There are many things that happen each day that are said to happen for a reason.  Sometimes, even good things come out of really bad things that happen.  These could be examples of Fate.  Do you think fate exists?

Optional:  You can explain why you think Fate exists or not.

Have fun, and include examples of when Fate happened to you if you want. :D:D:D

Well, as a famous spanish poet once said

[quote="Antonio Machado"]caminante, no hay camino se hace camino al andar.[/quote]
That in english it would be, more or less, trying to translate "poetically" too, with the intrinsec sence in the whole poem,
[i]Voyager, there is no fate, you do the fate as you travel[/i]. The fate is what we do from and for our lives. I wasn't born to be a mathematician, I chose to be one! I wasn't born to marry my wife, I chose to marry her, I wasn't born to be who I am right now, I chose to be who I am and not someone else!

Deep...
I thought more people would believe in Fate than not, but it is the other way around as the poll stands right now.  So you think we all choose and make our own decisions?

1)  Don't you think that there might be some force leading us on to a certian direction (that force may I add, would be Fate) and strongly pushing us that way? :)

2)   Do you think that it is possible that people die to keep balance with the people being born? :?

3)  Do you think we can ever get rid of all the problems in our world? ;)

All very thought- provoking questions.

[quote="lightspeed"]Deep...
I thought more people would believe in Fate than not, but it is the other way around as the poll stands right now.  So you think we all choose and make our own decisions?[/quote]Of course there are circumstances on the enviroment that affects the own decisions, but that does not mean that there is a fate. I honestly think that there are many variables when we talk about decisions, and some of the decisions are not even concius, but definitely not fate

[quote="lightspeed"]1)  Don't you think that there might be some force leading us on to a certian direction (that force may I add, would be Fate) and strongly pushing us that way? :) [/quote]Well, yes, I think there are some forces leading us on a certain path, but I don't call it Fate, I call it skills, talent, oportunity, preference, personal opinions, free will, decisions, etc. Not fate!

[quote="lightspeed"]2)   Do you think that it is possible that people die to keep balance with the people being born? :? [/quote]I think you are confusing motive with consecuence. The fact that people that dies open spots for those that are just being born is a consecuence of the fact that those people dies, but it is not the cause (otherwise we probably wold not have that much problem with overpopulation). People dies because of the oxidation in the different tissues on your body, that causes those tissues to get weaker and weaker and ventually not enough strong to mantain you alife (when we talk about death by natural causes), that could be acelerated for some kind of external factor such as a disease. Also, a strong trauma could cause different organs to stop working, that is why people dies, not because of the fate!

[quote="lightspeed"]3)  Do you think we can ever get rid of all the problems in our world? ;) [/quote] That would require that everyone would agree what are the problems in our world and all agree in the strategy to solve it (and all of us be willing to follow such strategy), what is sligly difficoult!

Often I think about Fate, and if it is true or not. Usually after some deliberation, I might do something out of the ordinary, and then see what happens (try it....).

But then again, would the consequences of that be considered still Fate, can I not change this invisible hand that guides me?!?!?! To believe that would be to believe that you are trapped in some ulterior motive. I say to hell with that, I want to be the decider of my future, and if what I do turns out to be my Fate, whatever.

I guess this ties in with religious views, I suppose believing in a supreme deity goes hand in hand with believing that you are part of some elaborate plan.

[quote="mcalderbank"]I guess this ties in with religious views, I suppose believing in a supreme deity goes hand in hand with believing that you are part of some elaborate plan.[/quote]
Not necessarly. Note that it is possible that such a deity created his world in such a way that it has some kind of control over it, but nevertheless, he cannot anticipate everything. Like when you creat a game in a computer, you usually create a "random" generator of conditions or responses of the game to some situations that gives you some uncertainty about the final resoult. Otherwise it could be too boring!

And as it has been discussed on other threads, the fate or destiny theories go against the free will of the humans, that is one of the core concepts in most of the religions.

[quote="mcalderbank"]right.....

Whom believes in Fate and does not believe in God(s)?[/quote]

i don't believe in god but i do believe in fate...

Could you elaborate?

For example, do you believe in Fate such that there is some 'invisible hand" that guides us through life.

[quote="Zratonx"]invisible hands...checking...not there

but i believe things r preset and unchangable...[/quote]So... you don't believe in free will?

[quote="Zratonx"]wut i am saying is that ur free will is governed by something...[/quote]then it is not free will! is it?

wut r u going free will

something up our head that help us make decision...how do u know wut u r doing is coincided with fate...

[quote="Zratonx"]wut r u going free will[/quote]WHAT???, could you please rephrase your statement? (and if possible, please, avoid chatspeak!)

[quote="Zratonx"]something up our head that help us make decision...how do u know wut u r doing is coincided with fate...[/quote]Well, probably if you believe in fate (therefore not in free will!) there is no way to know if it is or not coinciding with fate, but thus, it wouldn't be free will, because the decision was taken before you ever existed!

[quote="Zratonx"]sorry...i was not thinking...

i wanted to say is something like this

wut's ur idea of free will??[/quote]
What is my idea of free will? well, I would say that [b]free will is the power of making free choices that are unconstrained by external circumstances or by an agency such as fate or divine will[/b], even when the words are not originally mine, I agree with such definition. That is why fate and free will are incongruent one with the other!

well then since they are incongruent of each other this talk is useless due the fact we r calling the same thing by two different names....

different ppl believe in certain things due to the external expose they get in this world...

[quote="Zratonx"]well then since they are incongruent of each other this talk is useless[/quote]Why, you may believe in fate but not in free will or viseversa. Any way, Relativity and Quantum Mechanics are incongruent (I don't know that much about phisics, but I believe I heard that somewhere) but any way, people continue studying both of them. I don't see the futility of the discussion!

[quote="Zratonx"] due the fact we r calling the same thing by two different names....[/quote]What two different things are we calling by two different names?  :?

[quote="Zratonx"]different ppl believe in certain things due to the external expose they get in this world...[/quote] :huh: How is that related to our conversation?

[quote="Zratonx"]ppl were not born to believe in god or fate...

It is through their experience that they chose to believe in a certain thing...[/quote]Well, I still don't see what is the relation between that and our conversation. Any way, I do consider that childs are born without any belief at all, and it is experience what influences on that!


An array $ (a_1,a_2,...,a_{13})$ of $ 13$ integers is called $ tame$ if for each $ 1 \le i \le 13$ the following condition holds: If $ a_i$ is left out, the remaining twelve integers can be divided into two groups with the same sum of elements. A tame array is called $ turbo$ $ tame$ if the remaining twelve numbers can always be divided in two groups of six numbers having the same sum.
$ (a)$ Give an example of a tame array of $ 13$ integers (not all equal).
$ (b)$ Prove that in a tame array all numbers are of the same parity.
$ (c)$ Prove that in a turbo tame array all numbers are equal.

a) One example of a tame arraw: $ (1,1,1,1,1,1,1,1,1,1,1,1,11)$ (A generalization for odd $ n \equal{} 2k \plus{} 1$ instead of $ 13$ is: $ (1,1,\cdots,1,2k \minus{} 1)$)

b) Let $ S \equal{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{13}$ then $ 2|S \minus{} a_i$ for all $ i$ then it follows that all $ a_i$ have the  same parity.


logarithms
calculus
calculus computations

Determine if the following series are convergent, absolutely convergent, or divergent:

i)  $ \sum_{2}^{\infty}(\minus{}1)^n\log{(\frac{n\plus{}1}{n})}$

ii) $ \sum_{1}^{\infty}(\minus{}1)^{n\minus{}1}[e\minus{}(\frac{n\plus{}1}{n})^n]$

If you take away the alternating sign, in both cases the terms are limit comparable to a constant times $ \frac1n.$ Furthermore, in both cases, the absolute values of the terms decrease, so the alternating series test applies. Hence in both cases, the answer is conditionally convergent.

The first can be evaluated exactly; it's the logarithm of a [url=http://en.wikipedia.org/wiki/Wallis_product]Wallis product[/url].

[quote="Kent Merryfield"]If you take away the alternating sign, in both cases the terms are limit comparable to a constant times $ \frac1n.$ [/quote]

this is not so clear to me. :maybe:


I just am curious to know how many people are obsessed or atleast like Harry Potter. I know some of you may be thinking , " Who cares!", but for those avid Harry Potter fans, here's a poll you can enter... BTW, do you know why they changed the movie date from July 13th to July 11th?

I'm not allowed to read it?
You must have strict parents to not be able to read Harry Potter.

It's not as crazy as you might think:

http://www.educationworld.com/a_admin/admin/admin157.shtml

http://www.judyblume.com/articles/harry_potter_oped.html

http://www.speakout.com/activism/issue_briefs/1319b-1.html

http://www.factmonster.com/spot/banned-harry.html

Hmmm... My daughter just loved it. It was amazing to see a 7-year old girl reading a 200+ page book non-stop. I'm not so sure about myself. The first 3 volumes were fun to read but then it gradually became more and more boring (in my opinion). Still all children I know seem to enjoy it greatly.

Banning Harry Potter? Good luck! :rotfl:

You won't believe how many people I know who say they can't read Harry Potter because it's against their religion. That's why I added that option. By the way, I love Eoin Colfer. I especially adore the Artemis Fowl series. It's fantasy + logic + coolness all wrapped up into a book! :)

Harry Potter is my life. (Well, not really) I voted the first option obviously. Can't wait for July 21st (as my signature says). Read [u]Mugglenet's What Will Happen in Book 7, Who Lives Who Dies and Who Falls in Love and how will the Adventure Finally End [/u]if you like Harry Potter and care about theories. Shouldn't this be in Games & Fun Factory?

I think J.K. Rowling is a great writer, and I am excited to see the finale to this series. I like the series, but it's not like I camp out near a bookshop or a library to get the latest book. I wish the movies were more loyal to the books, however. Just a fun fact, the cover is gold with red lettering.

One thing I especially love about Harry Potter is when MAD magazine(basically the show on paper) spoofs it.

I'm a crazy fan of Harry Potter, but I haven't read any lately, as I have been instead anticipating for the movie and the book to come out. I wonder if the author will choose to have Harry win or Voldemort win for good. What do you all think?

[quote="tridecalogism_lover"]You won't believe how many people I know who say they can't read Harry Potter because it's against their religion.[/quote]

I read Staw Wars and used to read Bionicle(Different name for their god, Mata Nui, they call him the great spirit), but my parents won't let me read Harry Potter...

Also, don't read Bionicle. It was just a wast of time. You have to wait two months for a 5 minute comic.

I look at Harry Potter as a forms of entertainment. I don't have the seventh one on reserve, and I don't think I had the sixth one reserved either.... It's a great thing to just sit down with and flow some hours away into. However, I think its good to not get too obsessive with it. The theories are definitely a huge time sink. I've managed to stay away from theorizing about future installments for both HP and LOST, and I think that's saved me quite a bit of time :D

[quote="dbkarp"]I like Harry Potter, (I'll read all day to finish it but i do that with all books and they just fly by) but I think that people should consider the large variety of books that have similar themes to Harry Potter(fantasy basically). Some people are sooooooo obsessed with Harry Potter they never read another book and reread them over and over and over again. Other some other good authors are: Eoin Colfer, Angie Sage, and Garth Nix. Happy Reading![/quote]
I totally agree with you.  Some people will just read HP over and over and over again and if you tell them that there are hundreds of books just as good as HP, they won't even try to read another book.  It's really sad.

yeah...I'm one of the people who would choose the first choice, but you wouldn't see me reading HP every single day. I'd try to read variety..Atleast HP is promoting kids to read more(if you could say it that way.)..Also what gets me is that some people think it's ok to burn the harry potter books. No matter how much you are against a book or something, I  believe that you shouldn't burn books to show your cause. That's absurd!

[quote="Lepuslapis"]If I remember correctly, that Georgia mother trying by all means to have Harry Potter books removed from her children's school  library (claiming they were evil) was awarded the "Idiot of the Year" distinction.[/quote]

That must have made Rowling furious - she worked hard for it this year ...
Are the mods sleeping? What's this doing in the RT?

[i][color=green]Moderator responds: I move things (esp. polls) to F&G or elsewhere because they lack discussion or belong somewhere else.  This thread doesn't naturally belong elsewhere, and there is actual discussion taking place, so it's staying here for the time being.[/i][/color]

I keep getting this weird feeling that things are going to turn out much more unexpected..... What about Kreacher? He's an important character according to J.K. Rowling. What do you think about him?

[quote="7h3.D3m0n.117"][quote="ABC News"]
NEW YORK (Reuters) - Copies of the final Harry Potter book have already been shipped to customers by one U.S. online retailer, U.S. publisher Scholastic said on Wednesday, and purported copies of the novel have flooded the Internet.

Scholastic Corp. said it was taking legal action against book distributor Levy Home Entertainment and DeepDiscount.com for breaching an embargo preventing the seventh Harry Potter book from being sold in America before 12:01 a.m./0401 GMT on Saturday. The publisher said Levy delivered the books to the online retailer.

People started receiving copies of "Harry Potter and the Deathly Hallows" on Tuesday, but the number of copies shipped only made up "around one one-hundredth of one percent" of the 12 million copies due to go on sale, Scholastic said.

Photographs have also been posted on the Internet of what is claimed to be each page of the book, but Scholastic would not comment on whether they were real. Links to the pictures quickly flooded Web sites around the world.

Author J.K. Rowling appealed to fans to "ignore the misinformation popping up on the Web and in the press on the plot." The book is due to be released in the rest of the world when the clock strikes 12:01 a.m. Saturday in Britain (2301 GMT).

"I'd like to ask everyone who calls themselves a Potter fan to help preserve the secrecy of the plot for all those who are looking forward to reading the book at the same time on publication day," she wrote on her Web site, http://www.jkrowling.com. "In a very short time you will know EVERYTHING!"

Scholastic also issued a statement asking "everyone, especially the media, to preserve the fun and excitement for fans everywhere."

Andrew Moscrip, vice president of Infinity Resources Inc., which owns DeepDiscount.com, said the company was investigating the issue, adding: "Obviously we're taking this matter very seriously."
[/quote]

What do you guys think about the HP7 leak? I heard a bunch of people near Baltimore got some copies, and thankfully, they are keeping the secrets to themselves.[/quote]

It's just a couple more days.  Besides, if the public doesn't want to hear the secrets, they don't have to hear them.  There'll always be some idiots screaming out spoilers but I don't think that's such a big deal.

[quote="bookaholic"]At least two credible newspapers have already (claimed to have) gotten a hard copy of the book and reviewed it. They don't have any major spoilers other than what the Deathly Hallows are. (Okay, the second one does imply another spoiler at the end, but I thought it was obvious anyway.)
http://www.nytimes.com/2007/07/18/books/18cnd-potter.html?_r=1&oref=slogin
http://www.baltimoresun.com/entertainment/booksmags/bal-2potter0718,0,2741335.story?coll=bal_tab01_layout
There are lots of mutually contradictory spoilers online, too. While some of them are probably real, I think it's still best to assume everything is fake until proven otherwise on Saturday.[/quote]

Uh this is definitely gonna be a different book. Aw man we don't see Snape and Lupin for that long (they are my favorite characters even though they are so different) but oh well baltimore may be wrong. And yeah i was pretty sure that 784 pages won't suffice to explain the complexities of the wizarding world, but I hope the loose ends about all the characters are explained.

Edit: The first two paragraphs of the second page of NY times gives away something really big, read carefully

I'd better start rereading 5 and 6 so that I'll be ready to read 7 soon.  I want to finish it before people start publicly discussing the ending.  It's pretty insane to imagine...in a few days, we will know the end of the Harry Potter series!  Not that I was ever a Harry Potter fan, but this whole thing has been around for a long time...

[quote="archimedes1"][quote="sonny"]

Edit: The first two paragraphs of the second page of NY times gives away something really big, read carefully[/quote]
lol
I wish I knew what "blinkered" meant...[/quote]

um second page not first....

Hmmmm
oops

That's interesting.  Even Dumbledore?  And Dudley? How could she?

Well, of course they're complicated. I had that filed under "obvious".

Yes, but now...
they've [i]confirmed [/i]that they're complicated!

[quote="bookaholic"]I guess so :lol: But did anyone seriously think that Snape would just end up being another Bellatrix? JKR pretty much admitted, back in 1999, that he has a "redemptive pattern" going on. Dudley's complicatedness was confirmed by his reaction to the dementors in OotP and Dumbledore was already showing signs of complication in HBP. Also, I'm sort of confused because some people on Mugglenet claim that the epilogue Lazarus posted is real and makes much more sense in the context of one of the online versions of the book. Hmm.[/quote]

Yea um the baltimore newspaper's end suggests what Lazarus posted. But I don't really think JKR could really rearrange that last sentence to make the last word "scar" without preserving what it was supposed to mean.

Wow.  People are already talking.  I don't think I can safely leave my house until I've finished the book.  I should probably even cut off my internet connection.

Wow. I just read the epilogue again...it's NOT that bad. If it's JKR's work, it's not the best thing she's ever written, but it's not the worst thing she's ever written either, and it IS her style. So I believed it to be fake out of pure bias. Is my judgement really that subjective? Wow.

Okay, I still don't want to say for certain that that's the actual epilogue, but if it is, then I'm satisfied with it. Though I still don't understand the Headmaster thing. But we're almost there, we're almost there...

I'm officially bored with this subject.  Any further comments on Harry Potter should find themselves another forum.


find all $f(x): R\to R$ such that :
     $f(f(xy)+x+y)=f(x+y)+xy$

I thinh here is right  problem
find all continuous function $f(x): R\to R$ such that :
     $f(f(xy)+x+y)=f(x+y)+xy$ $\forall x,y\in R$

[quote="quangtrung"]I think here is right  problem
find all continuous function $f(x): R\to R$ such that :
     $f(f(xy)+x+y)=f(x+y)+xy$ $\forall x,y\in R$[/quote]
 thank you , quangtrung
     you are right


[img]http://img267.imageshack.us/img267/9905/problemsr1.png[/img]

Draw a parallel to $ r$ that passes through $ O$. It intersect the circle in $ A,B$. Let $ C\equal{}AP\cap C_1$ and $ O_1\equal{}OC\cap r$. $ O_1$ satisfies the conditions of the problem. The same with $ C'\equal{}BP\cap C_1$ that creates the center $ O_2$.


inequalities
function
calculus
derivative
inequalities proposed

Umm... I think you can't construct a triangle with sides $ 2$, $ 1$, $ 1$. ($ b\plus{}c>a$ isn't true).

[quote="mornik"]Umm... I think you can't construct a triangle with sides $ 2$, $ 1$, $ 1$. ($ b \plus{} c > a$ isn't true).[/quote]
Oh, no. In my definition a triangle is definited by the condition $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \ge 0$. ;)

[quote="mornik"]I guess it depends on what you call a triangle - it the case of triangle sides being $ 2$, $ 1$, $ 1$, you would only have a line, but I understand this is a degenerative case and as such, may be considered a triangle.

Anyway, even if you consider $ (2,1,1)$ a triangle, the statement is still not true - take any triple $ (2t,t,t)$ and all of its permutations and the equality will obviously still hold since the inequality is homogeneous.[/quote]
I think there is something misunderstanding between me and you, sorry for my poor English.
Actually, as you said, the equality holds when $ a\equal{}2t,b\equal{}c\equal{}t$ but I just want to write it as $ a\equal{}2,b\equal{}c\equal{}1$. I think it is nothing at all since the main important thing is my inequality. By the way, my inequality is true indeed. I am sure of it. ;)

Nah, there's no misunderstanding between the two of us and your English is great, by the way. I'm also sure that the inequality holds - it's just that I was a little confused because $ (2t,t,t)$ and $ (2,1,1)$ are not completely the same (they are in this problem because it's homogeneous... but you get the picture). I was also a little confused because $ (2,1,1)$ is not a triangle in the most obvious sense of the word, but I'm sure we got everything straightened out.

If I'll have the time, I will try to prove the inequality.

When (a,b,c)=(2,1,1) triangle--->line.
Nothing wrong in here. :)

I think it 's true by S.O.S
p/s can_hang:I hope you add nick me"deptrainhuthangxelai"
I really admire you  :blush:

Solution is here:
http://www.toanthpt.net/forums/showthread.php?p=66534#post66534
I think it' s not wrong :D

Using the substitutions $ a\equal{}y\plus{}z$, $ b\equal{}z\plus{}x$, $ c\equal{}x\plus{}y$ and $ x\plus{}y\plus{}z\equal{}p$, $ x^2\plus{}y^2\plus{}z^2\equal{}1$, the inequality becomes
$ \frac 1{1\plus{}2px\minus{}x^2}\plus{}\frac 1{1\plus{}2py\minus{}y^2}\plus{}\frac 1{1\plus{}2pz\minus{}z^2} \le \frac 9{3p^2\minus{}1}$.
We apply EV-Theorem to the function $ f(u)\equal{}\frac 1{u^2\minus{}2pu\minus{}1}$ for $ 0\le u<p$. Since $ g(x)\equal{}f'(x)\equal{}\frac{\minus{}2(x\minus{}p)}{(x^2\minus{}2px\minus{}1)^2}$ is convex, because
$ f"(x)\equal{}\frac{24(p\minus{}x)[(x\minus{}1)^2\plus{}p^2\plus{}1]}{(x^2\minus{}2px\minus{}1)^4}>0$,

it suffices to consider the cases $ x\equal{}0$ and $ 0<x\le y\equal{}z$. These cases are equivalent to $ a\equal{}b\plus{}c$ and $ b\equal{}c$, respectively.

Have you a proof without derivative, can_hang2007?

Thank you very much.

[quote="nguoivn"]Solution is here:
http://www.toanthpt.net/forums/showthread.php?p=66534#post66534
I think it' s not WRONG :D[/quote]

I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning.

WLOG we can assume $ a\plus{}b\plus{}c\equal{}3$ We put $ r\equal{}abc$ and $ q\equal{}ab\plus{}bc\plus{}ca$

Our inequality is equivalent to

$ 2\sum ab \sum a^4 \plus{}6\sum ab \sum a^2b^2 \leq 18a^2b^2c^2 \plus{}9\sum a^2b^2(a^2\plus{}b^2)$

which becomes after substitution

$ f(r)\equal{} 2q((9\minus{}2q)^2\plus{}q^2\minus{}6r) \minus{}9((q^2\minus{}6r)(9\minus{}2q)\minus{}r^2) \leq 0$

Now we use this fact

$ 6r \leq 5q\minus{}9$

that is $ \sum (b\plus{}c\minus{}a)(a\minus{}b)(a\minus{}c) \geq 0$

It is true if $ a,b,c$ are sidelenghts of a triangle, by Valenticu-Shur inequality.

Putting $ x\equal{}b\plus{}c\minus{}a$ , $ y\equal{}c\plus{}a\minus{}b$ and $ z\equal{}a\plus{}b\minus{}c$ it is sufficient to show that for $ a \geq b \geq c$

$ cz \geq by$ and this is true being equivalent at $ (b\minus{}c)(b\plus{}c) \geq a(b\minus{}c)$

Deriving $ f(r)$ we find that  $ f'(r)\equal{} 120(3\minus{}q) \plus{}126 \plus{}8r > 0$

Now we have

$ f(r) \leq f((5q\minus{}9)/6) \equal{} (q\minus{}3)^2(112\minus{}315)$ which is not least of zero as $ q leq 3$!!!!

So I think this solution is wrong. In the solution at Vietnamese forum I found a mistake in calculation as
$ f(r)$ is not equal at

 $ f(r)\equal{} 2q((9\minus{}2q)^2\plus{}2q^2\minus{}12r) \minus{}9((q^2\minus{}6r)(9\minus{}2q)\minus{}r^2) \leq 0$

in fact this expression is not equal to zero for $ q\equal{}3$ :wink:

Have you a simple solution can_hang2007?

[quote="manlio"][quote="nguoivn"]Solution is here:
http://www.toanthpt.net/forums/showthread.php?p=66534#post66534
I think it' s not WRONG :D[/quote]

I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning.

WLOG we can assume $ a \plus{} b \plus{} c \equal{} 3$ We put $ r \equal{} abc$ and $ q \equal{} ab \plus{} bc \plus{} ca$

Our inequality is equivalent to

$ 2\sum ab \sum a^4 \plus{} 6\sum ab \sum a^2b^2 \leq 18a^2b^2c^2 \plus{} 9\sum a^2b^2(a^2 \plus{} b^2)$

which becomes after substitution

$ f(r) \equal{} 2q((9 \minus{} 2q)^2 \plus{} q^2 \minus{} 6r) \minus{} 9((q^2 \minus{} 6r)(9 \minus{} 2q) \minus{} r^2) \leq 0$

Now we use this fact

$ 6r \leq 5q \minus{} 9$

that is $ \sum (b \plus{} c \minus{} a)(a \minus{} b)(a \minus{} c) \geq 0$

It is true if $ a,b,c$ are sidelenghts of a triangle, by Valenticu-Shur inequality.

Putting $ x \equal{} b \plus{} c \minus{} a$ , $ y \equal{} c \plus{} a \minus{} b$ and $ z \equal{} a \plus{} b \minus{} c$ it is sufficient to show that for $ a \geq b \geq c$

$ cz \geq by$ and this is true being equivalent at $ (b \minus{} c)(b \plus{} c) \geq a(b \minus{} c)$

Deriving $ f(r)$ we find that  $ f'(r) \equal{} 120(3 \minus{} q) \plus{} 126 \plus{} 8r > 0$

Now we have

$ f(r) \leq f((5q \minus{} 9)/6) \equal{} (q \minus{} 3)^2(112 \minus{} 315)$ which is not least of zero as $ q leq 3$!!!!

So I think this solution is wrong. In the solution at Vietnamese forum I found a mistake in calculation as
$ f(r)$ is not equal at

 $ f(r) \equal{} 2q((9 \minus{} 2q)^2 \plus{} 2q^2 \minus{} 12r) \minus{} 9((q^2 \minus{} 6r)(9 \minus{} 2q) \minus{} r^2) \leq 0$

in fact this expression is not equal to zero for $ q \equal{} 3$ :wink:

Have you a simple solution can_hang2007?[/quote]
Hello, my friend. :) My solution for it used the old way-SOS. Here is it :)

[quote="arqady"]Very nice inequality and very nive proof, can_hang2007! :lol:
Even, we can prove that $ S_b \plus{} S_c\geq0.$ :wink:
$ S_b \plus{} S_c \equal{} \frac {1}{a^2 \plus{} c^2} \plus{} \frac {2(b^2 \minus{} ac)}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \plus{} \frac {1}{a^2 \plus{} b^2} \plus{} \frac {2(c^2 \minus{} ab)}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} \frac {1}{a^2 \plus{} c^2} \plus{} \frac {2(c^2 \minus{} ab)}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {1}{a^2 \plus{} b^2} \plus{} \frac {2(b^2 \minus{} ac)}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} \frac {3c^2 \minus{} 2ab \plus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2ac \plus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)}\geq$
$ \geq\frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right) \equal{}$
$ \equal{} \frac {(b \minus{} c)^2(4a^2 \plus{} (b \minus{} c)^2)}{(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\geq0.$[/quote]
Nice, argady :)

[quote="manlio"][quote="arqady"]

$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right) \equal{}$
[/quote]

I think it is wrong. Can you please explain it to me, arqady.
Thank you very much.[/quote]
Yes, you are right, manlio! It can be
$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} (b \minus{} c)\left(\frac {3b \plus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \plus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right)$ and it's nothing.  :(
Thank you!

[quote=can_hang2007]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)[/quote]

only three lines with our trick!


If the roots of the equation

$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$

are equal,prove that

$a^3+b^3+c^3=3abc$

[hide]
rearrange and get

$3x^2-2(a+b+c)x+(ab+ac+bc)=0$

Discriminant = 0 , hence we obtain $(a+b+c)^2=3(ab+bc+ac)$ which is $a^2+b^2+c^2=ab+bc+ac$

But AM-GM tell us , $a^2+b^2+c^2\geq ab+bc+ac$ . Hence $a=b=c$

Hence $a^3+b^3+c^3=3abc$ is true since $a=b=c$
[/hide]

Rather than use AMGM, multiply both sides by $a+b+c$ and voila!

Another way:

[hide]rearranging the equation:

$3x^2-2(a+b+c)x+(ab+ac+bc)=0$ considering that $k$ is the  only root of the equation: by Viete's formula:

$2k=-\frac{-2(a+b+c)}{3}\implies k=\frac{(a+b+c)}{3}$ and $k^2=\frac{(ab+ac+bc)}{3}$ in consequence: (I) $k^3=\frac{(a+b+c)}{3}.\frac{(ab+ac+bc)}{3}$ but:

(II) $k^3={(\frac{(a+b+c)}{3})}^3=\frac{a^3+b^3+c^3+a(ab+ac)+b(ab+bc)+c(ac+bc)+2(a+b+c)(ab+ac+bc)}{27}$

$\Rightarrow k^3= \frac{a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-3.abc}{27}$

Equalizing (I) and (II):  $\frac{(a+b+c)}{3}.\frac{(ab+ac+bc)}{3}=\frac{a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-3.abc}{27}$

$\implies a^3+b^3+c^3=3.abc$[/hide]

[quote="seamusoboyle"][quote="shyong"]
But AM-GM tell us ,$a^2+b^2+c^2 \geq ab+ac+bc$ . Hence

[/quote]

Last time i checked this is true only for positive, real values. Being awfully assumptious arn't you? :)[/quote]

But for $a^2+b^2\geq 2ab$ , you dont need $a,b$ to be positive real though .
However , if a,b,c can take negative value then the equality sign deosnt neccesary to be $a=b=c$  :P

Dont forget about complex values either.


HI!
My first problem is to know the best way to know if a graph is bipartite.
I don't know any algorithm in particular to do it...and I think I can do it in the best way in O(n), or I hope so, but if someone knows a method or a good algorithm it would be nice to know it.
The second problem is this:
I have 1 graph, and I want to divide it in 2 diferent groups of nodes, in such a way that the amount of conected nodes between the first group and the second is maximum... I hope you understand my bad description.
In another way to see this problem, just think in a group of people where some of them are friends. And I want to divide it in 2 teams such that the amount of pairs of friends between 1 team and another is maximum.
thanks for any help

Color a node red. Color all nodes that connect to it blue. Then color all nodes that connect to those nodes red. When you get a contradiction stop.

The second problem is known as Max Cut. The problem is NP hard. So it cannot be solved exactly in polynomial time unless $P=NP$. The best known approximation algorithm is the Goemans-Williamson algorithm.

BTW, what is $n$ in your first problem?


Let $ a,b,c$ be integers with $ a>0$, and $ ac\minus{}b^2\equal{}P\equal{}p_1p_2...p_n$, where the $ p_i$'s are distinct primes. Let $ M(n)$ denote the number of pairs of integers $ (x,y)$ such that $ ax^2\plus{}2bxy\plus{}cy^2\equal{}n$.

Prove that $ M(n)$ is finite, and $ M(n)\equal{}M(P^kn)$, for all natural numbers $ k$.

It equavalent to $ z^2\plus{}Py^2\equal{}an$, were $ z\equal{}ax\plus{}by$. Therefore $ M(n)$ is finite.
If $ n\equal{}Pn_1$, then $ z\equal{}Pz_1$ and $ y^2\plus{}Pz_1^2\equal{}an_1$, therefore $ M(n)\equal{}M(n_1)$ (same equation). It prove $ M(P^kn)\equal{}M(n)$.

When you conclude $ M(n)\equal{}M(n_1)$ has the same number of solutions, how can you be sure?


1.  What is the least positive integer meeting each of the following conditions.

Dividing by 7 gives a remainder of 4.
Dividing by 8 gives a remainder of 5.
Dividing by 9 gives a remainder of 6.

( I already have the answer. If you can provide different solutions the better.)

2. Suppose that a metric clodk uses the following system.

  1 day = 10 metric hours
  1 metric hour = 10 metric minutes
  1 metric minute = 10 metric seconds
  1 metric seconds = 10 metric miniseconds

  What is the time in our system when the metric clock registers 4 hours, 5 minutes, 6 seconds and 7 miniseconds?

[quote="guile"]1.  What is the least positive integer meeting each of the following conditions.

Dividing by 7 gives a remainder of 4.
Dividing by 8 gives a remainder of 5.
Dividing by 9 gives a remainder of 6.

( I already have the answer. If you can provide different solutions the better.)
[/quote]
[hide="Solution"]
4 (+7), 11, 18, 25, 32, 39, 46, [b]53[/b],
53 (+56), 109, 165, 221, 277, 333, 389, 455, [b]501[/b]

So the answer is 501. There's most likely a far easier way to solve this, 'cept I don't know the method.
[/hide]

[quote="4everwise"]Couldn't you just use modular arithmetic?[/quote]

Of course you can solve it using modular arithmetic. How about other solutions? I was just wondering if there are other solutions.

[hide="peek if you want to"]The answer to Q1 is obviously  $-3$.. Ohh.. you don't want negative numbers  then just add $7 \cdot 8 \cdot 9 = 504 $ to it  to get $501$[/hide]

The first one is simple:

$x\equiv4(mod7)$
$x\equiv5(mod8)$
$x\equiv6(mod9)$

So:

$x+3\equiv0(mod7)$
$x+3\equiv0(mod8)$
$x+3\equiv0(mod9)$

so $x+3=2^3(3^2)(7)$, so $x=501$.

[quote="Iversonfan2005"][quote="236factorial"]Does anyone else want to try nubmer 2?[/quote]

To tell you the truth, no :P Too confusing and I'm way too lazy[/quote]

It isn't confusing, but it's a little long, and it took me about 2 minutes to figure it out with a calculator at hand.

[quote="236factorial"][quote="guile"]2. Suppose that a metric clodk uses the following system.

  1 day = 10 metric hours
  1 metric hour = 10 metric minutes
  1 metric minute = 10 metric seconds
  1 metric seconds = 10 metric miniseconds

  What is the time in our system when the metric clock registers 4 hours, 5 minutes, 6 seconds and 7 miniseconds?[/quote]

[hide]Percent of a day = 0.4567
0.4567*24 = 10.9068
= 10:54:24.48

No clue if that's close. Realize that 10:06:40.2 is incorrect.[/hide][/quote]

Actually the anwer is 10:57:38.88.  You must have erred somewhere. Sorry but im too busy to explain it now.

I must've read my calculator wrong...  :blush:
Oh well. Thanks for noticing it.


Summation from 1 to infinity of [8n^3 - 6n^5] / [12n^4 + 9n^5]

converges or diverges?

This one is too easy. you can do it yourself [hide](HINT: limit of the nth member!)[/hide]

[quote="Carcul"]$ \lim_{n \to \infty}\frac {8n^3 \minus{} 6n^5}{12n^4 \plus{} 9n^5} \equal{} \minus{} \frac {2}{3}$[/quote]

I didn't want to spoil his homework totally :)


Me gustaria q se colgaran por aca los nombres de los integrantes de los equipos q participaran en Ecuador 2006.

En El Salvador ya los selectivos fueron realizados, los resultados se conoceran al final de la semana.

Espero respuestas.
Chao.

 :ninja:  :ninja:

Lo mismo en Argentina. Ya se tomó la prueba, pero no están los resultados todavía. Cuando estén, los posteo.

Equipo Argentino:

1.Germán Sergio GIECZEWSKI - San Isidro - Buenos Aires
2.Pablo BLANC, E.E.T. N°1 "Brig. Gral. P. ECHAGÜE" - Concordia - Entre Ríos
3.Julián Martín EISENSCHLOS, Escuela Técnica ORT I - Cdad de Buenos Aires
4.Ignacio Agustín ROSSI, Instituto San Vicente de Paul - Escobar -  Buenos Aires

Equipo Salvadoreño:

1.  Gabriel Chicas       (IMO 2006)
2.  Otto Tang               (IMO 2006)
3.  Alfonso Alvarenga   (primera participación)
4.  Tatiana Vasquez     (primera participación)

Equipo Colombiano:
Diego Cifuentes (IMO 2006)
Felipe Ochoa (Centro 2005)
Nicolas Paez (Centro 2005)
Jose Alejandro Samper (Ibero 2005,IMO 2006) Yo!!

HOLA
EL EQUIPO DE PERU ES EL SIGUIENTE
CHEN SONNCO (IBERO2004;CONOSUR2005,2006;IMO 2005,2006)
FRANCO VARGAS(IBERO 2005, CONO SUR 2006, IMO 2005,2006)
JOSSY ALVA(IBERO 2005, CONO SUR 2005, IMO 2005,2006)
LUIS VELEZ.

El equipo chileno:

Simón Legassa
Héctor Pasten
Giancarlo Lucchini
Jaime Soza

Suerte para todos!!

[color=blue][b]Equipo Paraguayo[/b]

1. Michael Chan
2. Rafael Arias
3. Maurizio Caló
4. Gabriela Ratti[/color]

El equipo mexicano!!!!

1.- Pablo Soberon (Primer oro IMO de Mexico)
2.- Isaac Buenrostro
3.- Fernando Campos
4.- Ivan Joshua

Vamos Mexico!!
 :P


[quote="fjwxcsl"]Let $ {x,y,z}\ge{0},x + y + z = 1,f(x,y,z) = \sum{\sqrt {x^{3} + xyz},}$ prove
\[ \\
{f(\frac {1}{2},\frac {1}{2},0)}\le{f(x,y,z)}\le{f(1,0,0)}.\]
[/quote]
The left inequaliy can be proved by using the inequality
$ 2(x^3+y^3+z^3)^2+18x^2y^2z^2 \ge (x^2+y^2+z^2)^3.$
Indeed, from Minkowski's Inequality, one has
$ \sum \sqrt{x^3+xyz} \ge \sqrt{\left(\sum x^{3/2}\right)^2+\left(\sum (xyz)^{1/2}\right)^2},$
and thus, it is enough to prove that
$ (x^{3/2}+y^{3/2}+z^{3/2})^2+9xyz \ge \frac{1}{2}.$
This inequality is equivalent to
$ 2(x^{3/2}+y^{3/2}+z^{3/2})^2 +18xyz \ge (x+y+z)^3.$
Replacing $ x,y,z$ by $ x^2,y^2,z^2$, respectively, it becomes
$ 2(x^3+y^3+z^3)^2+18x^2y^2z^2 \ge (x^2+y^2+z^2)^3.$

For the right hand side, applying the AM-GM Inequality, we have
$ \sum \sqrt{x^3+xyz}=\sum \sqrt{x(x^2+yz)} \le \sum \frac{x+(x^2+yz)}{2} \le \sum \frac{x+(x^2+2yz)}{2}=1.$


During my high school years, I came to believe that academic success was more important than anything else. I planned on learning as much as possible in order to get accepted by a graduate school in the sciences. Upon learning that I could enter the university two years earlier than normal, I got excited about the prospect of obtaining the freedom to study whatever I wanted to sooner rather than later.
I believed that I could do anything if I could apply my will to it, that I would be capable of studying for longer amounts of time than I was ever able to – which was why I decided to live in the university dorms. I thought that I had all of the time in the world since I did not have extra-curriculars, a commute, too many friends, or any other potential sink of time. I also expected that I would not face the same problems that other college students face – like drugs or excess socialization, since I do not enjoy most social activities. Finally, I did not bring my computer games with me to university, eliminating my primary time sink in high school.
Nonetheless, this turned out to be too optimistic. I was not prepared to be a more independent learner, since I was still under the impression that the student did not need to read the textbook and that education was the responsibility of the institution more so than the responsibility of the student. Furthermore, college required a type of thinking different from that type that I was used to. Many of the math and physics problems seemed to be insurmountable and I became frustrated at my inability to meet my own expectations. Internet surfing, daydreaming, and aimlessly walking around campus replaced computer games as the main culprit of wasting my time, even though I did not engage in practices that waste the time of most other college students. Impulse control turned out to be much more difficult than I previously thought. I concurrently started to become fascinated by nihilism – the notion that life is pointless. My interest in astronomy, my agnosticism, my tendency to question authority, my inability to conform to social norms, my wide reading in the social and natural sciences, my lack of close friends in real life –all predisposed me to nihilistic thoughts. Such thoughts would provide a justification for me to no longer care about my grades, so that I could finally stop stressing myself out over my grades so much. Since I stopped trying to control my own impulses, my self discipline and performance deteriorated as the quarter progressed. I became aimless and disorganized, attracted to any lure of even remote interest. Eventually, I lost faith in my own ability to study, resulting in a self-fulfilling prophecy that caused me to give up hope.
As the novelty of nihilism wore off I had to start thinking about my future again. After talking with several people who had become nihilists in the past, I realized that they still yet continued to move on with their lives. Many created their own meaning for themselves. Furthermore the image of working on minimal wage jobs for the rest of my life is certainly an unattractive one. Whether they have a belief system or not, people still desire to seek happiness in their lives. While there is no intrinsic reason why academic pursuits are more important than other pursuits, I would probably be happier with academic pursuits, given my personality characteristics. So after my period of nihilism, I concluded that I would be happiest studying as much as possible for the next few years, but with a different approach than the one that I previously thought of. My experience had many parallels to the philosophy of Friedrich Nietzsche (1844-1900). He was known for propounding the theory that nihilistic beliefs prevail with the breakdown of religion and old values. But with an internal struggle with our own belief systems and by examining them from multiple perspectives, we can make ourselves deeper and stronger. We can then create new values that are independent of those of society.
Despite the damage that it did to my academic performance, my period of nihilism gave me some insight to life. My original plan involved the pursuit of science to the exclusion of all other fields. Now I know this is not the ideal route to go if I want to think about the relation of science both to other fields of knowledge and to myself. Nihilism has made me more open to new experiences that allow me to learn more about the world. I also now care more about learning than about my academic transcript, so that I am more inclined to study independently of class. Furthermore, I am now more open to questioning my own beliefs instead of stubbornly clinging onto them. Nihilism also tends to bring relativistic attitudes into the fray, and such attitudes impel me towards political apathy and greater tolerance towards different points of view. With tolerance comes a greater possibility of converting to a new belief system when the time comes, if it is convincing enough.
So even though I have not found a replacement, my thinking is no longer dominated by nihilism. While this period of nihilism had coincided with a major life change, which in turn impacted my grades and learning in a negative way, I am now more determined to try to control my impulses such that I can learn as much as possible, so that I can eventually land a position in graduate school to pursue my intrinsic love of knowledge. Though I fell somewhat behind, one year is not such a big deal. Unique as I am, I ultimately realized that my problems were not all that different from the problems that other college students typically face.

um a few points:
1.this is a MATH forum
2.this particular forum is devoted to people who want to procrastinate from their lives.
3.darn I can't think of anymore

If this is a college essay, then someone should please move this to the College forum.

or possibly other topics, as this doesn't pertain to math...


[i]Prove that there are infinite many triples of positive integers $ (x;y;z)$ such that $ gcd(x \ ; \ y \ ; \ z) \ \equal{} \ 1$

   and $ x^2 \plus{} y^2 \ \equal{} \ z^3$[/i]

[i]  Ps : It's my $ 200^{th}$ post  :lol: [/i]

[quote="nguyenvuthanhha"][i]Prove that there are infinite many triples of positive integers $ (x;y;z)$ such that $ gcd(x \ ; \ y \ ; \ z) \ \equal{} \ 1$

   and $ x^2 \plus{} y^2 \ \equal{} \ z^3$[/i][/quote]
If $ x^2\plus{}y^2\equal{}z^3$ with $ z$ odd and $ \gcd(x,y,z)\equal{}1$, then $ (x^2\minus{}y^2)^2\plus{}(2xy)^2\equal{}(z^2)^3$ and $ z^2$ odd and $ \gcd(x^2\minus{}y^2,2xy,z^2)\equal{}1$

And since $ 11^2\plus{}2^2\equal{}5^3$, we got infinitely many such triples.

[quote="nguyenvuthanhha"][i]  Ps : It's my $ 200^{th}$ post  :lol: [/i][/quote]
Congrats

[quote="Abdek"][quote="pco"][quote="nguyenvuthanhha"][i]Prove that there are infinite many triples of positive integers $ (x;y;z)$ such that $ gcd(x \ ; \ y \ ; \ z) \ \equal{} \ 1$

   and $ x^2 \plus{} y^2 \ \equal{} \ z^3$[/i][/quote]
If $ x^2 \plus{} y^2 \equal{} z^3$ with $ z$ odd and $ \gcd(x,y,z) \equal{} 1$, then $ (x^2 \minus{} y^2)^2 \plus{} (2xy)^2 \equal{} (z^2)^3$ and $ z^2$ odd and $ \gcd(x^2 \minus{} y^2,2xy,z^2) \equal{} 1$

And since $ 11^2 \plus{} 2^2 \equal{} 5^3$, we got infinitely many such triples.

[quote="nguyenvuthanhha"][i]  Ps : It's my $ 200^{th}$ post  :lol: [/i][/quote]
Congrats[/quote]

Nice solution  :D  But How did you find this identity  ?:)  est ce qu'il existe une methode pour arriver a cette identité ou c'est juste un tatonement  :lol:  Merci !![/quote]

$ (x^2\minus{}y^2)^2\plus{}(2xy)^2\equal{}(x^2\plus{}y^2)^2$ is a very classical identity uded to solve $ a^2\plus{}b^2\equal{}c^2$

[quote="FelixD"]@Abdek: Why did you write one part of your answer in english, the other one in frensh?  :P[/quote]

 Because I didn't Know how to write The second part  in English  :P  since my English is very poor  :( ^^

Thank you pco ^^  I'm just a beginner at the Diophantine equations  :)

Let $ z$ be a prime of the form $ 4k \plus{} 1$, then $ z^3$ is a sum of squares as it contains no prime divsors of the form $ 4n \plus{} 1$. The only way $ x$ and $ y$ could have a common factor is if $ x \equal{} za$ and $ y \equal{} zb$ since $ z$ is prime, so $ a^2 \plus{} b^2 \equal{} z$. It is well-known that this equation only has one solution. However, it is also known that $ z^3$ must be representable as a sum of squares in more than one way (actually $ 4$, see  [url]http://mathworld.wolfram.com/SumofSquaresFunction.html[/url]).Thus, there are infinitely many solutions to the original equation.


quadratics
Vieta
algebra
quadratic formula

$\alpha\beta\neq c$.

And what exactly do you want to prove? Either:

1. Given a,b,c derive the formula for the roots of the quadratic? That can't be done without factoring the equation.

or

2. Given the formula for the roots of a quadratic, show that those numbers satisfy Vieta's formulas, hence they ARE the solutions? That CAN be done, but you still need to be a HELLUVA guesser to "just guess" those expressions and plug them into Vieta's.

do you guys know the much easier form of the quadratic formula that is helpful when b is even?

[quote="13375P34K43V312"]I'm not completely sure that this works, but can you prove the quadratic formula using the fact that $\alpha\beta=c$ and $\alpha+\beta=-\frac{b}{a}$?[/quote]
But those formulas are derived from the quadratic formula.  Also, it should be $\alpha\beta=\frac{c}{a}$.

[quote="lingomaniac88"]But those formulas are derived from the quadratic formula.  Also, it should be $\alpha\beta=\frac{c}{a}$.[/quote]

They are derived from Vieta's formulas, which in turn follow from the Factor Theorem.  They are in agreement with the quadratic formula because they are all true.

The problem with your approach is that solving that particular system is equivalent to solving the original quadratic.

Oh, you can get them from Vieta's formulas, too?

They [b]are[/b] Vieta's formulas, for a quadratic.

They are also easy to prove to someone.
$yz=p$
$y=\frac{p}{z}$

$y+z=s$
$\frac{p}{z}+z=s$
$p+z^{2}=zs$
$z^{2}-zs+p=0$

THere you have it. P and s are constants.
Also, they can be proved with the quadratic formula pretty easily. Add and multiply
$\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac}}{2a}$
And you will have your answer.


all i know is that hong kong didn't really nail qn 1, but overall was quite good.


Let $a$ and $b$ be two integers such that $a$ and $b$ are congruent to eachother modulo every prime $p$. Show that $a=b$.

I thought you can show that $a-b = k_{1}p_{1}= k_{2}p_{2}\dots$ and by that show that $a-b$ must contain all existing prime factors and thus not be finit, but I dont really like it. Any other ideas?

Well, there are infinite primes, and none of them share any common factors (by definition)...

[hide="Solution"] If $a \equiv b \bmod p$ for all primes $p$ then all primes divide $a-b$, so $a-b = 0$. [/hide]

[hide]b-a is finite, choose p > b-a and pk = b-a so k = 0 and b = a..

[/hide]

though I suppose that's the same as torajiorou's


trigonometry
inequalities unsolved
inequalities

Find the smallest value of the expression (3+x)^0.5+(3+y)^0.5for the values of x and y which verify the relation x^2+y^2=9

If my mind works well , this problem is from Moldova 1999  ;)

Just wondering whether we can do some substitution like $x=3\sin \theta$ , $y=3\cos \theta$ ....


calculus derivative function algebra domain vector topology

Recently, swag posted a question about distributions (also known as "generalized functions"), so I thought I'd drag this up.

This is not a "problem." It's  a pedagogical exercise, or an illustrative example, and a rather simple one, at that. Still, I've always liked it.

Suppose that $(f_n)$ is a sequence of distributions on $\mathbb{R}$ and $f$ is a distribution such that $f_n\to f$ as $n\to\infty.$

Prove that $f_n'\to f'$ as $n\to\infty.$

Reconcile this statement with what you observe in the case $f_n(x)=\frac1n\sin(n^2x).$

What is "distributions"? And what is $f_n\to f$,if $f,f_n$ are distributions?
Please explain. :(

To define distributions, start with a space of "test functions." The most usual space of test functions is $\mathcal{D},$ the set of $C^{\infty}$ functions with compact support. (These are functions on a given domain. For simplicity, let's assume for now that that domain is $\mathbb{R}.$) There are other possible sets of test functions, but we'll  start with this case. $\mathcal{D}$ is a topological vector space. Now's not the time to go into great detail on the topology, but as a partial statement, $\varphi_n\to\varphi$ iff the supports of $\varphi_n$ are contained in a common compact set and each derivative of $\varphi_n$ converges uniformly to the corresponding derivative of $\varphi.$

The set of distributions, $\mathcal{D}',$ is the set of continuous linear functionals on $\mathcal{D}$ - that is, the set of continuous linear mappings from $\mathcal{D}$ to $\mathbb{R}.$ (Or $\mathbb{C},$ if you want complex-valued functions.)

So if $f\in\mathcal{D}',$ we write the action of $f$ on an arbitrary test function $\varphi$ as $(f,\varphi).$

Any locally integrable function $f$ defines a distribution by the action

$(f,\varphi)=\int_{-\infty}^{\infty}f(x)\varphi(x)\,dx.$

Another, very famous, example of a distribution is the Dirac delta, which is defined by

$(\delta,\varphi)=\varphi(0).$

If $f$ were actually a "nice enough" differentiable function, we would have:

$(f',\varphi)=\int_{-\infty}^{\infty}f'(x)\varphi(x)\,dx= -\int_{-\infty}^{\infty}f(x)\varphi'(x)\,dx=-(f,\varphi').$

(Note that the compact support of $\varphi$ causes the boundary terms in the integration by parts to vanish.)

We seize on this to make a [b]definition[/b], valid for all distribution. We define the derivative of $f$ to be the distribution with the action

$(f',\varphi)=-(f,\varphi').$

Call this the [b]weak[/b] derivative. As an example: let $H(x)$ be the unit step function which is $1$ for positive $x$ and $0$ for negative $x.$ Then, as a weak derivative, $H'=\delta.$

All distributions have infinitely many derivatives. Always.

I'm not going to show you the definition of $f_n\to f$ in $\mathcal{D}',$ but I'll leave you with two hints:

Hint #1: You really don't have much of a choice, when you consider what tools are available to you for defining that convergence.

Hint #2: Define $f_n$ by $f_n(x)=\begin{cases}n&0<x<\frac1n\\0&\text{ otherwise} \end{cases}.$

Then $f_n\to\delta$ in $\mathcal{D}'.$

I guess i'll try:

$(f_{n}',\phi) = -(f_{n},\phi') \rightarrow -(f,\phi') = (f',\phi)$

hence: $f_{n}' \rightarrow f'$

That's it. That's the whole argument. Think of how difficult it it to prove anything like that in a setting of classical analysis, and here it's nearly immediate from the definitions.

And as for my example, $f_n(x)=\frac1n\sin(n^2x):$

Since $f_n\to0$ uniformly, that's more than strong enough to imply that $f_n\to0$ in $\mathcal{D}'.$

$f_n'=n\cos(n^2x).$

The theorem above it true, and we must have that $f_n'\to0$ in the sense of distributions. (Call that "weak convergence.") This goes to show just how astonishingly weak this sense of convergence can be.


Hi,
Did anyone here got full score in the pizza problem of the usaco 2004 march contest?
My program always times out in the last 3 cases.
I would appreciate if anyone could help me by posting his code..
The problem statement can be found here (along with Bruce Merry's analysis):

http://ace.delos.com/MAR04.green.htm

Thanks in advance,

I know network flow algorithm, but I don't know how to use it in this problem.
If I will make all sequences with length K it will be many  :blush:.
Write here your algorithm and I'll try to solve it.

Here's the algorithm I use:
I use a find_path algorithm which tries to find a path in the bipartite graph (left = cows, right = pizzas) I have:
First I loop through the cows, and if I find a certain cow 'C' that has not already been assigned a pizza, I generate all the pizzas cow 'C' likes (using a recursive function). Each time I finish generating one pizza of those,  I check whether this pizza 'P' had already been assigned to another cow. If not, then I can go to the sink node, and I've found a path, if it had already been assigned, then in the residual graph, there is an edge from the pizza to the cow 'D' (which was assigned to pizza 'P'). So I take that edge, and makes the same thing for cow 'D' as for cow 'C'.

I keep calling find_path until I can't find any path.
Then I output the number of paths found.

Thanks,


Sides AB and AC of triangle ABC are extended to D and E such that BC and DE are parallel. If AB=4, BD=2, CE=3 and DE=9, find the perimeter of triangle ABC.

[hide]
$BC = \frac{DE \cdot AB}{AB+BD}= \frac{9 \cdot 4 }{ 4+2 }= 6$
$AC = \frac{BC ( AC+CE) }{DE}= \frac{6AC+18}{9}= 6$
Perimeter of triangle $ABC$ is $AB+BC+AC = 6+6+4 = 16$
[/hide]

yah.. that's what i got
[hide] you konw that triangle ABC and triangle ABE are similar so you can just set up a series of proportions to find the side lengths and find the perimeter...so... ... .. .. ... [b]16[/b][/hide]


AMC
AIME
algorithm
modular arithmetic
blogs
geometry
Pythagorean Theorem

This is a question more relevant to the AIME.

Alas, it occurs to me that I may be the worst at Arithmetic.  No kidding.  Questions that I can think up of the solution to in a few seconds, takes me minutes and minutes and minutes to compute.  I think one problem could be that I overanalyze my solutions too much, always conscious about whether or not I've made a "stupid mistake."  Past practise tests have indicated that I'm pretty accurate.  But I would like to be quick too.

Are there any specific tips on mental calculation that you guys might have?  Sometimes on the AIME...there are impossibly huge numbers that take quite some time to calculate; given the solution I discovered of course (and I don't think they're bad solutions, they're only bad because I'm pretty weak when it comes to arithmetic  :( ).

Also, should I be willing to trust myself more?  I don't want to risk going so fast that I'll miss something...but I usually get questions right anyways.  Plus...how do you guys multiply/divide/add numbers?  Are there special algorithms you guys have, or personally like to incorporate?

Another thing, I freeze a lot when it comes to arithmetic.  When I'm doing practise AIMEs, there's a subconscious voice in my head every time I'm calculating that says something like:  "6 times 4 is equal to...gosh darnit...hmmmm you know I've made good progress so far heh...24."  How should I go about so that Arithmetic is executed constantly...without thoughts of say...random things?

Psychological tips, Math tips...all would be appreciated.  Thanks :) .

Yeah, I personally have a problem with freezing on speed calculations.  How do you stop doing that...

and these are on competitions where speed is the entire purpose.

Uh, I don't know why you would trouble with things like 4*6, you've been doing those since 4th grade or before.

Well if you have something like $123 \times 456$, you can multiply it out, then check your last digit, estimate (bigger than 100*400, somewhere around 100*450), a cool trick is the sum of the digits of the first number times the sum of the digits of the second number should equal to the sum of the digits of the second number, modulo 9. In this example, $1+2+3\equiv 6 \pmod 9$ times $4+5+6 \equiv 6 \pmod 9$, so the sum of the digits of the product is congruent to $6 \cdot 6 \equiv 36 \equiv 0 \pmod 9$. Yeah other modulo tricks like that, if you're really unsure of your answer...

Oh and here's one of those cross multiplication examples for 2-digit numbers: 12*34 = 2*4 + 10(1*4 + 2*3) + 100(1*3)=8+100+300=408.

MCrawford's blog had a nice entry on this subject.

[quote="Silverfalcon"]Well.. 123 x 456.. Write it as (120+3)(450+6). That's a lot easier to do since 12 x 45 isn't that hard to do and rest is pretty easy calculation.

Think about how you can factor. I mean, if you see 45*68, think about (40+5)(70-2). This is a lot easier.[/quote]

$45 \times 68 = 90 \times 34$  :)

Personally, I like to factor heavily for multiplication and division questions.  As far as addition and subtraction, I've actually forgotten how to subtract the way they taught me in elementary school and I do it this way instead:

Suppose I wish to evaluate $1584367- 3869$.

$1584367 - 3869 = 1580000 + (4367 - 3869) = 1580000 + (4367 - 4000 + (4000 - 3869))$
$4000 - 3869 = 4000 - (3900 - 31) = 100 + 31 = 131$
$1580000 + (367 + 131) = 1580000 + (498) = 1580498$

I use a similar digit-grouping method for addition.  It's convoluted, but it gives me an easy way to do them in my head without keeping track of borrowing  :)

Instead of borrowing in addition, I give an extra +1 to the number that's doing the subtracting. For example, 98 - 79, I first see that 8-9 = 9, and then you do 9 - (7+1) = 1, because you looped and have to add one to the 7.

I guess I'm a generous person, haha.

Or you can just multiply it out on paper...

Yeah uhh... it's really not hard to multiply things out on paper ;). If your solution makes even that difficult, then maybe your solution isn't very good. For multiplication specifically, I always find that doing the corresponding division problem after your calculation is a good check.

[quote="probability1.01"]For multiplication specifically, I always find that doing the corresponding division problem after your calculation is a good check.[/quote]Or after subtraction, you can do the corresponding addition problem.  I find that my accuracy rate for addition is much higher than for subtraction, so that often helps me find errors.

Me too. I suck at substraction and division, so I check them with addition and multiplication. Especially substraction: I have only about a 50% success rate :P

[quote="lowest_reynolds"]Thanks for the suggestions so far guys.

Not hard to "do arithmetic out" on paper...maybe.  I just tend to freeze when performing tediously long calculations.  There are many great Math suggestions so far.  Anything psychological?[/quote]

Think like this: What would my calculator say?

And voila! You have your answer.

Honestly though, if you want to improve calculation speed, just keep doing more and more. You'll find great patterns.

often in geometry problems, using the difference of squares occurs with the pythagorean theorem,
ex. $\sqrt{174^2-26^2}=\sqrt{200*148}=20\sqrt{74}$
also, factoring can be a mess, but the euclidian algorithm usually helps, and also i use a separate my brute calculations from my drawings and such on different sheets of paper

mentally, you should have gotten enough sleep, etc

I should also mention that nowadays I add starting from the largest digit going down - I find it makes it easier to remember what I have left to add.  Easiest way to check your arithmetic, in my opinion, is last-digit check (basically, $\mod 10$).

Also, you might want to memorize squares.  It helps, especially in combinatorics problems or whatnot where you have to find the largest $n$, for example, such that $n(n+1) \le 1337$ (to provide an example).  The answer, if you have your squares memorized, should clearly be $n = 36$.   :)

A lot of times I don't do the multiplication till the end if the problem doesnt get too messy

for example, if you get x+98*47 = 3x+ 4 - 68*84
I leave it like that until I solve for x, then calculate the products
Of course this was a simple example, but this helps me do calculations all at once

And about squares... to guess and check anything that ends with a 5...
(x5)^2 = x*(x+1)attaced to 25
for example 45^2 = (4*5)25=2025


Hi, i need ALGEBRAIC EQUATIONS for these problems.   Please also give me a verbal model and the definitions of the variables if possible :) .  Ok onto the problems.

Fred can run at 11miles per second. Wilma can run at 13miles per second. How far ahead must fred be so that Wilma doesnt catch him within 15 seconds?

Sandy hikes up a road starting at 9 A.M.  She walks at an average rate of 4mph.   Salty starts after sandy and drives up the same road at 24mph.  If they both reach the top at 9:27 A.M.  how long was sandy hiking before salty started driving?

A gazelle can run 73 feet per second for several minutes.  A cheetah can run 88 feet per second but can only sustain its top speed for about 20 seconds before it is worn out.   How far away from the cheetah does the gazelle need to stay for it to be safe during the initial 20 seconds for the chase?

LAST ONE.

You and a friend are each driving 379 miles from LA to San Francisco. Your friend leaves first, driving at 52 mph.  She is 32 Miles from LA when u leave, driving at 60 mph.  After how many hours do you catch up to your friend?  How far are you from LA at that time?

THANK YOU VERY MUCH!

For number one:
Wilma runs 2 miles faster than Fred each second. (Very fast)
In 15 seconds, Wilma will be 30 miles ahead.
Therefore, Fred must start 30 miles ahead of Wilma.
The algebraic equation would be 11(15)+x=13(15), where x is the headstart needed and the speed and time is shown.
For number two:
Salty gets there six times as fast, so she started when Sandy was 5/6 the way there, or 9:22 and 30 seconds, so Sandy walked 22 1/2 minutes before Salty started to drive.
The algebraic equation would be 4*27=24*x, where the speeds of both and the time of the first is shown. Solving, Salty took 4.5 minutes. That means she started at 9:22 and 30 seconds.
For number three:
In twenty seconds, the cheetah will run 1760 feet. The gazelle will run 1460 feet. Therefore, the cheetah can be 300 feet away.
The algebraic equation would be 20(88)=20(73)+x, where the speed and time is shown and x is the allowed distance between them.

[quote="Ttocs45"]For number one:
Wilma runs 2 miles faster than Fred each second. (Very fast)
In 15 seconds, Wilma will be 30 miles ahead.
Therefore, Fred must start 30 miles ahead of Wilma.
For number two:
Salty gets there six times as fast, so she started when Sandy was 5/6 the way there, or 9:22 and 30 seconds, so Sandy walked 22 1/2 minutes before Salty started to drive.[/quote]

But i need equations. such as 11x15+x=13x15


Find a formula for the sum of the following sequence:
1(1!) + 2(2!) + 3(3!) + ... + n(n!)

[quote="Boridi"]Find a generalization (a small formula) for the following series:
1(1!) + 2(2!) + 3(3!) + ... + n(n!)[/quote]

Hint:
Notice that 1(1!)=2!-1!.

Now make a claim for n(n!), and figure out a way for telescoping.

[hide]
$n\cdot n!=(n+1)!-n!$.

Thus, the series telescopes to $\boxed{(n+1)!-1}$.[/hide]

[quote="Boridi"]Find a generalization (a small formula) for the following series:
1(1!) + 2(2!) + 3(3!) + ... + n(n!)[/quote]
[hide="or a proof by induction"]
Hypothesis=$(n+1)!-1$
The base case of 1 is clearly true.
$(n+1)!-1+(n+1)(n+1)!$
$(n+1)!(n+1+1)-1$
$(n+2)!-1$
and thus our induction is complete[/hide]


integration
real analysis
function
real analysis unsolved

SHow that if $ f \in L^1(I,\lambda)$ and if $ \int_If(x)dx\neq 0$ there is a Lebesgue mesureable function $ g$ such that \[ \int_I|f(x)g(x)| < \infty\] and \[ \int_I |f(x)\parallel{}g(x)|^2\equal{}\infty\]

The set $ A_k: \equal{}\{x\colon 2^k\le |f(x)|<2^{k\plus{}1}\}$ has positive measure for some $ k\in\mathbb Z$. Choose $ g\in L^1(A_k)\setminus L^2(A_k)$.


topology
real analysis
real analysis unsolved

Let f:X to Y  be a continuous biyective function. Prove  that if X is compact then the inverse of f is also
continuous.

$ f^{ \minus{} 1}$ is continuous iff $ (f^{ \minus{} 1})^{ \minus{} 1}(F)$ is closed for every closed $ F\subseteq X.$ But any closed subset of a compact space is compact, so $ F$ is compact. Then $ (f^{ \minus{} 1})^{ \minus{} 1}(F) \equal{} f(F)$ is the image of a compact set, and the continuous image of a compact set is compact. Since compact sets (in $ Y$) are closed, we have that $ f^{ \minus{} 1}$ is continuous.

Can you construct the counterexample that shows that we needed compactness? That is can you find a bijective continuous $ f: X\to Y$ with $ X$ not compact such that $ f^{\minus{}1}$ is not continuous?

Something is fishy here. Consider a set $ M$ with more than one element, and take $ X$ the set $ M$ with the discrete topology $ \mathcal{T} \equal{} \mathcal{P}(M)$, $ Y$ the same set $ M$ with the trivial topology $ \mathcal{T} \equal{} \{\emptyset, M\}$, and $ f \equal{} \textrm{id}_M$. Then $ f$ is continuous, while $ f^{\minus{}1}$ is not, even when $ M$ is finite, hence compact (compact sets in $ Y$ are not necessarily closed).
I think we should assume $ X, Y \subseteq \mathbb{R}$, or what?

Certainly you need Hausdorff. According to French usage, "Compact" includes "Hausdorff", so maybe this came from something French.

I was at least tacitly assuming Hausdorff. My request for a counterexample without compactness also assumes Hausdorff.

For counterexample, consider X an interval [0,2 pi) and Y a circle of radius 1, and somehow map X bijectively and continuously onto Y ...


[size=34][color=Red]can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx[/color][/size]

[img]http://i74.photobucket.com/albums/i249/wenlong104/scan.jpg[/img]

[quote="kansai"][size=34][color=Red]can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx[/color][/size][/quote]

can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx

Wow... you do complex stuff in gr12... or we do really really easy stuff (probably the latter) :D
I don't know how to do any of these :|

[hide="1"]

In cirlce $O_1$, $r_1=1$. Note that the radius of this circle is half of the diagonal of the inscribed square $S_1$. So the diagonal of $S_1$ is equal to 2, and the side is $\frac{2}{\sqrt{2}}=\sqrt{2}$. So the area of $S_1=2$. Now draw $O_2$. Clearly his radius is half of the side of $S_1$, so $r_2=\frac{\sqrt{2}}{2}$. But we said befor that the radius is half of the diagonal of the inscroibed square, so the side of $S_2$ is equal to 1, and so his area. In general, if $l_n$ is the lenght of the side of the $nth$ square, $l_n=\frac{1}{2}\cdot l_{n-1} \cdot \frac{2}{\sqrt{2}}=\frac{l_{n-1}}{\sqrt{2}}$, but we said that $l_2=\frac{l_1}{\sqrt{2}}$, so $l_n=\frac{l_1}{(\sqrt{2})^{n-1}}=\frac{1}{(\sqrt{2})^{n-2}}$, and

$S_n=l_n^2=\frac{1}{2^{n-2}}$[/hide]

[hide="5"]

I don't know if I understood question a) because my english isn't very good, but in general $a_{ij}$ means the element that is in row $i$ and in coloumn $j$, so in matrix $A$, $a_{12}=5$, $a_{22}=0$, $a_{31}=2$.

For question b), (first of all remember that the result of an operation between 2 matrix is another matrix), to calculate $3B$ just multiply each element of the matrix by 3 (simple!), and  to calculate $A-3B$, you have just to subtract at each element of the matrix $A$ the corresponding element of $3B$. So for example in matrix $3B$, $a_{11}=6$, and in matrix $A-3B$, $a_{11}=-5$.

Finding $A'$ is easy, but (for me) it's difficult to explain the method in english...

[/hide]

A' simply is the "opposite" or "transposed" matrix.

For an m x n matrix, one must transpose it to a n x m matrix where each value [g,h] turns to [h,g]... get it?

For example...

[A]=  1 2
         3 4

[A']= 1 3
         2 4

what about question 3,4,7,8?

can any one do them?????

I am pretty sure that I can solve Question3. Not sure how to post networks though. THink the answer is

Vertex A: 3

Vertex B: 4

Vertex C: 4

Vertex D: 3

Vertex E: 4

Vertex F: 2

Traceable, in fact it has an Euler Path, because it has exactly 2 odd degree vertex.

These are all based on the assumption that I drew the network right.[/u]

[quote="Franz Joseph"]I am pretty sure that I can solve Question3. Not sure how to post networks though. THink the answer is

Vertex A: 3

Vertex B: 4

Vertex C: 4

Vertex D: 3

Vertex E: 4

Vertex F: 2

Traceable, in fact it has an Euler Path, because it has exactly 2 odd degree vertex.

.[/u][/quote]

can u explain it how u got the answer? and would u mind do the work on BITMAP image....and upload it?
plz this is very improtant to me.......thnx

if any one know how to do more plz.....do so.....

buecase i need ur help.........so plz do what ever question u know.......plz

can any one do question 4?


induction
number theory open
number theory

Here's what I was wondering:

Let Ln be the Lucas sequence defined as usual: L0=2, L1=1 and Ln+1=Ln+Ln-1 for all n>=1. Let p be a prime, p>3.

Is it true that all prime divisors q of Lp are of the form kp+1? I managed to show that if (5/q)=-1 then p|q2-1, and I think that if (5/q)=1 then p|q-1.

Here's something stronger: are all prime divisors of Lp 1 or -1 (mod 5)?

Does ne1 have good Mathematica skills? I tried to investigate this, but I couldn't do it for large numbers for a number of reasons: if I define the sequence recursively, it takes a long time to compute even L30, and if I define it by un+vn where u and v are the roots of x2-x-1=0 then even if I put Ln=Simplify[un+vn], for n>=126 the term doesn't appear in its integer form, but in a form involving \sqrt5; even if it can be further simp-lified, the software doesn't do it properly. Any ideas as to what I might do to investigate this with Mahtematica?

I think I can show that all prime divisors of [tex]L_p[/tex] with prime p>3 are of the form 5k+1 or 5k-1. In order to do this we first prove by induction that [tex]L_n^2-5F_n^2=4(-1)^n[/tex], where [tex]F_0=0, F_1=1, F_{n+1}=F_n+F_{n-1}[/tex] (the Fibonacci sequence). Since p is odd, we have [tex]q|5F_n^2-4=5F_n^2-2^2=>(\frac5q)=1=>(\frac{q}5)=1=>q=5k\pm1[/tex] for all prime q s.t. [tex]q|L_p[/tex]. We need the condition p>3 in order to have [tex]L_p[/tex] odd, so that q<>2.

You can find more info here:

[url]http://mathworld.wolfram.com/LucasNumber.html[/url]


function
limit
integration
trigonometry
real analysis
calculus
calculus computations

Here I recall some well-known identities (suppose that $ f$ to be a continuous function)
\[ \mathop {\lim }\limits_{n \to \infty } \frac {1} {n}\int_0^n {f\left( {\sin x} \right)dx} \equal{} \frac {1} {{2\pi }}\int_0^{2\pi } {f\left( {\sin x} \right)dx} .
\]

\[ \mathop {\lim }\limits_{n \to \infty } \frac {1} {n}\int_0^n {f\left( {\left| {\sin x} \right|} \right)dx} \equal{} \frac {1} {{\pi }}\int_0^{\pi } {f\left( {\sin x} \right)dx} .
\]

\[ \mathop {\lim }\limits_{n \to \infty } \int_0^1 {xf\left( {\sin \left( {2\pi nx} \right)} \right)dx} \equal{} \frac {1} {{4\pi }}\int_0^{2\pi } {f\left( {\sin x} \right)dx} .
\]

\[ \mathop {\lim }\limits_{n \to \infty } \int_a^b {f\left( x \right)\cos \left( {nx} \right)dx} \equal{} \mathop {\lim }\limits_{n \to \infty } \int_a^b {f\left( x \right)\sin \left( {nx} \right)dx} \equal{} 0.
\]

[quote="bookworm_vn"]
\[ \mathop {\lim }\limits_{n \to \infty } \frac {1} {n}\int_0^1 {xf\left( {\sin \left( {2\pi nx} \right)} \right)dx} \equal{} \frac {1} {{4\pi }}\int_0^{2\pi } {f\left( {\sin x} \right)dx} .
\]
[/quote]
This one is clearly incorrect. For any $ f$ bounded on $ [0,1],$ the limit on the left has to be zero. I assume there's some typographical error. What did you mean to say here?

The fourth one is the well-known Riemann-Lebesgue Lemma.

The first and second are specific cases of the following theorem:

Suppose $ g(x)$ is a periodic function of period $ P$ and is absolutely integrable over each period. Then
\[ \lim_{y\to\infty}\frac1y\int_0^yg(x)\,dx \equal{} \frac1P\int_0^Pg(x)\,dx
\]

[quote="Kent Merryfield"][quote="bookworm_vn"]
\[ \mathop {\lim }\limits_{n \to \infty } \frac {1} {n}\int_0^1 {xf\left( {\sin \left( {2\pi nx} \right)} \right)dx} \equal{} \frac {1} {{4\pi }}\int_0^{2\pi } {f\left( {\sin x} \right)dx} .
\]
[/quote]
This one is clearly incorrect. For any $ f$ bounded on $ [0,1],$ the limit on the left has to be zero. I assume there's some typographical error. What did you mean to say here?[/quote]

Thanks KM, there is no $ \frac {1} {n}$ in the identity. Sorry for the misprint.

Here is an other usefull identity
\[ \mathop {\lim }\limits_{n \to \infty } \int_a^b {f\left( {nx} \right)dx}  \equal{} \frac{{b \minus{} a}}
{T}\int_0^T {f\left( x \right)dx}
\]
provided $ f$ is a continuous and periodic function of period $ T$, $ a<b$.


function
inequalities
logarithms
inequalities unsolved

When $x,y,z\rightarrow 0$, $A\rightarrow\infty$.
[edit] Sorry that I thought it was maximum.

The proper english terminology is 'value', not 'price'.  Actually, you often would omit that and refer only to the maximum or minimum.

Ok do you have a solution ?I think we should write the $8^x$ as a function of $e$ but I have not foind a solution.

[quote="silouan"]If $x,y,z>0$ find the minimum price of

$A=\frac{8^x+8^y+8^z}{(xy)^8+(yz)^8+(zx)^8}$[/quote]

Actually the problem seems very easy, but please correct me if I am wrong.

I will show that for any three positive reals x, y, z, the inequality

$\frac{8^x+8^y+8^z}{\left(yz\right)^8+\left(zx\right)^8+\left(xy\right)^8}\geq f\left(\frac{16}{\ln 8}\right)$

holds, where f is the function defined by $f\left(x\right)=\frac{8^x}{x^{16}}$, and that this inequality becomes an equality if and only if $x=y=z=\frac{16}{\ln 8}$.

In order to prove this, consider the function $f\left(x\right)=\frac{8^x}{x^{16}}$ for $x\in\mathbb{R}^+$. Since $f^{\prime}\left(x\right)=\frac{8^x}{x^{17}}\left(x\ln 8-16\right)$, it is clear that this function f(x) attains its maximum at $x=\frac{16}{\ln 8}$. Hence, for every $u\in\mathbb{R}^+$, we have $f\left(u\right)\geq f\left(\frac{16}{\ln 8}\right)$, with equality if and only if $u=\frac{16}{\ln 8}$. Since $f\left(u\right)=\frac{8^u}{u^{16}}$, this becomes $\frac{8^u}{u^{16}}\geq f\left(\frac{16}{\ln 8}\right)$, so that $8^u\geq f\left(\frac{16}{\ln 8}\right)\cdot u^{16}$. Applying this to the numbers x, y, z, we get

$8^x+8^y+8^z\geq f\left(\frac{16}{\ln 8}\right)\cdot x^{16}+f\left(\frac{16}{\ln 8}\right)\cdot y^{16}+f\left(\frac{16}{\ln 8}\right)\cdot z^{16}$,

with equality holding if and only if $x=y=z=\frac{16}{\ln 8}$. This transforms into

$8^x+8^y+8^z\geq f\left(\frac{16}{\ln 8}\right)\cdot\left(x^{16}+y^{16}+z^{16}\right)$.

But applying the well-known inequality $u^2+v^2+w^2\geq vw+wu+uv$ to the numbers $u=x^8$, $v=y^8$, $w=z^8$, we obtain $x^{16}+y^{16}+z^{16}\geq\left(yz\right)^8+\left(zx\right)^8+\left(xy\right)^8$ with equality if and only if x = y = z.

Hence, altogether, we get

$8^x+8^y+8^z\geq f\left(\frac{16}{\ln 8}\right)\cdot\left(x^{16}+y^{16}+z^{16}\right)\geq f\left(\frac{16}{\ln 8}\right)\cdot\left(\left(yz\right)^8+\left(zx\right)^8+\left(xy\right)^8\right)$,

so that

$\frac{8^x+8^y+8^z}{\left(yz\right)^8+\left(zx\right)^8+\left(xy\right)^8}\geq f\left(\frac{16}{\ln 8}\right)$,

with equality if and only if $x=y=z=\frac{16}{\ln 8}$. This completes the proof.

  Darij


calculus
integration
function
limit
parameterization
calculus computations

Can we integrate 1/x between -1 and 1 by considering a contour with a semi-circle around the origin?

Yes, but I wouldn't.  That's a really inconvenient integral to use contour integration on because unlike the case when one or both of the limits are $ \pm\infty$, you can't just make the other paths go to 0.  You need to actually evaluate the integral along three other paths (if you use the same path that I'm using which is an indented contour along the real axis, a line from $ 1\rightarrow1\plus{}\rho i$, a line from $ 1\plus{}\rho i\rightarrow \minus{}1\plus{}\rho i$, and one from $ \minus{}1\plus{}\rho i\rightarrow \minus{}1$.)

This is not a good problem for talking about complex tools of integration (Cauchy's Theorem, the Cauchy Integral Theorem, the Residue Theorem). Those results require a closed loop. In this case, not only do we have to deal with the singularity at zero, we are also faced with the ends of the interval at $ \minus{}1$ and $ 1.$ We'd have to connect those in some way to have a closed loop - and there are rather too many ways to do that.

You mostly see complex tools used for integrals that start and end as special places - singularities of the function, or $ \pm\infty.$ In the case of $ \pm\infty,$ you often see an argument estimating the size of the integral over a large semicircle - we'd really like to see that go to zero.

None of that applies here. It's really more likely that you'll use ordinary calculus to tell you something about a complex integral than the other way around.

Now, about that ordinary calculus:

If an integral is improper for more than one reason, we [i]must[/i] break it into pieces in which each piece isolates only one of the reasons the integral is improper. Then each and every such piece must converge in isolation for us to accept the improper integral as converging.

$ \int_{\minus{}1}^1\frac{dx}{x}$ is improper for two reasons: the left side of zero and the right side of zero.

Isolate those two reasons: $ \int_{\minus{}1}^0\frac{dx}x\equal{}\minus{}\infty$ and $ \int_0^1\frac{dx}x\equal{}\infty.$

Hence the improper integral diverges.

Now, there is an idea beyond that of improper integral: [b]principal value[/b]. If an integral is improper for two reasons and if there is a canonical way to "balance" the two pieces and to take the limit of the two partial integrals "at the same rate," and if the resulting limit exists, then we call that the principal value. In this particular case, the principal value would be:

$ \text{p.v.}\int_{\minus{}1}^1\frac{dx}x\equal{}\lim_{\delta\to 0}\left[\int_{\minus{}1}^{\minus{}\delta}\frac{dx}x\plus{} \int_{\delta}^1\frac{dx}x\right]\equal{}0.$

We can see that this is zero just by symmetry.

Again, I see no place for complex tools in any of this, and find it more likely that this real principal value result would inform some complex computation than the other way around.

[quote="AndrewTom"]Thanks JRav, could you show me how you've done it?[/quote]
As both Kent and I pointed out, it's really inconvenient.  If you want an idea...
We want the integral along the real axis, call it $ I$.  Also, since there are no singularities inside the contour, $ \Gamma$, the residue theorem yields
$ \oint_{\Gamma}\frac{dz}{z}\equal{}0$
The indented semi-circular contour yields $ \minus{}\pi i\text{Res}(f,0)\equal{}\minus{}\pi i$.  Now the annoying part: you have to parametrize each of the other three remaining contours (the path I mentioned above, $ \rho$ is more or less arbitrary, so long as it isn't 0).

If you want to make this calculation simple, simply close the contour with a semi-circle.  Let's make it of radius $ r$: Then $ z\equal{}re^{i\theta}$ and $ dz\equal{}iz\,d\theta$.  Hence
$ \int_{\gamma}\frac{dz}{z}\equal{}\int_{0}^{\pi}i\,d\theta\equal{}i\pi$
Combining this with the integral along the indent around the origin, we get $ I\equal{}0$, as desired.

Sorry, JRav (and Kent). I can see how the Cauchy principal value does it. but I cannot see how to parametrize the 3 line-segments (1 to 1 + rho i, etc.); nor can I see how these line-segments tie in with closing the contour with a semi-circle of radius r. Where is this semi-circle? The results -pi x i and pi x i seem OK, but I can't see how to deal with the rest of the contour, nor exactly where it is. Sorry to be a pain, but I'd like to get to the bottom of this so I can get on with other problems where I feel I'll have more success.

Ignore my first parametrization (the three lines), as it's more annoying.  The idea with that is to close the contour by forming an indented rectancle.  Remember, we have to close the contour to apply the residue theorem.  You have to parametrize each side separately, and then integrate along each line.

My second path is to simply close the contour with a semi-circle of radius 1.  Think of it like [url=http://img.tfd.com/wn/4B/6C995-semicircle.gif]this[/url], except with the indentation around the origin.  The integral we desire, $ I$, is the straight portion along the real axis (do you see why?).  The semi-circle contour is the upper half, which connects $ z\equal{}1$ and $ z\equal{}\minus{}1$.  The reason I chose a semi-circle is because it is only one contour, and it is simple to parametrize and integrate.  And it has radius 1 because that's how long our interval of integration requires.

Actually, I just noticed that previously, I said to use a semi-circle of radius $ r$, instead of 1.  By mistake, I actually showed that $ \int_{\minus{}a}^{a}\frac{dx}{x}\equal{}0$ for any $ a$, which is exactly what we would expect by symmetry.

Thanks JRav. That was the source of my confusion.


Hey does anyone have 2000-2003 school if you do I trade any of my stuff (I give you 2 items you give me 1).
This is my inventory:
School:
2008 sprint, target, countdown, answer key
2007 sprint, target, countdown, answer key
2006 sprint, target
Chapter:
2008 sprint, target, team, countdown, solution key, answer key
2007 sprint, target, team, countdown, solution key, answer key
2006 sprint, target, team, countdown, solution key, answer key
2005 sprint, target, team, countdown ppt, solution key, answer key
2003 sprint, target, solution key
2002 sprint, target, answer key
2001 sprint, target, answer key
2000 sprint, target, answer key
1999 answer key
1998 countdown
State:
2008 sprint, target, team, countdown, solution key, answer key
2007 sprint, target, team, countdown, solution key, answer key
2006 sprint, target, team, countdown, solution key, answer key
2005 sprint, target, team, countdown, solution key, answer key
2004 sprint, target, team, countdown, solution key, answer key
2003 sprint, target, solution key
2002 sprint, target, countdown ppt, answer key
2001 sprint, target, answer key
2000 sprint, target, answer key
1999 sprint, target, answer key
1996 sprint, target, team, countdown, answer key
1995 sprint, target, team, countdown, answer key
1994 sprint, target, team, countdown, answer key
1993 sprint, target, team, answer key
1992 sprint, target, team, answer key
National:
2008 sprint, target, team, countdown, answer key
2007 sprint, target, team, countdown, answer key
2005 sprint, target, team, countdown ppt (with answers)
2004 sprint, target, team
2003 sprint, target, answer key
2002 sprint, target, answer key
2001 sprint, target, team, answer key
2000 sprint, target, answer key, solution key
1999 sprint, target, team, answer key
1997 sprint, target, team, countdown, answer key
1996 target
1995 sprint, target
1999 sprint, target, team, answer key
1993-4 sprint, target
1991 sprint, target
Handbooks(full solutions):
2001-2
2004-5
2005-6
2007-8
2008-9
handbooks( no full solutions):
2004-5
2005-6
2006-7
2008-9
If anyone wants to trade pm me or e-mail.
My e-mail is    dalibi@verizon.net

in the future, please refrain from starting a new thread and just post in the big thread.


function
inequalities
algebra open
algebra

[b]Why function f(x)=log(x) is a concave function?[/b]
[color=green](concave function:function f:[a,b]*->R  is a concave function if for each numbers like x,y (that a<=x,y<=b) :
(( f(x)+f(y))/2)<=f((x+y)/2) )[/color]

 :lol:  :lol:

Because it has a negative second derivative.

Anyways, your condition is not enough, you need that midpoint inequality + continuity to conclude that a function is concave.


Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $64$ square units. What is the number of square units rounded to the nearest integer in the area of the largest circle which can be drawn in pentagon $MNAUQ$?

[img]7413[/img]

[hide]Let $O$ be the center of the circle and let $P$ be the foot of the perpendicular from $O$ to $MN$. Then $OP = r$ and $OU = r\sqrt{2}$. By the Pythagorean Theorem, $UP = 6\sqrt{2}$. So, $r+r\sqrt{2}= 6\sqrt{2}$. Then $r = \frac{6\sqrt{2}}{1+\sqrt{2}}$. Therefore, $r^{2}=216-144\sqrt{2}$. This makes the area $(216-144\sqrt{2})\cdot\pi \approx 39$.[/hide]

Question -- do we know for sure that the largest circle that can be drawn in the pentagon will be tangent to MN AU And QU as (almost) shown?

If you visualize a baby circle starting at U and then growing bigger and bigger while staying tangent to QU and AU, then it's logical that MN will eventually prevent its growth.

[quote="lifesapain"]lets see... [hide]could the answer be 25.1 units squared perhaps? [/hide][/quote]

This is what I got.. :)
Is it right?

BTW where did you get this problem?

there are many proofs of this, but I'm not in the mood to post 1. search for proofs about a triangle's incircle. (the circle tangent to the sides of a triangle)

[quote=".:Logic:."][quote="lifesapain"]lets see... [hide]could the answer be 25.1 units squared perhaps? [/hide][/quote]

This is what I got.. :)
Is it right?[/quote]

I got the same answer as DarkKnight (for the same reasons).  Perhaps you could post your reasoning and we can figure out if one of us went astray somewhere.

ok then.

[hide]the legs of the triangle have to be 4 units.
this makes the hypotenuse 4 root 2
the hypotenuse of the triangle is also the circles diameter
so the radius is 2 root 2
this makes the area of the circle 25 something
by rounding, i ended up with 25.1
see? i hope im right for once[hide]

[quote="lifesapain"]ok then.

the hypotenuse of the triangle is also the circles diameter
[/quote]

I don't think that is true.  Why do you think it is?

im pretty sure it just is but dont know how to explain it sry..

Well, I'm pretty sure that's where you went wrong.  It would be true if the center of the circle were directly below N, but that won't be the case here -- the center will be below and to the right of N. (So, if we call L the midpoint of MN, and O the center of the circle, triangle LNO is not necessarily isosceles -- angle LNO is not necessarily 45 degrees.)

Well, I can disprove lifesapain's theory.  Take the midpoints of the other two sides and draw another segment of identical length.  It is easy to see that it's not the diameter.

I got what DarkKnight got.

The circle is tangent to all 3 of those sides. Draw the tangents and the diagonal of the square. You can see that r (the radius) + r*rt2 + 2rt2 = 8*rt2.

r(rt2+1) = 6rt2

Solve for r: r = (12-6rt2)

Area = (12-6rt2)^2*pi = 38.8 = 39.


The numbers 1,2,3,...,n are written in a row. It is permitted to swap any two numbers. If 1007 such operations are preformed, is it possible that the final arrangement of numbers coincides with the original?

[hide="Group Theory"] The identity of the permutation group is an even permutation (having an even number of 'swaps') while making $ 1007$ swaps would make an odd permutation.[/hide]


Prove that A union B = A intersection B if and only if A = B.

I'm not sure how to start on this... could someone point me in the right direction?

[hide="Hint"]Use proof by contradiction.[/hide]

[hide="Solution"]First, we assume that the union of $ A$ and $ B$ is equal to the intersection of $ A$ and $ B$. Assume $ A \neq B$. We assume without loss of generality, via symmetry, that $ A$ has an element $ n$ that $ B$ does not have. However, this directly implies that the intersection of $ A$ and $ B$ does not contain $ n$, whereas the union does contain $ n$. Contradiction. Therefore, the union of $ A$ and $ B$ being equal to the intersection of $ A$ and $ B$ implies that $ A$ and $ B$ are equal.

Now, we must prove that $ A \equal{} B$ implies the union of $ A$ and $ B$ is equal to the intersection of $ A$ and $ B$. Assume the union of $ A$ and $ B$ contains an element $ k$ that the intersection does not contain, since the union must necessarily contain all the elements within the intersection. However, since $ A \equal{} B$, both $ A$ and $ B$ share that element, so it must be in the intersection of $ A$ and $ B$. Contradiction. Therefore, $ A \equal{} B$ implies the union of $ A$ and $ B$ is equal to the intersection of $ A$ and $ B$.[/hide]

I'm going to read more about proof by contradiction. That's such a beautiful way to solve the problem. Thanks for your help. :)

[hide="Or..."]
Let $ A\cup B \equal{} A\cap B$. If $ A$ is empty, then obviously $ B$ must be empty as well. Otherwise, consider an element $ x\in A$. Then $ x\in A\cup B$ and so $ x\in A\cap B$ implies $ x\in B$ as well, so $ A$ is a subset of $ B$. A similar argument shows that $ B$ is a subset of $ A$, and so $ A \equal{} B$, as desired.

Now let $ A \equal{} B$. Then (trivially) $ A\cup B \equal{} A \equal{} B \equal{} A\cap B$, since all four sets contain the exact same elements.
[/hide]

[hide="solution"]If: If $ A\equal{}B$, then by definition $ |A\cup B|\equal{}|A\cap B|$.

Only if: $ |A\cup B|\geq \max(|A|,|B|)\geq \min(|A|,|B|)\geq |A\cap B|$. Since $ |A\cup B|\equal{}|A\cap B|$, we have $ \max(|A|,|B|)\equal{}\min(|A|,|B|)$, or $ |A\cup B|\equal{}|A|\equal{}|B|\equal{}|A\cap B|$.

Now let us assume for the sake of contradiction that there is an element $ x$ that is in $ B$ but not in $ A$. Therefore $ |A\cup B|>|A|$. We have a contradiction. Therefore $ B$ is a subset of $ A$. We have a similar proof that $ A$ is a subset of $ B$. Therefore $ A\equal{}B$.[/hide]


integration
function
calculus
trigonometry
limit
logarithms
real analysis

Calculate :
  $ \int_{0}^{\infty}{\frac{sin^2{x}dx}{x^3}}$

This diverges, because it looks like $ \frac 1x$ near zero. Since it's a nonnegative function, the integral equals $ \plus{}\infty.$

[quote="Kent Merryfield"]This diverges, because it looks like $ \frac 1x$ near zero. Since it's a nonnegative function, the integral equals $ \plus{} \infty.$[/quote]
 Please Calculate :
 $ \int_{0}^{\infty}{\frac {sin^3{x}dx}{x^2}}$

[quote]Please Calculate :
 $ \int_{0}^{\infty}{\frac {sin^3{x}dx}{x^2}}$[/quote]
 Yes I have a mistake ! Sorry !!

what is your question??????????

if it is $ \int_{0}^{\infty}{\frac{sin^2{x}dx}{x^3}}$, then integral does not converge.
if it is $ \int_{0}^{\infty}{\frac {sin^3{x}dx}{x^2}}$ then ans= $ \frac{3 log(3)}{4}$


Hi, this is my first post in this wonderful forum.  :lol:

Can someone please give me a hand on proving the following proposition: "For any $ m,n\in N$, there exists some unique $ r_o,r_1,...,r_k \in N$ such that the following conditions hold:
(1) $ m\equal{}r_k*n^k\plus{}r_{k\minus{}1}*n^{k\minus{}1}\plus{}...\plus{}r_1*n\plus{}r_0$
(2) $ r_o,r_1,...,r_k\in \{0,1,2,...,n\minus{}1\}$

I have no clue on how to prove the part concerning the "uniqueness" of $ r_o,r_1,...,r_k\in$. Someone please help.

Thank you very much,  :)

Take another different set p0,p1,p2 and so on
such m = pk r^k + ...+p1*r+p0

Now equate both by finding m mod n, mod n^2 and so on.


1. In organic chemistry, naming alkanes and alkenes, there's something I'd like to be made clear. For example, with alkanes, lets say its butene with a ring structure, it would be named cyclobutene. With alkenes, in a butene with ring structure and a single double bond, it is also named cyclobutene. If asked to draw cyclobutene how would you know which to draw? The one with or the one without the double bond.

2. Naming alkanes with ring structures and lets say there's a methyl branch, is it 1-Methylcyclobutene or would it just be Methylcyclobutene where the former one is used if it is an alekene?

1.  Actually, the ending "-ene" in "butene" means there's a double bond, automatically (really the infix "-en-" if you want to be technical).  To name the straight chain alkane with four carbons, it would be called "butane" (note the ending "-ane" and its analogy to alkane).  So a four carbon cycloalkane would be named cyclobutane, and if there is a double bond it would be cyclobutene.  It's only one letter that makes the difference, but with practice you'll get used to it.

2.  Again, to denote an alkane vs. an alkene, you have to make sure you're using the right ending.  Also, you don't have to bother putting in numbers if there is only one way to put the structure together -- for example, in methylcyclobutane.  No matter where you draw the methyl, you always end up with the same compound.  So in this case, the 1 is unnecessary.

Hope that helps.


Let $ a,b,c\in \mathbb{R}, a,b,c>0$. Prove that $ \dfrac{a\plus{}b}{c}\plus{}\dfrac{b\plus{}c}{a}\plus{}\dfrac{c\plus{}a}{b}\geqslant 6$. help me. thank!

Consider $ x\plus{}\frac{1}{x}$ for $ x>0$.

Hint 1: We can write $ \frac{a\plus{}b}{c}\equal{}\frac{a}{c}\plus{}\frac{b}{c}$
Hint2: After writing, just use AM GM directly
Hint 3: Consider Kunny's hint, it too leads to a solution.

hello, another solution, we have $ \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\minus{}6\equal{}\frac{a^2b\plus{}ab^2\plus{}a^2c\plus{}ac^2\plus{}b^2c\plus{}bc^2\minus{}6abc}{abc}$. The nominator is nonnegative because we have
$ \frac{a^2b\plus{}ab^2\plus{}a^2c\plus{}ac^2\plus{}b^2c\plus{}bc^2}{6}\geq\sqrt[6]{(abc)^6}\equal{}abc$ by the AM-GM inequality.
Sonnhard.


If $ x\plus{}y\equal{}3$ and $ xy\equal{}4$, what is the value of $ x^4y\plus{}xy^4\plus{}x^3y^2\plus{}x^2y^3?$

It equals
[hide]$ xy(x^3 \plus{} y^3 \plus{} 3x^2y \plus{} 3xy^2 \minus{} 2x^2y \minus{} 2xy^2) \equal{} xy((x \plus{} y)^3 \minus{} 2xy(x \plus{} y)) \equal{} 12$
[/hide]There are many other ways to simplify it.


Prove that

$ \frac{a^2}{r_a^3}\plus{}\frac{b^2}{r_b^3}\plus{}\frac{c^2}{r_c^3}\leq\frac{4(R\minus{}r)}{3r^2}$

[quote="Ligouras"]Prove that

$ \frac {a^2}{r_a^3} \plus{} \frac {b^2}{r_b^3} \plus{} \frac {c^2}{r_c^3}\leq\frac {4(R \minus{} r)}{3r^2}$[/quote]

$ \Longleftrightarrow(b^3 \minus{} b^2c \minus{} b^2a \plus{} 4bca \minus{} ba^2 \minus{} bc^2 \plus{} a^3 \minus{} c^2a \minus{} ca^2 \plus{} c^3)(b^2 \minus{} ab \plus{} a^2 \minus{} bc \minus{} ac \plus{} c^2)\geq 0$

only prove

$ b^3 \minus{} b^2c \minus{} b^2a \plus{} 4bca \minus{} ba^2 \minus{} bc^2 \plus{} a^3 \minus{} c^2a \minus{} ca^2 \plus{} c^3\geq 0$

$ \Longleftrightarrow2z^2x \plus{} 2y^2x \plus{} 2yz^2 \plus{} 2y^2z \plus{} 2x^2z \plus{} 2x^2y \minus{} 4xyz\geq 0$

obvious true.


I'm not sure where this goes, so I'll post it here. I typed my commands on TeXinicCenter, but I don't know how I can get the symbols. I already read the page about LaTeX on AoPS website, but I can't follow the steps. Can anyone help me?

[quote="airelemental135"]I'm not sure where this goes, so I'll post it here.[/quote]This goes in the $\LaTeX$ forum. Follow the links in my signature to quickly learn how to use the symbols (there is a .pdf file attached in the first topic which contains the commands of the most common used symbols).


group theory
abstract algebra
advanced fields
advanced fields unsolved

Let $ G$ be connected topological group and $ U$ an arbitrary neighbourhood of $ e$ in $ G$. Show that $ G \equal{}\cup_{n \equal{} 1}^{\infty}U^{n}$.

obviously, we may assume that $ U$ is open and symmetric (otherwise take an open neighborhood $ W$ in $ U$ and consider $ W\cap W^{\minus{}1}$). then $ V: \equal{}\cup_{n}U^{n}$ is an open subgroup of $ G$, thus also closed. because $ G$ is connected and $ V$ is nonempty, it follows $ V\equal{}G$.


arithmetic sequence
number theory unsolved
number theory

Prove that there are no positive integers $ a$, $ b$, $ c$, $ d$ in arithmetic progression (consecutive terms of an arithmetic progression) such that the numbers $ ab \plus{} 1$, $ ac \plus{} 1$, $ ad \plus{} 1$, $ bc \plus{} 1$, $ bd \plus{} 1$ and $ cd \plus{} 1$ are all perfect squares.