In [url=http://www.geocities.com/mario_the_sheep/trig1.bmp][u]the figure[/u][/url]  P ( -4; 3 )  ,  Q ( -3, -4)
and R are points on a circle with its
centre at the origin.

$ \angle POX = B$ and reflex $ \angle QOX = A$

Determine, without the use of a calculator, the value of:
	1/	$ \tan B + \cot A$
	2/	$\csc 270^\circ  +\sin(360^\circ -B)$
         3/     Prove that  $ \angle POQ = 90^\circ$

(not referring to the diagram) It is given that  $\cot\Theta > 0$ , $\sin\Theta < 0$ .
If $\cot\Theta = p$      determine the following in terms of  $p$:
       1/	$\sin\Theta$
2/	$\sin\beta$  if $\Theta$  and $\beta$  are complementary

hi darkelve, since you already know how to get $\beta$ or $\Theta$ then for the trigonometric functions they follow the same principle. instead of sin $sin$ type \sin $\sin$. hope this helps your $\LaTeX$ tips

hmmm <: i don't know any other way atm.. i'll look around if i can find another method ;0

[quote="Chrnox"]hmm

i dun feel right when i see the way u present the answer in part 3).

erm.. i guess by using pythagoras theorem straight away does show that it is equal to PQ^2. however, u are [b]assuming[/b] that its 90 degrees [b]before[/b] proving.

hmm.. is there another way to show that its right angles besides using this? cos i feel that assumptions before proving and showing that is true doesn't really convince me. sorry.. i'm a little troublesome (asking for too much).. >.<.. haha..[/quote]

actually there is a theorem in geometry which says that if $a,b,c$ are the sides of a triangle such that $a^2+b^2=c^2$ then the angle opposite the hypotenuse is a right angle. so her answer is still correct without assuming 90 degrees


algebra
polynomial
function
combinatorics unsolved
combinatorics

Prove the following relations:
1.$ \sum_{k \equal{} 0}^{n}\binom{2k}{k}\binom{2(n \minus{} k)}{n \minus{} k} \equal{} 4^n$.
2.$ \sum_{i \equal{} 0}^{m}\binom{x \plus{} y \plus{} i}{i}\binom{y}{a \minus{} i}\binom{x}{b \minus{} i} \equal{} \binom{x \plus{} a}{b}\binom{y \plus{} b}{a}$.,where $ m \equal{} min(a,b)$.
3.For $ n\geq 0$.
$ \sum_{k \equal{} 0}^{n}\binom{n}{k}^2x^k \equal{} \sum_{j \equal{} 0}^{n}\binom{n}{j}\binom{2n \minus{} j}{n}(x \minus{} 1)^j$.
Pay attention to the name of the topic.I'm looking for the combinatorial solutions :wink:
[hide="Hint?"]
Using Combinatorial Nullstellensatz we can use that $ x,y\in N$.
[/hide]

i'm sure that for the three problems above there exist a combinatorial proof,but i can't find them...
Here two more similar problems(actually second is the generalization of the first,but i am uncertain whether combinatorial proof for the second problem is known):
1.Prove that if $ p\in\mathbb{P}$ then $ \binom{2p}{p} \minus{} 2$ is divisible by $ p^2$.
2.Prove that if $ p\in\mathbb{P},p > 3$ then $ \binom{2p}{p} \minus{} 2$ is divisible by $ p^3$.
This problems are not hard,especially the first one and i've already found one beautiful combinatorial proof for it.
Remember that i'm looking for only combinatorial proofs! :)
Best regards,Erken.

Prove $ \sum_{k \equal{} 0}^{n}\binom{x}{k}\binom{y}{n \minus{} k} \equal{} \binom{x \plus{} y}{n}$, it's not hard.
Use it. x=y=p and n=p
$ \sum_{k \equal{} 0}^{p}\binom{p}{k}\binom{p}{p \minus{} k} \equal{} \binom{2p}{p}$
$ \sum_{k \equal{} 0}^{p}\binom{p}{k}\binom{p}{p \minus{} k} \equal{} \sum_{k \equal{} 0}^{p}\binom{p}{k}^2 \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\binom{p}{k}^2 \plus{} 2$
$ \binom{2p}{p} \minus{} 2 \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\binom{p}{k}^2$
If $ k < p$ then $ \binom{p}{k}$ is divisible by p.
Then $ \binom{2p}{p} \minus{} 2$ is divisble by $ p^2$

Don't spam Ol. :rotfl: you know ,what i'm talking about!

[quote="JBL"]Pascual2005 outlined a combinatorial proof of the first [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316545#316545]here[/url] (but some other commenters weren't happy with it?).[/quote]
But what about 2nd and 3rd problems?Was they solved here?

well about the third...
before i start i apologise for my english!

we want to prove that two polynomials are equal, this is the same as proving that all coefficients are equal, the identity becomes
\[ {n\choose k}^2 = \sum_{i = k}^n{n\choose i}{2n - k\choose n}{i\choose k}( - 1)^{i - k}
\]
it's known that $ {n\choose i}{i\choose k} = {n\choose k}{n - k\choose i - k}$
so our identity becomes
\[ {n\choose k}^2 = {n\choose k} \sum_{i = k}^n{n - k\choose i - k}{2n - i\choose n}( - 1)^{i - k}
\]

\[ \Leftrightarrow {n\choose k} = \sum_{i = k}^n{n - k\choose i - k}{2n - i\choose n}( - 1)^{i - k}
\]
to prove this we want to prove that
\[ \sum_{j = 0}^a {a\choose j}{b + a - j\choose b}( - 1)^j = {b\choose a}
\]
(b>a) and then substitute $ j = i - k$, $ a = n - k$, $ b = n$.
here there are 2ways:
i) with generating functions we note that
\[ \{{b + h\choose b}\}_{h\geq 0}\leftrightarrow {(\frac {1}{1 - x})}^{b + 1}
\]
and
\[ \{{a\choose j}( - 1)^j\}_{j\geq 0}\leftrightarrow {(1 - x)}^a
\]
so
\[ \{\sum_{j = 0}^l {a\choose j}{b + l - j\choose b}( - 1)^j\}_{l\geq 0}\leftrightarrow {(\frac {1}{1 - x})}^{b + 1}\cdot\ {(1 - x)}^a = {(\frac {1}{1 - x})}^{b - a + 1}
\]

\[{ {(\frac {1}{1 - x})}^{b - a + 1}\leftrightarrow \{b - a + l\choose b - a}\}_{l\geq 0}
\]
and so
\[ \sum_{j = 0}^a {a\choose j}{b + a - j\choose b}( - 1)^j = {b\choose b - a} = {b\choose a}.
\]
ii) combinatorial way: consider sets $ A_{i\ (b < i\leq b + a)}$ of all the choices of $ b$ of the first $ b + a$ positive integers without integer $ i$. With PIE we count the total number of way to choose $ b$ of the first $ b + a$ positive integers without at least one integer $ i,\ b < i\leq b + a$.
By PIE this number is
\[ |A_{b + 1}\cup A_{b + 2}\cup\dots\cup A_{b + a}| = \sum_{k = 1}^b\sum_{b < i_1 < \dots < i_k\leq b + a}|A_{i_1}\cap\dots\cap A_{i_k}|( - 1)^{k + 1}
\]
but $ \sum_{b < i_1 < \dots < i_k\leq b + a}|A_{i_1}\cap\dots\cap A_{i_k}|( - 1)^{k + 1}$ are the way to choose $ b$ positive integers without $ k$ integers $ i_1,\ i_2,\ \dots,\ i_k$ between $ b$ and $ b + a$ and this number is $ {a\choose k}{b + a - k\choose b}$ (we choose $ k$ integers between $ b$ and $ b + a$ that we put out and between the other $ b + a - k$ integers we choose $ b$ integers). So by PIE our number is
\[ \sum_{k = 1}^a{a\choose k}{b + a - k\choose b}( - 1)^{k + 1}
\]
.

Now the total number of choices of $ b$ of the first $ a + b$ positive integers are $ {b + a \choose a}$, so the number of choices of $ b$ of the first $ b + a$ positive integers with all $ a + 1,\ a + 2,\ \dots,\ b$ are
\[ {a\choose 0}{b + a\choose a} - \sum_{k = 1}^a{a\choose k}{b + a - k\choose b}( - 1)^{k + 1} = \sum_{k = 0}^a{a\choose k}{b + a - k\choose b}( - 1)^{k}
\]
but obviously this number is also the number of choices of the other $ b - a$ integer (which are $ \leq b$) that is
\[ {b\choose b - a} = {b\choose a}.
\]

2 mattilgale:
At first thank you!I haven't checked your soution yet,but i'll do it later.
I'm not so good in english math.termins,so what is PIE?

i'm sorry, maybe PIE is an acronym used only in Italy ... it's Principle of Inclusion-Exclusion


Interesting proposition.  :)
[url]http://www.earthhour.org/[/url].

I saw that when I was Googling something. I'm definately going to join  :)

Looks awesome, but how do you join it...? I haven't read the full details.

Just turn off as much power-using things as possible at 8 pm tonight.

[quote="EarthHour.org"]
On March 29, 2008 at 8 p.m., join millions of people around the world in making a statement about climate change by turning off your lights for Earth Hour, an event created by the World Wildlife Fund.
[/quote]

Oh, me and 7h3.D3m0n.117, we were watching a movie.  :oops:  Shame on us...

Ah well everything was off except for the TV. I guess that counts  :) .

And the one light....

I wondered what ways any other AoPSer conserved energy.

[url]http://blackle.com/tips/[/url] has some good tips for saving energy. blackle itself is a energy saving version of Google. Pretty interesting...

I completely forgot about Earth Hour. I've been meaning to follow it, but just forgot.

I don't think having everyone turn off their appliances is necessarily a good idea, because if everyone actually did do that then the power plant's efficiency would go down dramatically, since they're designed to be more efficient at certain outputs...
I didn't even know about it, so I didn't do anything, but my family doesn't use much electricity in the first place so I don't see how that matters.

That's true. The conservation things are more for people who drive cars that get 1 mile/gallon, leave the faucet running while they brush their teeth, etc. etc. I still think the best solution is to go to an alternative energy source such as wind or solar.

However, in most regions, solar is the only way to go for a different energy source and that can get pretty expensive.

A comment on cars. I think driving habits/skill matters too. My mom has very good driving, its very comfortable sitting in the car when she drives, and she always averages near the advertised MPG (her car is a Honda Pilot, 18/24, she gets like 21 average). Then there's other people who buy Priuses and can't even break 30MPG because they don't have good control and step too hard on the gas, press too hard on brakes, step on gas when they see red light and want to roll over to the intersection faster for some odd reason, etc. Then you have people who buy fat American SUVs AND step super hard on the gas :O

I participated in this event, but one of my friends decided to do the eath hour at a more convenient time to him. Do you guys think that you should participate in an event like this at the given time, or whenever the time suits your need? Both ways save the same amount of money, but still....

[quote="jackdillon"]Do you guys think that you should participate in an event like this at the given time, or whenever the time suits your need?[/quote]

Turning off all appliances at the same time would be easier to see the dramatic change of energy use, much less monitor the energy use all day long. I would suggest that it would be at a very convenient hour, such as during the nighttime (maybe 11:00 pm :P) which is easier done than said.

Eh, I turned off some things.
Mostly all that was on was a TV that my parents were watching and my computer.

I would have joined..If i knew sooner, lol I think my family had 2 TV's all the computers(3) and like 3/4 of the lights running in our house. XD

[quote="serialk11r"]she always averages near the advertised MPG (her car is a Honda Pilot, 18/24, she gets like 21 average). Then there's other people who buy Priuses and can't even break 30MPG ...[/quote]
Or, in other words, because your mother drives an absurdly large vehicle, she uses 30% more gas per distance than even the worst driver in a Prius :P

More generally, I think that Earth Hour, while a nice idea, is somewhat off-point.  One needn't live in darkness to consume less energy, and individual choices to conserve will only be successful if they are complemented with large-scale societal actions (e.g. a carbon cap or tax that would help ensure that everyone reduces consumption regardless of personal morality).  That said, individuals do have pretty astounding environmental impacts.  I think there are basically three ways of mitigating them:

1) Eliminate pure waste.  This is consumption (of energy, plastic, water, or whatever) that offers you no benefits whatsoever.  Obvious examples are to turn things (computers, televisions, lights, faucets) off when you aren't using them -- I'm just shocked at the number of people who leave their computers on all the time, for example, when every modern operating system has some sort of sleep mode, or who leave the light on in their office when they go home for the day.  There is no legitimate case in favor of waste of this kind.

2) Increase efficiency.  Buy compact fluorescent lightbulbs, fuel-efficient cars, EnergyStar appliances, etc.  Eat organic or locally-raised foods.  Insulate your home.  These types of things tend to be "doubly green" (sorry, a US-specific notion) in that you save money and reduce your environmental impact (without changing your underlying behavior).

3) Reduce your consumption.  Choose activities that have less impact.  Eat vegetarian.  Go to sleep earlier (so you use less electric lighting), arrange your trips to minimize driving, bike or walk whenever possible, turn the thermostat down, use less hot water, live closer to your job (or vice versa), etc.  These are the most challenging, the ones that are probably best solved through collective action (by incorporating the environmental costs of activities into their dollar costs, i.e. by appropriate taxation), and probably the most important overall for avoiding environmental catastrophies (of which the global warming associated with increase CO$ _2$ levels is just one).

oh yeah!! my friend told me about that on monday, and I was like, WHA??

so i'm sorry.. but i'm probably gonna not waste any energy on Earth Day!!!


When try to edit  own message, I get "Ask the moderator to do it" type message.

Was able to do that in the past.

New feature or bug?

(Sorry if this was already reported).


function
floor function
inequalities
quadratics
number theory unsolved
number theory

Let $ p$ be a prime of form $ 4k\plus{}1$. Find the sum

$ \sum_{n\equal{}1}^{p\minus{}1}\lfloor \sqrt{np}\rfloor$ in explicit form.

Note that $ \lfloor \sqrt{np} \rfloor \equal{} m$ is equivalent to the inequality
\[ m^2 \leq np < (m\plus{}1)^2.\]
The number of $ n$ satisfying this inequality is exactly
\[ \left\lfloor\frac{(m\plus{}1)^2\minus{}1}{p}\right\rfloor \minus{} \left\lfloor\frac{m^2\minus{}1}{p}\right\rfloor,\]
implying that
\[ \sum_{n\equal{}1}^{p\minus{}1} \lfloor \sqrt{np} \rfloor \equal{} \sum_{m\equal{}1}^{p\minus{}1} m\left( \left\lfloor\frac{(m\plus{}1)^2\minus{}1}{p}\right\rfloor \minus{} \left\lfloor\frac{m^2\minus{}1}{p}\right\rfloor \right) \equal{} (p\minus{}1)^2 \minus{} \sum_{m\equal{}1}^{p\minus{}1} \left\lfloor\frac{m^2\minus{}1}{p}\right\rfloor.\]

Since $ p$ is prime, $ m^2\bmod p$ for $ m\equal{}1,2,\dots,p\minus{}1$ takes values equal quadratic residues modulo $ p$, each value two times. Therefore, for prime $ p$, we have
\[ \sum_{m\equal{}1}^{p\minus{}1} \left\lfloor\frac{m^2\minus{}1}{p}\right\rfloor \equal{} \sum_{m\equal{}1}^{p\minus{}1} \left\lfloor\frac{m^2}{p}\right\rfloor \equal{} \frac{\sum_{m\equal{}1}^{p\minus{}1} m^2 \minus{} 2S}{p} \equal{} \frac{(p\minus{}1)(2p\minus{}1)}{6} \minus{} \frac{2S}{p}\]
where $ S$ is the sum of the quadratic residues modulo $ p$. It is known that for prime $ p\equal{}4k\plus{}1$, $ S\equal{}kp$. Therefore,
\[ \sum_{n\equal{}1}^{p\minus{}1} \lfloor \sqrt{np} \rfloor \equal{} (p\minus{}1)^2 \minus{} \frac{(p\minus{}1)(2p\minus{}1)}{6} \plus{} \frac{p\minus{}1}{2} \equal{} \frac{(p\minus{}1)(2p\minus{}1)}{3}.\]

See also http://www.mathlinks.ro/viewtopic.php?t=150713 .

(Sorry for topic necromancy, but I'm just browsing indybar's posts for interesting and nontrivial problems.)

  darij


I was just looking on their website and it says that the 2005 convention is in Honolulu. That would be awesome to go, but I think attendance is going to be much lower. What do you think?

what is mu alpha thetat and how do you participate?

well, i think it would be more incentive for some ppl to go as well because they could stay there for vacation or somthin but i think overall attendence will drop just because east coast ppl will have a long ways to go (but is that really a bad thing? :D)

so the cost is high. why do people go at all?

Yeah, I'm in Alabama. Maybe we can go to the one in Huntsville this year, but there's about a .01% chance we will go to the Honolulu one.

is mu alpha teta by chance the problems which have a MAO right by it in the AOPS books?

Mu Alpha Theta = MAO = Communist *gasp* :D

Mu Alpha Theta (MAO) is a national mathematics honor society which has a national convention in a different city each summer.  The national convention consists of presentations on mathematical topics, competitions of various sorts, and tourist activities in the host city.  The conventions are a great deal of fun, and packed with great math.  I happen to be in charge of the test-writing effort for the 2005 Hawaii convention, and believe it will be one of the best in recent memory (many original problems and very few disputes), something that has often been said of the 2001 Denver convention for which I also led the test-writing effort.  Although it may be a long flight from the east coast, and possibly a little more expensive, I would encourage any school with students interested in extracurricular mathematics to attend.


induction
number theory proposed
number theory

The real sequence $ \{a_{n}|n\equal{}0,1,2,3,...\}$ defined $ a_{0}\equal{}1$ and \[ a_{n\plus{}1}\equal{}\frac{1}{2}\left (a_{n}\plus{}\frac{1}{3\cdot a_{n}}\right ).\]

Denote \[ A_{n}\equal{}\frac{3}{3\cdot a_{n}^{2}\minus{}1}.\]

Prove that $ A_{n}$ is a perfect square and it has at least $ n$ distinct prime divisors.

Here another solution:

Affirmation 1.- All $ a_n$ are rational numbers.
Proof.- By easy induction. It is left to the reader,

Let $ a_n\equal{} \frac{p_n}{q_n}$ with $ p_n$, $ q_n$ integers and $ (p_{n},q_{n})\equal{}1$

Affirmation 2.- For all $ n\geq 1$ is true $ 3p_{n}^{2}\minus{}q_{n}^{2}\equal{}3$
Proof.- For induction. To $ n\equal{}1 \rightarrow a_{1}\equal{} \frac{2}{3} \rightarrow 3.2^{2}\minus{}3^{2}\equal{}3$. If it is true for $ n$ then $ 3p_{n}^{2}\minus{}q_{n}^{2}\equal{}3 \rightarrow 3|q_{n}^{2} \rightarrow q_n\equal{}3q_{n}'$, then the induction hypothesis is written  $ 3p_{n}^{2}\minus{}9q_{n}'^{2}\equal{}3 \rightarrow p_{n}^{2}\minus{}3q_{n}'^{2}\equal{}1$. Of the recurrence relation $ a_{n\plus{}1}\equal{} \frac{3a_{n}^{2}\plus{}1}{6a_{n}} \implies a_{n\plus{}1}\equal{} \frac{3p_{n}^{2}\plus{}q_{n}^{2}}{6p_{n}q_{n}}\equal{} \frac{p_{n}^{2}\plus{}3q_{n}'^{2}}{6p_{n}q_{n}'}$.
Then $ 3(p_{n}^{2}\plus{}3q_{n}'^{2})^{2}\minus{}(6p_{n}q_{n}')^{2}\equal{}3(p_{n}^{2}\minus{}3q_{n}'^{2})^{2}\equal{}3$. End of induction.

By affirmation 2 we have $ A_{n}\equal{} \frac{3}{\frac{3p_{n}^{2}}{q_{n}^{2}}\minus{}1}\equal{}q_{n}^{2}$.                                                           $ \square$

Affirmation 3.- For all $ n\geq 1$ is true $ w(q_{n})\geq n$.
Proof.- For induction. If $ n\equal{}1  \rightarrow q_{1}\equal{}3 \implies w(q_{1})\equal{}1\geq 1$. We assume that is true for $ n$. Since $ (p_{n},q_{n})\equal{}1 \implies  (p_{n},q_{n}')\equal{}1 \implies (p_{n}^{2},p_{n}^{2}\plus{}3q_{n}'^{2})\equal{}1$. As $ p_{n\plus{}1}\equal{}p_{n}^{2}\plus{}3q_{n}'^{2}$ and $ q_{n\plus{}1}\equal{}6p_{n}q_{n}'$.
If $ 6|q_{n}' \implies w(6p_{n}q_{n}')\equal{} w(p_{n})\plus{}w(q_{n}')\geq {n\plus{}1}$. If $ 2|q_{n}'$ and $ 3\nmid q_{n}' \implies w(6p_{n}q_{n}')\equal{} w(3)\plus{}w(p_{n})\plus{}w(q_{n}')\geq {n\plus{}1}$. If $ 3|q_{n}'$ and $ 2\nmid q_{n}' \implies w(6p_{n}q_{n}')\equal{} w(2)\plus{}w(p_{n})\plus{}w(q_{n}')\geq {n\plus{}2}$. If $ 6\nmid q_{n}' \implies w(6p_{n}q_{n}')\equal{} w(6)\plus{}w(p_{n})\plus{}w(q_{n}')\geq {n\plus{}2}$. In total $ w(q_{n\plus{}1})\geq {n\plus{}1}$. End of induction.

By affirmation 3 we have $ w(A_{n})\equal{} w(q_{n}^{2})\equal{} w(q_{n})\geq n$.                                                                                $ \square$


inequalities
geometry
3D geometry
tetrahedron
octahedron
algebra
polynomial

Show that this inequality holds for every $ x,y,z,t,u,v\in (0,1) $ : $ xyz+uv(1-x)+(1-y)(1-v)t+(1-z)(1-u)(1-t)<1 $

[quote="bomb"]Can you symmetrize this inequality.

Bomb[/quote]
 :huh: ???

[quote="ehsan2004"]Show that this inequality holds for every $x,y,z,t,u,v\in (0,1)$ : $xyz+uv(1-x)+(1-y)(1-v)t+(1-z)(1-u)(1-t)<1$[/quote]

Nice one...

In the following, we will use the abbreviation |XYZW| for the volume of an arbitrary tetrahedron XYZW.

First recall a well-known lemma:

[color=blue][b]Lemma 1.[/b] If OABC is a tetrahedron, and A', B', C' are points on its edges OA, OB, OC, then

$\frac{\left|OA^{\prime}B^{\prime}C^{\prime}\right|}{\left|OABC\right|}=\frac{OA^{\prime}\cdot OB^{\prime}\cdot OC^{\prime}}{OA\cdot OB\cdot OC}$.[/color]

This lemma trivially follows from the linearity of the volume.

Now, construct a regular tetrahedron ABCD with edgelength AB = BC = CD = DA = AC = BD = 1, and let X, Y, Z, U, V and T be points on its edges AB, AC, DA, BD, BC and CD, respectively, such that AX = x, AY = y, AZ = z, BU = u, BV = v and CT = t. Then, of course, BX = AB - AX = 1 - x, CY = AC - AY = 1 - y, DZ = DA - AZ = 1 - z, DU = BD - BU = 1 - u, CV = BC - BV = 1 - v and DT = CD - CT = 1 - t. Hence, by Lemma 1, we have

$\frac{\left|AXYZ\right|}{\left|ABCD\right|}=\frac{AX\cdot AY\cdot AZ}{AB\cdot AC\cdot AD}=\frac{x\cdot y\cdot z}{1\cdot 1\cdot 1}=xyz$;
$\frac{\left|BUVX\right|}{\left|ABCD\right|}=\frac{BU\cdot BV\cdot BX}{BD\cdot BC\cdot BA}=\frac{u\cdot v\cdot\left(1-x\right)}{1\cdot 1\cdot 1}=uv\left(1-x\right)$;
$\frac{\left|CYVT\right|}{\left|ABCD\right|}=\frac{CY\cdot CV\cdot CT}{CA\cdot CB\cdot CD}=\frac{\left(1-y\right)\cdot\left(1-v\right)\cdot t}{1\cdot 1\cdot 1}=\left(1-y\right)\left(1-v\right)t$;
$\frac{\left|DZUT\right|}{\left|ABCD\right|}=\frac{DZ\cdot DU\cdot DT}{DA\cdot DB\cdot DC}=\frac{\left(1-z\right)\cdot\left(1-u\right)\cdot\left(1-t\right)}{1\cdot 1\cdot 1}=\left(1-z\right)\left(1-u\right)\left(1-t\right)$.

But |AXYZ| + |BUVX| + |CYVT| + |DZUT| < |ABCD| (in fact, the tetrahedra AXYZ, BUVX, CYVT and DZUT together form only part of the volume of the tetrahedron ABCD; the rest is the octahedron XYZTUV). Dividing this inequality by |ABCD|, we get

$\frac{\left|AXYZ\right|}{\left|ABCD\right|}+\frac{\left|BUVX\right|}{\left|ABCD\right|}+\frac{\left|CYVT\right|}{\left|ABCD\right|}+\frac{\left|DZUT\right|}{\left|ABCD\right|}<1$;

in other words,

$xyz+uv\left(1-x\right)+\left(1-y\right)\left(1-v\right)t+\left(1-z\right)\left(1-u\right)\left(1-t\right)<1$.

This completes the proof.

  darij

:o  :o  :w00t: it's a very good solution, thank very much darij grinberg

Very cool solution, Darij :thumbup:
Here is another approach:
We view the initial inequality as a linear polynomial f(x), 0<x<1. Thus f(x) attains the max/min when x->0 or x->1.
Putting x,y,z,u,v,t=1,0, we can check the max/min value, though one has to calculate f(x,y,z,u,v,t) for 32 (x,y,z,u,v,t)... :blush:


Prove that
$F_{2n}= F_{n}L_{n}$

where $F_{n}$ is a Fibonacci number of term $n$ and $L_{n}$ is a Lucas Number of term $n$ defined by $L_{n}= F_{n+1}+F_{n-1}, L_{1}= 1$

Backstory: From a book I'm learning number theory from. I got stuck on this problem.

You'll need to use strong induction.

http://en.wikipedia.org/wiki/Mathematical_induction#Complete_induction

Edit:  Actually, there's a better way.  Specifically, a generalization, which can be proven by regular induction.

[hide="Hint"] $F_{n+1}= F_{n}+F_{n-1}$
$F_{n+2}= F_{n+1}+F_{n}= 2F_{n}+F_{n-1}$
$F_{n+3}= F_{n+2}+F_{n+1}= 3F_{n}+2F_{n-1}$
$F_{n+4}= F_{n+3}+F_{n+2}= 5F_{n}+3F_{n-1}$ [/hide]

Oh, but Binet's is so... cheap.

[hide="By the way..."] That second method I outlined gives a suggestion for another method, which is to use regular induction, but on two expressions; the other is an expression for $F_{2n+1}$.  Can you conjecture it?

[hide="It is..."] $F_{2n+1}= F_{n}^{2}+F_{n+1}^{2}$ [/hide] [/hide]


limit
trigonometry
calculus
calculus computations

can we calculate $\lim_{n \to \infty}\cos{\frac{\pi}{4} \cos{\frac{\pi}{5} \cdots \cos{\frac{\pi}{n}}}}$

Could you clarify what the terms in the middle are?
Also, it's \infty you want in the limit.

thank you for reminding.
is it clean now? :)


calculus
integration
induction
modular arithmetic
number theory
number theory unsolved

Find all natural numbers a, b such that $ a^{n}\plus{} b^{n} \equal{} c^{n\plus{}1}$ where c and n are naturals.

I don't understand. So $ c$ and $ n$ are given, and we find $ a$ and $ b$ in terms of $ c$ and $ n$? That's weird. :maybe:

No we shouldn't find a and b in terms of c and n. (i.e. a=2 b=2 is a solution and we should find the solutions like this)

I think we should consider the cases.
1- a=b   2- a>b.
I found that in the first case only a=b=2 is a solution and in the second case i got a contradiction but I am not clear.

[quote="ridgers"]Find all natural numbers a, b such that $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$ where c and n are naturals.[/quote]

I dont know how is it possible to find $ a$ and $ b$ if $ c$ and $ n$ are fixed given values.

But here is a classical family of solutions : $ (u(u^n\plus{}v^n))^n\plus{}(v(u^n\plus{}v^n))^n\equal{}(u^n\plus{}v^n)^{n\plus{}1}$

[quote="pco"][quote="ridgers"]Find all natural numbers a, b such that $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$ where c and n are naturals.[/quote]

I dont know how is it possible to find $ a$ and $ b$ if $ c$ and $ n$ are fixed given values.

But here is a classical family of solutions : $ (u(u^n \plus{} v^n))^n \plus{} (v(u^n \plus{} v^n))^n \equal{} (u^n \plus{} v^n)^{n \plus{} 1}$[/quote]

$ c$ and $ n$ are not fixed value.

If $ c$ and $ n$ are not fixed.The equation $ a^n\plus{}b^n\equal{}c^{n\plus{}1}$ has infinitely many solutions in $ N$.For example,
$ a_k\equal{}1\plus{}k^n,b_k\equal{}k(1\plus{}k^n),c_k\equal{}1\plus{}k^n$,where $ k \in N$ satisfies the hypothesis.

[quote="SUPERMAN2"]If $ c$ and $ n$ are not fixed.The equation $ a^n \plus{} b^n \equal{} c^{n \plus{} 1}$ has infinitely many solutions in $ N$.For example,
$ a_k \equal{} 1 \plus{} k^n,b_k \equal{} k(1 \plus{} k^n),c_k \equal{} 1 \plus{} k^n$,where $ k \in N$ satisfies the hypothesis.[/quote]

Hummm, this is exactly my solution $ (u(u^n\plus{}v^n))^n\plus{}(v(u^n\plus{}v^n))^n\equal{}(u^n\plus{}v^n)^{n\plus{}1}$ where $ u\equal{}1$ and $ v\equal{}k$

 :)

The similar problems of Dorin Andrica and I think they are all very nice:
The following equations have infintely many intergral solutions $ \forall n \in N$:
i)$ x^n\plus{}y^n\plus{}z^n\plus{}t^n\equal{}a^{n\minus{}1}$
ii)$ x^n\plus{}y^n\plus{}z^n\plus{}t^n\equal{}a^{n\plus{}1}$

[quote="ridgers"]Find all natural numbers a, b such that $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$ where c and n are naturals.[/quote]The exact version is :
Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying  $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$.
And the unique result is $ a\equal{}b\equal{}2$ and then $ c\equal{}2$.

[quote="Allnames"]Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying  $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$.[/quote]

A nice one.
Lemma 1. If $ p \in \mathbb{P} \setminus\{2\},p^a \parallel{}a\minus{}b$ and $ p^b\parallel{}n$ then $ p^{a\plus{}b}\parallel{}a^n\minus{}b^n$.
Lemma 2. If $ p\equal{}2, p^{a\plus{}1}\parallel{}a^2\minus{}b^2, p^b\parallel{}n$ then $ p^{a\plus{}b}\parallel{}a^n\minus{}b^n$. (Both done by induction on b).
Let p>2 a prime s.t. $ \upsilon_p(a\minus{}(\minus{}b))\equal{}x \in \mathbb{N}_0$ then for all odd n, by lemma 1: $ n\plus{}1 \mid \upsilon_p(c^{n\plus{}1})\equal{}\upsilon_p(a^n\plus{}b^n)\equal{}x\plus{}\upsilon_p(n)$, contradiction because lhs is greater for enough large n. So a+b must be a power of 2, in particular 2|a-b.
Now if $ (a\minus{}b)^2>0 \text{ and }\upsilon_2{a^2\minus{}b^2}\equal{}x\plus{}1$ then for all odd n, by lemma 2: $ n\plus{}1 \mid \upsilon_2{a^n\plus{}b^n}\equal{}x\plus{}\upsilon_2(n)$, again contradiction. So a=b.
So $ 2a^n\equal{}c^{n\plus{}1}$: if 2<p|a then $ n\plus{}1|\upsilon_p(2a^n)\equal{}n\upsilon_p(a) \implies n\plus{}1 \mid \upsilon_p(a)$ again contradiction. So $ a\equal{}2^k$ for suitable choice of $ k \in \mathbb{N}$ and $ 2^{kn\plus{}1}\equal{}c^{n\plus{}1}$, in particular $ n\plus{}1|kn\plus{}1\minus{}k(n\minus{}1) \implies n\plus{}1 \mid k\minus{}1$ for all n, so a=b=c=2.

[quote="Allnames"][quote="ridgers"]Find all natural numbers a, b such that $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$ where c and n are naturals.[/quote]The exact version is :
Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying  $ a^{n} \plus{} b^{n} \equal{} c^{n \plus{} 1}$.
And the unique result is $ a \equal{} b \equal{} 2$ and then $ c \equal{} 2$.[/quote]

Thank you very much . This is the correct version :D  :D  :D

Another approach.
Lemma1. Denote $ \{c_i\}_{i > 0}$ the sequence of integer such that $ a^i \plus{} b^i \equal{} c_i^{i \plus{} 1}$. Then $ c_{i \plus{} 1} \ge c_i$. So it is definitively costant because of is bounded by (a+b).
Lemma2. We have $ 2 \equiv a^{p \minus{} 1} \plus{} b^{p \minus{} 1} \equal{} c^{p} \equiv c \pmod p$ for all p enough large, so c=2 definitively.
Then for all enough large n we have $ (2^{n \plus{} 2})^2 \equal{} 2^{n \plus{} 1} \cdot 2^{n \plus{} 3} \implies$ $ (a^{n \plus{} 1} \plus{} b^{n \plus{} 1})^2 \equal{} (a^n \plus{} b^n)(a^{n \plus{} 2} \plus{} b^{n \plus{} 2}) \implies a \equal{} b \equal{} c \equal{} 2$.

Another approach.
$ c$ tends to $ \max\{a,b\}$, so for enough large n we'll have $ a^n\equal{}(b\minus{}1)b^n \implies a\equal{}b\equal{}c\equal{}2$.

[quote="bboypa"]

Another approach.
$ c$ tends to $ \max\{a,b\}$, so for enough large n we'll have $ a^n \equal{} (b \minus{} 1)b^n \implies a \equal{} b \equal{} c \equal{} 2$.[/quote]
It was also my approach. I remember that  this nice problem belongs to Lautentiu Panatopol.

Russia Mo: a, b are two given natural numbers such that the sum of their n-th powers for any n is a (n+1)-th power: [code]a^{n}+b^{n}=c^{n+1}[/code]
Solution: Let p>a+b be a prime. [code]a^{p-1}+b^{p-1}=c^{p}[/code] By Fermat's little theorem it follows that p must divide c-2. So c=pk+2. If k>0 then [code]a^{p-1}+b^{p-1}\ge (p+2)^p\ge (a+b+2)^p[/code] which is a contradiction. So c=2 and it can be easily shown that a=b=2.

I guess there are infinitely many solutions. Because if     n = 1 then search natural a and b such that a +b equals
a perfect square. there are infinitely cases that a+b is a perfect square. so find sqrt(a+b) and you will find c.


inequalities
calculus
derivative
inequalities proposed

Let $x,y,z$ be positive real numbers satisfying  $xy+yz+zx=2$ . Prove that

$\frac{1}{x^2-yz+3}+\frac{1}{y^2-zx+3}+\frac{1}{z^2-xy+3}\ge\ 1$

By convexity of x->1/(x+3) (its second derivative being positive), you can apply Jensen's inequality:

1/(x^2-yz+3) + 1/(y^2-zx+3) + 1/(z^2-xy+3) >=
3/(1/3*(x^2+y^2+z^2-xy-yz-xz)+3) =
9/(x^2+y^2+z^2+7) >= 1 which is equivalent to:
2 >= x^2 + y^2 + z^2 which is true by the rearrangement inequality.

The last inequality is not true

$x^2+y^2+z^2\ge xy+yz+zx=2$

Strange  :roll: . I tried many method but ...

The inequalities 18 and 15 are the same. :lol:

I reduced the inequality to this one: \[ \frac{1}{x(x+y+z)+1}+\frac{1}{y(x+y+z)+1}+\frac{1}{z(x+y+z)+1}\geq 1. \]

opening the reduced inequality (see the post of cezar lupu),
we have to prove that $2+(x+y+x)^2 \ge xyz(x+y+z)^3$.
homogenizing and opening again, we have to prove that
$\sum {2x^4y^2 + 3x^3y^3} \ge \sum {2x^4yz + 3x^3y^2z}$.

this follows from bunching.  :)

$\sum \frac{yz}{x(x+y+z)+1} \ge \frac{x^2}{x(x+y+z)+1}$ which follows from rearrangement ineq for three sequence.


A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:

$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\
\textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\
\textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\
\textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\
\textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

[hide]
Let the side length of the equilateral triangle be s, and the perpendiculars be a, b, and c. The area can be expressed in two ways: $ \frac{s^2\sqrt{3}}{4}$ and $ \frac{as}{2}\plus{}\frac{bs}{2}\plus{}\frac{cs}{2}\equal{}\frac{s(a\plus{}b\plus{}c)}{2}$. So, setting them equal to each other, we have $ \frac{s^2\sqrt3}{4}\equal{}\frac{s(a\plus{}b\plus{}c)}{2}$. Simplifying, we get $ a\plus{}b\plus{}c\equal{}\frac{s\sqrt3}{2}$. Note that the altitude of an equilateral triangle is $ \frac{s\sqrt3}{2}$. $ \boxed{C}$.
[/hide]

[hide]Say the side length is x. When the point is the center of gravity of the triangle, then using 30-60-90 triangles, the sum of the perpendiculars is x*sqrt3/2. This is also the sum when the point is one of the vertices of the triangle=the altitude. So, the answer is C.[/hide]

:rotfl:  :rotfl:  :rotfl:

[hide = Solution(trivialized by Vivani's)]

[b]Vivani's Theorem[/b] states that for any point $P$ in an equilateral triangle, the sum of the distances from $P$ to each side length is equal to the triangle's altitude.

The answer is just $\boxed{\textbf{(C)}}$.

[/hide]


advanced fields
advanced fields theorems

First, I don't have a formal question. Actually, I am killing formalism.

It seems the axiom restricts what a set is. It seems a system with the axiom of constructibility has "fewer" sets than system where it is not valid. However, it implies the [i]negation[/i] of the Suslin's hypothesis, which asserts the [i]non[/i] existence of some sets. Is there any explanation for this? (or any idea that makes this result less disturbing to a layman?)

P.S.: I don't know if this is the best place to post this and I am sorry if it isn't.

The axiom of constructibility says that V=L. In layman terms, L is the smallest possible class that contains all ordinals and satisfies the axioms of ZFC (it's quite easy to show this with some absoluteness arguments). Anyway, although every member x of L is a member of V, it doesn't necessarily follows that x satisfies the same properties in both L and V. In the case of Suslin trees, although it's true that Suslin trees exist in L, there's no way to guarantee that the same tree will remain Suslin in V.

Ah, now I understand... Thank you.

http://en.wikipedia.org/wiki/Suslin_hypothesis

In this link, the conditions 3 and 4 may become invalid if $ V \not\equal{} L$, isn't it? Allowing more objects to be considered sets could cause the existence of a subset of $ R$ which does not have an infimum... Am I correct?

Well, this is still disturbing. It is very weird to accept a subset could suddenly emerge...

Later I am going to try to explain better what I am thinking. I need to review these concepts...


Asha adds all the odd integers from 1 through 101, inclusive, and
then subtracts all the even integers in that same range from her
sum. What result does she obtain?

there are $ 51$ odd integers from $ 1$ to $ 101$. The sum of the first $ n$ odd numbers is $ n^2$, $ 51^2 \equal{} 2,601$.
then there are $ 50$ even integers from $ 1$ to $ 101$. the first one is $ 2$ and the last one is $ 100$. the sum of those integers is $ \frac{(2 \plus{} 100)(50)}{2}$ evaluating this gives $ 2,550$. so the answer is $ 2,601 \minus{} 2,550 \equal{} 51$

You add an odd and then subtract an even.
When you perform the operations, you get something like this:
(1-2)+(3-4)+(5-6)... (99-100)+101

Each parentheses = -1 and there are 100/2 = 50 pairs
so -50+101 = [b]51[/b]

You can also use the formula for the sum of the first n consecutive odd integers which is n(n) and subtract that by the sum of the first n consecutive even integers which is n(n+1). Since you know that there are 51 odd integers and 50 even integers, you get this.

1. 51(51)-50(51)=?
2. 51(51-50)=?
3. 51(1)-?
4. 51=?


Let $a$ be real numbers. Given a matrix $A=\left(\begin{array}{cc} a & 1-a \\ 1-a & a \end{array}\right).$ Let $\left(\begin{array}{c} x_0 \\ y_0 \end{array}\right)=\left(\begin{array}{c} 1 \\ 0 \end{array}\right),$ and the sequences $\{x_n\},\ \{y_n\}$ are defined by the following equation.

\[ \left(\begin{array}{c} x_n \\ y_n \end{array}\right)=A \left(\begin{array}{c} x_{n-1} \\ y_{n-1} \end{array}\right),\ \ n=1,2,\ \cdots \]

then find the necessray and sufficient condition for $a$ such that the sequence $\{x_n\}$ converges.

$\left(\begin{array}{c} x_n \\ y_n \end{array}\right)=\left(\begin{array}{cc} a & 1-a \\ 1-a & a \end{array}\right)\cdot\left(\begin{array}{c} x_{n-1} \\ y_{n-1} \end{array}\right)$

So, we have:

$x_n = ax_{n-1} +y_{n-1} - ay_{n-1}$  and  $y_n = ay_{n-1}+x_{n-1}-ax_{n-1}$

Note that $1-x_n = ay_{n-1}-y_{n-1}-ax_{n-1}+1$, and this quantity is equal to $y_n$ iff $1-y_{n-1} = x_{n-1}$ and because $x_0 = 1$ and $y_0=0$ we can prove inductively that $1-x_n=y_n \Rightarrow 1-x_{n-1} = y_{n-1}$.  Substituting this into the original expression for $x_n$ gives:

$x_n = 2ax_{n-1}-x_{n-1}-a+1$

Unfortuately, this is as far as I could get, my guess would be that $-1<|A|<1 \Rightarrow 0<a<1$.

You're very close. Note that the case $a=1$ gives you the identity matrix, and with the identity matrix, the sequence, being constant, does converge.

A somewhat more direct method of addressing the stability of this process would be to diagonalize the matrix. This can be done fairly cleanly.

\[ A^n=\frac{1}{2}\left(\begin{array}{cc} 1+(2a-1)^n & 1-(2a-1)^n\\ 1-(2a-1)^n & 1+(2a-1)^n\\ \end{array}\right)  \]
\[ \; \; \; \; \Longrightarrow \; \; \; \; \; \left(\begin{array}{c} x_n\\ y_n \end{array}\right)= \frac{1}{2}\left(\begin{array}{c} 1+(2a-1)^n \\ 1-(2a-1)^n \end{array}\right)\; .  \]
Therefore $\lim\limits_{n\to \infty}x_n$ exists iff  $a\in (0,1]\; .$


Morgenvroeg neemt het Belgische team het in Zweden op tegen andere landenploegen in de Baltic Way competitie 2005 - tegenstanders zoals Polen, Sint-Petersburg (niet echt een land maar ja), Wit-Rusland ook geloof ik, enzovoort. Benieuwd hoe ze het er van af gaan brengen - sommige tegenstanders zijn vrij sterk, maar wij hebben ook zeker geen slecht team. En ook benieuwd wat de vragen gaan zijn, want de 6 vraagvoorstellen die België heeft ingediend komen allemaal van mij :) dus ik hoop dat er een of meerdere vragen van mij werden gekozen.

Succes aan het Belgische team, ook al kunnen ze dit nu niet lezen!

To be continued.

PS. De teamleden:

Guolong Li
François Gonze
Clara Mertens
Wim Vanrie
Augustin Lu An Khoa

Teamleider is Bart Windels (zoals gewoonlijk :D) en deputy is Ria Van Huffel.

Laat ons hopen dat ze't er goed van af brengen! :) met guolong in het team (en het is een team competitie) zou dat zeker moeten lukken. :)

[ en aangezien we special guest zijn, denk ik dat je ongeveer 100% kans hebt dat ze een of meer belgische vragen nemen :P ]

[quote="Peter VDD"]met guolong in het team (en het is een team competitie) zou dat zeker moeten lukken. :)[/quote]

Niet alleen Guolong is goed, ook François begint echt sterk te worden. De rest van het team is minder ervaren, maar blijkbaar is Augustin (die ik dus niet ken) zeker niet slecht en ik weet dat Clara en Wim hard werken (nu ja, ik ben dan ook hun "trainer").

Zo Arne, iedereen kijkt naar jou voor de eerste uitslagen. :)

Competitie is momenteel afgelopen. Ik duim voor de resultaten van het hele team, en hoop de ze niet teveel communicatieproblemen hebben ondervonden  ;) . Website is te vinden op [url=http://www.math.su.se/bw2005/]www.math.su.se/bw2005/[/url] overigens.

[quote="Arne"]De website ziet me er degelijk uit maar spijtig genoeg staat er nog niks "nieuws" op :([/quote]officiele websites lopen altijd een week achter :D

:? Ik voel me nu wel erg oud, ik ben totaal niet meer mee, in mijn tijd wist ik enkel van de IMO

Een Baltic competition in Zweden?  Wow, waarschijnlijk begonnen in de drie Baltische staten?
Hoeveel landen doen er zo aan mee?  Hoelang doet Belgie al mee?  Is het zoals bij fysica dat in de oneven jaar de Vlamingen eentje meer mogen sturen?  Is het voor zesdejaars alleen?

Euhm, elk jaar wordt er op de Baltic Way competitie (met deelnemers Sint Petersburg, Polen, Duitsland, Estland, Finland, Denemarken, Zweden, Noorwegen, Ijsland, ... nog landen, 10 à 12 in totaal denk ik) ook een gastland uitgenodigd (zoals in het verleden Israel, Wit-Rusland, enzovoorts). Dit jaar werd België uitgenodigd, dus dit is de eerste en enige deelname van België aan de Baltic Way competitie :)

En nee, er werden gewoon drie Vlamingen en twee Walen geselecteerd - weet niet juist waarom - en 2 van de 5 deelnemers zijn vijfdejaars.

't Is een teamcompetitie - de 5 deelnemers werken 4.5 uur lang in teamverband aan een set van 20 vragen. :)

De resultaten staan online!

Poland	95
Finland	75
St Petersburg	73
Germany	62
Latvia	60
Lithuania	53
Norway	48
Sweden	47
Belgium	39
Estonia	37
Iceland	30
Denmark	24
[url=http://www.math.su.se/bw2005/results.pdf](meer)[/url]

België is dus negende, misschien wat minder dan verhoopt, maar toch proficiat aan het Belgische team voor hun inzet!

[url=http://www.math.su.se/bw2005/bw2005_english.pdf]Vragen[/url] 1 en 16 werden overigens voorgesteld door Arne  :)

[quote="Stijn"]België is dus negende, misschien wat minder dan verhoopt, maar toch proficiat aan het Belgische team voor hun inzet![/quote]

Hm, laten we hopen dat de teamleden hier veel van opsteken voor de toekomst :)

[quote="Stijn"][url=http://www.math.su.se/bw2005/bw2005_english.pdf]Vragen[/url] 1 en 16 werden overigens voorgesteld door Arne  :)[/quote]

Waarom ze juist deze twee vragen hebben gekozen is me lichtjes een raadsel want ik had er minstens 2 die een stuk beter waren. Maar ja, tant pis :)

[quote="Arne"]Waarom ze juist deze twee vragen hebben gekozen is me lichtjes een raadsel want ik had er minstens 2 die een stuk beter waren. Maar ja, tant pis :)[/quote]Wees blij, nu heb die goeie nog over. ;) enfin ja, zij zullen die beter gevonden hebben zeker? [of mss kwamen die uit een ander gebied waarvan ze er al 5 hadden]
Proficiat met de 2 vragen anyway. :P

I've solved the first one, but I didn't write everything correctly, so we just get one point on it (when there are more thant 2008 terms, it's decreasing, so it's easy to prove the rest, but I didn't put everything in my solution :blush: )

(in fact, I've made a few little stupid thinks like this, and we could have had about 50 points without them :(  :(  :( )

Dommage. :)

I didn't consider the first problem as "difficult", not at all. But it is true that you need to be VERY careful when writing a solution for this problem. There are quite a few subtle points in my solution (which is exactly the "official" solution in the pdf document with the solutions, they didn't change anything to the solution).


Ok, I have to admit. I have been always a Nokia fan, ever since the 5110 model (the older folks remember the "brick" :) ), but recently I've switched from a 2-year old Nokia N95 (great phone!) to an iPhone. I did it for a couple of reasons: one, it's the only device that lets me [b]easily[/b] keep my calendar and contacts in permanent sync (via Apple's Push service, used with either Google or MobileMe -- I prefer Google as it's free) with my home and work computers. All other solutions I have tried (less WinMobile phones -- I just hate the hardware on those) haven't succeeded in providing that. The other main reason was the screen: it's big, and easy to handle. I had to give up a lot in the multimedia capabilities (read crappy camera) but I have come to the conclusion that I don't really use that much the camera in my phone (maybe 2-3% of my daily usage involved the camera?!) so if there's something I have to give up it's that.

Anyway, I've also grown to like some of the apps the iPhone comes with. Thus this topic. If you do have an iPhone (and I'm sure a fair number of you guys do, since we have thousands of pages viewed with iPhones every day on the site :) ) and (again this is pure fiction at this point) we would come up with an iPhone app, what would you imagine it would do?

Maybe allow users to play For the Win on the go? or something similar?

JRav's actually got a very good idea there. Have a keypad with buttons for pi, sqrt, etc. along with the standard numbers. The only issue would be the frequent typoes.

Similar to JRav's idea, perhaps have the application allow ease of forum access as well as Alcumus lessons?

P.S. Valentin, are you planning to jailbreak your iPhone?

[quote="tcjy"]There's a third version in june?![/quote]
Yes, there is. If you look at [url]http://www.apple.com[/url], you'll see the news.
Personally, I have a 3G and I plan to upgrade it when the new version comes out.

No, there's a new firmware coming up in June that's advertised on the Apple Website right now. Brings on a few minor improvements for all current phones. However, there's increase rumors and speculation that a new phone (hardware) is coming up soon ...

Anyway, until now we got

FTW (don't think it's possible due to the architecture of our server -- we need a flash client, and flash is not supported on the iPhone)
Alcumus
Ease of forum access (what exactly do you have in mind here)

Maybe just availability to your archives (amc, usamts, aime, usamo, imo, arml?)

Have you tried the old Palms? (Centro and older Treo) The ones that run on Palm OS sync perfectly with outlook AND palm desktop. In fact, my Palm IIIxe works perfectly after six years! (Not typical, but hey, I left it in my dad's car for most of the time.) Anyways, it's a pity that Palm will no longer make its own OS because that IS your best bet to store and synchronize data. Anybody interested in reviving Palm OS?

You're confusing syncing with a [b]local[/b] program (such as Outlook) a feature which [b]ALL[/b] my previous phones have had, and syncing OTA with PUSH technology, between 4 different devices (laptop, home desktop, work desktop, iPhone.)

[quote="Valentin Vornicu"][quote="n0vad3m0n"]P.S. Valentin, are you planning to jailbreak your iPhone?[/quote]I did that within the first hour of ownership. Can't stand not being able to put applications in background :)[/quote]

:D I wrote some of the software that automated the jailbreak process, and made quite a bit of money too.

I like the idea of accessing the classroom, because on a couple of occasions I have been unable to participate in the class because I'm on the go and I have to simply refresh the live transcript page continuously to follow along.

Though classes wouldn't be too effective unless you carry a stack of paper around (and a printer, for Olympiad Geo ;)).

[quote="Valentin Vornicu"]Classroom/game won't happen right now because we need flash to connect to our servers.[/quote]

A simpler task would be a nice interface to the forum (I use my iPhone to browse the forums a surprising amount).
I'm talking about a web interface, not necessarily an app on the App Store.

For reference, there are a lot of examples of iPhone-optimized sites here: http://www.apple.com/webapps/

Or at the very least, like Ihatepie mentioned, a simple interface for the resources section would be great.

err, although this may be considered reviving the topic by some people, but:

I heard that the new firmware is being postponed until september by apple since they don't have enough money to release/finish it.

I can't get an iPhone even if I wanted to. They don't sell it here because it doesn't work here. It is a "dead zone".

THANK YOU FOR BUMPING THIS i find this thread so FUNNY quite an antique isnt it


Find the value of the number C:

1/2(sin^2(x) + C = -1/4(cos2x)

In the textbook, x = theta.  Now, can you show how to solve for C?

things obviously must cancel out, just try and use trig identities for this problem

err where does that other parenthesis end?

is it 1/2(sin^2(x) + C) or 1/2(sin^2(x)) + C

Thank you all for your tips and efforts.


hello, by simplifying the left hand side and subtracting one we get:
$ \minus{}3\sqrt{3}x^2\minus{}x\sqrt{1\minus{}x^2}\plus{}6\sqrt{3}\minus{}10\sqrt{1\minus{}x^2}>0$.
The inequality above ist equivalent with
$ \left(7x^2\plus{}12x\plus{}8\right)\left(2x\minus{}1\right)^2>0$.
This is fulfilled for all $ x$ with $ \frac{1}{2}<x<1$.
Sonnhard.


Gauss
linear algebra
matrix
algorithm
absolute value
linear algebra theorems

Hey guys again, I'm coming back cause I really loved the way you answered me last time.

Again, first of all, thank you very much for any kind of answer you can give me.

I've searched the internet and wikipedia for the jacobi method cause I didn't fully get it when I was in class, the teacher wasn't there and there was a substitute that actually was a student... so....

My question is simple, I want to use jacobi method just for equations of 2 or 3 variables. No more, I know that I have to order the equations according to the Diagonals, and I know how to do it. I know that afterwards I have to leave the x alone in equation 1, and leave the y alone in equation 2. (Dunno what's the math term in english to "leave alone", sorry). Afterwards, I have to assign the 0 value to something, but I don't know to what, actually, I'm pretty lost there, and wikipedia's explanation is not enough.

After I post this, I will continue my research and try to find it myself using google or so. However, if you guys could give me a hand I would really appreciate it.

Thank you very much again! I hope one day I get much better at this so I can be the one that helps.

I've got to look a couple of things up in Burden & Faires (a numerical analysis textbook) to get in the right mood to answer this. Not today - maybe this weekend.

No problem!

I had to go to sleep so I couldn't get more research done, now I'm going to college... If I get my question answered there I will tell you guys.

Thanks btw.

Here's an example worked through.

We'd like to solve $ \begin{bmatrix}2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2\end{bmatrix}\begin{bmatrix}x \\
y \\
z\end{bmatrix} = \begin{bmatrix}1 \\
2 \\
3\end{bmatrix}.$

We can solve each equation for the variable on the main diagonal and wind up writing
\begin{align*}x & = \frac12(1 - y - z) \\
y & = \frac12(2 - x - z) \\
z & = \frac12(3 - x - y)\end{align*}
Now the idea is to turn this into an iteration. We suppose we already have $ (x_k,y_k,z_k)$ and we wish to compute $ (x_{k + 1},y_{k + 1},z_{k + 1}).$

For Jacobi iteration,we work all the way through this before we update any of the variables. We have:
\begin{align*}x_{k + 1} & = \frac12(1 - y_k - z_k) \\
y_{k + 1} & = \frac12(2 - x_k - z_k) \\
z_{k + 1} & = \frac12(3 - x_k - y_k)\end{align*}
The difference with Gauss-Seidel iteration is that as soon as any variable is updated, we use the new value rather than the old value. That is:
\begin{align*}x_{k + 1} & = \frac12(1 - y_k - z_k) \\
y_{k + 1} & = \frac12(2 - x_{k + 1} - z_k) \\
z_{k + 1} & = \frac12(3 - x_{k + 1} - y_{k + 1})\end{align*}
Let's run this, using $ (0,0,0)$ as a starting point.

Jacobi iteration does this:
$ (0,0,0)$
$ (.5,1,1.5)$
$ ( - .75,0,.75)$
$ (.125,1,1.875)$
$ ( - .9375,0,.9375)$
$ (.03125,1,1.96875)$
and so on. It doesn't converge. After some time, it becomes clear that we are approaching a limit cycle of alternating back and forth between $ ( - 1,0,1)$ and $ (0,1,2),$ neither of which is a solution to the original problem. The method did not work (we'll explore why later).

Here's the Gauss-Seidel iteration. I include these details mostly so you can set up your own version and confirm the arithmetic. Try to follow along.
$ (0,0,0)$
$ (.5,.75,.875)$
$ ( - .3125,.7188,1.2969)$
$ ( - .5078,,6055,1.4512)$
$ ( - .5283,.5386,1.4949)$
$ ( - .5167,.5109,1.5029)$

After 20 or 30 iterations of that (depending on the tolerance we set), we reach an acceptable approximation of $ ( - .5,.5,1.5),$ which is the solution to the problem.

[color=green][Continued in next post][/color]

Let's phrase this in matrix terms. We wish to solve $ Ax\equal{}b,$ and we write $ A\equal{}L\plus{}D\plus{}U$ where $ L$ is strictly lower triangular, $ D$ is diagonal, and $ U$ is upper triangular. In general, we need for $ D$ to be large in some sense compared to the other parts. In our particular problem above,
$ L\equal{}\begin{bmatrix}0&0&0\\1&0&0\\1&1&0\end{bmatrix},$ $ D\equal{}\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix},$ and $ U\equal{}\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}.$

Jacobi iteration is given by $ x^{(k\plus{}1)}\equal{}D^{\minus{}1}(b\minus{}(L\plus{}U)x^{(k)}).$

Gauss-Seidel iteration is given by $ x^{(k\plus{}1)}\equal{}(L\plus{}D)^{\minus{}1}(b\minus{}Ux^{(k)}).$

In each case, if we continue, a matrix eventually gets raised to a high power. We need for that power of a matrix to tend to zero, which in turn requires that the spectral radius be strictly smaller than one.

For Jacobi, that repeated matrix is $ \minus{}D^{\minus{}1}(L\plus{}U).$

For Gauss-Seidel, that repeated matrix is $ \minus{}(L\plus{}D)^{\minus{}1}U.$

In the particular problem I worked above, for Jacobi iteration, we see

$ \minus{}D^{\minus{}1}(L\plus{}U)\equal{}\minus{}\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}^{\minus{}1}\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}\equal{}\minus{}\frac12\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}.$

This last matrix has eigenvalues of $ \frac12,\frac12,$ and $ \minus{}1.$ That means the spectral radius is $ 1$ and the powers of the matrix don't tend to zero, which is why the Jacobi iteration fails to converge.

For Gauss-Seidel, we have:

$ \minus{}(L\plus{}D)^{\minus{}1}U\equal{}\minus{}\begin{bmatrix}2&0&0\\1&2&0\\1&1&2\end{bmatrix}^{\minus{}1}\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}\equal{}\frac18\begin{bmatrix}0&\minus{}4&\minus{}4\\0&2&\minus{}2\\0&1&3\end{bmatrix}$

The eigenvalues of this matrix are $ 0$ and $ \frac{5\pm i\sqrt{7}}{16},$ which gives a spectral radius of $ \frac{\sqrt{2}}4.$ Since this is comfortably smaller than $ 1,$ the Gauss-Seidel iteration converges in a reasonable amount of time.

A sufficient condition for the convergence of these iterations is that $ A$ be strictly diagonally dominant - that is, that each diagonal entry be strictly greater (in absolute value) than the sum of the absolute values of the other entries. You can see that we don't have that - at least not the strict inequality. And we paid the price for that in the failure of the Jacobi iteration to converge. The Gauss-Seidel iteration is more robust; it's easier for it to converge.

Sorry for answering so late. Seriously

I am really grateful, I *really* got it here, the difference between seidel and jacobi is perfectly explained. I have to do it with just 2 equations so it's way simpler than that and it takes 10 iterations or so long.

Thank you very much, sorry again for getting back to you so late, it's just I had a really rough time and I couldn't even go to college nor worry about studying.

Sorry and thank you again.

By the way, I rated you five stars cause you did a great work :) .But I stared for an hour before I realised you made a mistake when making the equations (we are humans, we all make mistakes, but I must point this so anyone that reads it can understand), you put three times 1/2 when it should've been 1, 1/2 and 1/3.
I must say, THANK YOU FOR THAT, cause I now know it better cause I didn't only have to learn it, but also I had to figure out what you did wrong and why. That gave me a priceless-understanding of these methods.

Thanks again

You're welcome. I think I've fixed the typo; not $ 1,\frac12,\frac13$ but rather $ 1,2,3$ on the right hand side rather than $ 1,1,1.$ The numerical stuff is probably OK.

This is all very intriguing to me, as I just learned some linear algebra. The first thing we learned in class was a basic algorithm (Gauss-Jordan elimination) to solve systems of linear equations. In what instances would the Jacobi iterations be useful (or have I completely not understood what they are?)? Thanks so much!

Gauss-Seidel Method is used in solving systems of algebraic linear equations..the process is in an iterative way...I first encountered this method in our Numerical Analysis for engineers course way back in college...the advantage of this method is you can easily create a simple program (JAVA, Matlab, etc.) of this...I even got a perfect score for creating such program in JAVA...


conics
projective geometry
geometry proposed
geometry

[color=blue]MacLaurin Theorem. [/color]If a quadrilateral is inscribed in a conic, the sections of two pairs of opposite sides are collinear with the section of the tangents at opposite vertices.

Is it a corollary of Pascal's theorem

José Carlos

This is a well known ( at least to me ) theorem, in the configuration of a cyclic quadrilateral.

     So, because of a conic may be considered as an appropriate projection of a circle, the proof of this theorem, is a consequence of the proof in the case of a cyclic quadrilateral.

     Kostas Vittas.


1. Given the equation 3a^4 - 2a^3 +7a^2 +11a - 9 = 0.

     E=the sum of the roots
     W=the sum of the reciprocals
     Y=the sum of the squares of the roots

  What is E + W + Y?

2. Find the polynomial whose roots are three more than those of the following polynomial:

     x^3 - 6x^2 +3x - 10

3. If 14/(x^2 + x -12) = A/(x-3) + B/(x+4), find (A+B)/2.

Thanks


algebra
polynomial
function
analytic geometry
inequalities

It depends on the actual functions; I know of no set rule.

For example, first degree functions on  xy coordinate system can at max intersect once, or they may not intersect at all.

Two second degree functions can at max intersect twice, or not at all.

To solve for the polynomials, set the two functiosns equal to each other and find out the solutions that satisfy.

For example, let's say our two functions are $y=x^2$ and $y = -x^2 +1$

We set $x^2 = -x^2 +1$

Then we get all the $x^2$ on one side

$2x^2 = 1$

$x^2 = .5$

In this case, x has two solutions, so they meet at two places.

You can do the same thing with $y=x^2$ and $y=-x^2$

Solving for this only yields one answer.

Therefore, it depends on the actual equations given.

[quote="riposte"]How do you solve a problem such as:
At how many points do the curves

     y = a specific polynomial and
     y = another polynomial

intersect?[/quote]Basically you're asking how to find the number of intersections of the graphs of $y_1=P(x)$ and $y_2=Q(x)$, where $P$ and $Q$ are polynomials in $x$. When they intersect you have $y_1=y_2$, or $P(x)-Q(x)=0$. So the number of intersection points is the number of zeros of the polynomial $p(x)=P(x)-Q(x)$. Really your question is asking for a general method of solving $p(x)=0$ for an arbitrary polynomial $p$, which you're probably not going to find (i.e. there isn't one).

[quote="riposte"]How do you solve a problem such as:
At how many points do the curves

     y = a specific polynomial and
     y = another polynomial

intersect?[/quote]

use a ti-83   :lol:

[quote="riposte"]How do you solve a problem such as:
At how many points do the curves

     y = a specific polynomial and
     y = another polynomial

intersect?[/quote]

It really depend on the actual polynomials you have, at least if you are talking on the real axis. If you are talking in the Complex Plane, it is a different story, but I don't think it is relevant right now.

In general, if you have $y=p(x)$ and $y=q(x)$ as the two curves, the most you can say in general is that $\#I\leq\deg (p-q)$ where $I$ is the set of intersections. In the complex plane, you have equality, if you count multiplicity.

Best,

[quote="ankur87"]djimenez

Can you explain the $\#I\leq\deg p+\deg q$ formula

I don't see how if I have two 2nd degree fumctions, I can have 4 intersection points.

$\#I\leq 2+ 2$

$\#I\leq 4$

?
Or am I interpreting the formula incorrectly?[/quote]

No, you are understanding the forumula well, I am the stupid who wrote it wrong (already edited). What I tryed to mean is that $\#I\leq\deg (p-q)$. I appologize. I better go to sleep before I write more stupidicities. Eitherway, the the original inequality is correct, but it is wrong that on the complex plane there is equality  :D .

Best,

[quote="ankur87"]Sorry to bother you again...but I still don't understand the formula...

$\#I\leq\deg (p-q)$

Let's again say that I have two 2nd degree functions

Does that mean that $\#I\leq (2-2)$

$\#I\leq\deg (0)$

Number of intersections is less than or equal to 0?
That means that there is no intersections in 2nd degree functions?[/quote]

You are understanding $\#I\leq\deg p -\deg q$, what, of course, it is in general false, specially because the number can become negative. Note that what it says is $\#I\leq\deg (p-q)$, then, the story is other. I mean, you have to substract the two terms, and after that, consider the degree. That is, if you have $y=2x^2-1=p(x)$ and $y=x^2+1=q(x)$, then this two curves have at most two (and in this case exactly two, even in the reals) intersection points, because $p(x)=q(x)=x^2-2$. It is easy to compute the intersections, that are $(\pm\sqrt2,3)$.

By the way, in most of the books, $\deg(0)=\infty$ and not zero. This is by definition, and basically to make sence of some results, like $\deg(pq)=\deg p+\deg q$ and so many others.

Best,

[quote="ankur87"]Sorry to bother you again...but I still don't understand the formula...

$\#I\leq\deg (p-q)$

Let's again say that I have two 2nd degree functions

Does that mean that $\#I\leq (2-2)$

$\#I\leq\deg (0)$

Number of intersections is less than or equal to 0?
That means that there is no intersections in 2nd degree functions?[/quote]$\deg(p-q)\ne\deg{p}-\deg{q}$

You have $\deg(p-q)=\deg(2-2)$, but $2$ and $2$ are the degrees of $p$ and $q$, not $p$ and $q$ themselves.

Oh, ok

I was getting simple concepts mixed up.

I get it now, thanks.


Δινονται οι θετικοι αριθμοι $ x,y,z$ ετσι ωστε:
$ 21xy\plus{}2yz\plus{}8zx \leq 12$
Να προσδιορισετε την ελαχιστη τιμη της παραστασης:
$ P\equal{}\frac{1}{x}\plus{}\frac{2}{y}\plus{}\frac{3}{z}$

Έπεσε χθες το μάτι μου σε άσκηση χωρίς απάντηση και είπα να τη δοκιμάσω...

$ a \equal{} 21xy, b \equal{} 2yz, c \equal{} 8zx$
Τότε:

$ x \equal{} \frac {84ac} {b}, y \equal{} \frac {21ab} {4c}, z \equal{} \frac {16bc} {21a}$

Επίσης έχουμε:

$ 12 \geq a \plus{} b \plus{} c \geq 3\sqrt [3] {abc} \rightarrow \frac {1} {\sqrt [3] {abc}} \geq \frac {1} {4}$

Τότε η ζητούμενη παράσταση παίρνει τη μορφή:

$ \frac {1} {x} \plus{} \frac {2} {y} \plus{} \frac {3} {z} \equal{} \frac {b} {84ac} \plus{} \frac {8c} {21ab} \plus{} \frac {63a} {16bc}$

$ \equal{} \frac {16 \cdot 21ab^3c \plus{} 8 \cdot 16 \cdot 84abc^3 \plus{} 63 \cdot 21 \cdot 84a^3bc} {84 \cdot 21 \cdot 16(abc)^2}$

$ \geq \frac {3 \sqrt [3] {16 \cdot21 \cdot8 \cdot 16 \cdot84 \cdot63 \cdot21 \cdot84(abc)^2} } {84 \cdot21 \cdot16abc}$

$ \equal{} \frac {3\cdot16\cdot21\cdot2\sqrt [3]{63}}{84\cdot21\cdot16\sqrt [3]{abc}}$

$ \geq \frac {\sqrt [3] {63}} {56}$

Το οποίο είναι και η ελάχιστη τιμή αν δεν έχω κάνει λαθος πράξεις... :P

Δεν είμαι σίγουρος, αλλά νομίζω ότι αυτό που βρίσκεις είναι ένα κάτω φράγμα αλλά μάλλον δεν είναι το καλύτερο δυνατό. Και το λέω αυτό γιατί πρέπει να ελέγξεις αν ισχύει η ισότητα.
Στην αρχή η πρώτη ανισότητα ισχύει ως ισότητα όταν $ a\equal{}b\equal{}c$  ενώ μετά με την ΑΜ-GM η ισότητα δεν ισχύει στο $ a\equal{}b\equal{}c$!

Συγχωρέστε με αν κάνω λάθος :oops:

[quote="bilot4"]Δεν είμαι σίγουρος, αλλά νομίζω ότι αυτό που βρίσκεις είναι ένα κάτω φράγμα αλλά μάλλον δεν είναι το καλύτερο δυνατό. Και το λέω αυτό γιατί πρέπει να ελέγξεις αν ισχύει η ισότητα.
Στην αρχή η πρώτη ανισότητα ισχύει ως ισότητα όταν $ a \equal{} b \equal{} c$  ενώ μετά με την ΑΜ-GM η ισότητα δεν ισχύει στο $ a \equal{} b \equal{} c$!

Συγχωρέστε με αν κάνω λάθος :oops:[/quote]

Όντως.. Κεκτημένη ταχύτητα... Όταν έχεις συνηθίσει η AM-GM να είναι με abc ξεχνάς τις σταθερές με τις οποίες πολλαπλασιάζονται...  :oops:

Πρόκειται για δύσκολο και ωραίο θέμα . Η λύση μου στο λινκ
http://www.mathlinks.ro/viewtopic.php?search_id=822077093&t=80026


function
complex analysis
algebra
polynomial
complex analysis unsolved

Here is one I cannot tackle at all:

Prove there is no one-to-one analytic map from $\mathbb{D}-\{0\}$ to $\mathbb{C}-\{0\}$, where $\mathbb{D}$ is the open unit disk.

Assume there is such a function, and let $f$ be its inverse. $f$ would then be a one-to-one holomorphic function from $\mathbb C\setminus\{0\}$ to $\mathbb D\setminus\{0\}$. There are now three cases:

[b]1)[/b] $0$ is an essential singularity of $f$.

By the [url=http://planetmath.org/encyclopedia/WeierstrassCasoratiTheorem.html]Weierstrass-Casorati Theorem[/url], the image through $f$ of any punctured neighbourhood of $0$ is dense in the complex plane, so $f(\mathbb C\setminus\{0\})$ cannot be a subset of $\mathbb D$.

[b]2)[/b] $0$ is a removable singularity of $f$.

We can regard $f$ as an entire function, but then it would be a bounded entire function, meaning that it must be constant, and we have a contradiction.

[b]3)[/b] $0$ is a pole of $f$ of order $n\ge 1$.

In this case, $f$ has the form $f(z)=\frac{g(z)}{z^n}$, where $g$ is an entire function which does not attain the value $0$. On the one hand, since $\frac{g(z)}{z^n}$ is bounded on the region $|z|>R$ for some $R>0$, $g$ must be a polynomial of degree $\le n$. On the other hand, however, $g$ does not attain the value $0$, and this contradicts the Fundamental Theorem of [b]Algebra[/b].

Edit:

Yeah, I had written "Fundamental Theorem of Arithmetic" :D.

Second edit:

Note that the problem as Julien stated it was not exactly correct. There are injective holomorphic functions from $\mathbb D\setminus\{0\}$ into $\mathbb C^*$ (see the exponential map $z\mapsto e^z$, for example), but there are no such functions which are also [i]onto[/i]. I implicitly used the fact that the function whose existence we assumed was onto, since I took its inverse to be defined on the whole punctured plane $\mathbb C^*$.

Of course there's a generalization.

Suppose $0< a<b$ and  $0< c<d$

Let $A_1=\{z\in\mathbb{C}: a<|z|<b\}$ and $A_2=\{z\in\mathbb{C}: c<|z|<d\}.$

Suppose there exists $f: A_1\mapsto A_2$ that is analytic, one-to-one, and onto.

Prove that $\frac ba=\frac dc.$


now by definition te change in internal energy as a system moves from one state to another is given by

$\Delta U$ $=$ $\Delta K_{cm}+\Delta I$                                  $(1)$

where
$\Delta U$ is the change in internal energy
$\Delta K_{cm}$ is the change in kinetic energy of the system as  seen from the c.o.m frame of reference and $\Delta I$ is the change in potential; energy of the system due to various conservative forces acting on the system .
Now by $1^{st}$ law of thermodynamics we conclude that
$\Delta U$ $=$ $W+Q$

Now [i]my question [/i] is
how to define $W$ and $Q$ in terms of variables in $(1)$


algebra
polynomial
vector
number theory
relatively prime
superior algebra
superior algebra unsolved

Let tau be in L(V) where V is a finite dimensional vector space.  If
p(x) in F[x] is irreducible and p(tau) is not one to one, prove that
P(x) divides the minimal polynomial of tau.

$ p(\tau)$ isn't one to one then $ \exists\lambda\in{spectrum(\tau)}$ s.t. $ p(\lambda)\equal{}0$. Thus $ p,m$ (where $ m$ is the minimal polynomial of $ \tau$) have a common divisor. $ p$ is irreducible, then $ p$ is a divisor of $ m$.

Without using eigenvalues-
Let $ q$ be the minimal polynomial of $ \tau$. Suppose that $ q$ and $ p$ are relatively prime. There are polynomials $ r,s$ in $ F[x]$ such that $ p(x)r(x)\plus{}q(x)s(x)\equal{}1$. Applied to $ \tau$, $ p(\tau)r(\tau)\equal{}I$, since $ q(\tau)\equal{}0$. Thus $ p(\tau)$ is invertible.
If $ p$ is also irreducible, it is relatively prime to $ q$ if and only if it doesn't divide $ q$.

This is the contrapositive, and the desired statement follows.

To run the argument with eigenvalues, you need to work with some algebraic extension of $ F$ in general.


limit calculus derivative ratio function parameterization quadratics

Find $ \lim_{r\to0}\frac{e^{rx}\minus{}1}{r}.$

The answer given by a plurality of the students in the class was not the correct answer.

Could we use L'Hopital's Rule?  $ \frac{d}{dr}\left(\frac{e^{rx}\minus{}1}{r}\right)\equal{}xe^{rx}$.  So by plugging in $ r\equal{}0$, we get the limit to be $ x$.  Is this right?

That's right, because you remembered that you were differentiating with respect to $ r.$ (Although I might take off a point for bad notation - you're not taking the derivative of the quotient.) Of those who attempted L'Hôpital, rather more of them wrote the derivative of the numerator as $ re^{rx},$ and then got an answer of zero.

Whew, I'm not as dumb as I thought. :lol:

L'Hopital's is kind of redundant here: this limit is the derivative of $ e^{rx}$ with respect to $ r$, no more and no less.

This is not worth opening another thread about, but my final exam had this multiple choice question:
How many fractions are in the partial fraction decomposition of $ \frac{4x^4\plus{}x\minus{}3}{(x^2\plus{}4)^3(x^2\minus{}3)^2}$?
(A) $ 2$  (B) $ 3$  (C) $ 5$  (D) $ 6$  (E) $ 7$

I got $ 6$ correct answers from the class of $ 70$ students. They would do much better by picking the answer at random.

Is L'Hopital's rule applicable here? It seems to be a circular reasoning.

From what I recall, doesn't partial fraction decomposition involve adding components?  Shouldn't it be factored like so:
$ \frac{A}{(x^{2}\plus{}4)^{3}}\plus{}\frac{B}{(x^{2}\minus{}3)^{2}}\equal{}\frac{4x^{4}\minus{}x\plus{}3}{(x^{2}\plus{}4)^{3}(x^{2}\minus{}3)^{2}}$
I apologize if my thought process is wrong, we don't use partial fractions here in multivariable calculus.  So isn't that 2?

but you also need a couple others like a
$ \frac {C}{x^2 \plus{} 4}$ term and a
$ \frac {D}{(x^2 \plus{} 4)^2}$ term etc.

[quote="mlok"]
How many fractions are in the partial fraction decomposition of $ \frac {4x^4 \plus{} x \minus{} 3}{(x^2 \plus{} 4)^3(x^2 \minus{} 3)^2}$?
[/quote]
Erm... How can I be sure that certain coefficients are non-zero without finding the actual decomposition? :?

Yes, I thought about that too (as a matter of fact, all numerators are nonzero). It would be better to ask how many undetermined coefficients one should use. But the students understood anyway that the question was about writing the general form of decomposition, with undetermined coefficients. They just did not know the correct general form.

Here are the series questions. #6b in particular didn't go well. How does one teach how to look at an algebraic expression and rapidly seeing what is truly important about it? I had so many students tie themselves into algebraic knots trying, badly, to apply the root or ratio test to it (and perhaps after spending all that effort, misremembering how to interpret the results.) How can you get people to just see it? Oh, and identification of the function in #7b - not good, either.

6. For each series: classify it as divergent, conditionally convergent, or absolutely convergent. Give reasons.

a) $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{\sqrt{n}}$

b) $ \sum_{n\equal{}1}^{\infty}\frac{2^n}{\sqrt{n}(3^n\plus{}n)}$

7. For each power series: [i]i[/i]) Find all $ x$ for which it converges and [i]ii[/i]) Identify the function to which it converges.

a) $ x\minus{}\frac{x^3}{6}\plus{}\frac{x^5}{120}\minus{}\frac{x^7}{7!}\plus{}\cdots$

b) $ x\minus{}x^3\plus{}x^5\minus{}x^7\plus{}x^9\plus{}\cdots$

(Several people simply rewrote the series in #7 in summation notation. I did not consider that to constitute "identifying the function.")

:roll: On 6b we must use the Ratio Test because there is something raised to power $ n$. So,
\[ \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac {2^{n \plus{} 1}}{\sqrt {n \plus{} 1}(3^{n \plus{} 1} \plus{} n \plus{} 1)} \frac {\sqrt {n}(3^n \plus{} n)}{2^n}
\]
Canceling the powers of $ 2$ and $ 3$, get
\[ \frac {2}{\sqrt {n \plus{} 1}(3 \plus{} n \plus{} 1)} \frac {\sqrt {n}(n)}{1}
\]
Now cancel $ n$ and observe that $ \sqrt {n \plus{} 1}/\sqrt {n}\to 1$:
\[ \frac {2}{3 \plus{} 1} \frac {1}{1} \equal{} \frac {1}{2}\ne 0,
\]
therefore the series diverges. [/sarcasm]

I put $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n}{2^n \plus{} 5n}$ on the final. The number of students who used the Ratio Test and got $ \frac17$ was somewhat disturbing.

My Method
[hide]
Can we use Cauchy tests on these?
[/hide]

mlok, that looks, uh, familiar. Have you been talking to any of my students?

madness - that will tie you up as well. Simplify your thinking.

[quote="Kent Merryfield"]Find $ \lim_{r\to0}\frac {e^{rx} \minus{} 1}{r}.$

The answer given by a plurality of the students in the class was not the correct answer.[/quote]

We should be obey the definition of $ \lim_{h\rightarrow 0} \frac{e^h\minus{}1}{h}\equal{}1.$

[quote="Kent Merryfield"]Find $ \lim_{r\to0}\frac {e^{rx} \minus{} 1}{r}.$[/quote]
We can also define $ u\equal{}\frac1{e^{rx}\minus{}1}.$

[quote="JRav"]So mlok, what is the official answer to your question?[/quote]
$ \frac {4x^4 \plus{} x \minus{} 3}{(x^2 \plus{} 4)^3(x^2 \minus{} 3)^2}$ $ \equal{}\frac{Ax\plus{}B}{x^2 \plus{} 4}\plus{}\frac{Cx\plus{}D}{(x^2 \plus{} 4)^2}\plus{}\frac{Ex\plus{}F}{(x^2 \plus{} 4)^3}\plus{}\frac{G}{x\minus{}\sqrt{3}}\plus{}\frac{H}{(x\minus{}\sqrt{3})^2}\plus{}\frac{I}{x\plus{}\sqrt{3}}\plus{}\frac{J}{(x\plus{}\sqrt{3})^2}$
So the correct answer is (E) $ 7$. The most popular choice was (C) $ 5$.

If you're going to count something in a partial fractions expansion, the most useful is the number of parameters- which doesn't depend on which field you're working over, or how you break up the pieces.
It's also very easy to count, since it's always equal to the degree of the denominator.

In other words, my answer would be (F)$ 10$.

[quote="jmerry"]it's always equal to the degree of the denominator.[/quote]
I did not realize that. :blush: I'll mention this next time I teach Calculus 2: if the number of coefficients you have is less than the degree of the denominator, something is missing from your decomposition.
I'm not worried about dependency on the field: in the context of Calculus 2 the field is $ \mathbb R$, and there are clear rules on what is a partial fraction and what is not.

Even within those "clear rules", what about terms like $ \frac{Ax\plus{}B}{x^2\plus{}4}$? You're going to break that one up when you integrate anyway.

If I'm thinking that way, I'll complete the square on all of the irreducible quadratics, and arrange the parameters accordingly. If I'm doing it myself rather than for a Calc 2 course, I'll evaluate at the complex roots to find the coefficients.

One can also write $ \frac{A}{x}$ as $ \frac{B}{x}\plus{}\frac{C}{x}$, etc. But I did not include answers of more than $ 7$.  :)

Yes, I know. I mostly just don't like it as a multiple choice question. Fortunately, the calculus classes where I am don't use multiple choice tests.

[hide="Hows this?"]
6a.$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n}{\sqrt {n}}$
$ \sum \frac {\sqrt {n \plus{} 1} \minus{} \sqrt {n}}{\sqrt {n^2 \plus{} n}}\approx\sum x^{ \minus{} 1.5}\approx\int x^{ \minus{} 1.5}$ which converges
I'm not sure what absolutely convergent means, but if you rearrange the terms, it diverges, so I'm going to guess conditionally convergent
6b. how could that possibly diverge? I think I'll go with absolutely convergent on that one

7a.sin(x) of course!
$ x\ e\ \mathbb{C}$
7b. $ \frac {x}{1 \plus{} x^2}$
$ |x| < 1$
[/hide]

I'd need more explanation on 6b). You can't just say it's obvious - you have to say why it's obvious. I would accept all of the others. Your answer to 6a) is unconventional (and a little notationally sloppy), but it is valid.

Just as a reminder: $ \sum a_n$ is called absolutely convergent if and only if $ \sum |a_n|$ converges. A sum is conditionally convergent whenever it is convergent but not absolutely convergent. There is a theorem that says that a real conditionally convergent series can be rearranged to converge to any sum whatsoever, or to diverge. What you wrote in 6a) looks like it references this theorem.

$ \sum_{n \equal{} 1}^{\infty}\frac {2^n}{\sqrt {n}(3^n \plus{} n)}$

$ 0 < \frac {2^n}{\sqrt {n}(3^n \plus{} n)} < \left(\frac {2}{3}\right)^n$

since all the terms are positive, it can't just alternate endlessly, and it has an upper bound of $ \sum_{n \equal{} 1}^{\infty}(2/3)^n \equal{} 2$

OK. I'll call that a use of the one-sided, or "simple" comparison test.

7 a) It converges for all $ x\in \mathbb R$ to $ \frac{e^x\minus{}e^{\minus{}x}}{2}\equal{}\sinh x$.

7 b) It converges for $ |x| < 1$ to $ \frac {x}{1 \plus{} x^2}$

[quote="0714446459923"]
6a.$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n}{\sqrt {n}}$
$ \sum \frac {\sqrt {n \plus{} 1} \minus{} \sqrt {n}}{\sqrt {n^2 \plus{} n}}\approx\sum x^{ \minus{} 1.5}\approx\int x^{ \minus{} 1.5}$ which converges
[/quote]

It is well known that if $ \lim_{n \rightarrow \infty} f(n) \equal{} 0$ then $ \sum ( \minus{} 1)^n f(n)$ converges!

Also, $ \sum^{\infty}_{n \equal{} 0}( \minus{} 1)^n \frac {x^{2n \plus{} 1}}{(2n \plus{} 1)!}$ is the basic Taylor series for $ \sin x$ and it can be shown using a simple differential equation $ y \plus{} y'' \equal{} 0$, $ y(0) \equal{} 0$ and $ y'(0) \equal{} 1$.

Yes, I wasn't expecting question to be so easy and I made a foolish mistake.
$ \sinh {x} \equal{} x \plus{} \frac {x^3}{3!} \plus{} \frac {x^5}{5!} \plus{} ...$
[quote="milin "]It is well known that if $ \lim_{n\rightarrow \infty}f(n) \equal{} 0$ then $ \sum ( \minus{} 1)^{n}f(n)$ converges!
[/quote]

It's not true. $ \lim_{n\rightarrow \infty}\frac {( \minus{} 1)^n}{n} \equal{} 0$  but  $ \sum \frac {( \minus{} 1)^n( \minus{} 1)^n}{n} \equal{} \sum \frac {1}{n}$ diverges.

$ f(n)$ has to be monotonous in order to satisfy that implication.

Has anyone done this the calc II way yet and said:

$ \sum\frac{(\minus{}1)^n}{\sqrt{n}}$ converges by the alternating series test ($ a_n \rightarrow 0$). It doesn't converge absolutely since the absolute value is the series $ \sum\frac{1}{\sqrt{n}}\equal{}\sum\frac{1}{n^p}$ which is known to diverge when $ p\leq 1$.

Well, yes, that's the answer I wanted to see.  I didn't see it often enough for various reasons, including getting the definition of absolute convergence backwards or getting the implication of $ p<1$ backwards.

[quote="Ognjen09"]
$ f(n)$ has to be monotonous in order to satisfy that implication.[/quote]

Yes, of course.. Sorry for not stating this..

About the initial question, see also [url=http://www.mathlinks.ro/viewtopic.php?t=124996]here[/url].

[quote="JRav"]I must admit I did not see all of those different factors.[/quote]

A decomposition with $ 2$ parameters spans only a two-dimensional subspace of the space of possible numerators.  The numerator can, of course, be any polynomial with degree strictly less than that of the denominator (or else you could simply divide out), whence jmerry's claim about the number of parameters.  :)

kent is that a class for math majors or some general requirement? i know quite a few who are (proudly) afraid of or hate math. or just plain satisfied and self-affirmed to fail it.

after two years of learning calculus i can't do more than the most routine series problems (applying tests, etc).  :oops:

The majority of the students are engineering majors, with significant numbers of science majors and math majors, and maybe a few people who are taking the course first and will figure out what it fits into later. These are people who mostly like math, or at least think they do. I think mlok was teaching essentially the same course this semester, to a very similar audience.

Yes, the only difference is that I had no math or science majors - only engineering majors and a few who are hoping to get into engineering. This semester's crop seemed to be weaker than usual, and "self-affirmed to fail" is a fitting description of some of them.   :|


[hide]$ 34^\circ \equal{} \arctan(\cos\theta)$

$ \tan 34^\circ \equal{} \cos\theta$

$ \theta \equal{} \arccos(\tan 34^\circ) \equal{} 47.58^\circ\implies\boxed{\text{D}}$[/hide]

If you're confused about the definition of arctan:

arctan of x = the angle that has a tangent value of x

For this problem, you have arctan (cos x) = 34 degrees.

Find tan 34 and then take the arccos of the value and you have your answer.

You can also think of the "arc" functions as the functions that "cancel out" trig functions.
Example: $ x\equal{}\tan\theta$.
Take the arctangent of both sides.
$ \arctan x\equal{}\arctan(\tan\theta)$.
But $ \arctan(\tan\theta)\equal{}\theta$ (like I said, they "cancel out").
So $ \arctan x\equal{}\theta$.


Basically, I'm trying to cleanup my view by hiding forums i don't frequent...

And one forum that I can't seem to touch is the Valley Math Circle community forum of Los Angeles. Not from that area, need a password to get into it anyways, and I can't hide it. -> problem


function
probability
integration
topology
expected value
real analysis
real analysis unsolved

Let $\Omega$ be a bounded open set in $\mathbb R^{n}$. Prove that for each point $x\in\Omega$ there is a Borel probability measure $\mu_{x}$ on the boundary $\partial\Omega$ of $\Omega$ such that for every continuous function $u: \overline{\Omega}\to\mathbb R$ which is harmonic on $\Omega$ we have $u(x)=\int_{\partial\Omega}u\ \mbox d\mu$.

One of the more entertaining approaches uses brownian motion. Send out a brownian motion from $x,$ and stop it when it hits the boundary. $u(x)$ will be the expected value of $u$ at the stopping point.

so $\mu_{x}$ is the probability measure on the boundary that describes the hitting point of a brownian motion starting from $x$ ?


From a point outside the circle, two tangents can be drawn to the ellipse. The chord of contact is the chord joining these two points of contact.

Let the point outside the ellipse be $ T(x_{0},y_{0})$, and assume the two tangents drawn from T contact the ellipse at $ P(x_{1},y_{1})$ and $ Q(x_{2},y_{2})$

Find the equation of TP and TQ then state why chord of contact has equation

$ \frac{xx_{0}}{a^{2}}$+$ \frac{yy_{0}}{b^{2}}=1$

ahh ok i understand now. Thanks for the descriptive proof appreciate it.

If there is another elegant way then please show me

thanks bye


algebra
function
domain
linear algebra
matrix
USAMTS
algorithm

Let $f: D\to {\mathbb R}$ , $f(x): =x\sqrt{x^2-1}$, where $D$$: =\{ x\in {\mathbb R}\; |\; f(x)\in {\mathbb R}\} .$ If $x_0=0$, compute $f(x_0) .$  It's true that  $x_0\in D$   ?

Whoa you sure this should be in the Round Table? :?

hmm...

it is correct do ask it in the round table since it is more a philosophical problem than a mathemetical one. however, extending it to a complex function, we will first get some problems because of the discontinuity of the square root, but even when arbritarily chosing the square root at each point, we always get that the limit at 0 is 0, thus it is very convenient to set $f(0)=0$, so $x_0\in D$.

This is a math problem, even if it is philosophical in nature.  I'm moving it to "Other Problem Solving Topics," which is where mathy discussions belong.

And I think f is poorly defined; you cannot define the domain of a function in terms of the function, because the function is not well defined unless you've chosen its domain.

[quote="ComplexZeta"]The question can be posed much better with a small modification.

Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=x\sqrt{x^2-1}$. Is 0 in the domain of $f$? My answer is yes -- any real $x$ satisfying $f(x)\in\mathbb{R}$ is in the domain of $f$, even if you have to leave the field of real numbers in order to compute it. Mainly, I don't think there should be any difference between the functions $x\sqrt{x^2-1}$ and $\sqrt{x^4-x^2}$.[/quote]

Hmm.  I'll weigh in, even though I'm a bit of a lightweight compared with some of you.  I think that it's not in the domain.

For example, if you want to define $ \mathbb{Z} $ from $ \mathbb{N} $ using an equivalence relation, the common way of doing that is to define:

(a,b)~(c,d) iff a+d=b+c.  Now, it's clear that this partitions N into pairs of integers having the same difference and we call each of these equivalence classes an integer.  It's obvious that this is the same condition as a-b = c-d, but we can't say this because it's entirely possible that a-b < 0, so we can't make sense of it.

Similarly, 0*x = 0 is an axiom, but x has to be a real number.  Why does 0 * (something that doesn't exist in R) have to compute to something?  For example, 0 * (a 2x2 matrix) is the zero 2x2 matrix, not 0.

So for me, having the expression undefined makes the most sense.

I didn't even realize what the problem was supposed to be until just now.  Simon, your definition of the function still isn't perfect because there are pieces of the domain, (-1, 0) and (0, 1), for example, that don't map to the range.  That point aside, I don't see why you would care.  The same function could be perfectly well defined by either assigning it that value at 0 or not assigning it that value at 0.  I am inclined to agree with Simon simply because the complex numbers are such a good extension of the real numbers.  If they didn't preserve all the really nice properties we want, I might be more skeptical, but there is almost never a problem with jumping into the complex numbers for a while and then jumping back out into the real numbers.

[quote="flip2004"]Let $f:D\to {\mathbb R} $ , $ f(x):=x\sqrt{x^2-1} $, where $D:=\{ x\in {\mathbb R}\; |\; f(x)\in {\mathbb R}\} .$ If $ x_0=0 $, compute $f(x_0) .$  It's true that  $ x_0\in D $   ?[/quote]

A difficulty related to this question occured in the Year 14 Round 4 of the USAMTS a year and a half ago. See the apology at the beginning of the solutions to Problem 3/4/14 at [url]http://www.artofproblemsolving.com/USAMTS/Tests/us1404s.pdf[/url]

There are two ways to figure out the domain of a function: the informal way and the formal way. Flip2004's odd phrasing above is an attempt to write up the informal way formally. The informal way is to just put together the algorithm for calculating the function, and then check which numbers yield a well-defined value when put through the algorithm. Since $f(0)$ has a good justification for equaling 0, we would say that 0 is in the domain.

The formal method for a composite function is to deduce the domain from the known domains of the pieces of the function. The pieces can be viewed as the binary multiplication fucntion $xy$, the square root function $\sqrt{x}$, and the polynomial function $x^2-1$. Neither multiplication nor the polynomial have restrictions on their domains, so since we are working in reals, their natural domains are all real numbers. The natural domain of the real-valued square root function is [0, :inf: ). That puts a restriction on the range of the polynomial function, $\mathstrut x^2-1\geq 0$, which yields a restricted domain (-:inf:, -1] :cup: [1,:inf:) for the composite function $\sqrt{x^2-1}$. Combining that restricted domain with the multiplication function to get $x\sqrt{x^2-1}$ still leaves the domain restricted, so the formal domain of $f$ is (-:inf:, -1] :cup: [1,:inf:). In this case, 0 is not in the domain.

Since formal and informal methods yield different answers, we have to decide which method is valid in our context. Are we more interested in getting as many answers as possible or are we more interested in being careful? It is a lot like the limit problems where we know that we can make the rational expression $\frac{x^2+x-2}{x-1}$ equal 3 when x=1 by cancelling the x-1 factor n the numerator and denominator, but we need to use limits to do it formally.

Erin Schram


algebra
polynomial
complex numbers
absolute value
number theory
Diophantine equation
number theory unsolved

Find  positive integers n and m such that :
$ 3^m \equal{}2.n^2 \plus{}1$

$ 3^m\equal{}2n^2\plus{}1\quad\Leftrightarrow\quad 1\plus{}3\plus{}9\plus{}\cdots \plus{}3^{m\minus{}1}\equal{}n^2.$  :maybe:

[quote="M@ri@n"]$ 3^m \equal{} 2n^2 \plus{} 1\quad\Leftrightarrow\quad 1 \plus{} 3 \plus{} 9 \plus{} \cdots \plus{} 3^{m \minus{} 1} \equal{} n^2.$  :maybe:[/quote]

ok.Can you continue?

$ 1\plus{}2n^2 \equal{} 3^m$

Lemma : the only elements of norm 3 in $ \mathbb{Q}[\sqrt{\minus{}2}]$ are $ \pm 1\pm \sqrt{\minus{}2}$. This is easy to see by looking at the equation $ a^2\plus{}2b^2\equal{}3$.

We have $ N(1\plus{}n\sqrt{\minus{}2}) \equal{} 3^m$. Hence $ 1\plus{}n\sqrt{\minus{}2}$ is of the form $ (\pm 1\pm \sqrt{\minus{}2})^m$. (it can't be a product including two types of these four elements because then 3 would divide $ 1\plus{}n\sqrt{\minus{}2}$ which it clearly doesn't). We can consider only the case $ 1\plus{}\sqrt{\minus{}2}$; all others are similar.

Now consider powers of $ 1\plus{}\sqrt{\minus{}2}$:

$ (1\plus{}\sqrt{\minus{}2})^2\equal{}\minus{}1\plus{}2\sqrt{\minus{}2}$
$ (1\plus{}\sqrt{\minus{}2})^3\equal{}\minus{}5\plus{}\sqrt{\minus{}2}$

I'm quite sure that past $ (1\plus{}\sqrt{\minus{}2})^5\equal{}1\minus{}11i$ there are no more powers whose real part is 1 in absolute value but I can't show it. I'm a bit tired. This method yields the solutions $ (n,m) : (1,1), (2,2), (11,5)$ and I'd be extremely surprised if there were others.

The only solutions $ (m,n)$ are indeed $ (1,1), \, (2,2)$ and $ (5,11).$

Following [b]brunojoyal[/b], if we let $ (u_n)$ be the sequence of the real parts of $ (1 \plus{} i \sqrt{2})^n,$ we see that it satisfies the recurrence relation $ u_{n\plus{}2} \equal{} 2u_{n\plus{}1} \minus{} 3u_n,$ with $ u_0 \equal{} u_1 \equal{} 1$ (nota : the minimal polynomial of $ 1 \plus{} i \sqrt{2}$ is $ X^2 \minus{} 2X \plus{} 3$).

We would like to show that the values $ \minus{}1$ and $ 1$ do not appear anymore in that sequence after $ n \equal{} 5.$

Fortunately, F. Beukers has done some work on such things, and his paper [url=http://www.numdam.org/numdam-bin/fitem?id=CM_1980__40_2_251_0]The multiplicity of binary recurrences[/url] destroys the problem at hand.
(in particular, cf. case ii) in Lemma 7 on page 9 of the PDF, and the 4th case of Theorem 2, on page 10)

A few side remarks :
$ \diamond$ That equation is a "cousin" of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=192321]Ramanujan-Nagell equation.[/url]
$ \diamond$ The way [b]M@ri@n[/b] rewrote it reminds the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=149570]Diophantine equation 1+x+...+x^m=1+y+...+y^n[/url]
$ \diamond$ Beukers also proved a result about [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26612]ternary recurrences.[/url]


parameterization
floor function
logarithms
algorithm
algebra
polynomial
combinatorics proposed

Let $ A_1,A_2,\cdots,A_m$ be $ n$-sets (i.e. $ |A_i|\equal{}n$). Suppose $ m < 2^{n\minus{}1}$. Prove that one can color the universal set $ \cup_{i\equal{}1}^mA_i$ with black and white such that non of $ A_i$s has the same color for all its members.

this is a very easy application of the probabilistic method.

immodestius being so modest, let us repair his terse statement. Also, in the statement, we obviously need $ n \geq 2$, and we may even take $ m \leq 2^{n \minus{} 1}$.

Consider all random (equiprobable) colorings of the elements of $ \bigcup_{1 \leq i \leq m} A_i$ with black and white. For each $ A_i$, let $ E_i$ be the event "all elements of $ A_i$ share a same color". Since there are exactly $ 2$ such colorings, we have $ \textrm{Prob}(E_i) \equal{} \left ( \frac {1} {2} \right )^{n \minus{} 1}$, so for $ m \leq 2^{n \minus{} 1}$ (as the events $ E_i$ are not disjoint)
\[ \textrm{Prob}\left ( \bigcup_{1 \leq i \leq m} E_i \right ) < \sum_{i \equal{} 1}^m \textrm{Prob} (E_i) \equal{} m\left ( \frac {1} {2} \right )^{n \minus{} 1} \leq 1.\]
Therefore there must exist a $ 2$-coloring such that no $ A_i$ is monochromatic.

Paul Erdos established a (not sharp) bound for $ m$, so that such colorings are impossible, namely $ m > n^2 2^n$, also using probabilistic methods. The lowest such $ m$ is only known for $ n \equal{} 2$ (when it is $ 3$), and $ n \equal{} 3$ (when it is $ 7$).

(Courtesy Proofs from the Book, by Aigner and Ziegler).

Thanks a lot for your elaborate answer. In fact the main problem that I was working on was to find the max $ m$ for which such a coloring exists. As you and immodestius mentioned, I found the bound $ 2^{n\minus{}1}$ easily through probabilistic method, but I had an intuition that $ m$ still can be larger. You mentioned Erdos's bound which is a big help for me. There is another similar problem (with some application in table coloring) which I posted in the section 'Combinatorics Open Questions'. For your convinced I state here:

Consider the following question: Given integers $ n$ and $ m$, is it true that for any $ m$ sets $ A_1,A_2,\cdots,A_m$ with $ |A_i|\equal{}n$ ($ i\equal{}1,2,\cdots,m$), there exists a black and white coloring of the set $ \cup_{i\equal{}1}^mA_i$ with at least one black member such that any $ A_i$ contains an even number of black members?
a) Come up with some value for the parameter $ m$ (in terms of $ n$), as large as possible, such that the above question has answer Yes. For example, an straightforward argument shows that $ m\equal{}\lfloor\log{n}\rfloor$ works. Can it be better?
b) Find an algorithm to answer the above question in polynomial time in terms of $ n$ (and preferebly $ m$ too)

I'm going to post a new topic in this section for the $ m\equal{}\lfloor\log{n}\rfloor$ bound as it is a solved problem.


geometry
3D geometry
modular arithmetic
number theory
prime numbers

1)Show that every cube number is of the form $9n$ or $9n(+-)1$

2)Show that the difference of the  squares of any two prime numbers greater than 6 is divisible by 24.

1)[hide]Half of it is easy and the other half by fermats theorem i suppose.[/hide]

2)[hide]not sure about this but see mod3 and mod8.[/hide]

2) Is pretty interesting. We know that $24=2^3\times8$. We can come up with the following pretty obvious facts:
$p_1\equiv p_2\equiv1\bmod{2}$
Therefore, $2|(p_1+p_2)$ and $2|(p_1-p_2)$. So we've found two of the three factors of 2.

Next, we know that
$p_1\equiv1\bmod{3}$ or $p_1\equiv2\bmod{3}$
$p_2\equiv1\bmod{3}$ or $p_2\equiv2\bmod{3}$
So no matter how you combine those possibilities, either $(p_1+p_2)$ or $(p_1-p_2)$ is going to be divisible by three. So we have the factor of three. All that's left is that last pesky two.

How can the last factor of $2$ be shown? So far all I've done is show that $12|(p_1-p_2)(p_1+p_2)$.

[hide="2"]it is pretty well known that a prime $>3$ is in the form $6k+1$ so that squared$(6n\pm 1)^2=36n^2\pm 12n+1$

$36(n^2-m^2)+12(\pm n \pm m)=12(3n^2-3m^2\pm m\pm n)$ if n and m are either both even, or both odd, there will be an extra factor of 2, so they will be divisible by 24

if one odd, one even, both parts are odd their sum will be even so an extra factor of 2, and div by 24[/hide]

[hide="Proof of #2"]Every prime number greater than $3$ can be written as $6m\pm 1$. Hence if $p_1=6k\pm 1$ and $p_2=6l\pm 1$, we have four cases for $p_1^2-p_2^2=(p_1-p_2)(p_1+p_2)$:

1) $p_1^2-p_2^2=12(k-l)[3(k+l)+1]$
2) $p_1^2-p_2^2=12(k+l)[3(k-l)+1]$
3) $p_1^2-p_2^2=12(k+l)[3(k-l)-1]$
4) $p_1^2-p_2^2=12(k-l)[3(k+l)-1]$

In any of these cases, if $k$ and $l$ are of equal parity, then $k\pm l$ is even, hence the whole expression is divisible by $24$. If $k$ and $l$ are of different parity, then $k\pm l$ is odd, hence $3(k\pm l)\pm 1$ is even, and the expression is again divisible by $24.$[/hide]

hmm does this mean we can check whether big numbers are primes by looking at whether or not they are $\pm 1 \bmod 6$?

No. Not ALL numbers $6m\pm 1$ are primes - for example $35=6\cdot 6-1$. But if number IS a prime, then it's $6m\pm 1$ and it can be a quick test if a number is composite.

oh right... darn so ig uess we cant use that trick haha

it is not really that useful a test because all that does is check if it is divisible by 2 or 3, and you can do that faster by the digit addition and even units digit...

[quote="matt276eagles"][hide="Solution to Part A"]$0^{3}=0\pmod{9}$
$1^{3}=1\pmod{9}$
$2^{3}=-1\pmod{9}$
$3^{3}=0\pmod{9}$
$4^{3}=1\pmod{9}$
$5^{3}=-1\pmod{9}$
$6^{3}=0\pmod{9}$
$7^{3}=1\pmod{9}$
$8^{3}=-1\pmod{9}$[/hide]

No words required! :D[/quote]

That isn't a solution to it at all.Could someone tell the 1st one bcos it isn't possible by fermats theorem...I'm not getting a solid solution by mods as well(the prime numbers are troubling me.)

can we go any further from here:-
$n^2-1$ is divisible by 3.So cubing or something :?

What was wrong with my solution?  I showed that every cube is equal to -1, 0, or 1 mod 9.

I don't think they understood the solution as it was stated. A few words would have helped, such as "since only the units digits affect modulo a digit and all units digits were tested, therefore modulo that digit will not vary outside of those given terms".

[quote="Hokkage"][hide="Number 1"]Any number can be written in the form $3n$, $3n+1$, or $3n+2$. The cubes of these numbers are, respectively, $27n^3 = 9(3n^3) = 9m$, $27n^3+27n^2+9n+1 = 9(3n^3+3n^2+n)+1 = 9m+1$, and $27n^3+54n^2+36n+8 = 27n^3+54n^2+36n+9-1 = 9(3n^3+6n^2+4n+1)-1 = 9m-1$ for some m in each case. Woohoo![/hide][/quote]

What's m in this case?$3n^3$?Then how come $(3n^3+3n^2+n)=m$?Or is m different in each case.
Any number can be written in the form $3n$, $3n+1$, or $3n+2$
According to the above,how do we write 2?

And for he modulo solution i as getting troubled by primes.Like $3^3$ congruentto 0(mod9).So how does this help for finding $23^3$?

I was just saying that the term (substituted with m) as a whole is a multiple of 9, so yes, m is different for each one. The important thing is that m is an integer, so therefore the whole thing is 9m, 9m+1, or 9m-1, which is what we wanted.

[quote="Hokkage"][hide="Number 2"]Any prime (>3) can be written in the form $6m \pm 1$.
[hide="Proof of this fact"]Since any other number is written in the form $6m$, $6m+2$, $6m+3$, or $6m+4$, we can show that all of those are composite because they are respectively divisible by 6, 2, 3, and 2.[/hide]
We may now assume that these two primes ($p_1$ and $p_2$ are in one of the two forms, either $6m + 1$ or $6m - 1$. Checking the cases:
$p_1 = 6m+1, p_2 = 6n+1 \Rightarrow p_1^2-p_2^2 = (6m+1)^2 - (6n+1)^2 = (6m-6n)(6m+6m+2) \equiv 0 (\bmod{24})$ since $(6m-6n) \equiv 0 (\bmod{12})$ (since m-n is even) and $(6m+6m+2) \equiv 0 (\bmod{2})$ and multiplying 2 and 12 yields $p_1^2-p_2^2 \equiv 0 (\bmod{24})$.
Second case: $p_1 = 6m-1, p_2 = 6n+1 \Rightarrow p_1^2-p_2^2 = (6m-1)^2 - (6n+1)^2 = (6m+6n)(6m-6m-2) \equiv 0 (\bmod{24})$ by the same argument.
The same result is achieved by swapping $p_1$ and $p_2$ and thus all cases are exhausted, so....
$\therefore \forall$ primes $p_1$ and $p_2$ ($p_1,p_2>6$), $p_1^2-p_2^2 \equiv 0 \bmod{24}$[/hide][/quote]

Almost exactly the same proof as mine (post #6)

Oh whoops. Sorry, didn't see your solution. Excuse my dupe post  :oops:

and what about the mod prrof?(What matt 276 eagles gave)how do we find 23^3 with it(In the fprm 9n+-1?