vector
logarithms
abstract algebra
combinatorics unsolved
combinatorics

A test includes 30 questions with YES/NO answers. With every attempt Michael gives his answers for the all questions. After that he can see the number of correct and incorrect answers that he have made. At the beginning he doesn't know any of the answers. Can Michael make such a attempts, so he can find out all correct answers if he makes:
a) 29 attempts (he should answer correct on the 30th attempt).
b) 24 attempts (he should answer correct on the 25th attempt)?

I don't think so: how can you distinguish between "all the evens are true, all the odds are false" and "all the evens are false, all the odds are true" with those questions?

First, let's see how we can solve $ 5$ questions with only $ 4$ attempts. The guesses are:

A. $ \{\}$ (all questions are guessed as "Yes")
B. $ \{1,2\}$ (questions $ 1$ and $ 2$ are guessed "No", the other three "Yes")
C. $ \{2,3\}$ (questions $ 2$ and $ 3$ are guessed "No", the other three "Yes")
D. $ \{1,3,4\}$ (questions $ 1$, $ 3$ and $ 4$ are guessed "No", the other two "Yes).

Notice that this protocol is "static" in the sense that Michael's guesses are independent of the answers he got on previous attempts. One could imagine that for a larger number of questions it might be possible to achieve better results with a "dynamic" protocol where Michael designs each attempt based on what he already knows. I don't know if this is indeed the case, but so far I found no case where a dynamic protocol beats the best static one.

Checking that this protocol works is a bit tedious but straightforward. There are some shortcuts possible: for instance, you may assume that there are zero, one or two "No" answers, since the other cases are obtained by reversing all the the answers which simply changes each response from $ k$ correct answers to $ 5\minus{}k$ correct ones.

Next, we see how to turn this protocol into $ 24$ attempts for $ 30$ questions. Our first 6 attempts are:

Z0: $ \{\}$ (all "Yes")
Z1: $ \{1,2,3,4,5\}$
Z2: $ \{6,7,8,9,10\}$
.
.
.
Z6: $ \{21,22,23,24,25\}$.

After these attempts, we know how many "Yes" answers there are overall, and we also know how many "Yes" answers there are in each consecutive block of $ 5$ questions. The last block doesn't need to be checked since it can be calculated from the total (Z0) and the other blocks.

We now proceed using the first protocol for each block of $ 5$ consecutive questions.

A1: $ \{1,2\}$
A2: $ \{2,3\}$
A3: $ \{1,3,4\}$

B1: $ \{6,7\}$
B2: $ \{7,8\}$
B3: $ \{6,8,9\}$

etc.

The responses we get for each attempt will be the number of correct answers in this block plus the number of "Yes" answers in the other blocks, but we know what that number is so we can subtract is and focus on the block at hand. After $ 6\plus{}3\times 6 \equal{} 24$ attempts we have the answers to all questions.

Just a few words about what this solution is really based on. Create a 30-dimensional vector $ u$ whose $ j - th$ component is $ + 1$ if the answer to the $ j$-th question is "yes" and $ - 1$ if it is "no". Now for each Mike's attempt create a similar vector $ v$ based on his answers. The output of the attempt can be easily recalculated into the scalar product $ \langle u,v\rangle$ and vice versa. Now, if we want the queries to be set up in advance, we need to choose the minimal set of $ \pm 1$ vectors $ v$ such that the mentioned scalar products determine $ u$ uniquely, i.e., that there is no vector in the orthogonal complement consisting of $ + 1$, $ 0$, and $ - 1$ only. The key idea of the above solution is to find $ 4$ vectors consisting of $ \pm 1$ in $ \mathbb R^5$ whose orthogonal complement is spanned by $ (1,1,1,1,2)$. These vectors are just $ (1, - 1, - 1, - 1,1)$, $ ( - 1,1, - 1, - 1,1)$, $ ( - 1, - 1,1, - 1,1)$ and $ ( - 1, - 1, - 1,1,1)$ and then to take the tensor product of this system with any spanning set of six $ \pm 1$ vectors in $ \mathbb R^6$ to get the system of $ 24$ vectors in $ \mathbb R^{30}$.

Now a few questions:

1) Can we do better than 25 with such scheme?
2) Can we do better than 25 with answer-dependent schemes?
3) If you have an $ n$-question quiz, approximately how many attempts are needed? (the above solution gives $ \frac 45n$ but I suspect that the answer is really $ o(n)$; also, since every attempt splits the set of possible vectors $ u$ into $ n$ classes, we cannot do better than $ n/\log_2 n$ no matter what).


is the 1994 AHSME especially easy or something?

i took it today for practice and got a 144

which is about 55 points higher than my score on this year's amc12a

it seems sort of strange that i would improve by so much in a few months time.

Yes, the 1994 AHSME was particularly easy.  There were more perfect scores that year than in all the previous 40-some years combined.

In contrast, if I remember correctly, the accompanying 1994 AIME was insanely difficult.

yeah, the 94 aime was in my opinion the hardest aime ever.


The question states:
What is the largest 2-digit prime factor of the integer 200C100?

And the given solution states:
200C100 = (200!)/(100!)(100!).
Therefore, our two digit prime p must satisfy 3p < 200.

I get until the (200!)/(100!)(100!) part, but I can't understand
where [b]3p<200[/b] comes from.
Please somebody give me a help! Thank you.[/b]

All 2-digit numbers appear twice in the denominator, once in each $100!$. Thus, it must appear in the numerator 3 times. The only way this will happen is if $p$, $2p$ and $3p$ are all in $200!$. Thus, we solve $3p<200$.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=328131#p328131


[quote="MithsApprentice"]Anyone have recommendations for computer security software? (Virus scan and removal, firewalls, adware and spyware blockers, etc...)

I need to buy new stuff for my computer...[/quote]
for adware/spyware blockers, spywareblaster(NOT to be mistaken with spy blaster) which is free.  If you meant spyware removal, get free ad-aware, spybot s&d, and spysweeper.  YOu should tryu not to by software without googling first( for example, stop sign from eacceleration seems good but googling gives you a whole bunch of faults such as it being adware)

I don't know what to recommend, so I'll just list my setup:

Antivirus: Norton Antivirus Corporate 7 (license from University of Toronto)
Anti-intrusion: [url=http://smb.sygate.com/]Sygate Personal Firewall[/url] (free)
Anti-spyware: [url=http://www.download.com/]Adaware and Spybot: Search and Destroy[/url] (free)
Anti-ads: [url=http://www.proxomitron.info]Proxomitron Naoko 4.5[/url] (free)

You should always try to keep your virus definitions up-to-date. It wouldn't hurt to update the spyware and firewall products once in a while, too.


inequalities
function
integration
logarithms
algebra
domain
complex analysis

Let $\phi(x)$ be the real valued functions,and integrabel in any finite interval of $[0,+\infty)$,Let
\[P=\int_{0}^{+\infty}e^{-(t+i\phi(t))dt},Q=\int_{0}^{+\infty}e^{-2(t+i\phi(t))}dt \]
Prove:
\[|4P^{2}-2Q| \le 3 \]
And find when does the equality hold?

${P=\int_{0}^{+\infty }e^{-(t+i \phi (t))}\, dt}$
${Q=\int_{0}^{+\infty }e^{-2 (t+i \phi (t))}\, dt}$

First Simplify

${e^{-(t+i \phi (t))}=e^{-t}e^{-i \phi (t)}}$
${e^{-2 (t+i \phi (t))}=e^{-2 t}e^{-2 i \phi (t)}}$

Substituing $\phi$ of anoter function to simplify

${\phi (u)=\log \left(f'(u)\right) i}$

${e^{-t}e^{-i \phi (t)}=e^{-t}f'(t)}$
${e^{-2 (t+i \phi (t))}=e^{-2 t}f'(t)^{2}}$

Substituing on ${P}$ and ${Q}$

${P=\int_{0}^{+\infty }e^{-t}f'(t) \, dt}$
${Q=\int_{0}^{+\infty }e^{-2 t}f'(t)^{2}\, dt}$

Remembering the laplace transform

${G(s)=\mathcal{L}_{t}[g(t)](s)=\int_{0^{-}}^{\infty }e^{-s t}g(t) \, dt}$

Since ${e^{-t}f'[t]}$ and  ${e^{-2 t}f'[t]^{2}}$ are integrable in the domain ${[0,+\infty \right) )}$

${P=\mathcal{L}_{t}\left[f'(t)\right](1)}$
${Q=\mathcal{L}_{t}\left[f'(t)^{2}\right](2)}$

${|4 \left(\mathcal{L}_{t}\left[f'(t)\right](1)\right)^{2}-2 \left(\mathcal{L}_{t}\left[f'(t)^{2}\right](2)\right) | \leq 3}$

How can i simplify the laplace transform of a product of 2 functions?

${P=\int_{0}^{+\infty }e^{-(t+i \phi (t))}\, dt\text{ }\land \text{ }Q=\int_{0}^{+\infty }e^{-2 (t+i \phi (t))}\, dt}$
${\left|4 P^{2}-2 Q\right|\leq 3}$

${e^{-(t+i \phi (t))}=e^{-t}e^{-(i \phi (t))}}$
${e^{-2 (t+i \phi (t))}=\left(e^{-t}e^{-(i \phi (t))}\right)^{2}}$

Substituing

${\phi (u)=\log (f(u)) i}$

${e^{-t}e^{-(i \phi (t))}=e^{-t}f(t)}$
${\left(e^{-t}e^{-(i \phi (t))}\right)^{2}=\left(e^{-t}f(t)\right)^{2}}$

Writing ${f}$ in Taylor Series

${f(t)=\sum_{i=0}^{\infty }\alpha_{i}t^{i}}$

${P=\int_{0}^{\infty }e^{-t}f(t) \, dt}$
${P=\mathcal{L}_{t}[f(t)](1)}$
${P=\mathcal{L}_{t}\left[\sum_{i=0}^{\infty }\alpha_{i}t^{i}\right](1)}$
${P=\sum_{i=0}^{\infty }\Gamma (i+1) \alpha_{i}}$

${Q=\int_{0}^{\infty }\left(e^{-t}f(t)\right)^{2}\, dt}$
${Q=\int_{0}^{\infty }e^{-t 2}f(t)^{2}\, dt}$
${Q=\mathcal{L}_{t}[f(t)](2)}$
${f(t)=\sum_{i=0}^{\infty }\alpha_{i}t^{i}}$
${f(t)^{2}=\left(\sum_{i=0}^{\infty }\alpha_{i}t^{i}\right)^{2}=\sum_{i=0}^{\infty }\left(\sum_{j=0}^{\infty }\left(\left(\alpha_{i}t^{i}\right) \left(\alpha_{j}t^{j}\right)\right)\right)}$
${Q=\mathcal{L}_{t}\left[\sum_{i=0}^{\infty }\left(\sum_{j=0}^{\infty }\left(\left(\alpha_{i}t^{i}\right) \left(\alpha_{j}t^{j}\right)\right)\right)\right](2)}$
${Q=\sum_{i=0}^{\infty }\left(\sum_{j=0}^{\infty }\left(2^{-i-j-1}\Gamma (i+j+1) \alpha_{i}\alpha_{j}\right)\right)}$

${\left|4 P^{2}-2 Q\right|\leq 3}$

${P^{2}=\left(\sum_{i=0}^{\infty }\Gamma (i+1) \alpha_{i}\right)^{2}=\sum_{i=0}^{k}\left(\sum_{j=0}^{k}\left(\left(\Gamma (i+1) \alpha_{i}\right) \left(\Gamma (j+1) \alpha_{j}\right)\right)\right)}$

${\left|4 \sum_{i=0}^{k}\left(\sum_{j=0}^{k}\left(\left(\Gamma (i+1) \alpha_{i}\right) \left(\Gamma (j+1) \alpha_{j}\right)\right)\right)-2 \sum_{i=0}^{\infty }\left(\sum_{j=0}^{\infty }\left(2^{-i-j-1}\Gamma (i+j+1) \alpha_{i}\alpha_{j}\right)\right)\right|\leq 3}$
${\left|\sum_{i=0}^{k}\left(\sum_{j=0}^{k}\left(4 \left(\left(\Gamma (i+1) \alpha_{i}\right) \left(\Gamma (j+1) \alpha_{j}\right)\right)\right)-\sum_{i=0}^{\infty }\left(\sum_{j=0}^{\infty }\left(2 2^{-i-j-1}\Gamma (i+j+1) \alpha_{i}\alpha_{j}\right)\right)\right|\leq 3\right.}$
${\left|\sum_{i=0}^{k}\left(\sum_{j=0}^{k}\left(4 \left(\left(\Gamma (i+1) \alpha_{i}\right) \left(\Gamma (j+1) \alpha_{j}\right)\right)-2 2^{-i-j-1}(\Gamma (i+j+1)) \alpha_{i}\alpha_{j}\right)\right)\right|\leq 3}$
${\left|\sum_{i=0}^{k}\left(\sum_{j=0}^{k}\left(\left(4 \Gamma (i+1) \Gamma (j+1)-2^{-i-j}(i+j+1) \Gamma \right) \alpha_{i}\alpha_{j}\right)\right|\leq 3\right.}$

no

i just show that its impossible to make it using taylor or laplace


would anyone happen to have and be able to sell this book by andreescu and feng to me
or know where i could get it thanks

On Amazon.com http://www.amazon.com/Problems-Algebra-Training-Team-Enrichment/dp/187642012X

 :P

thanks but i already tried they dont have it and i'm trying to get it before APMO for crash course in algebra lol :S

just google it!!!

[url]http://www.mathpropress.com/books/problemBooks/101ProblemsInAlgebra/[/url]

im not sure if it is reliable but its cheap!!!


Find all functions $ f: \mathbb R\rightarrow \mathbb R$ satisfying $ f(f(x)\plus{}y)\equal{}2x\plus{}f(f(y)\minus{}x)$ for every $ x,y\in \mathbb R$

We first prove the $ f(x)$ is injective. Suppose there is $ f(b)\equal{}f(c)\equal{}d$ with $ b\neq c$.
$ P(b,y): f(d\plus{}y)\equal{}2b\plus{}f(f(y)\minus{}b)$
$ P(c,y): f(d\plus{}y)\equal{}2c\plus{}f(f(y)\minus{}c)$.
So we have for any $ y$,
$ f(f(y)\minus{}b)\minus{}f(f(y)\minus{}c)\equal{}2c\minus{}2b$  ---------------(1)
$ P(x,b): f(f(x)\plus{}b)\equal{}2x\plus{}f(d\minus{}x)$
$ P(x,c): f(f(x)\plus{}c)\equal{}2x\plus{}f(d\minus{}x)$
So we have for any $ x$.
$ f(f(x)\plus{}b)\equal{}f(f(x)\plus{}c)$ ----------------------(2)
If we let $ u\equal{}f(x)\plus{}y$ and $ v\equal{}f(y)\minus{}x$, we have $ f(u)\minus{}f(v)\equal{}2x$. This implies we can always find $ x_0,y_0$ such that $ f(x_0)\minus{}f(y_0)\equal{}\minus{}b\minus{}c$. Plug this in (1), we have
$ f(f(y_0)\minus{}b)\minus{}f(f(y_0)\minus{}c)\equal{}f(f(x_0)\plus{}c)\minus{}f(f(x_0)\plus{}b)\equal{}2c\minus{}2b\neq0$, a contradiction with (2) when we plug in $ x_0$.
Assume $ f(0)\equal{}a$,
$ P(0,y): f(a\plus{}y)\equal{}f(f(y))$.
Since $ f(x)$ is injective, we have $ f(y)\equal{}y\plus{}a$.


Is there a sequence of sets $ A_n\subset X$ such that there is no convergent subsequence $ A_{n_k}$ if a)$ X \equal{} \mathbb N$, b)$ X \equal{} \mathbb N$ and $ A_n$ are finite sets for all $ n$?

I know the answer when   $ X\equal{} \mathbb R$ (sequence exists) and when $ X$ is a finite set (there is no sequence).

Some definitions.
Consider a sequnce of sets $ A_n \subset X$. Let $ A^*\subset X$ ($ A^* \equal{} \varlimsup A_n$) be a set whose members belong to infinite number of $ A_n$.  Let $ A_*\subset X$ ($ A_* \equal{} \varliminf A_n$) be a set whose members belong to $ A_n$ for all $ n$, except finite number of $ A_n$. If $ A^* \equal{} A_* \equal{} A$ we say $ A$ is a limit of this sequence $ A_n$.

I solved the problem. There is no subsequence. Thanks for all.


function
real analysis
calculus
derivative
integration
limit
topology

I just wanted to see how many nice examples would show up :).

Give an example of a measurable real set $E$ s.t. for each interval $I$ we have $0<m(E\cap I)<m(I)$.

Here is a sketch (I think the English here is very unclear)

Let $H_1$ be a Cantor-like set in $[0,1]$ s.t. $m(H_1)=1/2$.

In each $I_{1j}$ (which is an adjacent interval of $H_1$), make a cantor-like set $H_{1j}$ such that $m(H_{1j}) = |I_{1j}|/2^2$ and let $H_2 = \bigcup_{j=1}^\infty H_{1j}$

In each $I_{2j}$ (which is an adjacent interval of $H_1\cup H_2$), make a cantor-like set $H_{2j}$ such that $m(H_{2j}) = |I_{2j}|/2^3$ and let $H_3 = \bigcup_{j=1}^\infty H_{2j}$

...

In this way we obtain $\{H_m\}$. Let $E=\bigcup_{n=1}^\infty H_n$ then we have $0<m(E\cap I)<m(I)$ for all open interval $I$ in $[0,1]$.

Other nice constructions?
I was thinking in the same way as Li Yi.

I will not construct such a set, but I will deduce it's existence from a Baire category
theorem ( may be it is stupid, or also may be I have an error in my solution,
but nevertheless I write it here).

Denote by $ X $ the set of all measurable functions $ f : [0,1] \rightarrow \mathbb{R} $,
such that $ f(x) = 0 $ or $ 1 $ for all $ x \in [0,1] $ ( the functions which are
differ on a set of measure $ 0 $ are equal for us). Then we have an embedding
$ X \hookrightarrow L^{1}([0,1]) $, so on $ X $ we have a metric induced from
the metric on $ L^{1}([0,1]) $, and $ X $ is a closed subspace of $ L^{1}([0,1]) $,
so X is complete in this metric. Therefore $ X $ is complete metric space,
so it is of second category. We denote $ B_{r}(f) $ a ball of radius $ r $ around $ f $ in
$ X $ . Now for $ s,t \in [0,1] $, rational numbers, such that
$ s < t $ define $ Y_{s,t} \subset X $ as $ Y_{s,t} := \{ f \in X $, such that
$ f|_{[s,t]} \equiv 1 \}
\cup \{ f \in X $, such that $ f|_{[s,t]} \equiv 0 \} $.
Let us show that $ Y_{s,t} $ is nowhere dense in $ X $.
Indeed, consider a function $ f \in X $ , and $ r > 0 $.
Assume that $ r < t-s $. Take arbitrary open non-intersecting
intervals $ I,J \subset [s,t] $ of lengths equal to
$ r/3 $. Now define $ g \in X $ as : $ g(x) = 0 $ if $ x \in I $,
$ g(x) = 1 $ if $ x \in J $, otherwise $ g(x) = f(x) $. It is clear that
$ dist(f,g) < 2r/3 $. Therefore $ B_{r/4}(g) \subset B_{r}(f) $. Now I claim
that $ B_{r/4}(g) \cap Y_{r,s} = \emptyset $. Indeed , if $ h \in Y_{r,s} $ then
in the case $ h|_{[s,t]} \equiv 1  $ we will have that $ h|_{I} = 1 $ and
$ g|_{I} = 0 $ and length of $ I $ is $ r/3 $, and in the case of
$ h|_{[s,t]} \equiv 0  $ we will have that $ h|_{J} = 0 $ and
$ g|_{J} = 1 $ and length of $ J $ is $ r/3 $, so in any case we will have that
$ dist(g,h) \geqslant r/3 $.

So we have shown that $ Y_{s,t} $ is nowhere dense in $ X $.
Therefore $ Y := \bigcup_{r,s} Y_{r,s} $ is a set of first category in
$ X $, and $ X $ is a complete metric space, so it is of second category,
so $ Y \varsubsetneq X $. So if we take some $ f \in X $, which
is not in $ Y $, then the set $ A := f^{-1}(1) \subset [0,1] $ is a desired set.
If we want now such a set in $ \mathbb{R} $, we can take $ B \subset \mathbb{R} $,
such that $ x \in B $ if the fraction part $ \{ x \} \in A $.

I hope it is correct  :)  .

I think, it is an approach of a real mathematician.
Very nice!

Very nice indeed, Leva :).

Here's what I did (it's not really rigorous, it's the ugliest approach here so far, and I hope it's not wrong.. :)):

Consider the function on $I=[0,1]$ (it makes no difference), and let $(I_n)_{n\ge 1}$ be an enumeration of the intervals $\subseteq I$ with rational endpoints. Let $S_n$ be the set we have at step $n$ (which we keep modifying). At step $n$, take $I_n$ and add to $S_n$ a subset of $I_n$ with measure $m_n$ if the measure of $S_n\cap I_n$ is null, or delete from $S_n$ a subset of $I_n$ if $S_n\cap I_n$ has measure $m(I_n)$ (or do nothing otherwise). The $m_n$ can be chosen to be small enough so that for each $j$, the limit of the measures $m(S_n\cap I_j)$ as $n\to\infty$ exists and is strictly between $0$ and $m(I_j)$ (this is because at each step $n$ we have only considered $n-1$ intervals, i.e. a finite number, so we can adjust $m_n$ to be small enough for these).

Does it work?

[quote="grobber"]Does it work?[/quote]

Funny, but this "ugly" construction is exactly the same as Leva's "nice" one. The only difference is that Leva uses the result of the Baire theorem, while grobber's argument is more or less equivalent to the standard proof of the Baire theorem adjusted to this special case. :)

You can construct - and have constructed - a set $E$ such that $0<\frac{|E\cap I|}{|I|}<1$ for any interval $I.$

Nonetheless, the Lebesgue theorem on the differentiation of the integral implies that

$\lim_{|I|\to0,x\in I}\,\frac{|E\cap I|}{|I|} = 0\text{ or }1$ for almost every $x.$


topology
real analysis
real analysis solved

By Tychonoff's Theorem, the (direct) product of countably many compact sets is compact.

Could I see a proof for the case with two compact sets?

Thanks.

You have to use the so-called Tube Lemma:

Given the spaces $X,Y$, where $Y$ is compact, and an open subset of $X\times Y$ which covers $x_0\times Y$ for some $x_0\in X$, $N$ contains an entire "strip" (or "tube" :)) $V\times Y$, where $V$ is an open neighborhood of $x_0$.

Try to prove that, and I'm sure you can see how to apply it to your problem.

There's something I don't get: why mention countability? Tychonoff's Theorem states the compactness of the product of any family of compact spaces (it's not concerned with countable families). :?

You're right, it doesn't have anything to do with countability.

This mistake is because I really don't know much about topology  :blush:

heres the complete proof:
let us take the product of 2 compact sets X & Y.Let R be a cover of X x Y by basic open sets U x V. U open in X & V open in Y.We are to show that R has a finite subcover.Fix a belonging to Y.The slice X x {a} is compact.So there are finitely many sets of the form Un x Vn in R covering the slice.Now use the fact that Y is compact & use the above construction to construct an open cover for R.


Let $ abcd$  be a convex quadrilateral such that the sides  $ ab  ,ad,  bc$satisfy $ ab  \equal{}  ad  \plus{}  ac$. There exists a point $ p$  inside the quadrilateral at a distance  $ h$ from the line $ cd$ such that $ ap\equal{}  h  \plus{}  ad$ and $ bp  \equal{}h  \plus{}  bc$. Show that: $ \frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}\plus{}\frac{1}{\sqrt{BC}}$

Did you mean $ AB\equal{}AD\plus{}BC$?? Because that is how i interpreted the problem (it made the problem even simpler :D )

[geogebra]540c34d7fc1f5fe0f491a097f06b98a00a61b075[/geogebra]

[hide="Hint?"]The biggest circle has centre A. Go clockwise ABCD, so that the medium circle has centre B. The smallest circle has centre P. Note that the 3 circles are tangent to each other with radii AD, BC and h respectively. Cd is not necessarily tangent to the 2 circles.[/hide]

hey could u explain me a little more!
 :)  :D

Lets call AB=a, BC=b, CD=c, DA=d, AP=r , BP=s, PM=h

a=b+d
r=d+h
s=h+b

Now do those above equations look familiar?  :wink:


A bicycle tire has a diameter of 66 centimeter. Approximately how many times must the tire rotate to travel 1 kilometer?

(A) 5   (B) 10   (C) 48   (D) 480   (E) 960

[hide] The circumference of the tire is $ 66\pi$. The tire travels with the rest of the bike. since $ 1$ km=$ 100000$cm, we have $ \frac{100000}{66\pi}\approx480$. D.[/hide]


probability
Pascal\u0027s Triangle
combinatorics unsolved
combinatorics

I have a probability question I was wondering about, if anyone is interested.

Please excuse my lop-sided dots, but I hope it conveys the idea.

I seen an episode of the Price is Right where they played this game called Plinko

The contestant stands at the top and lets a disc randomly fall(they can release it anywhere, not a specific spot)

through the 10X10 grid. The arrow at the bottom is 10,000 dollars. They can win up to 5 disks. What is the

probability of actually landing in that spot 5 times?.

Mighty slim I would think.

Regardless, the two slots on either side of the arrow is where most land. That's 0 dollars, of course. I was just wondering, somewhere along the line someone may have actually calculated this probability somehow.

[img]http://img155.imageshack.us/img155/320/plinko5fk.gif[/img]

Seems the ball will go straight to the arrow if you put it in the middle

DUH. I drew it wrong. I revised the drawing. The peg below is in between the two above. Sort of like Pascal's triangle is arranged, only a square.


superior algebra
superior algebra solved

Let $R$ be the commutative ring, and let \[ N= \{x \in R | x^n=0 \text{ for n } \in \mathbb{Z} \} \]

Prove that $N$ is ideal of $R$.

is this problem correct?   :?

This was posted before.
The ideal is that one made of the nilpotent elements.
All properties of ideals are simple to show, except of that $a,b \in N \implies a+b \in N$. For that, let $m , n$ be integers with $a^m=0=b^n$. Then by binomial extension of $(a+b)^{m+n-1}$ one gets only summands that are $0$ (the summends are $\mathrm{blah} \cdot a^k b^l$ with $k+l = m+n-1$, so either $k\geq m$ or $l \geq n$ and it gets $0$ by the assumption), thus the sum is $0$.

Well, if it is for all $n$, then it's an ideal, too (the trivial one) ;)
But $n$ should be assumed as nonnegative everytime (or how is $x^{-1}$ defined¿).

Thanks you ZetaX,  looks like i miss that.  :)


[hide]43251[/hide]

i can explain later if you want, but this is an ASAP thign... so i decided to HURRY!

[hide]BC is equal to either 12, 24, 32, or 52. D=5. C+E+5 is divisible by 3.
$12453$ works too.[/hide]

chess64, not only is that another solution, but it is the unly solution...the original solution of 43251 by uglyman is incorrect because 251 is not a multiple of 3...

[quote="math92"]The digits 1, 2, 3, 4, 5 are each used once to compose a five digit number ABCDE.  The three digit number ABC id divisible by 4, BCD is divisible by 5, and CDE is divisible by 3.  Find the five-digit number ABCDE.[/quote]
[hide]
The tens digit is 5, since BCD is divisible by 5, the hundreds digit is 4, since ABC is divisible by 4.
5+4=9, So the last digit is 3. The answer is $\framebox{12453}$, since the last 2 digits have to be divisible by 4 in ABC and 14 is not divisible by 4.[/hide]

[hide]12453. hee hee, that's fun to type :D [/hide]

[quote="math92"]The digits 1, 2, 3, 4, 5 are each used once to compose a five digit number ABCDE.  The three digit number ABC id divisible by 4, BCD is divisible by 5, and CDE is divisible by 3.  Find the five-digit number ABCDE.[/quote]As Chess pointed out, since 4|ABC, at this point A can be any number, any multiple of 100 is divisible by 4. BC is a two digit number divisible by 4. Candidates are {12, 24, 32, 52}. Since 5|BCD, we know D = 5, because 0 is not on the list of digits. Since 3|CDE, $(C+5+E)  \equiv 0\mod{3} \Rightarrow (C+E) \equiv 1 \mod{3}$. If $C = 2$ then $E \equiv 2 \mod{3}$, however, the only two numbers in the list $\equiv 2\mod{3}$ have already been used.  Therefore BC = 24.  $C = 4 \Rightarrow E \equiv 0 \mod{3}$. The only number $\equiv 0 \mod{3}$ on the list is 3; therefore E =3. That leaves A = 1 as the only possibility.

$\boxed{ABCDE = 12453}$


Not own solution:
For any $i$ we have $(a_{i}+2)(a_{i}-17)\leqslant 0$. After expanding it becomes $a_{i}^{2}\leqslant 15a_{i}+34$. Then we simply sum up such inequalities for $1\leqslant i\leqslant 59$:
$\sum_{i=1}^{59}a_{i}^{2}\leqslant \sum_{i=1}^{59}15a_{i}+34= 15\sum_{i=1}^{59}a_{i}+59\cdot 34= 2006$.


inequalities
symmetry
inequalities unsolved

Prove the following inequality for positive real numbers $x, y, z:$

$(xy + yz + zx)(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2})      \geq \frac{9}{4}$.

This has been posted before:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=3547[/url]

This problem is nice,but it is simmetry inequality with three variable a,b,c.Almost ineq with symmetry var is nit difficult,for example in JAPAN math olympiad 1995,the problem:
a,b,c>=0,prove that
sum(sym):(b+c-a)^2/((b+c)^2+a^2 )>=3/5
it's quite as same as this problem(you can apply Shur's inequality-not elegant,or use CAUCHY -elegant and  easier)

The  well  known ineqs  are allready  discused in this  forum....long  time ago.


algebra
polynomial
function
vector
complex analysis
complex analysis unsolved

Let $ f\in\mathcal{O}_{\mathbb{C}^n,0}$ be irreducible. Then $ f\in\mathcal{O}_{\mathbb{C}^{n\plus{}1},0}$ is also irreducible?

Use the Weierstraß preparation theorem to work in the polynomial ring $ \mathcal O_{\mathbb C^n,0}[z_{n \plus{} 1}]$.

Can you give a more detail proof. I tried using the Weierstrass preparation theorem but I still can't solve it, because if I want to use  this theorem for a function $ g$, $ g$ must satisfy the condition $ g(0,z_{n \plus{} 1})\not\equiv 0$.

If your condition is not satisfied, then apply some appropriate linear transformation such that the condition is true under the new coordinates. Note that a linear transformation preserves the irreducibility of elements.

What does the $ 0$ in the notation mean?

... either zero in $ \mathbb C$ or the zero vector in $ \mathbb C^n$ or in $ \mathbb C^{n\plus{}1}$ - admittedly a sloppy notation, but a useful abuse I think. ;)

No no, I'm wondering about the notation $ \mathcal{O}_{\mathbb{C}^n, 0}$. Does it denote all holomorphic functions from $ \mathbb{C}^n$ to $ \mathbb{C}$ with $ f(0) \equal{} 0$?

Oh, sorry for this misunderstanding. $ \mathcal O_{\mathbb C^n,0}$ means the ring of convergent power series in $ n$ complex variables around $ 0\in\mathbb C^n$ which can be identified with the ring of holomorphic functions $ U\to\mathbb C$ on an open subset $ 0\in U\subset \mathbb C$ where functions are identified iff they coincide on a common open subset containing 0. So, e.g., $ f(0)\equal{}1$ is not excluded.

Ok, I understand the question and your suggested solution also now. But is it really necessary to use an analytic theorem to solve this problem? What about if we consider a more abstract ring of power series? Doesn't it seem "obvious" that you shouldn't be able to produce a factorization by introducing a new variable?

Hmm, this is the beginning of another track. But what theorems do we need in terms of regrouping, i.e. for comparing $ \mathbb C\{x,y\}$ and $ \mathbb C\{x\}\{y\}$- just boiling the problem down to $ n\equal{}1$ at first. Here we have to topologize the ring $ \mathbb C\{x\}$ to be able to assert a claim as far as I can see.


Suppose that a constant force acts on an object. The force does not vary with time, nor with the position or the velocity of the object. Start with the general definition for work done by a force \[ W = \int_{i}^j \textbf{F} \cdot \textrm{d}\textbf{r} \] and prove that the force is conservative.

I don't immediately see how to do this. However, this might be really easy :)

Hm, isn't that really obvious? Since F is constant, integration gives:

$W=\int_\textbf{i}^\textbf{j}\textbf{F}\cdot\text{d}\textbf{r}=\textbf{F}\cdot(\textbf{j}-\textbf{i})$.

But this is the definition of conservative force - the work done is path independent, it only depends on the two endpoints.

Maybe I just didn't understand the purpose of the question...

Or maybe it is just so trivial that I didn't see it since I expected it to be more difficult.

[quote="bus"]$W=\int_\textbf{i}^\textbf{j}\textbf{F}\cdot\text{d}\textbf{r}=\textbf{F}\cdot(\textbf{j}-\textbf{i})$.[/quote]

This may be a really stupid question but could you *prove* the above equality? I don't really have any experience with integrals involving vectors.

Integrals involving vectors are no different than integrals with scalars, only the way you multiply them is different. You probably know that in real numbers

$\int_a^bc\text{ d}x=[cx]_a^b=c(b-a)$

and with vectors this integration would be exactly the same. Perhaps one difference between real numbers and vectors is that there is only one way to get from real number $a$ to real number $b$ but many different paths from vector $\textbf{i}$ to vector $\textbf{j}$ - that's also why people usually write

$\int_C \textbf{F}\cdot\text{d}\textbf{r}$

instead of what you wrote where C is the path along which we integrate. To prove my statement.... it really depends on how you define integral, to me it seems just plain obvious that $\int_\textbf{a}^\textbf{b}\text{d}\textbf{x}=\textbf{b}-\textbf{a}$. Just imagine all those tiny little dxes how they add up to a vector connecting the beginning and ending point of the integration, which is exactly the vector $\textbf{b}-\textbf{a}$. And of course, multiplying this by a constant vector doesn't change anything because of the distributive properties of scalar multiplication - $\textbf{c}\cdot\int_\textbf{a}^\textbf{b}\text{d}\textbf{x}=\int_\textbf{a}^\textbf{b}\textbf{c}\cdot\text{d}\textbf{x}$ because $\textbf{a}\cdot(\textbf{b}+\textbf{c}+...+\textbf{n})=\textbf{a}\cdot\textbf{b}+\textbf{a}\cdot\textbf{c}+...+\textbf{a}\cdot\textbf{n}$.


Has anyone here read the whole Frank Herbert Dune series?  I am about half-way through and am really enjoying it.  What are your opinions on the story?

I'm also about halfway through... I should really get around to finishing it sometime. Good series. Erm, it's Matt, right? Matt Crawford? Anyway, you should probably have a more interesting signature. Just because you're an instructor doesn't mean that you need a math related quote... imo...

-Isaac

I prefer Mathew.  And I plan to keep the signature math related because, well, it's what I'm doing here :)

I've read most of the Dune books (not all because unfortunately my library doens't own all of them) but I liked the prequels better (ex. Dune: House Corrino).  I liked how they answered so many of the questions I had about what happened.

Of the House books, I have only read House Harkonnen.  I enjoyed it and how it answered a lot about the history of Dune, but I prefer Frank Herbert's writing to his son's.

Yep, that series rocked!!! Well, my opinion on it is that sometimes, F.H. makes it very thrilling and exciting and suspenseful. Oh yeah, a new trilogy is coming out starting when the Butlerian Jihad occured. The first book's already out, and I am waiting for the 2nd one. All in all, the series is AWESOME!!

More series... cool. If you want you can get the books online. Just do a Google search for "Heretics of dune.txt" or whatever.

-Isaac


I've read elsewhere that you can create .bat files to increase the Java Heap Space, however my question is how do I do it?

You can change the heap size simply by adding an extra parameter when running your program with the java command (from the command line):

java -Xms<initial heap size> -Xmx<maximum heap size>

Defaults are java -Xms32m -Xmx128m.

So if you wanted to double the maximum heap size, you'd run your program with java -Xmx256m program.java.

How would you do this if you were running it through an IDE?  Like textpad or NetBeans.

All IDEs will have a setting somewhere which has the 'java' command in it, where you can edit the parameters. I don't have NetBeans, but for example in Textpad, go into Configure->Preferences->Tools->Run Java Application, and you can change the Parameters field from $BaseName to -Xmx256m $BaseName, or whatever you want.


logarithms
algorithm
combinatorics proposed
combinatorics

If you are condemned to die in Sikinia you are put in to Death Row until the last day of the year .Then all prisoners from Death row are arranged in a circle and numbered 1,2,....,n .
Starting with #2 every second one in a double is shot until only one remains who is immediately set free .How do you find the place of the sole survivor ??

I like this kind of problems!  :)

Well, the first thing we should notice is that we have $n-1$ steps (in each step someone is killed).

This problem is like teams that are playing football!!

In every match, the first team (which is on the left) wins and continues playing. The second team loses and stops there.

If $n=8$, it is simple. The team #1 is the champion (because, at any level it is on the left - see the picture below)
Notice that we need $1+2+4+8=15$ boxes.

$\hline$

But what happens when n=9 or n=10?

Say, for example that $n=9$. As we saw above, we will have $n-1=8$ games. Each game has 2 teams, so we want $2\cdot 8=16$ boxes, plus one box on the top. So we want a total of $17$ boxes.

The first row (which contains only the winner) will be the row #0
The row with number $k$ can contain up to $2^{k}$ boxes.

We sum up the numbers $2^{k}$, starting from $k=0\Rightarrow$

$S_{k}= \sum_{i=0}^{k}{2^{i}}= 1+2+4+8+...$.
We are going to find how much rows are needed.

For $k=0,1,2,3,4$ we have $S_{k}= 1,3,7,15,31$ respectively.

We stop at $k=4$, because we got $S_{4}\geq 17$. Let $m$ denote this maximum value of $k$.

We need to put 17 boxes. As we can see above, $S_{3}=15$, which means that in the rows $0,1,2,3$ we can put 15 boxes. So we have 2 boxes more. This means that in the last row #4 we'll put the numbers $1,2$. So the winner is the number $3$.

$\hline$

Generally, we have to put $2(n-1)+1=2n-1$ boxes

What we need is the sum $S_{m-1}$ (that is, how many complete rows are there in the tree)

Notice that $S_{m-1}= 2^{m}-1$

a) If $n$ is a power of 2, then it's easy too see that the winner is the team #1 (it is just like the case of 8)

b) If not, then we have:

$S_{m-1}<2n-1<S_{m}\Rightarrow$

$2^{m}-1<2n-1<2^{m+1}-1\Rightarrow$

$2^{m}<2n<2^{m+1}\Rightarrow$

$2^{m-1}<n<2^{m}\Rightarrow$

$m-1 < \log_{2}{n}<m$

So we take $m=\{\log_{2}{n}\}+1,$

where $\{x\}$ is the integer part of $x$

Then, we calculate the sum until the $m-1$ row : $S_{m-1}= 2^{m}-1$

In the last row we will put $x$ boxes. We can find the value of $x$ by a simple substraction:
$x=(2n-1)-S_{m-1}$

$x$ is the last person in the last row, so the survivor is the number $x+1$.

$x = (2n-1)-S_{m-1}$
$x = 2n-1-2^{m}+1$
$\boxed{x = 2n-2^{m}}$

$\hline$

We can simply say that:

1) if $n$ is a power of $2$ then the winner is the person #1

2) else, if $m$ is the smaller natural number such that $n<2^{m}$, then the survivor is
$\boxed{s = 2n-2^{m}+1}$, where $m=\{\log_{2}{n}\}+1$

$\hline$

We took two cases, one when $n$ is a power of $2$ and another one, when it is not.
In the first case, the last person in the last row is $x=n$.

By the way, we can express the number of the survivor in a unique formula, since $n+1$ gives $1$ in modulo $n$:

$s = (x \mod n)+1$, where $\boxed{x = 2n-2^{m}}$ and  $\boxed{m=\{\log_{2}{n}\}+1}$

$\hline$

In the diagrams below we have cases for $n=8,9,10$.

With red are the persons who will be killed.

The blue is the winner.

Applying the above algorithm for the numbers $n=1,2, ... , 31$ we get the values:

1,
1,3,
1,3,5,7,
1,3,5,7,9,11,13,15,
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31...


AMC AIME geometry Gauss calculus USA(J)MO USAMO

Are olympiaders' math abilities, skills, creativities, etc. very much higher or better than normal math major students' at the end of the college year? How about after graduate school?

My concern is that..Actually, I think I'm much better than normal people in math. I am qualified for AIME as well..

And I really like math..Many hours in my day, I am thinking about math of physics stuff..So I'm going to major in math or physics..

But..am I late? I think I am pretty late..I heard that Einstein studied geometry himself in age of 15 or something. And many people here..even 4th or 5th graders...gets better score than me on AIME..actually 8 years old boy in my country, he is done with calculus..which I am struggling with..and attending college..

And also..I think I'm one of near genius..no, let me say one of the brilliant guy..but not real genius..or super genius like Gauss, Newton or Einstein..I heard that Newton started to study math after college, and Edward Witten studied History by age of 26 and started to study math and physics after that, got a Fields Medal in 13 years and became the best physicist in the world by string theory.

My question is..Does the IQ or gifted ability matter very much in math or physics? Equal to or maybe more than the hardworking?Can I catch up Einstein?..At least..Can I catch up the guys here like former international olympiad winners or equivalent people? I mean..I'm so depressed by this forum's people! Tooooooo smart..

Can anybody answer the question?

You can catch up enough to have a satisfying career. Here is a link to some [url=http://paulgraham.com/love.html]advice on finding work that you love[/url] that I discovered online just the other day. I think it is true that if you keep working steadily, you can become smarter, and you can become smart enough to do well at work you enjoy.

[quote="car202"]Does the IQ or gifted ability matter very much in math or physics? Equal to or maybe more than the hardworking?[/quote]

Talent is worth infinitely more.  In a sense, work ethic can be taught, but talent can't.  That doesn't mean that one cannot have a successful career or become a very good mathematician if they're not incredibly talented, but they probably will not achieve as much as someone truly brilliant, regardless of how smart they are.

[quote]Can I catch up the guys here like former international olympiad winners or equivalent people?[/quote]

I used to think the answer was yes, now I don't, at least in the vast majority of cases.  Many of the people who are winning IMO golds now could jump into courses at the graduate level instantly.  I think the gap only increases, personally.  Mathematical brilliance often manifests itself very early; look for example at people who have done very well in olympiads at a young age (Terence Tao, Noam Elkies, etc).  It's easy to say that people who are doing well in olympiads are just born bright and a lot of hard work is enough to catch up with them, but chances are that they are much brighter and harder working to boot.

But I think that trying to be an olympiad superstar is ultimately an unfulfilling goal.  Sure, it's nice for motivation (and I've used it, with very limited success, before) but at the end of the day, you have to like what you do, and be happy with whatever your accomplishments are.

Well i'm no pro but here's my 10cents,my 2cents are free...

When i came in this forum,I was like"8th graders are doing calculus and I'm sitting here doing nothing at all"  but i've heard this saying many times[b]"One who stops comparing himself with others and only compares himself with himself,he is god".[/b]

If you try to go ahead very fast,nor will you grasp anything at a higher level,nor will your basics be clear.You should start slowly and once the fire in you starts building up,you will realise that you've started doing so much stuff with no effort at all.Keep your mind busy at math and physics rather than thinking"he's done so much can I ever do that much...".This will  just create more anxiety in you.Your genius will build up automatically.(thooough some are born bright like blahblahblah says)

Just give your self a little push and the momentum will start building up :)

I wonder when I'll stop giving others advices and follow them :|

The more I think, the less I believe in talent. Not that I don't think it exists at all, but I think its effects are way less influential than most people might be inclined to believe. The problem is that other people can only guess that someone who is doing very well (say, winning olympiads) is talented, but such a person also has to work very diligently, and that's not generally observable. Therefore it is customary to attribute most of a person's success to talent. But actually it is clear to me that the people who win the IMO work a lot harder at winning the IMO than I ever did, and it is also clear to me that people who did not make the USAMO did not work as hard as I did. Now people say I'm talented because I take advanced classes, but I'm not. I just work harder than they do, and if they worked as hard I do, they would do just as well. I think talent is primarily an excuse for lazy people not to work hard.

Calvin agrees with ComplexZeta:

[img]http://images.ucomics.com/comics/ch/1995/ch950325.gif[/img]

(Unfortunately, I think this image will only last a week or so.)

I think Simon is mostly right. I would only add that there are more kinds of talent than just math that matter, such as tenacity and initiative. I believe that all people have a great deal of potential within them, and it's largely a matter tapping into it. Of course, at the highest levels, you start running into actual limitations, but most people have enough potential to make their work rewarding both in a sense of public recognition and personal fulfillment.

Hard work almost always leads to success, and inadequate work ethics mostly draws failure. Nevertheless, there have been far, far more hard workers than there have been the Einstein’s and the likes in history for me to believe that raw talent does not exist.

I know this has been said like.. a billion times, but whatever. repetition never hurt anyone (i think).

talent is an excuse for laziness. real talent is what your work ethic is. the better your work ethic, the more successful you will be in your endeavors.

anything is possible, if you try hard enough.


rotation
geometry
geometric transformation
number theory
relatively prime
combinatorics unsolved
combinatorics

Let's imagine that we have two coalescent regular [tex]n[/tex]-gones [tex]A_1A_2...A_n, B_1B_2...B_n[/tex] (and [tex]A_i = B_i[/tex]). Let the [tex]n[/tex]-gon [tex]A[/tex] be fixed and the [tex]n[/tex]-gon [tex]B[/tex] can rotate around its center so that after the first rotation [tex]B_1 = A_2, B_2 = A_3,..., B_n = A_1[/tex], after the second rotation [tex]B_1 = A_3, B_2 = A_4,..., B_n = A_2[/tex], etc. (similarly we can define [tex]m[/tex]-th rotation). Before the first rotation all vertices [tex]B_1, B_2,..., B_n[/tex] are coloured by black colour, but after each rotation the colour of each vertex can change from black to white and conversely according to this rule:

In each vertex of [tex]n[/tex]-gone [tex]A[/tex] there can be a converter or not. If (after some rotation) vertex [tex]B_i[/tex] get to some vertex [tex]A_j[/tex] with a converter, then the colour of [tex]B_i[/tex] will change to the contrary colour (white --> black, black --> white), if [tex]A_j[/tex] is a vertex without converter, then the colour of [tex]B_i[/tex] doesn't change.

Let [tex]m[/tex] denote the minimal possible number of rotations such that after [tex]m[/tex] rotations all vertices of [tex]n[/tex]-gone [tex]B[/tex] will be black again. Determine all possible numbers [tex]m[/tex] if the converters can be assigned to vertices [tex]A_i[/tex] arbitrarily.
note: obviously the colouring of [tex]n[/tex]-gone [tex]B[/tex] is "identical" (irrespective of the order of indexes of vertices [tex]B[/tex]) after [tex]p[/tex] and [tex]q[/tex] rotations if and only if [tex]p = q[/tex] (mod [tex]m[/tex]) - [tex]m[/tex] is a period.

My observation:
I can prove that maximal possible [tex]m[/tex] is [tex]2n[/tex] and I think (but only from some observation) that [tex]GCD(m,n) > 1[/tex] must work, but it's true that each [tex]m[/tex] satisfying [tex]GCD(m,n) > 1[/tex] doesn't have to be a period of previous rotations (I have found some contraexamples).

Could anybody give me some hint. Thanks...

Do you mean we calculate $m$ for each rotations, and you want to know what values it can take?
I think $m=2d$, $d\mid n$. :?

But it's not exactly true, let's take [tex]n[/tex] = 6 and let [tex]A_1,A_2,A_4,A_5[/tex] be the vertices of 6-gon [tex]A[/tex] that have a converter (and the remaining vertices [tex]A_3,A_6[/tex] don't have a converter), then it's easy to see that [tex]m[/tex] = 3 which doesn't satisfy your condition...

Number $m$ is good iff for all $i$ vertices $A_{i+1}$, $A_{i+2}$, ..., $A_{i+m}$ have even number of convertors. It follows that $A_{i+1}$ and $A_{i+1+m}$ have convertors or have not convertors in the same time. If we suppose $\gcd(m,n)=1$ then it yeilds ALL vertices have convertors.
Anyway, convertors are $m$-periodic.
It gives us that $m=d$ or $m=2d$, $d\mid n$. And we are to check what of these possibilities can be realized. All $m=2d$ work and $m=d\geq 3$ too.

Yes, that's similar to my further idea... But now it's the question how to exclude the other possible [tex]m[/tex] that aren't relatively prime with [tex]n[/tex], but don't satisfy that [tex]m = d[/tex] or [tex]2d[/tex] where [tex]d|n[/tex]. For concrete [tex]n[/tex] it isn't so hard to exclude those cases using the procedure you've written, because the location of all converters is always similar with some location we've obtain for some convenient [tex]m[/tex] (which is real period) (I've tried it for some little [tex]m[/tex]), but I don't know how to generalize it, maybe we can use that the distance between two neighbouring converters is [tex]GCD(m,n)[/tex]...

I don't understand you. In my opinion, I described all posibilities and pointed the solution.
Do you have couterexamples?

I'm sorry, I haven't read your message correctly...

But though I have a contraexample: let [tex]n = 12[/tex] and all converters are in vertices [tex]A_1,A_5,A_9[/tex], then [tex]m = 16[/tex] which doesn't satisfy your condition...

I'm sorry again... I'd better finish - your solution is complete.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Appologies if this has already been posted but I couldn't find it after
some hours of searching.  Prove:

$ \int_{\minus{}\infty}^{\infty}\sin(x)\sin(\frac{1}{x})\,\mathrm{d}x \equal{} \pi J_1(2)$

Where $ J_1(x)$ is the first order Bessel function of the first kind.
I found the result in a table of Fourier sine transforms published
decades ago by CalTech.  As you can see this integral is resistant to
contour integration because of the essential singularity at 0.

It seems to be true.  Or take half of it...
\[ \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( {x}^{ \minus{} 1} \right) {dx} \equal{} \frac {\pi}{2} \,{{\rm J}_1\left(2\right)}\]
According to Maple, both sides evaluate to 0.905917209663732742 ... pretty convincing, but not proof.

[u]Now, here is an argument that doesn't work right.[/u]

Write $ \sin(1/x)$ as a series:
\[ \sin\,\frac {1}{x} \equal{} \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k}{(2k \plus{} 1)! x^{2k \plus{} 1}}\]
This converges for all $ x \ne 0$.  Maple can do these integrals:
\[ E_k \equal{} \int _{0}^{\infty }\!{\frac {\sin \left( x \right) }{{x}^{2\,k\plus{}1}}}{dx
}\equal{}{\frac {{2}^{\minus{}2\,k\minus{}1}\sqrt {\pi }\;\Gamma  \left( \minus{}k\plus{}1/2 \right) }{
\Gamma  \left( 1\plus{}k \right) }}\]
So, assuming we can reverse the integral and the sum, our answer is
\[ \sum_{k\equal{}0}^\infty \frac{E_k (\minus{}1)^k}{(2k\plus{}1)!}\]
Plugging in the values for $ E_k$, we get another series that Maple can do.  The result is
\[ \frac{\pi}{4} \,{{\rm I}_1\left(2\right)}\plus{}\frac{\pi}{4} \,
{{\rm J}_1\left(2\right)} \equal{} 1.7022\]
Not the right answer.  It seems this is the correct sum of the series.  So this is a situation where interchange of the integral and the sum is not valid!

$ \int_0^{\infty}\frac {\sin(x) dx}{x^{2k \plus{} 1}}$ doesn't actually converge for $ k \geq .5$

It's not the sum and integral interchange that caused the problem.

Ok,
perhaps this one will help brew some interest in the original posting;  similar but
much easier:

$ \int_{\minus{}\infty}^{\infty}\left(1\minus{}x\sin(1/x)\right)^2\,\mathrm{d}x \equal{} \frac{\pi}{3}$

Ok, try the trick $ (1\minus{}a)^2\equal{}2(1\minus{}a)\minus{}(1\minus{}a^2)$ to express the integral:

$ \int_{\minus{}\infty}^{\infty}(1\minus{}x\sin(1/x))^2\,\mathrm{d}x \equal{} 2\int_{\minus{}\infty}^{\infty}(1\minus{}x\sin(1/x))\,\mathrm{d}x
\minus{}\int_{\minus{}\infty}^{\infty}(1\minus{}x^2\sin^2(1/x))\,\mathrm{d}x$

Another installment...

$ \int_{\minus{}\infty}^{\infty}\sin(x)\arctan(2/x)\,\mathrm{d}x
\equal{}\int_{\minus{}\infty}^{\infty}(1/x^2)\arctan(2x)\sin(1/x)\,\mathrm{d}x
\equal{}\pi(1\minus{}e^{\minus{}2})$

Why doesn't anyone seem to like these?

Here is my first approach to this problem, though it yielded no useful results at all...

\begin{eqnarray*} & & \int_{ - \infty}^{\infty} \sin x \sin \frac {1}{x} \, dx \\
& = & \int_{0}^{\infty} \sin x \sin \frac {1}{x} \, dx + \int_{0}^{\infty} \sin x \sin \frac {1}{x} \, dx \\
& = & \int_{0}^{\infty} \sin x \sin \frac {1}{x} \, dx + \int_{0}^{\infty} \sin x \sin \frac {1}{x} \, \frac {dx}{x^2} \qquad (x \mapsto 1/x) \\
& = & \int_{0}^{\infty} \left( 1 + \frac {1}{x^2} \right) \sin x \sin \frac {1}{x} \, dx \\
& = & \int_{0}^{\infty} \frac {1}{2} \left( 1 + \frac {1}{x^2} \right) \left[ \cos \left( x - \frac {1}{x} \right) - \cos \left( x + \frac {1}{x} \right) \right] \, dx \\
& = & \int_{ - \infty}^{\infty} \frac {1}{2} \left( \cos u - \cos \sqrt {u^2 + 4} \right) \, du \qquad (u = x - 1/x) \\
& = & \int_{0}^{\infty} \left( \cos u - \cos \sqrt {u^2 + 4} \right) \, du \end{eqnarray*}

Now put $ J(\alpha) = \int_{0}^{\infty} \left( \cos u - \cos \sqrt {u^2 + \alpha^2} \right) \, du$.

Then it is not hard to show that $ \alpha^2 J''(\alpha) - \alpha J'(\alpha) + \alpha^2 J(\alpha) = 0$. But the series solution to this equation yielded an unknown function, contrary to my expectation.

OKay, here is my solution to this integral. I have little time to write down my proof, so it is rather a sketch of my proof.

[url]http://sos440.springnote.com/pages/5461033[/url]

$ I: \equal{}\int_{\minus{}\infty}^\infty \left(1\minus{}x\sin\frac{1}{x}\right)^2 dx\equal{}2\int_0^\infty \left(1\minus{}x\sin\frac{1}{x}\right)^2 dx$

subst $ x \mapsto \frac{1}{x} \; : \quad I\equal{}2\int_0^\infty \left(1\minus{}\frac{\sin x}{x}\right)^2 \frac{dx}{x^2}$

$ \equal{}4\int_0^\infty \frac{1\minus{}\frac{\sin x}{x}}{x^2} dx\minus{}2\int_0^\infty \frac{1\minus{}\left(\frac{\sin x}{x}\right)^2}{x^2} dx\equal{}4\cdot \frac{\pi}{4}\minus{}2\cdot \frac{\pi}{3}\equal{}\frac{\pi}{3}$

For the other integral use the formula

$ \int_{\minus{}\infty}^\infty \frac{x \sin x}{t^2\plus{}x^2} \, dx\equal{}\pi\, e^{\minus{}t} \quad t>0$ .

Integrate with respect to $ t$ from $ 0$ to $ a$ :

$ \int_\infty^\infty \sin x \, \arctan\left(\frac{a}{x}\right) dx\equal{}\pi \left(1\minus{}e^{\minus{}a}\right)$


So just today my trusted and beloved ti-89 was stolen while I was at my university. I have had the thing for about 6 years, since iv been in middleschool. I am now a sophomore in College  double majoring in mathematics and computer engineering. I was going to look on ebay to get another old version of the ti-89 since I am used to and comfortable in it, but I ended up buying a TI-Nspire CAS from amazon and overnighting it to me (I have a test in my differential equations course on thrusday >< ). Anyways I suppose my question is, does anyone know if this calculator is good for mathematics majors? iv been lookind around and reading and I suppose it is still so new, no one has really evauluated it much, even on TI's nspire website they dont really list much about its capabilities. Though maybe it looks like its more intended for highschool level math courses, im not terribly sure. Anyone have any ideas?

I'm taking differential equations and haven't needed a calculator yet. In fact, I haven't needed a calculator in any of the math courses I've taken in college, except to check answers quickly in linear algebra. I have used Mathematica though.

The TI NSpire is a new model, and I hear that Texas Instruments is still issuing software upgrades for it. I can't imagine that it would be LESS useful for higher math than an 89, but I'm sure the marketing push is to get it into high school classrooms, so maybe that part of the feature set is better developed today. Does your college math department have advice on this issue?

[quote="asdf2345"]does this calculator do proofs?[/quote]

That's my LOL of the day.  :rotfl:

@tokenadult

Its not as stupid as it seems, there are programs which can prove simple geometry theorems.

[url]http://en.wikipedia.org/wiki/Automated_theorem_prover[/url]

Yeah, I know that, and I think my son has used three or four different online proving systems now (where he uses computer commands to show steps in proofs, which are then verified automatically). But I would really want to know what the proofs are like in a math course if someone tells me that his TI calculator can do the proofs for him.


MATHCOUNTS geometry perimeter AMC AIME ratio probability

rcv killed the other one, so it's time for a new one.

Problem 1

A circle is inscribed in a quarter circle with radius 8. What is the area of the inscirbed circle?

[quote="ragnarok23"]Problem 1

A circle is inscribed in a quarter circle with radius 8. What is the area of the inscirbed circle?[/quote]

[hide]$r(1+\sqrt{2})=8$
$r=8\sqrt{2}-8$

Answer: $(192-128\sqrt{2})\pi$[/hide]

[quote="daermon"]can you explain your reasoning?  I got a completely different answer  :huh:[/quote] [hide]You draw the radii of the smaller circle perpendicular to the radii of the large quarter circle. This forms a square. One of the quarter circle's radii consists of the diagonal of the square plus the radius of the small circle. That's where he got the equation. [/hide]

I still don't get it   :huh:   :help:

:ddr:

[quote="daermon"] $r=\frac{8}{1+\sqrt{2}}=8+4\sqrt{2}$[/quote] Try that again.  :wink:
$\frac{a}{b+c}\ne\frac{a}{b}+\frac{a}{c}$

[quote="daermon"]oops I was drawing a semicircle instead of a quarter circle :blush:  :rotfl:
thanks for the help guys  :wink:
now I think I have the answer [hide]$r+r\sqrt{2}=8 , r=\frac{8}{1+\sqrt{2}}=8+4\sqrt{2}, r^{2}\pi=(96+64\sqrt{2})\pi$[/hide][/quote]

lotrgreengrapes is right, the step to use from

$r=\frac{8}{1+\sqrt{2}}$ is to mutiply the top and bottom of the fraction by the conjugate of the denominator ($1-\sqrt{2}$)

Doing that gives you $r = 8\sqrt2-8$

[quote="lotrgreengrapes7926"]This should be familiar to many of you. :D

[size=134][b] Problem 2: [/b][/size]
In quadrilateral $ABCD, \angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD},$ $AB=18, BC=21,$ and $CD=14.$  Find the perimeter of $ABCD$.[/quote]

[hide]$AD^{2}=18^{2}+14^{2}+21^{2}$
$AD=31$

Answer: $18+21+14+31=\boxed{84}$[/hide]

[quote="usaha"]2006 aime #1.  I was surprised this was even an aime problem[/quote]

Yeah, it's rather easy and I would even expect it on the AMC 10.

The last questions of the AMC 10 are usually harder than the first question of AIME.

SplashD, you can post the next problem.

Problemo 3

A regular hexagon and a regular octagon have the same area. What is the ratio of their perimeters?

None posted anything in a while.....

Next problem  :)

$ b_1, b_2,b_3$ form a geometric sequence, $ b_1$ is the first term.
given that
$ b_1\times{b_2}\times{b_3} \equal{} 1728$
and
$ b_1 \plus{} b_2 \plus{} b_3 \equal{} 63$
find $ b$, and find the ratio.

[hide]$ 1728$ is equal to $ 12^3$, so a geometric sequence that would work without the sum constraint is 12, 12, 12. Since it is a geometric sequence, you divide the first number by a number and multiply the third by the same number.
12, 12, 12 Sum: 36
6, 12, 24 Sum: 42
4, 12, 36 Sum: 52
3, 12, 48 Sum: 63
$ b_1 \equal{} 3$
The ratio is equal to 4.
[/hide]
New problem:
Two teams are playing a series of games. The first team to win 3 games is the winner.  If team A has a 2/5 chance of winning any one game, what is the chance of team B winning the series, as a common fraction?

[quote="vamathletes"][hide]$ 1728$ is equal to $ 12^3$, so a geometric sequence that would work without the sum constraint is 12, 12, 12. Since it is a geometric sequence, you divide the first number by a number and multiply the third by the same number.
12, 12, 12 Sum: 36
6, 12, 24 Sum: 42
4, 12, 36 Sum: 52
3, 12, 48 Sum: 63
$ b_1 \equal{} 3$
The ratio is equal to 4.
[/hide]
[/quote]

is the ratio always an integer >1?

Fraenkel, a few notes about your solution:

The solution $ r\equal{}1$ is extraneous. It results in a zero denominator for $ b_{1}$. Also, the solutions 3 and 48 are analogous, they give the same sequence reversed.

Also, $ 4r^{3}\minus{}21r^{2}\plus{}21r\minus{}4\equal{}4r^{3}\minus{}4\minus{}21r^{2}\plus{}21r\equal{}4(r\minus{}1)(r^{2}\plus{}r\plus{}1)\minus{}21r(r\minus{}1)\equal{}(r\minus{}1)(4r^{2}\plus{}4r\plus{}4\minus{}21r)\equal{}(r\minus{}1)(r\minus{}4)(4r\minus{}1)$ would be the way to go for the factorization. Also, it can easily be seen that the sum of the coefficients is zero, so $ r\equal{}1$ must be a root, this can be factored out with synthetic division or whatever you want.

vamathletes, you are taking quite a dangerous path for your solution, you are assuming (a) there is only one possible answer, which is passable since the problem wording suggests this, and (b) you divide 12 by an integer. Why do you not check, for example, 8, 12, 18?

[quote="Fraenkel"][hide="$ 1^{st}$ attempt"]
$ b_1 + rb_1 + r^2b_1 = b_1(1 + r + r^2) = b_1\frac {1 - r^3}{1 - r} = 63$
$ b_1 = \frac {63(1 - r)}{1 - r^3}$
$ b_1\times b_2 \times b_3 = \left(\frac {63(1 - r)}{1 - r^3}\right)^3\times r^3 = 1728 = 12^3$
$ \left(\frac {(1 - r)r}{1 - r^3}\right)^3 = \left(\frac {12}{63}\right)^3$
$ \frac {(1 - r)r}{1 - r^3} = \frac {12}{63}$
$ 63r - 63r^2 = 12 - 12r^3$
$ 12r^3 - 63r^2 + 63r - 12 = 4r^3 - 21r^2 + 21r - 4 = 0$

I have no clue how to solve that for $ r$
Luckily I have a TI-89 that factors for me  :lol: .
$ 4r^3 - 21r^2 + 21r - 4 = (r - 4)(r - 1)(4r - 1) = 0$
$ r \in$ {$ 4$,$ 1$,$ \frac {1}{4}$}
plug it in $ b_1 = \frac {63(1 - r)}{1 - r^3}$ to find  $ b_1$.
$ b_1 = \frac {63(1 - r)}{1 - r^3} = \frac {63}{1 + r + r^2) = \frac {63}{1 + 1 + 1^2) = 21$
$ b_1 = \frac {63(1 - 4)}{1 - 4^3} = 3$
$ b_1 = \frac {63(1 - (\frac {1}{4}))}{1 - (\frac {1}{4})^3} = 48$
so in conclusion:
$ r \in$ {$ 4$,$ 1$,$ \frac {1}{4}$}
$ b_1 \in$ {$ 3$,$ 21$,$ 48$}

Im sure i must have done something wrong since I am tired.
I hope there is a much simpler method!
[/hide][/quote]
Yeah you got the answers but i didnt get r = 1 and b  = 21 in my solution..

[quote="vamathletes"][hide]$ 1728$ is equal to $ 12^3$, so a geometric sequence that would work without the sum constraint is 12, 12, 12. Since it is a geometric sequence, you divide the first number by a number and multiply the third by the same number.
12, 12, 12 Sum: 36
6, 12, 24 Sum: 42
4, 12, 36 Sum: 52
3, 12, 48 Sum: 63
$ b_1 = 3$
The ratio is equal to 4.
[/hide][/quote]
Yes,$ b = 3$,$ r = 4$ is one of the answers.
Here's my solution
[hide]

The sequence can be written as $ b, br, br^2...$
so we are given
$ b*br*br^2 = 1728$
$ b + br + br^2 = 63$
so
$ b^3\times{r^3} = 1728$
$ b(1 + r + r^2) = 63$

$ (br)^3 = 1728$

$ br = 12$
$ b(1 + r + r^2) = 63$

$ \frac {b(1 + r + r^2)}{br} = \frac {63}{12}$
so
$ 12 + 12r + 12r^2 = 63r$
$ 12r^2 - 51r + 12 = 0$
$ 4r^2 - 17r + 4 = 0$

$ (4r^2 - 16r) - (r - 4) = 0$
$ 4r(r - 4) - (r - 4)$
$ (4r - 1)(r - 4)$
Thus,

$ b = 3, r = 4$ or $ b = 48, r = \frac {1}{4}$

[/hide]

I dont know but $ r\equal{}1$ seems to work. But then $ b_1$ would just be the average of the sum $ b_1 \plus{} b_2 \plus{} b_3$
Thus,
$ \frac{b_1 \plus{} b_2 \plus{} b_3}{3}\equal{}\frac{63}{3}\equal{}21$

But I do not know whether by definition this is a geometric progression, so I am hesitant.
Although it is definitely a solution.

I think my teacher would call it [i][b]Trivial[/b][/i] and then roll his eyes  :P

[quote="Fraenkel"]I dont know but $ r \equal{} 1$ seems to work. But then $ b_1$ would just be the average of the sum $ b_1 \plus{} b_2 \plus{} b_3$
Thus,
$ \frac {b_1 \plus{} b_2 \plus{} b_3}{3} \equal{} \frac {63}{3} \equal{} 21$

But I do not know whether by definition this is a geometric progression, so I am hesitant.
Although it is definitely a solution.

I think my teacher would call it [i][b]Trivial[/b][/i] and then roll his eyes  :P[/quote]
Are you sure that works?

while it's true that $ 21\plus{}21\plus{}21 \equal{} 63$
but
$ 21\times21\times21 \neq 1728$

Okay, let's get this running again

I can't figure out the problem #

whatever

Consider the functions $ f(x)\equal{}2x^{2}\plus{}3x\minus{}5$ and $ g(x)\equal{}2x\plus{}3$. For what value(s) of $ x$ do(es) $ f(x)\equal{}g(x)$.

[hide]Set the functions equivalent and get $ 2x^{2} \plus{} 3x \minus{} 5 \equal{} 2x \plus{} 3$

Subtract on the left and get $ 2x^{2} \plus{} x \minus{} 8 \equal{} 0$

Applying the quadratic gives us $ \frac{\minus{}1 \pm \sqrt{65}}{4}$[/hide]

[quote="legodude1050"]Next problem:
Mr. Tedious starts at point $A$. He wants to travel to a river (a line $l$ that does not pass through $A$) and then to another point $B$ that lies on the same side of the river as $A$. However, he wants his path to intersect the river at exactly one point $P$, and he wants the angle formed by $A$, $P$, and $l$ to be one-fourth the size of the angle formed  angle formed by $B$, $P$, and $l$. How would he make a ruler-and-compass construction of his path?[/quote]
[hide]First, he would reflect $B$ over line $l$ to make $B'$. Then, he would construct a circle that has center $B'$ and is tangent to line $l$. Then he would construct a line tangent to that circle. Call the point of tangency $x$. Then he would  construct a circle with center $x$ tangent to $l$. Then he would construct a line tangent to circle $x$. This line intersects $l$ at point $M$. His path would then be $AMB$.[/hide]

It's been nearly 6 years since this was last posted in.

Well, aren't you glad that legodude found out this thread and revived it?


geometry
geometric transformation
homothety
geometry unsolved

Suppose that we have an equilateral triangle,in which there is a grasshopper and a flower .Every second the grasshopper chooses a vertex of the triangle and jumps toward that vertex a distance equal to $\frac{1}{10}$ of its distance from that vertex before the jump .Prove that disregarding to the original position of the grasshopper it can choose such a succession of jumps that after some time the distance between it and the flower will be less than $\frac{1}{1000}$.

This is my solution... I hope it is correct. Sorry if I can't speak English... I try.

Lemma: let P be a point inside the triangle. Then there is a vertex (let's call it A) such that $PA \le \frac{9}{10}h$ where $h$ is the altitude of the triangle.
Proof: if this is not true, the circles with centers in A,B,C and radius $\frac 9{10}h$ do not cover the triangle. But it is sufficient to take radius = $\frac{2}{3}h$ and they do.

Let us consider a circle centered at the flower with radius $\frac{1}{ 1000}$. Let $X$ be a vertex such that the distance from X to the flower is less than $\frac9{10}h$. Now, the circle $C_{1}$ obtained with a homotety centered at X with factor $\frac{10}9$ gives us a locus of points from which you can go sufficiently near the flower. Its centre is inside the triangle.
We continue to take the nearest vertex from the last centre, and do homoteties to find a succession of circles $C_{1},C_{2},C_{3},\ldots$ with centres inside the triangle. From each point inside one of these circles, I can go sufficiently near the flower.

But they radii get bigger and bigger, so one of these circles will cover the triangle. (Or even an arbitrary point in the plane!). Then, wherever the grasshopper is, he can go near the flower.

[size=75][edited: 910 ---> 9/10[/size]

Congratulations [b]edriv[/b]!Very nice solution! :)
In fact this problem was unsolved during the mentioned competition!
Thank you! :wink:

[quote="edriv"] Let $X$ be a vertex such that the distance from X to the flower is less than $9{10}h$. Now, the circle $C_{1}$ obtained with a homotety centered at X with factor $\frac{10}9$ gives us a locus of points from which you can go sufficiently near the flower.[/quote]

I think you mean ''the distance  from X to the flower is less than $\frac{9}{10}h$   and not $910h$ .
Also why did you take factor of homothety  $\frac{10}{9}$  ?

[quote="silouan"]
Also why did you take factor of homothety  $\frac{10}{9}$  ?[/quote]
Notice that the circle $C_{1}$,which obtained after homotety,is the locus of point from where the grasshopper can get near enough to the flower jumping towards the vertex $X$!
When the grasshopper jumps towards a vetex $X$,its transmition can be discribed as a homotety with center $X$ and coificent $\frac{9}{10}$.


Prove that among any four positive integers greater than or equal to $1$ there two, say $a$ and $b$, such that $\frac{\sqrt{(a^{2}-1)(b^{2}-1)}+1}{ab}\geq\frac{\sqrt{3}}{2}$.

Try It

[hide]${f(a,b)=\frac{\sqrt{\left(a^{2}-1\right) \left(b^{2}-1\right)}+1}{a b}}$

${f(20,1)==\frac{1}{20}<\frac{\sqrt{3}}{2}}$[/hide]

[quote]
Prove that among any four positive integers greater[/quote]

Four!? i didnt get this.... are two no?!

its just greater, no!?  if it is equal to one its untrue


Since that Russian guy proved it two months ago or something, I think the Poincare Conjecture change to the Poincare Theorem.
I think you admins should change the rank thingy (when you have two stars and 100 posts) to the PT instead of keeping it on Poincare "Conjecture".

I think there was already a discussion about that...
One could still argue that the ranks are the conjectures they were when the Clay institute offered their prices.
At least I'm against any rank that sounds as it is already proven ;) :D

PS: He didn't prove it two months ago, but earlier (a good time earlier).


Let $x_{1}, x_{2},...,x_{n}$ be positive reals, find the greatest possible $k$ such that \[{\left(\sum_{i=1}^{n}x_{i}\right)^{n}}-n^{n}{\prod_{i=1}^{n}x_{i}}\geq k\left(\sum_{i=1}^{n}x_{i}\sum_{i=1}^{n}\frac1{x_{i}}-n^{2}\right){\prod_{i=1}^{n}x_{i}}.\]

For those who don't want to calculate what the best constant might be, I guess it's $k=\frac{1}{2}n^{n-1}$. At least, I can prove that $k\leq \frac{1}{2}n^{n-1}$ setting $x_{i}=1+\epsilon_{i}$ and calculating everything up to terms of second degree in the $\epsilon_{i}$.

Edit: This is spam, as always. Setting $x_{2}=...=x_{n}=1$ and letting $x_{1}\to 0$, we see that $k\leq (n-1)^{n-1}$ which is then best possible.

Have a look here http://www.mathlinks.ro/Forum/viewtopic.php?t=14906 :)

actually we can put a stronger bound on $k$
if one of the $x_{i}$ goes near $0$ and the other $n-1$ equal 1 then $k\le (n-1)^{n-1}\le \frac{n^{n-1}}{2}$ with equality in the last only for n=2
and for the proof:
(n-1)EV reduces this problem directly to seeing the case when n-1 variables are equal (to 1 for instance since it's homogenuous)
then we are left with:
$n(x+n-1)^{n}-n^{n}x-(n-1)^{n}(x-1)^{2}\geq 0$

but if you set $x_{i}=\frac{1}{n}$, the left side of the equation becomes $0$. But the factor $\sum_{i=1}^{n}x_{i}\sum_{i=1}^{n}\frac1{x_{i}}-n^{2}$ is also $0$. Thus, k can be any number, because the product is always $0$. I.e., you can easily choose a k which is greater then the k that Peter has mentioned. Or am I making some terrible kind of mistake? Please correct me, if I am wrong!

[quote="Blender"] Please correct me, if I am wrong![/quote]The ineq with $k$ has to hold for $any$ choice of the $x_{i}$  :lol: Doing a particular choice gives you an upper bound each time, not more  :wink: So you've just proved the upper bound is $\infty$ :rotfl:


A car travels due east at $ 2/3$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $ 51$ miles, moves southeast at $ 1/2 \sqrt2$ mile per minute. At time $ t \equal{} 0$, the center of the storm is $ 110$ miles due north of the car. At time $ t \equal{} tsub1$ minutes, the car enters the storm circle, and at time $ t \equal{} tsub2$ minutes, the car leaves the storm circle. Find $ 1/2(tsub1 \plus{} tsub2)$

Please bear with my inability to use Latex, or something. I'm still fairly new to this site.

All right then, so what I thought of doing was creating a diagram in which the storm is $ 110$ miles directly above the car. Therefore, the edge of the storm is only $ 59$ miles above the car at this point. Because its moving $ 1/2 \sqrt2$ mph, it must be moving $ 1/2$ miles southward every minute.

Okay, so if its $ 59$ miles above, then it should take $ 118$ minutes to reach the car right?

In that time, the storm will have also moved $ 59$ miles east, and the car will have moved $ 78 \plus{} 2/3$ miles east. So, then I just calculated how much time it would take for the car to move atleast $ 51$ miles east of the storm so that it cannot be inside it (because the radius of the storm is 51).

Well, the overtake speed of the car is $ 1/6$ miles per minute ($ 2/3 \minus{} 1/2$). When the car enters the storm, the car will have already progressed $ 78 \plus{} 2/3 \minus{} 59 \equal{} 19 1/3$ miles east of the storm's center. It needs only to go $ 51 \minus{} 19 \plus{} 1/3 \equal{} 31 \plus{} 1/3$ miles more. At the overtake speed, this will take $ 188$ minutes. Therefore, if the car stays in the storm for $ 188$ minutes, I find that the answer is $ 94$

Well, if I did anything wrong or assumed something (and I'm sure I did), I hope someone will tell me. Thanks for the help in advance.

[quote="veezbo"]Therefore, the edge of the storm is only $ 59$ miles above the car at this point. Because its moving $ 1/2 \sqrt2$ mph, it must be moving $ 1/2$ miles southward every minute.[/quote]
I haven't thought about this question carefully, but the point of the storm closest to the car is not the same thing as the southernmost point of the storm.

In the future, keep all discussion confined to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=393642#p393642]original thread.[/url] That way, if another person has the same problem in the future, they need only look at one thread.


inequalities
group theory
abstract algebra
superior algebra
superior algebra theorems

Not hard but slightly cute... Prove:
If $f$ is an endomorphism of a finite group $G$, then
\[|f(G)|^{2}\le |G|\cdot|f(f(G))|\]
where $|S|$ denotes the number of elements of a finite set $S$.

there's an obvious surjective map of left-cosets

$G / f(G) \to f(G) / f(f(G))$

so that the inequality follows from $|G/U|=|G|/|U|$, where $U$ is a subgroup of the group $G$.

in the same manner, we get more generally

$|f(G)| \cdot |U| \leq |G| \cdot |f(U)|$

where $U$ is a subgroup of the group $G$.


calculus integration function parameterization derivative logarithms trigonometry

Compute the following: integral from 0 to  \pi /4 from the function
f(x)=ln(cosx)/cosx

Seems that this integral is not that simple.
But it appears in the 12th grade manual!

First of all, we cant express the primitive of ln(cosx)/cosx by elementary functions (it reduces at ln(x)/x-1 which appears in M Smoleanski table of primitives as incalculable)

I think it can be made as an parameter integral(we shall introduce a parameter and then use the derivation under integrals).

[quote="xxxxtt"]
I think it can be made as an parameter integral(we shall introduce a parameter and then use the derivation under integrals).[/quote]

Are you thinking something like
$g(t)= \int _{0}^{\pi /4}(cosx)^t dx$
$g'(t)= \int _{0}^{\pi /4}ln(cosx).(cosx)^t dx$ and $g'(-1)$

Do I miss something or it is possible to write it as x1ln(cosx)/cosx and simply integrate by parts?

The function is ln(cosx)/cosx not xlncosx/cosx.

Have anyone tried to put this integral into a symbolic program like Mathematica or Maple or Matlab? The result is that the program used complex variables and some 'dilogs' and 'Catalan's constant'. In other words, not only is this problem hard but the exact answer is insanely big. Not in our lifetime would this problem appear on any test, it's ugly even for a homework question. This is just as hard as finding an elementary function whose derivative is (sin x)/x.

I tried that too, and I got something using the elliptic integrals, so, basicly, it still uses an integral. I don't think the problem is suitable for work in the classroom :).

Use powers series

For $|x|<\frac{\pi}{2}$

$ln(cosx)=-\sum_{k\geq 1}\frac{2^{2k-1}(2^{2k}-1)|B_{2k}|}{k.(2k)!}x^{2k}$

$\frac{1}{cosx}=\sum_{k\geq 0}\frac{|E_{2k}|}{(2k)!}x^{2k}$

Use Cauchy product for powers series to get an expansion of $f(x)$ and get

$\int_{0}^{\frac{\pi}{4}}\frac{ln(cos(x))}{cosx}dx$

I found (after some calculation with Maple)$
\int _{0}^{1/4\,\pi }\!{\frac {\ln  \left( \cos \left( x \right)
 \right) }{\cos \left( x \right) }}{dx}=-1/2\,\ln  \left( 2 \right)
\ln  \left( \sqrt {2}-1 \right) +5/8\, \left( \ln  \left( 2 \right)
 \right) ^{2}+1/2\,{\it polylog} \left( 2,-1/4\,\sqrt {2}+1/2 \right)
-1/4\, \left( \ln  \left( 2+\sqrt {2} \right)  \right) ^{2}-1/24\,{
\pi }^{2}
$
where $
{\it polylog} \left( 2,x \right) =\sum _{n=1}^{\infty }{\frac {{x}^{n}
}{{n}^{2}}}
$
But ${\it polylog} \left( 2,-1/4\,\sqrt {2}+1/2 \right)$ is not a known value of the polylog of order 2. So there is no simple value for this integral


calculus
integration
trigonometry
logarithms
Euler
calculus computations

[quote="prithwijit"]1) Evaluate $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$.

2) Evaluate $\int_{0}^{\pi}\frac{dx}{(1+\cos\alpha\cos x)^{2}}$.

 :lol:[/quote]
Solution to [b]problem (1)[/b]:

First note that $\sqrt{\frac{1-\sin x}{1+\sin x}}= \frac{\cos x}{1+\sin x}.$ So,

$I = \int \frac{\cos x}{1+\sin x}\sqrt{\frac{2-\sin x}{2+\sin x}}\,dx$. Now make the substitution $y=\sin x$. We then have

$I = \int \frac{1}{1+y}\sqrt{\frac{2-y}{2+y}}\,dy = \int \frac{(2-y)\,dy}{(1+y)\sqrt{4-y^{2}}}=-\int \frac{(1+y)-3}{(1+y)\sqrt{4-y^{2}}}\,dy$
$I =-\int \frac{\,dy}{\sqrt{4-y^{2}}}+3\int \frac{\,dy}{(1+y)\sqrt{4-y^{2}}}=-{\sin}^{-1}\left( \frac{y}{2}\right)+I_{1}$ (say)

Make the substitution $y = 2\sin\theta$ in $I_{1}.$ We have
$I_{1}= 3\int \frac{\,d{\theta}}{1+2\sin\theta}= 6\int \frac{\frac12{\sec}^{2}\frac{\theta}{2}}{{(\tan\frac{\theta}{2}+2)}^{2}-{(\sqrt{3})}^{2}}\,d{\theta}$ ... [making use of the identity $\sin\theta = \frac{2\tan\frac{\theta}{2}}{1+{\tan}^{2}\frac{\theta}{2}}$]
$I_{1}= \frac{6}{2\sqrt{3}}\ln \Big| \frac{ \tan\frac{\theta}{2}+2-\sqrt{3}}{ \tan\frac{\theta}{2}+2+\sqrt{3}}\Big|$

Thus, $I =-{\sin}^{-1}\left( \frac{\sin x}{2}\right)+\frac{1}{\sqrt{3}}\ln \Big| \frac{ \tan\left( \frac{{\sin}^{-1}\left( \frac{\sin x}{2}\right)}{2}\right)+2-\sqrt{3}}{ \tan\left( \frac{{\sin}^{-1}\left( \frac{\sin x}{2}\right)}{2}\right)+2+\sqrt{3}}\Big|+C$

[quote="prithwijit"]1) Evaluate $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$.

2) Evaluate $\int_{0}^{\pi}\frac{dx}{(1+\cos\alpha\cos x)^{2}}$.

 :lol:[/quote]

I got #2: Let $\beta = \cos\alpha$ so that our integral is $K: =\int_{0}^{\pi}\frac{dx}{(1+\beta\cos x)^{2}}$

Notice that since $\left| \beta\cos x\right| <1$ for $0<x<\pi$, the following power series expansion is valid:

$\frac{1}{(1+\beta\cos x)^{2}}= \frac{1}{\beta\sin{x}}\cdot\frac{d}{dx}\left( \frac{1}{1+\beta\cos x}\right) = \frac{1}{\beta\sin{x}}\cdot\frac{d}{dx}\left( \sum_{k=0}^{\infty}(-\beta)^{k}\cos^{k}x\right)$
$=-\frac{1}{\beta\sin{x}}\sum_{k=1}^{\infty}(-\beta)^{k}k\cos^{k-1}x\sin x = \sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\cos^{k}x$

hence we have

$K=\int_{0}^{\pi}\left(\sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\cos^{k}x\right) dx = \sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\int_{0}^{\pi}\cos^{k}x dx$
$= \sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\left( \int_{0}^{\frac{\pi}{2}}+\int_{\frac{\pi}{2}}^{\pi}\right) \cos^{k}x dx$
$= \sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\left(\int_{0}^{\frac{\pi}{2}}\cos^{k}x\, dx+(-1)^{k}\int_{0}^{\frac{\pi}{2}}\sin^{k}x\, dx\right)$
$= \frac{1}{2}\sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\left(\mbox{B}\left( \frac{k}{2}+\frac{1}{2},\frac{1}{2}\right)+(-1)^{k}\mbox{B}\left( \frac{k}{2}+\frac{1}{2},\frac{1}{2}\right)\right)$
$= \frac{1}{2}\sum_{k=0}^{\infty}(-\beta)^{k}(k+1)\left(1+(-1)^{k}\right) \frac{\Gamma\left(\frac{k}{2}+\frac{1}{2}\right)\Gamma\left( \frac{1}{2}\right) }{\Gamma\left(\frac{k}{2}+1\right)}$
$= \sum_{k=0}^{\infty}(-\beta)^{2k}(2k+1)\frac{\Gamma\left( k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) }{\Gamma\left( k+1\right)}= \sum_{k=0}^{\infty}\beta^{2k}(2k+1)\frac{\Gamma\left( k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) }{\Gamma\left( k+1\right)}$

recall that $\Gamma\left( k+\frac{1}{2}\right)\Gamma\left( \frac{1}{2}-k\right) = \frac{\pi}{\sin\left(\frac{\pi(2k+1)}{2}\right)}=\pi (-1)^{k}$ and that $\Gamma (x+1)=x\Gamma (x)$ to get

$K= \sum_{k=0}^{\infty}\beta^{2k}(2k+1)\frac{\Gamma\left( k+\frac{1}{2}\right)\Gamma\left( \frac{1}{2}-k\right)\left(-\frac{1}{2}\right) \Gamma\left(-\frac{1}{2}\right) }{\Gamma\left( \frac{1}{2}-k\right)\Gamma\left( k+1\right)}$
$=-\frac{\pi}{2}\sum_{k=0}^{\infty}(-1)^{k}\beta^{2k}(2k+1)\frac{\Gamma\left(-\frac{1}{2}\right) }{ \left(-\frac{1}{2}-k\right) \Gamma\left(-\frac{1}{2}-k\right)\Gamma\left( k+1\right)}$
$= \pi\sum_{k=0}^{\infty}(-1)^{k}\beta^{2k}\frac{\Gamma\left(-\frac{1}{2}\right) }{ \Gamma\left(-\frac{1}{2}-k\right)\Gamma\left( k+1\right)}= \pi\sum_{k=0}^{\infty}(-1)^{k}\left(\begin{array}{c}-\frac{3}{2}\\k\end{array}\right) \beta^{2k}= \frac{\pi}{(1-\beta^{2})^{\frac{3}{2}}}$

where $\left(\begin{array}{c}-\frac{3}{2}\\k\end{array}\right)$ is a generalized binomial coefficient, and since $\beta = \cos\alpha$ we have $1-\beta^{2}= 1-\cos^{2}\alpha = \sin^{2}\alpha$ so that the required value of the integral is

$\int_{0}^{\pi}\frac{dx}{(1+\cos\alpha\cos x)^{2}}= \frac{\pi}{(1-\beta^{2})^{\frac{3}{2}}}= \frac{\pi}{|\sin\alpha |^{3}}$

-Ben

Here are some other ways to deal with integrals like No2.
n is a positive integer ,a and b are real numbers.

1.The substitution cos$\theta$ =$\frac{acosx+b}{a+bcosx}$ ( with appropriate restrictions on the values of and b)transforms

$\int_{0}^{\pi}{(a+bcosx)^{-n}dx}$ into

${(a^{2}-b^{2})^{-n+{\frac{1}{2}}}}$.${\int_{0}^{\pi}(a-bcos\theta)^{n-1}d\theta}$

2.We can find a reduction formula for$U_{n}$ =$\int(a+bcosx)^{-n}dx$ by differentiating

$\frac{sinx}{(a+bcosx)^{n-1}}$

The reduction formula is

$(n-1)(a^{2}-b^{2})U_{n}$ =$\frac{-bsinx}{(a+bcosx)^{n-1}}$ +$(2n-3)aU_{n-1}-(n-2)U_{n-2}$

Another solution to #1:

$I \, = \, \int \sqrt{\frac{2-\sin x}{2+\sin x}}\, \frac{\cos x \cdot dx}{1+\sin x}$

Now, with the substitution (similar to Euler 3rd one)

$y^{2}\, = \, \frac{2-\sin x}{2+\sin x}$,

$\sin x \, = \, 2 \frac{1-y^{2}}{1+y^{2}}$,

$1+\sin x \, = \, \frac{3-y^{2}}{1+y^{2}}$

and

$\cos x \cdot dx \, = \,-8 \frac{y \cdot dy}{\left( 1+y^{2}\right)^{2}}$

so

$I \, = \,-8 \int \frac{y^{2}\cdot dy}{\left( 3-y^{2}\right) \left( 1+y^{2}\right)}$

which is an easy rational integral


quadratics
number theory unsolved
number theory

let $m,n,a$ be natural numbers find nessecery and suffient condition such that if $a$ be a quadratic residue mod$m$ and quadratic residue mod$n$ . then $a$ be a quadratic residue mod$mn$.

For $a$ being a quadratic residue $\mod k$, its necessary and sufficient that it is a quadratic residue $\mod$ all prime powers dividing $k$. But for odd primes, $a$ is a quadritic residue $\mod p^m$ iff it is a quadritic residue $\mod p$, so there will no problems occure.
So $a$ is a quadratic residue $\mod mn$ iff it is a quadratic residue $\mod$ $m$ and $n$ and $\mod$ the highest power of $2$ dividing $mn$.


geometry
geometric transformation
reflection
circumcircle
IMO Shortlist
geometry proposed

Let $ ABC$ be a triangle. Let $ \mathcal C(O, R)$, $ H$ be its circumcircle and orthocenter, respectively. Denote by $ A_1$, $ B_1$, $ C_1$ the reflection points of $ A$, $ B$, $ C$ across the sides $ BC$, $ CA$, $ AB$, respectively. Prove that $ A_1$, $ B_1$ and $ C_1$ are collinear if and only if $ OH \equal{} 2R$.

[b]Remark.[/b] $ OH\equal{}2R\Longleftrightarrow a^2\plus{}b^2\plus{}c^2\equal{}5R^2$ a.s.o. (metrical proof !)

This is problem from IMO Shortlist 1998 and it was also posted at:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=122241#122241[/url]

and solution can be found in this topic.

[color=darkblue]Thank you ,[b] Delegat[/b], for the lower address ! Here is a very nice [b]Grobber[/b]'s proof.[/color]
http://www.mathlinks.ro/Forum/viewtopic.php?p=122241#122241 [quote="grobber"][color=darkred][b][u]Proof.[/u][/b] Let $ X,Y,Z$ be the projections of the nine-point center $ \mathcal E$ on $ BC,CA,AB$ respectively. If $ M,N,P$ are the midpoints of $ BC,CA,AB$ respectively and $ D,E,F$ are the feet of the perpendiculars from $ A,B,C$ on the opposite sides, then $ X,Y,Z$ are the midpoints of $ MD,NE,PF$ respectively. [b]Use vectors to show that[/b] $ XY\parallel A_1B_1$, $ YZ\parallel B_1C_1$ and $ ZX\parallel C_1A_1$. This means that $ A_1,B_1,C_1$ are collinear $ \Longleftrightarrow$ $ X,Y,Z$ are collinear $ \Longleftrightarrow$ the projections of $ \mathcal E$ on the sides of $ \triangle\ ABC$ are collinear $ \Longleftrightarrow \mathcal E$ lies on the circumcircle of $ ABC$ (using reciprocally the [u]Simson's line[/u]) $ \Longleftrightarrow$ $ \frac {OH}{2} \equal{} R$, i.e. $ OH \equal{} 2R\ .$[/color][/quote]
[color=darkblue]I"ll show now only what is missing from the [b]Grobber[/b]'s proof

(for lack of time or because it is simply),  i.e. $ A_1B_1\parallel XY\ .$

Indeed, since the nine-point center is the middlepoint of the segment $ [OH]$, obtain

$ \left\{\begin{array}{ccccc} 2\cdot\overline {\mathcal EX} & \equal{} & \overline {HD} \plus{} \overline {OM} & \equal{} & \overline {HD} \plus{} \frac 12\cdot\overline {AH} \\
 \\
2\cdot \overline {\mathcal EY} & \equal{} & \overline {HE} \plus{} \overline {ON} & \equal{} & \overline {HE} \plus{} \frac 12\cdot\overline {BH}\end{array}\right\|$ $ \implies$

$ \frac {\overline {\mathcal EX}}{\overline {\mathcal EY}} \equal{}$ $ \frac {\overline {HD} \plus{} \left(\overline {AH} \plus{} \overline {HD}\right)}{\overline {HE} \plus{} \left(\overline {BH} \plus{} \overline {HE}\right)} \equal{}$ $ \frac {\overline {HD} \plus{} \overline {AD}}{\overline {HE} \plus{} \overline {BE}} \equal{}$ $ \frac {\overline {HD} \plus{} \overline {DA_1}}{\overline {HE} \plus{} \overline {EB_1}} \equal{}$ $ \frac {\overline {HA_1}}{\overline {HB_1}}$ $ \implies$

$ \frac {\overline {\mathcal EX}}{\overline {\mathcal EY}} \equal{}$ $ \frac {\overline {HA_1}}{\overline {HB_1}}$, i.e. $ \triangle\ X\mathcal EY\sim\triangle\ A_1HB_1$. In conclusion, $ A_1B_1\parallel XY\ .$[/color]

[color=darkred][b]Repeat, I like very much the Grobber's proof.[/b][/color]


I'm given a figure of a right triangle with base of 12, leg of 5 and hypotenuse of 13.
I'm supposed to find the exact trig value of the function in 2 ways using power reducing formulas.

The 2 problems are:

sin^2 (x)

and

cos^2 (x)

where angle x is opposite of leg (5)

I know the answer is 25/169 for sin^2 and 144/169 for cos^2, but I don't know how to use power reducing formulas to get there--I just know the answer from the figure. My teacher said I'd be using double angle formula, but I don't know where.

The power reducing formulas are:
sin^2 (x) = [u]1- cos 2x[/u]
........................2

and

cos^2 (x) = [u]1+ cos 2x[/u]
.........................2

please help me if you can.

Why would you use the power-reducing formulas?  To calculate $ \cos 2x$ you'd need to calculate either $ \sin^2 x$ or $ \cos^2 x$ anyway...

i don't really know, but my teacher told me to use the power reducing formulas to solve for it, so i didn't know if anyone knew why they'd be useful because it seems to me you could just get the answers from the diagram.  i posted this to see if i was missing some major concept or if my teacher is perhaps just crazy. jk--well i don't know--just checking for input

The power-reducing formulas aren't useful at all here for the reason I've already stated.

The thing is

the power reducing formulas are derived from the double angle formula for cosine

If you're going to use the power reducing formula and then using the double angle formula right after, then essentially, you've just wasted two steps.


trigonometry
trig identities
Pythagorean Theorem
geometry
Law of Cosines

[b][u]Derive:[/u][/b] $ \sin (a \plus{} b) \sin (a \minus{} b) \equal{} (\sin a \plus{} \sin b) (\sin a \minus{} \sin b)$

All methods are welcome, especially geometric ones :) (just please don't cite similar trig identities)

[i]I like this one because I find it particularly easy to remember[/i]

Can we use the sine addition/subtraction identities?

You can use whatever you want. The idea is that deriving trig formulas is the best way to learn trig and/or familiarize oneself with it.

I just mean like, don't use the Law of Cosines to prove the Pythagorean Theorem type of thing -- nothing that makes it a non-exercise.

[hide="Derivation"]
\begin{align*} \sin \left(a + b\right)\sin \left(a - b\right) & = \left(\sin a \cos b + \cos a \sin b\right)\left(\sin a \cos b - \cos a \sin b\right) \\
& = {\left(\sin a \cos b\right)}^2 - {\left(\cos a \sin b\right)}^2 \\
& = \sin^2 a \cos^2 b - \cos^2 a \sin^2 b \\
& = \sin^2 a \left(1 - \sin^2 b\right) - \sin^2 b\left(1 - \sin^2 a\right) \\
& = \sin^2 a - \sin^2 a \sin^2 b - \left(\sin^2 b - \sin^2 b \sin^2 a\right) \\
& = \sin^2a - \sin^2a \sin^2b - \sin^2 b + \sin^2 a \sin^2 b \\
& = \sin^2 a - \sin^2 b \\
& = \boxed{\left(\sin a + \sin b\right)\left(\sin a - \sin b\right) }\end{align*}
[/hide]

[hide="Solution using identities."]
$ sin(a \plus{} b)sin(a \minus{} b)$
$ \equal{} [ sin(a)cos(b) \plus{} sin(b)cos(a) ][ sin(a)cos(b) \minus{} sin(b)cos(a) ]$
$ \equal{} sin^2(a)cos^2(b) \minus{} sin^2(b)cos^2(a)$
$ \equal{} sin^2(a)[ 1 \minus{} sin^2(b) ] \minus{} sin^2(b)[ 1 \minus{} sin^2(a) ]$
$ \equal{} sin^2(a) \minus{} sin^2(a)sin^2(b) \minus{} sin^2(b) \plus{} sin^2(a)sin^2(b)$
$ \equal{} sin^2(a) \minus{} sin^2(b)$
$ \equal{} [ sin(a) \plus{} sin(b) ][ sin(a) \minus{} sin(b) ]$

Oops, too late.
[/hide]


induction
number theory unsolved
number theory

Prove that there are infinitely many positive integers which cannot be expressed by the arithmetic mean of different nonnegative powers of 2.

Let a positive integer $ n$ be good if it is the arithmetic mean of some pairwise distinct power of $ 2$, and bad otherwise.

First, we prove that $ n$ is good if and only if $ 2n$ is good.

- If $ n$ is good then multiply by $ 2$ each of the power of $ 2$ which appear in the arithmetic mean, then we deduce that $ 2n$ is good.
- If $ 2n$ is good then the numerator of the mean equals to $ 2n$ is even. But $ 2^0$ is the unique odd power of $ 2$ and the power of $ 2$ appearing in the numerator are pairwise distinct, therefore $ 2^0$ does not appear in the numerator. This ensures that we may divide each of these power of $ 2$ by $ 2$, which gives that $ n$ is good.

It follows that we only have to find one bad number.

Now, we prove that $ 13$ is bad :
Assume that $ 13 \equal{} \frac 1 k (\sum 2^i)$ where the sum is over $ k$ pairwise distinct nonnegative integers $ i$.
Thus $ 13k \geq 2^0 \plus{} 2^1 \plus{}\cdots \plus{} 2^{k\minus{}1} \equal{} 2^k \minus{}1$.
But an easy induction shows that, for $ k  \geq 7$, we have $ 13k < 2^k \minus{} 1$.
Thus, we have $ k \leq 6$.
Direct checking gives that the binary decomposition of $ 13k$ does not use exactly $ k$ digits $ 1$ for $ k\leq 6$. A contradiction.
Thus, $ 13$ is bad and, from above $ 13\cdot 2^n$ is bad for each nonnegative integer $ n$.

A question (open for me) : if $ n$ is good, is the decomposition unique or not? I do not see why it would be unique but...

Pierre.

[quote="pbornsztein"]
A question (open for me) : if $ n$ is good, is the decomposition unique or not? I do not see why it would be unique but...[/quote]
It is not unique, here comes an example:
$ 27 \equal{} \frac {2^0 \plus{} 2^4 \plus{} 2^6}{3} \equal{} \frac {1 \plus{} 16 \plus{} 64}{3}$

$ 27 \equal{} \frac {2^2 \plus{} 2^3 \plus{} 2^5 \plus{} 2^6}{4} \equal{} \frac {4 \plus{} 8 \plus{} 32 \plus{} 64}{4}$

A question: Is it $ 13$ the minimum bad number?


quadratics
algebra
quadratic formula
AMC

Help with this simplifying fraction. Thanks. :D

The fraction \[\frac {2(\sqrt 2 + \sqrt 6)}{3(\sqrt {2 + \sqrt {3})}}\]

I tried my best but I just can't take the bottom square root off easily.

 ;)

You say you are having difficulty rewriting the denominator in a usable form.  One method is to guess the form of the result.
[hide]
Let us guess the form of the denominator is:

$\sqrt{a}+\sqrt{b}$

If our guess is correct, we should be able to set this equal to the denominator, and solve for a and b:

$\sqrt{a}+\sqrt{b} = \sqrt{2+\sqrt{3}}$

Square both sides:

$a+b+2\sqrt{ab}=2+\sqrt{3}$

$(a+b) + \sqrt{4ab} = 2+\sqrt{3}$

Then equate the integer part and the radical part:

$a+b = 2$
$4ab = 3$

Now, a little algebra, followed by the quadratic formula:

$a+b=2$
$a=2-b$

$4ab=3$
$4(2-b)b=3$
$8b-4b^2=3$
$4b^2-8b+3=0$

$b=\frac{8\pm\sqrt{8^2-4(4)(3)}}{2(4)}=\frac{8\pm4}{8}=\frac{2\pm1}{2}$

$a=\frac{2\mp1}{2}$

$\sqrt{2+\sqrt{3}}=\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}$

Almost done.  Can you finish the problem?

It is important to note that if we had guessed poorly, this method might not have worked so well.  [For example, the radicand in the quadratic formula might not have been a perfect square.]  When guessing the form of result, expect some trial and error.  Your guesses should improve with experience.[/hide]


MIT college Stanford Harvard Yale Princeton articles

[url=http://www.thes.co.uk/worldrankings/]The Times Higher Education Supplement[/url] has recently posted its world ranking of universities.

[b]General:[/b]

[color=blue][i]- The World's top 200 universities[/i][/color]

[b]Hard Subjects:[/b]

[color=blue][i]- The world's top 100 science universities
- The world's top 100 technology universities
- Top non-university institutions in science
- Top non-university institutions in technology[/i] [/color]

[b]Soft Subjects:[/b]

[color=blue][i]- The world's top arts and humanities universities
- The world's top social science universities
- The world's top non-universities in social science[/i] [/color]

[b]Biomedicine:[/b]

[color=blue][i]- Top 100 biomedicine universities
- Top non-university institutions in biomedicine [/i][/color]

[b]Geographical Excellence: [/b]

[color=blue][i]- Top 50 European universities
- Top 50 North American universities
- Top 50 universities in the rest of the world [/i][/color]

Wrt my interest in technology (technology universities) the TOP 5 is as follows: MIT, Berkeley, IIT, Stanford and Imperial College London.  :)

http://www.ccer.pku.edu.cn/ss/world-rankingsUnis.pdf
http://www.tuwien.ac.at/pr/news/doc_news/times_world_ranking_2005.pdf
http://www.its.caltech.edu/~xuanluo/random/documents/worldrankings2005.pdf
http://cgi.cse.unsw.edu.au/~ksho/public/worldrankings2005.pdf
http://www.comp.nus.edu.sg/~wangdeli/worldrankings2005.pdf
http://www.epfl.ch/soc/etudes/pdf/world-rankingsUnis.pdf
http://www.jeffooi.com/archives/THES-RANKS2005.pdf

Well, I see there is little discussion about whether the ranking is correct wrt one's own feelings and opinions. Anyway here is the TOP-5 of the [b]richtest US-Universities[/b]:

1.) Harvard: $\$$  22,587,305,000

2.) Yale: $\$$ 12,740,896,000

3.) Princeton: $\$$ 10,251,952,000

4.) Stanford: $\$$ 9,922,041,000

5.) MIT: $\$$ 5,869,800,000

Well, what to do with all the money? [url=http://www.admissionsconsultants.com/college/ivy_league_table.asp]Havard[/url] for example provided more scholarships. Thus the acceptance rate dropped below 10% as more smart people of the poor pool applied. In 1991 Jack Meyer took over the Harvard endownment fond which was worth $\$$ 4.7 billions back then. By means of his superior strategy with an average return of 15.9% he managed to accomplish that within 14 years. It should be noted that his portfolio included 13% hedge fonds. But Yale has even more: 26%. Due to Harvard student's criticism about their salaries Jack Meyer and his crew will retire from this job and will start their own hedge fund - with an unbelievably high seed capital of $\$$ 4 billion dollar. Student's criticised the last year's brokerage of Jack Meyer, David Mittelman and Maurice Samuels of $\$$ 34.1 million and $\$$ 35,1 million, respectively. Well I don't get the students' issue. They should be happy that scientific standard due to scholarships of Harvard will probably rise even higher and control their envious feelings.

I would have to agree with the idea that if Harvard has a good fund manager, that is good for Harvard.

As long as his increased return covers his salary and then some :-).

I think they're just jealous the Yale guy takes home much less. :p Fortune did an article about them a couple of months back, and I must say I am fairly impressed that these two aren't in the private sectors on their own.

But on another note, how did MIT get rich? I mean the first 4 were always up there but I didn't recognize that MIT had quite a few billion for itself.

It's primarily concerned with grad schools and not undergrad, which makes a world of difference where Berkeley and several other universities (Michigan, Chicago, etc) are concerned.

[quote="aznphatso"]Interesting that Berkeley is so high. From my high school (which is in SF bay area) ~50 kids get into Berkeley each year, ~3 get into Stanford, and the top maybe 1 or 2 opt for the East coast.[/quote]

Well I didn't browse for the exact facts but Berkeley can admit at least twice as many students as Stanford.

um the richest school part is WRONG

I know that the university of texas at austin has a little over 12 billion in assets

[quote="blahblahblah"]Endowment and assets are not the same.  Endowment refers mostly to the money they have invested, e.g, which is free to be spent.[/quote]
I know this

how does that change the fact that it is one of the "richest"

i am assuming when they say "richest" schools they mean the ones with the most money regardless of where it came from

Read between the lines.  Those numbers quoted refer to endowments.


Each valve $ A$, $ B$, and $ C$, when open, releases water into a tank at its own constant rate.  With all three valves open, the tank fills in $ 1$ hour, with only valves $ A$ and $ C$ open it takes $ 1.5$ hour, and with only valve $ B$ and $ C$ open it takes $ 2$ hours.  How long will it take to fill the tank with only valves $ A$ and $ B$ open?

[hide="ANSWER"]$ \boxed{1.2 \ \text{hours}}$[/hide]

Thanks in advance!

[hide]
$ \frac1{A} \plus{} \frac1{B} \plus{} \frac1{C} \equal{} 1$

$ 1.5\left(\frac1{A} \plus{} \frac1{B}\right) \equal{} 1$

$ 2\left(\frac1{B} \plus{} \frac1{C}\right) \equal{} 1$
Then solve $ t(\frac1{A} \plus{} \frac1{B}) \equal{} 1.$
[/hide]

this just MIGHT be faster that $ \frac{\pi^2}{6}$'s way
to find b, consider valves a and c as 2 fused valves *valve ac*
now valve ac fills the tank in 1.5 hrs and valves ac and b fill the tank in 1 hour
valve ac did $ \frac{1}{1.5}$ or $ \frac{2}{3}$ of the filling and valve b did $ 1\minus{}\frac{2}{3}$ or $ \frac{1}{3}$
which means valve b can fill the tank in 3 hrs
doing the same for valve bc and a, ytou find a fills the tank in 2 hrs and then $ \frac{3\cdot2}{3\plus{}2}\equal{}1.2$


Roy, the babysitter, can clean the entire house in 30 minutes. The children can mess up the entire house in 55 minutes. If the entire house is a mess and Roy begins cleaning the house while the children are messing it up, how long will it take Roy to clean the entire house?

[quote="quach.jim"][hide]$(\frac{1}{30}-\frac{1}{55})(t) = 1$
$\frac{1}{66}(t) = 1$
$t=66$ minutes[/hide][/quote]
[hide="An explanation of the above:"]
Imagine that the cleanness of the house (or the house itself) is on a number line, from $0$ (completely messy) to $1$ (completely clean). Then, the kids are pushing the house towards the dirty end at a rate of $\frac{1}{55}$ and Roy is pushing the house towards the dirty end at a rate of $\frac{1}{30}$, since if you remember your rate formula we have $d=rt \Longrightarrow r=\frac{d}{t}$.
$d=1$ in this case, and $t$ is either $30$ or $55$ minutes. The two rates will partially cancel each other out, but the push towards the clean end is greater. Thus the rate towards the clean end is equal to $\frac{1}{30}-\frac{1}{55}=\frac{1}{66}$. Then, since the house starts at the $0$ end (i.e. the house starts completely messed up), the time will be $t=\frac{d}{r}=\frac{1}{\frac{1}{66}}=\boxed{66 \text{ minutes}}$.[/hide]

[hide]in one minute, the babysitter cleans up $\frac{1}{30}$ of the mess.  in one minute, the kids create $\frac{1}{55}$ of a mess.  so, in every minute, $\frac{1}{30}-\frac{1}{55}=\frac{1}{66}$ of a mess is cleaned up and it will take $\boxed{66}$ minutes for the babysitter to clean up.[/hide]


Hello,
[quote="hieuchuoi@"]Let $ABC$ be a triangle with three angles $A,B,C$ such that $sin(nA)+sin(nB)+sin(nC)=0$ where n is a positive integer. Prove that there is an angle of triangle ABC which equals to $\frac{\pi}{n}$[/quote]

It seems wrong :
$n = 8$, $A=\frac{\pi}{4}$, $B=\frac{\pi}{4}$, $C=\frac{\pi}{2}$

$sin(nA)+sin(nB)+sin(nC) = sin(8A)+sin(8B)+sin(8C)=sin(2\pi)+sin(2\pi)+sin(4\pi) = 0$
and neither A, B or C is $\frac{\pi}{8}$

--
Patrick


What is the remainder when $ 2007$ is divided by $ 81$?