geometry
geometric transformation
reflection
rotation

Let A, B, C, D, E be five points lying on a circle k in the given order. Assume that < ABE = < CEB = < ECD = 45°. Let S be the area of the circle k. Find the sum of the areas of the triangles ABE and CDE and of the circular segments intercepted from the circle k by the chords AE and DE. (These areas are red in the attached figure.)

Since the angles $\measuredangle ABE = \measuredangle CEB = \measuredangle ECD$, the chords $EA = BC = DE$ are equal and since these angles are equal to $45^o$, the chords $EA, BC, DE$ have lengths equal to the side length $a$ of a square inscribed in the circle $(K)$. Let $F$ be the reflection of the point $E$ in the circle center $K$, i.e., $EF$ is a diameter of the circle $(K)$. Rotate the triangle $\triangle ECD$ counter-clockwise around the center $K$ of the circle $(K)$ by the angle of $90^o$ and denote the rotated triangle $\triangle E'C'D'$. The rotated vertex $E'$ coincides with the original vertex $D$, the rotated vertex $D'$ coincides with the point $F$ and the rotated vertex $C'$ coincides with the point $B$. Since $AEDF$ is a square inscribed in the circle $(K)$, $EA \parallel DF$. Hence, the triangles $\triangle AEB$ and $\triangle E'C'D \equiv \triangle DBF$ have the common vertex $B$ and parallel sides $EA \parallel DF$, which are also the sides of the inscribed square $AEDF$. Let $G, H$ be the feet of altitudes of the triangles $\triangle AEB$ and $\triangle E'C'D \equiv \triangle DBF$ from their common vertex $B$ to the sides $EA, DF$, respectively. The distance between the lines $EA, DF$, equal to the sum $BG + BH$ of these altitudes, is also equal to the side lenght $a$ of the inscribed square. Consequently, sum of the areas $\triangle_1, \triangle_2$ of the triangles $\triangle AEB$ and $\triangle E'C'D \equiv \triangle DBF$ is

$\triangle_1 + \triangle_2 = \frac 1 2 BG \cdot EA + \frac 1 2 BH \cdot DF = \frac{a}{2}(BG + BH) = \frac{a^2}{2}$

The areas of the circle sections bound by the chords $EA, DE$ and the minor arcs $EA, DE$ are equal to each other and to

$\triangle_3 = \frac 1 4 (\pi r^2 - a^2)$

where $r$ is the radius of the circle $(K)$. The total area in question is

$\triangle_1 + \triangle_2 + 2\triangle_3 = \frac{a^2}{2} + \frac 1 2(\pi r^2 - a^2) = \frac 1 2 \pi r^2 = \frac 1 2 S$

Sorry to revive this thread, but I think it's worth of!
Let $M=AD\cap BE, N=AD\cap CE, P=Ac\cap BD$. We shall prove that $[ABM]+[DCN]=[MEN]\ (\ 1\ )$.
We see that triangles $AMB, NME, NDC$ are similar so, to prove $(1)$ we need to show $AM^2+ND^2=MN^2\ (\ 2\ )$.
Easy angle chase gives $AC\bot BE, CE\bot BD$; with $BE$ bisector of $\angle ABD, CE$ bisector of $\angle ACD, P$ is both the symmetrical of $A$ through $BE$ as well as of $D$ through $CE$. $\angle APM=\angle MAP, \angle NPD=\angle NDP, \angle PAD+\angle ADP=45^\circ$, so $\angle APM+\angle DPN=45^\circ$; with $\angle APD=135^\circ$ we infer $\angle MPN=90^\circ$ and $(2)$ has been proven.

Best regards,
sunken rock


[quote="xholydivinity"]How would CO2 absorbed from the atmosphere affect the molarity of a solution? Thanks[/quote]
What do you mean "affect the molarity of a solution"
the molarity of CO2 will increase, the other will remain the same

had CO2 reacted with something in experiment it will change it.
if you burn something and need really correct results it's better to get the oxygen from K2MnO4 or KClO3.


I want to self-study macroeconomics, would you please introduce some text (book or pdf) to me?
Moreover, I have the "[i]Macroeconomics[/i]" (18th ed) by P. A. Samuelson and W. D. Nordhaus. How do you think about it? And is there any supplement (in pdf) to that book?

i highly recommend Dr. Mankiw's book that covers both macro and micro economics extensively, and assumes virtually no prerequisite of any type of economics. Perfect for any ap/ib exam (covers more than ib however) and any intro micro/macro college course as well i presume.

[quote="basted"]MIT OCW can help. Apart from lecture notes they have the textbooks they use for a class.[/quote]

I personally thought the MIT OCW notes are a little too mathematics oriented (and way too much emphasis on the Solow Growth model, but I guess for good reasons  :rotfl: )

Personally, I found the Berkeley Podcast on Macroeconomics are pretty good. The Economics lectures by Professor Rimothy Taylor (via The Teaching Company *hint bt) are really good as well.

In terms of textbooks, most of them are really similiar. So I won't go all picky.

你好！


group theory
abstract algebra
superior algebra
superior algebra unsolved

let < x > denote the generator for a Group G. and x^-1 (x inverse)

Prove

If G = < x > , then G = < x^-1 >

[b]Proposition:[/b][i]Let $ \left\langle g\right\rangle\equal{}G$ and assume that $ \left|g\right|\equal{}n<\infty$.  Then $ G\equal{}\left\langle g^{a}\right\rangle$ if and only if $ \gcd(a,n)\equal{}1$.[/i]
[i]Proof.[/i]  It is a fact that $ g^{a}$ generates a subgroup of $ G$ of order $ g^{a}$.  This subgroup equals all of $ G$ if and only if $ \left|g^{a}\right|\equal{}\left|g\right|$.  It is also a fact that this is true that if and only if $ \frac{n}{\gcd(a,n)}\equal{}n$, i.e. if and only if $ \gcd(a,n)\equal{}1$.  You can check the infinite case.

Since $ \gcd(\minus{}1,n)\equal{}1$ regardless of $ n$, $ g^{\minus{}1}$ generates $ G$.

Here's another way to see it in the case of a finite group:

You know that $ \{1,g,g^{2},\ldots,g^{n\minus{}1}\}$ has to contain the inverses of all its elements.  So $ \{1,1\}$, $ \{g,g^{n\minus{}1}\}$, $ \{g^{2},g^{n\minus{}2}\}$, and so on.  In the set $ \{1,g^{\minus{}1},g^{\minus{}2},\ldots,g^{1\minus{}n}\}$, the set of inverses are $ \{1,1\}$, $ \{g^{\minus{}1},g^{1\minus{}n}\}$, $ \{g^{\minus{}2},g^{2\minus{}n}\}$, and so on.  So $ 1\equal{}1$, $ g\equal{}g^{1\minus{}n}$, $ g^{2}\equal{}g^{2\minus{}n}$, and so on.

[quote="JRav"]It is a fact that $ g^{a}$ generates a subgroup of $ G$ of order $ g^{a}$.[/quote]
What did you actually mean to say here?

If you don't want to split into finite and infinite cases, just show that both groups contain each other.


modular arithmetic
number theory proposed
number theory

prove that if $ p$ is a prime $ >3$ and $ m\equal{}\frac{4^p\minus{}1}{3}$ then $ 2^{m\minus{}1} \equiv 1 \pmod {m}$.

$ 3m\equal{}4^{p}\minus{}1$  $ \implies$  $ 3m\equiv 3\pmod{p}$  $ \implies$ $ m\equiv 1\pmod{p}$
$ m\equiv 1\pmod{2}$  $ \implies$  $ 2p|m\minus{}1$
$ 2^{2p}\equiv 1\pmod{m}$  $ \implies$  $ 2^{m\minus{}1}\equiv 1\pmod{m}$   :wink:

Double reference and already posted: VJIMC 2002 (see [url=http://vjimc.osu.cz/]here[/url]) and British Mathematical Olympiads 1993 (see [url=http://www.bmoc.maths.org/home/bmolot.pdf]here[/url])  :wink:


If $ (AB,CD) \equal{} \minus{} 1$ prove that $ AC,AB,AD$ are in Geometrical Progression...
this doesn't look right though

$ (AB,CD) \equal{} k \Rightarrow \frac {\vec{AC}}{\vec{CB}}: \frac {\vec{AD}}{\vec{DB}} \equal{} k$
where $ A,B,C,D$ are collinear points


function
probability
expected value
combinatorics unsolved
combinatorics

$n\in{Z^{+}}$ and $A={1,\ldots ,n}$. $f: N\rightarrow N$ and $\sigma: N\rightarrow N$ are two permutations, if there is one $k\in A$ such that $(f\circ\sigma)(1),\ldots ,(f\circ\sigma)(k)$ is increasing and $(f\circ\sigma)(k),\ldots ,(f\circ\sigma)(n)$ is decreasing sequences we say that $f$ is good for $\sigma$. $S_\sigma$ shows the set of good functions for $\sigma$.
	a) Prove that, $S_\sigma$ has got $2^{n-1}$ elements for every $\sigma$ permutation.
	b)$n\geq 4$, prove that there are permutations $\sigma$ and $\tau$ such that, $S_{\sigma}\cap S_{\tau}=\phi$
	.

This got bumped by a troll post that was deleted, but it seems like it's worth answering anyhow.  First, let me rewrite it:

Fix a positive integer $n$, and let $[n] = \{1, \ldots, n\}$.  Given a permutation $\sigma: [n] \to [n]$, we say that the permutation $f: [n] \to [n]$ is [i]good[/i] for $\sigma$ if there exists some $k \in [n]$ such that $f(\sigma(1)) < f(\sigma(2)) < \ldots < f(\sigma(k)) > f(\sigma(k + 1)) > \ldots > f(\sigma(n))$.  Let $S(\sigma)$ be the set of permutations of $[n]$ that are good for $\sigma$.
a) Prove that $|S(\sigma)| = 2^{n - 1}$.
b) For $n \geq 4$, prove that there exist permutations $\sigma$ and $\tau$ such that $S(\sigma) \cap S(\tau) = \varnothing$.

Part (a) is straightforward: first we count permutations $w: [n] \to [n]$ such that there exists $k$, $w(1) < \ldots < w(k) > \ldots > w(n)$.  Each such permutation corresponds to the subset $\{w(1), \ldots, w(k - 1)\}$ of $[n - 1]$, and in fact this correspondence is easily seen to be a bijection.  Thus, the number of such permutations is $2^{n - 1}$.  For each such permutation $w$, we there exists a permutation $f = w \circ \sigma^{-1}$ that is good for $\sigma$.  Since the set of permutations of $[n]$ forms a group, these $f$s are all distinct, and every $f$ good for $\sigma$ must be of this form.

Part (b) seems trickier; maybe someone else can come up with a nicer argument.  Consider choosing a permutation $\sigma$ at random.  Because of the nice group structure, a given permutation $w$ is in $S(\sigma)$ with probability exactly $\frac{|S(\sigma)|}{n!} = \frac{2^{n - 1}{n!}$.  Thus, if we choose two permutations $\sigma$ and $\tau$ independently at random, the expected number of permutations in $S(\sigma) \cap S(\tau)$ is precisely $n! \cdot \left(\frac{2^{n - 1}}{n!}\right)^2 = \frac{2^{2n - 2}}{n!}$.  For $n \geq 7$, this number is less than 1; if the expectation is less than 1 then there must be actual values less than 1, and so in particular there is some pair $\sigma, \tau$ where the two sets have empty intersection.  For $4 \leq n \leq 6$, this leaves us to find a pair of permutations by hand.  We may as well take $\sigma$ to be the identity permutation; for $n = 4$ take $\tau = 2143$, for $n = 5$ take $\tau = 21435$, and for $n = 6$ take $\tau = 214356$.

Edit: okay, I see, the probabilistic argument is not really necessary at all: we can just show that the pair $12\cdots n$ and $2143 56\cdots n$ works for any $n \geq 4$.  But the probabilistic argument shows something more, namely, having $S(\sigma)$ and $S(\tau)$ intersect is in fact an extremely rare coincidence.


Find all x > 0 such that (13 + \sqrt x)1/3 + (13 - \sqrt x)1/3 is an integer.

[i]unknown source[/i]

This problem is easy, but I think every1 scares to write sln.

Let a = (13 +  sqrt(x))^1/3 , b = (13 -  sqrt(x))^1/3 then a+b is integer and a^3 + b^3 = 26. We also have
       26 = a^3 + b^3 = (a+b)^3 - 3ab(a+b)
=>  ab = ((a+b)^3 - 26)/(a+b)
But a.b = (13 +  sqrt(x))^1/3.(13 -  sqrt(x))^1/3 = (169-x)^1/3 < 6
=> a+b = 1, 2, 3 and respectively we have a.b = -25, -9, 1/3, x=15794, 898, 4564/27.

I really dont like such kind of problem.

Namdung


induction inequalities trigonometry geometry vector logarithms conics

So the exam's tomorrow...and I thought that we ought to have a Post-Exam Discussion Thread...
Also Best of luck to all!!

if i'm not wrong u'r in delhi naa rijul...best of luck BTW mate..i think RMO'09 will be seriously tough seeing the level of '08 :) :)

@all
post ur sols,I wud like to check them with mine [hide]provided I get any  :( [/hide]

First of all
Congrats to all for faring well in RMO 2009-10 :)

How did you write the exam?
How many questions did you attempt?

I think RMO paper may be different for different regions
In AP we got the one prepared by NBHM only-no separate paper.

well! once again it went terrible atleast for me  :(

i cud perfectly solve only 2 questions (Q4) and (Q6)

i cud have done correctly (Q1)

but alas i didnt try

this time , i think cutoffs gonna increase !!!!

plzzz post solutions and tell how you all fared!!!!

Got 3 correct. Proved another one, but my method was totally wrong and might just be able to confuse the examiner into a few marks. Last question......EFFFF I imagined tearing page numbers instead of pages!!! Also attempted another one very lil, got nowhere.

I cudn't do well........I did only 2....i.e Q2 & Q4
Wont make it :(

i attempted all .....got 5 perfectly right and one i didnt complete.......

i could provide solutions if at all anyone wants..........

can anybody please upload the questions ??? i did not qualify for the RMO , but i dont think anyone would mind me trying .

Right if at all ur posting solutions please hide them .

Well, 4 questions completely correct... :lol: .though I doubt the combinatorics one...full of calculations...
and a partial attempt on the polygon's first part...
So summary goes like this....
1) Question 1 - Filled this one up with trigo....but neatly...and then got the answer to be 90

2) Question 2 - Had very little clue about this one at first....so went crazily brutal....showed that only a $ \equiv$ 10 (mod 17) satisfies $ 17|a^2 \minus{} 3a \minus{} 2$ out of 0 to 16 (so wasted 17 lines for getting just this) and then ended up with a contradiction because $ 17|a\minus{}10 \Longrightarrow 289|a^2 \minus{} 20a \plus{}100 \Longrightarrow 289|\minus{}17a \plus{} 119 \Longrightarrow 17|\minus{}a \plus{} 7 \Longrightarrow 17| a\minus{}7$ which is a contradiction with $ 17|a\minus{}10$....

3) Question 3 - Got no clue about this one....but banged my head when I saw the Sophie-Germain answer ......I could have easily done this one.... :(

4) Question 4 - Changed the question into $ Sum \ of \ all \ three \ digit \ nos. \minus{} sum \ of \ the \ three \ digit \ nos. \ with \ the \ parity \ of \ all \ three \ nos. \ the \ same...$ and so got something like  $ 43xxxx$  I don't remember exactly...

5) Question 5 - Tried this one right at the end....using induction....I used the theorem that the line segment joining any two sides of a triangle is always less than or equal to the greatest side of the triangle....so first took no. of sides as 3 and then by induction tried to extend it to n+1 sides...but ran out of time and couldn't complete it.....
6) Question 6 - This one I'm sure about the answer....3 pages..did this one at the start....though I found that those who had tried the question x of rmo 199x could do this easily..
Thanks to all those who completely read my post.... :D

I'm posting the  Q1&2

[hide="Q1"]
$ Q1.$ Let in $ \Delta ABC$, $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle[BAC]$
[16M]
[hide="Sol."]
$ AI \equal{} (s \minus{} a)sec \frac {A}{2}$
$ a \equal{} c \plus{} (s \minus{} a)sec\frac A2$
That gives me
$ a(1 \plus{} \frac {sec \frac A2}{2}) \equal{} c(1 \plus{} sec \frac {A}{2})$
Put $ a \equal{} 2RsinA$
$ c \equal{} 2RsinC$ $ \implies c \equal{} 2R cos \frac {A}{2}$
You get $ \angle[BAC] \equal{} \frac {\pi}2$
[/hide]
[/hide]

[hide="Q2"]
$ Q2.$ Show that there is no integer $ a$ such that $ a^2 \minus{} 3a \minus{} 19$ is divisible by 289.
[15M]
[hide="Soln"]
Let $ a \equiv k(mod17)$
Then for$ 17 \mid a^2 \minus{} 3a \minus{} 19$    $ a \equiv 10(mod17)$
So let $ a \equal{} 17k \plus{} 10$
Then $ a^2 \minus{} 3a \minus{} 19 \equal{} 289a^2 \plus{} 289a \plus{} 51$
That ends the proof
[/hide]
[/hide]

I accidentally reached the answer after things canceled out after 1.25 pages :P

@Layman
http://www.isid.ac.in/~rbb/inmosol_07.pdf

http://www.isid.ac.in/~rbb/rmoqpapers.html

[quote="mihir0492"]i just realised that i gave CBSE group maths olympiad(GMO) instead of RMO...though the paper was same as rmo, but only 30 students all India get selected from gmo for Inmo.....anybody has any idea when the rsults are declared?[/quote]
Yeah...its quite alright mihir...you have the [b] same chances [/b]as RMO in GMO...even though here 30 are selected from all over INDIA
And the results are not yet out...you can contact antiparticle (another user here) as he has also given the GMO...

Also @all 11th standard Mathlinkers---The INMO is the same date (and for some also the same time) as the KVPY Interviews....even though I'm not in 11th ...still I think somethings gotta be done for this...Any views on this one?

This was initially happening last year as well
But we asked Amitabha sir to help and he did so
Actually he could help bcos of 2 reasons
1) KVPY date was only tentative and not officially declared(unlike this year)
2) He himself is 1 of the interviewer at KVPY and could not allow it to clash with INMO
He told us in camp that be relieved that at least for Delhi region(in reference to KVPY) KVPY will not clash with INMO
But yes he can't do anything for other regions of KVPY

[quote="Swatigupta"]CBSE GMO and KV GMO olympiad is out.
From where i can download previous year papers of CBSE GMO?
RMO papers are available but I guess they have different paper from RMO?
Swati[/quote]
No Swati the GMO papers are exactly the same as the RMO...
BTW did Antiparticle (Archit) get selected?

[quote="Layman conjecture"]so exactly how many problems do you have to do in the inmo to be sure that you get selected?[/quote]
1 if you are a genius and invent something completely new.
2 if you are a flawless smartass AND very lucky
3 if you are good at proofwriting AND lucky
4 if you aren't that perfect
5 if you suck
6 you CANNOT do all :P

Disclaimer: Just my opinion.

[quote="abhitayal"]Yes
antiparticle got selected
congrats antiparticle as well as other AOPSians and mathlinkers who got selected[/quote]

Thanks! I got lucky, GMO went pathetic as the marks suggest.

By the way, how do you know me?

Hey does anyone kno from where I can get the RMO 2009 OFFICIAL SOLUTIONS? Please don't give me the solutions FIITJEE, etc. have posted...  :)  I really need them... :o

[quote="fbaggins"]Hi!!!!

EVERYONE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I got a call today from the MP Co-ordinator :lol:

He said that I QUALIFIED FOR RMO WITH A RANK 10 IN THE STATE!!!!!!!!!!!!!!! :D

Hurray!!! feeling on top of the world today!!!!

Thanks all aopsians specially Agr_94 math (Aravind) for encouraging me and Pocha too !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Still cant believe that only 2 problems can get u a rank 10 in the state :rotfl:

Anyways

On to INMO NOW!!!!!!!!!!![/quote]

@fbaggins... do you know how many marks you had scored in rmo? is your true name tanmoy?

@Faros Amora:- Yes, his name is Tanmoy, I think.


absolute value
complex analysis
complex analysis unsolved

I have a midterm tomorrow, and this appeared on a past test.

Let $ p(z)\equal{}az^4\plus{}bz\plus{}1$ for $ a\in [1,\pi]$ and $ b\in [2\pi \minus{} 2, 7]$. Find the maximum number of roots in the annulus $ \{z: 1<|z|<2\}$.

It seems to me that the natural approach is to use Rouche to minimize the number of roots in $ |z|<1$ and maximize the number in $ |z|<2$. Then the difference of these would be the max in the annulus.

It just seems like there are too many variables floating around to effectively use Rouche, though. Any other suggestions?

We can check that all four zeros always lie in $ |z| < 2$ by comparing with $ az^4$:

On $ |z| = 2$ we have

\begin{eqnarray*} |p(z) - az^4| & = & |bz + 1| \\
& \leq & b|z| + 1 \\
& = & 2b + 1 \\
& \leq & 15 \\
& < & 16a \\
& = & |az^4| \end{eqnarray*}

So by Rouche, $ p(z)$ and $ az^4$ have the same number of zeros inside $ |z| < 2$, which is four.

Now the goal is to find a choice of $ a$ and $ b$ with the given restrictions, that has none of the zeros inside $ |z| < 1$, try comparing with 1:

On $ |z| = 1$

\begin{eqnarray*} |p(z) - 1| & = & |az^4 + bz| \\
& = & |z\parallel{}az^3 + b| \\
& \leq & a|z|^3 + b \\
& = & a + b \end{eqnarray*}

Thus if there is a choice of $ a$ and $ b$ such that $ a + b\leq |az^4 + bz + 1|$ on $ |z| = 1$, we will be done. I don't see if this is possible or not, though. I feel like it is quite possible that there will always be a zero inside $ |z| < 1$...

Edit: I just checked the case $ a=1$ and $ b=2\pi-2$ on maple, and found that there is a mistake in what I wrote, since it gives 2 roots outside of $ |z|<2$.

[quote="HilbertThm90"] I don't see if this is possible or not, though. I feel like it is quite possible that there will always be a zero inside $ |z| < 1$
[/quote]
What can you say about the product of your zeroes? ;)
[quote]
Edit: I just checked the case $ a \equal{} 1$ and $ b \equal{} 2\pi \minus{} 2$ on maple, and found that there is a mistake in what I wrote, since it gives 2 roots outside of $ |z| < 2$.
[/quote]
No, Hilbert, your argument showing that all roots are less than 2 is correct. If Maple thinks otherwise, it is a shame to its developers. Trust your proof in such questions, not the software :).

Maple 10 says
[quote]> a:=1; b:=2*Pi-2;
> evalf(solve(a*z^4+b*z+1=0));
 0.8857070733 + 1.411896051 I, -0.2341731756, -1.537240971, 0.8857070733 - 1.411896051 I [/quote]
All roots are in the disk $ |z|<2$. I guess [b]HilbertThm90[/b] forgot to take the square root when computing the absolute value of complex roots.

:blush:  Ah. That is precisely what I did.

The product of the roots is $ \frac{-1}{a}\leq 1$. Hmm, so at least 1 is in $ |z|<1$. Maybe compare with $ bz$ then?

On $ |z|=1$:

\begin{eqnarray*}
|az^4+bz+1-bz| & = & |az^4+1|\\
  & \leq & |a\parallel{}z|^4 + 1\\
  & \leq & a + 1\\
  & \leq & 2\pi - 2 \\
  & \leq & |bz| \\
\end{eqnarray*}

So there is always 1 zero in $ |z|<1$, meaning that there are 3 in the annulus.

Ack. If I only thought to check the last option. I was semi-convinced I could find one with no zeros in the inner disk, though... I find this problem strange, since I didn't have to find the maximum. According to this, any $ a$ and $ b$ that satisfy those conditions will always have 3 in the annulus.


a forensic biologist conducts a simulation aimed at determining how the growth rate of maggots increases with time, in order to investigate post mortem death calculations. Six cultures are set uo and growth rates recorded over time. Data obtainded is summarised below:

CULTURE                              A      B      C       D       E      F
t=Time period (hours)           2.9    6.1    0.7    6.9    4.8    8.9
R=Growth rate (units/hour)   10.0  13.1   8.1   13.4   10.8  14.5

the relationship over this period is approximated by a straight line relationship of the form
         R=A+Bt    ....[1]

where time (t) is plotted on the horizontal axis and R (growth rate) on the vertical axis.

(1) Use the linear regression method to determine the constants A and B

(2) Determine the correlation coefficient r

(3) Determine the coefficient of determination, r squared, and interpret its meaning

(4) What is the growth rate after 10 hours, assuming the linear relationship (given by equation [1] above) holds?

[hide]The independent variable is  and the dependent variable is , as the goal is to write  as a function of .

(1) To determine the value of the constants  and , recognize

and

Input the values to obtain  and

(2) The linear correlation coefficient  is determined by the equation

Input the values to obtain

(3)

(4) The equation with the constants  and  is

Because  is in hours,  is substituted, yielding

[/hide]


A pair of fair six sided dice is rolled. What is the probability the positive difference between the two numbers is a factor of one of the numbers?

The probability of getting a positive difference between the numbers on the dice [hide="is"]5/6 because in 6/36=1/6 of thre roll both dice will show thw samwe number.[/hide]
In those cases there are only [hide="hidden"] 15[/hide]
possible pairs of dice readings. We can then use straight brute force or a some brute force after [hide="a little algebra and divisibility"] If m and n are the numbers shown on the dice, with m>n,
m=k(m-n) with k>1 and a natural number, or
n=p(m-n) with p a natural number[/hide]
Solving those equations we get [hide="hidden"]the same five possibilities
2n=m
3n=2m
4n=3m
5n=4m
6n=5m
There are 3, 2, 1,1,and 1 (brute-forced) solutions respectively, adding to 8 of the possible 15 sets.
This occurs in 8/15 of the 5/6 of rolls with different numbers on the dice, or
(8/15)(5/6)=4/9 of all dice rolls.[/hide]
Or did I mess up?


[i]Αυτή την άσκηση την έχω καιρό στην επιφάνεια εργασίας , αλλά δε μπορούσα για τεχνικούς λόγους να τη στείλλω. Δεν θυμάμαι καν που τη βρήκα. Νομίζω ότι κάπου είχα και λύση , αλλά την έχω επίσης χάσει. Καιρός λοιπόν να την τακτοποιήσουμε!(Μάλλον είναι από κάποιο CRUX)[/i]

Ωραία ασκησούλα .  :)

Έστω ότι η $ EF$ τέμνει την $ BC$ στο $ S$ . Τότε $ (DS,DF,DA,DE)\equal{}\minus{}1$ και αφού $ NM//BS$ θα είναι $ AN\equal{}AM$

Άρα $ AQ//DE$ και $ AP//DF$ επομένως $ \angle{EDF}\equal{}\angle{AFE}\equal{}\angle{APE}$ άρα το $ P$ ανήκει στον περιγεγραμμένο κύκλο του $ AEIF$ . Όμοια και για το $ Q$ so QED  :)

[quote="silouan"]Ωραία ασκησούλα .  :)

Έστω ότι η $ EF$ τέμνει την $ BC$ στο $ S$ . Τότε $ (DS,DF,DA,DE) \equal{} \minus{} 1$ και αφού $ NM//BS$ θα είναι $ AN \equal{} AM$

Άρα $ AQ//DE$ και $ AP//DF$ επομένως $ \angle{EDF} \equal{} \angle{AFE} \equal{} \angle{APE}$ άρα το $ P$ ανήκει στον περιγεγραμμένο κύκλο του $ AEIF$ . Όμοια και για το $ Q$ so QED  :)[/quote]

Σιλουανέ , μιας και έχεις το σχήμα στο αρχείο , δεν κάνεις τις γραμμές λίγο πιο έντονες , για να μπορώ και γω να μελετήσω τη λύση σου που μου φαίνεται εξαιρετική; Η όραση όσο περνούν τα χρόνια , κάνει νάζια  :) !
 Σε ευχαριστώ προκαταβολικά.

       Μπάμπης

[quote="silouan"]Ωραία ασκησούλα .  :)

Έστω ότι η $ EF$ τέμνει την $ BC$ στο $ S$ . Τότε $ (DS,DF,DA,DE) \equal{} \minus{} 1$ και αφού $ NM//BS$ θα είναι $ AN \equal{} AM$

Άρα $ AQ//DE$ και $ AP//DF$ επομένως $ \angle{EDF} \equal{} \angle{AFE} \equal{} \angle{APE}$ άρα το $ P$ ανήκει στον περιγεγραμμένο κύκλο του $ AEIF$ . Όμοια και για το $ Q$ so QED  :)[/quote]

Silouane, poli wraia lisi. Apla gia na deikseis oti AM=AN tha mporouses na peis oti gonia(AME)=gon(EDC)=gon(AEM) opote kai AM=AE. Omoia AN=AF opote to MNFE einai eggrapsimo me kentro to A. Tora gia na fanei akoma pio poli oti auti i askisi einai poli xalari (opws px exoume tis xalares anisotites) mporeis na sinexiseis me ton tropo pou piges(diladi mia akoma sxesi) i na pareis to theorima Casey kai na deikseis oti to P anikei ston kiklo AEIF. Pragmati pare to Casey gia to P gia tous kiklous (MEFN) kai (DEF) kai efarmose auti ti sxesi stin isotita pou sou dinei to Casey gia to P gia tous kiklous (EFNM) kai (AEF) opote deixneis oti to P exei mideniki dinami os pros ton (AEF). Fusika to :"$ \angle{EDF} \equal{} \angle{AFE} \equal{} \angle{APE}$ άρα το $ P$ ανήκει στον περιγεγραμμένο κύκλο του $ AEIF$" einai poli pio kompso kai grigoro, apla to parapanw einai mia poli bati efarmogi tou Casey.

Ωραίες λύσεις παιδιά... Μετά απ' το ΑΜ=ΑΝ (το οποίο κι εγώ το βρήκα όπως ο Στέφανος) αρκεί να αποδείξουμε ότι το $ AFQIPE$ είναι ο κύκλος Euler του $ DMN$. Αρα αρκεί να αποδείξουμε ότι $ MF, NE$ ύψη του τριγώνου αυτού. Για το $ F$ έχουμε, απ' το ισοσκελές $ AFM$ $ \angle{AFM}\equal{}90\minus{}\frac{FAM}{2}\equal{}B/2\equal{}IFD$. Επειδή όμως $ IFA\equal{}90\iff MFQ\equal{}90$. Ομοια για το $ E$ και τέλος :)

Ισως θα μπορούσε να γίνει και μια πολύ ωραία λύση με αντιστροφή με κέντρο το I κι ακτίνα αυτή του εγγεγραμμένου. Τότε θα έπρεπε να αποδείξουμε ότι τα $ Q', P'$ ανήκουν στην $ EF$ αλλά δεν έχω πολύ χρόνο να κοιτάξω αυτόν τον τρόπο...


trigonometry
AoPSwiki
geometry
trig identities
Law of Sines
Law of Cosines

Can someone please explain the Law of Sines and the Law of Cosines?  What's the difference?

Law of Sines is a statement about arbitrary triangles in the plane. If the sides of the triangle are a, b and c and the angles opposite to those sides are A, B and C, then the law of sines states:

$ \frac{a}{\sin{A}}\equal{}\frac{b}{\sin{B}}\equal{}\frac{c}{\sin{C}}$

Law of Cosines states that for a triangle (same one as above)

$ a^{2}\equal{}b^{2}\plus{}c^{2}\minus{}2bc\cos{A}$

both the law of sines and the law of cosines have entries on AoPSWiki (go to AoPSWiki by clicking the link on the left panel)

[url]http://www.artofproblemsolving.com/Wiki/index.php/Law_of_cosines[/url]
[url]http://www.artofproblemsolving.com/Wiki/index.php/Law_of_sines[/url]

Also, when solving triangles, the Law of Sines is used when you have ASA, SSA :) , SAA.

Law of Cosines is used when you have SSS, SAS, and occasionally, SSA :) .

As fishythefish pointed out, the law of cosines is frequently more useful if we are look for a missing side or a missing angle when he have everything else.  Also note, the if we have a right triangle, and $ \angle C\equal{}90^{\circ}$, then the law of cosines reduces to the pythagorean theorem.


When I try to raise something to a power that takes more than one symbol to make, I get something like this: $ x^12$, when I mean x^12.

How do you fix it so the 2 also becomes part of the exponent?


[color=blue]
Let $a, b, c>0$ such that $abc=1$. Prove that the following inequality holds true:

                                        \[ 5+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq (1+a)(1+b)(1+c).  \]
 [/color]

Equivalent to (sums are cyclic)

$\sum \frac{a}{b} + 3 \geq \sum ab + \sum a$
$\sum x^3+3xyz \geq \sum xy(x+y)$ upon setting $a = x/y$ etc... which is just Schur's


Hey everyone...I really need help...the school textbook doesn't have probability till 12th grade but I wanna learn so I borrowed a 12th grade worksheet..I don't know how to do it now!!
So can u guys please...just show your working for these sums and then I'll understand how to do the rest of them? Thanx a lot!
1)A card is drawn from a pack...what is the probability that it is a  King?(no jokers)
2)there are 5 white balls and 5 black ones in a bag...three are drawn out...what is the probability that a)all are black and b)all are white?
3)Tim forgot the code of his safe but he knows that there are 3 dials and each dial has numbers from 0-9. How many different possibilities are there and what is the minimum one he has to try?
4)Lisa has to attempt any 5 out of 6 questions in a paper.What is the probability that she skips Question 3?
5)A coin has to be flipped thrice. The first time, it's heads.The second time,it's tails.What is the probability that it will be a)heads and b)tails the second time?
6)A number is drawn from the set of natural numbers less than ten thousand.What is the probability that it's first and last digits are the same?
7)How many different ways are there to paint the 6 faces of a cube in 4 different colours?
Thanks a lot,
Maya.
And sorry if this is too easy to be in High School..I really don't know where to put it so since I'm in 8th grade i thought High School was ok... :maybe:

[hide="1"]
There are 52 cards in the deck, and exactly four of them are kings. Therefore, the probability is $\frac{4}{52}= \frac{1}{13}$
[/hide]
[hide="2"]
First ball being black is $\frac{5}{10}$. After drawing there are 9 left, 4 black and 5 white, so the probability that the second one is black is $\frac{4}{9}$. By similiar logic, the probability that the third one is black is $\frac{3}{8}$. The probability that all are black is the dependent probability that these events happen in order, or $\boxed{\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}}$ = $\boxed{\frac{1}{12}}$

The second part is exactly the same, replace "black" with "white" in my solution.
[/hide]
[hide="3"]
For the first, second, and third dials, there are 10 positions.
Together, there are 10^3 possible orders.
My problem with this question is that it doesn't specify whether the order of dialing matters, or if you just need the right three numbers. I'll let someone else answer more completely.
[/hide]
[hide="4"]
There are $\binom{6}{5}$, or 6 ways to pick the questions to answer. The element three is in 5/6 the sets, so the probability is 1 - 5/6 = $\boxed{\frac{1}{6}}$
[/hide]
[hide="5"]
This one is a conceptual, rather than computational. See if you can figure it out on your own. Hint: try flipping a coin 1,507,534 (or insert large random number here) times. Then flip it once more. Does the outcome of the first 1,507,434 times affect the outcome of the 1,507,535th time?
[/hide]
[hide="6"]
We'll use some case work. There are 10^3 elements in the set.
For digits 1-9, all work. (This is 9)
For digits 10-99,
11,22,33,44,55,66,77,88,99 work. (This is 9 more)
For digits 100-999,
1?1, 2?2, 3?3, 4?4, 5?5, 6?6, 7?7, 8?8, 9?9 (This is 9 more)
As a general case, you can insert whatever digits you want in between, and you'll have 9 correct numbers for each power of 10.
Thus, we have 9 + 9 + 9 + 9 correct numbers out of 10000 possible, for a probability of $\frac{36}{10000}$.
[/hide]
[hide="7"]
This turns ugly if you account for rotational symmetry. I'll assume you can only orient the cube one way, because that seems to be the level of these problems.
Also, I'm not sure if the question REQUIRES that you use exactly four colors (i.e. can we have a solid colored cube?) or not.
The first face can be any of four colors. This determines 4 choices. For each choice of the first, there are four choice for the second. Now we're at 4 * 4 choices. For the third, there are 4 more choices for each of the 16 previous ones, making the total 64 choices. Now do this process for 6 faces, and we end up with our answer: $\boxed{4^{6}}$
[/hide]

Wheeew!! Got most of them. I had a few clarification questions on some though.

i think order matters for the code one.

also for the coin flipping one, (a) is 0 and (b) is 1 because its asking for the secodn flip which already happened.

for the last one, its not 4^6

[quote="Art of Owna"]DiscreetFourierTransform, I dont think number 6 is right..to begin with, isnt the demoninator 9999-not 10000?  and i dont think the numerator is right..[/quote]

[hide]there are 9*9*9 numbers that have the same first and last digits

$\frac{9^{3}}{9999}=\frac{81}{1111}$[/hide]

Yeah, I misread the coin flip one.
Ermm, yeah, I messed #6 up. Here's the augmented solution.
1-9 = 9 correct
11-99 = 9 more correct
101 - 999 = 9 * 9 (9 choices for each pair of first and last, times 9 choices for the middle digit per choice)
1001 - 9999 = 9*9*9 correct

And yes, the denominator should be 9999. I thought it said less than or equal to 10000, but really it's just less than.

So I get a revised answer of $\frac{9+9+81+729}{9999}= \frac{19}{1111}$

Is THIS right?

Actually I think there are 9+9+9*10+9*10*10 numbers with the first and last digit the same less than 10000, so tahat is 1008, 1008/9999=112/1111 (this is also saying that single digit numbers have the first and lsat digit the same)

oops...i messed up my solution :mad:

art of owna's solution is right though

and I messed up BOTH of mine. I agree, art of Owna is correct.


function
symmetry
inequalities open
inequalities

Let a,b,c be none negative numbers.prove that:

a^2+b^2+c^2>=2(a-1/a+1+b-1/b+1+c-1/c+1)+a+b+c

[quote="Amirhossein"]Let a,b,c be none negative numbers.prove that:

$a^{2}+b^{2}+c^{2}\geq 2(\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1})+a+b+c$[/quote]

Is this what you mean?

I rewrite it better with LaTeX...I hope I've understood the text.

$a^{2}+b^{2}+c^{2}\geq 2 ({\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}})+a+b+c$

$a^{2}-a+b^{2}-b+c^{2}-c \geq 2 ( \frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1})$

$a(a-1)+b(b-1)+c(c-1) \geq 2\frac{(a-1)}{(a+1)}+2\frac{(b-1)}{(b+1)}+2\frac{(c-1)}{(c+1)}$

$a(a-1)+2\frac{(a-1)}{(a+1)}+b(b-1)+2\frac{(b-1)}{(b+1)}+c(c-1)+2\frac{(c-1)}{(c+1)}\geq 0$

Take the single ones

For example prove that $a(a-1)+2\frac{(a-1)}{(a+1)}\geq 0$

In fact we have  $(a-1) ({ a-\frac{2}{a+1}) \geq 0}$

Now you have two cases since $a, b ,c \geq 0$

$(1)$ $1 > a \geq 0$

Then $(1-a) \geq 0$ and we can rewrite the expression

$-a (1-a)+2 \frac{(1-a)}{(1+a)}\geq 0$

$\frac{2-2a-a(1-a)(1+a)}{1+a}\geq 0$

$\frac{a-1}{a^{2}+a-2}\geq 0$

$\frac{a^{3}-3a+2}{a+1}\geq 0$

$\frac{(a-1)(a^{2}+a-2)}{a+1}\geq 0$ which is true

since $1 > a \geq 0$ [ $a-1 < 0$ and $a^{2}+a< 2$]

$(2)$ if $a \geq 1$ then it is clearly true

since $(a-1) \geq 0$,

and $a-\frac{2}{a+1}= \frac{(a+2)(a-1)}{a+1}\geq 0$

The same can be proved for $b$ and $c$.

[hide](Next time be more precise when you write a title for the new topic...hard it's too generic  :wink:
and try to learn how to use LaTeX... :) )[/hide]

I denote $E(a,b,c)=\sum a^{2}-2\sum \frac{a-1}{a+1}-\sum a$
Let $f: [0,\infty)\to \mathbb R$ ,$f(a)=E(a,b,c)$ , and we have to prove that $f(a)\geq 0$ $,\forall a,b,c \geq 0$
But $f$ is convex because is a sum of convex functions so it has a minimum.
Because $f'(1)=0$ $\Rightarrow$ $f(a)\geq f(1)$ ,$\forall a\geq 0$
Because of the symmetry $E(a,b,c)\geq E(1,1,1)=0$ ,$\forall a,b,c \geq 0$


If we solve the problems and send it then what happened?

What do you mean? You wait :)

PS I got your email.

[quote="samangharib"]I mean this program have prize or something? :stink:[/quote]Not really, no financiar incentive is offered, mainly to prevent cheating :D


This math concept/question has been bothering me for a while. I've looked around for a while now but I haven't found much about it: What is a difference function?

Any help would be awesome! Thank you!!!

eeehhh...wow...thats alot of red lines  :D
sorry, that didnt clear it up at all. but thanks for trying  :D

Try "divided difference" in Google. It leads to discussions on more elementary and much more important topics than the very specialized "prime difference function".


quadratics
LaTeX
algebra
quadratic formula

In the equation $ \frac {x(x \minus{} 1) \minus{} (m \plus{} 1)}{(x \minus{} 1)(m \minus{} 1)} \equal{} \frac {x}{m}$ the roots are equal when

$ \textbf{(A)}\ m \equal{} 1 \qquad\textbf{(B)}\ m \equal{} \frac {1}{2} \qquad\textbf{(C)}\ m \equal{} 0 \qquad\textbf{(D)}\ m \equal{} \minus{} 1 \qquad\textbf{(E)}\ m \equal{} \minus{} \frac {1}{2}$

[hide]

Cross-multiplying and expanding the equation gives $ mx^2\minus{}mx\minus{}m^2\minus{}m\equal{}mx^2\minus{}mx\minus{}x^2\plus{}x$. Simplifying, we have $ m^2\plus{}m\equal{}x^2\minus{}x$. If we treat m as a constant, then we can apply the quadratic formula to $ x^2\minus{}x\minus{}(m^2\plus{}m)\equal{}0$. The roots are equal when the discriminant is zero, so we must find m such that:

$ 1\plus{}4m^2\plus{}m\equal{}0$

Hence, $ m\equal{}\minus{}\frac{1}{2}$, so $ \textbf{ (E) }$

[/hide]

[quote="lifeisacircle"][hide]$ 1 \plus{} 4m^2 \plus{} m \equal{} 0$
[/hide][/quote]Minor latex fix:
[hide]$ 1 \plus{} 4m^2 \plus{} 4m \equal{} 0$[/hide]


modular arithmetic
number theory solved
number theory

Find all natural numbers $n$, for which exists a prime number $p$, such that the numbers
$p+2, 2^n+p, 2^n+p+2$ are primes too.
Please post only full solutions or nice ideas to this interesting problem. Thank you.

For $n=1,p=1;n=2,p=1;n=3,p=3;n=4,p=1$. So we have to proove that for $n>4$ there is no solution, but how to do that?

Looking $\mod 3$ just gives that when $p \neq 3$, then at least one of the numbers is divisble by $3$, thus not prime.
So the numbers $2^n+3$ and $2^n+5$ have both to be prime. So how do you procede¿

e.lopes, if you can solve the problem solve it else leave it. I tried (mod3) , but it didn`t work for n odd.

$p+2=p-1(mod3)$
Let $n$ be even: $2^n+p=p+1(mod3), 2^n+p+2=2(2^{n-1}+1)+p=p(mod3)$
so one of the numbers $p+2, 2^n+p, 2^n+p+2$ is not prime ($p-1,p,p+1$)
Now let $n$ be odd: $2^n+p=p-1(mod3), 2^n+p+2=2(2^{n-1}+1)+p=p+1(mod3)$, but since $p$ is prime then one of the numbers $p-1,p+1$ is $=0(mod3)$.

But why then for $n=1,2,3,4$ there exist a prime $p$ such that $p+2, 2^n+p, 2^n+p+2$ are primes? Where do we use that $n>4$?

No, there is no such prime for $n=2;4$.
It's correct to look $\mod 3$ to get:
the only possibility for $p$ is $p=3$ (you excluded this, why¿). And then simple checking gives that we are left to find all $n$ such that $2^n+3$ and $2^n+5$ are both primes. This happens for $n=1;3$, and I think there are no other solutions.

For $n=2$, $p+2,p+4,p+6$ are all primes when $p=1$ so there exist a prime?
Similary for $n=4$

Your last post only implies that n is odd and p is divisible by 3.  Since 1 is not actualy a prime, this corresponds to the only known solution n=p=3.
So the three numbers are
$5,2^n+5,2^n+3$
we know that n is odd
order of 2 mod 7 is 2,4,1,2,4,1 which yields n=6k+3=3(2k+1).
So the task is to find odd odd k such that $8^{2k+1}+5$ and $8^{2k+1}+3$ are both primes
order of 8 mod 5 is 3,-1,-3,1
order of 8 mod 13 is 8,-1,-8,1
so if k was odd, $8^{2k+1}+3\equiv0 (mod 5)$ (so it cannot be prime since 8>5), else
$8^{2k+1}+5\equiv 0 (mod 13)$ (so it can be prime only for k=0).
So k=0 which yields n=3 is the onlz solution.

I must be sleepy today, but Kondr what about when n=1 and p=3.
And I`ll be thankful if you write the full solution ( word by word) becouse i`m only $14\frac{1}{2}$ years old and i want to understand the problem.

[quote="Kondr"]
order of 2 mod 7 is 2,4,1,2,4,1 which yields n=6k+3=3(2k+1).
[/quote]
This is probably the piece of my solution that is not detailed enogh.
It means, that the residues of $2^n$ periodicaly change with period 3. As you see, for n not divisible by 3 we have $2^n+5\equiv 2+5\equiv 0(mod 7)$ or $2^n+3\equiv 4+3\equiv 0(mod 7)$. There i lost the solution n=1, p=3 because 2+5 is divisible by 7 although it's a prime.
Because of that n is divisible by 3. Thus n is odd, n canot be 6k, so n=6k+3.
Than I'm using order of 8 mod 5 and mod 13 in similar way, so I hope you will understand it.

2^n+p=p-1(mod3), 2^n+p+2=2(2^{n-1}+1)+p=p+1(mod3), but since p is prime then one of the numbers p-1,p+1 is =0(mod3). -beat

Your last post only implies that n is odd and p is divisible by 3 -Kondr

why p=3, what about if p=2?

Kondr wrote:

order of 2 mod 7 is 2,4,1,2,4,1 which yields n=6k+3=3(2k+1).

and why that?

Ok [i]Beat[/i], I shall try to explain it as detailed as possible. (After all, we were all beginners once)

Firstly, we immediately rule out $p=2$ since $2$ is obviously the only even prime and $p$ & $p+2$ have the same parity.

Thus $p \geq 3$ and since $p+2, p+2^n, p+2^n+2$ are primes greater than $3$, they must be $\not\equiv 0 \pmod{3}$. We then observe that when $p \equiv 1 \pmod{3}$, we have $p+2 \equiv 0 \pmod{3}$ (Contradiction), and when $p \equiv 2 \pmod{3}$, we have $p+2^n+2 \equiv 0 \pmod{3}$ when $n$ is odd and $p+2^n \equiv 0 \pmod{3}$ when $n$ is even, since $2^n \equiv 2 \pmod{3}$ when $n$ is odd, and $2^n \equiv 1 \pmod{3}$ when $n$ is even.

This leaves us with the only possibility of $p=3$. (Remember that $3$ is the only prime divisible by $3$)

Substituting in, we have $2^n + 3$  and $2^n+5$ must be primes. We then observe that taking $\mod 7$ we have $2^n \equiv 1, 2, 4 \pmod{7}$ for $n \equiv 0, 1, 2 \pmod{3}$ respectively. Note that $2^n + 3 > 7$ for $n > 2$, thus $\forall n >2$ we must have $2^n + 3, 2^n+5 \not\equiv 0 \pmod{7}$ for these two integers to be primes. As for $n \leq 2$, we check the finite cases to obtain the solution $n=1$ (which yields $p, p+2, p+2^n, p+2^n + 2 = 3, 5, 5, 7$ respectively). So we assume $n>2$.

When $n \equiv 1 \pmod{3}, 2^n \equiv 2 \pmod {7} \Rightarrow 2^n+5 \equiv 0 \pmod{7}$ (Contradiction).
When $n \equiv 2 \pmod{3}, 2^n \equiv 4 \pmod{7} \Rightarrow 2^n+3 \equiv 0 \pmod{7}$ (Contradiction).
Thus $n \equiv 0 \pmod{3}$.

We also know that $n$ must be odd. This follows from the fact that when $n$ is even, we have $2^n \equiv 1 \pmod{3} \Rightarrow 2^n + 5 \equiv 0 \pmod{3}$, yet we know that $2^n + 5$ is a prime greater than $3$. (Contradiction)

It then follows from $n$ is odd and $n$ is divisible that $n \equiv 3 \pmod{6}$. Letting $n = 6k+3$ for some non-negative integer $k$, we have $2^n+3 = 2^{3(2k+1)} + 3 = 8^{2k+1} + 3, 2^n+5 = 2^{3(2k+1)} + 5 = 8^{2k+1} + 5$ are both primes.

We then note that taking $\mod 5$ we have $8^m \equiv 3, -1, -3, 1 \pmod{5}$ for $m \equiv 1, 2, 3, 0 \pmod{4}$ respectively, and that taking $\mod 13$ we have $8^m \equiv 8, -1, -8, 1 \pmod{13}$ for $m \equiv 1, 2, 3, 0 \pmod{4}$ respectively.

Finally, we see that when $k$ is odd, $8^{2k+1} \equiv -3 \pmod {5} \Rightarrow 8^{2k+1} + 3 \equiv 0 \pmod{5}$ (Contradiction as $8^{2k+1}+3 \geq 11$ is prime and so cannot be divisible by $5$), and when $k$ is even, $8^{2k+1} \equiv 8 \pmod{13} \Rightarrow 8^{2k+1} + 5 \equiv 0 \pmod{13}$. Since $8^{2k+1} + 5 > 13$ for $k>0$, the only way for $8^{2k+1} + 5$ to be prime is when $k=0$, which in this case yields another solution $n=3$, with $p, p+2, p+2^n, p+2^n+2 = 3, 5, 11, 13$ respectively.

Therefore $n=1, 3$ are the only solutions.

Our motivation behind this solution is that we first guess that $n =1,3$ are our only solutions, which yields the possible primes $3, 5, 7, 11, 13$. Thus to prove that there only finitely many solutions, we can try modulo $3, 5, 7, 11, 13$ and use the fact that if $p, q$ are primes such that $q \equiv 0 \pmod{p}$, then $p=q$. Indeed, considering modulo $3, 5, 7, 13$ is sufficient to solve this problem.

[i]dblues[/i] thanks a lot for the detail post and thaks to e.lopes, ZeTaX and Kondr as well.


Given any set of 10 consecutive positive integers, prove that there is at least one of the 10 numbers which is coprime to the other 9 numbers.

. There are 5 numbers divisible by 2.
. There are at most 3 or 4 numbers divisible by 3, of which at most 2 are not even.
. There is 2 number divisible by 5 of which one is even.

Prime factors >= 7 are not to consider :
. There is at most 1 number divisible by p prime >= 7.

5+3+1 < 10 so there is a number not divisible by 2 or by 3 or by 5 which is prime with the others.

Hope it's correct now.


Alright I'm back on a drink break and apparently some moron has deleted my thread before I saw the answer.  Once again,
I'm on a drinkbreak right now while the rest of my class is taking a midterm. I really need somebody to tell me this answer so I won't fail. The question is from Integrated 2 - Discovery and Exploration published by Prentice Hall - Chapter 3.6 Problem 17. When does x^2 reach its minimum value? I know this is a trick question because I can't figure it out. Answers within the next 5 minutes are appreciated.

I'm the "moron" who deleted your thread.

1. Your question has nothing to do with the AMC.
2. You're claiming to ask about a midterm question, which means either (a) you're cheating or (b) you're making a joke, neither of which are appropriate for this forum.

If you want to post a [i]problem[/i] that people can [i]discuss[/i], fine.

If you want to post a question that you "need" the answer for in the next 5 minutes, that's not the purpose of this forum.


What is the value of the expression $ (37 \plus{} 12)^2 \minus{} (37^2 \plus{}12^2)$?

@cgyao15:
It's a form of (a+b)^2. It generally goes that:

$ (a \plus{} b)^2 \equal{} (a \plus{} b)(a \plus{} b) \equal{} a^2 \plus{} b^2 \plus{} 2ab$

Thus here it would be:

$ (12 \plus{} 37)^2 \equal{} 12^2 \plus{} 37^2 \plus{} 2(37)(12)$

Hope I helped to answer your question. Also thanks for the solution, LaTex!


monovariant
combinatorics unsolved
combinatorics

[quote="Johan Gunardi"]A deck of 50 cards contains two cards labeled $ n$ for each $ n \equal{} 1, 2, . . . , 25$. There are 25 people seated at a table, each holding two of the cards in this deck. Each minute every person passes the lower-numbered card of the two they are holding to the right. Prove that eventually someone has two cards of the same number.[/quote]

First consider the position of the two '25's.  Either one person has both--in which case we are done--or the two '25's are held by different people and will never move.

Next consider the position of the two '24's.  If at any stage one person has both, we are done--and if not, we keep iterating until both '24's are held by people not holding a higher valued card.  From this point onwards, neither the '25's nor the '24's will ever move again.

We continue this process until either someone has had a duplicate or all cards valued between between 14 and 25 are held by different people and will never move again.  At this point there will be exactly one person--call him/her 'A'--not holding a card valued 14 or above.  Let's think about what happens to the '13's.  Both '13's must keep moving until 'A' acquires one of them.  At that point the '13' that A has stops moving but the other '13' keeps moving until it is also acquired by A--and we are done.

[quote="Johan Gunardi"]That's a nice solution. But I'm looking for a solution using monovariant.[/quote]

Hmmm...I'm a bit confused because if you are looking for a monovariant solution then you must have had a specific solution in mind--meaning this problem was already solved from your perspective when you posted it.  I thought this forum was for problems where the poster did not know the solution.

In any event I suspect what you are looking for is something along the following lines.  Let the values of the two cards that person $ i$ has at iteration $ t$ be given as $ A_{i}(t)$ and $ B_{i}(t)$.  Then the sum $ S(t) \equal{} \sum_{i \equal{} 1}^{25}|A_{i}(t) \minus{} B_{i}(t)|$ will never decrease as $ t$ increases--it must always increase or remain constant.  Since $ S(t)$ is always an integer and is clearly bounded above, eventually $ S(t)$ will reach a point where it remains constant permanently.  Moreover, when someone passes their lower card and it becomes the strictly higher valued card in the hand of its new owner, $ S(t)$ strictly increases and does not remain constant.  Therefore, once the point is reached where $ S(t)$ is permanently constant, the same set of 25 cards must be passed at each interation and the same set of 25 cards must remain fixed.  Since 25 is an odd number, there must be at least one card denomination $ i$ where one of the two cards is fixed and the other is moving.  Those two cards of denomination $ i$ must therefore eventually meet and the problem is solved.


[quote="ln(dx/dy)"]Shortcut please...

If $(x+y+z)^{4}$ is multiplied out and like terms are combined, what is the coefficient of the $x^{2}yz$ term?
(This was a warm-up problem, so you can't use calculators.)[/quote]

[hide]

If you look at this as $[x+(y+z)]^{4}$ you can consider (x + y+ z) as a binomial instead of a trinomial and then use binomial expansion to get :

$x^4 + 4x^3(y+z) + 6x^2(y+z)^2 + 4x(y+z)^3 + (y+z)^4$

Now since you want the coefficient of the $x^{2}yz$ term, you only really need to use the $6x^2(y+z)^2$ term since its the only one that will have $x^2$ in it.  Expanded, it looks like this :

$6x^2(y^2 +2yz + z^2)$  You can distribute but you only need the term that has y and z, so you get $12x^2yz$.  The coefficient is $12$.[/hide]

[hide]You are counting the number of ways to pick two of the four x's for inclusion in the term, times the number of ways to pick one of the two y's that are still valid after the x's have been chosen. C(4,2)*C(2,1)=12.[/hide]

Think about it:
[hide]$\frac {4!}{2! \ 1! \ 1!}$

General formula for cofficient of  $x^m y^n z^k $ in expansion of $(x+y+z)^{m+n+k} $ is

 $\frac {(m+n+k)!}{m! \ n! \ k! } $

Another way to Think about  it  is to see it as: $(x+y+z)(x+y+z)(x+y+z)(x+y+z)$
 How many  ways can you pick two terms (to get x^2)  (4 choose 2) (=6)
and then  two ways to choose  y and z from rest of the two...so the answer is  (or 6*2)[/hide]

[quote="bob123"][hide]Well, you can easily find out that (x+y+z)2=x2+y2+z2+2xy+2xz+2yz. Then, you see that you can only get x2yz by multiplying x2 and 2yz, so then you will do that twice, thus making the coefficient 2+2=4.
Sorry I don't know LaTeX. hehe I'm not sure I'm right, either.[/hide][/quote]
Your approach is OK, but you missed a case:
[hide]$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$
You will get $x^2yz$ terms from $(x^2)(2yz)$ or from $(2xy)(2xz)$.  Each occurs twice, so the coefficient is $(2+4)\times 2=12$.[/hide]


Prove that for each natural n, 2^n is not divisible by 23

Errrr..... 2^n is not divisible with 23? Come on man, what did you really mean? :D

O,sorry, i wanted to say 2^n+1 is not divisible by 23!

let k be the smallest integer such that 2^k= -1 mod 23
then 2^2k =1 mod 23 and hence 2k is divisible by order of 2 mod 23
= 11.
hence k is a multiple of 11.but then 2^k+1=2 mod 23. contradiction


algebra
polynomial
Rational Root Theorem

problem we got in algebra class today......

it was on the test today and i couldnt get it, so i was wondering how this problem was solved...

the problem:

On the planet ZOG, colored coins are used for money. Suppose 12 white, 9 red, 8 yellow, 4 blue, and 0 green coins can be exchanged for 2 white, 1 red, 0 yellow, 1 blue, and 1 green coin. Also, suppose 1 green = n blue, 1 blue = n yellow, 1 yellow = n red, and 1 red = n white.

What is the whole number rate of exchange of these coins? Is only one exchange rate possible?

Whites are the smallest denomination, so from this point everything is in units of whites.  Specifically, a yellow is $ n^{2}$ whites, a blue is $ n^{3}$ whites, and a green is $ n^{4}$ whites.  We are also given

$ 12\plus{}9n\plus{}8n^{2}\plus{}4n^{3}\equal{} 2\plus{}n\plus{}n^{3}\plus{}n^{4}\Leftrightarrow$
$ n^{4}\minus{}3n^{3}\minus{}8n^{2}\minus{}8n\minus{}10 \equal{} 0$

By the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational Root Theorem[/url], the only possible positive rational (hence, integer) roots are $ n \equal{}\pm 1, 2, 5, 10$.  Checking, we find that $ \boxed{n \equal{} 5}$ is a solution.  Factoring out,

$ (n\minus{}5)(n^{3}\plus{}2n^{2}\plus{}2n\plus{}2) \equal{} 0$

The cubic factor clearly has no positive root, nor a rational one, so $ n \equal{} 5$ is both the only positive and the only integer exchange rate possible.

cant you put everything in terms of n as white chips?

oh, thanks everybody!

yay, i somehow got it right. go random guessing!
i had no idea and i put that white equaled five (a nice, round number) and worked out the rest....

guess i got that question right.

thanks again.


how do u define the molecular mass of a mixture(gaswous liquid solid)

Perhaps by $ M_{T}\equal{}\frac{\sum m_{i}}{\sum n_{i}}$.


manu  thanks for the problems i  have a question about inversion i  know  that  inversion  preserves angles  but  what  is the angle  of a line  and a circle?

More generally:
You can define an angle between any two curves $f,g$ at an intersection point $X$ when they are differantiable in $X$ (and meet $X$ just once).
To define the angle, just take the tangents $t_f , t_g$ through $X$ on $f$ and $g$. The angle between $t_f$ and $t_g$ is by definition the angle between the curves $f,g$ at this point.

So for $f$ being a line and $g$ being a circle, the angle of intersection is the angle between the line $f$ itself (the tangent to a line is the line itself) and the tangent on $g$ through the intersecton point (this angle is the same for both points of intersection).
The same works also for $f,g$ both being circles: then the angle between them is the angle between their tangents at an intersection point.


What is the maximum product that can be obtained by multiplying two distinct numbers form the given set? Express  your answer as a common fraction. \[ \{ \frac7{48},\frac3{16},\frac16,\frac5{24}\}\]

First, we order the numbers from least to greatest:

$ \{\frac {7}{48}, \frac {1}{6}, \frac3{16}, \frac5{24} \}$

The largest product is the largest number multiplied by the second-largest number.

We have $ \frac3{16} \times \frac5{24}$, or $ \boxed{\frac {15}{384}}$.

One must reduce 15/384
Divide the numerator and denominator by three
get 5/128


there is a box with sides a and b in a corridor of width c.find the conditions in which it can be pushed through

a door of width d.

Here are the linksssss: ;)  ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=58006

http://www.mathlinks.ro/Forum/viewtopic.php?t=58019

i think the links dont help us....

(ba ba amir joon narin be ma digeh :( )


Let $ f: \mathbb{A} \to \mathbb{A}$ where $ \mathbb{A} \equal{} ( \minus{} 1,\infty)$ and
$ f(x \plus{} f(y) \plus{} xf(y)) \equal{} y \plus{} f(x) \plus{} yf(x)$ $ \forall x,y \in \mathbb{A}$
be such that $ \frac{f(x)}{x}$strictly increasing for $ x\in ( \minus{} 1,0)\cup(0,\infty)$. Find all Such $ f$.

[quote="Mashimaru"]Put $ x \equal{} y$ we have: $ f(x \plus{} (x \plus{} 1)f(x)) \equal{} x \plus{} (x \plus{} 1)f(x),\forall x\in ( \minus{} 1; \plus{} \infty)$

Let $ u \equal{} x \plus{} (x \plus{} 1)f(x)$ then $ f(u) \equal{} u$.

For each $ u$ satisfies $ \exists x: u \equal{} x \plus{} (x \plus{} 1)f(x)$, substitute $ x$ by $ u$ we have: $ f(u^2 \plus{} 2u) \equal{} u^2 \plus{} 2u$.

Thus, function $ f$ has two sets of immovable points (i.e. $ \{x\in\mathbb{A}|f(x) \equal{} x\}$): $ u$ and $ u^2 \plus{} 2u$. We will consider the following cases:

1. $ \minus{} 1 < u < 0\Rightarrow \minus{} 1 < u^2 \plus{} 2u < 0$:
We have: $ \frac {f(x)}{x}$ strictly increases on $ ( \minus{} 1;0)$ then there exists only one $ x_0$ such that $ \frac {f(x_0)}{x_0} \equal{} 1$, thus $ u \equal{} u^2 \plus{} 2u \Leftrightarrow u \equal{} 0 \vee u \equal{} \minus{} 1$, contradiction.

2. $ u > 0 \Rightarrow u^2 \plus{} 2u > 0$, similarly we got $ u \equal{} \minus{} 1 \vee u \equal{} 0$, contradiction.

Since $ u > 0$ or $ \minus{} 1 < u < 0$ lead to contradictions, we must have $ u \equal{} 0$, since then $ f(0) \equal{} 0$, thus $ f(x)$ has only one immovable point: $ 0$.

This implies $ x \plus{} (x \plus{} 1)f(x) \equal{} 0$ or $ f(x) \equal{} \frac { \minus{} x}{x \plus{} 1}$, which is the only function satisfies our condition.[/quote]

Thank you for this great solution. :)
My second question is if the increasing condition were not given , what would have been the solution set? :)

[quote="ith_power"][quote="Mathias_DK"][quote="Mashimaru"]Put $ x \equal{} y$ we have: $ f(x \plus{} (x \plus{} 1)f(x)) \equal{} x \plus{} (x \plus{} 1)f(x),\forall x\in ( \minus{} 1; \plus{} \infty)$

Let $ u \equal{} x \plus{} (x \plus{} 1)f(x)$ then $ f(u) \equal{} u$.

For each $ u$ satisfies $ \exists x: u \equal{} x \plus{} (x \plus{} 1)f(x)$, substitute $ x$ by $ u$ we have: $ f(u^2 \plus{} 2u) \equal{} u^2 \plus{} 2u$.

Thus, function $ f$ has two sets of immovable points (i.e. $ \{x\in\mathbb{A}|f(x) \equal{} x\}$): $ u$ and $ u^2 \plus{} 2u$. We will consider the following cases:

1. $ \minus{} 1 < u < 0\Rightarrow \minus{} 1 < u^2 \plus{} 2u < 0$:
We have: $ \frac {f(x)}{x}$ strictly increases on $ ( \minus{} 1;0)$ then there exists only one $ x_0$ such that $ \frac {f(x_0)}{x_0} \equal{} 1$, thus $ u \equal{} u^2 \plus{} 2u \Leftrightarrow u \equal{} 0 \vee u \equal{} \minus{} 1$, contradiction.

2. $ u > 0 \Rightarrow u^2 \plus{} 2u > 0$, similarly we got $ u \equal{} \minus{} 1 \vee u \equal{} 0$, contradiction.

Since $ u > 0$ or $ \minus{} 1 < u < 0$ lead to contradictions, we must have $ u \equal{} 0$, since then $ f(0) \equal{} 0$, thus $ f(x)$ has only one immovable point: $ 0$.

This implies $ x \plus{} (x \plus{} 1)f(x) \equal{} 0$ or $ f(x) \equal{} \frac { \minus{} x}{x \plus{} 1}$, which is the only function satisfies our condition.[/quote]
But $ f(x) \equal{} \frac { \minus{} x}{x \plus{} 1}$ is strictly DEcreasing.. ;)[/quote]
Yes it is. But THe problem wants $ f(x)/x$ to be increasing and it is. :)[/quote]
Oh.. :blush:  :blush:  :blush:


quadratics
number theory theorems
number theory

Is there some list of all quadratic residues (Legendre symbols) online?

I know some, like (-1/p) and (2/p), but other (a/p) with proof might come in handy too

What speaks against computing them yourself using quadratic reciprocity¿


[hide]Let $\frac{x}{2} = \alpha$.

$2 + \frac{\sin 2\alpha\cos \alpha}{\cos 2\alpha\sin \alpha} + \frac{\cos 2\alpha\sin \alpha}{\sin 2\alpha\cos \alpha} = 0$

$2 + \frac{2\cos^2 \alpha}{2\cos^2 \alpha - 1} + \frac{2\cos^2 \alpha - 1}{2\cos^2 \alpha} = 0$

Let $2\cos^2 \alpha = a$. Suppose $a \neq 0$ and $a \neq 1$.

Then $(2a - 1)^2 = 0 \Rightarrow a = \frac{1}{2} \Rightarrow \cos^2 \alpha = \frac{1}{4} \Rightarrow x = \frac{2}{3}\pi + 2k\pi \vee x = \frac{4}{3}\pi + 2k\pi$.

Now we see if $a = 0$ and $a = 1$ are solutions.
$2\cos^2 \alpha = 0 \Rightarrow x = \frac{\pi}{2} + 2k\pi$, $2\cos^2 \alpha = 1 \Rightarrow x = \frac{\pi}{2} + 2k\pi \vee x = \frac{3}{4}\pi + 2k\pi$.
Finally the solutions are $x = \frac{2}{3}\pi + 2k\pi \vee x = \frac{4}{3}\pi + 2k\pi$.[/hide]

[quote="inom"]$2+tgxctg(\frac{x}{2})+ctgxtg(\frac{x}{2})=0$
Solve for x?[/quote]

I can't quite understand your equation.  What does tgxctg mean as well as ctgxtg?

i think tg=tan and ctg=cot...by the solution that's what it seems like...

Oh yeah. THAT'S WHAT SOMEONE TOLD ME WHEN I WAS IN INDIA.


Find the number of integer solutions to the equation $ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}28$ where $ 3\leq x_{1}\leq 9$ and $ 0\leq x_{2}\leq 8$ and $ 7\leq x_{3}\leq 17$

[quote="guile"]Find the number of integer solutions to the equation $ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{} 28$ where $ 3\leq x_{1}\leq 9$ and $ 0\leq x_{2}\leq 8$ and $ 7\leq x_{3}\leq 17$[/quote]

[hide]$ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}18$

where $ x_{1}\leq 6$, $ x_{2}\leq 8$ and $ x_{3}\leq 10$.

The max sum is $ 24$ so we need to take out $ 6$.

$ \therefore{24\choose 6}$.[/hide]

[quote="paladin8"]Somehow I doubt that, seeing as there are at most $ (9\minus{}3\plus{}1)(8\minus{}0\plus{}1)(17\minus{}7\plus{}1) \equal{} 693$ possibilities without the sum condition :wink:.

[hide="hint to error"] We are selecting 6 from 24, but they're not all distinguishable. [/hide][/quote]

oh rite...

im tired

use the balls and walls method so there are 6 balls and 2 dividors so $ {8\choose 2}\equal{}28$

[quote="Altheman"]If we expand the generating function, we get 28...I don't see how it suffices to do $ \binom{8}{2}$...[/quote]

Balls and urns on the six $ 1$'s (i.e. the $ 24\minus{}18$ in the first reply) we need to take away from the three variables.


Syntax error in ....asy, absence of ...asy in the working directory and standard directories, or some other reason. In short, [color=red]give a minimal example, please![/color].


Prove that $ \left[\frac{\sqrt{a^{4}\plus{}b^{4}}\minus{}\sqrt{b^{4}\plus{}c^{4}}\minus{}\sqrt{c^{4}\plus{}a^{4}}}{2}\right]^{2}\plus{}\frac{(|ab|\minus{}|bc|\minus{}|ca|)^{2}}{2}\geq{c^{4}}$ IF $ \sqrt{a^{4}\plus{}b^{4}}\minus{}\sqrt{b^{4}\plus{}c^{4}}\minus{}\sqrt{c^{4}\plus{}a^{4}}>0$ and $ |ab|\minus{}|bc|\minus{}|ca|>0$ and $ a,b,c\in \mathbb{R^*}$

@Canhang2007: It is the same inequality that i told you via the pm but i forgot to say that the bases must be positive!. Now try it  :D

$ \sqrt{a^{4}\plus{}b^{4}}\minus{}\sqrt{b^{4}\plus{}c^{4}}\minus{}\sqrt{c^{4}\plus{}a^{4}}>0$!!!!!!!
is that a spam  :rotfl:
sorry but you must review you inequality

[quote="Γιώργος"]Prove that $ \left[\frac {\sqrt {a^{4} \plus{} b^{4}} \minus{} \sqrt {b^{4} \plus{} c^{4}} \minus{} \sqrt {c^{4} \plus{} a^{4}}}{2}\right]^{2} \plus{} \frac {(|ab| \minus{} |bc| \minus{} |ca|)^{2}}{2}\geq{c^{4}}$ IF $ \sqrt {a^{4} \plus{} b^{4}} \minus{} \sqrt {b^{4} \plus{} c^{4}} \minus{} \sqrt {c^{4} \plus{} a^{4}} > 0$ and $ |ab| \minus{} |bc| \minus{} |ca| > 0$ and $ a,b,c\in \mathbb{R^*}$

@Canhang2007: It is the same inequality that i told you via the pm but i forgot to say that the bases must be positive!. Now try it  :D[/quote]
Sorry, my friend. But I think we can't have $ \sqrt {a^{4} \plus{} b^{4}} \minus{} \sqrt {b^{4} \plus{} c^{4}} \minus{} \sqrt {c^{4} \plus{} a^{4}} > 0$. Because it is clear that
$ \sqrt{a^4\plus{}c^4}\plus{}\sqrt{b^4\plus{}c^4} \ge a^2\plus{}b^2 \equal{}\sqrt{a^4\plus{}b^4\plus{}2a^2b^2} \ge \sqrt{a^4\plus{}b^4}.$


number theory
relatively prime
number theory proposed

Let $N = 6^{n}$, where $n$ is a positive integer, and let $M = a^{N}+b^{N}$, where $a$ and $b$ are relatively prime integers greater than $1. M$ has at least two odd divisors greater than $1$ are $p,q$. Find the residue of $p^{N}+q^{N}\mod 6\cdot 12^{n}$.

It' s enough to prove that $p^{6^{n}}\equiv 1\ (\mbox{mod }6\cdot 12^{n})$, for prime divisors $p$ of $M$. Note that $\varphi(6^{n+1})=2\cdot 6^{n}$ and thus $p^{6^{n}}$ is congruent $1$ or $-1$ mod $6^{n+1}$. So it's enough to prove that $p^{6^{n}}\equiv 1\ (\mbox{mod }2^{2n+1})$ and this is true if we prove $p\equiv 1\ (\mbox{mod }2^{n+1})$. Since $p\mid A^{6^{n}}+1\mid A^{2\cdot6^{n}}-1$ where $A=ab'$ and $bb'\equiv 1\ (\mbox{mod }p)$, we see that $o_{p}(A)\mid 2\cdot 6^{n}$, $o_{p}(A)\nmid 6^{n}$ and $o_{p}(A)\mid p-1$. Therefore from first $o_{p}(A)=2^{k}3^{l}$, from the second $k=n+1$ and from the third $2^{n+1}\mid p-1$.


Hi,

I am a die hard math fan, but also a movie fan.  Music and art has been declared the movie place( not games and fun) and I will act accordingly :

What are really the worst movies.

Wikipedia claims these :[url]http://en.wikipedia.org/wiki/Films_considered_the_worst_ever[/url]

"Manos : Hands of fate" and "Plan 9 from Outer Space" sound really really bad.

These are a bit more modern films that I have seen and I found them proposterous :

[url]http://www.imdb.com/title/tt0120847/[/url] : 'Substitute 2 : School's out'
About a man whose brother has been killed, and he poses as a teacher in a very bad school to find the killers.  The movie deals with delinquent kids.  It also tackles the subtle racial issues in education.
In the end it teaches you one or two things :
1) When there is trouble raising kids, laser guided guns and explosives are the best way!
2) You get nowhere when you just give up on the African American youngsters.  Asians are bad too!  :|
Really this movie is hopeless : all the baddies are black, and there is this one Asian guy that turns out be the top very bad guy (I think all in all there is just one non white in this movie that helps the principal character!)

[url]http://www.imdb.com/title/tt0116055/[/url]
'Deadly heroes' : about a soldier whose wife is kidnapped, and he and a super elite team have to get her back.
Amazingly, one car chase scene is just exactly the same footage as in 'Killing streets', in which the same two actors happened to play :o

[url]http://www.imdb.com/title/tt0245562/[/url]
John Woo disappoints here.  It is about Indians in WOII using a secret code against the Japanese, and the US soldiers can't let them be captured.  Doesn't sound really bad, but it is worked out very badly :
The acting is superficial.  Real low quality black and white footage of the war is used.  The racism is black and white : some soldiers just beat up the indians for no reason, other people stop whatevery they are doing to gaze at the indians and their magnificent music.  There is no plot : it's just one scene after the other with people getting blown up.  In the end the code is only used for an order that was not secret at all.

[url]http://en.wikipedia.org/wiki/Azumi[/url]
Azumi is a movie adaption from a comic book series.  It doesn't sound bad : ten young kids trained by a master to assassinate warlords who threaten peace.  But for the rest the scenario is really bad and empty.
Which is worse : they used young kids (which is fine, they don't look too old like in American films) but they don't act very well.  Aya Ueto may be a popular young celebrity, but you have to act well...

[url]http://www.imdb.com/title/tt0190138/[/url]
This thing drags on.  It's just Matthew Perry standing there surprised and scared all the time, about how slick and cool Bruce Willis' character is, and how he talks in a nonchalant way about his profession.

So share your thoughts!  Disagree with me! Wanna add other movies?

I have "Plan 9 from Outer Space" on CD, and have seen it. Yeah, it's bad, but unintentionally funny for all of that.

[quote="fredbel6"][quote="shadysaysurspammed"]Some of the "jason lives:friday 13th" series are dumb[/quote]
Can't say I heard a lot of good things about Freddy versus Jason  :roll:[/quote]

Didn't I say $some$ :roll:

[quote="Captain Sugar"]i saw this one movie that looked really boring. it was "killer clowns from outer space." its scary description was "in space you can't taste your ice cream." :huh:[/quote]

Well I hadn't heard of that but the title seems to come from
Plan Nine from Outer Space, and that tagline comes from "In space no one can hear you scream"from Alien.
That doesn't sound too good.

[url]http://en.wikipedia.org/wiki/Ernest_Scared_Stupid[/url]
Ernest scared stupid is really an incredibly dumb film, well I call these movies "wacky comedies" and when the poster of the film says 'wacky' itself.... :|

[url]http://en.wikipedia.org/wiki/2001:_A_Space_Travesty[/url]
2001: A Space ....Travesty  :huh: .
Well what can you possibly expect....except a crime against millennia of human evolution.

there was another one i was forced to watch when i was about 8. it was called mumphie. it starts off mumphie the elephant never gets any mail. so he went on a quest to get some mail.  he meets a wooden guy and they go to find some mail. somehow there is a twist in the story line so that, instead of looking for mail, he is saving some princess from an underwater fortress thing. luckily "the wooden man could breathe underwater, because he was made of wood, and mumphie could breathe underwater because he was a very special elephant." (that is almost a direct quote.) at the end, the princess is saved, and the bad guys turn into good guys because they "have seen the error of their ways." (that is also a direct quote.)  it was a horrible movie. i think i fell asleep during part of it.

the da vinci code: see it only because of the book

[quote="José"][quote="fredbel6"][quote="José"]I don't remember the name, but this one new about Hitler (In spanish:La caída)[/quote]

You mean der Untergang = Downfall?[/quote]

Yes that.... I couldn't finish it, I just couldn't, it was so bad for me. However, maybe once I'll try to finish it[/quote]

Well I really think you should watch it again.  As I said, I know a tiny bit of German, maybe that's why it had a different impact on me...you didn't watch a dubbed version did you??? :huh:

I will watch Da vinci code eventually, this summer...or later, we will see...

[quote="fredbel6"][url]http://en.wikipedia.org/wiki/Magic_Adventures_of_Mumfie[/url]

Was it a show or a movie.
Wow that sounds really bad.[/quote]

It was a movie. it hurt my brain.

[quote="José"][quote="fredbel6"][quote="José"]I don't remember the name, but this one new about Hitler (In spanish:La caída)[/quote]

You mean der Untergang = Downfall?
Why?  It is one of the best movies I've seen in the past year?  It is the first German speaking film about Hitler, and (I know a little bit of German) I think it was extremely well done.[/quote]

Yes that.... I couldn't finish it, I just couldn't, it was so bad for me. However, maybe once I'll try to finish it[/quote]
WHY???
I thought it was the GREAT movie. One of the BEST movies about WWII.

I am proudly voting for 'Lady in the Water' as the worst movie of the year.

very,very bad. :lol:
imagine a little kid(around 8years old)  predictiing what will happen, by looking at a cereal box.

Time I add another movie to my own topic : :wink:

[url=http://www.imdb.com/title/tt0116341/]Forest Warrior (1996)[/url] starring Chuck Norris

A bunch of evil men want to chop down a forest.... but a bunch of kids who love to play there AND Chuck Norris.... who happens to transform into an eagle, bear or wolf from time to time cannot let that happen. :|

Are there any similarities among the movies in this topic?

Honestly, I select very careful the movies that I watch before. Sometimes I don't watch any movie because I'm not satisfied about what I found (doo'h .. I think I'm exagerating a bit) As my brother said one of this days: I couldn't live whitout IMDb :).

Generally I don't like [b]sweet movies[/b] as: A Cinderella Story, 10 things I hate about you, The Sweetest thing, Legally Blonde etc because they are all the same, so when you see one of them your brain doesn't need to participate in order to understand the plot  :D
Don't get me wrong but I like [b]true romantic[/b] moives, when a couple (in some movies) is speaking all the time about all kind of things (Before Sunset, Before Sunrise).

So far I think the genre I like is Thriller/Horror/Mystery (a little the last one) Once I saw [i]Decoys[/i] .. not good.. mmm...actually bad bad movie..

I'm sorry (but happy in the same time, you'll get it why) because that's all I have to say here since the movies that I watch are selected by the rating (>6.4 or .5) and of course by the number of votes.  :)

I'm starting to get tire of HP:GOF because i watch it every night. The worst movie currently is shark boy and lava girl-a babyish movie i can't believe i went to see.

[quote="funcia"][quote="kyyuanmathcount"]Anything that you see on the SciFi channel is probably bad, except once in a while when they show The Mummy series.[/quote]

Take that back! The Bone Cruncher(was that the name?) was really good.[/quote]

Ok, fine, maybe that was a stereotype.  How about any movie that you haven't heard of, and is on the scifi channel is bad.

[quote="thatin"]I think that HPGOF is not a good movie... too short and fast for my opinion... the magic is gone...[/quote]
yeah...not sticking to the book as much as they used to. they just should have made 2 parts or something. :|
Hope the 5th HP is better

me too as i love the books...but the movies are okay i guess. GOF was good at the DE scene with LV but not at all good for the quid.(i'm a quid freak i've attempted to play it in my backyard...) i was hoping the world cup'd be exciting and that theyd have scenes with the other teams b/c i love the Germans.

Do you think that the DE scene with LV is good? Did you noticed the color of LV's eyes? It is [color=blue]BLUE[/color]!!!

I know there are many problems in it(where was the missing spots in the circle? or is it me) but I like the script...if you watch CoS you can tell the sameness(though it would be weird if you couldn't) of LV and TMR. "It's funny what a drop of your blood can do eh Harry?"

Anyway...Poseidon wasn't that great. I didn't like the people in it...

I wonder how OoTP will turn out...I hope Bella is good.

[quote="metehan"]I watched a film about a mystery.The film contained only a room and 12 people in this room.The name of this film maybe "12 angry man".(It was black-white film.So too old.)[/quote]
I have heard of that film but I have not seen it.  Nominated for three oscars apparently, but that does not guarantee quality.
You do know that this is a topic about "Worst movies"?

[i]Twelve Angry Men[/i] is a movie about a jury in a courtroom trial in America. It is not at all a bad movie.

What about [url=http://www.imdb.com/title/tt0799949/]Epic Movie[/url]? It's really one of the worst movies... ever :first:
I will never understand how I managed to watch it until the end :blush:

[i]Death Ring[/i], starring Mike Norris (Chuck's son), is pretty bad.

what about that movie about Glenn Gould's interpretation of the 32 inventions by bach or something? like, there's 32 clips...


function
topology
real analysis
real analysis solved

let $(E,d)$ be a metric space such that any $k$-lipschitzian,$0<k<1$,  function has a fixed point. ($k$ is not fixed).
Does it follows that $E$ is complete ?

Very nice problem ! :)
The solution I found is hidden
[hide="here"]

The answer is "No". The simplest construction I could come up with is the following. Let $a$ be a "distinguished point". Let $[a_n,b_n]$ be intervals of length $1-\frac 1n$ ($n\geqslant 2$) with the usual distance between the points of the same interval and initially huge distance between the points of different intervals. Now identify all points $b_n$ to one point $b$ and make the distance from $a$ to $a_n$ equal $\frac1n$. This will create a complete metric space $X$ with the following properties: $a$ is a limit point of $X\setminus\{a\}$, $X\setminus \{a\}$ is path connected, the distance from each point of $X$ to $a$ is not greater than $1$ and the intersection of any ball centered at $a$ of radius less than $1$ has no path connected component except the one-point set $\{a\}$ that comes arbitrarily close to $a$. Now put $E=X\setminus\{a\}$. If $f: E\to E$ is any contraction, then it extends to the entire space $X$ by continuity. Hence, $f$ has a fixed point in $X$. But this fixed point cannot be $a$ because otherwise $f(E)$ would be path connected, come arbitrarily close to $a$ and be contained in the ball centered at $a$ of radius strictly less than $1$. The only possibility for this to happen is to have the entire $E$ mapped to the point $a$ but we must have $f(E)\subset E$, which renders this impossible.

[/hide]


Nice one .By AM-GM and then the famous $(a^{2}+b^{2}+c^{2})^{2}\geq 3(a^{3}b+b^{3}c+c^{3}a)$ applied for the numbers $\sqrt{a}$ , $\sqrt{b}$, $\sqrt{c}$
we have
 $\frac{a^{2}b}{a+b}+\frac{b^{2}c}{b+c}+\frac{c^{2}a}{c+a}\leq \frac{\sqrt{a^{3}b}+\sqrt{b^{3}c}+\sqrt{c^{3}a}}{2}\leq \frac{1}{6}(a+b+c)^{2}$

Many people have asked this before.. see http://www.mathlinks.ro/Forum/viewtopic.php?t=6026

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[\frac{a^{2}b}{4a+7b}+\frac{b^{2}c}{4b+7c}+\frac{c^{2}a}{4c+7a}\leq\frac{(a+b+c)^{2}}{33}\]

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[\frac{a^{2}b}{4a+7b}+\frac{b^{2}c}{4b+7c}+\frac{c^{2}a}{4c+7a}\ge\frac{5(ab+bc+ca)-2(a^2+b^2+c^2)}{33}\]


There are 4 quarts in a gallon. One dollar is worth 4.5 francs. A
quart of gasoline costs 2 francs. How many dollars does a gallon of
gasoline cost? Express your answer rounded to the nearest cent.

First let's find how much one gallon of gasoline costs in francs, and then find out how many dollars that is.

So 1 quart is just 2 francs. There are 4 quarts in a gallon, so 1 gallon is 8 francs($ 4 \times 2 \equal{} 8$). To find out how many dollars this is, we can set up a proportion.

$ \frac{1}{4.5} \equal{} \frac{x}{8}$

where $ x$ is the amount of dollars one gallon costs. Now we cross-multiply:

$ 4.5x \equal{} 8$

$ \frac{4.5x}{4.5} \equal{} \frac{8}{4.5}$

$ x \equal{} 1.7777...$

We need to round to the nearest cent, so our answer is $ \boxed{\$1.78}$


If $a,b,c,d,e$ are positive numbers such that $abcde=1$, then
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}-4 \geq \sqrt[4]{a+b+c+d+e-4}$.

My standard solution uses Maclaurin's Inequality (2), too.
Moreover, we can apply EV-Theorem.

Vasc, can you please post a full solution to this nice inequality of yours? :)  ;)


A number line has its 0 fixed at (1,0) on a unit circle.  What central angle, in radians, is formed by the radius to (1,0) and the radius to the point that corresponds to 5.7 on the number line when it is wrapped around the circle?

It's $ 2\pi\minus{}5.7$ radians.  The unit circle has circumference of $ 2\pi$, which is the same number of radians in a circle.  Therefore, one unit on the number line corresponds to one radian, and so 5.7 units is 5.7 radians.  Then the angle is in the other direction so just subtract.


The question is $ \ce{C6H5 \sbond CHO} \plus{} \ce{NH3} \Delta$
now the obviuos produvt is Benzaldimine .But then the book say sit condenses that is this Benzaldimine attacks another Benzaldehyde .Then that attacks another and finally we have Hydro Banzanamide as the final Product.
First is this a Name reaction?
if no then how does one get to know (not in th Lab but in the exam Hall) that the reaction doesn't stop in the Benzaldimine stage

Well the product has structure $ \ce{Ph \sbond CH \dbond N \sbond C(H)(Ph) \sbond N \dbond HC \sbond Ph}$

[quote="Pardesi"]if no then how does one get to know (not in th Lab but in the exam Hall) that the reaction doesn't stop in the Benzaldimine stage[/quote]

We cannot know if the problem is incorrectly stated, as it seems to be. First, if what you have described is the expected reaction, then your book is wrong, because for that to happen the correct statement would be

$ 3C_6H_5CHO \plus{} NH_3 \longrightarrow ?$

that is, we cannot have an equimolar mixture of both reagents. Second, although the [i]intramolecular[/i]attack of an imine nitrogen on a carbonyl carbon is quite a common reaction in heterocyclic chemistry, I have never heard a usefull example of this outside heterocycles and intermolecular. If I propose this problem, as you stated, to any of my Organic Chemistry students then the answer I would expect to get (and the one I would also gave) is "benzaldimine", which is a far more important reaction than what your book expects to be answered. From the point of view of teaching/learning, that problem is almost "garbage" (its almost something like "what product will be obtained when acetylene is treated with a metal catalyst and heated?"). So, two questions need to be answered at this stage:

1) What is the purpose of that problem?

2) Why are you studying that book, specially since you are not an university student?

The problem is i hate the kind of problems like metal+acetylide...this one...
but the question paper setters in our country(most) love such questions .


Oduvek me je zanimalo sta matematicari vole da čitaju ...pa evo prilike da pitam direktno :)
Mislim na knjizevnost, ne na zbirke  :lol:   Poezija, proza, bilo sta ...

Ja sam kao mali jako puno čitao dječje romane, kasnije su me neki književnici kao Ferić ili Jergović oduševili ali zbog mnogo drugih interesa nisam puno čitao. Sad razmišljam da se vratim književnosti, prvenstveno da čitam modernu književnost jer mi je ona najbolja.

Ali pokušao sam i pisati književnost. Ta me ideja okupirala u djetinjstvu kada sam započeo bar 20 romana i na svakom odustao. Danas imam nekoliko sitnih radova, uglavnom nedovšenih. Shvatio sam koliko je zapravo to teško i pitam se kako se to može naučiti. Jer, matematika me je naučila da se izražavam jasno, koncizno, sažeto, a za književnost mi treba malo drugačiji stil.

[quote="djuro"]Ja sam kao mali jako puno čitao dječje romane[/quote]

This.

[quote="djuro"]Ali pokušao sam i pisati književnost. Ta me ideja okupirala u djetinjstvu kada sam započeo bar 20 romana i na svakom odustao. Danas imam nekoliko sitnih radova, uglavnom nedovšenih. [/quote]

This too.

[quote="djuro"]Jer, matematika me je naučila da se izražavam jasno, koncizno, sažeto[/quote]

Mene (još) nije. Mislim da predanih četrdesetak stranica na IMOu dobro govore o tome. :D

Ja na svakom putovanju čitam istu knjigu - Balkanski špijun Dušana Kovačevića (jest zbirka, ali zbirka drama  :lol: ). To mi je nekako kultna knjiga i o njoj me možete usred noći pitati što god hoćete. (Ostavit ću mobitel kraj kreveta ako netko želi isprobati ovu moju tvrdnju.  :P )Osim toga, ne čitam puno - nažalost, više se družim s najboljim prijateljem, kompjuterom. :lol: Volim Krležu. Recimo. (Ne Povratak Filipa Latinovicza, god no!, ali recimo Hrvatski bog Mars (antiratne novele) i Glembajeve).

Ovaj thread me podsjetio na http://www.mathlinks.ro/viewtopic.php?t=76903  :lol: i na diskusiju o Selimovicu, Poeu... Ah dobra stara vremena...

Prepoznao sam se u djurinim zapocetim romanima (ja sam poceo 3  :lol: ), ali jos uvijek pisem... I to neku vrstu techno-postmoderne: koga zanima, neka me PMuje  :lol:

Sto se citanja tice, uvijek i iznova mogu citati Orwella, Selimovica, Andrica, Dostojevskog i Gogolja. Sto se modernije literature tice: Hemon i Jergovic.

Ali ipak, omiljena knjiga: Surely you're joking Mr. Feynman.

A ove jeseni pocinjem aBd sa Godel-Escher-Bach...

hsiljak, drago mi je da si jos uvek ovde  :) . Osim toga, moram priznati da se slazem sa tobom-bio je to zanimljiv topic. Ja i dalje voooolim da citam i to ne samo zbirke zadataka  :wink: .

Citah Stivena Kinga- Mracnu kulu. Jedva sam docekala poslednji deo. Serijal je fantastican, ne samo zato sto je mesavina fantastike, sf-a, vesterna, nego zato sto je tako inteligentno napisan, likovi su veoma dobro izgradjeni, obicaji originalni, jezik i izrazi zarazni.

U planu je Den Simons-Hiperion, Pad Hiperiona, Endimion...Cini mi se da je trenutno najjveci naucnik medju piscima i da u knjigama ne samo da pominje razne naucne teorije, nego i zna o cemu prica.

Trenutno Citam Slikare i modele od Suzan Vriland. Knjiga me licno zanima. Donekle je pokusaj da razumem prirodu slikara i njihovodnos prema zivotu, zeljma, potrebama, vaznim i nevaznim stvarima...

Malo je razocaravajuce pomisliti da matematicari ne vole da citaju. Buduci da smo u stalnom traganju za istinom, kako cemo saznati istinu o svetu ako ne zelimo da ga upoznamo? Matematika jeste precizna,ali nije uskogruda ni u kom pogledu. Osim toga ne govori nista o emocijama, iako ih izaziva u prilicnoj meri. Cini mi se da ne bi trebalo ici iskljucivo u jednom smeru. Biti matematicar ne znaci i biti odvojen od umetnosti i zivota.

Pozdrav, Nienna! Dugo nisi pisala... (tacnije, godinu i pol)

Vidim da jos uvijek imas dobar knjizevni ukus: kad smo kod Simmonsa, ne znam da li si citala (da li su i ostali citali) Simona Singha. Njegov Code Book je pravi udzbenik kriptografije, napisan u formi pitkog i razumljivog stiva-sto nimalo ne trivijalizira temu, ali je cini pristupacnijom i totalnom laiku-a ostavlja dio izazova i za upucenije u tu oblast. Fermat's Last Theorem istog autora takodjer zasluzuje najmanje prolaznu ocjenu...

Sto se mene tice: nema promjena u odnosu na prethodni post: uz cinjenicu da jos uvijek citam GEB i planiram da, po djurinoj preporuci, pocnem sa djelima Zorana Ferica.

[quote]Malo je razocaravajuce pomisliti da matematicari ne vole da citaju. Buduci da smo u stalnom traganju za istinom, kako cemo saznati istinu o svetu ako ne zelimo da ga upoznamo? Matematika jeste precizna,ali nije uskogruda ni u kom pogledu. Osim toga ne govori nista o emocijama, iako ih izaziva u prilicnoj meri. Cini mi se da ne bi trebalo ici iskljucivo u jednom smeru. Biti matematicar ne znaci i biti odvojen od umetnosti i zivota.[/quote]

Odlicno receno. Ni slikar nije samo slikar, ni knjizevnik nije samo knjizevnik-ni matematicar ne smije ostati samo u svijetu matematike (kakva lijepa njemacka rijec, Fachidiot  :)  ). Ljepotu i sklad koju nadje u matematici moze vidjeti i u umjetnosti, naci je u pisanoj rijeci-a ako je ne nadje-moze pristupiti kreativnom cinu i sam je preinaciti u svoj vlastiti diskurs ;)

Opet ja poceo filozofirati...  :blush:

A sad, moram da podijelim sa svekolikim pucanstvom MathLinksa svoje odusevljenje romanom Ljubav u doba kolere G.G. Marqueza. Prije 4 godine sam citao Pukovniku nema ko da pise, i zavolio Marqueza. Onda sam prije dvije godine citao Sto godina samoce: nije to bio onaj isti Marquez, ali sam opet bio fasciniran.

A prije 5 dana dobio sam na poklon Ljubav u doba kolere... to je treci Marquez  :lol:  isto ime na koricama, ista biljeska o piscu, ali taj stil - ne nalikuje prethodnim, a opet osvaja i privlaci... Sjajno.

Nakon što prođe neko vrijeme ja imam za reći: ozbiljno bih htio biti pisac. Naime toliko stvari, misli, raznih emocija se u meni rađa i šteta je sve to ne pretočiti u umjetničko djelo.

Kako Borges lijepo rece: "Svako od nas ima nekoliko trenutaka u zivotu u kojima je sposoban napisati najljepsu pjesmu na svijetu" (neka me neko ispravi ako ima pri ruci Tom 1 sabranih djela, posto citat vadim iz glave  :)  ) Tako da svaki onaj ko odluci da iskoristi te svoje trenutke ima moju punu podrsku  :wink:

Proza, poezija, drama, nije bitno-dok je god to iz srca, nepatvoreni sentiment. A i mene si uspio natjerati u razmisljanje- da li ostati medju Citateljima, ili preci u drugi tabor.  :)

Ma mislim ovo je nevjerojatno. Koja blasfemija.

Nađete se na matematičkom forumu i onda pričate kako treba postati pisac...  :lol:

Kad bi svi postali pisci, tko bi osvajao zlata na Olimpijadi?! :P

Osvojis zlato na IMOu, i napises knjigu o tome  :P

A salu na stranu, matematika je samo jedan segment zivota svakog od nas, nekom veci, nekom manji. Ostatak prostora u zivotnom spektru popunimo svako na svoj nacin. Ukoliko neko odluci da matematiku komplementira knjizevnoscu /i krene stopama Lewisa Carrolla/ to niposto nije iznenadjujuce-jer inventivnost i masta koju posjeduje matematicar natjecatelj svakako ima potencijal da se ispolji i u nekoj umjetnickoj formi.

Kad smo vec kod citanja, da li obracate paznju na tekstove pesama koje slusate? Mene je poceo da odusevljava Nick Cave. Preslusala sam neke pesme uz tekstove i ostala prijatno iznenadjena. Murder ballads mozda jeste najpoznatiji album, ali No More Shall We Part ima mnogo kompleksnije tekstove. U pitanju je zaista poezija koja u skopu sa muzikom daje mnogo jaci utisak. Ako to povezete sa njegovim licnim zivotom, dobijate potpuniju sliku. Nesto nalik na autoportret umetnika. Nije za zanemarivanje ni njegov pogled na religiju, boga, ljubav.... Cini mi se da zaista prvi rekapitulaciju dogadjaja od kojih mnogi nisi ni prijatni ni dobri, ali svakako cine zivot stvarnijim.

Osim Kejva, Fernando Pessoa:

     Nothing with nothing around it
     And a few trees in between
     None of wich very clearly green,
     Where no river or flower pays a visit.
     If there be a hell, I've found it,
     For if ain't here, where the Devil it is?

Citala sam nekad Sto godina samoce, ali se knjige vise ne secam. Cim zavrsim sa ispitima, vracam se citanju necega sto nije zbirka pesama.   :yup:

Nasla si mi slabu tacku: Nick Cave. Oduvijek sam spadao u red onih koji gotovo samo i iskljucivo slusaju tekstove pjesama a ne pjesme kao melodijske cjeline ritma i rijeci (valjda sam zato zavolio Serbedziju, Stulica, Cavea, Adama...). I valjda zato nikad nisam slusao mainstream muziku...
Konkretno Cave: nekad se zamislim da li je covjek lirski ili epski pjesnik-jer je sentiment uvijek uokviren tako sjajno (i slikovito) ispricanom pricom o dogadjaju, o licima i mjestima...
A kad smo vec na toj temi: vecina nas ovdje voli Iron Maiden  :) I u tom kontekstu moram reci da je poprilican broj Dickinsonovih (i Harrisovih) tekstova umjetnicko djelo.

Odoh slusati Arsena.

ja se moram pohvaliti da sam i strastveni slušač muzike, i ne okupira me tekst koliko glazba. tekstove ponekad primijetim, pogotovo kod domaćih izvođača, ali muzika, ta genijalna muzika koja je stvarno umjetnost nad umjetnostima, mi je na prvom mjestu.

Izgleda da trenutno na ovom forumu, matematičari pričaju samo o književnosti.

Osim imo kompendiuma, u zadnje vrijeme sam čitala Derviš i smrt. I knjiga me baš prijatno iznenadila.
I bacila u duboko razmišljanje.
A tekstovi Iron maidena su zaista prave pjesme.

Pink Floyd, jesi li ti možda Jelena? Ako jesi onda lijep pozdrav :)

Ajde, kad jadnom djuri nitko ne odgovara (Ili je to bilo retoričko pitanje?)...

Hm, djuro... veliki informatičar, a ne potrudi se pogledati nečiju mail adresu ispod profila... :D Eto, sigurno si je toliko razočarao svojim pitanjem da ti ne želi ni odgovoriti... :P

E, da, u vezi književnosti i umjetnosti, djuro, pozovi nas na otvorenje svoje prve izložbe... ili promociju svoje prve knjige... ili na proslavu svojeg povratka matematici... :P

[size=67]Napomena: Stavovi koje je iznio autor u svojem uratku nisu službeni stavovi Unije izumitelja tople vode, Stranke vječne opozicije, Savjeta za ekonomsku analizu propasti, kućnog savjeta njegove zgrade i/ili daljnjih udruga čiji je autor član. Autor zadržava pravo da zaniječe svaku povezanost s bilo kojim svojim napisom koji jest ili nije uopće napisao. Sve transakcije koje autor čini su završne i na njih nema reklamacije, osim na autorove autograme ili službene biografije, koje mogu i/ili ne moraju biti točne. Prije čitanja, posavjetujte se s liječnikom, ljekarnikom ili barem visokopozicioniranim političarem optuženim za korupciju. Baterije nisu ni uključene niti isključene. Kad otvorite kutiju, ostvarujete pravo na popust prilikom vašeg sljedećeg otvaranja iste kutije. Ovaj tekst je napisan na ekološki prihvatljivom virtualnom prostoru, ali preporučujemo printanje na ekološki neprihvatljivom bijelom papiru. Smisao za humor nije uključen, ali autor rado prima donacije.[/size]

Vec godinama uopste ne slusam domacu muziku. Na neki nacin mi je opterecujuca. Valjda i sama imam dovoljno slovenske sentimentalnosti da mi je tesko da to pronalazim i u muzici. Jedino sto me je zaista odusevilo u zadnje vreme je Laibach. Od starih stvari, tu su Azra, EKV...

[quote]Izgleda da trenutno na ovom forumu, matematičari pričaju samo o književnosti.[/quote]

Zar nije dosadno pricati samo o matematici?

[quote="Nienna"]
[quote]Izgleda da trenutno na ovom forumu, matematičari pričaju samo o književnosti.[/quote]

Zar nije dosadno pricati samo o matematici?[/quote]

Predlažem da napravimo Mathlinks revoluciju. Otvarali bismo sve više i više tema o umjetnosti sve dok ne bi one matematičke postale rijetkost. Za to nam naravno treba internacionalna potpora, ali od Balkana se počinje :) Dakle, tko je uz mene?

Zar smo Pink Floyd (djelomično, vidim da se i ona nešto bavi lijepom književnošću) i ja jedini glasovi razuma na ovom forumu?

Ljudi, čitajte što više, ozbiljno. Posebno preporučujem 102 Combinatorial Problems od Titu Andreescua i Zuming Fenga.

A i djuro, znaš da sam ja (gotovo) uvijek zainteresiran za revoluciju. Ali pobogu, istjeravati matematiku s matematičkog foruma... people, to je kao da mene probate naučiti da sviram gusle. Ne ide.

 :D

Sad ću si ja dati vremena danas-sutra, pa otvoriti onako baš pravu matematičku temu, a ne da se priča o ko-zna-kakvim tu piskaralima i zelenim, crvenim i žutim vragovima (tko zna koga parafraziram, zna, tko ne zna, nije bitno - može se smatrati sretnim).  :)

Hajd' da si rekao da je problem nauciti svirati violinu, ili klavir... ali gusle...  :lol: Ne moze biti toliko tesko guditi na jednoj zici  :)

102 CP je dobar, dapace sjajan... mada ja citam Teoriju grafova od Diestela :)

P. S. Nisam te srece da ne prepoznam duh dr. F. T. u posljednjoj recenici  :)

Nevjerojatno nešto ali opet mi je matematika ispred književnosti. Ko bi rekao. Je li tome kriv dozlaboga dosadni Šenoa, shvaćanje koliko je jako jako jako teško dobro pisati ili prodike roditelja, ne bih znao.

Melkiore, ako misliš da to treba proslaviti, mogu te počastiti pićem.

Ne mislim ja da to treba proslaviti... moje loše šale su čisto zabavljačke prirode i najozbiljnijim mogućim tonom uvjeravam te da stojim iza onoga što sam već i prije rekao: U mojem shvaćanju života (za koje shvaćam da nije u potpunosti prihvatljivo svim, pa čak ni većini ljudi), čovjek se treba baviti onime što ga najviše veseli. I to je to - bez daljnje filozofije. Veseli te - radi to.

Mislim da će te, ako si već spreman raditi (bez pritiska, ako ostaješ pri nedolasku na natjecanja, tvoja stvar), Dražić raširenih ruku primiti. Naravno, i ja i svi ćemo pomoći koliko možemo.

Welcome back, Adriane.


"Two cars 150 miles apart are traveling toward each other along the same one-lane road. The first car goes 90 miles per hour; the second car rushes along at 60 miles per hour. A bee is hovering just above the nose of the first car. It buzzes from the first car to the second car, turns around immediately, flies back to the first car, and turns around again. It goes on flying back and forth between the two cars until they collide. If the bee's speed is 120 miles per hour, how far will it travel?" (Source:Found on Dr. Math Forum)

Is the bee travelling 120 mph relative to the train?

oh sorry. the first car/train is traveling at 90 mph not 150 mph and second car/train 60 mph. fixed.

@VenDiddy: I believe you may have misinterpreted. the 2 cars are actual, separate cars, not cars on a train.

Anywho, here's [hide="my solution."]The 2 cars are traveling towards each other with a total velocity of 150 mph. This means that they will meet after traveling for 1 hour, because they started 150 miles apart. Hence the bee has also flown for 1 hour. Since it flew for 1 hour at a constant speed of 120 mph (flies constantly, turns are instantaneous), the bee covered 120 miles.[/hide]

I agree.  [hide="My Answer"]The speed is $90+60$ is $150$.  They total distance is $150$, so  it takes 1 hour.  Therfore, the fly travels 1 hour at 120 mph, resulting in $120$ $miles$.[/hide]

@lasagnaman: I was just confused because earlier it said that the first train was travelling at 150 mph, so it would have gone faster than the 120 mph bee and collided before the bee would have even gotten a chance to touch the nose of the other train.


complex analysis
complex analysis theorems

[quote="Lang In  [i]Complex Analysis 3rd Edition[/i]"]
[b]Theorem[/b]

[i]Let $\sum a_{n}z^{n}$ be a power series, and let $r$ be its radius of convergence. Then
\[\frac{1}{r}= \lim \sup |a_{n}|^{1/n}\]
If $r=0$, this relation is to be interpreted as meaning that the sequence $\{|a_{n}|^{1/n}\}$ is not bounded. If $r = \infty$, it is to be interpeted as meaning that the $\lim \sup |a_{n}|^{1/n}=0$[/i]

[i]Proof:[/i] Let $t=\lim \sup |a_{n}|^{1/n}$. Suppose first that $t \not= 0, \infty$. Given $\epsilon>0$, there exist only a finite number of $n$ such that  $|a_{n}|^{1/n}\geq t+\epsilon$. Thus for all but a finite number of $n$, we have
\[|a_{n}| \leq (t+\epsilon)^{n}, \]
Whence the series $\sum a_{n}z^{n}$ converges absolutely if $|z| < 1/(t+\epsilon)$, by comparasion with geometric series. Therefore the radius convergence $r$ statisfies $r \geq 1/(t+\epsilon)$ for every $\epsilon > 0$, whence $r \geq 1/t$.

  Conversely, given $\epsilon$ there exist infinitely many $n$ such that $|a_{n}|^{1/n}\geq 1-\epsilon$ and therefore
\[|a_{n}| \geq (t-\epsilon)^{n}\]
Hence the series $\sum a_{n}z^{n}$ does not convergence if $|z|=I/(t-\epsilon)$, because its $n$-th term does not even tend to $0$. Therefore the radius convergence $r$ stastisfies $r \leq 1/(t-\epsilon)$ for every $\epsilon>0$, whence $r\leq 1/t$. This conclude the proof in case  $t \not = 0, \infty$.
[/quote]

what i can't understand is this step:
[quote]
 Given $\epsilon>0$, there exist only a finite number of $n$ such that  $|a_{n}|^{1/n}\geq t+\epsilon$. Thus for all but a finite number of $n$, we have
\[|a_{n}| \leq (t+\epsilon)^{n}, \]
[/quote]

after saying that for [i]finite[/i] number $n$ then $|a_{n}|^{1/n}\geq t+\epsilon$,   why it is also  $|a_{n}| \leq (t+\epsilon)^{n}$ for all but finite $n$ ??? how can $\geq$ becomes $\leq$ here???
 :blush:
thanks before

Edit: Typo (not the book but me. :blush: )

[quote="bolpoin"] $|a_{n}| \leq (1+\epsilon)^{n}$ for all but finite $n$ ???
 [/quote]

Exists $n_{0}$ such that $|a_{n}| \leq (1+\epsilon)^{n}\forall n\geq n_{0}$, i think  :wink:

Not "for finite number $n$" but "for finitely many numbers $n$".

i've  change "for all but finite..."  to "for all but a finite..."

still don't understand... :blush:  maybe my english is so bad...  :blush:

Those $1+\epsilon$ should be $t+\epsilon$.

For each $n$, either $|a_{n}|\ge (t+\epsilon)^{n}$ or $|a_{n}|<(t+\epsilon)^{n}$. The first is true for only finitely many $n$, so the second is true for all others- all but finitely many.

It's important to understand what happens to quantifiers when we negate statements.

:blush:  that's my typo... and i made my self confuse of it, whoaaaa that's really stupid  :blush:  :blush:  (actually the book is in the library while i type my first post, its from handwriting paper ) really sorry and thanks for you all help, but still confuse

@jmerry why all but finitely many?? you true that it is true for all other...