calculus inequalities limit quadratics algebra polynomial function

My book is somehow grabbing the variable five and proving this limit. There is a serious lack of descriptive language and detailed logic. The logic wayyy too implied for me.

Book:

[quote]
Use the epsilon-delta definition of limit to prove that

lim x^2 = 4
x -> 2

To find an appropriate delta, begin by writing |x^2 - 4| = |x -2||x+2|. For all x in the interval (1,3), you know that |x+2| < 5. So, letting delta be the minimum of epsilon/5 and 1, it follows that, whenever 0 < |x - 2| < delta, you have...

|x^2 - 4| = |x-2||x+2| < (epsilon/5)(5) = epsilon[/quote]

The parts I'm not understanding:

[quote]
Use the epsilon-delta definition of limit to prove that

lim x^2 = 4
x -> 2

To find an appropriate delta, begin by writing |x^2 - 4| = |x -2||x+2|. For all x in the interval (1,3)[/quote]

Here is one of my questions...

ok, where is it getting this idea of (1, 3)???
Why is (1,3) being used?

[quote]...For all x in the interval (1, 3), you know that |x+2| < 5. So, letting delta be the minimum of epsilon/5 and 1, it follows that, whenever 0 < |x - 2| < delta, you have...

|x^2 - 4| = |x-2||x+2| < (epsilon/5)(5) = epsilon[/quote]

Uh, where is it getting this five? I know that (1,3) applied to x would be less than 5, but why are we doing it this way? Why are we even using (1,3) and applying it to 5?

I don't really the logic of this all.

I mean, it looks like it's grabbing numbers out of air that just work. There seems to be a large understanding of calculus on the author's part of how these choices were determined, but none of them were explained. So, possibly one of you could?

On another website, it used a different kind of method for a different but similar limit equation:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION%205

It sadly had implied language too, which made me assume this about that website's problem...

Solution 5.

Since negative two is the least it could be. Therefore, less than 3
Since positive one is the most it could be. Therefore, greater than 1

[quote="Genecks"]My book is somehow grabbing the variable five and proving this limit. There is a serious lack of descriptive language and detailed logic. The logic wayyy too implied for me.

Book:

[quote]
Use the epsilon-delta definition of limit to prove that

lim x^2 = 4
x -> 2

To find an appropriate delta, begin by writing |x^2 - 4| = |x -2||x+2|. For all x in the interval (1,3), you know that |x+2| < 5. So, letting delta be the minimum of epsilon/5 and 1, it follows that, whenever 0 < |x - 2| < delta, you have...

|x^2 - 4| = |x-2||x+2| < (epsilon/5)(5) = epsilon[/quote]

The parts I'm not understanding:

[quote]
Use the epsilon-delta definition of limit to prove that

lim x^2 = 4
x -> 2

To find an appropriate delta, begin by writing |x^2 - 4| = |x -2||x+2|. For all x in the interval (1,3)[/quote]

Here is one of my questions...

ok, where is it getting this idea of (1, 3)???
Why is (1,3) being used?

-->> We want the lim x^2 as x --> 2, so we interested how x^2 behaves in an interval which includes 2. We choose of the interval (1,3)  because it is a simple one. We can do the same choosing every open interval which includes 2.
Try to make the same for the interval $ \left[{{3\over2},\,{5\over2}}\right]$.<<--

[quote]...For all x in the interval (1, 3), you know that |x+2| < 5. So, letting delta be the minimum of epsilon/5 and 1, it follows that, whenever 0 < |x - 2| < delta, you have...

|x^2 - 4| = |x-2||x+2| < (epsilon/5)(5) = epsilon[/quote]

Uh, where is it getting this five? I know that (1,3) applied to x would be less than 5, but why are we doing it this way? Why are we even using (1,3) and applying it to 5?

-->> For every $ \epsilon$ we already find a $ \delta \equal{} max\left\lbrace{{\epsilon\over5},1}\right\rbrace$ and because  |x+2| < 5,  we prove that $ |x^2 \minus{} 4| < \epsilon$, as we want. <<--

I don't really the logic of this all.

I mean, it looks like it's grabbing numbers out of air that just work. There seems to be a large understanding of calculus on the author's part of how these choices were determined, but none of them were explained. So, possibly one of you could?

On another website, it used a different kind of method for a different but similar limit equation:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION%205

-->> The method it's exactly the same!
Try to focus on the meaning of definition of limit and the role of $ \epsilon$ and $ \delta$ in this definition.
Don't forget the quantifiers! "$ \forall$ (for every) $ \epsilon > 0$, we must find or (there exists) $ \exists$ a $ \delta > 0$, such that if $ |x \minus{} 2| < \delta$, then $ |x^2 \minus{} 4| < \epsilon$.
And, if you are a beginner, be patient! <<--

It sadly had implied language too, which made me assume this about that website's problem...

Solution 5.

Since negative two is the least it could be. Therefore, less than 3
Since positive one is the most it could be. Therefore, greater than 1[/quote]

they pick (1,3)=(2-1,2+1) because they use $ \delta\equal{}min(1, \frac{\epsilon}{5})$

you want to be able to control |x+2||x-2| for all x near 2

they decided that they would easily be able to calculate it for when "near" meant within 1 at the most.

but if they picked within $ \theta$ at the most then the problem would be to control it in $ (2\minus{}\theta, 2\plus{}\theta)$

watching the term |x-2| in this range, we see that the biggest it can be is $ \theta$
watching the term |x+2| in this range, we get anything in $ (4\minus{}\theta, 4\plus{}\theta)$

so alltogether we want to pick a value for theta that makes solving for delta easier.

we want a delta such that when $ |x\minus{}2|<\delta$ then we get $ |x\minus{}2\parallel{}x\plus{}2|<\epsilon$

using our thetas first, we want  $ |x\minus{}2\parallel{}x\plus{}2|\le |x\minus{}2|(4\plus{}\theta)\equal{}\delta (4\plus{}\theta)$

so we pick our delta to be $ \frac{\epsilon}{4\plus{}\theta}$  the issue here is that this value might be bigger than theta, which would ruin our inequality, so we just set it $ \delta\equal{}min(\frac{\epsilon}{4\plus{}\theta},\theta)$

this is what your book did, with the easiest value of $ \theta\equal{}1$

interestingly, we could have solved this without a min.
if we just used our thetas

$ |x\minus{}2\parallel{}x\plus{}2| \le \theta (4\plus{}\theta)\equal{}\theta^2\plus{}4\theta < \epsilon$ iff $ \theta^2\plus{}4\theta\minus{}\epsilon <0$ iff

$ \left[ \theta\equal{}\frac{\minus{}4 \pm \sqrt{16\plus{}4\epsilon}}{2}\equal{}\minus{}2\pm \sqrt{4\plus{}\epsilon}\right]$

iff $ \theta \in (\minus{}2\minus{}\sqrt{4\plus{}\epsilon},\minus{}2\plus{}\sqrt{4\plus{}\epsilon})$  since we only care about positive thetas we look towards the top of this range

and set $ \delta\equal{}\sqrt{4\plus{}\epsilon}\minus{}2$  now $ |x\minus{}2|<\delta$ gives us $ |x\minus{}2\parallel{}x\plus{}2|<\epsilon$

hope i didnt make things infinitely worse!

First, do you understand how to prove that $ \lim_{x \to a} x \equal{} a$ using epsilon-delta?  It is easy to choose delta here because we can take $ \epsilon \equal{} \delta$; in other words, $ \frac{x \minus{} a}{x \minus{} a} \equal{} 1$, so the error in the output is exactly the error in the input.

When we move from linear to quadratic polynomials, things get hairier because the value of, say, $ \frac{x^2 \minus{} a^2}{x \minus{} a} \equal{} x \plus{} a$ locally depends on $ a$ as well as on $ x$; in other words, the error in the output depends locally on the [i]value[/i] of the input as well as on its error.

(I suppose it would be meaningful to define what I mean by "error."  If we have a function $ f(x)$ and we input a value $ a$ into $ x$ is that is off by an error $ \epsilon$, the we can write $ f(a \pm \epsilon) \equal{} f(a) \pm \delta$ where $ \delta$ is the corresponding error in the output.  In other words, if we are performing calculations on data which we know to have some error $ \pm \epsilon$ then we would like to provide bounds on what the corresponding error $ \pm \delta$ in our final calculations is.  This is one way to think about the epsilon-delta definition of a limit.)

Now, if $ |x \minus{} a| < \epsilon$ then what can we say about the value of $ |x^2 \minus{} a^2|$?  Well, it must be less than $ \epsilon |x \plus{} a|$, so now how do we bound $ |x \plus{} a|$?

The author does this by looking in a neighborhood $ x \equal{} a \pm 1$ and bounds $ |x \plus{} a|$ by $ |1 \plus{} 2a|$, but as kenn4000 noted $ 1$ can be replaced by any positive real.  The point is that if $ x$ is "around" $ a$ then $ |x \plus{} a|$ will be "around" $ 2a$.


A game is played with five cards numbered 3, 6, 2, 5, and 1. Face down, the cards are shuffled and then three are turned face up. Your score is the sum of the three numbers showing. What is the probability that your score will be 9?

There are $ \binom53\equal{}10$ ways to choose the cards. Of those, only 1,2,6 and 1,3,5 work. So it's $ \frac{2}{10}\equal{}\boxed{\frac15}$


I would like to see IMO profiles of ALL contestants this year .  ;)

[b]All it needs are a room where you can take pictures properly and some people who type the answers of the contestants![/b]

[url=http://imo.wolfram.com/contestants/ARG5/]Martín Miranda[/url] had the following questions/answers in his profile:

[i]Email: martin@strike.com.ar
Home town: Buenos Aires

School: Instituto Industrial Luis A. Huergo

Age: 17

Native Language: Castellano

Handedness: right

Math Relatives:

School subjects(s) most liked: Electronics Subjets

School subjects(s) least liked: Is not the subject, is the profesor who teach it.

Hobbies/Interests: Chat.

Special achievement(s):

Favorite book(s): Lord of the Rings and other works of JRR TolkienDune (the entire saga) Frank Herbert

Favorite movie(s): Star Wars.Lord of the Ringsd (not yet, ;( )

Planned college subjects: Electronics.

Future career plans: Electronics and Informatics.

Favorite math fact, theorem or formula: I dont have any

Why math is great: Because it is Perfect

Personal Webpage: http://

People to thank:

Final message: Argentina is the Best, the SAPO have the power. [/i]

[size=125][b]What would you ask the contestants? [/b][/size] :D

[quote="orl"]

[size=125][b]What would you ask the contestants? [/b][/size] :D

[/quote]
I would also ask their musical instruments. Most math people seems to be able to play one.

We were given a questionaire somewat like this. Is that going to be a profile? Are they going on the internet? Does anyone know anything about what those questions were for?

This is a question to which I also wish for answers. For my questionnaire did flow with the divine milk of comedy.


Since when can the signature contain no external URLs?  The only reason that my signature contains a URL is that I entered it a while ago.


How do you make your text into like that hexagon thing on the first page of this document?

Could be PSTricks which is not for beginners. Ask them for the code.


real analysis
calculus
integration
logarithms
inequalities
function
real analysis solved

[quote="liyi"]$f:[0,1]\to[1,\infty)$ is a measurable function.
Show that
$\left(\int_{[0,1]} f(x)dx\right)\left(\int_{[0,1]} \ln f(x)dx\right) \leq \left(\int_{[0,1]} f(x) \ln f(x)dx\right)$[/quote]

 Using Jensen inequality for convex ( and concave ) functions , this problem seems to be quite straightforward.

$\int_{[0,1]} f(x) \ln f(x)dx \geq \left(\int_{[0,1]} f(x)dx\right)\ln \left(\int_{[0,1]}f(x)dx\right) \geq \left(\int_{[0,1]} f(x)dx\right)\left(\int_{[0,1]} \ln f(x)dx\right)$

 The first inequality is due to $u\ln u$ is convex, the second is due to $\ln u$ is concave

 Isn't it??

Yes. dickchimney's proof is completely correct, and shows that the original stipulation on the range of $f$ could be weakened to $f:[0,1]\to(0,\infty),$ as long as all of the integrals in dickchimney's proof converge absolutely.

[quote="dickchimney]
$\left(\int_{[0,1]} f(x)dx\right)\ln \left(\int_{[0,1]}f(x)dx\right) \geq \left(\int_{[0,1]} f(x)dx\right)\left(\int_{[0,1]} \ln f(x)dx\right)$
[/quote]
How to obtain this?

I think he did mention that it follows from the concavity of $\ln$.

thanks.
i realized it soon after posting it....
//shamed


probability
linear algebra
matrix
geometry

[quote="JeffreyJones88"]What I meant to say in my last subject post was this: The following is a link which proves *MATHEMATICALLY* that the Bible is true.if there is a 1/1000000 chance that the Bible was true, you would owe it to good sense to check it out. As it turns out, there is a 1/200000000 chance that the Bible is WRONG. Anyway, check it out:
http://video.google.ca/videoplay?docid=8648337797409966990&q=Professor+and+the+Prophets[/quote]
[b]Please do not double-post! The same post is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111656]here[/url]![/b]

What's more, the topic is completely irrelevant. And even without watching it, I have some doubts about your figures- specifically, they seem to be made up.

So basically, you are saying that there is a 1/200000000 possibility of the earth NOT being made in 7 days, of 6/7 of the surface of the earth NOT being land, etc.?

Probably it is the case that there is that probability that everything in the Bible is false.  :D

Ah if that means that there is something in the Bible that is true, I accept it gladly, and let the matter rest.

[quote="wobster109"]the bible cannot be true.  to give an example used in the scopes-monkey trial (the trial of a teacher for teaching evolution in a state where evolution was banned), the bible claims god made the sun and stars on the third morning/day, or something similar.  since one cannot have days and nights without the sun, the bible cannot be true.[/quote]

Haha, very funny.

That isn't reason for the Bible to be false, and really, there isn't much which we can use to disprove the Bible, but many of the things in there just seem so improbable.  Not to discredit the Bible, but I honestly think it's better to follow a logical conclusion rather than simply follow a book like a sheep. (this sentence is directed at the creator of the thread)

You can't prove a whole book is 'true' unless you prove every single statement in it.  I haven't looked at your link but I'm sure it doesn't prove everything in the bible.

Moreover, we cannont say that something in the Bible has a probability. For example, if it states that something happened, it wasn't a random event and we can't define probability.

I agree with Rzeszut: how on earth did they arrive at a probability of something being true? It's not exactly like this is a replicable science experiment of some sort.

I would like to remind anyone posting on this thread, however, that the religious sensitivities of others shouldn't be ignored. There are plenty of people on this forum who are Christian (Catholics included here) and accept the Bible as divine.

Also, I'm moving this to the Round Table. Please remember in the future that F&G is reserved for threads that are basically completely pointless. Anything that might merit a thoughtful discussion belongs elsewhere.

To keep on the pointless discussion, and have some fun in the meantime, I don't know about Genesis in other languages, but surely similar to spanish where it goes something like 'And God made the stars and the sun and he saw it was good'.  What?! What do you mean he SAW it was good?! Wasn't God supposed to know it WAS GOING to be good in the first place?
But of course, all of this is nonsense if you think of it as an argument against the Bible or God.

[color=red]Edit: I forgot to mention about the practical impossibility of disproving the Bible versus the practical possibility of convincing most people of its veracity. Because if tomorrow God makes its appearance and miracles start to happening, well we'd have a hard time denying them. We could (claiming we're on a Matrix for instance) but we would had a hard time[/color]

Ursula

Can god create a rock that he cannot move? :rotfl:

[quote="mustafa"]Paradoxes like that annoy me.

What is the plural of Paradox? Paradoxes? Paradoxs? Paradoxi?[/quote]
Paradoxes.

[quote="mustafa"]Paradoxi sounds much cooler[/quote]
I agree, but somewhere I found that the plural is paradoxes. I think it only becomes an "i" ending when the singular is "us," like cactus (cacti), and radius (radii).

[quote="mustafa"]Who wrote this stupid language? They should have some common sense, and pluralize paradox the best way.[/quote]
Is pluralize a word? If it is, then I agree, this is a very strange language.

I would say that's more of a contradiction than a paradox.  What you're saying is along the lines of "can god have all the power but not have some of that power at the same time?".  No, he can't.  Being unable to move the rock is the lack of a power, not a power in itself.

And the God is unable to answer the question about the rock.

Yeah, he can't answer this question either:

True or false:  find the area of a 3-4-5 right triangle.

There are very many such questions: `prove that $0=1$`, `Am I lying?`, etc.

Who's to say God can't make two contradictory things simultaneously true?  If we were accept that God makes logic, then that means God can define "true".

[quote="mustafa"]And if you annoy God with stupid paradoxi, God can smite you.[/quote]

Lol I'd love to call that spam, delete, it and call it a day. Unfortunately, it is appropriately humorous.

Anyway, about the rock. I think this is more a case of God being able to create such a rock and it being immovable not because God is unable to move it, but rather because he created he to be...well...not moved. Thus he created something for the very purpose of being unmovable and God, with his infinite power, will not move it (which is different from not being able to move it). And that in turn means that the rock is actually not movable, but God could move it...and so on and so on and so on.

Honestly, there's better questions to waste your time contemplating. Reimann's for one.

[quote="mustafa"]Who wrote this stupid language?[/quote]

In this context (the "us" endings pluralizing as "i"), the Romans.

In general, the Romans, Greeks, Germanic people, and probably quite a few more.


superior algebra
superior algebra unsolved

Let $R$ be a commutative infinite ring with identity. Suppose that every proper ideal of $R$ has smaller cardinality than $R$. Prove that $R$ is a field.

[quote="harazi"]Hmm, you know you should not post problems for which solutions are still expected![/quote]
Sorry. I thought the april edition was already over, since it's june. Do the problems keep running untill someone solves them? Or how does it work?

You may close the thread if you prefer :)

is it allowed to discuss it this month yet? I'm very curious for this beatiful result

Since it hasn't been locked up yet, and since it's such a cute and simple problem (I'm sure it must have been solved right away in AMM anyway :)), it's a pity to just leave it lying here like this. Here's a proof:

Let $ (x)$ be the ideal generated by an arbitrary element $ x$ of $ R$, and let $ \mbox{Ann}~x$ be the annihilator of $ x$ in $ R$, i.e. the set of elements $ a\in R$ such that $ ax=0$. Now consider the short exact sequence $ (*)\ 0\rightarrow\mbox{Ann}~x\rightarrow R\rightarrow(x)\rightarrow 0$, the next-to-last arrow representing multiplication by $ x$, i.e. the map $ a\mapsto ax$. $ (*)$ provides a bijection of $ R$ onto the product $ \mbox{Ann}~x\times(x)$. Because $ R$ is infinite, this means that at least one of the ideals $ \mbox{Ann}~x$ and $ (x)$ has cardinality equal to that of $ R$, and hence, by hypothesis, coincides with $ R$. There are thus two cases: either $ \mbox{Ann}~x=R$ and $ x=0$, or $ (x)=R$ and $ x$ is invertible. In other words, $ R$ is a field.


Stanford college articles AMC USA(J)MO USAMO AIME

Hey, Gunn people. Who's taking 51H next year?

Doesn't look like too many people...I think ccy is doing it too. Argh I want to skip analysis really really bad next year...
Speaking of classes, who is planning to enroll in advanced problem solving III/IV? I signed up already.

[quote="bookaholic"]I thought 51H referred to the entire sequence...anyway, well, I meant all three parts.[/quote]
hmm i dont know you mean like linear algebra and stuff right

APS i cant take, 52H or 51H or whatever is going to make me really late to the APS class

Crap I'm really hoping that class will have enough people enrolled. Lynelle go help us solicit to your paly folk and tell them to sign up for APS3/4!!!

[quote="bookaholic"][url]http://math.stanford.edu/undergrad/50seq.html[/url]
Oh...so 51H is just the first trimester.

I'll try to promote APS to math club, but I can't take it.

Taylor, what's the rest of your schedule like? Is Gunn also moving to a late start next year?[/quote]
i have no idea
hm im going for 3 aps + stanford + eng H + java/living skills + prep (YESH)

You two can be guinea pigs :P if any of you die next year then I might reconsider my plan for junior year (AP physics C, calc hopefully stanford although that doesn't look possible, AP bio, English H, whatever history ap or honors, hopefully not Chem AP as I hope to be able to somehow convince the teacher into letting me into that next year, AP chinese (actually it'll be my "prep" since I'm 100% sure I can sleep everyday and ace tests), AP music theory/stat/some other BS AP?)
I don't think you two will die though, since apparently there's a good deal of people who took a ton of AP crap junior year and they weren't exceptionally smart or anything and didn't die.

I am taking 3 APs--APUSH, physics, Spanish--but my fifth Paly class is art spectrum, which is said to be only slightly more demanding than a prep. And physics<bio when it comes to homework, too.

[quote="serialk11r"][quote="junggi"][quote="bookaholic"]Whoa cool! Now when someone tries to scare me about my plan for next year, I can be like, I know someone whose schedule is at least as hard and probably harder! Yay. Will one of those APs be APUSH?[/quote]
whoa youre not taking even 3 aps?
luckyyy

apush, ap bio, ap spanish lang

[quote="serialk11r"]AP chinese[/quote]
i think i said this before, but im sure some colleges suggest/require 3 years of the same language
anddddd lotsa chinese tried to take the chinese lang class and got kicked out in the first week.
andddd the classes obviously arent designed for native speakers, and my guess is that the colleges will realize that and dock points off of your application
so i strongly do suggest you continue with spanish...[/quote]
But Spanish is so boring...and I'm still in Spanish 1. I don't want to be stuck with the retards in my class right now, so going to chinese 3 next year will help that, since chinese 3 has people who actually get better than a D in their classes usually.[/quote]
not really
the "retards" have to stay in the languages to get the graduation requirements

I can't imagine any Chinese class being less boring than a Spanish class for someone who already speaks a basic amount of Mandarin. Spanish class never stops being boring, but in my opinion the language itself is kind of fun at higher levels. Have you considered maneuvering to skip Spanish 2 (and thus avoiding the type of people who either quit after fulfilling the requirement or go into 3 rather than 3H)? Incoming freshmen get into 3H after some tutoring, and I don't see why incoming sophomores shouldn't be able to do that too.

[quote="DiscreetFourierTransform"]I'll have to disagree with all of you and claim, that infact, the only instrument worth playing, is the little plastic guitar from Guitar Hero.

I'm a rock GOD.[/quote]
1) i bet i can beat you at guitar hero
2) no guitar hero guitar isnt an instrument
3) get a real guitar
yeah
real electric guitars are awesome
if you really are awesome at guitar hero you should improve fast enough

Guitar hero is awesome! Except I suck because I almsot never play it.
I like FFR though

http://www.vgcats.com/comics/?strip_id=232

This is aimed at the guitar heroing youtubers.

Has ubemaya lost weight since starting ddr.

LOL don't hate on mozart.
itg > ddr > guitar hero.

Rock Band is for noobs. The guitar is way easier and it is lame without 4 people (or at least 2)

Dude, Beat Boxing with the mike
nuf said

but seriously Rock Band is sweet.

:huh: um what the original topic was about some random test, now it's all about musical instruments

i'm impressed at how you people manage to change topics so quickly in what, 70 posts?

take your non spamming somewhere else like oklahoma.

or arkansas.

but you guys, guitar hero/rock band is lame. 4-panel dancing games = win.

[quote="serialk11r"][quote="junggi"][quote="serialk11r"]WTF piano*10^-(indefinitely large positive number)>violin*10^(indefinitely large positive number)[/quote]
GO SPREAD YOUR PROPAGANDA ELSEWHERE[/quote]
HELL NO
(btw, why would you care? its not like you play violin. I didn't mention guitar YET)[/quote]
i see violins and pianos as equal
violins get extra points here
[url]http://youtube.com/results?search_query=electric+violin&search_type=[/url]


Let $ n \in N$, $ n\geq 2$. Find maximum :
                    $ A \equal{} \prod_{i \equal{} 1}^{n} (1 \minus{} x_i)$
where $ x_i > 0$ $ \forall i\in N\bigcap [1,n]$ and $ \sum_{i \equal{} 1}^{n} x_i ^2\equal{} 1$

[quote="conan_naruto236"]Let $ n \in N$, $ n\geq 2$. Find maximum :
                    $ A \equal{} \prod_{i \equal{} 1}^{n} (1 \minus{} x_i)$
where $ x_i > 0$ $ \forall i\in N\bigcap [1,n]$ and $ \sum_{i \equal{} 1}^{n} x_i ^2 \equal{} 1$[/quote]
Nobody?  :(


induction
function
number theory proposed
number theory

[quote="iura"]For a positive integer denote by $f(n)$ the largest integer $k$ with $2^k|n$.
Let $a_0=0$ consider the recurrence
$\frac 1{a_n}=1+2f(n)-a_{n-1}$.
Prove this sequence contains every positive rational exactly once[/quote]
Rewrite your problem because is BAD-defined, because, by definition, $f(0)=+\infty$ so $a_1=0=a_0$, contradiction with your afirmation, so maybe you need to say something about $f(0)$!.

Best Regards.
RX the Crazy Math

[quote="Myth"]You don't need $f(0)$ at all. Please, read the problem more carefully![/quote]
WHere my student was watching  :blush:  :blush: , ejem, ejem , I mean, sorry I made a mistake  :?  :what?:

RX

Let $p_{n}$,$q_{n}$ the two sequences of non negative numbers such that $a_{n}=\frac{p_{n}}{q_{n}}$,$p_{n}$ and $q_{n}$ coprime. Then $p_{n+1}=q_{n}$ and $q_{n+1}=(1+2f(n+1))q_{n}-p_{n}=(1+2f(n+1))q_{n}-q_{n-1}$ for each $n\geq 1$. As the numerator of $a_{n+1}$ equals the denominator of $a_{n}$, we have:$a_{1}\dots a_{n}=\frac{1}{q_{n}}$. So $a_{n}=\frac{q_{n-1}}{q_{n}}$. Moreover, we can prove by induction that
$q_{2n}=q_{n-1}+q_{n}$
$q_{2n+1}=q_{n}$
Indeed, we have $a_{0}=0$,$a_{1}=1$,$a_{2}=\frac{1}{2}$,$a_{3}=2$, so $q_{0}=q_{1}=1$,$q_{2}=2$,$q_{3}=1$, so both relations are satisfied for $n=1$.
Suppose they are true for $n\geq 1$.
$q_{2n+2}=(1+2f(2n+2))q_{2n+1}-q_{2n}$
$q_{2n+2}=(1+2(f(n+1)+1))q_{n}-q_{n}-q_{n-1}$
$q_{2n+2}=(1+2f(n+1))q_{n}-q_{n-1}+2q_{n}-q_{n}$
$q_{2n+2}=q_{n+1}+q_{n}$
And:
$q_{2n+3}=(1+2f(2n+3))q_{2n+2}-q_{2n+1}$
($f(2n+3)=0$)
$q_{2n+3}=q_{2n+2}-q_{2n+1}$
$q_{2n+3}=q_{n+1}+q_{n}-q_{n}$
$q_{2n+3}=q_{n+1}$
This proves that if $a_{n}=\frac{p}{q}$, then $a_{2n}=\frac{p}{p+q}$ and $a_{2n+1}=\frac{p+q}{q}$ (*).
We now define a function $i$ fom the rational numbers to the natural integers.
Let $\frac{p}{q}$ be a rational number. We apply a succession of elementary transformations according to the following rule:
if $p \geq q$, then $\frac{p}{q}$ becomes $\frac{p-q}{q}$.
if $p < q$, then $\frac{p}{q}$ becomes $\frac{p}{q-p}$. It is clear that after a finite number of operations, we find $0$. Now take $n=0$ and apply the following transformations to $n$:
We take the sequence of transformations applied on $\frac{p}{q}$ and invert it. If the transformation is
$\frac{p}{q}\rightarrow \frac{p-q}{q}$, then $n\rightarrow 2n+1$.
if $\frac{p}{q}\rightarrow \frac{p}{q-p}$, then $n\rightarrow 2n$. $i(\frac{p}{q})$ is the final $n$ we get. Maybe it is clearer on a concrete example :
 :blush:
$\frac{12}{7}\rightarrow \frac{5}{7}\rightarrow \frac{5}{2}\rightarrow \frac{3}{2}\rightarrow \frac{1}{2}\rightarrow \frac{1}{1}\rightarrow \frac{0}{1}=0$
So corresponding operations on $n$ give:
$0 \rightarrow 1 \rightarrow 2\rightarrow 5 \rightarrow 11\rightarrow 22\rightarrow 45$ and we have $i(\frac{12}{7})=45$. Because of (*), $i(a_{n})=n$ and $a_{i(\frac{p}{q})}=\frac{p}{q}$.
This means that $a$ is a bijection and all rationals appear exactly once in the sequence. I would be very interested to know Iura's solution :-)

Iura's solution is actually the same:
We can prove by straightforward induction that $a_{2n}=\frac {a_n}{a_n+1}$ and $a_{2n+1}=1+a_n$.
From now on, by a simple finite descent method we can go backwards from any rational to reach $0$.
So I can't gladden your heart, Xell, it's the same.
Note that this problem doesn't look easy to me, but it's solution is very simple: just you must observe the two relations.


Find all the positive integers $a,b$ such that the number $\frac{\sqrt{2}+\sqrt{a}}{\sqrt{3}+\sqrt{b}}$ is [b]Rational[/b]

[quote="maskman"]why don't you try to look for a more  interesting problem ? this one has no value and shouldn't be here[/quote]
This problem doesn't deserve this kind of criticism, especially from someone so new to the forum.  If you don't like it, then just ignore it, or contribute your own "interesting" problems.


Prove that:
$ F_1F_3\plus{}F_2F_6\plus{}F_3F_9\plus{}...\plus{}F_nF_{3n}\equal{}F_nF_{n\plus{}1}F_{2n\plus{}1}$

This problem is a thing thought about in 1969 by Recke :lol:

by PMI

Since $ F_1F_3 \equal{} 1 \cdot 2 \equal{} 2$ and $ F_1F_2F_3 \equal{} 1 \cdot 1 \cdot 2 \equal{} 2$, the given statement is clearly true when $ n\equal{}1$.
Now we assume it is true for an arbitrary positive integer $ k$ : $ \sum_{i\equal{}1}^kF_iF_{3i}\equal{}F_kF_{k\plus{}1}F_{2k\plus{}1}$.
Then
$ \sum_{i\equal{}1}^{k\plus{}1}F_iF_{3i}$
$ \equal{}F_kF_{k\plus{}1}F_{2k\plus{}1}\plus{}F_{k\plus{}1}F_{3k\plus{}3}$
$ \equal{}F_{k\plus{}1}(F_kF_{2k\plus{}1}\plus{}F_{3k\plus{}3})$
$ \equal{}F_{k\plus{}1}(F_kF_{2k\plus{}1}\plus{}F_{(2k\plus{}1)\plus{}(k\plus{}2)})$
$ \equal{}F_{k\plus{}1}(F_kF_{2k\plus{}1}\plus{}F_{k\plus{}2}F_{2k\plus{}2}\plus{}F_{k\plus{}1}F_{2k\plus{}1})$
$ \equal{}F_{k\plus{}1}((F_k\plus{}F_{k\plus{}1})F_{2k\plus{}1}\plus{}F_{k\plus{}2}F_{2k\plus{}2})$
$ \equal{}F_{k\plus{}1}(F_{k\plus{}2}F_{2k\plus{}1}\plus{}F_{k\plus{}2}F_{2k\plus{}2})$
$ \equal{}F_{k\plus{}1}(F_{k\plus{}2}(F_{2k\plus{}1}\plus{}F_{2k\plus{}2}))$
$ \equal{}F_{k\plus{}1}(F_{k\plus{}2}(F_{2k\plus{}3}))$
$ \equal{}F_{k\plus{}1}F_{k\plus{}2}F_{2k\plus{}3}$
$ \equal{}F_{k\plus{}1}F_{(k\plus{}1)\plus{}1}F_{2(k\plus{}1)\plus{}1}$
Thus the formula works for $ n\equal{}k\plus{}1$.
So, by PMI, the statement is true for every integer $ n \ge 1$.


parameterization
geometry
trigonometry
analytic geometry
complex numbers

(I actually have no idea what level this is since I can't solve it). $ABC$ is a triangle. A perpendicular $BD$ to $BA$ is constructed such that $BD=r\cdot BA$. Similarly, a perpendicular $CE$ to $CA$ is constructed such that $CE=r\cdot AC$. Find the locus of the mid point of $DE$ for all values of the parameter $r$.

So I did the case for an equilateral triangle. I set $A$ and $C$ at $-\frac{1}{2}$ and $\frac{1}{2}$ on the $x-$axis and $B$ at $1$ on the $y-$axis. Using basic geometry I was able to find that the midpoints of $DE$ are $\left\{(x,y): \left(\frac{1-\sqrt{2}r}{4}, \frac{2+\sqrt{2}r}{4}+\frac{r}{2}, ~r\in\mathbb{R}\right)\right\}$. But this is a line, there isn't really a locus for a line is there? This problem seems fairly straightforward, does anyone get anything different?

P.S.-I've been told that the answer involves complex numbers, if that helps you're doing better than me...

Or maybe I'm misinterpreting the problem...?

are you sure that you have stated the problem correctly? cause maybe D should be on AC and BD perpendicular to AC, etc..
anyway if that is true, you can WLOG put AC on the x-axis and A on the origin so that $AD = AB\cos{A}$ and $AE=AC\cos{A}$. so if $A=(0,0)$ and $C = (b,0)$, then the coordinates of $E=(b\cos{A}\cos{A},b\cos{A}\sin{A})$ and $D=(c\cos{A},0)$.
you can then say the the midpoint of $DE$ has coordinates $\left(\frac{b(\cos{A})^{2}+c\cos{A}}{2},\frac{b\cos{A}\sin{A}}{2}\right)$ and i think here somebody else can bust out the complex numbers or whatever

I definitely agree that the problem may be mistated, but what about the part with $r\cdot$ stuff? That can't be a typo like the perpendicular part because that is the parameter that should be varying. Does anyone have a guess as to what this problem is supposed to say? Thanks.


1) Let $ A,B,C \in M_n{\mathbb{(C)}}$ are Hemitian matrices and positive such that $ A \plus{} B \plus{} C \equal{} \pi I_n$
Prove that $ 1 < \det(sin\frac {A}{2} \plus{} sin\frac {B}{2} \plus{} sin\frac {C}{2}) < (\frac {\pi}{2})^n$
2) Let $ A_1,A_2,...,A_m \in M_n{\mathbb{(C)}}$ are Hemitian matrices and $ x_1,x_2,...,x_m$ are $ m$  real number
Prove that :
 $ (x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_m^2)(A_1^2 \plus{} A_2^2 \plus{} ... \plus{} A_m^2)\geq (x_1A_1 \plus{} x_2A_2 \plus{} ... \plus{} x_mA_m)^2$  (Same[b] inequality cauchy -schwarz[/b])
 [hide="A small example "]for $ A,B,C \in M_n{\mathbb{(C)}}$ are Hemitian matrices such that $ 3A \minus{} 4B\plus{}C \equal{} 5I_n$ then use it we have $ (3^2 \plus{} 4^2\plus{}1^2)(A^2 \plus{} B^2\plus{}C^2) \geq (3A \minus{} 4B\plus{}C)^2$ or $ A^2 \plus{} B^2\plus{}C^2 \geq \frac{25}{26}I_n$ then $ \det(A^2 \plus{} B^2\plus{}C^2) \geq (\frac{25}{26})^n,Tr(A^2) \plus{} Tr(B^2)\plus{}Tr(C^2) \geq \frac{25n}{26}$ ......[/hide]

Let $ A,B,C \in M_n{\mathbb{(C)}}$  are Hemitian matrices and semipositive define such that $ AB \equal{} BA,BC \equal{} CB,CA \equal{} BC$ and $ A \plus{} B \plus{} C \equal{} \pi I_n$
 then $ 0 \leq Sin(A) \plus{} Sin(B) \plus{} Sin(C) \leq \frac {3\sqrt {3} I_n}{2}$, $ Cos(A) \plus{} Cos(B) \plus{} Cos(C) \leq \frac {3I_n}{2}$ ,.....
 More .... I hope that you are interested it  :(


algebra
polynomial
function
limit
calculus
derivative
real analysis

It is given a sequence of polynomials with real coefficients
\[ \phi_n(x)\equal{}\sum_{k\equal{}0}^na_{n,k}x^k\qquad|a_{n,k}|\leq 1,\]
which, for $ n\to\infty$, pointwise converge to a function $ f(x)$, for all $ x\in I: \equal{}(0,1)$. Prove that

(a) $ f$ is $ C^\infty$ on $ I$.
(b) $ f$ is analytic on $ I$.
(c) the convergence $ \phi_n\to f$ may be non uniform, even if you take the stronger hypotesis $ \sum_{k\equal{}0}^n|a_{n,k}|\leq 1$ on the coefficients.

The trick is to observe that the family of all such polynomials is a normal family in the unit disk $ \mathbb D$ (it is dominated by $ \frac 1{1\minus{}|z|}$). Therefore, any sequence contains a subsequence converging uniformly on every compact subset of $ \mathbb D$. Applying this fact to our given sequence, we get (a) and (b) immediately. As to (c), just consider the sequence $ x^n$.

[quote="smullyan"]It is given a sequence of polynomials with real coefficients
\[ \phi_n(x) \equal{} \sum_{k \equal{} 0}^na_{n,k}x^k\qquad|a_{n,k}|\leq 1,
\]
which, for $ n\to\infty$, pointwise converge to a function $ f(x)$, for all $ x\in I: \equal{} (0,1)$. Prove that

(a) $ f$ is $ C^\infty$ on $ I$.
(b) $ f$ is analytic on $ I$.
(c) the convergence $ \phi_n\to f$ may be non uniform, even if you take the stronger hypotesis $ \sum_{k \equal{} 0}^n|a_{n,k}|\leq 1$ on the coefficients.[/quote]
Here's a real variable proof:

Note that $ \lim_n a_{n,k} \equal{} b_k$ exists for all $ k$. Indeed, suppose $ K$ is the smallest index for which
\[ \limsup_n a_{n,k} \equal{} \alpha > \liminf_n a_{n,k} \equal{} \beta\]
and set $ f_K(x) \equal{} \lim_n \sum_{k \equal{} 0}^{K\minus{}1}a_{n,k}x^k$, $ f_K(x) \equal{} 0$ if $ K\equal{}0$. Then for any $ 0 < x < 1$
\[ f(x) \equal{} f_K(x) \plus{} \alpha x^K \plus{} \lim_n \sum_{k \equal{} K\plus{}1}^na_{n,k}x^k \\
\ge f_K(x) \plus{} \alpha x^K \minus{} \lim_n \sum_{k \equal{} K\plus{}1}^nx^k \\
\equal{} f_K(x) \plus{} \alpha x^K \minus{} \frac{x^{K\plus{}1}}{1\minus{}x} \]
and similarly
\[ f(x) \le f_K(x) \plus{} \beta x^K \plus{} \frac{x^{K\plus{}1}}{1\minus{}x} \]
resulting in
\[ \alpha \minus{} \beta < \frac{2x}{1\minus{}x} \]
for all $ x > 0$. Hence in fact $ \alpha \equal{} \beta$.  Hence $ \lim_n a_{n,k} \equal{} b_k \in [\minus{}1,1]$ for all $ k$.

Then by a standard argument, $ f(x) \equal{} \sum_{k\equal{}0}^\infty b_kx^k$, and $ f$ is analytic. The convergence is uniform (together with all derivatives) on any compact subinterval of $ (\minus{}1,1)$, but not necessarily up to $ x\equal{}1$, as fedja noted.


geometry
incenter
circumcircle
angle bisector
complex numbers
geometry proposed

let ABC be a triangle.find the locus of M,wich is in the interior of ABC,,,inwich:

angle MAB+angle MBC+angle MCA=90

(u can also find it for an equaliraterall triangle )

For an equlateral triangle, finding the locus is trivial - all points of the altitudes $\equiv$ angle bisectors $\equiv$ medians ...

Let the $\triangle ABC$ be equilateral, D, E, F the feet of the altitudes from the vertices A, B, C and P an arbitrary point, say on the altitude AD. The triangles $\triangle PAB \cong \triangle PAC$ are congruent by SAS (AB = AC, PA is common and $\angle PAB = \angle PAC = 30^\circ$). Hence,

$\angle PAB + \angle PBC + \angle PCA = 30^\circ + \angle PBC + \angle PBA = 30^\circ + 60^\circ = 90^\circ$.

For an arbitrary triangle $\triangle ABC$, we can identify several points of the locus:

[color=blue][b](1)[/b][/color] the incenter I:

$\angle IAB + \angle IBC + \angle ICB = \frac{\angle A + \angle B + \angle C}{2} = \frac{180^\circ}{2} = 90^\circ$

[color=blue][b](2)[/b][/color] the orthocenter H:

$\angle HAB + \angle HBC + \angle HCB = 90^\circ - \angle B + 90^\circ - \angle C + 90^\circ - \angle A =$

$= 270^\circ - (\angle A + \angle B + \angle C) = 270^\circ - 180^\circ = 90^\circ$

[color=blue][b](3)[/b][/color] the feet of the altitude isogonals $D' \in BC, E' \in CA, \F' \in AB$, for example, the point F'. Let $F \in AB$ be the foot of the C-altitude and $M \in AB$ the foot of the angle bisector CI:

$\angle F'AB = 0,\ \angle F'BC = \angle B$

$\angle F'CA = \angle FCM + \angle MCF + \angle FCA =$

$= 2 \left (\frac{\angle C}{2} - (90^\circ - \angle A) \right) + 90^\circ - \angle A = \angle C + \angle A - 90^\circ$

$\angle F'AB + \angle F'BC + \angle F'CA = 0 + \angle B + \angle C + \angle A - 90^\circ = 180^\circ - 90^\circ = 90^\circ$

[color=blue][b](4)[/b][/color] the circumcenter O, identical with the concurency point of the altitude isogonal conjugates AD', BE', CF', i.e., the isogonal conjugate of the orthocenter H:

$\angle OAB + \angle OBC + \angle OCA = \frac{180^\circ - 2 \angle C}{2} + \frac{180^\circ - 2 \angle A}{2} + \frac{180^\circ - 2 \angle B}{2} =$

$= \frac{3 \cdot 180^\circ - 2 \cdot 180^\circ}{2} = 90^\circ$

But the general shape and/or equation of the locus is ugly. Let the vertices A, B, C and the locus point P be represented by complex numbers a, b, c, z, respectively. WLOG, let a = -1, b = +1 be real. the angles $\angle PAB, \angle PBC, \angle PCA$ are given by

$\angle PAB = \text{Arg}\frac{z - a}{b - a},\ \angle PBC = \text{Arg}\frac{z - b}{c - b},\ \angle PCA = \text{Arg}\frac{z - c}{a - c}$

$\angle PAB + \angle PBC + \angle PCA = \text{Arg}\left(\frac{z - a}{b - a} \cdot \frac{z - b}{c - b} \cdot \frac{z - c}{a - c} \right) =$

$= \text{Arg}\left(- \frac{z + 1}{2} \cdot \frac{z - 1}{c - 1} \cdot \frac{z - c}{1 + c} \right) = \text{Arg} \frac{(z^2 - 1)(z - c)}{2(1 - c^2)}$

If the argument of this product is $90^\circ$, the product must be purely imaginary. Thus the equation of the locus is

$\text{Re}\frac{(z^2 - 1)(z - c)}{2(1 - c^2)} = 0$

or after multiplying by the complex conjugate of the denominator $2 (1 - \overline{c}^2)$

$\text{Re}\left[(z^2 - 1)(z - c)(1 - \overline{c}^2) \right] = 0$

Substituting $z = x + iy$ yields the locus equation in x-y coordinates. It is an ugly curve of the 3rd degree in both x and y, not in one piece,  passing through the locus points identified above (and through the triangle vertices A, B, C, where one of the angles $\angle PAB, \angle PBC, \angle PCA$ is undefined).

Have a look at
http://perso.wanadoo.fr/bernard.gibert/Exemples/k003.html


number theory
prime numbers
number theory unsolved

PLEASE DELETE THIS POST... SORRY ABOUT THAT
PLEASE DELETE THIS POST... SORRY ABOUT THAT

[quote="baz"]Just suppose $ p_{1}\neq p_{2}$ and reach contradiction.[/quote]This can be also the start of the problem. The finish is not obvious from here.

[quote="mathpk"]Prove that for odd prime numbers $ p_{1},q_{1},p_{2},q_{2},P$ such that
\[ \frac{p_{1}^{q_{1}}\minus{}1}{p_{1}\minus{}1}\equal{}\frac{p_{2}^{q_{2}}\minus{}1}{p_{2}\minus{}1}\equal{}P\]
,
then $ p_{1}\equal{} p_{2}$.[/quote]
(p1^q1-1/p1-1)mod q1 = (1+p1+p1^2+...+p1^(q1-2)+p1^(q1-1)) mod q1 = 1 mod q1 => P=1modq1
Similarly p2^q2-1/p2-1 mod q2=1mod q2 => P=1mod q2
p1^q1-1/p1-1 = 1 mod q2
=> p1^q1-1 = (c*q2+1)(p1-1)
Upon simplification we get
p1*(p1^(q1-1)-c*q2-1)=-c*q2
q2 does not divide p1, so p1^(q1-1)-1 mod q2 = 0
By Fermat's Little Theorem, q1-1 is a divisor of q2-1 --(1)
Similarly considering the other equation, we get q2-1 is a divisor of q1-1.--(2)
From these two statements q1=q2.


Consider the positive integers 121 + 19n where n = 0, 1, 2, ...., 3000. The number of the numbers which are the squares of integers is

a. 12
b. 13
c. 19
d. 25
e. 2002

Thanks! :)

I'll give an explanation if I turn out right. Is the answer [hide="suspected answer"]D.[/hide]?  I probably did it too quickly and made a mistake.


consider tringle ABC and its excircles: one of them touches the side AC at K and touches lines AB and BC at L and L. the other touches side AB at P and touches the lines AC and BC at Q and R. prove thet the intersection point X of the lines LM and QR lies on the altitude (passing throut A) of tringle ABC.

It's old problem.
See here: http://www.mathlinks.ro/viewtopic.php?p=928863#928863


geometry
3D geometry
tetrahedron
linear algebra
matrix
analytic geometry
LaTeX

I've seen so many problems dealing with the volume of a tetrahedron without you knowing the hieght, although many of the sides are known. I was wondering if there was some technique to find the area other than using a huge matrix complex.

use three dimensional drawing with coordinates. draw it out. it becomes easier to see. also, looking at the so-called center of the tetrahedron helps as well.

i would suggest one of two things:
1) find the height and then calculate it based on a bunch of right tetrahedra

2)find the coordinates in some 3d coord system, and then use a matrix, i dont remember exactly how to do this, but i think it looks something like this. (im to lazy to put htis in latex, so hopefully it comes out nice)

| 1      1     1     1 |
| x1    x2   x3   x4 |*(1/6)
| y1    y2   y3   y4 |
| z1    z2    z3   z4 |

that is, the determinant of the matrix times 1/6, but dont hold me accountable, im not positive thats right.

If the tetrahedron is $ABCD$ and we take vectors $AB, AC, AD$, the volume is

$\frac{1}{6}\cdot [(AB \times AC) \cdot AD]$.

Yes.  Also, if we write the vectors a, b, and c in component form, we can then do:

$V = \frac{1}{6}\parallel \begin{pmatrix}a_{1}& a_{2}& a_{3}\\ b_{1}& b_{2}& b_{3}\\ c_{1}& c_{2}& c_{3}\end{pmatrix}\parallel $

Where the inner set of lines denotes the determinant, and the outer denotes the absolute value function.

It is also useful to notice that we can use this to determine if 3 vectors are coplanar.  If the determinant of that is 0, then the area of the tetrahedron is 0, and thus the vectors are coplanar.


[quote="gaussintraining"]I'm not certain, but I'm fairly sure that the $ 6$ face is opposite the $ \boxed{3}$ face.[/quote]
Yes, it is correct because if you visualize it, the $6$ face will fold up to the side of the $4$ face and so will the $3$ face but on the opposite side.  So the number on the face opposite the number $6$ face is $\boxed{3}$.

This isn't related, but how do you delete your posts?


limit
logarithms
real analysis
real analysis solved

hi, i haven't been able to figure this one out:

Lim       ( x / cos(x-1) ) ^ (x / (1-x^3) )
x -> 1

??

thanks!!

-Trevor

thanks, that makes perfect sence!!

-Trevor


geometry
analytic geometry
graphing lines
slope
puzzles

This puzzle [url=http://en.wikipedia.org/wiki/Image:Missingsquare.png]here[/url] is pretty fun. See how long it takes you!  :lol: [/url]

[hide]What's the puzzle? The areas of the quadrilateral and the concave octagon are clearly the same.[/hide]

[hide="Solution"] The thing is, the hypotenuses of the red and the blue triangles do not make a straight line (just find the slope of them). Then in the first figue they create a concave quad, and in the bottom one -- a convex one, but it seems it is just a triangle, which resluts in the illusion.[/hide]

[quote="box of rocks"]rem,

"very easy?"

Well, no, if you have to give THAT much description about what the situation is to adequately cover a (correct or not correct) solution, then you have contradicted yourself with "very easy."  Best to avoid said phrase.[/quote]
OK sorry. I will follow your advice.  :)


[color=cyan]This is a problem which was given to me by a teacher/friend of mine:

Consider four spheres of equal radius r, each tangent to the other three.  There exists a regular tetrahedron containing the three spheres, with each of its faces tangent to three of the spheres.  Compute, in terms of r, the volume of this tetrahedron.[/color]

Ok, Joel, you know this is one of those questions that would usually skip over, but I tried it anyway. So...

Hopefully, the length of each side of the tetrahedron is 2r(1+sqrt[3]).

Sooooo, with that formula we derived in class, the volume is...
 (20*Sqrt[2]*r^3)/3 + 4*sqrt[6]*r^3.

Ugh, but then again, I'm wrong, as I am usually with, uh, any type of problem in fact.

[color=cyan]How did you figure out you were wrong?  And how did you approach the problem to begin with?[/color]

to begin with, i stacked up 4 tennis balls, just to see what it would look like. But then, i made the erroneous assumption that each face of the tetrahedron is an equilateral triangle with three circles of radius r, each tangent to 2 of the sides and the other 2 cirlces. of course, that is wrong, so, my answer is wrong.


I've encountered quite a few stupid rules at school.  What's your experience?
Stupidest one at mine: You can't wear a hood or hat, because it reminds people of gangsters. (!?)

I don't think this [i]really[/i] belongs in Round Table, but...
Similar to your experience, you can't make any "gang signs" in school photos. This, however, includes the stereotypical asian fob peace sign. I can see a point for this rule, but I live in a safe town with no gang activity (look at my location!), so I think it is pretty pointless.
Also, me and a few friends of mine have gotten into trouble for (in my opinion) trivial things at middle school. My friend, for instance, once flipped a school computer's monitor with ctrl+alt+down arrow. Following this, the class was given a lecture on the wrongs of hacking, with a specific mention to monitor flipping.

Rather than "stupid", you mean "frustrating".

At my previous public school, there was a rule that you need a signed hall pass to go from one place to another. Which is totally legitimate, except it became a pain in the bum for me. After orchestra, I wanted to go to the library during study hall. Study hall instructors did not have the authority to give out library passes, so I had to get one from the math director. But i couldn't go there without getting a pass from my math teacher. So I got a pass from my study hall teacher to go to my math teacher to go to the math director to go to the library. The whole time I was like, why the ƒ~ can't I just go to the library?

Fortunately in the real world, rules are easier to bend. Of course, you have to find the right people  :wink:

our school ended up having bomb threats written on the bathroom wall.

the school was evacuated and we had to leave all our stuff in the classrooms and lockers.

and that happened about five times.
yes. five bomb threats.

so after the bomb threats, for the rest of the year, the 2nd floor bathrooms were locked (and all my classes were 2nd floor), and we had to have hall passes signed by teachers. You can't go to the bathroom between classes, you can't go to the bathroom during lunch, and you couldn't go to the bathroom at the end of the day.
And if you try getting your pass signed by your teacher during class, he or she will get very angry at you for interrupting the class and probably won't let you go.

so what's the point of having bathrooms in the building??

and it was the very end of the year when this happened, and most people burned their hall passes in massive bonfires (like I did), so I couldn't go during class anyway...

and also, this isn't at my school, but at some schools you can't wear jeans.
and at Exeter, you can't wear shirts with writing on them.
Which I think is pretty stupid, but I feel worse for the boys. They have to wear nice button-down shirts and ties. They tried to remove the rule, but the girls liked it, so they kept it. :)

[quote="kimmystar94"]Which I think is pretty stupid, but I feel worse for the boys. They have to wear nice button-down shirts and ties. They tried to remove the rule, but the girls liked it, so they kept it. :)[/quote]

GASP

what about shirts with no writing? They exist! :)

One of the really dumb rules that my school gave was that I'm not aloud to smoke on the bus.

1. So we can smoke inside the school?
2. I don't even smoke.
3. I've never seen anybody smoke on the bus.
4. (Just Plain Steeeeewwwpid)

[quote="lex95wiz"]Of all the things in america, school is the most un-democratic.  I mean, schools can punish you for cursing while the first amendment says freedom of speech.[/quote]

one of the articles under htat says that free speech is limited if you are under the protection/watched over by an administration. then that administration can limit your free speech. when you become a school newspaper writer, you will quickly find that out ;]

i really hate the No Spaggetti Straps one. i LUV spagetti straps. they're so much mor comfortable! and the No Short Shorts rule, which states ur shorts cant be above ur fingertips when u put ur arms down at ur sides. honestly, i dont have any shorts that DONT break the rul! my arms are linger than most peoples', for some reason. and i get hot easily, so knee-lengnth pants just dont do around the end of school. in fact, i dont like most of the school dress code...

oh yeah and i also dont like the no hoods rule. and my school doesnt let me wear sunglasses :dry:

There really aren't any rules in high school that I'm too frustrated at.

We also use hall passes for restrooms but my biology teacher just let's me go and says "Pretend I signed it."

No gum chewing in class but some teachers don't mind.

The dress code is practically "Just don't come to school naked."

One time we had a fad in our school where you leave the price tags on your shirts/sweaters on so that you can brag how expensive your clothes are and how small sizes you wear. lol it was very cocky but very fun. then the principle banned it and gave us a long speech on taking the price tags off.

[quote="1=2"]what about shirts with no writing? They exist! :)[/quote]

that wasn't what I was meaning.
like even if they have the brand name on it, i can't wear it.
and that means nearly 50% of my shirts aren't allowed.

but I'm bringing them anyway cuz that rule is often disregarded by teachers.

who keeps rating everyone's posts here? :maybe:

some1's being biased..

[quote="7h3.D3m0n.117"]

[quote]
[u]Conesquences:[/u]
1. Teachers are free to make fun of you
2. Teachers are free to give dirty looks
[/quote]

This happens anyway  :P[/quote]

Man, I can't even count how many times either of these have happened to people at my school.

Ah. I don't think I explained it enough. By teachers making fun, I mean, say the Spanish teacher saw you doing some monkey business. Then, when you went to Spanish class, the first 10 minutes would be a series of jokes at your expense. Then, maybe he told the math teacher, so then the first 10 minutes of math would also be a series of embarrassing remarks at your expense. Stuff like that.

rule:

no hats

i wonder why the even have that rule...

They're distracting and people can hide drugs or cheat-sheets in them. It's also disrespectful to the teacher.

Because the teacher wants to make sure you are paying attention.

@chenhsi:
It doesn't work anyway...
there are people in my classes who use their cellphones during class and get away with it. And they sit in the front.
I think TZF is right.

my school is really crappy, i think i even saw a kid with a bottle of beer on the bus and he threw it away and the bus driver didn't even care! but, i was really sleepy so i don't know... but the quality of education and the people there is dreadful. I HATE MY SCHOOL, ALL I WANT TO DO WITH IT IS SEE IT IN MY REAR VIEW MIRROR :mad:

oh i got one!! no costumes on halloween. so some seniors dressed in white for some bull fight after school and they wrote down all their names cuz they MIGHT be involved...

In response to someone: Haha my school is tiny too, all the teachers and students (mostly) know you. My classes, usually 40 minutes long, rarely are very content-filled. Especially Latin/Greek--I absolutely LOVE our classics department. We learn a lot pretty quickly and goof off for the whole class.

[quote="Poincare"]i absolutely hate my school. The people that go there aren't the best kind. and, the lvl of education is not very high. i wish i had never moved.[/quote]
i wish i never moved from my downtown chicago apartment either! :mad:

Entirely off topic: Its nearly impossible to get a separate house in big cities like Chicago....

[quote="Poincare"]i absolutely hate my school. The people that go there aren't the best kind. and, the lvl of education is not very high. i wish i had never moved.[/quote]

I was pretty good at math in elementary school, then I went to my middle school and got really good at math, (at least compared to the other kids at my school), and then I came to this new school that isn't magnet, and now I am in a math class with 9th and 10th graders, (my school is every grade up to 12th grade) when I am just a 7th grader. Don't get me wrong, the highschoolers are pretty cool.  :lol: It is just a little awkward, because, I began the school year in an 8th grade math class, and finally when we started to get to know eachother, I got moved up.  :D


geometry
integration
symmetry
analytic geometry
calculus
calculus computations

Find the area between the curves

$ y \equal{} x^2 \minus{} a$
$ y \equal{} \minus{}x^2 \plus{} a$
$ x \equal{} y^2 \minus{} a$
$ x \equal{} \minus{}y^2 \plus{} a$

[quote="mathwizarddude"]Find the area between the curves

$ y \equal{} x^2 \minus{} a$
$ y \equal{} \minus{} x^2 \plus{} a$
$ x \equal{} y^2 \minus{} a$
$ x \equal{} \minus{} y^2 \plus{} a$[/quote]
It will be very helpful if there is a restriction for $ a$. Is there any?

first assume $ a > 0$ else we get no intersection

then we look to the intersection in the first quadrant and get

$ x \equal{} \minus{} y^2 \plus{} a$ and $ y \equal{} \minus{} x^2 \plus{} a$

$ x \equal{} \minus{} ( \minus{} x^2 \plus{} a)^2 \plus{} a \to x \equal{} a \minus{} x^4 \plus{} 2ax^2 \minus{} a^2$

$ x^4 \minus{} 2ax^2 \plus{} x \plus{} (a^2 \minus{} a) \equal{} 0, (x,y) \equal{} (\alpha, \beta)$

$ \alpha \equal{} \minus{} \beta^2 \plus{} a$, $ \beta \equal{} \minus{} \alpha^2 \plus{} a$

$ A_1 \equal{} \int_0^\alpha \sqrt {a \minus{} x} dx \plus{} \int_\alpha^{\sqrt a} a \minus{} x^2 dx$

$ \equal{} \frac {2}{3}[a^{\frac {3}{2}} \minus{} (a \minus{} \alpha)^{\frac {3}{2}}] \plus{} \frac {2}{3}a^{\frac {3}{2}} \minus{} a\alpha \plus{} \frac {1}{3}\alpha^3$

$ \equal{} \frac {1}{3} \left( 4a^{\frac {3}{2}} \plus{} \alpha(\alpha^2 \minus{} 3a) \minus{} 2 (a \minus{} \alpha)^{\frac {3}{2}} \right)$

$ \equal{} \frac {1}{3} \left( 4a^{\frac {3}{2}} \minus{} \alpha (\beta\plus{}2a) \minus{} 2 \beta^3 \right) \equal{}$

symmetry tells us that the total area is then

$ \frac {4}{3} \left( 4a^{\frac {3}{2}} \minus{} \alpha (\beta\plus{}2a) \minus{} 2 \beta^3 \right)$

where calculating alpha and beta is something doable

[quote]irst assume $ a>0$ else we get no intersection [/quote]
what about $ \frac{1}{4}\leq{a}<0$?
is the picture in the attachment wrong?

And secondly for $ a > 0$ we have the picture below. For which area we are talking about?

[quote="kenn4000"]how do you make those graphs, is it a diff program or something? maybe a website i can peruse?[/quote]
I've made them with a program named just "graph".
"Graph is an open source application used to draw mathematical graphs in a coordinate system."
You can download it from here [url]http://www.padowan.dk/graph/[/url], although it is for Windows platforms only.


1. Find the value of tan^2 1+ tan^2 3+...tan^2 (2n-1)...+tan^2 89.

2.Find the ordered pairs x and y where x and y are real numbers such that x and y satisfy the two equations below.

          (1/sqrt x)+(1/sqrt y)=(x+3y)(3x+y)
          (1/sqrt x)-(1/sqrt y)=2 (y^2-x^2)

[b]#1.[/b]

I’ll use a trick I’ve used before; it’s rather clumsy and inefficient, but it works. :P

Note that $ \tan^2{x^\circ} \equal{} \cot^2{(90 \minus{} x)^\circ}$; thus the sum is the same as $ \cot^2{1^\circ} \plus{} \cot^2{3^\circ} \plus{} \ldots \plus{} \cot^2{89^\circ}$. I will sum this instead – or rather, I will sum $ \cot^2{\frac {\pi}{180}} \plus{} \cot^2{\frac {3\pi}{180}} \plus{} \ldots \plus{} \cot^2{\frac {89\pi}{180}}$, working in radians.

Let $ \frac {x \plus{} \mathrm{i}}{\sqrt {x^2 \plus{} 1}} \equal{} \cos{\theta} \plus{} \mathrm{i}\sin{\theta}$, where $ x\in\mathbb{R}$.

Then $ \frac {(x \plus{} \mathrm{i})^{90}}{(\sqrt {x^2 \plus{} 1})^{90}} \equal{} \cos{90\theta} \plus{} \mathrm{i}\sin{90\theta}$.

$ \cos{90\theta} \equal{} 0\ \Rightarrow\ 90\theta \equal{} \frac {\pi}{2},\frac {3\pi}{2},\ldots,\frac {89\pi}{2}$

$ \color{white}.\hspace{17mm}\color{black}\Rightarrow\ \theta \equal{} \frac {\pi}{180},\frac {3\pi}{180},\ldots,\frac {89\pi}{180}$

$ \color{white}.\hspace{17mm}\color{black}\Rightarrow\ x^2 \equal{} \cot^2{\theta} \equal{} \cot^2{\frac {\pi}{180}},\cot^2{\frac {3\pi}{180}},\ldots,\cot^2{\frac {89\pi}{180}}$

Now $ x$ is a root of $ \mathrm{R}(x) \equal{} \mathrm{Re}\left((x \plus{} \mathrm{i})^{90}\right) \equal{} x^{90}\minus{}4005x^{88}\plus{}\ldots\minus{}1 \equal{} 0$. Since $ \mathrm{R}(x)$ has only even powers of $ x$, $ x^2$ is a root of $ \mathrm{R}(\sqrt {x}) \equal{} x^{45} \minus{} 4005x^{44} \plus{} \ldots \minus{} 1 \equal{} 0$. Hence the sum of the roots $ x^2$ is equal to the negative of the coefficient of $ x^{44}$, i.e.

$ \cot^2{\frac {\pi}{180}} \plus{} \cot^2{\frac {3\pi}{180}} \plus{} \ldots \plus{} \cot^2{\frac {89\pi}{180}}\ \equal{} \ 4005$

That’s the best (or worst) I can do so far. All right, someone please put me out of my misery. Someone please show me the relevant telescoping formula that should be used in this problem which I can’t for the life of me discover. :oops:


Circle $ C_{1}$ is internally tangential to circle $ C_{2}$, touching it at $ S$. The tangent $ UTV$ to $ C_{1}$ cuts $ C_{2}$ at $ U$ and $ V$ and is tangential to $ C_{1}$ at $ T$. Show that angle $ UST \equal{}$ angle $ TSV$.

[hide="Hint"]

Produce $ UTV$ to meet tangent through $ S$ at $ W$.

[/hide]

[hide=" Further hint"]

Then triangle $ STW$ is isosceles.

[/hide]

Let $ X$ be a point on the tangent $ WS$ on the other side of $ S$ to $ W$.

angle $ WSV$ = angle $ SUV \equal{} \alpha$, say (alternate segment theorem).

Let angle $ SVT \equal{} \beta$, angle $ VST \equal{} \gamma$ and angle $ UST \equal{} \delta$.

angle $ XSU \equal{}$ angle $ SVT \equal{} \beta$ (alternate segment theorem).

$ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} 180^\circ$ (straight line $ XSV$). (*)

Triangle $ WST$ is isosceles (equal tangents $ WS$ and $ WT$).

Therefore angle $ STV \equal{} \alpha \plus{} \gamma$.

Therefore $ \alpha \plus{} \gamma \plus{} \beta \plus{} \gamma \equal{} 180^\circ$ (angle sum in triangle $ SVT$). (**)

Therefore, from (*) and (**), we have $ \gamma \equal{} \delta$, that is, angle $ UST \equal{}$ angle $ TSV$.

Is there a neater way of doing this?

This is what the figure looks from your information.
[geogebra]8cb79ce7b4f9cdfd054a4c0bad2c898f1193c4ad [/geogebra]

I have been trying to somehow show that $ \frac {SU}{UT} \equal{} \frac {SV}{VT},$ if that is true, then by the angle bisector theorem, $ \angle UST \equal{} \angle TSV.$ Or alternately, we can also show that arc UD = arc DV.

Invert about $ S$. This gives a diagram of a point $ S$ on a circle with $ C_1$ a line tangent to the circle at $ T$ and $ C_2$ a line parallel to $ C_1$ intersecting the circle at $ U$ and $ V$. Show that $ \angle UST \equal{} \angle VST$.

Draw the perpendicular line from $ T$ to $ C_2$. Let the point where this line meets $ C_2$ be $ P$. Since $ C_1$ and $ C_2$ are parallel, this line is also perpendicular to the tangent at $ T$ and so passes through the centre of the circle. This means that the chord $ UV$ is bisected by the perpendicular. Now it is a simple matter to show that triangles $ UTP$ and $ VPT$ are congruent and so $ UT\equal{}VT$. Since equal chords subtend equal angles we have the result.

Thanks limac and KrazyFK.

KrazyFK, your proof looks nice. Unfortunately I don't know anything about inversion but I think I'd better find out.

Here's a site with plenty of information:

http://whistleralley.com/inversion/inversion.htm

Thanks KrazyFK. I've also got a copy of Gerry Leversha's Crossing the Bridge, so have no excuse not to look at chapter 13: Inversion.


geometry
geometric transformation
reflection

If the upper-right hand corner of a $8\frac{1}{2}\times 11$ piece of paper is folded onto the lower-left hand corner, how far does the upper-left hand corner travel?

Well, it depends how fast you're running while you do it ...

[hide="Hint"]"Folding" is just a reflection across a perpendicular bisector.[/hide]


calculus
integration
function
calculus computations

Let $ f$ be a continuous function on the interval $ [0,k]$ such that $ f(x)\plus{}f(k\minus{}x) \ne 0$ for all $ x \in [0,k].$
Find $ \int_0^k \! \frac{f(x)}{f(x)\plus{}f(k\minus{}x)} \, dx.$

$ \int_0^k \frac{f(x)}{f(x)\plus{}f(k\minus{}x)}dx\equal{}\int_0^k \frac{f(k\minus{}y)}{f(k\minus{}y)\plus{}f(y)}dy$

$ \int_0^k 1dx\equal{}k$ so we get $ \frac{k}{2}$

[quote="-Elixir-"]Let $ f$ be a continuous function on the interval $ [0,k]$ such that $ f(x) \plus{} f(k \minus{} x) \ne 0$ for all $ x \in [0,k].$
Find $ \int_0^k \! \frac {f(x)}{f(x) \plus{} f(k \minus{} x)} \, dx.$[/quote]

$ \boxed{\int_0^k \! \frac {f(x)}{f(x) \plus{} f(k \minus{} x)} \, dx \equal{} \int_0^\frac {k}{2} \! \frac {f(x)}{f(x) \plus{} f(k \minus{} x)} \plus{} \frac {f(k\minus{}x)}{f(k\minus{}x) \plus{} f(k \minus{}(k\minus{}x))} \ dx\equal{}\int_0^\frac {k}{2} \! 1 \ dx\equal{}\frac {k}{2}}$


function
real analysis
real analysis unsolved

Find the form of the function f , if

           $f '( \frac{x+y}{2}) = \frac{f(x) - f(y)}{x - y}$ , for all  $x, y$ with x different from y.

  I suspect it is a two degree polynomial(or less) but how?

[u]Babis[/u]

Really many thanks !!! Very nice solution !

         [u]Babis[/u]

This problem was on some competition....
New-York 1977 i think...


induction
graph theory
algebra proposed
algebra

The even positive integers.

If $ n$ is good then there exists  a permutation $ \sigma$ with the desired property.
Consider the graph of $ \sigma$ that is the bipartite graph whose sets of vertices are the positive integers from $ 1$ to $ n$, two being connected, say $ a$ and $ b$, iff $ \sigma (a) \equal{} b$.
The edges connecting two integers with the same parities use as many even (resp. odd) integers in both sets. This there are the same number of even integers $ x$ such that $ \sigma (x)$ is odd than odd integers $ x$ such that $ \sigma (x)$ is even.
It follows that the numebr of $ x$ such that $ x$ and $ \sigma (x)$ have opposite parities is even.
Thus, the sum of all the $ |\sigma(x) \minus{}x|$ is even, so that $ n$ is even.

In the other hand, if $ n\equal{}2p$ then consider the permutation $ \sigma$ such that $ \sigma(1)\equal{}p\plus{}1$ and $ \sigma(p\plus{}1)\equal{}1$ and $ \sigma(x) \equal{}x$ in all the other cases. It is easy to verify that $ \sigma$ has the desired property so that $ n$ is a solution of the problem.

Pierre.

Ooops... :blush:  I think I misread the problem, I tought there was a sum....

Pierre.

[quote="pohoatza"]Let $ n$ be a nonzero positive integer. Find $ n$ such that there exists a permutation $ \sigma \in S_{n}$ such that
\[ \left| \{ |\sigma(k) \minus{} k| \ : \ k \in \overline{1, n} \}\right | = n.\]
\]
[/quote]
just an idea, not the complete solution. :roll:
if $ n$ is solution then $ 3n\plus{}1$ is solution so $ 3^{p}n\plus{}\frac{3^p\minus{}1}{2}$ is solution.
$ \exists f\in S_{n}: \ \{ |f(k) \minus{} k| \ : \ k \in \overline{1, n} \}\equal{}\{0,1,2,...,n\minus{}1\}$
we take $ F\in S_{3n\plus{}1}$ such that $ \forall x\in\{n\plus{}2,n\plus{}3,...,2n\plus{}1\}: \ F(x)\equal{}f(x\minus{}n\minus{}1)$ and $ F(x)\equal{}x$ in all the other cases
we take $ g\equal{}Foh\in S_{3n\plus{}1}$ such that $ \forall x\in\{n\plus{}1,n\plus{}2,...,2n\plus{}1\}: h(x)\equal{}x$ and:
$ h(x)\equal{}(3n\plus{}1)\minus{}x\plus{}1$ if $ x\le n$
$ h(x)\equal{}(3n\plus{}1)\minus{}x\plus{}2$ if $ x\ge 2n\plus{}2$
$ h(n\plus{}1)\equal{}1$
$ \{ |g(k) \minus{} k| \ : \ k \le n\plus{}2 \}\equal{}\{n,n\plus{}2\plus{}2k: \ k\equal{}0...n\minus{}1\}$
$ \{ |g(k) \minus{} k| \ : \ k \ge 2n\plus{}1 \}\equal{}\{n\plus{}1\plus{}2k: \ k\equal{}0...n\minus{}1\}$
$ \{ |g(k) \minus{} k| \ : \ k \in\{n\plus{}2,...,2n\plus{}1\} \}\equal{}\{0,1,2,...,n\minus{}1\}$
then $ |\{ |g(k) \minus{} k| \ : \ k \in\{1,2,...,3n\plus{}1\} \}|\equal{}|\{0,1,2,...,3n\}|\equal{}3n\plus{}1$

for exemple $ n\equal{}4$ is solution for $ f\equal{}(1\ 4\ 2)$ then $ n\equal{}3^{n}4\plus{}\frac{3^n\minus{}1}{2} is solution$

I hope I do not misread again...

More precisely, we must have $ n \equal{} 0,1 \mod[4]$.

First, let $ \sigma$ be a permutation with the desired property.
Let $ d_i \equal{} |\sigma (i) \minus{} i |$ for each $ i$.
Note that $ 0 \leq d_i \leq n\minus{}1$, thus the $ d_i$'s give exactly the values $ 0,\cdots, n\minus{}1.$
It follows that $ \sum_i d_i \equal{} \frac{n(n\minus{}1)}2$.
We want to prove that this sum is even.

- If $ n$ is even, there are $ \frac n 2$ even integers and $ \frac n 2$ odd integers in $ \{1,\cdots , n \}$.
Let $ p$ be such that there are $ \frac n 2 \minus{} p$ even integers $ i$ such that $ \sigma(i)$ is odd. Since there are $ \frac n 2$ odd integers $ \sigma(i)$, it remains $ p$ odd integers $ i$ such that $ \sigma(i)$ is also odd. It follows that there are $ \frac n 2 \minus{} p$ odd integers $ i$ such that $ \sigma(i)$ is even and $ p$ even integers $ i$ such that $ \sigma(i)$ is even.
It follows that $ \sum_i d_i \equal{} n\minus{}2p \equal{} 0 \mod[2].$

- If $ n$ is odd, there are $ \frac {n\minus{}1} 2$ even integers and $ \frac {n\plus{}1} 2$ odd integers in $ \{1,\cdots , n \}$.
As above, let $ p$ be such that there are $ \frac {n\minus{}1} 2 \minus{} p$ even integers $ i$ such that $ \sigma(i)$ is odd. Then there are $ p\plus{}1$ odd integers $ i$ such that $ \sigma(i)$ is also odd. It follows that there are $ \frac {n\plus{}1} 2 \minus{} (p\plus{}1)$ odd integers $ i$ such that $ \sigma(i)$ is even.
It follows that $ \sum_i d_i \equal{} (\frac {n\minus{}1}2\minus{}p ) \plus{} (\frac{n\plus{}1}2 \minus{}(p\plus{}1))\equal{}n\minus{}1\minus{}2p\equal{} 0 \mod[2].$

Now, for $ n\equal{}4$ we may consider the permutation $ 4,1,3,2$.
And for $ n\equal{}5$, we have the 'good' permutation $ 5,2,4,1,3$.
I think, we may probably construct good permutations in each cases by induction, as done by aviateurpilot...

Pierre.

hi [b]pbornsztein[/b], you konw that $ x\equiv |x|(mod\ 2)$
so $ \frac{n(n\minus{}1)}{2}\equal{}\sum_{i}d_i\equiv \sum_{i} f(i)\minus{}i (mod\ 2)\equiv \sum_if(i)\minus{}\sum_ii(mod\ 2)\equiv 0(mod\ 2)$

[quote="shfdfzhjj"][quote="pbornsztein"]I hope I do not misread again...

I think, we may probably construct good permutations in each cases by induction, as done by aviateurpilot...

Pierre.[/quote]

Can you complete your proof？
I don't think the left is easy. :)[/quote]

yes , I think so .

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=43&t=448845&p=2526198#p2526198]This link[/url] redirects here, which is locked.

This was the problem of RMO 2011, which held today only ( 4 december according to indian time)

The user [b]solo1729[/b] is a cheater...

Sorry for reviving, but does anyone has a inductive construction? I‘ve tried but found it hard to accomplish.


Mafia
Support
rotation
videos
blogs
algebra
binomial theorem

There are currently too many RPG-style games going on, and this forum is legendarily overflowing. Thus, we're probably going to need to start limiting the number of forum games occurring at one time. From now on, if you wish to moderate a game, put your name on this waiting list, and you will be notified when a previous game has ended so you can start your game.

Currently running games:
-The Legend
-The Legend of the Drakon
-The Legend of the Black Dragon
-ATTOS
-Synthetica
-AntiRPG
-Quest of a Wisp
-Kill Code II
-HCGTOC
-TLOAVP (pending)
-WOTH (pending)

[quote="PhireKaLk6781"]Currently running games:
-The Legend (isabella2296)
[b]-The Legend of the Drakon (Butler)[/b]
[b]-The Legend of the Black Dragon (piegraph)[/b]
-ATTOS (PhireKalk6781)
-Synthetica (james4l)
-AntiRPG (Eggylv.999)
-Quest of a Wisp (meewhee009)
-Kill Code II (dragon96)
-HCGTOC (abcak)
[b]-TLOAVP (Butler)[/b]
-WOTH (musicnmath)
[b]-Clans of Arthera (piegraph)[/b][/quote]

Organized.

Forum Game starters: Please put the entire name of the Forum Game in the main Title and any abbreviations in the description.

I'd personally rather not start another CTC in the Games Forum. Any other possible alternatives?

Honestly, the RPG trend will die out itself, in my opinion.

hey there is another RPG you forgot.
It is Clans of Arthera.

To be fair, Synthetica is only 1/5th RPG, and contains dozens of other elements. (it's also somewhat of a satire of some browser games and some AoPS forum games). It is also pending to be started

Also, perhaps we limit people to 1 game-mod at a time.

[quote="james4l"]Also, perhaps we limit people to 1 game-mod at a time.[/quote]

I think dragon96 was hinting at that with the bold.  :wink:

I'm starting today, for TLOAVP. its full name is The Legacy On Alkan Vichi Preedator

[quote="piegraph"]well i am modding 2 games.
clans of arthera hasn't reallly started so can i put a dead line for that and quit legend of the black dragon on that same day?[/quote]

That sure makes people currently in LOBD happy...:P

hmmmmmmmmmm
once it dies out.
and all the people in legend of the black dragon are in clans of arthera bro.
*beats up his bro*

um whats wrong with having too many games? the more the merrier

Remember what happen to ahirbhairav before he was banned? I suggest doing something similar to this "hoidori".

It was so bad I actually heard about it on a blog first. ( ;) mathman ).

Currently  there are actually too many games for me to keep up with. I can't put everything onto the list now. I have to watch it for a while. [i] Most [/i] of the games are good though :wink:

[quote="goodguy888"]I will post what I said in "Territorial Conquest" here, as I think all AoPsers should have a say in the future of the Civ game. I'm sorry if you read the same thing twice.

Hello. I'm goodguy888. You might know me from somewhere, be it video games to AoPS. Anyway, on with these Civilization games.

Personally I'm not against another story chain. I know that I cannot be the sole proprietor of the map-based games. I wouldn't mind playing in such a game myself... (May soon. DunDunDuh...) [i][b] However, [/b][/i] My advice is this:

1) The first of this class of games has not been completed yet. [b][i]We don't know how well the reception will be or if this class of game will even take off.[/i][/b]
2) [i][b]These are extremely complex games[/b][/i]. A number of people have thought about starting such a game, I feel the urge to say something. [b] No Offense [/b] , But what some are suggesting, is a game that is even more complex than Civ 5 in some ways, which, if you will allow me to say this, is a very [i] interesting [/i] decision. I feel the need to dissuade you from such a goal. Why? Not because I want to control the Games Forum. It is because I see how busy it is to mod a slightly more complex game such as mine, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=535&t=564363]Paradyce[/url]. I can forsee a very busy job for the mod in the endgame, which is why I'm recruiting real-life-friends to help me out. Now, Already I am spending at least 30 minutes a day trying to keep up with the game, PMing players, etc. This will increase much. What you are suggesting will result in an overwhelming workload each day, or a very slow-paced game. Just some friendly advice here: If you want to be very complex, [i] save it till summer. [/i] Do you have about 1 to 2 hours [b] every [/b] day to mod this complex of a game? For most people, the answer would be no.
3) [i][b] Three of this class of game running at the same time would be suicide for all the games. [/i][/b] From what I see, there is a limited player base for Civ games on AoPS. There will not be enough players for 3 games to run at the same time. Therefore, it isn't very wise to start 3 of them at the same time.

[i][b]What is my suggestion?[/b][/i] Well, considering the situation, I suggest one single chain of Civ game, where they run in turn, one at a time. Those interested in modding such a game could take turns being co-mod and mod for a complete game. This way, not only will we reduce the workload of mods, we will also avoid imminent death by player base. If Paradyce succeeds, I am planning another game in the storyline, albeit not exactly a Civ game. We'll see what happens to Paradyce. And we will chat again then.

goodguy out.[/quote]
1) Agree.
2) This is all correct( for me, the answer is yes since I'm homeschooled :rotfl: but these are still hard to mod)
3) "Suicide" is not the best term. It's just that the Games forum would crash, especially with TDS and TDC going on right now.

Suggestion) This sounds too much like there should be a whole forum for these.

[quote="fz0718"]Perhaps one could automatically process the players' actions into the game?[/quote]
Players would have to send in exact posts that could be read by a program.
[quote="chaotic_iak"]Clearly the best action is to stop making too many similar games running at once.[/quote]
Exactly. Already there are almost 30 games in the games forum currently, an all-time-high for my career as game-list-keeper.

[quote="goodguy888"][quote="fz0718"]Perhaps one could automatically process the players' actions into the game?[/quote]
Players would have to send in exact posts that could be read by a program.[/quote]Precisely my point. Or you could just automatically update the civ-board :wink:

[quote="goodguy888"][quote="chaotic_iak"]Clearly the best action is to stop making too many similar games running at once.[/quote]
Exactly. Already there are almost 30 games in the games forum currently, an all-time-high for my career as game-list-keeper.[/quote]Your [b]short[/b] career as game list keeper...

[quote="fz0718"][quote="goodguy888"]
Players would have to send in exact posts that could be read by a program.[/quote]Precisely my point. Or you could just automatically update the civ-board :wink:
[/quote]
The civ-board would have to be in the program also. Additionally, I do not own any of this software needed (Though my friend can get it for me :wink:)

Precisely my point. Create a program to simulate a civ-board. Simple enough and saves you tons of time.

Then again, AoPS is based on a forum site. And if we want the games forum to be like this, ask the developer, or Paul H. Ryu.


[quote="phuc_Math_4ever"]Let's x,y that : 2(x^2+y^2)=xy + 1
Prove that:
18/25 <= 7(x^4+y^4)+x^2*y^2 <= 70/33.
Thanks all of you very much.[/quote]
Try $ x\equal{}\frac{1}{\sqrt5}$ and $ y\equal{}\minus{}\frac{1}{\sqrt5}.$


Let $ y \equal{} \sqrt {2x^2 \plus{} 5}$ (actually, go ahead and replace the $ 2x^2 \plus{} 5$ with any expression in x, it's just an example). To solve for x in terms of y, we would first square both sides to get $ y^2 \equal{} 2x^2 \plus{} 5$, when y is nonnegative. Instead of writing all that, would it be useful (I may be totally wrong on this) to just use something that indicates that we are squaring only the principal root?

For example, we could say $ y^{*2}$ means only the positive/principal y is squared, to avoid confusion in domain/range problems, so we would get $ y^{*2} \equal{} 2x^2 \plus{} 5$. After we finish solving, (it ending up being $ x \equal{} \pm \sqrt {\frac {y^{*2} \minus{} 5}{2}}$), when we examine the domain (in a domain/range problem, eg. to check an answer), we won't get confused with all the range stuff because the $ ^*$ indicates y has to be positive. Later on, if we get a negative y, we can toss it away.

(For example, geometry is limited to positive units, although that comes intuitively).

Just wondering... would this be useful to anyone?

I think it's easiest just to state which solution you want at the end, and why. There's no standard notation for this.


Iran 1997/3:
"Suppose f: R+->R+ is a decreasing continuous function such that for all xeR+,
f(x+y) + f(f(x)+f(y)) = f(f(x+f(y))) + f(y+f(x))
Prove that f(f(x))=x."

A solution is given here.
http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html
on the 1997-1998 file.

However, the problem seems to be inconsistent.

(1) Suppose there exists a such that a=f(a).
Put a=x=y:
f(a+a) + f(f(a)+f(a)) = f(f(a+f(a))) + f(a+f(a))
f(2a) + f(2f(a)) = f(f(a+a)) + f(a+a)
f(2a) + f(2a) = f(f(2a)) + f(2a)
So f(2a) = f(f(2a)).
f is decreasing, so it is surjective.
2a = f(2a)
But f(2a)<f(a) since f is decreasing, so 2a<f(a)=a. Contradiction.

(2) So for all x, f(x)<x.
So f(1)<1
f(f(1))<f(1), so f(f(1))<1

Thus we have found a counterexample.
(It is probably possible to show that f(f(x)) does not exist in positive reals)

Yes,It is our's.And your counterexample has a little mistake,See here for two other proves.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=260589#p260589

PS>i Still hold my idea about ,Being hard.
Ali`s solution doesnt make sense to me.


Given four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice.

:huh:
just divide them into 2 groups, and weight twice!! ;)

[hide="Proof"]

Let the four objects be $a,ar,ar^2,ar^3$ where $r$ is the common ratio greater than 1 (since the problem has the heaviest object).

There are 3 cases to investigate.

Case 1: $[a,ar], [ar^2,ar^3]$

$[x], [y]$ means that $x$ is on the left side of the balance and $y$ is on the right side.

$a+ar \ ? \ ar^2+ar^3$

$1+r \ < r^2+r^3$ since $r > 1$.

So, after choosing right side and using balance, it is easy to determine which is heavier by comparing $ar^2$ and $ar^3$. So, case 1 satisfies the problem.

Case 2: $[a,ar^2], [ar,ar^3]$

$a+ar^2 \ ? \ ar+ar^3$

$1+r^2 \ ? \ r+r^3$

$1 < r$ which is what problem stated. Then using the balance one more time gives the largest one. So, this case is proven.

Case 3: $[a,ar^3], [ar,ar^2]$

Assume that $a+ar^3 < ar+ar^2$. Then dividing by $a$ yields $1+r^3 < r+r^2$. Moving $r^3$ to RHS and factoring $r$, we get $1 < r(r+1-r^2)$. But since $r > 1$, the quantity will be negative and the assumption is wrong. So, $a+ar^3 > ar+ar^2$. From here, use one balance one more time to get the largest one. [/hide]

Yeah, that's basically what I did.

[hide="brief outline"]
Let the four weights be $a$, $ar$, $ar^2$ and $ar^3$.
For our convenience, assume $a>1$ and $r>1$.
(This makes $ar^3$ be the heaviest one, but we can use the same reasoning for other cases.)

Then, we can show that $a+ar^3>ar+ar^2$ for all $a$ and $r$ such that $a>1$ and $r>1$:
$a+ar^3=a(r^3+1)=a(r+1)(r^2-r+1)>ar+ar^2=ar(r+1)$

This implies that the pair with the heaviest weight has to be heavier than the other pair. (If $ar^3$ and a weight that is heavier than $a$ are paired up, then obviously the weight of the pair is more than $a+ar^3$, so it has to be heavier than the other pair.)

So, the method is simple: we divide them into 2 pairs and compare their weights. The heavier pair must contain the heaviest weight, so we weigh the heavier pair and we're done.
[/hide]
I didn't take much time to check, so correct me if I made any mistake.


The parenthesis actually arent needed, so we have 3x+5x+7-8 or 8x-1


Let n be a positive integer, and let p be a prime number.Prove that  p^p|n! imlies that  p^(p+1)|n!

If $p^p$ divides $n!$ then there are at least $p$ multiples of $p$ less than or equal to $n$.  But the $p$-th multiple of $p$ is a multiple of $p^2$.  So $p^{p+1} \mid n!$.

Oops I didn't see that before answering in the other forum...

Please do not post the same problem several times!!!!

Pierre.


Let $ a,b,c$ be [b]real[/b] numbers. Prove that
$ 6abc\plus{}2\sqrt{(a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ca\plus{}a^2)} \ge (a\plus{}b)(b\plus{}c)(c\plus{}a).$
When does equality hold?

It's ture because
\[ 4(a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ca\plus{}a^2)\minus{}\left((a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}6abc\right)^2\]
\[ \equal{}3(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\]
Equal holds when $ a\equal{}b$ or$ b\equal{}c$ or $ c\equal{}a$.

[quote="can_hang2007"][quote="Tourish"]It's ture because
\[ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2\]

\[ \equal{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\]
Equal holds when $ a \equal{} b$ or$ b \equal{} c$ or $ c \equal{} a$.[/quote]
Dear friend,
What do you think about a proof without expanding?[/quote]

I didn't expand it at all.I just thought like the following way:
First I let $ a\equal{}b$,then I found $ LHS\equal{}RHS$,and as the inequality is symmetrical,we can deduce that $ a\equal{}b$ or $ b\equal{}c$ or $ c\equal{}a$ can make the equal hold.Hence
$ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2$must heve the factor such that $ (a\minus{}b)(b\minus{}c)(c\minus{}a)$.Moreover,
the inequlity holds for all $ a,b,c$ as you wanted to prove,so I thought the factor $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$ is needed.The following we should do is to find $ K(a,b,c)$ such that
\[ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2\]
\[ \equal{}K(a,b,c)(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\]
But notice that $ deg(LHS)\equal{}6$,so $ K(a,b,c)$ should be a constant.Then I let $ a\equal{}0,b\equal{}1,c\equal{}2$,I find $ K(a,b,c)\equal{}3$,which ended the all prove.
If my thought is wrong,please point it out.And I also want to see a nice solution with only $ C\minus{}S$ and $ AM\minus{}GM$ :)

[quote="Tourish"][quote="can_hang2007"][quote="Tourish"]It's ture because
\[ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2\]

\[ \equal{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\]
Equal holds when $ a \equal{} b$ or$ b \equal{} c$ or $ c \equal{} a$.[/quote]
Dear friend,
What do you think about a proof without expanding?[/quote]

I didn't expand it at all.I just thought like the following way:
First I let $ a \equal{} b$,then I found $ LHS \equal{} RHS$,and as the inequality is symmetrical,we can deduce that $ a \equal{} b$ or $ b \equal{} c$ or $ c \equal{} a$ can make the equal hold.Hence
$ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2$must heve the factor such that $ (a \minus{} b)(b \minus{} c)(c \minus{} a)$.Moreover,
the inequlity holds for all $ a,b,c$ as you wanted to prove,so I thought the factor $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$ is needed.The following we should do is to find $ K(a,b,c)$ such that
\[ 4(a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2) \minus{} \left((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 6abc\right)^2\]

\[ \equal{} K(a,b,c)(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\]
But notice that $ deg(LHS) \equal{} 6$,so $ K(a,b,c)$ should be a constant.Then I let $ a \equal{} 0,b \equal{} 1,c \equal{} 2$,I find $ K(a,b,c) \equal{} 3$,which ended the all prove.
If my thought is wrong,please point it out.And I also want to see a nice solution with only $ C \minus{} S$ and $ AM \minus{} GM$ :)[/quote]
Your thougts are very nice, dear friend. The official solution of mine is to use the Cauchy-Schwarz Inequality. You can view it here (the solution of leedt26 is the same as mine): http://ddbdt.co.cc/forum/showthread.php?t=134


You probably mean Asymptote which allows you to draw diagrams both on the forum and in your LaTeX dcuments. Full details on the  [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]AoPS Wiki[/url].

i downloaded ghostscript but it is a winrar file and i don't know what to do after that. on opening gsview, it says ghostscrript is not yet downloaded on my pc...can u help me?

Assuming you are running Windows (32 bit) go to [url=http://pages.cs.wisc.edu/~ghost/doc/GPL/gpl860.htm]here[/url] and download the self-extracting file labelled
[i]gs860w32.exe, GPL Ghostscript 8.60 for 32-bit Windows (the common variety)[/i]
just under the Microsoft Windows heading near the bottom of the page, then run the file.

[quote=hell_ever]I see many geometry diagrams on this site (generated by users)
can anyone tell me what is this software and where can i get it?

thanks for the help[/quote]

Hi,

I'm not sure where you can get it but I want to share with you a geometry software which I recently use.
It's a Geometry for Mac OS X https://itunes.apple.com/us/app/geometrie/id972453094?mt=12 and it works really well.


Triangle ABC is a right triangle in which AB=66 and BC=77. If hypotenuse AC>50 and AC is expressed in simplified form $ x\sqrt{y}$, then x + y =

a. 13
b. 24
c. 85
d. 96

Is the hypotenuse AC? Then the AC > 50 is unnecessary, since AC must be greater than AB and BC.
AC = sqrt(66^2+77^2) = sqrt(10285) = 11*sqrt(85).

11+85 = 96

D

just to rewrite everything Riders on the Storm wrote, in $ latex$,
Use Pythagoreans therom.
$ 66^{2}+77^{2}=(x\sqrt y)^{2}$

$ %Error. "arrow" is a bad command.
\sqrt{66^{2}+77^{2}}=x\sqrt y$  We can square both sides because $ AC>0$

$ %Error. "arrow" is a bad command.
\sqrt{10285}=x\sqrt y$

$ x\sqrt y=11\sqrt85$

$ 11+85 = 96$

$ D$

whats with the D at the end and yeah the AC>50 part is unessacary. just use the pythagorean theorem. sorry i cant spell

the d is the answer choice. look at the original post.


Consider $ \triangle{ABC}$. Take a point $ H$ on $ \overline{AC}$ such that $ \overline{BH}\perp\overline{AC}$, $ \measuredangle~CBH=3x$ and $ \measuredangle~ABH=5x$. Take a point $ D$ on the altitude such that $ \measuredangle~CAD=x$ and $ \measuredangle~ACD=2x$. Compute $ x$

I've already found a solution with trigo, actually $ x=10^\circ$, but I'd like to see if someone can show a synthetic solution.

Thanks a lot.

[color=darkred][b]Sorry, I am also a trigonometrical proof ![/b][/color]

$ \left\{\begin{array}{c}m(\widehat{BAD})=90-6x\\\ m(\widehat{BCD})=90-5x\end{array}\right\|$ $ \implies$ $ \cos 6x\sin 3x\sin 2x=\sin x\sin 5x\cos 5x$ $ \implies$ $ \cos 6x(\cos x-\cos 5x)=\cos 5x(\cos 4x-\cos 6x)$ $ \implies$ $ \cos 6x\cos x=\cos 5x\cos 4x$ $ \implies$ $ \cos 7x+\cos 5x=\cos 9x+\cos x$ $ \implies$ $ \cos 7x-\cos 9x=\cos x-\cos 5x$ $ \implies$ $ \sin 8x\sin x=\sin 3x\sin 2x$ $ \implies$ $ \sin 8x=2\sin 3x\cos x$ $ \implies$ $ \sin 8x=\sin 4x+\sin 2x$ $ \implies$ $ 2\sin 4x\cos 4x=2\sin 2x\cos 2x+\sin 2x$ $ \implies$ $ 4\cos 2x\cos 4x=2\cos 2x+1$ $ \implies$ $ 2(\cos 6x+\cos 2x)=2\cos 2x+1$ $ \implies$ $ 2\cos 6x=1$ $ \implies$ $ \cos 6x=\frac{1}{2}$ $ \implies$ $ 6x=60$ $ \implies$ $ \boxed{x=10}$.

Thanks Virgil Nicula

I actually have the same proof.

I hope you or somebody can get a synthetic proof, it'd be great.

Best regards.

P.S.: you can use $ \implies6x=60$, the "implies" shows entirely  :)

[hide][img]http://img412.imageshack.us/img412/3701/wdff7.png[/img]
[b]Pure geometry[/b]
Take simetry of ABE triangle according to [AB] and change the place of m(EAB) &
m(DEB) according to ceva theorem.Now we try to solve AEC triangle.
[img]http://img299.imageshack.us/img299/6228/t45h45j65juv8.png[/img]
1-Extend [EA] such that EFC is isosceles.If we write the angles we see that AFC is isosceles.
2-Draw height [EG].And [EG] divides [FC] into equal two parts.
3-m(DGC)+m(DBC)=180 therefore BDGC is a cyclic.Then draw [BG] so m(DBG)=4a.
4-Now draw the centroid of ABC triangle and join H and G points.ABC is right angled triangle and AFC is isosceles.Therefor |BH|=|HC|=|GC|.
5-m(HCG)=2a, m(HBG)=a |HC|=|GC| so B,H,G points are on the same circle.C is center of this circle.Therefor  |BC|=|HC|=|GC| are radius of this circle.Finally we find that HBC is equaliteral triangle.6a=60 a=10.
[img]http://img410.imageshack.us/img410/8185/f43f34gkt8.png[/img][/hide]


function
Support
real analysis
parameterization
integration
topology
abstract algebra

Hi,

this is a rather technical question, so I will try to explain it :?

Let $\mathcal{D}(\mathbb{R}^n)$ be the set of infinitely differentiable functions with compact support in ${\mathbb{R}^n}$

Now testfunctions are linear continuous maps to $\mathbb{C}$ on these testfunctions

Suppose on needs the convolution $S*T$

now this is defined in this way $<S*T,\phi> =<S_x,<T_y,\phi(x+y)>>$

Now my question is , the function
$\psi(x,y)=\phi(x+y)$ does not have compact support, even if $\phi$ has

how to get rid of this technical difficulty?

My professor claims all can be solved if $[S]$ or $[T]$ are compact supports (just one of them suffices)

The main difficulty is assuring yourself that the resulting convolution is in fact still a distribution : linearity is always easy of course but boundedness isn't

Click here for definition of boundedness of linear functional $T$ on $\mathcal{D}$

[hide]for every compactum $K$ there are constants $C$ and $k$ such that for
all $\phi\in \mathcal{D}(K)\subset\mathcal{D}$ one has :

$|<T,\phi>| \leq C sup_{x\in K}sup_{\alpha\leq k} |D^{\alpha} \phi(x)|$[/hide]

Sorry for the very technical question, it is NOT homework but it is examrelated, I just don't think the proof in my syllabus really solves all technicalities. :oops:  :ninja:

That $\phi(x+y)$ should be viewed as a function of one variable at a time. That $\langle T_y,\phi(x+y)\rangle$ is $T(\phi_x)$, where $\phi_x(y)$ is $\phi(x+y)$. In all of this, $x$ is a constant parameter, and each $\phi_x$ is compactly supported.
Then, we read $T(\phi_x)$ as a function of $x$. If $T$ is compactly supported, this function of $x$ will be compactly supported, and will be a valid input for $S$.

But there is still a problem. One cannot define the convolution of two arbitrary distributions. One can define the convolution is at least one [i]of the distributions[/i] itself has compact support.

Suppose that $T$ has compact support. Then $\langle T_y,\phi(x+y)\rangle$ defines a function of $x$ that has compact support - and we can argue that that function is smooth, hence a test function.

But we can't just convolve any two distribution. Here's a ridiculously simple example: let $f(x)=1$ on $\mathbb{R}.$ That's a perfectly good distribution on $\mathbb{R}.$ Its action is given by $\langle f,\phi\rangle=\int_{\mathbb{R}}\phi.$

But $f*f$ makes no sense.

OK, now I read your post again and your professor was asking for the correct result.

So what jmerry says seems to work , but Kent Merryfield, are you saying you disagree now with jmerry?

Anyway, the weird thing is that he said, the support of $S\otimes T$ is $[S]\times [T]$

Then he said, if you have a general distribution $V$ and a testfunction $\phi$ such that their supports intersect in a compactum, you can still define $<V, \phi>$ by creating a  testfunction $\alpha$,with a compact support, and 1 in an open set around the compact intersection of $V$

Then you define $<V, \phi> = <V,\alpha \phi>$

Advantage of this : you do not (strictly)need  that one of them has a compact support, in facht, if for every compactum $K\subset\mathbb{R}^n$ this is true :

$[S]\cap (K-[T])$ is compact and $[T]\cap( K-[S])$ is compact

then you can do it too (which immediately implies you can convolve the heaviside distributions)

Disadvantage : what a weird definition?  Suppose I would want to check continuity (with the definition I gave in the first post here), well, what to do with that $\alpha$
It changes, it depends on $\phi$
How can I estimate now? :ninja:

Wait wait!

let us agree that for every set $K\subset\mathbb{R}^n$
$K^c=\{(x,y)|x+y\in K\}$

Could it be that  , for any (closed )sets $[S]$ and $[T]$ this is equivalent? :

1. for every compactum $K$    $[S]\cap (K-[T])$ is compact
2. for every compactum $K$   $[T]\cap (K-[S])$ is compact

3. for every compactum $K$ : $K^c \cap ([S]\times [T])$ is compact

So if I assume this, and not that one of them has compact support (which is stronger), could I solve everything now, and still define a distribution?

Howver, at one point in the process I have to apply a distribution with possible noncompact support on a infinitely differentiable function that is possibly without compact support

 :?

[quote="fredbel6"]So what jmerry says seems to work , but Kent Merryfield, are you saying you disagree now with jmerry?[/quote]
There is no conflict. Look at my last line- this process requires $T$ to have compact support. If $S$ has compact support instead, it still works- we're just using a larger space of test functions in that step (or that construction involving $\alpha$).

The tensor product is completely different- I don't see what it has to do with this.

? I don't understand.  As I said in my last post, in the most general theory, you don't assume one of them has compact support, you use a weaker assumption.

In our class of functional analysis, the convolution was actually defined by tensor products? :?

$<S* T, \phi > = < S_x \otimes T_y, \phi(x+y) >$


geometry
inequalities
inequalities unsolved

H  is  the  orthocenter  of  a  triangle  ABC .Prove  that
a 2 + b 2  + c 2   \geq (AH+BH+CH) 2

We have:AH=2RCosA;BH=2RcosB;CH=2RcosC and a=2RsinA;b=2RsinB;
c=2RsinC.So you inequality is equal to:
(cosA+cosB+cosC) 2  \leq sin 2 A+sin 2 B+sin 2 C which can be done by Cauchy. :)