hence $ K/JK \equal{} 0$. Since $ P$ is projective, the first sequence splits and $ K$ is a finitely generated $ R$-module, such that again Nakayama implies $ K \equal{} 0$.[/hide]


A large sphere has a volume of $ 288\pi$ cubic units.  A smaller sphere has a volume which is $ 12.5\%$ of the volume of the larger sphere.  What is the ratio of the radius of the smaller sphere to the radius of the larger sphere? Express your answer as a common fraction.

[hide="Click here for solution"]

The formula for the volume of a sphere is $ \frac{4}{3}\pi r^3$.

So, for this problem, we have:

$ \frac{4}{3}\pi r^3\equal{}288\pi \Rightarrow 4\pi r^3\equal{}864\pi \Rightarrow r^3\equal{}216 \Rightarrow r\equal{}6$

For the new sphere, the volume is just $ \frac{1}{8}$ of the volume of the original one meaning that it is $ 36\pi$.

So, now we have:

$ \frac{4}{3}\pi r^3\equal{}36\pi \Rightarrow 4\pi r^3\equal{}108\pi \Rightarrow r^3\equal{}27 \Rightarrow r\equal{}3$

Therefore, the ratio of the smaller radius to the larger radius is $ \boxed{\frac{1}{2}}$.

[/hide]

A faster way:
If you know your decimal equivalents 12.5% is equal to 1/8.
Then since you are calculating volume, you find the cube root of 1/8 which is [b]1/2[/b].[/b]


Read the title,

And what are you doing for the holiday?

I'm going to a barbecue.

1'm r34d1n6 4b0u7 c0n1c5 1n 1n732m3d1473 4l63b24 4nd c0mpl371n6 4 phy51c5 3x4m. h0w fun!

In other words:

I'm reading about conics in intermediate algebra and completing a physics exam. How fun!

Celebrating Birthday (today yay!)

BBQ ribs, wings, and drummettes.

FIREWORKS

@bluecarneal: happy birthday!

I'm gonna be at home, chatting on gmail and playing Dance Dance Revolution. (I am DETERMINED to beat "Super Samurai"!)

The fireworks were AMAZING!
Right now I'm waiting for my b-day on wednesday.


For $m,\ n$ be positive numbers, let $x_{n}$ be the greatest number of the form $\frac{m}{10^{n}}$ among the numbers less than $\sqrt{2}$.
Prove that the rational number $(x_{n})_{n\geq 1}$ makes Cauchy sequence.

[quote="kunny"] greatest number among the numbers less than$\sqrt{2}$.[/quote]
Seems like nice problem, but I dont understand this part of the sentence..  :(

Notice that replacing $\sqrt{2}$ by any other real number would make absolutely no change in the problem. This is not about any special property of that number.

[quote="jmerry"]It's probably a minor mistake in translation. Here's what I think it should say:

... let $x_{n}$ be the greatest number of the form $\frac{m}{10^{n}}$ among the numbers less than $\sqrt{2}$.[/quote]

Thank you! :)

$\sqrt{2}$ is an upper bound for $\{x_{n}\}$. Suppose there exists another distinct upper bound $\alpha$. Then exactly one the following is true; $\alpha>\sqrt{2}$ or $\alpha<\sqrt{2}$.
Suppose $\alpha<\sqrt{2}$. Then there exists no $x_{n}$ between $\alpha$ and $\sqrt{2}$. But this contradicts our assumption that $x_{n}$ can take the greatest number of the form $\frac{m}{10^{n}}$ less then $\sqrt{2}$. Indeed there exist such numbers between $\alpha$ and $\sqrt{2}$. Thus $\alpha>\sqrt{2}$. Hence $\sqrt{2}$ is a least upper bound for $\{x_{n}\}$.
Suppose $x_{n}>x_{n+1}$ i.e. $\frac{m}{10^{n}}>\frac{m'}{10^{n+1}}$. Then $10m>m'$. Since $10^{-n}>10^{-n-1}$ we have to have $m>m'$ for $x_{n}>x_{n+1}$ to hold.
But $10m>m'$. Contradiction. Thus $x_{n}\leq~x_{n+1}$. Since $x_{n}$ is a motonically increasing sequence which is bounded above, it converges to its supremum $\sqrt{2}$.
Given an $\epsilon>0$ we can find positive integers $N_{1},N_{2}$ s.t. $|x_{m}-\sqrt{2}|<\frac{\epsilon}{2}$ and $|x_{n}-\sqrt{2}|<\frac{\epsilon}{2}$ for all $n\geq~N_{1}$ and $m\geq~N_{2}$. Let $N=max(N_{1},N_{2})$.
Then $|x_{n}-x_{m}|\leq~|x_{m}-\sqrt{2}|+|x_{n}-\sqrt{2}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ for all $n,m\geq~N$. Hence $\{x_{n}\}$ is a Cauchy sequence.


Given that Cos(theta)=3x and 0<theta< pi/2, find 4(theta)+sin(2theta)

Ok Theta would be arccos(3x) so 4(theta) would = 4arccos(3x)

cos(theta)= 3x/1 so the sin(theta) = square root of 1-9x^2.

Sin(2Theta) = 2 Cos Sin and we now know all value of cos and sin so the final answer would be

4arccos(3x)+6x(square root of 1-9x^2) yes?

[quote="Farcus"]Given that $\cos\theta=3x$ and $0<\theta< \frac{\pi}{2}$, find $4\theta+\sin(2\theta)$.

Ok $\theta$ would be $\arccos(3x)$ so $4\theta$ would $= 4\arccos(3x)$.

$\cos\theta= 3x/1$ so the $\sin\theta = \sqrt{1-9x^{2}}$.

$\sin(2\theta) = 2 \cos \sin$ and we now know all value of $\cos$ and $\sin$ so the final answer would be

$4\arccos(3x)+6x\sqrt{1-9x^{2}}$ yes?[/quote]
Since they give you $0<\theta<\frac{\pi}{2}$, they probably want a numerical answer.
EDIT: Never mind.  :blush:


Did you try reading the lines right after that?  I think that they explain why.

[/quote]

If $a = \frac{1 + \sqrt 5}{2}$ and if we define the sequence $\{g_n\}$ as $g_n = a^n$, then

$g_n + g_{n-1} = a^n + a^{n-1} = a^{n-1} \left( a + 1 \right) = a^{n-1} \left( 1 + \frac{1 +\sqrt 5}{2} \right) =$

$= a^{n-1} \frac{3 + \sqrt 5}{2} = a^{n-1} \left( \frac{1 + \sqrt 5}{2} \right)^2 = a^{n-1} a^2 = a^{n+1} = g_{n+1}$

which is the Fibonacci recurrence relation. Similarly for the sequence $\{h_n\}$ defined as $h_n = b^n$, where $b = \frac{1 - \sqrt 5}{2}$. To get an integer sequence $\{f_n\}$, we can define $f_n = g_n + h_n$, the odd powers of $\sqrt 5$ disappear and we get Lucas numbers (Fibonacci recurrence formula with the seed $(2, 1)$ or $(1, 3)$). Alternately, we can define $f_n = \frac{1}{\sqrt 5} (g_n - h_n)$, the even powers of $\sqrt 5$ disapper and we get Fibonacci numbers (Fibonacci recurrence formula with the seed $(0, 1)$ or $(1, 1)$).


trigonometry
inequalities
trigonometric substitution

Prove that for every $x \in R$ we have:
\[ \left| \frac{x(1-x^{2})}{(1+x^{2})^{2}} \right| \leq \frac{1}{4} \]

Let $x=\tan A$. Then the inequality is equivalent to $\frac{|\sin4A|}4 \le \frac14$ which is obvious.


Dear Problem Solvers,

I am compiling a list of quotes that related to mathematical problem solving.  If you have any favorite mathematical problem solving related quotes, especially quotes that are inspiring for the problem solver, would you be so kind as to e-mail the quotes to me at < yu250248@mathstat.yorku.ca >.  Please include the name of the person who the quote is attributed to and where it may be from (example, a book).  My apologies for cross-postings.

Yours truly,

Mathfreak

Hi Problem Solvers,

[quote="Peter VDD"]?
you mean funny quotes, or quotes that have a deeper truth or so? Or just general hints for problem solving? or what?[/quote]

All of the above: funny quotes, quotes with a deeper meaning, general hints for problem solving, inspirational quotes, etc.  My request is really broad in that I'm not being very specific about what types of quotes I'm looking for other than that they be related to mathematical problem solving and solvers.   Again, if you would be so kind, please e-mail your favourite such quotes to me at < yu250248@mathstat.yorku.ca >, or post them here to share with everyone on MathLinks.

Yours truly,

mathfreak

there is a very good site which gathers a lot of mathematician and physician's quote and it is in english !
l don't remember the address, but l'll check my 'favorites sites'

arthur

This is funny, it's from MOSP website. :D

http://moppers.kaseorg.com/01/quotes/

"Most graph theory problems don't actually come out and say they are graph theory problems  they just say 'Alice is friends with Bob'... And then there are planar graph theory problems which say 'Alice and Bob live in a plane, and Alice is friends with Bob..."

alex s.


Η άσκηση αυτή είναι κατασκευή του αδελφού μου Βασσίλη που είναι Χημικός !

Αν  p,q είναι πρώτοι αριθμοί , να αποδείξετε ότι $ pq|n( pn^{q \minus{} 1} \plus{} qn^{p \minus{} 1} \minus{} p \minus{} q )$  ... κατασκευή   05/2004.

Η απόδειξή της με κούρασε και αποτελείται από δυο σελίδες ... :(
Στηρίζεται βέβαια σε πρωταρχικές έννοιες. Το φοβερό είναι ότι πρόσφατα το έφερε η κουβέντα γύρω από αυτή την άσκηση και δυστυχώς ο αδερφός μου δεν θυμόταν με ποιές υποθέσεις κατέληξε στο παραπάνω συμπέρασμα.
 Είμαι χαρούμενος που μέσω του forum μου δίνεται η δυνατότητα μοιράζομαι κάποιες ασκήσεις μαζί σας. Πολύ καλύτερο από το να κάθονται στον υπολογιστή μου.

Η άσκηση βγαἰνει σε δύο γραμμές . Είναι άμεση συνέπεια του θεωρήματος Fermat
Για p πρώτο pou δεν διαιρεί τον n $ n^{p\minus{}1}\equiv 1(\mod p)$  τελος ...... :D

Ημουν σίγουρος ότι θα συμβεί κάτι τέτοιο ...  :)
Πάντως χωρίς να εφαρμόσεις το θεώρημα του Fermat έχει μεγάλο ενδιαφέρον η λύση της  :)   Χρησιμοποιώντας μόνο πράγματα που βρίσκονται στο βιβλίο κατεύθυνσης της Β΄Λυκείου ...

Το θεώρημα Fermat είναι κάτι εντελώς στοιχειώδες όμως και αποδεικνύεται πολύ εύκολα   :wink:

Επισης ισχυει οτι $ pqn/n(pn^{q \minus{} 1} \plus{} qn^{p \minus{} 1} \minus{} p \minus{} q)$.

Η λύση με το θεώρημα του Fermat:
για $ p,q \equal{}$ πρώτους είναι:
$ n^{p\minus{}1} \equiv 1 (mod p) \Leftrightarrow qn^{p\minus{}1} \equiv q (mod pq) \Leftrightarrow$
$ qn^{p\minus{}1} \minus{} q \equiv  0 (mod pq)$
ομοίως
$ pn^{q\minus{}1} \minus{} p \equiv 0 (mod pq)$
προσθέτοντας κατά μέλη έχουμε
$ pn^{q\minus{}1} \plus{} qn^{p\minus{}1} \minus{} p \minus{} q \equiv 0 (mod pq)   \Leftrightarrow n(pn^{q\minus{}1} \plus{} qn^{p\minus{}1} \minus{} p \minus{} q) \equiv 0 (mod npq)  \Leftrightarrow$
$ npq |  n(pn^{q\minus{}1} \plus{} qn^{p\minus{}1} \minus{} p \minus{} q)$

(Δεν μπορούσα να γράψω τόσο καιρό γιατί είχα κόψει το χέρι μου.) Δεν είναι δύσκολη αλλά παρουσιάζει ενδιαφέρον σαν θεωρία.


Ring Theory
superior algebra
superior algebra unsolved

Prove or disprove that a subring of $ Z[x]$ that is generated by $ 3$ and $ x^2\plus{}1$ is an ideal in $ Z[x]$. Thanks in advance.

[quote="t0rajir0u"]Look through the definition of an ideal.  Is every condition satisfied?[/quote]
I don't know, please help me and describe it to me, thanks once again.

The definition of an ideal has a couple conditions.  Why don't you walk us through them and tell us where you get stuck?


trigonometry
ratio
limit
geometric sequence

[hide="Problem 1"]
I don't know if this is what he wants...

Let $S= r+2r^2+3r^3+...nr^n$.
$\frac{S}{r}=1+2r+3r^2+...+nr^{n-1}$.
$S\left(\frac{1}{r}-1\right)=1+r+r^2+r^3+...+r^{n-1}-nr^n$
$S\left(\frac{1-r}{r}\right)=\frac{r^n-1}{r-1}-nr^n$
$S=\left(\frac{r^n-1}{r-1}-nr^n\right)\left(\frac{r}{1-r}\right)$
[/hide]

[hide]1. Not sure how to make it more simple, but I got it down to this...
$r+2r^2+3r^3+...+nr^n=\frac{nr(r^n-1)}{r-1}-\sum_{k=1}^{n-1} \frac{k(k^n-1)}{k-1}$  I hope i made some dent in this one...
EDIT: nvm, looks like towersfreak got to it before i did...nice solution!

2. Looks messy...i'm not good with trig functions...

3. $\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$ is a factorization.  I know that for all the odd numbers, the n+1 and n-1 will cancel out, leaving you with something like 2/n.  Same thing with the even numbers, but this time 1/n.  In the other term, plugging $n+1$ into $n^2-n+1$ for n yields $n^2+n+1$.  So you'll have a huge cancellation there.  You'll end up with something like $\frac{n^2+n+1}{2^2-2+1}=\frac{n^2+n+1}{3}$ with n->infinity.  If we let n->infinity, then that expression tends to $\frac{n^2}{3}$.  We can completely disregard the 1 and the n as well because they're infinitesimal next to $n^2$ as n->infinity.  If we multiply, then this gives us
$\frac{2}{3}$
Hope that was clear...and correct  :D [/hide]

[quote="pkothari13"]I came across these problems in one of the math books I have, and I was stumped.

1. Find a formula for the sum $r+2r^2+3r^3+...nr^n,$ and generalize.

2. Evaluate the product $\displaystyle\prod_{k=0}^n \cos(2^k\theta)$

3. Evaluate the infinite product $\displaystyle\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}$[/quote]

The math book you are referring to is the Art and Craft of Problem Solving, right :D ?  I recognize the second two from there.

[quote="towersfreak2006"][hide="Problem 2"]
$\displaystyle P=\prod_{k=0}^n \cos(2^k\theta)=\cos(\theta)\cos(2\theta)\cos(4\theta)...\cos(n\theta)$
$P\sin(\theta)=\sin(\theta)\cos(\theta)\cos(2\theta)\cos(4\theta)...\cos(n\theta)$
$P\sin(\theta)=\frac{1}{2}\sin(2\theta)\cos(2\theta)\cos(4\theta)...\cos(n\theta)$
$P\sin(\theta)=\frac{1}{4}\sin(4\theta)\cos(4\theta)...\cos(n\theta)$
$P\sin(\theta)=\frac{1}{2^{n+1}}\sin(2^{n+1}\theta)$
$P=\frac{\sin(2^{n+1}\theta)}{2^{n+1}\sin(\theta)}$
[/hide][/quote]

You made some typos.  I believe you meant $\cos(2^n\theta)$ instead of $\cos(n\theta)$ for all those terms on the ends of the lines.

Oops. It's fixed now.  :blush:

Yeah, I think they're all from the algebra section. Here's another one:
Let $f(n)$ be the integer closest to $\sqrt[4]{n}$. Find $\displaystyle\sum_{k=1}^{1995}\frac{1}{f(k)}$.

I did 3) in a similar manner as K81o7 and got got the same answer, but theres a part that i don't quite understand.  Part of the solution involves finding $\prod_{n=2}^\infty \frac{n^2+n+1}{n^2-n+1}$. If you do the algebra, you see that the numerator of every term cancels out with the denominator of the next term, and we find the product to be $\frac{1}{2^2-2+1}=\frac{1}{3}$.  But clearly every term in this product is greater than 1, yet the product comes out to be less than one.  Doesn't this break sum kind of rule?  (It seems kind of like finding the sum of an infinite geometric progression with common ratio greater than one, which often gives a negative answer.)

The part on the top is always one ahead of the bottom...So in this case, rather than canceling out, it tends to a certain value.  That value would be $\frac{n^2+n+1}{3}$ with n going to infinity, because you can say that it cancels out for every term, but then the term after each one is left.  So n must go to infinity.  This can be simplified to $\frac{n^2}{3}$.  Then the other part yields $\frac{2}{n}\frac{1}{n}=\frac{2}{n^2}$ so then you can multiply to obtain $\frac{2n^2}{3n^2}=\frac{2}{3}$

hey i cancelled these terms $n+1=(n+2)-1$ and $(n+1)^2-(n+1)+1=n^2+n+1$ and what remains from product is $\frac{(2-1)(3-1)}{2^2-2+1}=\frac23$

Yeah, that's exactly what i did.  As you said, the n+1 in the denominator cancels with the n-1 in the numerator.  However, because it is a product, you can't just ignore the second degree terms (well you can here, because there are two, one on the top and one on the bottom, so they cancel).  Suppose you have $\displaystyle\prod_{n=2}^\infty\frac{n-1}{n+1}$?  Then, for odd numbers, you'd get a big cancellation, and you'd end up with $\frac{3-1}{n+1}$, while for even numbers, you'd get $\frac{2-1}{n+1}$.  Then you can let n go to infinity, and you would get 0 as a limit.  There are ways to write that expression which make this more obvious.  I suppose then you would have to write it as $\displaystyle\lim_{n \to \infty}\prod_{n=2}^\infty\frac{n-1}{n+1}$...no wait you wouldn't...only if it was one expression...don't quite know i guess either way works.


function
LaTeX
abstract algebra
real analysis
real analysis unsolved

As we have known: function $ f: \mathbb{R}-----\to \mathbb{R}$ is continuous on $ (a,b)$ then it doesn't generally follow that $ f$ is differential on $ (a,b).$ My question is if $ f$ is a.e differential on $ (a,b).$

Please, help me.

Thanks

Actually Weierstrass gave an example of a function that is continuous everywhere and differentiable nowhere.  Take a look at:

 [url]http://pirate.shu.edu/projects/reals/cont/fp_weier.html[/url]

Thank you very much. My friend asked me to give him an encouter example but I couldn't find out, and I wondered like that.

In the formula, there is $ \bigg(\frac{3}{4}\bigg)^{''}$. What does it mean, edoo? (It means time-number present?)

Thanks.

I just selected that link randomly.  The formula on that webpage should read as follows in $ LaTeX$ :

$ \sum_{n=0}^{\infty}\left( \frac{3}{4}\right)^{n}| \sin(4^{n}x)|$.

It of course is not the only such function, I believe Gelbaum and Olmstead has some more examples.

Put
$ f(x)=\sum_{n=0}^{\infty}\left( \frac{3}{4}\right)^{n}| \sin(4^{n}x)|$.

It's clear that $ f$ is bounded in $ \mathbb{R}.$

[u]First, $ f$ is continuous in $ \mathbb{R}:$[/u]

Let $ \{x_{m}\}\subset \mathbb{R}: x_{m}--\to  x, m--\to+\infty.$ We then have
 \[ |f(x_{m})-f(x)|=\bigg|\sum_{n=0}^{\infty}\left( \frac{3}{4}\right)^{m}| \sin(4^{n}x_{m})|-\sum_{n=0}^{\infty}\left( \frac{3}{4}\right)^{n}| \sin(4^{n}x)|\bigg|\]
\[ =\bigg|\sum_{n=0}^{\infty}\left( \frac{3}{4}\right)^{n}\big| | \sin(4^{n}x_{m})|-|\sin(4^{n}x)|\big|\bigg|\]

Applying Lagrange's theorem, we obtain the result.

[u]First, $ f$ is not differenriable in $ \mathbb{R}:$[/u] ?????????? Help me.

[quote="NguyenThanh_545"]Put

[u]First, $ f$ is not differenriable in $ \mathbb{R}:$[/u] ?????????? Help me.[/quote]

To be honest, I have not been able to prove this although it's right.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=150273]here[/url]. My post includes a version of the Weierstrass function, with proof that it works. The rest of the thread is also worth reading, with several different proofs of the existence of nowhere differentiable functions.

Dear jmerry,

Thank you very much.

Best wishes


ratio
induction
geometric sequence
arithmetic sequence

1. In the geometric sequence a sub n, Sn is the sum of the first n terms. If S sub3 = 9/2, S sub6 = 14/3, what is "a sub1" (the first term) and q, the constant ratio?

2. If the explicit formula for a sequence is "a sub n" = 2^n+2n-1, what is the formula for the sum of its first n terms?

3. There are four numbers. The first three numbers form an arithmetic sequence, and the last three numbers form a geometric sequence, and the sum of the first and fourth terms is 37, the sum of the second and third terms is 36... what are the four numbers?

4. Known: a,b,c,d form a geometric sequence (constant ratio = q)
    Prove: (a-d)^2=(b-c)^2+(c-a)^2+(d-b)^2

5. Known: {a sub n} is a geometric sequence, and S sub n is the sum of its first n terms, a sub1, a sub7, a sub4 form an arithmetic sequence
   Prove: 2S sub3, S sub6, S sub12 - S sub6 form a geometric sequence

PLLEEEEEEASE HELP IF YOU CAN. THEY ARE THE LAST PROBLEMS IN THE BOOK!!!

[quote="sush"]1. In the geometric sequence a sub n, Sn is the sum of the first n terms. If S sub3 = 9/2, S sub6 = 14/3, what is "a sub1" (the first term) and q, the constant ratio?[/quote]

$ S_{6}\minus{}S_{3}\equal{} q^{3}\cdot S_{3}$

$ \frac{14}{3}\minus{}\frac{9}{2} \equal{} q^{3}\cdot\frac{9}{2}$

$ q^{3}\equal{}\frac{1}{6}\cdot\frac{2}{9}$

$ q \equal{}\frac{1}{3}$

Now,

$ a_{1}\plus{}q\cdot a_{1}\plus{}q^{2}\cdot a_{1}\equal{}\frac{9}{2}$

$ a_{1}\cdot(1\plus{}q\plus{}q^{2}) \equal{}\frac{9}{2}$

$ a_{1}\cdot\frac{13}{9}\equal{}\frac{9}{2}$

$ a_{1}\equal{}\frac{81}{26}$

[quote]4. Known: a,b,c,d form a geometric sequence (constant ratio = q)
Prove: (a-d)^2=(b-c)^2+(c-a)^2+(d-b)^2 [/quote]

$ a\equal{}1\cdot a$
$ b\equal{}q\cdot a$
$ c\equal{}q^{2}\cdot a$
$ d\equal{}q^{3}\cdot a$

$ a^{2}\cdot(1\minus{}q^{3})^{2}\equal{}a^{2}\cdot(q\minus{}q^{2})^{2}\plus{}a^{2}\cdot(q^{2}\minus{}1)^{2}\plus{}a^{2}\cdot(q^{3}\minus{}q)^{2}$

$ q^{6}\minus{}2q^{3}\plus{}1\equal{}(q^{4}\minus{}2q^{3}\plus{}q^{2})\plus{}(q^{4}\minus{}2q^{2}\plus{}1)\plus{}(q^{6}\minus{}2q^{4}\plus{}q^{2})$

$ q^{6}\minus{}2q^{3}\plus{}1\equal{}q^{6}\minus{}2a^{3}\plus{}1$

Problem: Given $ a_{n}\equal{} 2^{n}\plus{}2n\minus{}1$, generalize $ S_{n}\equal{} a_{0}\plus{}a_{1}\plus{}\dots\plus{}a_{n}\equal{}\sum_{i \equal{} 0}^{n}{a_{i}}$. Note: While it is not stated, it doesn't matter whether or not $ a_{0}$ exists since $ a_{0}\equal{} 0$.

Solution:
[hide]
$ S_{n}\equal{}\sum_{i \equal{} 0}^{n}{a_{i}}\equal{}\sum_{i \equal{} 0}^{n}{2^{i}\plus{}2i\minus{}1}\equal{}\sum_{i \equal{} 0}^{n}{2^{i}}\plus{}\sum_{i \equal{} 0}^{n}{2i}\plus{}\sum_{i \equal{} 0}^{n}{\minus{}1}$

$ \equal{} (2^{n\plus{}1}\minus{}1)\plus{}[n(n\plus{}1)]\minus{}(n\plus{}1) \equal{}\boxed{S_{n}\equal{} 2^{n\plus{}1}\plus{}n^{2}\minus{}2}$
[/hide]

For details on the jump from summations to the explicit formula:
[hide]
Lemma 1: $ \sum_{i \equal{} 0}^{n}{2^{i}}\equal{} 2^{n\plus{}1}\minus{}1$
Proof of Lemma 1, by induction:
Base case n=0 is trivial. Then assuming $ \sum_{i \equal{} 0}^{n}{2^{i}}\equal{} 2^{n\plus{}1}\minus{}1$, we find $ \sum_{i \equal{} 0}^{n\plus{}1}{2^{i}}\equal{}\left(\sum_{i \equal{} 0}^{n}{2^{i}}\right)\plus{}2^{n\plus{}1}\equal{} 2^{n\plus{}1}\minus{}1\plus{}2^{n\plus{}1}\equal{} 2(2^{n\plus{}1})\minus{}1 \equal{} 2^{n\plus{}2}\minus{}1 \equal{}\sum_{i \equal{} 0}^{n\plus{}1}{2^{i}}$, and the induction is complete.

Lemma 2: $ \sum_{i \equal{} 0}^{n}{2i}\equal{} n(n\plus{}1)$.
Proof of Lemma 2, by induction:
Base case n=0 is trivial. Then assuming $ \sum_{i \equal{} 0}^{n}{2i}\equal{} n(n\plus{}1)$, we find $ \sum_{i \equal{} 0}^{n\plus{}1}{2i}\equal{}\left(\sum_{i \equal{} 0}^{n}{2i}\right)\plus{}2(n\plus{}1) \equal{} n(n\plus{}1)\plus{}2(n\plus{}1) \equal{} (n\plus{}1)(n\plus{}2) \equal{}\sum_{i \equal{} 0}^{n\plus{}1}{2i}$ and the induction is complete.
[/hide]

[hide="Problem 3"]
$ [a\plus{}d\minus{}1 \equal{} b\plus{}c],\; [a\plus{}c \equal{} 2b]\implies d \equal{} 2c\minus{}b\plus{}1$
$ db \equal{} c^{2}\implies d \equal{} c^{2}/ b$

$ \implies 2bc\minus{}b^{2}\plus{}b \equal{} c^{2}\iff b\plus{}4bc \equal{} (b\plus{}c)^{2}\iff b\plus{}4b(36\minus{}b) \equal{} 36^{2}$
This gives $ b \equal{} 16\;\text{or}\; b \equal{}\frac{81}{4}\implies c \equal{} 20\;\text{or}\; c \equal{}\frac{63}{4}$

We can now fill in the sequence:
$ (a,b,c,d) \equal{}\boxed{\left( 12,16,20, 25\right)}\;\text{or}\;\boxed{\left(\frac{99}{4},\frac{81}{4},\frac{63}{4}, 49\right)}$[/hide]


integration
calculus
calculus computations

[quote="kunny"]Is $ m$ integer or any real number?[/quote]
Sorry $ m$ positive integer number  :blush:

All right.

I have no time to solve it now, I will try it later.

My idea: Rewrite the integrand as

$ (x \plus{} my)(x \minus{} y)^m \equal{} \{x \minus{} y \plus{} (m \plus{} 1)y\}(x \minus{} y)^m \equal{} (x \minus{} y)^{m \plus{} 1} \plus{} (m \plus{} 1)y(x \minus{} y)^m$.

Integrate this with respect to $ x$, ...

Sorry I am lazy ,too . :(  It is  :blush:  ....

For any integer m$ \geq 0$

$ \int_{0}^{1}\int_{0}^{1}(x\plus{}my)(x\minus{}y)^{m}dxdy$

$ \equal{}\int_{0}^{1}\int_{0}^{1}x(x\minus{}y)^{m}dydx\plus{}m\int_{0}^{1}\int_{0}^{1}y(x\minus{}y)^{m}dxdy$

$ \equal{}\int_{0}^{1}\left[ \minus{}\frac{x}{m\plus{}1}(x\minus{}y)^{m\plus{}1}\right] _{y\equal{}0}^{1}dx\plus{}m\int_{0}^{1}\left[ \frac{y}{m\plus{}1}(x\minus{}y)^{m\plus{}1}\right] _{x\equal{}0}^{1}dy$

$ \equal{}\minus{}\frac{1}{m\plus{}1}\int_{0}^{1}\left[ x(x\minus{}1)^{m\plus{}1}\minus{}x^{m\plus{}2}\right] dx\plus{}\frac{m}{m\plus{}1}\int_{0}^{1}\left[ y(1\minus{}y)^{m\plus{}1}\minus{}(\minus{}1)^{m\plus{}1}y^{m\plus{}2}\right] dy$

$ \equal{}\minus{}\frac{\left( \minus{}1\right) ^{m\plus{}1}}{m\plus{}1}\int_{0}^{1}x(1\minus{}x)^{m\plus{}1}dx\plus{}\frac{1}{\left( m\plus{}1\right) \left( m\plus{}3\right) }\plus{}\frac{m}{m\plus{}1}\int_{0}^{1}y(1\minus{}y)^{m\plus{}1}dy\minus{}\frac{m(\minus{}1)^{m\plus{}1}}{\left( m\plus{}1\right) \left( m\plus{}3\right) }$

$ \equal{}\minus{}\frac{\left( \minus{}1\right) ^{m\plus{}1}}{m\plus{}1}B(2,m\plus{}2)\plus{}\frac{1}{\left( m\plus{}1\right) \left( m\plus{}3\right) }\plus{}\frac{m}{m\plus{}1}B(2,m\plus{}2)\minus{}\frac{m(\minus{}1)^{m\plus{}1}}{\left( m\plus{}1\right) \left( m\plus{}3\right) }$

$ \equal{}\minus{}\frac{\left( \minus{}1\right) ^{m\plus{}1}}{\left( m\plus{}1\right) \left( m\plus{}2\right) \left( m\plus{}3\right) }\plus{}\frac{1}{\left( m\plus{}1\right) \left( m\plus{}3\right) }\plus{}\frac{m}{\left( m\plus{}1\right) \left( m\plus{}2\right) \left( m\plus{}3\right) }\minus{}\frac{m(\minus{}1)^{m\plus{}1}}{\left( m\plus{}1\right) \left( m\plus{}3\right) }$

$ \equal{}\frac{1}{\left( m\plus{}1\right) \left( m\plus{}2\right) \left( m\plus{}3\right) }\left( m\minus{}\left( \minus{}1\right) ^{m\plus{}1}\right) \plus{}\frac{1}{\left( m\plus{}1\right) \left( m\plus{}3\right) }\left( 1\minus{}m(\minus{}1)^{m\plus{}1}\right)$

$ \equal{}\frac{1\minus{}m(\minus{}1)^{m\plus{}1}}{(m\plus{}1)(m\plus{}3)}\left[ 1\minus{}\frac{(\minus{}1)^{m\plus{}1}}{m\plus{}2}\right]$

I'm not sure if it's the same as what IndoChina wrote, but I got: $ \frac {2 \plus{} ( \minus{} 1)^m (m \plus{} 1)}{(m \plus{} 2)(m \plus{} 3)}$

Work:
Let $ u \equal{} x \minus{} y$ so that:
$ \int_0^1\int_0^1 (x \plus{} my)(x \minus{} y)^m dxdy \equal{} \int_0^1\int_{ \minus{} y}^{1 \minus{} y} (u \plus{} (m \plus{} 1)y)u^m dudy$
$ \equal{} \int_0^1\int_{ \minus{} y}^{1 \minus{} y} u^{m \plus{} 1} \plus{} (m \plus{} 1)yu^m dudy \equal{} \int_0^1 \frac {u^{m \plus{} 2}}{m \plus{} 2} \plus{} y u^{m \plus{} 1} \Bigg|_{ \minus{} y}^{1 \minus{} y} dy$
$ \equal{} \int_0^1 \left( \frac {(1 \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} y (1 \minus{} y)^{m \plus{} 1} \right) \minus{} \left( \frac {( \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} y( \minus{} y)^{m \plus{} 1} \right) dy$
Let $ v \equal{} 1 \minus{} y$ for the second term of this integrand to get:
$ \equal{} \frac { \minus{} (1 \minus{} y)^{m \plus{} 3}}{(m \plus{} 2)(m \plus{} 3)} \plus{} \frac {v^{m \plus{} 2}}{m \plus{} 2} \minus{} \frac {v^{m \plus{} 3}}{m \plus{} 3} \minus{} \frac {( \minus{} 1)^{m \plus{} 2}y^{m \plus{} 3}}{(m \plus{} 2)(m \plus{} 3)} \minus{} \frac {( \minus{} 1)^{m \plus{} 1}y^{m \plus{} 3}}{m \plus{} 3} \Bigg|_0^1$
$ \equal{} \frac {1}{(m \plus{} 2)(m \plus{} 3)} \plus{} \frac {1}{m \plus{} 2} \minus{} \frac {1}{m \plus{} 3} \minus{} \frac {( \minus{} 1)^m}{(m \plus{} 2)(m \plus{} 3)} \minus{} \frac {( \minus{} 1)^{m \plus{} 1}}{m \plus{} 3}$
$ \equal{} \frac {1 \plus{} (m \plus{} 3) \minus{} (m \plus{} 2) \plus{} ( \minus{} 1)^{m \plus{} 1} \plus{} ( \minus{} 1)^m(m \plus{} 2)}{(m \plus{} 2)(m \plus{} 3)}$
$ \equal{} \frac {2 \plus{} ( \minus{} 1)^m (m \plus{} 1)}{(m \plus{} 2)(m \plus{} 3)}$

That's correct answer.

$ (x \plus{} my)(x \minus{} y)^m \equal{} \{x \minus{} y \plus{} (m \plus{} 1)y\}(x \minus{} y)^m \equal{} (x \minus{} y)^{m \plus{} 1} \plus{} (m \plus{} 1)y(x \minus{} y)^m$,

$ \therefore \int_0^1 (x \plus{} my)(x \minus{} y)^m\ dx \equal{} \left[\frac {(x \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} (m \plus{} 1)y\cdot \frac {(x \minus{} y)^{m \plus{} 1}}{m \plus{} 1}\right]_0^1$

$ \equal{} \frac {(1 \minus{} y)^{m \plus{} 2} \minus{} ( \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} y\{(1 \minus{} y)^{m \plus{} 1} \minus{} ( \minus{} y)^{m \plus{} 1}\}$

$ \therefore \int_0^1 \int_0^1 (x \plus{} my)(x \minus{} y)^m\ dxdy$

$ \equal{} \frac {1}{m \plus{} 2}\int_0^1 \{(1 \minus{} y)^{m \plus{} 2} \minus{} ( \minus{} y)^{m \plus{} 2}\}dy \plus{} \int_0^1 y\{(1 \minus{} y)^{m \plus{} 1} \minus{} ( \minus{} y)^{m \plus{} 1}\}dy$

$ \equal{} \frac {1}{m \plus{} 2}\left[ \minus{} \frac {(1 \minus{} y)^{m \plus{} 3}}{m \plus{} 3} \plus{} \frac {( \minus{} y)^{m \plus{} 3}}{m \plus{} 3}\right]_0^1$

$ \plus{} \left[\left\{ \minus{} \frac {(1 \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} \frac {( \minus{} y)^{m \plus{} 2}}{m \plus{} 2}\right\}y\right]_0^1 \minus{} \int_0^1 \left\{ \minus{} \frac {(1 \minus{} y)^{m \plus{} 2}}{m \plus{} 2} \plus{} \frac {( \minus{} y)^{m \plus{} 2}}{m \plus{} 2}\right\}dy$

$ \equal{} \frac {1 \minus{} ( \minus{} 1)^m}{(m \plus{} 2)(m \plus{} 3)} \plus{} \frac {( \minus{} 1)^m}{m \plus{} 2} \minus{} \frac {1}{m \plus{} 2}\left[\frac {(1 \minus{} y)^{m \plus{} 3}}{m \plus{} 3} \minus{} \frac {( \minus{} y)^{m \plus{} 3}}{m \plus{} 3}\right]_0^1$

$ \equal{} \frac {1 \minus{} ( \minus{} 1)^m}{(m \plus{} 2)(m \plus{} 3)} \plus{} \frac {( \minus{} 1)^m}{m \plus{} 2} \minus{} \frac {1}{m \plus{} 2}\cdot \frac {( \minus{} 1)^m \minus{} 1}{m \plus{} 3}$

$ \equal{} \boxed{\frac {2 \plus{} (m \plus{} 1)( \minus{} 1)^m}{(m \plus{} 2)(m \plus{} 3)}}$

[quote="gauss202"]I'm not sure if it's the same as what IndoChina wrote, but I got: $ \frac {2 \plus{} ( \minus{} 1)^m (m \plus{} 1)}{(m \plus{} 2)(m \plus{} 3)}$[/quote]

IndoChina's answer $ \equal{}\frac{1\minus{}m(\minus{}1)^{m\plus{}1}}{(m\plus{}1)(m\plus{}3)}\left[1\minus{}\frac{(\minus{}1)^{m\plus{}1}}{m\plus{}2}\right]$

$ \equal{}\frac{\{1\plus{}m(\minus{}1)^m\}\{m\plus{}2\plus{}(\minus{}1)^m}{(m\plus{}1)(m\plus{}2)(m\plus{}3)}$

$ \equal{}\frac{m\plus{}2\plus{}\{1\plus{}m(m\plus{}2)\}(\minus{}1)^m\plus{}m}{(m\plus{}1)(m\plus{}2)(m\plus{}3)}$

$ \equal{}\frac{2(m\plus{}1)\plus{}(m\plus{}1)^2(\minus{}1)^m}{(m\plus{}1)(m\plus{}2)(m\plus{}3)}$

$ \equal{}\frac{2\plus{}(m\plus{}1)(\minus{}1)^m}{(m\plus{}2)(m\plus{}3)}$ is also correct.

Thanks very much ! [b]Kunny ,gauss202 ,IndoChina[/b]


Did anyone take any CTY distance courses in humanities?  How did the courses worked out?  Do they help you on the SATs?  I'm thinking of taking crafting the essay or crafting the essay bilingual.  Is the biligual one for ESL students?

Thanks

I've taken Crafting the Essay and Analysis/Persuasion.  They were very good courses and I think they would help you on the SAT Essay.  I don't know about the bilingual version.

according to the description, the bilingual one is for esl students.

does any one have any recomendations for the santa cruz summer CAA courses?

[quote="#H34N1"]Its a scam. I've done it before.[/quote]

I disagree.  I am taking Crafting the Essay Webased right now and I have taken Honors Algebra II, Honors Precalculus and am currently taking AP Calculus AB from JHU CTY.  If you get a good teacher (I have great, very helpful teachers, my english teacher was a Fulbright scholar and my math teacher is wonderful--she has a B.S. and M.S. from Northwestern too)  If you want more info please pm me.

~lyra

I wasn't particularly impressed by the CTY courses either, actually.

i do NOT like the CTY distance education courses, but i LOVED their summer programs...

i hope to return this year at Lancaster Session 2 at Franklin and Marshall's College...
it the BEST!!!!!!

i definitely recommend that..


trigonometry geometry calculus integration

How do you find the rule for x for the equation sin(x)>cos(x)?  This was on math test a few months ago and [hide]I drew the graphs of sine and cosine and showed which was greater where, but I never found an actual rule.

The next step would be to find the intersections of the graphs, by solving sin(x)=cos(x), tan(x)=1, and then x = pi/4 + kpi are the intersections.

From this you can see that it is the intersections that are the areas where sin(x)>cos(x).  From k=0 to k=1, then from k=2 to k=3, then from k=4 to k=5 and so on.  How do you state this though?  I mean you using something like 3<x<4 or (- :inf: , 5]U[10,+ :inf: ) <--just random examples.[/hide]

Please tell me if there is any mistakes in what I have already done.

That looks right to me.  What you really want are the regions between every even number and the odd number directly following it, right?  One could write that like this:
[tex]\bigcup_{m\in\mathbb{Z}}(\pi/4+2m\pi,\pi/4+(2m+1)\pi)[/tex].  However, your way of describing it is perfectly valid -- it's okay to use words to say that, and it is possibly clearer that way, too.

Wow JBL thanks for the quick response.  Thats really cool how you got it to represent only even,odd.  Sortof simple but I coulden't think of it.  One quick question, what does UmeZ.  I know U=union, e=element of,Z probably represents all integers (not sure on this one), and m I have no ideal.  Can you tell me how to use this notation?  Thanks!

In the notation [tex]\displaystyle\bigcup_{m\in\mathbb{Z}}[/tex], the U-like symbol means union, m is an index variable (i.e. we just let it run over all integers, from -:inf: up through :inf:).  You may have seen a similar notation used for summation

[tex]\displaystyle\sum_{m=1}^\infty a_m[/tex],

which means that we let m run over all positive integers. An equivalent way of writing this would be

[tex]\displaystyle\sum_{m\in\mathbb{Z}^+} a_m[/tex],

where [tex]\mathbb{Z}^+[/tex] is used to denote all positive integers.

Okay, sure.  You got the basic idea of what I'm doing.  The point is that if x is in an interval which looks like [tex]\displaystyle (\pi/4 + 2\pim, \pi/4 + (2m + 1)\pi)[/tex], where m is any integer, then our conditions will be satisfied.  The notation I was using, [tex]\displaystyle \bigcup_{m\in\mathbb{Z}}[/tex], means literally "The union over all integers m."  m is acting as a dummy variable here -- we take m = 0, 1, 2, ..., -1, -2, ..., and each of those gives us an interval.  Then, we take the union of all of those intervals.  This is very similar to the use of dummy variables in definite integrals or in summation (sigma) notation.  For example, have you ever seen a formula like this:

[tex]\displaystyle \sum_{i=1}^ni=\frac{n(n+1)}{2}[/tex]?

All the i tells you is that you plug in the values 1, 2, 3 and so on all the way up to n, and the sigma tells you to add them all up.  Similarly, in what I wrote, the m tells you to plug in the values of all of the integers, and then the U tells you to take their union.

Really, this is just notation -- you solved the problem perfectly, and this is just about finding a short-hand way so that math people will understand what you mean easily without too much writing.  Another way that might seem more familiar to you is to say, "sin(x)>cos(x) if and only if there exists some integer m such that [tex]\displaystyle \pi/4 + 2m\pi < x < \pi/4 + (2m + 1)\pi[/tex]."  This sentence, while a bit wordier, is probably more clear and has exactly the same meaning.

:Edit: I see Simon was thinking along the same lines as me, and is also better at Texing than I :)

Thanks again.  Isn't the difference between these two notations that U means a solution set and sigma means a sum?

Right -- the U versus the :Sigma: changes the operation you are carrying out -- U for union, :Sigma: for addition -- but the meaning of the little notations underneath is the same, namely that we keep plugging in values for the "dummy variable" and then carry out the appropriate operation (either unioning or adding, in this case) on all the results.

Oh, my bad.  I knew that! =)   The UmeZ looked sortof foreign so I didn't think that m was the value you plug in.  Also, nvm my last question.  I see the similarity between U and sigma, not that they mean the same thing though.


plz solve this probability question.......

ya i know that......but not getting.....can u plz solve it... :)


topology
real analysis
real analysis unsolved

A Peano space $E$ is a compact, connected, locally connected metric space. Prove that a Hausdorff space is the continuous image of the unit interval $[0;1]$ if and only if it is a Peano space.

One implication, namely image of $[0,1]\ \Rightarrow$ Peano space, can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hahn&t=34104]here[/url], in post #14 (or post #4 counting from the end of the page; it's easier that way :)).

At some point I thought I had a proof for the other implication as well, but while trying to type it in that topic I decided it was too lengthy, so I gave up :).


Which is greater,

$ m^n \quad \text{or}\quad n! \cdot \binom{n\plus{}m\minus{}1}{n}$?

$ m,n\in\mathbb{N}$.

The first one represents the number of ways to distribute $ n$ distinguishable objects to $ m$ distinguishable objects.

The second one represents the number of ways to distribute $ n$ indistinguishable objects to $ m$ distinguishable objects. However, it then distinguishes every permutation of the $ n$ indistinguishable objects in their positions.

Now, note that for every arrangement of the first one, once the objects are selected to be in certain positions, they can't be permuted. But in the second one, they can be permuted, so $ n!\binom{n\plus{}m\minus{}1}n>m^n$ as long as $ n\ne1$.

For the second expression, it could also be interpreted that the n objects are distinguishable while the number of permutations of k ($ 0\le k\le n$) objects being distributed to any of the m objects are also counted, but all the number of permutations are only counted once in the first expression.


conics
parabola
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14344[/img]

 :(

Easy to show that $ \angle BAH_a\equal{}\angle OAC$, so $ AT_a$ is the bisector of $ \angle OAH_a$. Then, OA touches the circle whose center Ta and radius TaHa. The perp line from Ha to HaTa meets the tangent from O of the circle $ (T_a; T_aH_a)$ at A. Draw a circle with center O and radius OA meets the line HaTa at B,C. Done!

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14373[/img]

 :thumbup:

Since $ AT_a$ bisects the angle $ \angle HAO \equal{} \angle H_aAO,$ $ AT_a$ is tangent to a parabola with focus $ O$ and directrix $ h_a \perp T_aH_a$ at $ H_a.$ The parabola vertex tangent $ v \perp T_aH_a$ goes through the midpoint of $ OH_a.$ $ v$ is locus of the feet of perpendiculars from the focus $ O$ to all parabola tangents. Circle with diameter $ OT_a$ cuts $ v$ at $ F_1, F_2$ and $ T_aF_1, T_aF_2$ cut $ h_a$ at $ A_1, A_2,$ the 2 possible solutions. Circles with center $ O$ and radii $ OA_1, OA_2$ then cut the line $ T_aH_a$ at $ B_1, C_1$ and $ B_2, C_2,$ respectively.


geometry
variety
analytic geometry
algebra
polynomial
advanced fields
advanced fields unsolved

suppose $ h \equal{} (h_{1}\cdots h_{n})$ is a regular function from $ k^{m}\rightarrow k^{n}$ (k is algebra closed) $ V \subseteq k^{m}$ is a variaty.
Give an example that $ h(V) \equal{} W$ is not a variety in $ k^{n}$.

I think there isn't a such example.  :roll: How to prove it?

What is your meaning of variety? Anyway, this maybe what you are looking for
{xy = 1} in k^2 maps to k by projecting onto first coordinate.

by "variety" I mean algebraic variety, the common zero points of a set of polynomials. My question is, for example, $ k\equal{}C$ m=1 n=2 $ h\equal{}(t, t^2)$ this map maps $ C$ into the "whole" zero points of $ x^2\minus{}y$. But is there an example, in which $ h$ maps $ k^m$ into a "partial" algebraic variety.

So this maybe what are looking for:
$ \{(x,y): xy \equal{}1 \} \subset \mathbb{C}^2$ mapping to the first coordinate, i.e. $ (x,y) \rightarrow x$. The image is $ \mathbb{C} \minus{} \{0\}$, which is not a subvariety of $ \mathbb{C}$.

but $ (x,y)\rightarrow x$ isn't regular. I need $ h\equal{}(h_1\cdots h_n)$ where all $ h_i$ is polynomial ( resp affine space) or polynomial of same degree (resp proj space). Besides, $ C$ is itself a variety, I need the image not being a variety.

And what leads you to the conclusion that $ (x,y) \mapsto x$ isn't a polynomial¿

sorry for this mistake. But, still, $ C$ is variety, not a "partial" variety. I think I get a kind of examples. $ h: k^2\rightarrow k^2$  $ h\equal{}(yf(x),y^n)$ whose image is $ k^2\bigcup {(0,0)}\minus{}k*{0}$. It now seems trivil. Maybe I haven't make my statement clear.

by the way, I don't know how my "direction" change from PvsNP to Hodge Conjecture, and forget how to reset it, somebody would like to help?  :oops: (thanks :) )


Why isn't the original equilibrium of a chemical reaction altered if any of the species ,which is solid, is added?

I think its just that the effects are so negligible they can be ignored.  I might be wrong though.

When we write equilibrium expressions we consider the concentration of each species. Think about it, the amount of solid in a specific volume is its density and hence a constant. Therefore it doesn't matter how much solid you add it's going to be a constant and thus not affecting the actual equilibrium.

The more accurate reason is that the activity of solids do not change.


inequalities
function
calculus
derivative
calculus computations

How would one prove the Arithmetic Mean-Geometric Mean Inequality using Lagrange Multipliers?

Let $ a_{i}$ be a nonnegative sequence $ a_{1}, a_{2},\ldots, a_{n\minus{}1}, a_{n}$.

Let $ f(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n}) \equal{} (a_{1}*a_{2}*\cdots*a_{n\minus{}1}*a_{n})^{\frac{1}{n}}$ be the function to be minimized.

Let $ g(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n}) \equal{}\frac{a_{1}\plus{}a_{2}\plus{}\ldots\plus{}a_{n\minus{}1}\plus{}a_{n}}{n}$ and $ g(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n}) \equal{} c$, where $ c$ is some number be the constraint.

How would I proceed?  Or is this the wrong approach?  My work has only led to dead ends... maybe I'm not seeing some trick?

I know one has to find $ (a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n})$ such that:

$ \nabla f(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n}) \equal{}\lambda\nabla g(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n})$ and $ g(a_{1}, a_{2}, ..., a_{n\minus{}1}, a_{n}) \equal{} c$

Is there supposed to be a maximum of $ f$ when all the terms of the sequence $ a_{1}, a_{2},\ldots, a_{n\minus{}1}, a_{n}$ are equal and only when they are all equal to each other?  How would I show that?

Thanks for any help!

As an operational hint, it's a little easier if you define $ f\equal{}\prod_{k\equal{}1}^{n}a_{k}\equal{}a_{1}\cdot a_{2}\cdots a_{n}.$ That is, leave off the $ n$th root. After all, you maximize the product if you maximize the $ n$th root of the product. Similarly, I wouldn't bother dividing by $ n$ in the definition of $ g.$ Just let $ g\equal{}a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}.$

Now let's look at the partial derivative with respect to $ a_{k}:$

$ a_{1}a_{2}\cdots a_{k\minus{}1}a_{k\plus{}1}\cdots a_{n}\equal{}\lambda$

Multiply by $ a_{k}$ to get

$ a_{1}a_{2}\cdots a_{n}\equal{}\lambda a_{k}$

And now note that this is true for all $ k,$ and that that left hand side is the same for all $ k.$ Hence the only critical point we can find (within the positive numbers, which forces $ \lambda > 0$) has $ a_{1}\equal{}a_{2}\equal{}\cdots\equal{}a_{n}.$


Topic.  I have a class in a few minutes, so how do I open up the window?  Please help!

The classroom does not open until 10 minutes before the class is scheduled to start.  Try again in a few minutes.

Okeydoke, but which page doth it open up in, My Classes > Intro to Geometry or Classroom?

Click on "Classroom" on the left side of the screen.

It should be open right now.


geometry proposed
geometry
U sviti matematyky

Squares $ BCC_1B_2,$ $ CAA_1C_2,$ $ ABB_1A_2$ are constructed from the outside at sides of triangle $ ABC$ and $ O_A, O_B, O_C$ are the centres of these squares. Let $ A_0, B_0, C_0$ be the intersection points of the straight lines $ A_1B_2$ and $ C_1A_2,$ $ A_1B_2$ and $ B_1C_2,$ $ C_1A_2$ and $ C_1A_2$ respectively. Prove that the straight lines $ O_AA_0,$ $ O_BB_0$ and $ O_CC_0$ are concurrent.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme vol. 5  p.17
Sincerely
Jean-Louis


Let  a,b,c,A,B,C   be  positive   real  numbers  and   a+A=b+B=c+C=k
Show  that     aA + bB + cC < k^2

This is a direct consequence of the Arithmetic-Geometric Mean inequality.

We can change $ k^2$ to $ \frac{3k^2}{4}$, right?

By AM-GM, $ k\equal{}a\plus{}A \ge 2 \sqrt{aA}, so \frac{k^2}{4} \ge aA$.
Adding this up cyclically, $ \frac{3k^2}{4} \ge aA\plus{}bB\plus{}cC$.


I have two questions how can I put constant text in the footer.  And how can I decide how to flush it to the center, left or right. I am trying to do the following:

blabla ... current page number / number of pages in the document

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]Give this a try.[/url]


AMC AIME logarithms algebra polynomial function geometry

Which chapter in AoPS2 doesn't cover in the AIME test?
I feel like there are many useless concept in the AoPS2 for AIME.

Please note this post comes from someone who does not own AOPS2, just looked at the table of contents posted online.

The following chapters are covered by AIME tests, unless their name is misleading.
1. Logarithms
2. Not Just for Right Triangles
3. More Triangles
4. Cyclic Quads
6. Polynomials
7. Functions
9. Complex Numbers
12. Analytic Geometry
13. Equations and Expressions
14. Inequalities
15. Combinatorics
16. Sequences and Series
17. Counting in the Twilight Zone
19. Probability
23. Number Theory
24. Diophantine Equations

The following are either (1) probably going to rarely be helpful, but it probably would help to learn this stuff or (2) I'm not exactly sure. (i.e. I dumped a lot of geometry stuff in here because I suck at geometry and still am not exactly sure what's useful)
10. Vectors and Matrices
11. Cross Products and Determinants
18. Again and Again
20. Find it and Make it
21. Collinearity and Concurrency
22. Geometry Tidbits

That ended up being the majority of the chapters. That said, there are probably sections in each of those chapters that won't help.

The most important thing to note is that [b]all[/b] of AOPS2 is probably worth reading. There's a lot more to math than the AIME and Olympiads, and I think it is a mistake to learn math only for them. In fact, you probably will do better by broadening your horizons.


limit
modular arithmetic
induction
algebra unsolved
algebra

1)Let $ a_0 \equal{} 0,a_1 \equal{} 1, a_n \equal{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 3}$ ($ n > 1$)
Prove that $ 2^k|a_n\Leftrightarrow 2^k|n$
2) Let $ a_1\equal{}a,a_2\equal{}b, a_{n\plus{}2}\equal{}\frac{a_{n\plus{}1}\plus{}a_n}{2}$ ($ n>1$)
Find $ \lim_{n\to\infty }a_n$

First, from $ x_{n}\equal{}2x_{n\minus{}1}\plus{}x_{n\minus{}2}$ we have $ x_{n} \equiv x_{n\minus{}2} \pmod {2}, x_0\equal{}0, x_1\equal{}1 \Rightarrow x_n \equiv n \pmod {2}.$
We will show by induction after $ k$ that $ x_{n\plus{}k}\equal{}x_{k\plus{}1}x_{n}\plus{} x_{k}x_{n\minus{}1}$ for $ n \in N$. The equality holds for $ k\equal{}1$. Now suppose it's true for  $ i \le k$, then $ x_{n\plus{}k\plus{}1}\equal{}2x_{n\plus{}k}\plus{}x_{n\plus{}k\minus{}1}\equal{}2(x_{k\plus{}1}x_{n}\plus{}x_{k}x_{n\minus{}1}) \plus{}x_{k}x_{n}\plus{}x_{k\minus{}1}x_{n\minus{}1} \Rightarrow x_{n\plus{}k\plus{}1}\equal{}(2x_{k\plus{}1}\plus{}x_{k})x_{n}\plus{}(2x_{k}\plus{}x_{k\minus{}1})x_{n\minus{}1}\equal{} x_{k\plus{}2}x_{n}\plus{}x_{k\plus{}1}x_{n\minus{}1}.  q.e.d.$
Now if we choose $ k\equal{}n$ it follows that $ x_{2n}\equal{}x_{n\plus{}1}x_{n}\plus{}x_{n}x_{n\minus{}1}\equal{} x_{n}(x_{n\plus{}1}\plus{}x_{n\minus{}1})\equal{}x_{n}(2x_{n}\plus{}2x_{n\minus{}1})
\equal{}2x_{n}(x_{n}\plus{}x_{n\minus{}1})$. But $ x_{n}\plus{}x_{n\minus{}1}\equal{}2m\plus{}1, m \in N$, so $ x_{2n}\equal{} 2(2m\plus{}1)x_{n}$. Now if $ n\equal{}2^{k}a$, with $ 2 \nmid a$, we obtain by recursion that $ x_{n}\equal{}2^{k}(2t\plus{}1)x_{a}$, but $ 2\nmid a \Rightarrow 2\nmid x_{a} \Rightarrow 2^{k} \mid x_{n}$, and
$ 2^{k\plus{}1} \nmid x_{n} \Rightarrow 2^{k} \mid x_{n} \Leftrightarrow 2^{k} \mid n$ :cool:

[quote="kunny"]$ \lim_{n\to\infty} a_n \equal{} \minus{} (a \plus{} 2b)$.[/quote]
It's wrong. Try $ a_1\equal{}a_2\equal{}1$. We will have $ a_n\equal{}1$ for all $ n$, and $ \lim_{n\to\infty} a_n \equal{} 1 \neq \minus{}(1\plus{}2)$

You are right. The right answer is $ \lim_{n\to\infty} a_n\equal{}\frac{2a\plus{}b}{3}$. :D

1) You should write $ n\geq 3$.
2) You should write $ n\geq 1$.

2) Answer. $ \lim_{n\to\infty} a_n=\frac{a+2b}{3}$.

$ 2a_{n + 2} - a_{n + 1} - a_n = 0\Longrightarrow a_{n + 2} + \frac {1}{2}a_{n + 1} = a_{n + 1} + \frac {1}{2}a_n$.

$ \therefore a_{n + 1} + \frac {1}{2}a_n = a\2 + \frac {1}{2}a_1 = \frac {a + 2b}{3}$.

$ \therefore a_{n + 1} - \frac {a + 2b}{3} = - \frac {1}{2}\left(a_n - \frac {a + 2b}{3}\right)$.

$ \therefore a_n - \frac {a + 2b}{3} = \left( - \frac {1}{2}\right)^{n - 1}\left(a\1 - \frac {a + 2b}{3}\right)\longrightarrow 0\ (n\to\infty)$.


function
limit
calculus
derivative
topology
set theory
real analysis

Assume $ f(x)$ is continuous in the open set $ (a,b)$,

for $ \forall$ $ x \in (a,b)$

we have: $ \lim_{\overline{h \rightarrow 0^\plus{}}}\frac{f(x\plus{}h)\minus{}f(x\minus{}h)}{h}\geq0$

show that: $ f(x)$ is monotonic-increasing on the open set $ (a,b)$

If you can, i'd suggest you to read

Boas R.P.: "A primer of real functions" The
Carus mathematical monograph, number thirteen, The mathematical association of America,
John Wiley and Sons Inc. New York 1960, pp.122-123

and

Erdos P.: "Some remarks on Set Theory" The Annals of Mathematics, Second Series,
Volume 44, Issue 4, (Oct.,1943), 643--646

Otherwise I can try to provide a proof taken from the given refernces.

[quote="Xingyuan"]Assume $ f(x)$ is continuous in the open set $ (a,b)$,

for $ \forall$ $ x \in (a,b)$

we have: $ \lim_{\overline{h \rightarrow 0^ \plus{} }}\frac {f(x \plus{} h) \minus{} f(x \minus{} h)}{h}\geq0$

show that: $ f(x)$ is monotonic-increasing on the open set $ (a,b)$[/quote]

[quote="Arrigo-Sacchi"]If you can, i'd suggest you to read
Boas R.P.: "A primer of real functions" The
Carus mathematical monograph, number thirteen, The mathematical association of America,
John Wiley and Sons Inc. New York 1960, pp.122-123

and

Erdos P.: "Some remarks on Set Theory" The Annals of Mathematics, Second Series,
Volume 44, Issue 4, (Oct.,1943), 643--646

Otherwise I can try to provide a proof taken from the given refernces.

[/quote]

Thank you very much :lol:
But I am at home,so i don't have a library .These books I can't find.... :( ...Could I download them from internet? (I have searched in "Google",but can't find the book that can be download)
Is the books your mentioned above discuss the problem like I ask?

[quote="Arrigo-Sacchi"]actually the theorems I alluded for are based on the Dini-derivatives

Boas at page 123 says

[i] If $ f$ is continuous and one of the Dini derivatives is nonnegative except for a countable number of points, it follows that $ f$ is nondecreasing[/i]

I apologize for my inaccuracy in the previous post.
[/quote]

Yes, It's Dini derivatives.....thanks very much...

[quote="yzymat"]I  have  wrote   down  a  proof   some  months   ago ,  which  was  simpler   than  SCIbird's

http://old.math.org.cn/[/quote]

haha......I can't find it,  your ID is?

When that bbs closed, I come to here.... :D

[quote="yzymat"][quote="Arrigo-Sacchi"]actually the theorems I alluded for are based on the Dini-derivatives

Boas at page 123 says

[i] If $ f$ is continuous and one of the Dini derivatives is nonnegative except for a countable number of points, it follows that $ f$ is nondecreasing[/i]

I apologize for my inaccuracy in the previous post.

[quote="Xingyuan"]Assume $ f(x)$ is continuous in the open set $ (a,b)$,

for $ \forall$ $ x \in (a,b)$

we have: $ \lim_{\overline{h \rightarrow 0^ \plus{} }}\frac {f(x \plus{} h) \minus{} f(x \minus{} h)}{h}\geq0$

show that: $ f(x)$ is monotonic-increasing on the open set $ (a,b)$[/quote][/quote]

P123  ?  I can't  find  the  theorem.

the book  can  be  download   from    http://ifile.it/5dztca/a_primer_of_real_functions_088385029x.djvu

to  Xingyuan:  my  ID  is  he_2100[/quote]

Thank you for the site, very interesting.  The book downloadable is the enlarged version of H.P.Boas (the son)  while the book in my post was of the father (dead) R.P.Boas. In any case the theorem is at page 150 of the enlarged version and 123 of the old version.


function
probability
inequalities
probability and stats

[quote]A bank accepts rolls of pennies and gives 50 cents credit to a customer without
counting the contents. Assume that a roll contains 49 pennies 30 percent of
the time, 50 pennies 60 percent of the time, and 51 pennies 10 percent of the
time.[/quote]
Using the distribution function, of X for each roll, m(-1) = 0.3, m(0) = 0.6 and m(1) = 0.1:
E(X) = -1*0.3 + 0*0.6 + 1*0.1 = -0.2
V(X) = 0.36

Now: Estimate the probability that the bank will lose more than 25 cents in 100 rolls.

Let $ S_{100}$ be the sum of 100 rolls. We want to estimate
$ P(S_{100} < - 25) = P( - 101 < S_{100} < - 25)$ $ = P \left(\frac { - 101 + 20}{6} < \frac {S_{100} + 20}{6} < \frac { - 25 + 20}{6}\right)$$ \Large{\textasciitilde}$ $ \Phi( - 5/6) - \Phi( - 13.5)$

Is this correct? |-13.5| = 13.5 feels like a too large number.

I wouldn't bother with the left inequality: the normal approximation doesn't really hold that far from the mean anyway. Just take the entire left tail and that's it. On the other hand, it is completely harmless because $ \Phi(\minus{}13.5)\equal{}0$ for all practical purposes. The rest seems correct, provided that you made no error in computation of $ E$ and $ V$, which I was too lazy to check :).


i'm assuming these are standard 6 sided dice, non-weighted, numbered in the usual manner, etc.

[hide]the only possible equivalent fractions are 1/2=2/4=3/6, 1/3=2/6, and 2/3=4/6, and their opposites (2/1=4/2=6/3, etc.). The only other way to have equivalent fractions are to actually have the same fraction. Ok, the probability that we have the exact same fraction on both sides is 1/36, since the top dice and the bottom dice on both sides must be equal. Now, the probability to have four specific numbers will be 1/6^4, so we multiply that 1/6^4 by 20. This comes from seeing that we can have 3 possible equivalences from 1/2=2/4=3/6, plus 2 more from the other equations. Now, we can turn these backwards (2/4=1/2 for ex.) or upside down (2/1=4/2) or both, making us multiply by 4. Alright, so the final answer should be 20/6^4+1/6^2=56/6^4=7/162. Correct?[/hide]

[color=#F090F0]This question is a little ambiguous:  do you get to roll the four dice and then choose which go where in the fractions?  Because that was how I interpreted it, but Mystic did something else.[/color]

oh. um, i assumed that the dice were already fixed in a set place as to which one was the numerator/denominator of the right/left hand side.

[color=#C0A0FF]Well, following Mystic, we can just list all quadruples that work.
1/2 = 2/4 = 3/6
We can have (1, 2, 2, 4), (1, 2, 3, 6), or (2, 3, 4, 6).  The first gives us 12 possibilities, the second gives us 24 and the third gives us 24.

1/3 = 2/6 gives us (1, 2, 3, 6) and 2/3 = 4/6 gives us (2, 3, 4, 6), both of which we've already counted.

Now we are left with all the rolls (a, a, b, b).  If a = b, there are 6 possibilities.  If a != b, we have 6 choices for a and 5 for b.  Then we must multiply by 6 for choosing which dice get which numbers, and divide by 2 for interchanging a and b.  This gives us 90 arrangements here.  All told, we have
(12 + 24 + 24 + 6 + 90) = 156 ways that work out of 6^4 total.  Dividing, we find the simplified probability of rolling numbers on four dice that can be arranged into two equal rational numbers to be 13/108.[/color]


Due to that I can't seem to add the Latex to my website now, I'll post the problems here since some questions need Latex format.

Just for this round, you MAY send the answers by Art of Problem Solving. If you pass the round, I'll send you out that you have been passed with the ID and password for the my website. If you did not pass, well, you won't get registered by me. And you're out of the run.

1.

$\frac {\sqrt 2}{2} + \frac {\sqrt 3}{3} + \frac {\sqrt 4}{4} + \frac {\sqrt 5}{5} + \frac {\sqrt 6}{6} + \frac {\sqrt 7}{7} + \frac {\sqrt 9}{9}$ is equal to what? Write your answer in two terms in which first term consist a simplified fraction with radical.

2.

$4^2*3^2*2^2 = 576 $
$5^2*4^2*3^2*2^2 = 14400$

Then what's:

$8^2*7^2*6^2*5^2*4^2*3^2*2^2$

3. Factor this:

$(x^2+2x+1)(x^2+5x+6)$

4. A polygon has sum of angles 17460. What is the sum of its exterior angles?

5. Calculate:

$\frac {(99999-99998)(99999+99998)}{(99998-99999)(99998+99999)}$

Calculator is not allowed on any of the following five questions.

Good luck!

Question 3 is a bad question, e.g., it doesn't make sense nor is the solution you're looking for (I assume a-bi) necessarily correct.

It is OKAY to send me answers through AoPS now.

If you have answers but didn't send them in, check and do everything you want and please send.

I really appreciate it.

Silverfalcon

Oh, and for the multiplication question, it's going to be ugly, because you neither of the quantities you posted have any factors of 7 in them.

I don't see any rule saying you can't just multiply with pencil and paper.

That's not the point I was trying to make in that question.

Look this way:

as you numbers get larger, you'll going to have a larger product. For operation, 1^2*2^2 --> you can just do 1^2 = 1 and 2^2 = 4 and then do it.

But what if you have like this?

45^2*44^2.......1^2? Would you do 45^2, 44^2 then do the same thing till 1?

No, there is an easier way and since it's round 1, I'll post it.

$a^2*b^2*c^2*d^2 = (a*b*c*d)^2$

Plus, there is no reason why you can't do with hand.

Wait. I'm sort of confused.

[quote]Yes, but 8^2*7^2*6^2 != 4(4^2*3^2*2^2)[/quote]

Explain please.

What I mean is that there's no easy way to get to the answer from what you posted.

So, you gave the value of $2^2\cdot 3^2 \cdot 4^2\cdot 5^2$.  However, there is no easy way to find the value of $2^2\cdot 3^2 \cdot 4^2\dotsc 7^2\cdot 8^2$, because you need to multiply by 6^2,7^2, and 8^2 (e.g., none of the quantities you gave have any factors of 7 in them, but the last one does)

I see.. But look here... In that question, concept was what I was aiming at, not the easy way to solve the problem. I know it will be a lot calculating but well, I guess you have to go through it. I try not to make a problem that in future or if I do, I'll let the calculator allowed.

I'm not still letting calculator because it's easily solve with calculator even you don't understand the concept.

Thanks for pointing out, though. ;)  I appreciate your point.

Silverfalcon

Yes. You need to send the solutions.
-----------------------------------------------
Since the number one's most tricky, I'll give some approaches.

We know this:

$\frac {\sqrt a}{a} = \frac {\sqrt a}{\sqrt a * \sqrt a}  $

Which leads us to:

$\frac {1}{\sqrt a}$

And by getting this, you should be able to have common denominators and that should give you two kinds of terms.

 :) I believe this is fairly good question.

[quote]How simple does number one have to be? Will two fractions work as long as they're both common?[/quote]

Do you mean if they're simplified? If that's what you're asking, yes.

 :cool: I think you're getting the answer, RC-7th.

ANSWER MUST BE TURNED TO ME BY THIS FRIDAY!!!!

I ONLY RECEIVED ONE!!!!

IF I DON'T GET ANY MORE, THE ONE WHO SENT ME WILL BE THE WINNER!

EASILY~~~

PLUS, FOR ONES DON'T KNOW WHERE THE QUESTION IS, IT'S IN THIS THREAD!!!

GO TO FIRST PAGE!

Did you receive mine? I PM'd it to jwec05 on SilverFalcon's forum on Wednesday.

Yes. I just checked it and received your answer.

So, I have only two solutions now. MOre!!

DEADLINE IS TOMORROW!!!


Let's call an arbitrary triangular prism $ABCDEF$. There are real numbers $a_1, a_2, a_3, b_1, b_2, b_3$ assigned to each vertex of the prism. If any one of those numbers is equal to the arithmetic mean of the three numbers in the adjacent vertices, prove that $a_1 = a_2 = a_3 = b_1 = b_2 = b_3$

[hide="Hint"]Consider the number that is greater than or equal to the rest of the numbers...

Also, consider the number that is less than or equal to the rest of the numbers...[/hide]

[hide="Different Hint"]A graph also trivializes this[/hide]

i had a different hint in mind...

[hide]
actually write down equations and solve it out
[/hide]


四边形ABCD外切于圆O,BD交圆O于P,Q两点,M为PQ中点,求证:$\angle{AMD}=\angle{CMD}$
其实是凭印象打的,奉劝各位仁兄:最好在证之前用几何画板验证一下(本人不会用几何画板 :blush:  :oops: )

dingdongdog, 想借你這個topic問個問題: 如何證明(空間)五點共球面?

[quote="dingdongdog"]四边形ABCD外接于圆O,BD交圆O于P,Q两点,[/quote]
BD都在圆O上，怎么BD又交圆O于PQ两点呢？    :mad:  :?:

这是蝴蝶定理的配极命题.


The answer would obviously be yes as long as they pick some # from every state :P

Woah there are a lot of nsf topics on the state forums now!

I mean as in an AoPSer... If any one is coming to nats... I'll be the indian kid with silver rimmed glasses, nicely combed black hair, a growing mustache, decent height, (probably) with my hands in my pockets, and I made it for vocab....(sadly not math)

Hm..im in NSF..I'm going to nats too!
If I dont get first in vocab I'm gonna kill myself!
Hopefully I'll make top 5 in math..

But I dont live in Virginia


The probability of it raining any given day during June in Mexico City is $ \frac{1}{10}$, what is the probability that it will rain on at most two days during June in Mexico City?

I think you misinterpreted the question a bit...
I don't think that's right.

[quote="ZhangPeijin"]Ummmm why?[/quote]

It can rain on 0 days.


geometry
circumcircle
symmetry
geometric transformation
rotation
geometry unsolved

Let triangle $ABC$ , $H$ is the orthcenter. Let $M$ is arbitrary point in circle $(O)$ is the circumcenter of triangle $ABC$ . Let $M_{1},M_{2},M_{3}$ are the symmetry point of $M$ through $BC,CA,AB$,respectively.
Prove that $M_{1},M_{2},M_{3},H$ is collinear

Let $MM_{3}$ intersect $AB$ in $M_{3}'$ (similiar for $M_{1}'$ and $M_{2}'$). Then the points $M_{1}'$, $M_{2}'$ and $M_{3}'$ form the Simson line. Because $M_{i}M_{j}$ is parallel to $M_{i}'M_{j}'$, the points $M_{1}$, $M_{2}$ and $M_{3}$ are also collinear. Thus it remains to prove that $H$ is also on this line. And that is clear, since the projectivity with the center in $H$, taking $HM_{1}$ to $HM_{2}$ (combined from the projectivities which map $M$ into $M_{1}$ and $M_{2}$ into $M$), is clearly a rotation for the angle $\pi$ (check the three obvious cases - $M=A$, $M=B$ and $M=C$), and therefore $H$ truly is on the Simson line for every point $M$.

This is an obvious result of a very well known theorem (I'm quite sure it has been proved many times on mathlinks, although I'm not able to find it now):
The simson-line of $M$ wrt $ABC$ bisects $HM$.

[quote="Amir.S"]your solution is wrong.

You proved H lies on two distint lines parallel two each other(simson line and it's homothec wrt M) :roll:[/quote]
??? I proved that $H$ lies on $M_{1}M_{2}$, and because $M_{1}$, $M_{2}$ and $M_{3}$ are collinear, $H$ is on that line. What's wrong?

[quote="Erik."] ...and therefore $H$ truly is on the Simson line for every point $M$.[/quote]

I didn't  read your solution completely, I just read the Quote.
you said that $H$ is on simson line of point $M$


Hi.

The Australian Open this year is January 18-31, 2010.

I would like to discuss players and predictions!

Imo:

Roger Federer has won a few before, and of course he's done well in the other Australians, but he's getting toward 30. Also, Nadal and Djokovic have won here before so I think they definitely have chances.

Sadly, however, Murray hasn't ever really done well at Australian Open, so I'm not looking to him for a win.

It's too early to start making predictions.  Someone could come up with an injury, which ruins everything.  The past few years, the Australian Open as seen underdogs in the final against Federer (excluding last year, of course).  For example, the previous three years were Fernando Gonzalez, Marcos Baghdatis , and Jo-Wilfried Tsonga.  So someone surprising could emerge.

You mean last 3 against Fed? or last 3 winners.

Well the last 4 [b]distinct[/b] winners were Nadal, Djokovic, and Federer, and Safin.  The Australian Open has a recent history of little-known underdogs making very big gains.  Hasn't turned out well for them but they can't be counted out.

My point is that it's too early to make predictions.  Wait until a week or before before the tournament.

[quote="tennis123"]Why is everyone listed twice?[/quote]

IDK, a poll error thing. Just vote for the person with votes already

[quote="royalflower"]Don't you think it's odd that Murray beat Federer, and Roddick beat Murray who beat Federer, but Roddick hasn't been able to beat Federer himself? I'm looking forward to see the matches.[/quote]

Transitive property does not work for sports :lol:

By the way, if you vote "other" I would like to know who that is and what player.

Also, I think Federer pwns. I agree also with Nadal, and probably Djokovic.

Del Potro will do well, but I'm not yet going for his win. Also, other?? who is that? and which player

From the start of the tourney, I picked a Fed-Murray final.

As for the final, it's hard for me to pick. It's difficult for a defensive player such a Murray to win the final of a GS against an offensive player like Roger, as it means the offensive player is playing very well.

However, I want Andy to win this title, so I'm gonna say Murray in 5.   :blush: :D

Ok, my pick was a bit off.

Note to self: Next time, go with the head, not the heart.

(1) Roger Federer [SUI] def. (5) Andy Murray [GBR], 6-3 6-4 7-6(13-11).

Congratulations to Fed on his continuing march deeper into the history books. My pick was off too.....

[quote="robbiez"][quote="piegraph"]i want nadal to win.

(Hey awesometoad, can you stop making so many of those tennis posts?  Didn't you have one already?)[/quote]

Well, I can tell you're not a tennis fan...[/quote]
heheh...

Murray loses to Federer in 3 sets, 6-3, 6-4, 7-6(13-11)

Mirka looked sort of bored during the first and second sets.

And in the last set, Murray had about 5 chances to close the set....

I feel sorry for murray... he really wanted to be the first British in 73 years to get a slam.

@piegraph: I think Nadal is out for sometime. He isn't going to be #1 anytime soon.