Note that the inequality that needs to be proved is symmetric w.r.t. $a$ and $b.$ So WLOG, we assume $a \geq b.$

There are three cases we need consider here.

[b]Case 1[/b]: $c \geq a \geq b.$
Then the given inequality
$\Leftrightarrow (a-b)+(c-b)+(c-a) \geq 2\cdot \frac{c^{2}-ab}{a+b}$
$\Leftrightarrow (c-b)(a+b) \geq c^{2}-ab$
$\Leftrightarrow ca+bc-ab-b^{2}\geq c^{2}-ab$
$\Leftrightarrow c(a+b)+(c^{2}-b^{2}) \geq 0$, which is true.
[b][Edit[/b]: This is wrong and I am working on it now. Thanks [b]towersfreak2006[/b] for pointing out the error! I guess this is where we use the given equality... I still haven't been able to fix this one. I will leave it as it is until someone fixes it or comes up with another solution. ]
[[b]Edit[/b]: Finally, [b]Kurt Gödel[/b] pointed out the error below in the given problem. No doubt, I had such a hard time trying to prove what's not possible. Thanks Kurt! ]

[b]Case 2[/b]: $a \geq c \geq b.$
Then the given inequality
$\Leftrightarrow (a-b)+(c-b)+(a-c) \geq 2\cdot \frac{c^{2}-ab}{a+b}$
$\Leftrightarrow (a-b)(a+b) \geq c^{2}-ab$
$\Leftrightarrow a^{2}-b^{2}-c^{2}+ab \geq 0$
$\Leftrightarrow (a^{2}-c^{2})+b(a-b) \geq 0$, which is again true.

[b]Case 3[/b]: $a \geq b \geq c.$
Then the given inequality
$\Leftrightarrow (a-b)+(b-c)+(a-c) \geq 2\cdot \frac{c^{2}-ab}{a+b}$
$\Leftrightarrow (a-c)(a+b) \geq c^{2}-ab$
$\Leftrightarrow a^{2}+ab-ca-bc \geq c^{2}-ab$
$\Leftrightarrow (a^{2}-c^{2})+a(b-c)+b(a-c) \geq 0$, which is again true.

In the first case, you didn't change the sign of $c^{2}$ when you moved it to the LHS.

[quote="Lovasz"]Let $a,b,c>0$ satisfy $a^{2}+b^{2}+c^{2}=1$. Prove that: \[|a-b|+|b-c|+|c-a| \ge 2\cdot\frac{c^{2}-ab}{a+b}.\] [/quote]

I could be wrong, but this inequality does not seem to hold. Choose $a=b=\frac{1}{2}$ and $c=\frac{1}{\sqrt{2}}$ for example. In fact, whenever $c>a=b$, the statement is false. Because we would have to prove that $2(c-b)\geq \frac{c^{2}-b^{2}}{b}$, which is equivalent with $b\geq c$, contradicting the assumption $c>a=b$.


Let a,b,c be positive real numbers. and a+b^2+c^3=1, Prove that the following inequality holds
$ (1\minus{}a)(1\minus{}b^2)(1\minus{}c^3) \geq 8ab^2c^3$
BQ

[quote="xzlbq"]Let a,b,c be positive real numbers. and a+b^2+c^3=1, Prove that the following inequality holds
$ (1 \minus{} a)(1 \minus{} b^2)(1 \minus{} c^3) \geq 8ab^2c^3$
BQ[/quote]

It can be easily solved by AM-GM :)

[quote="xzlbq"]Let a,b,c be positive real numbers. and a+b^2+c^3=1, Prove that the following inequality holds
$ (1 \minus{} a)(1 \minus{} b^2)(1 \minus{} c^3) \geq 8ab^2c^3$
BQ[/quote]

Let $ x\equal{}a,y\equal{}b^2,z\equal{}c^3$, ineq becomes $ x\plus{}y\plus{}z\equal{}1,\,\,(1\minus{}x)(1\minus{}y)(1\minus{}z)\ge 8xyz$, i.e. $ (x\plus{}y)(y\plus{}z)(z\plus{}x)\ge8xyz$, i.e.... :wink:

open the brackets. it becomes  $ ac^3  \plus{} ab^2  \plus{} b^2 c^3  \ge 9ab^2 c^3$  or  $ {1 \over a} \plus{} {1 \over {b^2 }} \plus{} {1 \over {c^3 }} \ge 9$  which is clear from Cauchy-Schwarz(in Engel form).


integration
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the 2 curves are...

y = x^5 - 2ln(x+5)

and

y = x^3 - 2ln(x+5)

now i think the point of intersections are x= 0 and 1
so i would do the antiderivative from 0 to 1 of the top curve - bottom curve,,, but wwhat is the top curve and what is the bottom curve?

You don't need to know which the top curve is.
$ x_1 \equal{} \minus{} 1,x_2 \equal{} 0,x_3 \equal{} 1$
$ S \equal{} \int_{ \minus{} 1}^{1}\left|x^5 \minus{} x^3\right|dx \equal{} \left|\int_{ \minus{} 1}^{0}\left(x^5 \minus{} x^3\right)dx\right| \plus{} \left|\int_{0}^{1}\left(x^5 \minus{} x^3\right)dx\right|$
$ \equal{} 2\left|\int_{0}^{1}\left(x^5 \minus{} x^3\right)dx\right| \equal{} \frac {1}{6}$.

The second curve is on top. Because you're only dealing with differences, the natural logarithms are there just to annoy/confuse people.

You do need to know which is on top- or at least realize that it's not the same on the two halves.

The first curve is on top from $ x_1 \equal{} \minus{} 1$ up to $ x_2 \equal{} 0$.
The second curve is on top from $ x_2 \equal{} 0$ up to $ x_3 \equal{} 1$.
But calculating this area without "which is the top curve?". We note  $ y\equal{}x^5\minus{}x^3$  is an odd function.


linear algebra
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probability and stats

This is a Markov chain {Xn, n >= 0} with state space E = {1,2,3} The transition matrix P is
.2 .5 .3
.2 0 .8
.3 .1 .6
The initial distribution vector is p(0) = (0,1,0).
How do I calculate these?
Pr{X3 = 3 given that X1 = 1}
Pr{X3 in {2,3} given that X1 = 1}
Pr{X2 = 1, X3 = 2}

It depends a little bit on your notation. If the entry $ P_{ij}$ denotes the probability to arrive at state $ j$ conditional upon starting at state $ i$, then we get the following:

$ \mathbb{P}(X_3\equal{}3|X_1\equal{}1)\equal{}(P^2)_{13}$

$ \mathbb{P}(X_3\in\{2,3\}|X_1\equal{}1)\equal{}1\minus{}(P^2)_{11}$

$ \mathbb{P}(X_2\equal{}1,X_3\equal{}2)\equal{}\mathbb{P}(X_3\equal{}2|X_2\equal{}1)\cdot\mathbb{P}(X_2\equal{}1)\equal{}P_{23}\cdot (P^2)_{21}$

(For the last identity note, that we start at the zero step always in state $ 2$. Apparently, I have left the matrix calculations for you.)


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I think it should be legal. They make it illegal because of religion, but religion isn't supposed to be controlled by the government.

Lol. I don't have a problem with same-sex marriages. It seems like the only reason not to have same-sex marriages is because of religion, and what happened to the separation between church and state?

[quote="13375P34K43V312"]MA stands for Massachusetts, genius[/quote]
They're both on the east coast... Close enough...

[quote="joml88"] ... I think it's certainly within the government's power to promote natural, healthy lifestyles.[/quote]

Natural, healthy lifestyles??! Sorry gotta go now but I'll comeback to this, In the meantime, if you please clarify, what is a natural, healthy lifestyle??

Ursula

[quote="joml88"]I'm not very religious, but I maintain that marriage should be between a man and a woman.  The government should not acknowledge same-sex marriage.  Neither should they punish such couples.  I see marriage as the beginning of a pact between two people that they will join together as one, reproduce, and raise a family.  Two people of the same sex cannot do this by themself.  I mean, I think nature made it pretty obvious that a couple should be between a man and a woman...  I think it's certainly within the government's power to promote natural, healthy lifestyles.[/quote]

OK, you start giving arguments in the line where you go

[quote="joml88"] ..I see marriage as the beginning of a pact between two people that they will join together as one, reproduce, and raise a family...[/quote]

Well, of course you can see marriage in any way you like, as for myself, I don't feel this respect you seem to follow for the institution of marriage. Your viewpoint is in my opinion, at least naive, probably obsolete. In the first place, while the idea of everlasting love and a pact and all that is, in my opinion very beautiful indeed, you cannot expect everybody in the world shares this particular liking of ours and act accordingly. You cannot expect that every couple that decides to engage is doing so obbeying to such noble feelings, nor you can expect it to be their intention in the first place. It doesn't seem that you can also deny them the possibility of getting married, so your idilic idea of marriage in modern society stumbles in the first step.

And about the supposed fiancés ultimately reproducing and raising families, jo, you simply cannot base your argument in that because having children is a decision that corresponds to the couple and I'm pretty sure you as much as I too often hear about couples nowadays who do not intend at all to have any children. When a homosexual couple decides to marry, or for what matters, to enter a long relationship, obviously they have decided not to have children in the way you call 'natural'...just as the natural couple next door has.

So for my point of view, marriage is not, in principle, the begginning of any bond or commitment to maintain our species since this last thing is no further an issue. It might be\become this bond or it can end with divorce in three months. Marriage is in other respects, a deprecated legal status, which we inherit from times when we were deeply tied to religious ideas. I understand our modern society does need something like marriage but the whole implication of religious instances is out of time.

As for the 'natural' argument, man has overcome nature. So there are many things that do not apply to mankind as you apply them to chimpanzees or whatever. If the chimpanzees do not reproduce they die as a species, if every person of the world turns all of sudden homosexual I seriously doubt about our extinction. A Los Alamos kind of biotechnological effort I'm pretty sure would garantee us in no time the biological possibility of surviving, and even that the children me and my gay couple are going to have will look just like us cause they'll have our genes. Man has overcome nature in many other respects, we can argue this if you like but for instance, man in driven more by society and by learned knowledge than by instincts(this by itself is a long discussion with many important followers in both sides). It has not overcome N in the fact that it still depends on it, thank goodness! because it might be a reason not to destroy it. So man is in the position and in the duty right now of taking care of nature, not of following its commandments.

And I just don't see in which way a gay lifestyle is.. unhealthy?

Ursula

Perhaps "healthy" isn't the best adjective for what I'm saying...let's say "procreative".

However, I still think that homosexuality indicates an unhealthy mind.  I see it as a disorder...not natural.  This is a much more debatable opinion, so let's leave it out of the argument.

[quote="joml88"]
However, I still think that homosexuality indicates an unhealthy mind.  I see it as a disorder...not natural. [/quote]

But what is exactly a healthy mind? I have many homosexual friends and I do not see anything unusual in their behavior except for the fact they're gay. Perfectly normal beings in every other respect for life and even in the part that concerns love, because I see beautiful love relationships between people of the same sex that would be the envy of many heterosexual couples. So they're basically abnormal merely in the fact that they're getting laid with someone of the same sex. And this is what we are judging them diifferent for?

Ursula

[quote="13375P34K43V312"]What are your opinions on same-sex marraige? I live in the only state in which it is legal (MA), but I don't really have a problem with it.[/quote]

"state refers to set of governing institutions with sovereignty over a definite territory".  So Belgium and the Netherlands are states too, making your claim about living in the only state in which it is legal false.
But I don't know where you live?  Montenegro?  Liechtenstein?  Or maybe Djibouti?

Allowing them to marry allows them to take better care of matters like inheritance....

I never understood what people mean when they say "I THINK a marriage is something between a man and a woman" :huh:
Then what is a marriage : getting married, divorcing a year later, fighting over every dollar/euro you have to pay for a kid, forcing the kid to walk from one car to the other on some parking while the parents don't look at each other?
Let's face it people, marriage between heterosexuals isn't that sacred anymore either....

He meant the only state in the United States.

Fred, what you're saying only leads me in the direction of doing away with marriage for everyone.  Arguing that since a large percentage of heterosexual marriages end up in divorce has nothing to do with homosexuals having the right to a marriage.

Actually, the whole idea of doing away with marriage in the political sense seems fine to me.  If people want to get married, they can do so in their church but tying this to politics is sure to go wrong.  I don't foresee such a change happening anytime soon though...

And I haven't forgot about ursula's post.  It will require more time on part to articulate my views more precisely and address his points.  But overall, I seem to get the feeling that we different on a fundamental premise which means we'll probably end up agreeing to disagree since I doubt either of us will change our stance on this premise.  I'll re-evaluate and get back later. (classes and such now...)

[quote="Boy Soprano II"]Same-sex couples [i]can[/i] have children (modern scientific advances, I suppose).  I know a same-sex couple who have had a child.  Furthermore, what is the point of denying them the right to be legally married?  It looks like discrimination (which it is, if from nothing but a taxpayer's point of view).  As for marriage outside of the church, it definitely serves a purpose.  Many people wish to be married, but do not belong to a church (or other religious entity; pardon my loss of generality), or belong to two differing churches which cannot be reconciled.[/quote]
Well I think that is quite different in the USA.
In Belgium you have to go to City Hall to get married legally.  Then you go to church for a (usually catholic) wedding.  Only the first is strictly necessary. (I don't plan on ever doing the second.)
Am I correct that two procedures are not necessary in the USA?

If everyone was homosexual and stuck to it, the human race would die and we would not be having this debate.  :rotfl:

Otherwise, I actually find it unnatural in the sense that if we were in the jungle and living by the laws of [i]nature[/i] then homosexuality would be a hinderance to creating survival of the fittest since, it is possible, that theoretically, a homosexual could have that single gene to bolster the immunity of an entire population, thus saving its future.

But, we are not abiding by the laws of the jungle, and hence, based on the separation between church and state, I say we just keep giving them a civil union; a marriage is what people define it as and until the definition changes, well...yea.

I guess until someone comes and changes the US definition of marriage, it is civil unions for any homosexuals.

I still find it disturbing that there are homosexuals, but it is another person's choice and I don't think anyone can really interfere with it.

wow... quite a debate..

I think same-sex marriages should be legal. If it is legal, people actually have a marriage "freedom" in my opinion.

[quote="bookaholic"]This debate is confusing. The first confusing thing about it is that some of you seem to think that what the bible says matters at all when we're talking about legal issues. It doesn't and shouldn't. The second confusing thing is that some of you seem to think that what you consider right and natural should be everyone's definition of right and natural. I don't personally know any homosexuals, but I would guess that they find the idea of heterosexual sex just as unnatural and disgusting as some people here find homosexual sex. Why should your opinion be counted over theirs? Because they're a minority?[/quote]

LOL.

"I would guess that they find the idea of heterosexual sex just as unnatural and disgusting as some people here find homosexual sex."

LOL LOL LOL

I really don't see the big problem that people have with homosexuals.... It doesn't affect you at all.

I've always been fine with homosexuals; they don't really bother me or anything

i mean if they started banging each other in the middle of the street i might be offended, but not because they're gay, but because i don't believe in blatant display of sexual behavior

It doesn't matter how educationally unbiased all of you people sound, it boils down to this:

If you saw a hot lady and a hot man having sex in the street, you'd turn to look.

If you saw two hot men banging each other in the street, you'd turn away.

I agree that I might glance out of simple curiosity whether it was two gay men or a man and a woman, but I wouldn't blatantly stare as if interested in what they're doing. I'd probably turn around and get on with my business (no joke intended). Private acts should probably be kept private, whether or not it involves homosexuality. Intercourse is just one of the many facets a relationship that has to be evaluated; I'm sure that many people are ok with the concept of same sex marriages but opposed to the idea of sexual relationships of the same nature. As ignorant as I might seem discussing the concept of sex in relation to homosexuality and its offensiveness, nevertheless it still holds value in the discussion.

and then again, the fact that i recognize the sexual aspect of the relationship as opposed to simply talking about the more platonic issues demonstrates more that i'm willing to address all parts of the topic as opposed to creating a specific side bias

[quote="bookaholic"]Erm...isn't heterosexuality/homosexuality defined by who you want to have intercourse with? I don't get it. What else do you want me to address? I was just responding to the "icky" argument. I haven't posted my complete opinion or anything like that.

And I'm pretty sure I wouldn't want to look whichever gender the two people were, but if I did have a bias, so what? Are you going to ask homosexual people if they have a bias? They probably do. Do you think a heterosexual person's bias is more important than a homosexual person's bias?[/quote]

Not necessarily intercourse, I would say rather an attraction towards a particular gender, that may or may not lead to a desire for intercourse. I'm not saying that a heterosexual person's bias is more important than a homosexual's; I'm just arguing that some people prefer not to witness public displays of sexual intimacy whether it's homosexual or not, and that group of people probably includes gays and nongays.

I wasn't talking to you :P Sorry. But I was responding to the two posts above yours, sorry for not making that clear. Yeah, I didn't phrase that (the definition of sexual orientation) right. I used the word "intercourse" because that's what Ignite168 used. Note to self: the quote button is your friend.

Disclaimer: I'm not trying to say by my syllogistic form of argument that what I say must be correct. I'm stating it in the same way someone gives a "proof" that 0=1 - the arguments all seem correct to me, so I'm posting it here for people to find holes in it.

Let me start with an axiom which I believe to be true: An action is good if it increases the happiness of one person without decreasing the happiness of
anyone else. This belief forms the basis for my argument, and it seems simple enough to consider it an axiom, although I understand some people disagree with it. This is, I think, the definition of an ethically good action, as opposed to a morally good action, which is imposed by a higher power. I won't consider moral beliefs here, because absolute morals are the result of religion or can be considered religions themselves, and we're discussing this from a legal, not religious, point of view.

Thus, to show that allowing homosexual marriage is good, I should show that (1) homosexuals benefit from marriage, and (2) no one else is harmed by homosexual marriage. That homosexuals would benefit from being allowed to marry, I feel, is pretty obvious. Why? Because they want to. They know what they're getting into, but they want to marry a same-sex partner anyway. That implies that allowing them to marry would make them happier. Not in all situations, but then, the same could be said about heterosexual marriage.

Before I address the second concern, there is the question: what is marriage? Marriage comes with two parts, the reasons for marrying and the benefits of marrying. I see, in this discussion, two possible reasons being mentioned: love and procreation, as well as two different benefits: legal benefits and religious benefits.

Starting with the benefits, I feel that there should be two kinds of marriage, legal and religious. If your religion does not condone homosexual marriage, then why should it allow them? If the reason for marriage is to be married in the eyes of God, but your religion believes homosexuals cannot be married in the eyes of God, then you shouldn't marry someone of the same sex. Thus, I feel that religious organizations should have the right to choose how they want to treat homosexual marriage. When it comes to legal marriage, however, the couple seeks equal legal rights as heterosexual couples, and they should not be denied these rights, as it would detract from their happiness without increasing anyone else's happiness. The argument for this will come later.

Next, I want to address the two reasons for marriage. If the reason is love, then there is nothing wrong with homosexual marriage, because people of the same sex can love each other just as much as people of the opposite sex. This is even true regardless of sexuality. If, on the other hand, the reason for marriage is procreation, then homosexuals should not be allowed to marry, as homosexuals are unable to procreate, and allowing them to marry would go against the definition of marriage. However, then people who are infertile should not marry either, nor people who do not wish to have children. Since I don't think anyone is arguing this, I won't discuss it further.

Finally, I want to say why I think homosexual marriage is harmless to others. This is a bit harder, since the only way I can show this is by countering all arguments that it is harmful, and I don't really understand some of them. A common claim is that it destroys the sanctity of marriage; however, I have never heard this argument in different terms and really have no clue what it's supposed to mean. Another argument is that it would encourage other homosexuals to marry (but I'm trying to show that this isn't bad), or even that it would encourage others to become homosexual, in particular their children. The only counterargument I have to this is my belief that sexual orientation is determined at birth. As has been argued many times before, there's no reason for someone to choose to be homosexual (that, and the fact that most homosexuals would argue that they did not choose to be so).

That's all I have to say about homosexuality, however there is one somewhat unrelated thing I'd like to point out, which is that I seem to disagree with about everyone who's posted here on one point (which I briefly mentioned in my argument above): I believe that anyone can love anyone else regardless of sexuality. First, I should qualify this. What do I mean by love? Of course it's true that two heterosexual friends or brothers can love each other, if you're not using love to mean "relationship" love. I guess what I'm trying to say is that I don't feel that "relationship" love is anything special beyond regular love, it just has the additional component of physical/sexual attraction. Any love can have all the important elements: trust, compassion, commitment, intimacy, etc. So, if marriage is based on real love, without the physical component, sexuality shouldn't matter. But why add in sex? What's so important about sex, besides its use as procreation? Obviously procreation is important, but there's no reason for the requirement that both biological parents be married to each other. Just wondering what your opinions are about this.

One last thing: in response to:
[quote="BoesFX"]Well, Jesus doesn't meet homosexual couple (at least from the Bible it is not told) or have said something about homosexual [or He have implicitly] (And that doesn't mean that Jesus supporting homosexual). We won't know what He will say about homosexual (unless He somehow tell me or you). But just for reference:  [url=http://www.catholic.com/library/gay_marriage.asp]click here[/url] (for catholics, but I think it is okay too for non-catholic to read :wink: and this article may be can explain much better than I can what I mean)[/quote]
I don't get it. You say that Jesus hasn't said anything about homosexuality, and you give a link for more information. The link is to a site which is not only horribly biased against homosexuals, it is very insulting, and practically goes out of its way to include every possible logical fallacy it can. If you actually believe what this site says, I can argue against specific points, but I won't bother if I don't have anyone to argue with.

[quote="Ignite168"]The fact you guys keep bringing up the intercourse portion shows your inexperience and lack of knowledge about the issue.[/quote]

To classify as homosexual a relationship has to be sexual.  The definition on wikipedia is "[h]omosexuality refers to sexual and romantic attraction between two individuals of the same sex."  (yes, I'm citing wikipedia *sigh*).  To hammer the point home, I can have really close friendships with other males but not be gay.  I can love my friends and not be gay.  It's the sexual/romantic aspects that draw the line between straight friends and gay friends.

Some relevant news: the New Jersey legislature will have to [url=http://www.cnn.com/2006/LAW/10/25/jersey.samesex.ap/index.html?eref=rss_topstories]rewrite marriage laws[/url] so that homosexual and heterosexual couples are afforded the same rights.

[quote="Justice Barry T. Albin"]Although we cannot find that a fundamental right to same-sex marriage exists in this state, the unequal dispensation of rights and benefits to committed same-sex partners can no longer be tolerated under our state Constitution.[/quote]
This quote from the article (taken from the court's opinion in the case) represents pretty much exactly what I think should be the government's opinion on the topic. The government should not issue moral edicts, but there is a responsibility to afford the same rights to everyone. This leaves everyone free to have their own beliefs about what "marriage" truly means, and at the same time homosexual couples are no longer at a legal or economic disadvantage due to the actions of the government.


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So this is a standard marathon, but restricted to defining English words or phrases. You can do it as either "hard phrase" or "This is a sentence using hard phrase. What is another attribute of something in the sentence related to the hard phrase" - or of course you can do just one word. Each post must define or answer the word coming before it, and must pose a new question. It is preferred that you don't just copy-and-paste a dictionary definition for the word, but actually try to distill the definition down or rephrase it non-trivially.. Feel free to mention interesting etymologies, origins, etc.

#1: Define 'linn'  [happens to be my favorite word.]

linn - rushing water or waterfall

Define Defenestrate [happens to be my favorite word]

I actually know this word! Hooray for Latin!

defenestrate=throw out the window (Latin de=down from, fenestra=window, -te=verbal action suffix)

Define 'cwm'.

Basically a crater in the side of a mountain?

'limn' [!=linn]

limn=outlining in detail (comes from illuminate)
Define pneumonoultramicroscopicsilicovolcanoconiosis

pneumonoultramicroscopicsilicovolcanoconiosis=black lung disease
i don't think that you really need more info

define math(wow, this is hard)

fuddy-duudy= a person who doesn't follow fashion and dresses poorly

argy-bargy= a minor arguement

albumin= the white part of an egg

 :D =me

Here's a funny one(the pronunciation)
 Define and pronounce marmennill.

i remember that word! i spelled it correctly in a spelling bee(barely)

My favorite loan word into English is the German word "ersatz."

My favorite Chinese compound is "chechuang," which means "lathe" but which might be misinterpreted by an unwary Western reader as "cart bed."

The word poij taken from rolling your fingers across a keyboard

"a really horrible problem that is not easy to fix and most usually more awkward than necessary." (Taken from urbandictionary.com)

Coruscating and etinescent are two of my favourite words.

In German, I like Dingsbums, Überhängmandate and Kauderwelsch.

Coruscating means flashing I believe. I don't know about the second word.

define aglets

Yes, coruscating means glittering or sparkling, from the Latin coruscare.

Etinescent means beginning to shine forth and comes from the Latin etinesco (I begin to shine forth).

An aglet is a metal or plastic tube fixed tightly around the end of a shoelace, from the French aiguillette, a small needle.

More Latin :D!

effluvium-an unpleasant smell, as of decaying matter

[size=150]
abaculus[/size]

Tile of glass used in a mosaic.

Engastrimyth

Ventriloquist.

Next word:
[b]matutolypea[/b]

The state of being grumpy or irascible.

New word:
LEUCIPOTTOMY

leucipottomy / leucippotomy (LYU-si-puh-toh-mee) - n., the craft of cutting white horses on hillsides.

next: audaciousness

audciousness- fearless (reckless) daring, boldness

next: absquatulate

absquatulate- to leave promptly, run away, flee

New word: voussoir

voussoir- One of the wedge-shaped stones forming the curved parts of an arch or vaulted ceiling.
Next word: sepulchral

It's been a while...

sepulchral--of or relating to a tomb or burial; gloomy, melancholy, or dismal

Next word: anfractuous

Anfractuous-- Something extremely roundabout or circuitous.
I'm so glad to see this isn't (totally) dead!

Next word: [b]Indefatigable.[/b]
One must be indefatigable to traverse the anfractuous path to fulfillment.

[quote=GeronimoStilton]The state of not being able to be fatigued.

Jubilate.[/quote]

To express or feel happiness.

Paradoxical.


In a new work, after processing, I met the error "underfull \hbox", badness 1540 (if I remember well). I tried to rewrite the lines 404-406 but could do anything.
Here are the 404-406 lines, the first three only:

[quote]\bibitem{Lorenz}Lorenz, W.H, Nonlinear dynamical economics
and chaotic motion, Springer-Verlag, 1993
\bibitem{Miron}Miron, R., The geometry of higher order
Lagrange spaces. Applications to Mechanics and Physics.
Kluwer Academic Publishers, FTPH no. 82, 1997
[/quote]

What can I do?

Jean

It's not possible to tell from so little information since on its own in a document it doesn't produce any warnings. Please submit a complete [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] which shows the warning (it's not an error) about underfull boxes.


$1.$
A 45kg bicyclist climbs a hill at a constant speed of 2.5m/s by applying an average force of 85 N. Approximately how much power does the bicyclist develop ?

$2.$
How much heat must be removed from 0.5 kg of mercury at -39°C to change it from a liquid to a solid ?

http://en.wikipedia.org/wiki/Power_(physics)

$P = 2.5 * 85$

and for the second

http://en.wikipedia.org/wiki/Heat  -> Changes of phase

we need some constants about mercury. since i dont know :)


Still nobody?

I don't think it's that hard...

i think i've seen this. it's from japan or somewhere.

i don't know solution either.

Wow, this one is still unsolved after such a long time...


Does anyone know when the AMC state statistics be published on the AMC site?

In about a month.  We have many other, much higher priority, things to attend to right now.

Steve Dunbar
MAA Director, American Mathematics Competitions


calculus
integration
function
algebra
domain
logarithms
calculus computations

Let $f(x)=\frac{12(e^{3x}-3e^x)}{e^{2x}-1}$ with the domain of $x>0.$

(1) Prove that a function $y=f(x)\ (x>0)$ has the inverse function with the domain of the whole of real numbers, that is to say, prove that there exist single $x>0$ such that $f(x)=a$ for any real numbers $a$.

(2) Let $y=g(x)\ (-\infty <x<\infty)$ be the inverse function determined by the last question. Evaluate the definite integral $\int_8^{27} g(x) dx$.

(1)

$f(x)=\frac{12(e^{3x}-3e^x)}{e^{2x}-1}$
is a strictly increasing function.......

We have to prove that the function $f(x)$ is bijective. To simplify the calculations one can prove that $h=\frac 1 {12} fo\ln$ is bijective, and then since $fo\ln o h^{-1}=1_{\mathbb{R}}$ we have that $\ln o h^{-1}$ is an inverse for $f$.
Now it is easy to differentiate $h$ and remark that $h'>0$ and to see that $h$ has limits $\pm \infty$ at $\pm \infty$ and $h$ is continuous . Hence, $h$ is increasing continuously from $-\infty$ to $\infty$, hence it is bijective.
For the second point it may be used Young identity: $\int_a^bf(x)dx+\int_{f(a)}^{f(b)}g(x)dx=bf(b)-af(a)$.


geometry
3D geometry
prism
calculus
integration

I need help making a net with a specific volume and surface area. I don't
know how to do it.

Work Done:

Volume=80
surface area=144

Well, if it's a rectangular prism, given the side lengths are $x$, $y$, and $z$, you can write the following equations:

$xyz = 80$
$2(xy+yz+zx)=144$

Doesn't look like it has nice integral solutions, though. Oh well.

:huh:  Do not double post http://www.mathlinks.ro/Forum/viewtopic.php?t=150220.


If the sum $ 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} K$ is a perfect square $ N^2$ and if $ N$ is less than $ 100$, then the possible values for $ K$ are:

$ \textbf{(A)}\ \text{only }1 \qquad\textbf{(B)}\ 1\text{ and }8 \qquad\textbf{(C)}\ \text{only }8 \qquad\textbf{(D)}\ 8\text{ and }49 \qquad\textbf{(E)}\ 1,8,\text{ and }49$

[hide="Solution"]It's well known that the closed form to this is $ K(K\plus{}1)/2$.
Simply throwing in answer choies, $ 1$ works, $ 8$ works, and $ 49$ all work (they sum to $ 1^2$, $ 6^2$, and $ 35^2$ respectively).
So it's $ \boxed{\textbf{(E)}}$.[/hide]


Let real sequence {an} n=0,1,2... $\infty$ và a(n+1)=2^n -3an (n >=0)
Calculate (an) follow n and ao;
 a0=? such that a0<a1<a2<...

[quote="Rust"]\[ a_{n+1}=2^n-3a_n=2^n-3*2^{n_1}+3^2a_{n-1}=...=2^n[1-x+x^2-...+(-x)^n]+(-3)^{n+1}a_0=(-3)^{n+1}(a_0-\frac 15)+\frac{2^{n+1}}{5}  \][/quote]
Your message contains only a quote. Please add some content to the message before submitting <-- this is what you get when you just want to correct it...


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G=Sym(n). Suppose that $ \theta \in Hom(C, G)$ where C is a cyclic group. Then how to show that $ \gamma(ker \theta)\equal{}ker \theta$ for $ \gamma \in Aut(C)$.

It suffices to assume $ C$ to be a finite cyclic group, otherwise we factor out the subgroup $ Ker\theta\cap\gamma (Ker\theta )$ of finite index in $ C$. Therefore, for $ m\equal{}|Ker\theta |$ we notice that $ Ker\theta$ is the unique subgroup of order $ m$ in $ C$ and since $ m\equal{}|\gamma (Ker\theta )|$ holds, we must have $ Ker\theta\equal{}\gamma (Ker\theta )$.


linear algebra
matrix
arithmetic sequence

In most elementary or fundamental terms, when does an inverse of a matrix exist (meaning, no determinants or anything like this please) ? At first I thought that you should be able to find an inverse if the matrix is a square matrix but I've been proven wrong. If you were just looking at a matrix first glance, would you be able to tell whether or not it is invertible? (say it was a block matrix)

 thanks a lot

[quote="jmerry"]In order to answer the question, you must do work equivalent to finding a determinant; row-reduce until either it's upper triangular or there's a zero row. If you did this in a sensible way, you can also just write down the determinant at the end of the process.[/quote]

 so there's no "common property" amongst all invertible matrices so that when you see a matrix that is like ____ you can say, "oh yeah since it's ____, it's invertible"? A property I mean would be something like (just examples of common properties) "it is a square matrix" or "every column is different"... et c. I hope you know what I mean.

Thanks for replying

There are certain families of matrices for which you can say at a glance if they are, or not, invertible. For example, diagonal matrices and triangular matrices are non-singular iff all diagonal elements are non-zero. On the other hand, there are also some matrices for which it is easy to see that they are singular (besides the above two cases with at least one diagonal zero element): skew-symmetric matrices of odd order, and matrices where you can see that there is a row (column) which is a scalar multiple of another row (column). There are, of course, other more "specialized" cases: for example, if all elements of each row (or column) of a square matrix of order $ \geq 3$ are in arithmetic progression, with equal or different common differences, then the matrix is singular. In almost all other types of matrices, you have to do some kind of calculation (for example, Gaussian elimination) in order to determine if a matrix is, or not, invertible.


With reference to problems 36-40 find the coordinates of the point equidistant from r s and t.

[img]http://img.photobucket.com/albums/v603/Slippy6/MATHaops.jpg[/img]

As $ R$, $ S$ and $ T$ are over a circunference, we need to find the coordinates of its center.
Using the general form of a circle in plane: $ x^2\plus{}y^2\plus{}ax\plus{}by\plus{}c\equal{}0$(*)
We stablish three equations by sustituting the couples $ (x,y)$:
$ 1\plus{}9\minus{}a\minus{}3b\plus{}c\equal{}0\Leftrightarrow\minus{}a\minus{}3b\plus{}c\equal{}\minus{}10$
$ 4\plus{}4\plus{}2a\plus{}2b\plus{}c\equal{}0\Leftrightarrow2a\plus{}2b\plus{}c\equal{}\minus{}8$
$ 25\plus{}9\plus{}5a\minus{}3b\plus{}c\equal{}0\Leftrightarrow5a\minus{}3b\plus{}c\equal{}\minus{}34$
Wich can be reduced to:
$ 3a\plus{}5b\equal{}2$
$ \minus{}3a\plus{}5b\equal{}26$
Hence, $ a\equal{}\minus{}4,b\equal{}\frac{14}{5},c\equal{}\minus{}\frac{28}{5}$
Sustituting this in (*) and completing squares we get: $ (x\minus{}2)^2\plus{}(y\plus{}7)^2\equal{}\frac{273}{2}$

So the center(equidistant point) is $ (2,\minus{}7)$, unless I made a computing mistake but this is the idea.

 :)


EPGY does offer Physics B, actually. They call it AP Introductory Physics.

[quote="xxreddevilzxx"]where do i take AP commputer programming/science course or AP Physics classes online, besides EPGY?

I know they offer Physics, but they do not offer the AP Physics B, and certainly they dont offer computer programming or computer science in AP.[/quote]

If you are a homeschooler, here is a good place to take some reasonably priced APO classes. http://www.pahomeschoolers.com/courses/index.html

I am not homeschooled, does that mean i cannot do it?

also, is it possible for you to find another site?

I am not homeschooled, does that mean i cannot do it?

and are there any other than EPGY that is open to everyone?


$ 8^6  \plus{}  3^4$

If you are told that the number resulting from the above expression has four prime factors, two being same, less than $ 10$, one between $ 10$ to $ 100$ and the other between $ 500$ and $ 1000$ and you are asked to find out a quick methosd to find out the largest prime factor, can you do it ?

[hide="Hint"] Write $ 8^6 \equal{} 2^{18} \equal{} 4^{9}$ and use [url=http://en.wikipedia.org/wiki/Sophie_Germain#Contributions_to_number_theory]Sophie Germain's identity[/url]. [/hide]


Quite often (but not always), when I try to do a search and the results occupy more than two pages, I can see the first page but when I attempt to view any other page, I get the "No topics or posts meet your search criteria" message. Is it a bug or just some problem with my computer? I'm using the Mozilla browser run under Windows-95.

I often have this problem, too. The same thing happens when I go to the list of my old posts from the profile section. I use AOL browser with Windows XP.

EDIT- I have just tried it again and it seems that it's working properly now... Was this fixed?

It's not yet clear to me what happens but it might be an issue with the browser (if it were on the server, it would not work at all).


calculus
videos
integration
calculus computations

So most of the students at our school skip AP Calculus AB and head on to AP Calculus BC, at least those who are capable of taking it.
I am wondering if skipping part A is going to be a problem. Is there a website or free online videos or a simple "cliffnotes" version of A?

EDIT: I haven't taken regular calculus, but have some background knowledge of limits and integrals

[quote="JBL"]$ \text{AP Calc AB} \subset \text{AP Calc BC}$[/quote]
I thought A, B, and C were all separate sections... Are they not?
Or does BC review some of A?

[quote="jmerry"]This depends on how the class is taught. BC is about a year's worth of college classes, and certainly many people can learn it in a year. This is faster-paced than most high school classes, including the usual way AB is taught. Some versions of the BC class do the whole thing in a year, while others limit the class to students who already have AB and spend more time on the new content.

JBL is correct as to the content of the tests. Everything that is on the AB test is also on the BC test, and you get an AB subscore. Think of it as a three-quarter sequence covering single-variable calculus: AB is the first two quarters, while BC is all three.[/quote]

Eh.. I'm not so sure of how college classes work, but I think I get an idea of what you are saying.
Thanks :)

Just reiterating what was already said: the "A", "B", and "C" no longer stand for anything specific. The Calculus AB course is roughly equivalanet to 60% of a two semester college Calculus course, while the Calculus BC course is roughly equivalant to two full semesters of a college Calculus course. In fact, 60% of the questions which appear on the AB exam for a given year are duplicated on the BC exam for that year. Those questions are used to give BC students an AB subscore in case they don't pass the full BC exam, but could pass the AB portion of the material.


function
trigonometry
calculus
calculus computations

Hi, all :D i have a problem
find the solution of this diff equation
\[ (x^2)y'' - 2x y' +(x^2+2)y= x^3  \]

One "obvious" solution is $y=x$. Unfortunately, I do not see any way to find the general solution of the corresponding homogeneous equation except by using the power series method. It is close to the equation for the Bessel function but is not exactly it.

that was I got in exam. I subtitute $y=x^m$, $y'=mx^{m-1}$ and $y'' = m(m-1)x^{m-2}$ then the given ODE will becomes.
\[ m(m-1)- 2m +2= x^{3-m} -x^2  \]
obviously $m=1$ statisfies the equation. so $y=x$ is the solution.
can you show me your Bessel/power series solution?

[quote="kunny"]Since $y=x$ is solution, set $y=xu,$ we have $y'=u+xu',\ y''=2u'+xu''.$ Plugg these into the given equattion, yielding $x^3(u''+u-1)=0.$ Solve the differential equation of Simple Harmonic Equation-typed, $u''+u-1=0.$
Rewrite this as $(u-1)''=-(u-1)$ we obtain $u-1=A\cos x+B\sin x.$ Thus the desired solution is $y=x(A\cos x+B\sin x+1).$[/quote]
that's it :) I know someone will have posted  it before me, i'm late.


Where can I find full past AMC and AIME tests?

I think AIME in on the WikiAOPS, but what about AMC 12's?

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm for AMC 12s.
The Contests section *points up* contains all AIMEs. http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45

It also includes a lot of the more recent AMC problems that aren't on the site he gave:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44

you could also buy that cd that holds all the contests from 2000.

[quote="anirudh"]What about AMC-10? I mean besides 2000 and 2001.[/quote]
practice with AMC 12. they're about the same difficulty except 12 has harder problems in the end. When you're done with them, do AHSME I guess.

[quote="lotrgreengrapes7926"][quote="bpms"]The archives only have one of the AIMEs from 2000-2003 for each year...[/quote] They started giving 2 AIMEs a year in 2004...I think. :D[/quote]
No, they definitely started in 2000. If you look at the pages from 2000-2003 it will either say AIMEI at the top, then AIME II with the questions, or it will have AIME I at the top and at the bottom it will say AIME II without the questions.

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

That link doesn't work? I can access the page but the links to the contest porblems do not work.

Some libraries have "The Contest Problem Book" (I-VI). These are old AHSME exams. ISBN of No.VI is 0-88385-642-5.
Sometimes you will be surprised at what you can find at your local library.
I was looking for "Mathematics of choice" by Niven, found it plus all the Contest problem books.
Use the library website catalog search option and see what other local libraries you can borrow from. In CT one can borrow books from any CT library.


calculus
integration
trigonometry
function
logarithms
complex analysis
complex analysis solved

[quote="Fermat -Euler"]Prove that
$I_n = \int_0^{ + \infty } {\frac{{dx}}{{1 + x^n }} = \frac{\pi }{{n\sin (\frac{\pi }{n})}}}$[/quote]
This can be done with residues, i.e. it really belongs in the section on Complex Analysis.

Let $R>1$ and let $\gamma = \gamma_1 \cup \gamma_2 \cup \gamma_3$ consist of the paths $\gamma_1 = \{z | z = t, \, 0\le t \le R\}$, $\gamma_2 = \{z | z = Re^{i t}, \, 0\le t \le 2\pi/n\}$, $\gamma_3 = \{z | z = (R-t)e^{2 \pi i/n}, \, 0\le t \le R\}$. That is, $\gamma_1$ starts at the origin and follows the x-axis out to $z=R$, $\gamma_2$ is one n-th of a circle of radius R, starting at $z=R$ to $z = Re^{2\pi i/n}$, and $\gamma_3$ returns from that point to the origin along a straight line. The function $f(z) = \frac{1}{1+z^n}$ has a single pole at $z_0 = e^{i\pi/n}$ inside $\gamma$. The residue of $f$ at that point is $Res(f,z_0) = \frac{1}{nz_0^{n-1}} = -\frac{z_0}{n}$.
Now by the residue theorem
\[ \int_{\gamma} f(z) dz = 2 \pi i \, Res(f,z_0) = -2\pi i \frac{z_0}{n}\, .  \]
Then
\[ \int_{\gamma_1} f(z)dz = \int_0^R \frac{1}{1+t^n}dt \to \int_0^\infty \frac{1}{1+t^n}dt\, ,  \]
\[ \left|\int_{\gamma_2} f(z)dz \right| \le \int_0^{2\pi/n} \frac{R}{R^n - 1}dt \to 0 \, ,  \]
\[ \int_{\gamma_3} f(z)dz = - e^{2\pi i/n}\int_0^R \frac{1}{1+t^n}dt \to - e^{2\pi i/n}\int_0^\infty \frac{1}{1+t^n}dt \, .  \]
Consequently$(1 - z_0^2) \int_0^\infty \frac{1}{1+t^n} dt = -2\pi i \frac{z_0}{n}$ with $z_0 = e^{\pi i/n}$.
Since $\frac{z_0}{1-z_0^2} = \frac{1}{z_0^{-1} - z_0} = \frac{1}{\bar{z_0} - z_0} = \frac{1}{-2 i \sin \pi/n}$, we obtain
\[ \int_0^\infty \frac{1}{1+t^n} dt = -2\pi i \frac{1}{-2 i n \sin \pi/n} = \frac{\pi/n}{\sin \pi/n} \, .  \]
[i]See Cartan's book III.6 exercise 23.[/i] :lol:

[quote="Omid Hatami"]Posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91258]here[/url] before.[/quote]

no i posted it before you see the date

[quote="Omid Hatami"]OK :lol:
But you can see another solution there.[/quote]
in coputitions and utorials we have find integral of this
$\int\frac{dx}{1+x^{n}}$

Can anyone find the answer to the integral of
natural logarithm log e

$ \int \ln x \, dx \equal{} x \ln x \minus{} x \plus{} C$, but really, did you have to revive a 3 year old topic?

With a modified contour which cuts out the origin, $-\gamma_4 = \{z | z = \epsilon e^{i t}, \, 0\le t \le 2\pi/n\}, \epsilon\to0$, and which of course modifies $\gamma_3$ to join with this small semicircle, one can obtain

$$\int_0^{\infty}\frac{\ln(x)}{1+x^n}\,dx=-\frac{\pi^2}{n^2}\csc\frac{\pi}{n}\cot\frac{\pi}{n}$$


Problem: Prove that if $x,y,z\geq0$ then
\[ (x+y+z)^3\geq 3(xy+yz+zx)\sqrt{3(x^2+y^2+z^2). }\]
Solution: Since the desired is homogenous, we can suppose WLOG that $x+y+z=3$. Set $w=xy+yz+zx$, now all we need to prove is
\[ 27\geq w^2(9-2w). \]
This is true since $w\leq3$.

it's just am-gm... :D

$(a+b+c)^2 =3 \frac{(a^2+b^2+c^2) + (ab+bc+ca)+ (ab+bc+ca)}3 \geq 3 \sqrt[3]{a^2+b^2+c^2} \sqrt[3] {(ab+bc+ca)^2}$

Now taking the square root and cubing we get the result!

Just a quick remark here:

 [b] Problem. [/b]

       Let $a, b, c$ three non-negative real numbers such that $a+b+c=3$. Prove the following inequality

             \[ \frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{1}{a^{2}}\geq a^{2}+b^{2}+c^{2}.  \]

  [i] Vasile Cartoaje, Romanian TST 2006 [/i]

   [b] Solution. [/b]

     We will use Thuan's lemma:  \[ (a+b+c)^{3}\geq 3(ab+bc+ca)\sqrt{3(a^{2}+b^{2}+c^{2})}.  \]
     Using $a+b+c=3$, we get \[ 27\geq (ab+bc+ca)^{2}(a^{2}+b^{2}+c^{2}).  \]
     Now, if we prove that \[ \frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{1}{a^{2}}\geq\frac{27}{(ab+bc+ca)^{2}},  \] we are done. This last inequality reduces to

        \[ (ab+bc+ca)^{2}\left(\frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{1}{a^{2}}\right)\geq 27.  \]
     But, this follows immediately from these two trivial inequalities:

    \[ ab+bc+ca\geq\sqrt{3abc(a+b+c)}  \] and \[ \frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{1}{a^{2}}\geq\frac{a+b+c}{abc}  \].

The original post by arqady is here
[url]http://www.artofproblemsolving.com/Forum/topic-78236.htm[/url]


limit
real analysis
real analysis unsolved

Prove that for any $ a > 0$ we have $ \lim_{n\rightarrow \infty}\sum_{k = 1}^{n}{\frac {1}{\sqrt {n^{2} + kn + a}} = 2\left(\sqrt 2 - 1\right)}$ .

I get that the limit is $ 2(\sqrt{2}-1)$ but maybe I'm wrong. Here is the solution given by me:

$ \sum_{k = 1}^{n}{\frac {1}{\sqrt {n^{2} + kn + a}}<\sum_{k=1}^{n}{\frac {1}{\sqrt{n^2+nk}}}}$

Denote by $ x_n=\frac {1} {\sqrt{n}} (\frac {1} {\sqrt{n+1}}+...+\frac {1} {\sqrt{2n}})$ Form Stolz-Cesaro $ \Rightarrow \lim_{n\rightarrow \infty}x_n=\lim_{n\rightarrow \infty}\frac {\frac {1} {\sqrt{2n+1}}+\frac {1} {\sqrt{2n+2}}-\frac {1} {\sqrt{n+1}}} {\sqrt{n+1}-\sqrt{n}}$

$ \Rightarrow \lim_{n\rightarrow \infty} x_n=2(\sqrt{2}-1)$

Now, for $ n$ sufficiently large, we have $ n>a$ and $ \sum_{k = 1}^{n}{\frac {1}{\sqrt {n^{2} + kn + a}}>\sum_{k = 1}^{n}{\frac {1}{\sqrt {n^{2} + kn + n}}}}$

Denote by $ y_n=\frac {1} {\sqrt{n}}(\frac {1} {\sqrt{n+2}}+...+\frac {1} {\sqrt{2n+1}})$ Using again Stolz-Cesaro we get:

$ \lim_{n\rightarrow \infty} y_n=\lim_{n\rightarrow \infty} \frac {\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+3}}-\frac{1}{\sqrt{n+2}}}{\sqrt{n+1}-\sqrt{n}}=2(\sqrt{2}-1)$

Finally, we obtain that the limit is $ 2(\sqrt{2}-1)$


function
number theory proposed
number theory

Find all multiplicative functions $ f$ such that $ f(n^{2})=\sum_{d|n}\mu^{2}(d)f(\frac{n}{d})$ and $ f^{2}(n)=\sum_{d|n}f(d^{2})$ for all $ n$.

What do you mean by $ f^{2}(n)$? The pointwise product $ f(n) \cdot f(n)$ or the composition product $ f(f(n))$?

let $ \mathbb{P}=\{primes\}$
$ \forall x\in \mathbb{N}: \ f(n)=0$ is solution.
if $ \exists a: \ f(a)\neq 0$ then $ f(a)=f(a)f(1)$ gives $ f(1)=1$
and we have $ {\forall p\in\mathbb{P}: \ f(p^{2})=f(1)+f(p)=f(p)+1=f(p)^{2}}$ then $ f(p)=x\ or\ y$ such that $ \varphi^{2}-\varphi-1=\overline{\varphi}^{2}-\overline{\varphi}-1=0\ and\ \varphi\neq \overline{\varphi}$
we suppose that $ n=\prod_{i=1}^{m}p_{i}^{a_{i}}$ such that $ p_{1},p_{2},...,p_{m}\in S_{x}$ distinct primes and $ a_{1},a_{2},...,a_{n}>0$ so
$ {f(n^{2})=\prod_{i=1}^{m}(f(p_{i}^{2}))^{a_{i}}=\prod_{i=1}^{m}(f(p_{i})+1)^{a_{i}}}$
$ NOW$ we take $ S_{h}=\{p\in\mathbb{P}: \ f(p)=h\}$ and $ S(h,n)=\sum_{p_{i}\in S_{h}}a_{i}$
then $ (f(n))^{2}=f(n^{2})=x^{2S(x,n)}y^{2S(y,n)}$ so $ f(n)=\overline{\varphi}^{S(\overline{\varphi},n)}\overline{\varphi}^{S(\overline{\varphi},n)}$
now i must use $ f^{2}$ to find more information for $ f$.

[quote="msceok"][quote="aviateurpilot"]
...then $ (f(n))^{2}=f(n^{2})=x^{2S(x,n)}y^{2S(y,n)}$ so....[/quote]
Why $ (f(n))^{2}=f(n^{2})$ is true? I think $ f(n)$ is [b]multiplicative[/b] function so it cannot be the direct result from the problem...[/quote]
$ f(n)$ is [b]multiplicative[/b] function $ \iff$ $ f(ab)=f(a)f(b)$ by definition?

[quote="aviateurpilot"]
$ f(n)$ is [b]multiplicative[/b] function $ \iff$ $ f(ab)=f(a)f(b)$ by definition?[/quote]
Actually,
$ f(n)$ is a [b]completely multiplicative[/b] function $ \iff$ $ f(ab)=f(a)f(b)$ for all $ a,b\in \mathbb{N}$
and
$ f(n)$ is a [b]multiplicative[/b] function $ \iff$ $ f(ab)=f(a)f(b) \mbox{~whenever~}(a,b)=1$

thanks $ Baz$
[quote="msceok"][quote="aviateurpilot"]
...then $ (f(n))^{2}=f(n^{2})=x^{2S(x,n)}y^{2S(y,n)}$ so....[/quote]
Why $ (f(n))^{2}=f(n^{2})$ is true? I think $ f(n)$ is [b]multiplicative[/b] function so it cannot be the direct result from the problem...[/quote]
$ f$ multiplicative gives $ f(ab)=f(a)f(b)$ where $ gcd(a,b)=1$
now we can suppose that $ n=\prod_{i=1}^{m}{p_{i}}^{a_{i}}$ where $ p_{i}$ are primes.
then $ f(n^{2})=\prod_{i=1}^{m}{f(p_{i})}^{2a_{i}}=(\prod_{i=1}^{m}{f(p_{i})}^{a_{i}})^{2}=(f(n))^{2}$


Clearly i think: $f(2m) = 2m^{2}-1$, $f(2m+1) = 2m^{2}+2m-1$, can you post your complete solution$?$

I think it is :

$\frac{n^{2}-2}{2}$  if n is even  and
$\frac{n^{2}-3}{2}$  if n is odd and it is from Balkan 1994.


What data structures do you guys think are useful in IOI?
I'm asking this because CLRS has some data structures which are very complicated, such as red-black trees. I may be wrong, but I think these data structures which are too complicated are not useful in IOI.
What do you guys think?
I'll start the list:

- Queues
- Stacks
- Binary heap

Thanks for your help.

[quote="mahbub"]i think red black tree is useful. last year a problem came  which needed use of binary search tree (though it could be build in different way). but you know, here it means balanced one. and so red black would come into play![/quote]

Thank you. Apart form the ones already said, which data structures do you think are important? Hash tables, maybe?

well... i don't think hash table would be that much important. if you know STL map i think that will be enough for that!

besides some times, deque is also useful. you may learn the "set" of stl. i think that will be enough for IOI.


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

\[ \lim_{n \rightarrow \infty} { \left ( \frac{(2n+1)!}{(n!)^2} \right )^2 \int \int_Q { (xy(1-x)(1-y))^n f (x, y) dx dy }} = ? \]

where we integral over $Q = [0, 1] \times [0, 1]$ and $f: Q \rightarrow R$ is continuous function.

$\left ( \frac{(2n+1)!}{(n!)^2} \right )^2 = O(4^n \sqrt{n})^2=O(16^n n)$ by stirling formula.

we need now to estimate $I_n=\int_0^1 x^n (1-x)^n dx$.

$\int_0^1 x^n (1-x)^n dx=\int_{-\frac 1 2}^{\frac 1 2 } (\frac 1 4-x^2)dx=2 \int_0^{\frac 1 2} (\frac 1 4-x^2)dx$
setting $x=\frac 1 2 \sin t$ we obtain  $I_n=2 \frac 1 {4^n}\int_{0}^{\frac \pi 2} sin^{n+1} dx$.
hence $I_{2n}=\frac{(2n-1)!!}{(2n)!!}\frac \pi 2$ and $I_{2n+1}=\frac{(2n)!!}{(2n+1)!!}$.

$\int\int_Q x^ny^n(1-x)^n(1-y)^ndxdy=(\int_0^1 x^n(1-x)^n dx)^2$
Using the fact that $f$ is continuous on a comapct, hence bounded we get that the limit is 0.

The solution is wrong , becouse the answer is $f(\frac{1}{2},\frac{1}{2})$.

The problem is that estimate for $I_n$. $I_n$ is actually equal to $\frac{(n!)^2}{(2n+1)!}$. To prove this, integrate by parts repeatedly:
$\int_0^1 x^n(1-x)^n\,dx=\left[\frac1{n+1}x^{n+1}(1-x)^n\right]_0^1+ \int_0^1\frac{n}{n+1}x^{n+1}(1-x)^{n-1}\,dx=$
$\dots=\int_0^1 \frac{n\cdot(n-1)\cdot(n-2)\cdots 1}{(n+1)\cdot(n+2)\cdots (2n)}x^{2n}\,dx=\frac{(n!)^2}{(2n+1)!}$
The boundary terms are zero at every step.

Sorry  :blush:  Can you psot a full solution ?

My solution is like this (and it will be good if someone tells if it is without any obvious mistake)

Since we have \[ { \left ( \frac{(n!)^2}{(2n+1)!} \right )^2 =\int \int_Q { (xy(1-x)(1-y))^n dx dy}}  \]

So \[ J_n={ \left ( \frac{(2n+1)!}{(n!)^2} \right )^2 \int \int_Q { (xy(1-x)(1-y))^n f (x, y) dx dy }} = \frac{\int \int_Q { (xy(1-x)(1-y))^n f (x, y) dx dy}}{\int \int_Q { (xy(1-x)(1-y))^n dx dy}} \]

Which means we can interprate $J_n$ as a weighted mean of $f$ on whole $Q$ with is compact, so $\lim_{n \rightarrow \infty}J_n=f(x_1,y_1)$ for some $x_1,y_1\in[0,1]$ since $f$ is continuous.

Because weights other than the one for $x=1/2, y=1/2$ tends to 0 and the one is not. So the mean goes closer to the walue of $f(1/2,1/2)$ at any increase of $n$. Therefore the limit must be $f(1/2,1/2)$


Asymptote
combinatorial geometry
real analysis
real analysis unsolved

Given some Cantor set $ C$ on the unit circle, consider the set $ A$ of all chords between points of $ C$.
1) Prove that $ A$ is compact.
2) Is $ A$ convex?

1) I started off by saying that $ A$ is a subset of the unit disk, which is compact. So, if we can show that $ A$ is closed then it follows that it is compact. To this end, suppose that $ x$ is a limit point of $ A$; claim is that it is contained in $ A$. There exists some sequence $ (x_n)$ of points of $ A$ which converges to $ x$. But, each element $ x_n$ of the sequence must lie on a chord between two points $ c_n$ and $ d_n$ of the Cantor set $ C$. What I want to do here is define sequences $ (c_n)$ and $ (d_n)$, show that they have some convergent subsequences $ (c_{n_k})$ and $ (d_{n_k})$ with limits $ c$ and $ d$ by sequential compactness, and then show that $ x$ has to lie on the chord between $ c$ and $ d$. Something tells me I have to do a little more work here though. Any suggestions or am I ok with what I have?

2) I have no idea. The author of the book lists this question as one to which he does not know the answer and I'm losing sleep over it.

Your idea for the proof of 1) is fine. A more "high-tech" proof is: the product $ C\times C\times [0,1]$ is compact, and $ A$ is its image under the continuous map $ F(c,d,t) \equal{} (1 \minus{} t)c \plus{} td$.

2) is not true for general Cantor sets (defined as compact subsets of the circle with no interior points and no isolated points). Imagine an equilateral triangle inscribed into the circle. (No, I don't want to use Asymptote  :P ). Let $ C$ be contained in three short arcs positioned near each vertex of the triangle. Then the convex hull of $ C$ contains the center of the disk, but $ A$ does not.

That much is clear. The question is "Can it ever be convex?" ;)

Hm. I can't even tell if $ A$ can be of second category.  :maybe:


A line segment of $15+5\sqrt{3}$ is folded to create a 30-60-90 right triangle. Two of the triangles are put togther to form an equilateral triangle. Four of these equilateral triangles are put together to form a tetrahedron. Two of these tetrahedrons are out together to form a hexahedron. What is the volume of the hexahedron?

[hide]The ratio of the sides is $1: \sqrt{3}: 2$, so the sides of the triangle are $5, 5\sqrt{3}, 10$. Putting two together results in an equilateral triangle with sidelength 10. Usint the formula for volume of a regular tetrahedron (too lazy to derive it now, might be wrong), we have $\frac{25\sqrt{3}}{3}$ as the volume of the tetrahedon, and $\frac{50\sqrt{3}}{3}$ as the volume of the hexahderon.[/hide]

[hide]Is the answer $\frac{500\sqrt2}{3}$?[/hide]

[quote="sonny"]Is the answer $\frac{500\sqrt2}{3}$?[/quote]
Yes, I just figured that out, I got that too.


counting
distinguishability
combinatorics theorems
combinatorics

I was told that nHr is the combination with repetition, and nHr = (n+r-1)Cr, but actually what is nHr?
Thank you very much! :P

Let $M=\{\infty .a_{1},\infty .a_{2},...,\infty .a_{n}\}$. The number $H_{r}^{n}$ of $r$-element multi-subsets of $M$ is given by $H_{r}^{n}=C_{r+n-1}^{r}$.

Sorry, I am only a schoolgirl, who is not capable to understand the abstract set idea of nHr, can you please put it in a statement form.
For instance, "nCr is the number of unordered arrangements of r objects selected from a collection of n objects." :)

Thank you very much! :P

I've never seen the notation you're using before; however, the sentence you're looking for is that, "$nHr$ is the number of unordered arrangements of $r$ objects selected from a collection of $n$ objects, where each object may be selected repeatedly."

nHr=(n+r-1)Cr=$\binom{n+r-1}{r}$ where
$\binom{n}{r}$ is Newton's symbol (called "n choice r" or shortly nCr as you have written)
\[\binom{n}{r}\ =\ \frac{n!}{r!(n-r)!}\]
where $k!$ is defined as a product of natural numbers less/equal $k$ :
\[k!=1\cdot 2\cdot\ldots\cdot k \]
Assume that you have $n$ different balls, for example with numbers $1,2,\dots,n$,
 and you choose $r$ of them.
If the order of choosen balls (the order of numbers on them) isn't important for you,
then the number of all possible choices is $\binom{n}{r}$.

For example
$\binom{4}{3}=4$ because you can have exactly 4 possible results : $1,2,3$ and $1,2,4$ and $1,3,4$ and $2,3,4$;
$\binom{4}{2}=6$ because you can have exactly 6 possible results : $1,2$ and $1,3$ and $1,4$ and $2,3$ and $2,4$ and $3,4$.

I have seen this symbol in some Taiwanese books before.
It is called "combination with repetitions."

[b]Problem:[/b]
Let [i]r[/i] and [i]n[/i] be any two positive integers. There are [i]r[/i] indistinguishable balls of the same colour, and [i]n[/i] distinct cells. In how many ways can we place the r balls into the n cells if each cell can hold unlimited number of balls?

[i][b]The answer to this problem is denoted by [/b][/i][b]nHr.[/b]

[b]Solution[/b]
Use 0 to denote a ball and 1 1 to denote a box.
A sequence of 0's and 1's is used to represent a situation in which balls are placed into boxes.
*************************************************
[color=brown]Example[/color]
11 <--> One cell with no balls.
1001<-->One cell holding two balls.
101001<-->1 ball in the 1st cell and then 2 balls in the 2nd cell.
1010001001 <--->There are 1 ball in the 1st cell, 3 balls in the 2nd cell and 2 balls in the 3rd cell
*************************************************
After building up the 1-1 correspondence between the set of arrangements and the set of specified binary numbers, we may then easily solve the original problem:

[b]The required answer is the C(n+1+r-2,r),
             that is, C(n+r-1,r).[/b]

[i]Reason: n+1 comes from the number of 1's, for there should be n+1 walls for n cells. Note that there are r 0's. BUT, the leftmost 1 and the right most 1 are unchanged forever, so we don't need to count them. Finally, the problem is equivalent to counting the number of binary numbers of length n+r-1 in which r 0's appear in each sequence. [/i]


how many relation you can define on a set with N member that :
1-the relation is sequential compeletly
2- the relation is squential not compeletly
3- the relation is equivalant
4- the relation is reflective
5- the relation is asimetric
...

nvm, I go it. Thanks to people who considered answering this post. :)


geometry
inequalities
inequalities unsolved

Let $ ABC$ be triangle with sides $ a, b, c$ and area $ S$. Further, let $ A_1B_1C_1$ be triangle with sides $ \sqrt{a}, \sqrt{b}, \sqrt{c}$ and area $ S_1$. Prove that
\[ {S_1}^2 \geq \frac {S\sqrt{3}}{4}.\]

It is Finsler-Hadwiger inequality, isn' t it?

This is my full proof with some key hints for you to find out by yourself! :)
With the proof of the identity, I think everyone can take it easily
By the way, I rewrite the inequality after using the identity as followed:
$ \sum_{cyc}{(a \minus{} b \plus{} c)(a \minus{} c \plus{} b)} \ge 4\sqrt {3}S$
Set $ a \plus{} b \minus{} c \equal{} x .....$
and it implies to prove the following inequality which is familiar to us
$ ab \plus{} bc \plus{} ca \geq \sqrt {3abc(a \plus{} b \plus{} c)}$

and what about the identity?(how to achieve it?)

[quote="great math"]It is implied from one of the best well known inequalities in triangle. That is
$ 2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2 \ge \frac {S\sqrt {3}}{4}$
Actually, we can prove the following identity
$ S_{1}^2 \equal{} 2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2$[/quote]
This is not true,my friend.you made a mistake.
This is true:
$ 2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2 \ge 4\sqrt {3}S$(Finsler-Hadwiger inequality).
and remember that:
$ S_{1}^2 \equal{} \frac{2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2}{16}$
Q.E.D

I have already corrected it in the following post! Thanks for pointing out the mistake!
About the identity, all you need to do is to open the bracket ( expansion term by term)! Have fun!


For a map $ h: \mathbb{R}\to\mathbb{R}$, define $ Disc(h) \equal{} \{x: h$ is not continuous at $ x\}$.
My book asserts that $ Disc(h)$ is always measurable even if $ h$ is not.
How can i prove this?

Consider the function $ f(x)\equal{}\inf_{r>0}\sup_{y\in (x\minus{}r,x\plus{}r)} h(y)$. Note that if $ f(x)<t$, then there exists an interval $ (x\minus{}r,x\plus{}r)$ and $ t'<t$ such that $ h<t'$ on $ (x\minus{}r,x\plus{}r)$ and, therefore, $ f(y)<t$ for all $ y\in(x\minus{}r,x\plus{}r)$. Thus $ \{f<t\}$ is open for all $ t$ and $ f$ is measurable. Similarly, $ g(x)\equal{}\sup_{r>0}\inf_{y\in (x\minus{}r,x\plus{}r)} h(y)$ is measurable. But the set in question is just $ \{f>g\}$.

If you want a more difficult exercise in the same spirit, show that the set of points where $ h$ is differentiable is measurable.


limit
trigonometry
calculus
integration
function
logarithms
real analysis

Prove that
$\lim\limits_{n\to\infty} |\cos nx|^{1/n} = 1, \mbox{ a.e. } x\in \mathbb{R}$ (*)

=============================================

The following problem maybe is helpful (prove by integral)
Suppose $f$ is a measurable function with period of $T>0$. $A=\int_0^T |f(x)|dx < \infty$. Show that $\lim\limits_{n\to\infty} n^{-2}f(nx) = 0, \mbox{ a.e. } x\in\mathbb{R}$.

=============================================

I took logarithm of the function in (*) then tried the apply the conclusion above, but failed.

That doesn't work! Take for exemple $x = \frac{\pi}{2} $
 :cool:

hey, notice there is an "a.e." right? It is "almost everywhere"...In french, i think you say "p.p."

I think we can use some simple in nature but hard to write (:)) geometric arguments (working on the unit circle) to show that the set $A_n=\{x\in[0,\pi)||\cos nx|<(1-\varepsilon)^n\}$ has measure going to $0$ as $n\to\infty$. Since our $x$'s in $[0,\pi)$ are in the intersection of infinitely many $A_n$, it means that the set they form has null measure. This also applies to the bad $x$'s in every interval of the form $[n\pi,(n+1)\pi)$, for integers $n$, and when we take the countable union, we still get null measure.

[quote="grobber"] the set $A_n=\{x\in[0,\pi)||\cos nx|<(1-\varepsilon)^n\}$ has measure going to $0$ as $n\to\infty$. Since our $x$'s in $[0,\pi)$ are in the intersection of infinitely many $A_n$, it means that the set they form has null measure. [/quote]

Just having the measures of  $A_n$ tending to $0$ seems not enough: take, say,
$A_n=[\sqrt n, \sqrt{n+1}] \mod \pi$. Then $|A_n|=\sqrt{n+1}-\sqrt n\to 0$
but each point $x\in[0,\pi)$ belongs to infinitely many $A_n$...

I'll see what I can do. Sorry about that.

Ok, we should try to show that the series $\sum_{n\ge 1}m(A_n)$ is convergent. If we do that, we can employ the following theorem:

If $(A_n)_n$ are measurable sets of a measurable spae $X$ s.t. $\sum_{n\ge 1} m(A_n)<\infty$, then almost all $x\in X$ belong only to finitely many $A_n$.


vector
complex analysis
complex analysis unsolved

Hi,
suppose
$ z_1, z_2\in\mathbb C$,
$ L\equal{}\mathbb Z z_1\plus{}\mathbb Z z_2\subset \mathbb C$ is a lattice.
How to proof the number of the shortest non-zero vectors in $ L$ is not bigger than 6?

[code]    o-o-o-o-o
   o-c-c-c-o
  o-c-z-c-o
 o-c-c-c-o
o-o-o-o-o[/code]
how to show a circle $ K$ centered at $ z$, with $ K^o$ contains no other points than $ z$, doesn't touch any o's? (is this even true anyway?)

Both you questions are not well stated. The first is wrong without any bounds on the $ z_i$. In the second one, are you asking if such a circle with some positive radius exists¿

Hi,
$ L$ is a lattice, according to my definition that is $ \frac{z_1}{z_2}\notin\mathbb{R}$
The second of my questions was just my try to the first problem.
I thought I have to show a circle that contains no non-zero point inside, cannot cut any o-points in my diagram.
(so that it can only cut at most six of those c-points(for the angle between $ z_1$ and $ z_2$ is $ \frac\pi 3$, for example)).
It's easy to show that 2, 4 and 6 can be the number in my first question. Just why not 8?

And what about $ z_1 \equal{} 1337$ and $ z_2\equal{}481i$¿

[quote="ZetaX"]And what about $ z_1 \equal{} 1337$ and $ z_2 \equal{} 481i$¿[/quote]
hi,
I think in this case, there're 2 shortest vectors, $ v_1 \equal{} (0,481), v_2 \equal{} (0, \minus{} 481)\in\mathbb R ^2$

Indeed, whenever $ |z_1|\neq |z_2|$, the number must be 2.
 [geogebra]be4fc7ac55f6d09d76196fda7eb9cc2d5ee5d3e3[/geogebra]
and for the angle between $ z_2$ and $ z_1$ greater than $ \frac\pi 3$,lesser than $ \frac {2\pi}{3}$, $ |z_1| \equal{} |z_2|$, there are 4 such vectors.

Ah, sorry, I misread "number" for "distance"...
Well, then it is also not so difficult: If you have $ n$ vectors of smallest distance $ r$ from $ 0$, their pairwise distances needs to be at least $ r$, too. Draw a picture of a circle with radius $ r$ and try to put at least $ 7$ points on it that have pairwise distance $ r$. More formally, you should add the corresponding angles.
And no, you can have $ |z_1| \neq |z_2|$ and still have $ 6$ smallest distance vectors.

[quote="ZetaX"] If you have $ n$ vectors of smallest distance $ r$ from $ 0$, their pairwise distances needs to be at least $ r$, too. [/quote]

thanks, I think I got it.  :lol:

 [geogebra]ac2ba8888c31c52c2c62f7e28cee2809468e8d53[/geogebra]

 :blush: and also thanks for this.


integration
function
inequalities
real analysis
calculus
derivative
real analysis unsolved

Hi,
I have some difficulty with the following question, wonder if anyone can help. Thanks alot!!

Suppose $ f$ is of bounded variation on $ [a,b]$, then show that
$ \int_a^b \vert f' \vert \leq T$. where $ T$ is the total variation of $ f$ on $ [a, b]$

For each partitiion $ a\equal{}x_0 x_1<\cdots<x_n\equal{}b$ we choose $ \xi_i$ such that
\[ f'\left( {\xi _i } \right) \equal{} \frac{{f\left( {x_i } \right) \minus{} f\left( {x_{i \minus{} 1} } \right)}}
{{x_i  \minus{} x_{i \minus{} 1} }}
\]
where $ i\equal{}1,2,..,n$. Now
\[ \sum\limits_{k \equal{} 1}^n {\left| {f'\left( {\xi _k } \right)} \right|\left( {x_k  \minus{} x_{k \minus{} 1} } \right)}  \equal{} \sum\limits_{k \equal{} 1}^n {\left| {f\left( {x_i } \right) \minus{} f\left( {x_{i \minus{} 1} } \right)} \right|}  \leqslant T.
\]
Letting $ n \to \plus{}\infty$ we obtain
\[ \int_a^b {\left| {f'\left( x \right)} \right|dx}  \leqslant T.
\]

I am wondering the existence $ \xi_i$ in the above argument.

The existence of $ \xi_i$ is not guaranteed: consider $ f(x)\equal{}|x|$ on $ [\minus{}1,1]$.

[quote="mlok"]The existence of $ \xi_i$ is not guaranteed: consider $ f(x) \equal{} |x|$ on $ [ \minus{} 1,1]$.[/quote]
$ f(x)\equal{}|x|$ is not derivable on $ [ \minus{} 1,1]$, so that's out of subject    :D
It's trivial : the existence of $ \xi_i$ is guaranteed by the "mean value theorem for derivable functions"

Diogene: that's not out of subject. The stated hypothesis was that the function was of bounded variation; such functions need not be differentiable everywhere.

Suppose $ f$ were $ C^1.$ Then we'd have $ \int_a^b|f'|\equal{}T$ and we wouldn't need the inequality. We also have equality for $ f(x)\equal{}|x|,$ say, on $ [\minus{}1,1].$ In general, we would have equality for piecewise $ C^1$ functions, which is what $ |x|$ is.

So why an inequality at all? Because of functions whose variation has a singular part. For instance: let $ f$ be the Cantor-Lebesgue "devil's staircase" function. $ f'$ exists on the complement of the Cantor set (which is almost everywhere) and $ f'(x)\equal{}0$ wherever it exists. Hence $ \int_0^1|f'|$ makes sense as a Lebesgue integral and equals zero. But the total variation for a monotone nondecreasing function is just $ T\equal{}f(b)\minus{}f(a),$ and in this case $ T\equal{}f(1)\minus{}f(0)\equal{}1.$

Note: if $ f$ is of bounded variation, $ f'$ will exist almost everywhere - that's why there's something to integrate.

Of course Kent Merryfield , but the initial statement wasn’t clear for me about the definition of $ f'$ and I believed that's the usual derivative
 :cool:

A function of bounded variation can be represented as a sum of two monotone functions ($ f \equal{} g \plus{} h$) in such a way that $ \mathrm{var}\,(f) \equal{} \mathrm{var}\,(g) \plus{} \mathrm{var}\,(h)$. Thus we may assume that $ f$ is increasing, in which case the inequality reduces to $ \int_a^b f'(x)\,dx\le f(b) \minus{} f(a)$. But I can't think of a neat elementary proof right now...  :maybe:

By the way, $ f(x)\equal{}\mathrm{sign}\,(x\minus{}1/2)$ is a simpler example of strict inequality than the staircase.

[quote="mlok"] But I can't think of a neat elementary proof right now...  :maybe:
[/quote]
There is no neat elementary proof of the fact that the derivative exists a.e. but, if you assume that, all you need to do is to split the interval into $ n$ equal subintervals, to construct the functions $ f_n$ that are linear on each subinterval and coincide with $ f$ at the division points, to notice that $ |f_n'|\to |f'|$ at every point where $ f'$ exists that is not one of countably many division points and to apply Fatou's lemma.


I wonder how important are general gre scores compared vs subject test scores when applying to a graduate school? I performed quite badly on the general test therefore am thinking about retaking it...should I?

& just curious what would be reasonable standartized test scores of ppl thinking about the following

berkeley (I guess max?:))
chicago
yale
cornell
pennsylvania state
u washington
rice ?

thank you so much!

As I understand it, the general GRE is basically a filter: if you get above a certain score, they'll read your application; if you get below that score, they won't bother. The subject test is a bit different: it serves as a sort of equalizer. That is, it's hard to tell if your professor who writes that you're smarter than Gauss is telling the truth or not. But if you get 40th percentile on the subject GRE and your professor says you're smarter than Gauss, then something funny is going on.

By the way, don't retake it. Avoid giving the people at ETS any more money than is absolutely necessary.

I'd love not to retake, but would 790/410 filter through? :) ...I'm afraid that's too bad for most of the places, isn't it? If I retake it, I'm pretty sure I can manage 800/450 at least. Would that increase the chances significantly?

The GRE verbal section is (imo) genuinely very hard.  I've always sucked at analogies, and I managed to pull off the same score as I did on my SAT I verbal (730) even though I didn't study for it, but I felt like the GRE verbal was genuinely very difficult, and I had to guess at many questions (and suspect I got lucky at least a couple of times, since the questions got very hard very quickly), whereas on the SAT I I felt like I should've done better.

The subject test was also quite hard in comparison with the sample test they give; I imagine I answered incorrectly/left blank something like 6-8 questions, even though the sample test took me about 90 minutes and I didn't leave a single question blank.

[quote="Kent Merryfield"](I've never once used the phrase "smarter than Guass" in a letter, of course. Simon was exaggerating with that. Letters do have to be grounded in reality - but his point stands about the subject exam being a checkup on just how grounded in reality they are.)[/quote]

I was definitely exaggerating. But I think people understood the point.

[quote="Kent Merryfield"]I've heard rumors - third-hand, unconfirmed - that some places like to look at the GRE general verbal score. The GRE general math is pretty much worthless to them, because it's too low-level and too easy (for a math major), so it fails to distinguish people. But there's the idea floating around that the verbal says something about mental agility or general intelligence. (Get a bunch of the brightest math nerds together and play purely verbal games - Scrabble, anagrams, crossword puzzles, and so on - and the idea might not seem so far-fetched.)

Of course, because of the premium placed on vocabulary, that puts those whose native language is not English at a large disadvantage - which is OK, they understand that and make allowances for it. But once they know your native language is not English, then your application is in a different pile, the one for which one of the questions is "If we make this guy a TA, will disgruntled freshmen and their parents be pounding on the chair's door complaining about instructors who don't speak English?" In that case, at least one of your letters needs to say something reassuring about your communication skills.

Again, this is rumor and speculation; don't treat it more seriously than that.

(I've never once used the phrase "smarter than Guass" in a letter, of course. Simon was exaggerating with that. Letters do have to be grounded in reality - but his point stands about the subject exam being a checkup on just how grounded in reality they are.)[/quote]

Thankfully ETS is changing the general test next September. They claim that analogies and antonyms will be completed removed. I've tried some of the new sample questions (available on their website). I really like the feel of the new test, and I think I'm going to wait to take that one. It places much less emphasis on pure vocabulary knowledge and lot more emphasis on verbal/analytic reasoning.

thank you for the comments!
still, i wonder if there are some strong suggestions 800/450 (say) would/wouldn't make a significant difference instead of 790/410?:)

That takes you from the 40th percentile to the 50th percentile on the verbal; I don't think it will make that big of a difference at all.

I don't know anything about admission personally, but conventional wisdom goes like this (for *math* grad school only):

1.  Math subject GRE is important, sort of.  Top schools use it to filter out people who didn't get the highest possible score (because almost all of their serious applicants have the highest possible score).  Second-tier schools and lower care a lot about it.
2.  General scores almost worthless.  But common sense dictates that someone who blows the general math GRE probably shouldn't become a mathematician.
3.  Verbal might be important for international students, but it's not nearly as important as the TOEFL.

My (fairly dated) understanding is that the top mathematics graduate schools expect you to do very well on both the General and Subject GREs.  The General is necessary to be admitted to a top graduate school in general, and the Subject to be admitted to the department.  (Yes, there are two separate admissions processes, and you must be admitted by both graduate school and department in order to matriculate.)

so if the grad school accepted you and the department did not, does that mean you can still go to that school with a different major?

Usually the department decision is decisive for grad school.


limit
factorial
induction
algebra unsolved
algebra

Let $ a_1,a_2,a_3,...$ be an infinite sequence of positive reals. Prove that there are infinitely many $ n$ such that

$ n(1\plus{}a_n)>(n\plus{}1)a_{n\minus{}1}$

Suppose the contrary, then there is an $ N\in \mathbb{N}$ such that $ a_{n \plus{} 1}\leq \left(1 \plus{} \frac {1}{n \plus{} 1}\right)a_{n} \minus{} 1$ for all $ n\geq N$.  Hence the sequence $ \{b_n\}_{n\geq N}$ defined by $ b_N \equal{} a_N$ and $ b_{n \plus{} 1} \equal{} \left(1 \plus{} \frac {1}{n \plus{} 1}\right)b_n \minus{} 1$ satisfies $ a_n\leq b_n$.  We can show that regardless of the value of $ b_N$, $ \lim_{n\to \infty}b_n \equal{} \minus{} \infty$ (we can show this by, for example, solving the difference equation in terms of the digamma function).  Therefore, there is an $ n$ such that $ a_n\leq b_n \leq 0$, which contradicts the assumption $ a_n > 0$.

[quote="JoeBlow"] (we can show this by, for example, solving the difference equation in terms of the digamma function).  [/quote]

It seems you're using high-level knowledge. Is there more elementary way?

It's not any advanced concept, I was just using unnecessary terminology, my apologies.  Instead of [url=http://mathworld.wolfram.com/DigammaFunction.html]'digamma function'[/url] I really should have just used the more familiar term [url=http://mathworld.wolfram.com/HarmonicNumber.html]'harmonic numbers'[/url]; they're essentially the same thing at integer values, just as the 'Gamma function' is closely related to factorials at integer values.

So all I meant is that by induction we can show:
\[ b_n \equal{} (n \plus{} 1)\left(\frac {a_N}{N \plus{} 1} \plus{} H_{N \plus{} 1} \minus{} H_{n \plus{} 1} \right) \equal{} (n \plus{} 1)\left(\frac {a_N}{N \plus{} 1} \minus{} \sum_{j \equal{} N \plus{} 2}^{n \plus{} 1}\frac {1}{j}\right)
\]
where $ H_k$ is the $ k$th harmonic number; $ H_k \equal{} \sum_{j \equal{} 1}^k\frac {1}{j}$.  This shows that $ b_n\to \minus{} \infty$ and hence a contradiction.


To show experimentally that it can be converted into something that is already recognized as a "form of energy". Actually, I do not know any other way to demonstrate it ;).

one nice experiment would be to heat water with a lid on it and after sometimes the lid starts bouncing...Actually this is a classical result and this is how it was first found that heat is indeed a form of energy

When object moves on a surface with some friction, part of its kinetic energy is changed to heat.So heat and energy are related to each other.

[quote="aahmadi"]What is the easiest way to explain that heat is a form of energy?[/quote]

The only way to demonstrate, prove, that heat is a form of energy, as others have already said, is to perform an experiment whereby heat is transformed into work. This because energy is defined as the capacity to do work.

However, if you merely want to explain it as such, start by pointing out that adding heat to a body raises its temperature. Then suggest that temperature is a measure of molecular motion, thermometer/speedometer analogy, i.e. the higher the temperature the faster the molecules are moving. And finally point out that speed is a measure of kinetic energy. Close the circle by stating, adding heat increases (molecular) kinetic energy; therefore, heat is (a kind of) energy.

Thank you very much guys.  Now if you feel like maintaining your kindness, please check out the newest topic I just brought up.

[quote="Dr. No"][quote="Carcul"][quote="Dr. No"]that adding heat to a body raises its temperature[/quote]

Not always: be carefull with that argument.[/quote]

Okay, adding heat raises temperature if the substance is not undergoing a phase transition. But who would want to make that distinction to a ten-year-old?[/quote]
Dr. NO ok that may not be so also what about isothermal processes ... :wink:

I really hate to spam but [b]t0rijir0u[/b], that was awesome!  :clap2:

Yeh, Seeing is believing! :lol:

t0rajir0u's explanation is the best.

[quote="pardesi"]Dr. NO ok that may not be so also what about isothermal processes ... :wink:[/quote]

In an isothermal process, if you add heat then entropy increases (disorder at the molecular level), but again, if explaining that heat is energy to fifth graders (ten year olds) my inclination would be to avoid that discussion.

[quote="pardesi"]then what the point is there is heat added without change in temp.....
well u surely can't teach wrong things to a 5th grader....
u may not tell him but whatever u tell has to be  right[/quote]

I am certainly not going to communicate false information to anyone intentionally. Nonetheless, it is not necessary to exhaust every possible case when giving an introductory lesson on a particular topic.

The usual circumstance, that can be demonstrated by a simple experiment is that adding heat to an object increases its temperature. The general rise in temprature stops once a substance reaches its melting point. It continues again when the entire sample has melted; and stops once more when the substance reaches its boiling point, and then continues once the entire sample has vaporized.

These plateaus in the general trend of temperature increase do not invalidate the explanation that heat is energy because the addition of it raises temperature and thereby kinetic energy.

Furthermore, isothermal (constant temperature) processes are special cases that occur mostly, if not only, in gases. That the temprature does not increase during such processes again does not suggest that heat is not energy. It is merely that it is increases entropy rather than temprature for this process. [b] I do not intend to explain entropy to children.[/b]


The sequence $a_{1},a_{2},\dots$ of natural numbers satisfies

$GCD(a_{i},a_{j})=GCD(i,j)$

for all $i \neq j$. Prove that $A-i=i$ for all $i$.

[hide="what i have done"]so far all i have shown is that $k|a_{k}$ but am having trouble moving on from this point.
[/hide]

[hide]
assume $a_{k}=k*p$ where $p>1$. then look at $GCD(a_{k},a_{kp})$ an derive a contradiction from your own conclusion...;)[/hide]

well to make typing easier im just going to type $(a,b)$ instead of $GCD(a,b)$

ok so since i already have $k|a_{k}$ i let $a_{n}=ns$ for some integer $s$.
$(a_{n},a_{nr})=(ns,nr)=n$ so $r$ and $s$ are relatively prime.
now
$(a_{n},a_{nrs})=(n,nrs)=(ns,nrs)=ns=n$ so $s=1$ thus $a_{n}=n$


How do you do this problem?
In the expansion (a+b+c)^3 what is the coefficient of the "abc" term?

We expand:

$ (a \plus{} b \plus{} c)(a \plus{} b \plus{} c)(a \plus{} b \plus{} c)$

Let's take care of the first two terms first and then multiply by the third term.

$ (a \plus{} b \plus{} c)(a \plus{} b \plus{} c) \implies a^2 \plus{} ab \plus{} ac \plus{} ab \plus{} b^2 \plus{} bc \plus{} ac \plus{} bc \plus{} c^2$

$ \implies a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc$

Now we do:

$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc)(a \plus{} b \plus{} c) \implies$

$ a^3 \plus{} ab^2 \plus{} ac^2 \plus{} 2a^2b \plus{} 2a^2c \plus{} 2abc \plus{} a^2b \plus{} b^3 \plus{} bc^2 \plus{} 2ab^2 \plus{} 2abc \plus{} 2b^2c \plus{} a^2c \plus{} b^2c \plus{} c^3 \plus{} 2abc \plus{} 2ac^2 \plus{} 2bc^2$

When we simplify, we find that the coefficient of the $ abc$ term is $ \boxed{6}$.

......

way easier solution

(a+b+c)(a+b+c)(a+b+c) so we can choose a from either first, second or last group

and we can choose b from other 2 groups since you cant multiply two numbers which are in same parenthese

and we can choose c from last 1 group.

3*2*1=6

answer : 6

similarly, coefficient of abcd in $ (a\plus{}b\plus{}c\plus{}d)^4$ is 24.

hey fantasy are you sure that works for every situation possible it looks like it was just luck

no, cause he just counted all the solution for 3 and that would work for all real n

generally

coeffiecient of $ a_1a_2...a_n$ in $ (a_1\plus{}a_2\plus{}...\plus{}a_n)^n$ is $ n!$.

Is there an overall "rule" that expanded expressions in the form $ (a_1 \plus{} a_2 \plus{} a \plus{} 3 \plus{} \dots \plus{} a_n)^r$ follow, depending on the values of $ n$ and $ r$?

Edit: I know this might get messy, but can anyone post it? There seems to be a pattern I cat give name to, maybe something to do with binomial theorem.


conics
parabola
analytic geometry
graphing lines
slope
geometry

For the origin $ O$, given two points $ A,\ B$ such that $ \angle{AOB} \equal{} 90^\circ$ on a parabola $ y \equal{} x^2$. Suppose that the line $ AB$ has the slope 2.

(1) Find the equation of the line $ AB$.

(2) Find the area of $ \triangle{AOB}$.

(3) When the point $ P$ moves in the line $ AB$, find the minimum value of the length $ OP$.

[hide]$ A(a,a^2),B(b,b^2)$ Calculating dot product $ ab\plus{}(ab)^2\equal{}0\implies ab\equal{}\minus{}1 (ab\neq0)$
$ 2\equal{}(a^2\minus{}b^2)/(a\minus{}b)\equal{}a\plus{}b$ Solving $ x^2\minus{}2x\minus{}1\equal{}0\implies x\equal{}1\pm\sqrt2$

$ (1\plus{}\sqrt2)^2\equal{}2(1\plus{}\sqrt2)\plus{}k\implies k\equal{}1$ So the equation of the line is $ y\equal{}2x\plus{}1$

$ [AOB]\equal{}1/2\cdot AO\cdot BO\equal{}\sqrt2$

$ OP\equal{}2\sqrt2/AB\equal{}1/\sqrt5$[/hide]

That's correct.

(1) AB: $ y\equal{}(a\plus{}b)x\minus{}ab$

(2) $ [OAB]\equal{}\frac 12|ab(a\minus{}b)|\equal{}\frac 12 |ab|\sqrt{(a\plus{}b)^2\minus{}4ab}$


[quote="mathgeniuse^ln(x)"]If interest were compounded annually, what annual interest rate would (most nearly) double my money in 2 years? [Give your answer to the nearest $0.1\%$]

No Calculators[/quote]
[hide]
Let 100x=interest rate

$(1+x)^{2}=2$
$x=0.414$

Answer: $\boxed{41.4\%}$[/hide]

Think about it.  If you are only compounding it by 41.4%, you won't quite reach double, but another value will surpass double.  So 41.4% is incorrect.

[quote="mathgeniuse^ln(x)"][hide]Here is a solution

"Let's look at the money that's there at each stage.  First, we have $\$x$.  Next year we have $k(\$x)$.  The second year we have $k[k(\$x)]>\$2x$, so $k^{2}>2$ and $k>\sqrt{2}=141.4+\%$.  Since the principal is $100\%$, the least rate of interest must be $41.5\%$

$41.5\%$"[/hide][/quote]
That's mean.
Well, the answer is $(100\sqrt2-100)\%$, rounded to the nearest $0.1\%$, like it states, is $41.4\%$.

I got 41.4% too, because $1.414^{2}$ nearly equals 2.

[quote="mathgeniuse^ln(x)"]Don't ask me, that is what Math League said the answer was, and I think that was their solution as well. so there you have it.

I hope this doesn't happen on later Math Contests.[/quote]

Most nearly indicates that it's as close as possible. 41.4gets it a heck of a lot closer than 41.5

[quote="mathgeniuse^ln(x)"]I don't know, but that is what I got first, but then I realize the asnwer is 41.5, not 41.4

There you have it.[/quote]
Why did you realize the answer is 41.5?

That is also what the Math League said, so you read its mind. :lol:

[quote="fingerfood"]well i support the answer of 41.5% because the question says most nearly double, where double is used so you must atleast double. therefore you must round up. rounding down you dont get the full double minimum[/quote]

Do you know what most nearly means?
It doesn't mean most near value that is greater than what's wanted....

[quote="mathgeniuse^ln(x)"]Yeah, I thought about reporting the error, but I think their solution is correct, so no need to protest.

Here is another solution, in case you're wondering.

"Let the rate of interest be r, and let the original principal be P.  After the 1st year, the principal becomes P+rP.  The second year, the principal becomes (P+rP)+r(P+rP)>2P.  The question is meaningless unless P doesn't equal 0, so we may divide both sides of this inequality by P to get $r^{2}+2r-1>0$.  The least positive solution of the inequality is given by $r=\sqrt{2}-1$, which rounds [i]up[/i] to $r=41.5\%$."

So that is another solution.[/quote]

But that's a [i]wrong[/i] solution.
You're assuming that (P+rP)+r(P+rP)>2P, but actually, what you should be doing is (P+rP)+r(P+rP)=2P, and then getting the [i]closest[/i] value.

Did you miss this question when you were doing this test?

Yes, I did since I said $41.4\%$, but then I still was in shock that it said $41.5\%$.

But, I still agree with all of you that is why I posted this question.

I just hope this doesn't happen in future contests.

They said that because 41.4% doesn't double it.  But 41.5% surpasses x2, so that is why that is the answer sort of.

I've never heard of Math League, but w/e it is, it must not have very good solution checkers...

Now that I think about it, I think I saw this question before. Maybe it was during the year that I got a 39 on my math league, hmm, i wonder. :ninja:

Um, this is a High School MathLeague, so it is out of 36, so you didn't really see it.  Unless you meant 35.

I know I saw it somewhere before. Maybe in a practice test or the real thing, I forget.

Sorry, about this problem, if anyone is getting upset by it.


Let a,b,c be real numbers. Prove
$ \frac{(a\minus{}b)^2}{(b\minus{}c)^2}\plus{}\frac{(b\minus{}c)^2}{(c\minus{}a)^2}\plus{}\frac{(c\minus{}a)^2}{(a\minus{}b)^2} \ge 5$

hello, your inequality
$ \frac{(a\minus{}b)^2}{(b\minus{}c)^2}\plus{}\frac{(b\minus{}c)^2}{(c\minus{}a)^2}\plus{}\frac{(c\minus{}a)^2}{(a\minus{}b)^2}\minus{}5\equal{}$
$ \frac{(a^3\minus{}2a^2b\minus{}ab^2\plus{}b^3\minus{}ca^2\plus{}6abc\minus{}2b^2c\minus{}2c^2a\minus{}b^2c \plus{}c^3)^2}{(a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2}\geq0$
and this is true.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your inequality
$ \frac {(a \minus{} b)^2}{(b \minus{} c)^2} \plus{} \frac {(b \minus{} c)^2}{(c \minus{} a)^2} \plus{} \frac {(c \minus{} a)^2}{(a \minus{} b)^2} \minus{} 5 \equal{}$
$ \frac {(a^3 \minus{} 2a^2b \minus{} ab^2 \plus{} b^3 \minus{} ca^2 \plus{} 6abc \minus{} 2b^2c \minus{} 2c^2a \minus{} b^2c \plus{} c^3)^2}{(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}\geq0$
and this is true.
Sonnhard.[/quote]

Does exist any easier way ? this is complicated  :blush:

How can you find that such expression? . I doubt that there must be a programme set up to factorize or cancel out some complicated expressions. Anyway, this is an era in which technology is leading the way.

oh, no, there is a way to do that by hand  :D

let $ \frac{b\minus{}a}{b\minus{}c} \equal{}x$... and so on. We have

$ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx\plus{}2$

It is equivalent to prove that

$ (x\plus{}y\plus{}z\minus{}1)^2 \ge 0$ which is true.

this problem deverves a position in IMO like IMO2008


Prove that among 39 sequential natural numbers there always is a
number with the sum of its digits dividable by 11.

Look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=6301&highlight=39

Pierre.

Ok, and about this: find the smallest counterexample for 38 instead of 39.....It appeared in a Romanian TST i think....