vector quadratics trigonometry conics hyperbola AMC AIME

Suppose an alpha patricle is not aimed at the center of a lead nucleus, but had a nonzero angular momentum (with respect to the lead nucleus) of magnitude L=pb, where p is the magnitude of the initial momentum of the alpha and b=1.00*10^-12m. What is the distance of closest approach? ( you may use the mass and the charge of the alpha particle, I don't know)

hey is it aimed at the centre or not...if not where is it aimed at  :?:

[quote="maomao13"]It is not aimed the center; it has a nonzero angular momentum.[/quote]
then where is it aimed at  because i could aim it at any direction and ofc all would giev me different ansewers :)

I am sorry. In this problem, we know nothing else but L(angular momentum)=pb, in which p is linear momentum, and b is a constant.

[quote="maomao13"]If you can get different answers, please show me.[/quote]
i ma assuming that the velocity vector makes an angle $ 179.99$ degree with the line joiningit to nucleus...
and another case when it makes $ 0.0001$ degress...and u expect teh answer to be same :huh:

As $ b$ is the 'moment arm' (or whatever you call it), the particle is aimed at a point at a distance $ b$ from the lead nucleus. However, you do need to know where the particle is thrown from. Is it from infinity?

EDIT: I was putting this up at about 2:25GMT but my network got disconnected.

[hide]
So I'll assume its from infinity
Also, I'm assuming the mass of the Lead nucleus is much larger than that of the $ \alpha$ particle.

At its closest approach, the particle's velocity will be perpendicular to its radius vector (obviously putting the lead nucleus at the origin).
As force is central, angular momentum is conserved, i.e. $ p_f r_0 \equal{} pb$
Where $ r_0$ is minimum distance and $ p_f$ is final momentum.

Using energy conservation we get the quadratic equation $ p^2 r_0^2 \minus{} \frac {q_{\alpha}q_{Pb}}{4\pi\epsilon_0}r_0 \minus{} p^2 b^2 \equal{} 0$ which we can solve to get $ r_0$.
[/hide]

@Shreyas
without the direction data the proof won't hold true....see if i throw the particle at $ 179.99$ degress from say $ \infty$ it will never reach the nucleus that close that is there [b]does not exist[/b] a time $ t$ when it's velocity is perpendicular to radius vector...and ur entire proof depends on the fact that "if"
 the velocity at a time $ t$ becomes perpendiculat to the radius vector "then" if the first one doesn't happen the second won't :)

$ \vec{L}\equal{}\vec{p}\times\vec{r}$
Hence, $ |\vec{L}|\equal{}|\vec{p}\parallel{}\vec{r}|\sin\theta$

But we know that $ |\vec{r}|\sin\theta\equal{}b$ is constant, even as $ |\vec{r}|\rightarrow\infty$

Hence, the particle is aimed at a point $ P(b,0)$ (say).

i get that but the problem is the partcile if releaseed in arbitrary direction [b]from infinity[/b] won't reach the nucleii...say for example the direction i gave ...the key word being infinity :)

[quote="maomao13"]L(angular momentum)=pb, in which p is linear momentum, and b is a constant.[/quote]

The particle is NOT released in an arbitrary direction. $ b$ is constant implies that it is thrown towards a point at that distance from the nucleus.

I don't know if I can prove it, but the particle will move in a hyperbolic orbit with the nucleus as the focus and $ b$ is the distance between its focus and vertex.

why not since $ r$ is pretty unknow and $ b < < <1$ and if we consider $ r > > > > > > > 1$ then clearly we can have $ \theta \approx 180$ :wink:

Consider Coördinate system with the nucleus as origin and the initial direction of the particle's momentum (both constant) as positive Y axis.
The Locus of points having angular momentum $ |\vec{L}|\equal{}|\vec{p}|b$ where $ b$ is constant is the line $ x\equal{}b$

We are assuming that the particle was thrown from infinity along this line. My gut feeling is that the particle will travel along (one half of) the hyperbola $ (x\minus{}b)(y\plus{}b)\equal{}k$, $ k$ is constant.

In any case, you need only to use conservation of energy and angular momentum to get the minimum distance.

actually i have no problem with ur proof it's the question that is 'wrongly' stated...it shud have been like this
"Suppose the $ \alpha$ particle is aime not toward the centre of lead nucleus but at a certain distance away but towards it with say an impact parameter $ b_{0}$...."
then it becomes logical....further no distance is specified so u r left with two choices
1)thrown from $ \infty$ then u need the direction...because of the reasons stated above
2)if not infinity then where's the distance it is thrown from :)

I'm trying to justify that the statement $ L\equal{}pb$ alone suffices to fix the direction of propogation. (as being along $ x\equal{}b$)

i ma deleting ur above post ...i posted ur other problem in a [url=http://www.mathlinks.ro/viewtopic.php?t=189609]seperate thread[/url]


function
trigonometry
number theory
calculus
calculus computations

Today On the big break at university I and my classmate together drank tea and talked about theorem Dirichlet. Me has interested: How Theorem Dirichlet  by Gammaf Function is proved?
Theorem Dirichlet:
In number theory, Dirichlet's theorem, also called the Dirichlet prime number theorem, states that for any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $a+nd$, where $n > 0$, or in other words: there are infinitely many primes which are congruent to $a$ modulo $d$. Moreover, the sum of the reciprocals of such primes diverges.

[quote="mlok"]I don't know what $\Gamma$ has to do with Dirichlet's theorem, but you can find an accessible proof in [url=http://www.princeton.edu/~ppollack/notes/notes.pdf]Pollack's notes[/url], pages 71-89.[/quote]
Thanks. It achieved by $\zeta$ function. I don't remember somewhere was
$\zeta(x)=2^{x}\pi^{x-1}\sin{\frac{\pi{s}}{2}}\Gamma(1-x)\zeta(1-x)$ thereore it's can be achieved (proof) by $\Gamma$.Thanks again


abstract algebra
superior algebra
superior algebra unsolved

In the book Algebra of Lang I read the following property : Let $ A$ be a ring, $ M$ a free $ A$-module, $ I$ a nonzero set and $ (x_i)_{i \in I}$ a basis of $ M$. If $ N$ is an $ A$-module and $ (y_i)_{i \in I}$ a subset of $ N$, there exists an unique homomorphism $ f : M \rightarrow N$ such that $ f(x_i)\equal{}y_i$ for all $ i$ of $ I$.
I would like use this property to prove that for all $ A$-module $ M$ there exists a free $ A$-module $ F$ and an epimorphism $ f$ from $ F$ to $ M$. But How to do it ?

First, there's some messiness in your phrasing: $ I$ is not "nonzero" but "nonempty", and $ (y_i)_{i\in I}$ is not a subset but a list. (the elements can all be the same, for example)

As for your question, let $ I$ be $ M$ itself.

Thanks you for your answer, if I have good understanding I form the free module $ F$ on $ M$ generated by $ M$ and the map $ f : F \rightarrow M$ with $ f(x)\equal{}x \in M$, therefore $ f$ is an epimorphism?

Yep!

... (need to make these messages longer, apparently)


linear algebra
matrix
complex numbers
complex analysis

This is a [i]very[/i] basic question but I cannot find any other forum where discussing complex numbers seems appropriate. (Please move if there is one)

My question is about the definition of $ i$ . My textbooks define it such that $ i^2 \equal{} \minus{}1$ . However, I have also seen it written as $ i \equal{} \sqrt{\minus{}1}$ . Surely it should be written as $ i \equal{} (\sqrt{\minus{}1} \lor \minus{}\sqrt{\minus{}1})$ - i.e. the disjunction of the negative and positive roots? This hardly seems like a number to me, and I have never seen it written like that in any textbooks; however, I think there are some well-known proofs that start from $ i \equal{} \sqrt{\minus{}1}$ and work to a contradiction. I am confused.  :( This whole issue has been bothering me for a while, so can anybody clear it up for me?

The formal way to construct complex numbers is to define them as pairs $ (a,b)$ of reals with the addition law $ (a,b)\plus{}(c,d)\equal{}(a\plus{}c,b\plus{}d)$, multiplication law $ (a,b)\cdot(c,d)\equal{}(ac\minus{}bd,ad\plus{}bc)$ and the imbedding of the real field given by $ x\mapsto (x,0)$. Then $ i\equal{}(0,1)$, period. This may seem a bit too formal but it  eliminates all conversations about which square root of $ \minus{}1$ to take, what is it, etc.

Another construction - entirely equivalent to the one fedja gave - is to construct the complex numbers as the set of matrices of the form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}.$

In this version, $ i\equal{}\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}.$

This is a specific case of the construction of "numbers of the form $ a\plus{}b\sqrt{m}$" as

$ \begin{bmatrix}a&mb\\b&a\end{bmatrix}.$

It actually doesn't matter "which" square root to take as long as you're consistent:  in other words, $ \mathbb{R}[i]$ is isomorphic to $ \mathbb{R}[ \minus{} i]$.

A third way that I've seen to define the complex numbers formally is as $ \mathbb{R}[X]/(X^2 \plus{} 1)$; here $ i \equal{} X$.  Of course, this definition of the complex numbers obscures entirely the fact that they're algebraically closed, but the point here is that $ X$ is just a formal variable with the property that $ X^2 \equal{} \minus{}1$.

[quote="Drone"]however, I think there are some well-known proofs that start from $ i \equal{} \sqrt { \minus{} 1}$ and work to a contradiction. [/quote]

Such proofs are based on assumptions about the square root that do not hold in general.  In particular $ \sqrt {ab} \equal{} \sqrt {a} \sqrt {b}$ is not true in general.

Thank you for the replies. I've seen Fedja and Kent's definitions before (in Linear Algebra) but was thinking about how complex numbers are introduced to beginners. Are you suggesting it might be best to stop trying to reconcile that with what I know now, and focus on the formal definitions?

(t0rajir0u's definition is also nice  :) )


MATHCOUNTS
geometry
FTW
AoPS Books
number theory

Is it better to prepare with good books like AoPS or with problems (if both, what should I do more of)?
Are there any topics that the AoPS books don't cover that are important for MC?

You should definately read the aops books. They give you lots of problems while also telling you [i]how[/i] to do them. Volume 1 is the best, start with that. It has basically every topic MC will ask you about.

If you have time, use the Introduction series (first take the post-tests found on the separate pages of the books, and find your weakest subjects, then buy those books that you're weak in). But yeah, Volume 1 is the best. Read through it really carefully.

Also, I reccommend doing lots of probability/counting and geometry problems. These are usually the two subjects everyone is weakest in and on MATHCOUNTS tests, they are usually the hardest.

[quote="math154"]Sometimes number theory can be a little tricky too...[/quote]
Yes. I agree. I often get stuck on what method to use. That's why number theory is so hard for me. :(

I'm best at number theory :D
This is when I wish I hadn't lost Volume 1...

hey guys... i think we're getting a bit off topic

drill old cds i guess?
playing ftw also helps, until you start memming

Isn't Tiny Tim that kid in [u]A Christmas Carol[/u] who was ill?

Yeah...but isn't that kind of random? :P

You've posted a lot of random things.
I'm just going to do the 2000-2003 Nats and check wrong answers.


binomial coefficients
Pascal\u0027s Triangle

Is there a way to sum up $ \binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+...+\binom{n-k}{k}$ until $ k>n-k$

can someone help or is there no generalized formula?

It's a Fibonacci's sequence.
$ f(0)=1$
$ f(1)=1$
$ f(n)=\sum_{k}\binom{n-k}{k}=f(n-1)+f(n-2)$
You can prove the recurrence using the following property of the binomial coefficients (remember Pascal's triangle):
$ \binom{a+1}{b}=\binom{a+1}{b}+\binom{a+1}{b}$
and applying it to
$ f(n-1)=\sum_{k}\binom{(n-1)-k}{k}=\sum_{k}\binom{n-k-1}{k}$
$ f(n-2)=\sum_{k'}\binom{(n-2)-k'}{k'}=\sum_{k}\binom{n-k-1}{k-1}$ (with the shift $ k'=k-1$)


Prove that the only square numbers in the Lucas sequence are $ 1$ and $ 4$.

I am not sure what you are asking.  Maybe some clarification is needed.
The Fibonacci sequence is an example of a Lucas sequence and moreover $ 144\equal{}12^2$ is a term in this sequence which is known to be the largest square in this Lucas sequence.

Hi there, [b]baz[/b]!

I think that [b]Nemion[/b] is specifically referring to the sequence $ 2, 1, 3, 4, 7, \ldots$ [b]That[/b] one is the [i]mother[/i] of all Lucas sequences.

Regards,

[b]Coquitao[/b] is right, I am referring to those: $ 2,1,3,4,7,\dots$

$ a_n\equal{}(\frac{1\plus{}\sqrt{5}}{2})^{n}\plus{}(\frac{1\minus{}\sqrt{5}}{2})^{2}$
$ a_{2n}\equal{}(\frac{1\plus{}\sqrt{5}}{2})^{2n}\plus{}(\frac{1\minus{}\sqrt{5}}{2})^{2n}$
$ a_{2n}\equal{}(a_n)^{2}\minus{}2(\minus{}1)^{n}$
Hence $ a_{2n}$ can't be a square (the residue modulo 4 is either 2 or 3)


[quote="Lovasz"]The sequence $(x_n)$ defined by $x_n=a\sqrt{n^2+1}$.

$a)$ Find the value of a such that $(x_n)$ converges.

$b)$ Find the value of a such that $(x_n)$ is monotone increasing.[/quote]
Maybe you had in mind anything like this $x_{n+1}=a\sqrt{x_n^2+1}$ else problem looks rather strange.

I think the original problem is \$PAM because only for $a=0$ it would be convergent and for $a>0$ monotone increasing.
A solution for the second problem cuold be for $a\leq k$ convergent and for $a>k$ monotone increasing, where $k$ is the solution for $x=k\sqrt{x^2+1}$, so $\lim_{x\rightarrow \infty}k^{2}=1$, so it diverges for $a\geq 1$. But I think the "convergent" problem is a bit more complicated...

Naphthalin

[quote="spix"]Define $y_n=x_n^2$ so $y_{n+1}=a^2y_n+a^2$ and from here you can easily find the general term of $y_n$ and so forth.[/quote]

I haven't been learnt any way to find general term of $y_{n+1}=a^2y_n+a^2$ :|


Support
analytic geometry
graphing lines
slope
geometry
geometric transformation
reflection

This is a very sensible topic. I hear it being debated everywhere, so I decided to post it for you guys to discuss. Especially people in CA. And please, respect others' views. Thanks.

I'd be highly offended that this was even proposed if I lived in CA. Any attempt to write discrimination into a constitution is highly unethical and immoral. According to its exact wording it will "eliminate the right of same-sex couples to marry." Any time I hear the phrase "eliminate the right of..." fill in the blank, I shudder to think that I actually live in a world where people pass measures such as these of their own free will. Just put yourself in the place of the person that has their rights being eliminated. You are quite heartless if you still vote for it.

I can't vote against it since I don't live there now, but I've done the next best thing and donated a little money towards that cause.

[quote="HilbertThm90"] "eliminate the right of..." [/quote]

people to torture small mammals

I feel like you probably meant something slightly more specific than any possible fill-in-the-blank I could come up with . . .

Ha. Good point. You know what I mean.  :P

I live in California...

NO on Proposition 8!

ya, the "fill in the blank thing is a little bit broad.

eliminate the right to murder
hmm . . .

i'm no on 8 and i live in CA (woot!)

The problem is that the sentence "eliminate the right to murder" doesn't make sense, since murder is not a civil right. You can't take it away. Of course I should have been more specific and said, "eliminate the right of *fill in a group of people* to *fill in civil right that all others have*."

Geez. It is like you guys are mathematicians or something. You don't let anyone get away with imprecise language.  :wink:

NO ON PROP 8!

I've been hearing all sorts of pessimism about how it's probably going to pass. Can someone find contrary evidence? Please? Pretty please?

[url=http://www.latimes.com/news/politics/la-me-gaymarriage23-2008oct23,0,65703.story]Some evidence?[/url]
I don't know, I've heard that yes is actually leading over no.

This discussion is pretty useless when only one side is presented. And it seems that there is a majority of people against 8 on AoPS.

I also have major problems against Prop 8, and really hope it doesn't pass. But I would like to hear the points of the other side... (Don't worry about flaming, this is AoPS, not 4chan?)

[quote="HilbertThm90"]
Geez. It is like you guys are mathematicians or something. You don't let anyone get away with imprecise language.  :wink:[/quote]

 :)

Well, if you wonder if there is [i]anybody[/i] on AoPS, who supports proposition 8, I'm in favor of it, though I don't live in California either, so my vote doesn't really count. As to why, just consider me a conservative old timer for now (the arguments put forward against it in this thread are hardly any more elaborate so far ;)). By the way, the words "eliminate the right" are very imprecise: they tacitly assume that it has been a constitutional or whatever right up to the moment, while, as far as I know the question is far more complicated and the idea of allowing something like that is fairly recent, so one should rather say "not institute the right" (for more details, look [url=http://en.wikipedia.org/wiki/Same-sex_marriage_in_California]here[/url]).

According to the [url=http://www.latimes.com/news/local/la-me-gaymarriage16-2008may16,0,6182317.story]California Supreme court[/url], it is a right. It's right there in the language of the decision, which is the basis of state law right now.

[quote="fedja"]As to why, just consider me a conservative old timer for now (the arguments put forward against it in this thread are hardly any more elaborate so far ;)).[/quote]  Marriage gives societal recognition to individuals in committed relationships, providing especially a large number of rights (e.g., hospital visitation rights, parental rights) that are difficult to secure through other means.  In addition, it's clear that it matters a great deal to a great many people that they be able to take part in this particular societal tradition, and it's utterly unclear what benefit would derive to anyone from forbidding them from doing so.  (Indeed, this was made quite clear by the Massachusetts experience: after gay marriage had been enshrined for a year or so with no fire and brimstone raining from the heavens, opponents couldn't get even 25% of the legislature to agree to attempt to revoke it.)

Moreover, society has now broadly recognized homosexual relationships as legitimate -- while there are obviously a large number of social conservatives fighting this trend, acceptance of homosexuality is larger in each age cohort than in the one that preceded it, and (as courts and legislators in various states have now implicitly recognized) this trend has no sign of abating.  If you like, there might be enough "conservative old timers" around today to pass Proposition 8 (or maybe not), but in 20 years a great many of you will have died out :P.

[quote="JBL"]Marriage gives societal recognition to individuals in committed relationships, providing especially a large number of rights[/quote]

Yes, we both understand that. The whole thing is, indeed, about whether the society should encourage or discourage homosexual relationships. I'm in favor of discouraging them by "economic" means though I do not want to go as far as directly outlawing them.

[quote]after gay marriage had been enshrined for a year or so with no fire and brimstone raining from the heavens, opponents couldn't get even 25% of the legislature to agree to attempt to revoke it.[/quote]
No, there will not be any immediate "fire and brimstone from the heavens". Just the border between what is acceptable and what is not will be moved a little bit further. The next step will be to allow polygamy (so far directly outlawed). And so on.

[quote]
Moreover, society has now broadly recognized homosexual relationships as legitimate -- while there are obviously a large number of social conservatives fighting this trend, acceptance of homosexuality is larger in each age cohort than in the one that preceded it.
[/quote]
Perhaps. The public acceptance of such things is characteristic for societies that get older and richer. The idea was played with in Ancient China, Ancient Greece and the Roman Empire (see [url=http://en.wikipedia.org/wiki/History_of_same-sex_unions]here[/url]), so it is nothing new and radical and definitely not a result of an unstoppable development towards better and fairer societal structures :).

[quote]If you like, there might be enough "conservative old timers" around today to pass Proposition 8 (or maybe not), but in 20 years a great many of you will have died out[/quote]

These are the words the younger generation always tells the older. I'm young enough to remember saying them and old enough to hear them now. What really happens is that young progressives grow into old conservatives, so I'm not really afraid of my breed dying out anytime soon :P.

[quote="worthawholebean"][quote="polishcalculator"]Begs the question: why do governments incorporate it in the first place?[/quote]Historical reasons. Up until quite recently (excluding some prechristian attitudes), homosexuality was considered an abomination because of Christian doctrine.
[/quote]
I don't think you should settle for 'historical reasons' - ever in your analysis. Everything has an origin and most things have a motive. Why did Christians pick it up? Surely the leaders wouldn't have adopted doctrine that would have turned people away. Anyway, most of the world hasn't been and isn't Christian.

Homosexuality was prevalent in the Roman Empire during the height of their supremacy. Perhaps our hedonistic ways come out during times of extreme affluence.

[quote="worthawholebean"]
[quote="polishcalculator"]There may be another less objective reason folks voted for 'Yes' on Prop 8. It is a selfish but innocently selfish notion that they want their children to follow in their footsteps.[/quote]I don't think Prop 8 would have much effect on a gay person with respect to the closet.[/quote]
I think most parents want to teach their kids to pursue a heterosexual partnership and do not wish to expand the definition of their own husband and wife relationship. Ask your own parents. In an age, in which we have accepted all kinds of relationships and lifestyles, I don't feel it should be too painful to allow a majority of folks to hold on to a little semantic heritage.

[quote="JBL"][quote="polishcalculator"]Governments (as well a religious institutions) both have vested interests to produce the next generation.[/quote]  And the evidence that also recognizing homosexual marriages interferes with this goal is where?  You seem awfully willing to let this background assumption go by without any supporting documentation.[/quote]

JBL - no doubt you are brilliant, but I would love to see your math proofs some day.

Anyway, I was trying to get at the government motive for hijacking the religious custom of sanctioning the bond between a husband and wife. It was in the context of does marriage have any place in government jurisdiction.

It's more to the point that there is no motivation to incorporate homosexual marriages if (and big IF) the objective is to encourage family units with aspiration of giving birth to new tax payers, soldiers, etc.. The status quo has been marriage is between a male and a female.

I think I laid out why as a parent this continues to be attractive and why this tradition even exists. But let me ask you this, why should I change my views? What's in it for me? I've heard the equal rights arguments but that doesn't immediately benefit me. I'm open ears - why should I vote 'No' when this comes around again? Get me to change my vote.

I see only one reason that can (and will) change [i]my[/i] (hypothetical, because I'm not a California resident) "Yes" to "No": the established public image of a typical homosexual couple as a pair of useful members of society living up to acceptable standards that have some family duties preventing them from working, getting income, etc. simultaneously.

Alas, as I said before, the current statistics just does not support this image. JBL is, no doubt, a very clever person who can argue really well but his problem seems to be that he likes to imagine other people to be just like him, only having different personal preferences. As I said, I would be in favor of allowing homosexual marriages, carring firearms in public places, paying as much income tax as the taxpayer finds appropriate himself, and reducing the whole legal code of the USA to "live the life in the way you consider decent and respect opinions and interests of others" [b]if[/b] all people were, indeed, like him. It is a great and noble thing to see one's equal in everyone and ask for the same rights and priviliges for them but, alas, for some unknown reason, this nobility manifests itself exclusively in the people who are far better than average, i.e., exactly when it is unjustified.

[quote="polishcalculator"]I would love to see your math proofs some day. [/quote]  With a little bit of cleverness, I'm sure you can find one or two of them on your own.  Perhaps also you can find a less insulting way to suggest that you don't believe my point was on-topic  :wink: .  In fact, though, I believe my comment is very much to the point.  The state of the law at the time of Proposition 8 was that gay marriage was legal.  As such, [i]status quo[/i] bias favors a no vote -- the government already was sanctioning marriages between consenting adults of the same gender, and proponents of Proposition 8 were not merely preventing the granting of a civil right but actually removing one already granted.  (Of course, I also oppose other similar legislation in places where gay marriage is not currently legal, where this argument doesn't apply.  But it [i]does[/i] apply in the case under consideration.)  So "no motive to incorporate homosexual marriage" isn't an argument in favor of Prop 8, while "no motive not to incorporate it" [i]is[/i] an argument against it.

[quote="polishcalculator"]I think I laid out why as a parent this continues to be attractive and why this tradition even exists. But let me ask you this, why should I change my views? What's in it for me? I've heard the equal rights arguments but that doesn't immediately benefit me. I'm open ears - why should I vote 'No' when this comes around again? Get me to change my vote.[/quote]  Rather than take up your challenge, I thought I'd just ramble a bit about what thoughts it inspires.  Basically, I read this and I think, "Okay, this is fine.  You basically agree with me that your position is entirely one of personal disgust, and as long as that's what the other side is, I can just sit back and [url=http://www.fivethirtyeight.com/2009/04/will-iowans-uphold-gay-marriage.html]let time[/url] [url=http://www.fivethirtyeight.com/2009/04/gay-marriage-by-numbers.html]win this argument for me[/url]."
The expansion of civil rights rarely benefits everyone in a society.  This is a case in which opposition to it strikes me as particularly strange, since there isn't even a group of people harmed by the change.  (This is unlike, say, the Civil Rights Movement, in which some people were clearly benefiting from the denial of civil rights to other people.)  Opponents are left with saying things like, "I don't want my kids to get gay married" or "acceptance of homosexuality is a sign of hedonism in society and is a harbinger of the fall of Western civilization" or "If we let homosexuals get married, soon you'll have to raise your children in a world where the union of man and box turtle has the same footing as the union of man and wife" or "What's in it for me to let gays marry?"  (None of these are meant to be understood as exact quotes of anyone taking part in this conversation.)  And those aren't statements that are likely to convince anyone who, for example, has a good friend who turns out to be gay and is raising a child with his or her life partner.

Finally, one off-topic point: when did it become a standard rhetorical tactic to suggest that the person you disagree with is intelligent or a competent debater as part of the assertion that he or she is wrong?  :huh:  It is evident in both of the posts that precede mine, and (while I'm always happy to accept compliments) that rather disconcerts me.   I mean, I'd be happy if all three of us could get together and agree that I'm clever, brilliant, charming, morally unimpeachable, and correct (and also perhaps if we could get me a tenured professorship somewhere), but what's with asserting the first four things as part of an explanation of how I'm completely wrong?  :maybe:

[i][size=85]Edit for minor typo.[/size][/i]

JBL, I agree.  And it's been said a sick number of times already in this thread alone.  I'd bet so much polishc didn't read the thread.
If you don't want your kids to know, but kids tend to be more accepting, are you REALLY doing it for your kids?  Or are you trying to impose your own views and fears?  Are you trying to avoid letting them think for themselves?

Remember desegregation?  Leaving everything alone FAILS.  I mean, at that, even look at the economy.  It was left alone and people took advantage and pulled off scandals.  Care to tell me that issues left on their own simply die off?  The only problems that die after time are the leaves, the snow, and adolescent angst.

Let me go with a few things that are apparently axioms.  These are some of the main ideas discussed so far for those who don't wish to read through what would be 40 or so printed pages.
Axiom: Being constantly prejudged for being part of a group (which has not collectively done anything immoral **)  , not as an individual is the definition of prejudice.  Unwarranted prejudice is unwarranted.
Axiom: Telling homosexuals that they cannot be married, or must be "separate but equal" gives the message there is something wrong, different with them, to everyone.  By literally saying that they must be looked at differently by the law.
Axiom: If homosexuality exists in nature, then homosexuality is nature-al.  Natural.  And it does.

Care to debate these point-by-point anyone?  So we don't keep jumping back in circular arguments?

polishc, if you indeed read this, say so.  Also, please, remember: you are making it so we have to prove it is moral.  Why not try to explain  why you feel homosexual unions are wrong?  Do you find something innately wrong with homosexuality?

@fedja: As usual, I feel a little sick to my stomach whenever I read one of your posts about this topic. Just assuming that your reason is legitimate (which I don't think it is, since it has to do with "earning" civil rights, an oxymoron in my mind), it still doesn't hold water. As someone that circulates with gay people all the time and actually knows gay couples, exactly what you described is what I see. Don't rely on weird statistics to formulate the "public image", actually formulate an opinion based on experience.

Most homosexual couples are "useful members of society living up to acceptable standards."

[start rant]
[quote="polishcalculator"]The distinction being that a male and a female are likely to produce children, or rather a new generation. Governments (as well a religious institutions) both have vested interests to produce the next generation.[/quote]Are you saying that we should invalidate all marriages when one is sterile? Do you think that every time two people get married, the government should look through and make sure that they are doing it to have offspring? The bottom line is that it's not the government's job to decide why people get married.

[quote] It's for this reason, they make distinct and promote what we have known as the traditional family structure. Consequently, there are a lot of social traditions and laws to facilitate the traditional family structure, I believe for this reason. And once children come into the picture, I haven't seen anything that bests a family structure. Once in place, a father and mother benefit and influence a son or daughter in very different but complementary ways.[/quote]

[quote]The concept of marrying for love is relatively recent in the history of human civilization. Arranged marriages were the norm and still happen in some cultures today.[/quote]As are monogamous relationships. Polygamy was the norm for thousands of years.

[quote]It's the idea that we are products of both nature and nurturing. And while we can't control the nature part (not very well at least), we will sure try hard on the nurturing part. So while a sympathetic majority doesn't want to deny any tangible rights to a gay couple, at the same time they do not want to cloud the distinct expectations they have for their children.[/quote]Legalizing gay marriage won't make more people straight or gay. It'll just make people happier as they can finally express what they truly want.

[quote]I don't think you should settle for 'historical reasons' - ever in your analysis. Everything has an origin and most things have a motive. Why did Christians pick it up? Surely the leaders wouldn't have adopted doctrine that would have turned people away. Anyway, most of the world hasn't been and isn't Christian.[/quote] In fact, gay marriage was only condemned twice in the [b]entire[/b] Bible and Jesus never mentioned it. This doctrine was picked up by the church, not the Bible. As for a reason, I would assume that it was for power. The more kids of the religious, the more people the Church has control of. In addition, condemning homosexuality could also give them something to blame when something else went wrong.

[quote]I think most parents want to teach their kids to pursue a heterosexual partnership and do not wish to expand the definition of their own husband and wife relationship. Ask your own parents. In an age, in which we have accepted all kinds of relationships and lifestyles, I don't feel it should be too painful to allow a majority of folks to hold on to a little semantic heritage.[/quote]Here's how this argument would've went with the Civil Rights Movement. "I think most parents don't want to explain that blacks are just as good as whites. They don't want to expand on their definition of a human. In an age in which we have given women rights, accepted the Irish, and let many immigrants come into our country, I don't feel it should be too painful to allow a majority of folks to hold on to a little semantic heritage." Do you still think it's a good argument?

So here's the deal. Homosexuals are humans, that we all agree on. Therefore, they cannot be discriminated upon due to their sexual orientation. We have heterosexual relationships, so we should allow homosexual ones.

Being gay is not a choice. In fact condemning it has cause gay suicides to be miles above the percentage of straight suicides. By not giving them equal rights you are basically saying: "You're not good enough. You don't deserve everything that we get." But they do deserve it. [/end rant]

P.J. O'Rourke has nice food for thought on this (he's probably more for gay marriage than opposed; though I do not think he appreciates the children side of the debate):

"I mean, say, for instance, the issue of gay marriage. Now marriage has got entirely two entirely separate aspects to it. One of them is religious, if indeed, you believe in a religion. It is the spiritual bonding of two souls into one, etc. And I’m a Catholic, I go along with that.

But the other side, on a political side, marriage is a contract. And why any two people shouldn’t have equal contract rights, I can’t imagine. Now, of course, part of the problem is prejudice, it’s bigotry, it’s short-sightedness on the part of political conservatives. But part of the problem is also the activists. Because activists have not, in general, been satisfied with getting just the contractual rights. They say, no, no, we won’t be fully equal in this society unless it is called and regarded as a marriage.

Well, it isn’t actually within the capability of political society to make that change in people’s minds. That is a social, a religious, an extra-governmental decision, whether two gay people, who intend to stay together forever, whether they are, in fact, married, is something that people have to decide for themselves. It’s not something the government can decide. The government only covers contractual law.

Now, I would say most of the fault here is on the part of the body of people are political conservatives, not all of whom are social conservatives. But the social conservatives within that body of political conservatives bear most of the blame for the situation. But there is some blame in the insistence common to activists in any field that, no, we don’t want a reasonable solution to this that people can accept. We will not be satisfied with anything short of perfection."

Personally, I find that to be a perfectly reasonable argument.  Let me just ask you this: would you allow a gay couple who have a civil union to say they are "married" ?

This is just a sidenote.  Nothing to do really with much of anything besides curiosity.

Two things:

1. Many states don't hold civil unions equal to marriage legally. Married people in those states can get much better benefits than a gay couple.
2. Marriage isn't partly religious. Even if you think it is, you must agree that it doesn't belong to just the Christians. It would belong to the Jews, the Muslims, the Buddhists, etc. And in addition, if two atheists don't get married in a church, it doesn't give them a civil union. It gives them a full marriage, with all the benefits [b]and[/b] the name.
So to answer your question, yes I would allow them to say they are "married."

Btw, as I already mentioned in this thread, gay marriage is specifically condemned only [b]twice[/b] in the entire Bible and [b]never[/b] by Jesus. I don't see what the big fuss is about when many more mentioned things are just passed over by the Church.

Well maybe conservative Christians just want to fuss over those? I am atheist, but if the Bible indeed specifically condemns it, unlike how radical religious groups bend religious texts into interpreting them as "God says so and so should die because such and such", then they have a good reason. However what I've been hearing on the news or whatever is that the Bible doesn't ever mention gay marriage...

[quote="serialk11r"]if the Bible indeed specifically condemns it[/quote]  Yes, in the section after it bans clothes of mixed fabrics but just before the exhortation to always vote Republican.  (The explanation of how evolution and [url=http://scienceblogs.com/islandofdoubt/2009/04/another_reminder_of_just_how_f.php?utm_source=sbhomepage&utm_medium=link&utm_content=channellink]global warming[/url] are conspiracies by atheist scientists is a few books earlier.) :wink:

[hide="Yes yes I know"] :spam: [/hide]

[quote="Ihatepie"]Two things:

1. Many states don't hold civil unions equal to marriage legally. Married people in those states can get much better benefits than a gay couple.
[/quote]

I think it would forward the discussion if you spoke with specific examples. Prop 8 dismissal wouldn't fix that anyway.

If we christen everything as 'marriage', should a male-male relationship get the same benefits (notice I didn't say rights) as a female+X relationship? Clearly, one relationship will not result in a partner going through a pregnancy. They are really not equal since the components are not equal (QED! ;-)). The ramifications of all this should be thought out. Does the government have a vested interest in couples that contain a woman that they be promoted to have a children? If not, wouldn't it be best for all that there isn't a concept of a civil marriage in the books?

I honestly don't find it a religious issue. I believe most religions are trying to promote children and healthy families and have adopted the issue. And to some extent, they justify it by reading very heavily between the lines in their ancient religious text. If good Christian and Muslim people want to delegate their decision on this to their religious councel, it is their prerogative.

I also find the Christan/Bible a bit cowardly and doubt very much the same folks would take open first shots at a Muslim/Koran.

[quote="polishcalculator"]I also find the Christan/Bible a bit cowardly and doubt very much the same folks would take open first shots at a Muslim/Koran.[/quote] This sentence is not quite coherent, and my best guess as to what it means is moronic.  Christians are a majority religious group in this country.  Prominent Christian leaders spent lots of effort to get their followers to provide votes or money to support Proposition 8.  If Islam were the religion of more than about 2% of the population, and their religious authorities were responsible for an elimination of civil rights based on interpretations of 1400-year-old fantasy novels, I would be directing my ire at them.  Perhaps you should avoid calling people cowardly out of your own ignorance.  (If this isn't the comment you were making, perhaps you could fix the grammar of your sentence so its content is clearer.)

I agree with JBL that history will decide who is right.  I bet that forty years from now, those now writing anti-gay marriage arguments will either (a) look back on these arguments with embarrassment, or (b) they'll still hold these views even though mainstream society finds them offensive.  But maybe that's just wide-eyed optimism.  After all, kids and teens continue to call each other "gay" and "fag" all the time.  And the pro-life movement is still going strong despite decades of legal abortions.

On a tangent: I do believe that progressive ideals will continue to march on.  Sixty years ago no one would have even dreamed of a "gay marriage," and even a very progressive and thoughtful person would have thought it ridiculous.  Michael Kinsley asked the question of whether today's progressives have opinions that will be considered offensive to the next generation's progressives, but they just don't realize it yet.


Can anyone tell me when APMO 2008 result be released? m looking forward to seeing result... the result seem so late-release this year...

Last year they were released in Mid-June, so the results this year will probably be released in a few weeks.

The results are up now.

[url]http://www.kms.or.kr/competitions/apmo/data/20th%20APMO.htm[/url]


[quote="Fierytycoon"]By the way, just out of curiosity...who IS that in your avatar picture?[/quote]

I answered that question just a few days ago. Look at the avatar thread in F&G.

[spoiler]BoA[/spoiler]

[quote="MithsApprentice"]For some reason, my avatar doesn't appear as a selection in list in the classroom. This hasn't been a problem before.[/quote]

Im having the same problem, except for me it NEVER appeared in the list.

I asked about this once in the classroom, and I think someone tried to PM me about it, but I accidentally closed the window.

I believe that avatars were not showing up as icons for a period, but that the problem has been corrected.


Hi to all!!!

This is my first post here in mathlinks.ro hope answer the following: (this is the Philippine’s Regional Screening Test for the National’s Mathematical Olympiad.)

2. If 2A99561 is equal to the product when 3 x (523 + A) is multiplied by itself, find the digit A.

3. The perimeter of a square inscribed in a circle is p. What is the area of the square that circumscribes the circle?

4. The sum of the first ten terms of an arithmetic sequence is 160. The sum of the next ten terms of the sequence is 340. What is the first term of the sequence?

5. It is given that triangle CAB is congruent to triangle EFD. If AC = x+y+z, AB = z+6, BC = x+8z, EF = 3, DF = 2y-z, and DE = y+2, find x^2+y^2+z^2.

6. Container A contained a mixture that is 40% acid, while container B contained a mixture that is 60% acid. A chemist took some amount from each container, and mixed them. To produce 100 liters of mixture that is 17% acid, she needed to pour 70 liters of pure water to the mixture she got from containers A and B. How many liters did she take from container A?

Please do not triple post the same topic in multiple forums. ONE is good enough, kay? You didn't have to post in here or Pre-Olympiad if you already posted in HSB forum

[hide="2"]because 2A99561 must be divisible by 9, A=4.[/hide]

[hide="3"]the diamter of the circle is the diagonal of the inscribed square, which is $ \frac {p\sqrt2}{4}$, which is also the side of the largest square. so the area of the square is $ \frac {p^2}{8}$.[/hide]

[hide="4"]$ x \plus{} x \plus{} d \plus{} x \plus{} 2d \plus{} ... \plus{} x \plus{} 9d \equal{} 160.$
$ x \plus{} 10d \plus{} x \plus{} 11d \plus{} ... \plus{} X \plus{} 19d \equal{} 340$

subtracting the first from the second,
$ 100d \equal{} 180\rightarrow d \equal{} \frac {9}{5}$

then if we know d, then we can find x. x=7.9[/hide]

sorry just wanna make sure others will reply...thanks for the answers...

[hide="Alternate Solution for 2"]Since that number must be a perfect square, the only value of $ A$ that works is $ A\equal{}4$.[/hide]

Again geoffrels, you had been posting the same thing way too many times all in THREE forums! PLEASE stop doing that all the time; it could possibly get you banned, like many others say!


vector
linear algebra
linear algebra unsolved

If S and T are linear operators on V so that STST = 0, is it necessarily true that TSTS = 0?

I believe that it is, and my proof up to now is as follows: ST and TS have the same eigenvalues, and since ST is nilpotent, all vectors in V are generalized eigenvectors corresponding to the eigenvalue 0. Hence we know V = null[(ST)^2] = null[(ST)^(dimV)] and since ST and TS have the same eigenvalues, null[(ST)^(dimV)] = null[(TS)^(dimV)]. Noting that (TS)^3 = TSTSTS = T((ST)^2)S, we see null[(TS)^3] = null[(TS)^(dimV)]. Now if I could only show that null[(TS)^3] = null[(TS)^2] I would be in business. I can't see it though---am I going about this all wrong? I also thought about using the characteristic polynomial...but that didn't really get me anywhere.

One more restriction: we're trying to develop the notion of determinants and traces and the like from the abstract theory of vector spaces, so if you can find a way that doesn't involve either of those two it would be greatly appreciated!

In three dimensions, let $ S$ be the shift $ S(x_1,x_2,x_3)\equal{}(0,x_1,x_2)$ and let $ T(x_1,x_2,x_3)\equal{}(0,x_2,x_3)$.


algebra
polynomial
integration
calculus
calculus computations

I know that the answer is $\frac{1}{2n+1}$ but I haven't been able to figure out how. I tried integrating by parts but couldn't succeed. Can anyone help me?

Start with

$\int_{0}^{1}\left[\left(\frac{d}{dx}\right)^{n}(x^{n}(1-x)^{n})\right] \left[\left(\frac{d}{dx}\right)^{n}(x^{n}(1-x)^{n})\right] \,dx$

Integrate by parts $n$ times. All of the boundary terms evaluate to zero, and we wind up with:

$(-1)^{n}\int_{0}^{1}\left[\left(\frac{d}{dx}\right)^{2n}(x^{n}(1-x)^{n})\right] x^{n}(1-x)^{n}\,dx = (2n)!\int_{0}^{1}x^{n}(1-x)^{n}\,dx.$

This is now a beta integral. It is known that

$\int_{0}^{1}x^{j}(1-x)^{k}\,dx=\frac{j!k!}{(j+k+1)!}$

So our integral is now $\frac{(2n)!(n!)^{2}}{(2n+1)!}$

So far in setting this up I have omitted the division by $n!$ that was part of the definition of $P_{n}.$ Include that, and we get

$\int_{0}^{1}[P_{n}(x)]^{2}\,dx=\frac{(2n)!}{(2n+1)!}= \frac1{2n+1}.$

This problem is about the normalization of the Legendre polynomials. The Legendre polynomials are usually defined on $[-1,1];$ these are a variant with the definition shifted to $[0,1]$ but the are still essentially the Legendre polynomials.


Hi, I have been looking at this [url=http://www.rotorbrain.com/foote/interactive/hacks/ConformalMapping.html]Conformal Map[/url], but I don't fully understand it.

In the first map, $ f(z)\equal{}z \plus{} \frac{1}{z}$, there are two points which the map gravitates to. I think they are the points $ 2\plus{}0i$ and $ \minus{}2\plus{}0i$ and I can sort of see why, but map 4, $ f(z) \equal{} \log(z) \plus{} z^2$ there are also two points that which the map gravitates two. Would someone please explain what there points represent?
Thanks

For the first map, take a look here:

[url=http://math.fullerton.edu/mathews/c2003/JoukowskiTransMod.html]Joukowski Transform[/url]


Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \frac{ab}{9\minus{}c^2}\plus{}\frac{bc}{9\minus{}a^2}\plus{}\frac{ca}{9\minus{}b^2} \le \frac{3}{8}\]
:)

I have a proof using expansion and $ pqr$....
But I think Can's proof is much more consize and beautiful.....mine is messy!! :(

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {ab}{9 \minus{} c^2} \plus{} \frac {bc}{9 \minus{} a^2} \plus{} \frac {ca}{9 \minus{} b^2} \le \frac {3}{8}
\]
:)[/quote]

It's equivalent to
$ \sum\frac {ab}{(a\plus{}b\plus{}c)^2 \minus{} c^2}\le \frac {3}{8}\Leftrightarrow \sum\left(\frac1{8}\minus{}\frac {ab}{a^2\plus{}b^2\plus{}2\sum ab} \right)\ge0$
$ \Leftrightarrow \sum\left(\frac {(a\minus{}b)^2\plus{}2c(a\plus{}b)\minus{}4ab}{a^2\plus{}b^2\plus{}t}\right)\ge0$, where $ t\equal{}2\sum ab$

Due to $ \sum\left(\frac {(a\minus{}b)^2}{a^2\plus{}b^2\plus{}t}\right)\ge0$, ought to we only need to prove $ \sum\left(\frac {2c(a\plus{}b)\minus{}4ab}{a^2\plus{}b^2\plus{}t}\right)\ge0$
$ \Leftrightarrow \sum\left(\frac {b(c\minus{}a)\minus{}a(b\minus{}c)}{a^2\plus{}b^2\plus{}t}\right)\ge0$
$ \Leftrightarrow \sum c(a\minus{}b)\left(\frac1{b^2\plus{}c^2\plus{}t}\minus{}\frac1{c^2\plus{}a^2\plus{}t}\right)\ge0$
$ \Leftrightarrow \sum\left(\frac{c(a\plus{}b)(a\minus{}b)^2}{(b^2\plus{}c^2\plus{}t)(c^2\plus{}a^2\plus{}t)}\right)\ge0$ , which is obviously true.

This completes the proof! :)

Besides, we heve $ \frac {ab}{9 \plus{}c^2} \plus{} \frac {bc}{9 \plus{}a^2} \plus{} \frac {ca}{9 \plus{}b^2} \le \frac {3}{10}$   for all $ a,b,c>0: a\plus{}b\plus{}c\equal{}3$

Sorry,I first go to here,and I'm not good at English.

Expand this one,$ a\plus{}b\equal{}X,b\plus{}c\equal{}Y,a\plus{}c\equal{}Z$
We should proof:$ (ab\plus{}bc\plus{}ac)((sumxy)^2\plus{}6xyz)<\equal{}9xyz(x\plus{}y)(x\plus{}z)(y\plus{}z)/8
sumab<\equal{}3(a\plus{}b)(b\plus{}c)(a\plus{}c)/8\equal{}3xyz/8
Then9(x\plus{}y)(x\plus{}z)(y\plus{}z)/4>\equal{}(xy\plus{}yz\plus{}xz)^2,because36\equal{}(x\plus{}y\plus{}z)^2>\equal{}3(xy\plus{}yz\plus{}xz),6(xy\plus{}yz\plus{}xz)\equal{}(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}xz)\minus{}(x\plus{}y)(x\plus{}z)(y\plus{}z)\equal{}xyz<\equal{}(x\plus{}y)(y\plus{}z)(x\plus{}z)/8.
3(x\plus{}y)(x\plus{}z)(y\plus{}z)>\equal{}6xyz$
Proof.

The following is also true:

If $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$ and $ p$ is a real number such that $ 0<p\le\frac94$ then \[ \frac{ab}{p\plus{}c^2}\plus{}\frac{bc}{p\plus{}a^2}\plus{}\frac{ca}{p\plus{}b^2}\le\frac9{4p}.\]

It's nice but not easy. Can you show your solution Can_hang??

[quote="bimbo"]The following is also true:

If $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$ and $ p$ is a real number such that $ 0 < p\le\frac94$ then
\[ \frac {ab}{p \plus{} c^2} \plus{} \frac {bc}{p \plus{} a^2} \plus{} \frac {ca}{p \plus{} b^2}\le\frac9{4p}.
\]
[/quote]
Yes, nice, bimbo. There is a nice solution for it :) and the case $ p \equal{} \frac {9}{4}$ is known as Hungkhtn's Inequality. :)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {ab}{9 \minus{} c^2} \plus{} \frac {bc}{9 \minus{} a^2} \plus{} \frac {ca}{9 \minus{} b^2} \le \frac {3}{8}
\]
:)[/quote]
My solution is the following
\[ \frac{ab}{9\minus{}c^2}\equal{}\frac{ab}{(a\plus{}b)(a\plus{}b\plus{}2c)} \le \frac{ab}{2(a\plus{}b)\sqrt{(a\plus{}c)(b\plus{}c)}} \le \frac{(ab)^{3/4}}{2\sqrt{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\]
It suffices to show that
\[ \frac{8}{9}((ab)^{3/4}\plus{}(bc)^{3/4}\plus{}(ca)^{3/4})^2 \le (a\plus{}b)(b\plus{}c)(c\plus{}a)\]
By Cauchy Schwarz Inequality, we have
\[ ((ab)^{3/4}\plus{}(bc)^{3/4}\plus{}(ca)^{3/4})^2 \le (ab\plus{}bc\plus{}ca)(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}) \le (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\]
It suffices to show
\[ \frac{8}{9}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \le (a\plus{}b)(b\plus{}c)(c\plus{}a)\]
which is true by AM-GM. :)

[quote="bimbo"]The following is also true:

If $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$ and $ p$ is a real number such that $ 0 < p\le\frac94$ then
\[ \frac {ab}{p \plus{} c^2} \plus{} \frac {bc}{p \plus{} a^2} \plus{} \frac {ca}{p \plus{} b^2}\le\frac9{4p}.
\]
[/quote]
ah, of course, without EV ;)

[quote="bimbo"]The following is also true:

If $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$ and $ p$ is a real number such that $ 0 < p\le\frac94$ then
\[ \frac {ab}{p \plus{} c^2} \plus{} \frac {bc}{p \plus{} a^2} \plus{} \frac {ca}{p \plus{} b^2}\le\frac9{4p}.
\]
[/quote]
Note that it suffices to prove this inequality for $ p\equal{}\frac{9}{4}$, then it is equivalento the following: http://www.mathlinks.ro/viewtopic.php?t=57558&start=20
:)

Oh, sorry, I didn't know it was repost  :oops: . (I found it setting $ b \equal{} c$ while trying to find the best constant for Vasc's inequalities -which I didn't get by hand-. Then I noted that for little values of $ p$ it got the maximum at $ a \equal{} b \equal{} \frac32$ so I conjectured, proved and posted it, never considering that it could be so old and well-known. Moreover, It has a name  :oops:  so I should have known it.)

I didn't use EV Theorem, but not because I don't like it. In fact, I love that theorem and I think that is one of the greatest breaktroughs in the field of elementary inequalities, but until now I didn't find the time to prove it so I can't use it.
I proved the inequality by $ p,q,r$ technique (extremely ugly) using can_hang's theorem: $ 27r\le (p \minus{} s)^2(p \plus{} 2s)$, for $ s^2 \equal{} p^2 \minus{} 3q^2$, $ s\ge0$.

I really like your proof, can_hang! It's unbelievable.
Thanks for sharing!

[quote="Vasc"][quote="dduclam"]
Besides, we have $ \frac {ab}{9 \plus{} c^2} \plus{} \frac {bc}{9 \plus{} a^2} \plus{} \frac {ca}{9 \plus{} b^2} \le \frac {3}{10}$   for all $ a,b,c > 0: a \plus{} b \plus{} c \equal{} 3$[/quote]
Actually, we have for all $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$:

$ \frac {ab}{4 \plus{} c^2} \plus{} \frac {bc}{4 \plus{} a^2} \plus{} \frac {ca}{4 \plus{} b^2} \le \frac {3}{5}$ ,

$ \frac {ab}{7 \plus{} 2c^2} \plus{} \frac {bc}{7 \plus{} 2a^2} \plus{} \frac {ca}{7 \plus{} 2b^2} \le \frac {1}{3}$ ,

$ \frac {ab}{10 \plus{} 3c^2} \plus{} \frac {bc}{10 \plus{} 3a^2} \plus{} \frac {ca}{10 \plus{} 3b^2} \le \frac {3}{13}$ .[/quote]

The best constant for $ \sum\frac{ab}{k\plus{}c^2}\le\frac3{k\plus{}1}$ is $ k\equal{}3,26903...$.

Besides, we have a nice result (well known): $ \sum\frac{ab}{3\plus{}c^2}\le\frac{11\sqrt{33}\minus{}45}{24}$ :)


Find the number of functions defined on positive real numbers such that $ f\left(1\right) \equal{} 1$ and for every $ x,y\in \Re$, $ f\left(x^{2} y^{2} \right) \equal{} f\left(x^{4} \plus{} y^{4} \right)$.

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ \text{Infinitely many}$

[hide]Substitute $ a \equal{} x^2,b \equal{} y^2$ to get $ f(ab) \equal{} f(a^2 \plus{} b^2)$ for all positive reals $ a,b$. Letting $ a \equal{} y/k$ and $ b \equal{} k$; we see that $ f(1) \equal{} f(k^2 \plus{} y/k^2)$ for all positive reals $ k,y$; the expression inside the brackets can range over the reals; hence $ f(x) \equal{} 1$ for all positive reals $ x$ and so we have the answer as $ B$.[/hide]

the substitution wasn't really necessary, but I did it anyways.

A fix of Temperal's solution:

[hide="Solution"]Select $ x^2 \equal{} k$, $ y^2 \equal{} 1/k$ for $ k > 0$.
Then, $ 1 \equal{} f(1) \equal{} f(k/k) \equal{} f(x^2 y^2) \equal{} f(x^4 \plus{} y^4) \equal{} f(k^2 \plus{} k^{ \minus{} 2})$.
The range of $ k^2 \plus{} k^{ \minus{} 2}$ is the interval $ [2, \infty)$

Now, let $ x^2 \equal{} 2 \cos t$ and $ y^2 \equal{} 2 \sin t$ for appropriate $ t$.
$ 1 \equal{} f(4) \equal{} f(x^4 \plus{} y^4) \equal{} f(x^2 y^2) \equal{} f(4 \sin t \cos t) \equal{} f(2 \sin 2t)$
The positive elements in the range of $ 2 \sin 2t$ form the interval $ (0,2]$

Combining the above, we see that $ f$ must be identically $ 1$ on the interval $ (0, \infty)$[/hide]

I don't know what I misunderstood,But if we put $ x\equal{}\frac{a}{b}$ and $ y\equal{}\frac{b}{a}$ then $ f(x^{2}y^{2})\equal{}f(1)\equal{}f(x^{4} \plus{}y^{4})$.Since there are infinitely many numbers $ x,y$ such that $ x\equal{}\frac{a}{b}$ and $ y\equal{}\frac{b}{a}$ so the number of functions should be infinitely many or $ \boxed C$

Sorry about that; I am not very adept at thinking late at night.


A massive disc rotates about a vertical axis at angular velocity $ \omega_0$. A smaller disc of mass $ m$ and radius $ r$, whose axis is strictly vertical, is lowered on the first disc. The distance between the axes of the discs is $ d\;\;(d>r)$, and the co-efficient of friction is $ \mu$.

Determine the steady state angular velocity $ \omega$ of the smaller disc. The radius of the larger disc is $ R>d\plus{}r$. Friction at the axes should be neglected.

One question
Is the axis of the smaller disc fixed  ?

Axis of the smaller disc is [b]fixed[/b] (immotile) on the larger one.

Don't you think it would be better to end the discussion on your former riddle (with the same name), before  this one ?

[hide="Assuming the bigger disc maintains his initial angular velocity (it is massive)"]
Anyhow, even I should wait for someone else to try:
\[ \omega_0\] [/hide]

[quote="Immanuel Bonfils"]
\[ \omega_0
\]

[/quote]

Correct!

I insist you explain. Atleast provide some clue. I believe I can carry further without help. I'll need just a little push.

More or less like the moon around the earth (but the answer should be better hidden, he he he...).

[quote="Rajiv"]A massive disc rotates about a vertical axis at angular velocity $ \omega_0$. A smaller disc of mass $ m$ and radius $ r$, whose axis is strictly vertical, is lowered on the first disc. The distance between the axes of the discs is $ d\;\;(d > r)$, and the co-efficient of friction is $ \mu$.

Determine the steady state angular velocity $ \omega$ of the smaller disc. The radius of the larger disc is $ R > d \plus{} r$. Friction at the axes should be neglected.[/quote][hide]

I believe one ought to approach this problem by conservation of angular momentum.

The moment of inertia of the large disc about its center of mass is $ I_1 \equal{} \frac12 MR^2$.

The moment of inertia of the small disc about the center of mass of the large disc, due to the parallel axis theorem, is $ I_2 \equal{}\frac12 mr^2 \plus{} md^2$.

By the conservation of angular momentum: $ I_1\vec{\omega}_0 \equal{} (I_1\plus{}I_2)\vec{\omega}$.

Thus $ \vec{\omega} \equal{}\frac{I_1}{I_1\plus{}I_2}\vec{\omega}_0$.

This implies,

\[ \boxed{\vec{\omega} \equal{}\frac{MR^2}{MR^2 \plus{} mr^2 \plus{} 2md^2}\;\vec{\omega}_0}\][/hide]

This problem appeared in a mock test for IIT-JEE of a coaching institute. Later, I found it in [b]Science for Everyone[/b] by [b]S. S. Krotov[/b]. Luckily, detailed solutions are provided at the back, and as you'll soon discover, a little too much in detail.

I have photographed the three pages of the solution.

Seems  the problem you've found the text has others questions besides the angular velocity.

For [b] your question[/b], with a massive disk, you can simple consider Dr. No's result with M>>m (even this wouldn't be necessary since it's obvious that at steady state the smaller rotates with the bigger one...).

[quote="Rajiv"][quote="Immanuel Bonfils"]....at steady state the smaller rotates with the bigger one....[/quote]

I think you are mistaken, the smaller disc rotates with $ \omega_0$ about its own axis besides going about the axis of the larger disc with $ \omega_0$.[/quote]

Does it? I don't believe that is indicated in the problem statement.

To my understanding of the term "steady state," i.e. after all transient effects have extinguished, Immanuel is correct. Furthermore, angular momentum is conserved: there are no external torques applied to the system. Therefore, unless I neglected some term in writing the conservation equation, the result I obtained [b]is[/b] the answer.


Am I wrong or this reduces to :
 \sum (1 \leq i<j \leq n) (x_i + x_j + 1/2) 2   \geq  0 ?

Pierre.

Use the fact that 2sum x_ix_j=(sum x_i)^2-sum x_i^2. The inequality becomes (n-2)/(n-1)sum x_i^2+sum x_i+1/n-1(sum x_i)^2+n/8>=0. Since sum x_i^2>=1/n(sum x_i)^2 , the inequality reduces after some computations to ( sum x_i+n/4)^2>=0.


having the equation $2x^3-2x^2-2x-3=0$
find $\lim_{n->\inf} x_1^n + x_2^n + x_3^n$
i tried with the role sequence to get to the roots but i only can
find that there's no roots but in the interval $(\frac{9}{5},2)$
so what can i do to calculate the other 2 roots
and how can i solve the limit ?

You can check that the modulus of the complex roots $z,\bar z$ are $<1$.
Hence  $z^n\to 0$ and $(\bar z)^n\to 0$

why should i check modulus of complex roots if i don't even know those complex roots.
please give me a more readable solution to this problem if you can please.

but the root between (9/5,2) will go to infinity

according to vieta's theorem,
$xz\bar{z} = \frac{3}{2}$
where $x$ is the real root and $z,\bar z$ are complex roots.
hence
$|z|^2 = z\bar{z} = \frac{3}{2x} \in [\frac{3}{4},\frac{5}{6}]$
thus
$|z|<1$.

ok now i understand very well what you are saying about the roots.
now i know that $x_1 \in R x_1^n -> \inf$ if i'm right ?
and the limit of those other two complex roots i can't be sure of.
what is their limit ?

$|z| < 1 \Rightarrow |z^n| = |z|^n \to 0 \Rightarrow z^n\to 0$.

i understood now thaks.
what is the sturm sequence and how could i use it here to find the roots or what is it
supposed to do ?


Prove that in an ellipse, the sum of the distances from a point to the two foci are a constant.

Stick two tacks in a piece of paper, take a string, tie it around the two tacks, put a pencil in the string and draw around the tacks. Voila! An ellipse. Therefore, each point on the ellipse is the same distance (the length of the string) from the two foci (add the distance from each focus).

You can probably do this mathematically, but I'm not sure how. :P

I thought that the ellipse was defined as the locus of points such that the sum of the two distances from two fixed points, the foci, is a constant. If this is correct, then there's no need to prove it because the ellipse is defined that way.
If you want to use the general ellipse equation, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, you would still have to assume that your assertion is true anyway, in order to use $a^2=b^2+c^2$ because this is also based on the definition of the ellipse.

@JesusFreak: How do you know what you drew is an ellipse? It's obvious that from the construction, that the curve you drew has the property that the sum of the distances from any point on the curve to the two fixed points is constant, but what makes the figure you drew an ellipse? You have to be careful with circular logic.

[quote="Bictor717"]@JesusFreak: How do you know what you drew is an ellipse? It's obvious that from the construction, that the curve you drew has the property that the sum of the distances from any point on the curve to the two fixed points is constant, but what makes the figure you drew an ellipse? You have to be careful with circular logic.[/quote]

What makes a compass draw a circle? It's the same principle. A compass is a tool to draw circles, and that method is a tool to draw ellipses. You can't "prove" that the sum of the distances from the foci to the ellipse is constant by drawing one using that tool any more that you can draw a circle using a compass and say that that proves that a circle is equidistant from a point.


integration
Euler
calculus
calculus computations

evaluate
$ I\equal{} \int \frac{dx}{x\plus{} \sqrt{x^2\plus{}x\plus{}1}}$

hello
We can also use [b]Euler substitution[/b]
[hide="Solution"]
put $ \sqrt {x^2 \plus{} x \plus{} 1} \equal{} ux \minus{} 1$
 $ I \equal{} \int \frac {dx}{x \plus{} \sqrt {x^2 \plus{} x \plus{} 1}}$
$ \equal{} \int \frac { \minus{} 2u^2 \plus{} 2u \minus{} 2}{(u)(u \minus{} 1)(u \plus{} 1)^2} du$
$ \equal{} \int \frac {2du}{u} \minus{} \int \frac {du}{2(u \minus{} 1)} \minus{} \int \frac {3 du }{2(u \plus{} 1)} \minus{} \int \frac {3du}{(u \plus{} 1)^2}$
now i believe u can proceed by ur self. :)
[/hide]
thank u


abstract algebra
group theory
topology
advanced fields
advanced fields unsolved

A colleague of mine just asked the following question:

Let $ G$ be a compact abelian group and let $ m$ be the Haar measure on $ G$. Let $ U$ be an open set in $ G$ such that $ m(U)<1$ and $ m(\partial U)\equal{}0$ (as usual, $ \partial U$ is the boundary of $ U$). Let $ S$ be any subset of $ G$. Then, obviously, the set $ V\equal{}SU$ is open. Suppose that it is dense in $ G$. Does it follow that $ m(V)\equal{}1$?

Any ideas?

Alas, this is false even on the circle :(. Still, it is a good exercise to construct a counterexample, so I'm moving it to Unsolved and Proposed :P.

Take $ S^1 \equal{} [0, 1]/\sim $, $ S$ as the identity element, and construct $ U$ by enumerating all rationals between $ 0$ and $ 1$ as some sequence $ \{r_n\}$ and taking the union of a ball of radius $ \frac{\varepsilon}{2^n}$ around each $ r_n$ ?

Nope. You miss the $ m(\partial U)\equal{}0$ condition this way.

The boundary of the infinite union more often than not is [b]not[/b] the union of the boundaries. ;)

:blush: Alright. I'll buy that, but I haven't done anything with measures in over a year. Is there an easy way to see that it isn't measure 0?

The closure is the entire space (of measure $ 1$), the set itself is of measure  $ \varepsilon$ at most, so the boundary is of measure $ 1\minus{}\varepsilon$ at the very least. :)


geometry
circumcircle
inequalities unsolved
inequalities

Let $G$ be centroid and $R$ the circunradius of a triangle $ABC$. The extensions of $GA,GB,GC$ meet the circuncircle again at $D,E,F$. Prove that:

$\frac{3}{R} \leq \frac{1}{GD} + \frac{1}{GE} + \frac{1}{GF} \leq \sqrt{3} \leq \frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}$

[quote="M4RI0"]Let $ G$ be centroid and $ R$ the circunradius of a triangle $ ABC$. The extensions of $ GA,GB,GC$ meet the circuncircle again at $ D,E,F$. Prove that:

$ \frac {3}{R} \leq \frac {1}{GD} \plus{} \frac {1}{GE} \plus{} \frac {1}{GF} \leq \sqrt {3} \leq \frac {1}{AB} \plus{} \frac {1}{BC} \plus{} \frac {1}{CA}$[/quote]
$ Ga;GB;GC$ cut $ BC; CA; AB$ at $ M; N; P$,respectively
Put $ AM \equal{} m_a;BN \equal{} m_b;CN \equal{} m_c$
We have $ AM.MD \equal{} BM.MC$=>$ MD \equal{} \frac {a^2}{4m_a}$
=>$ GD \equal{} GM \plus{} MD \equal{} \frac {m_a}{3} \plus{} \frac {a^2}{4m_a}$
By AM - GM we have:$ GD \geq \frac {a}{\sqrt {3}}$
so $ \frac {1}{GD} \leq \frac {a}{\sqrt {3}}$= >$ \frac {1}{GD} \leq \frac {\sqrt {3}}{BC}$
Similar we have$ \frac {1}{GD} \plus{} \frac {1}{GE} \plus{} \frac {1}{GF} \leq \sqrt {3} \leq \frac {1}{AB} \plus{} \frac {1}{BC} \plus{} \frac {1}{CA}$
Otherhand $ \frac {GA}{GD} \equal{} \frac {8m_a^2}{4m_a^2 \plus{} 3a^2} \equal{} \frac {2b^2 \plus{} 2c^2 \minus{} a^2}{a^2 \plus{} b^2 \plus{} c^2}$
So $ \frac {GA}{GD} \plus{} \frac {GB}{GE} \plus{} \frac {GC}{GF} \equal{} \frac {3a^2 \plus{} 3b^2 \plus{} 2c^2}{a^2 \plus{} b^2 \plus{} c^2} \equal{} 3$
Note that$ \frac {AD}{GD} \plus{} \frac {BE}{GE} \plus{} \frac {CF}{GF} \equal{} 6$
so $ \frac {1}{GD} \plus{} \frac {1}{GE} \plus{} \frac {1}{GF} \geq \frac {3}{R}$

hello, there is a typo in this problem it must be
$ \frac{3}{R}\leq\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}\leq\sqrt{3}\left(\frac{1}{AB}+\frac{1}{BC}+\frac{1}{CA}\right) $
therefore the proof is also wrong.
Sonnhard.

for this problem we need a nice solution!!!


[quote="The Zuton Force"]Let $p(x)=x^{6}-6x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+1$ be a polynomial in $x$ with at least 5 positive real roots. Find, with proof, all possible values of $a+b+c+d$.[/quote][hide]Because there are at least five real roots, the last root must also be real, as complex roots come in conjugates. Since the product of the roots is $1$, then the polynomial has six positive real roots. By AM-GM, \[\frac{r_{1}+r_{2}+r_{3}+r_{4}+r_{5}+r_{6}}{6}\ge\sqrt[6]{r_{1}r_{2}r_{3}r_{4}r_{5}r_{6}},\] with equality if and only if \[r_{1}=r_{2}=r_{3}=r_{4}=r_{5}=r_{6}.\] Since the sum of the roots is $6$ and the product is $1$, we see that equality does indeed hold, so the roots are all $1$ and the polynomial is \[(x-1)^{6}=x^{6}-6x^{5}+15x^{4}-20x^{3}+15x^{2}-6x+1.\] The only possible value of $a+b+c+d$ is $\boxed{4}$.[/hide]

It is =)
It took me a while to figure out that I can just use AM-GM when I got the problem, because I usually don't come across polynomial problems involving AM-GM.

That's a special case of a series of inequalities known as Maclaurin's Inequalities.

Given a monic polynomial $P(x) = x^{n}-a_{1}x^{n-1}+a_{2}x^{n-2}\pm ... \pm a_{n}$ with all real positive roots (the implication here is that $a_{i}\ge 0$), let $s_{i}= \frac{a_{i}}{{n \choose i}}$.  Then

$s_{1}\ge \sqrt{s_{2}}\ge \sqrt[3]{s_{3}}\ge \sqrt[4]{s_{4}}\ge ...$

AM-GM is just the statement $s_{1}\ge \sqrt[n]{s_{n}}$.


Let $ S$ be the set of nonnegative real numbers. Find all functions $ f: S\rightarrow S$ which satisfy $ f(x\plus{}y\minus{}z)\plus{}f(2\sqrt{xz})\plus{}f(2\sqrt{yz})\equal{}f(x\plus{}y\plus{}z)$ for all nonnegative $ x,y,z$ with $ x\plus{}y\ge z$.

$ P(0,0,0)$ implies $ f(0)\equal{}0$.
We can compare $ P(x,y,z)$ and $ P(0,x\plus{}y,z)$ and get
$ f(2\sqrt{xz})\plus{}f(2\sqrt{yz})\equal{}f(2\sqrt{xz\plus{}yz})$ for $ x\plus{}y\ge z$.
It is not hard to prove $ f(\sqrt{x})\plus{}f(\sqrt{y})\equal{}f(\sqrt{x\plus{}y})$ for $ x,y\ge0$
$ g(x)\equal{}f(\sqrt{x})$ has cauchy equation and is increasing.
So $ f(x)\equal{}g(x^2)\equal{}cx^2$, where $ c>0$

[quote=arshakus]hi,
in this function if I get $f(z-x)=f(x-z)$, what it means??[/quote]

It means the function is even.


group theory
abstract algebra
function
superior algebra
superior algebra solved

Show that if H and K are two subgroups of finite index in G then $H \cap K$ has also finite index

Define a function $f: (G/H\cap K)_{\ell}\to(G/H)_{\ell}\times(G/K)_{\ell}$, where $(G/H)_{\ell}$ is the set of left cosets of $H$ in $G$. We define $f$ like this: $f(x(H\cap K))=(xH,xK)$.

Let's show it's a function: if $x(H \cap K)=y(H \cap K)$, then $y^{-1}x\in H \cap K \Leftrightarrow y^{-1}x\in H$ and $\in K\Leftrightarrow xH=yH$ and $xK=yK=(xH,xK)=(yH,yK)$. Now we show that it's injectiive: if $f(x(H \cap K))=f(y(H \cap K))\Leftrightarrow (xH,xK)=(yH,yK)=>xH=yH$ and $xK=yK\Leftrightarrow y^{-1}x\in H$ and $\in K\Leftrightarrow y^{-1}x\in H\cap K\Leftrightarrow x(H \cap K)=y(H \cap K)$. This shows that we can define an injective function from a set with $|G:H\cap K|$ elements to a set with $mn$ elements, where $m=|G:H|$ and $n=|G:K|$, which means that $|G:H\cap K|\le mn$.


Maybe I'm saying something stupid :? , but shouldn't the problem be :

If a, b, c > 0 then \sum (a + 2b + 4c)/bc \geq 21 ?

Probably you will all laugh, but what's wrong with this ? :?

\sum (a + 2b + 4c)/bc =  \sum (a + 2ab + 4ac)/abc

Now we have
(1) a + b + c \geq 3abc
(2) 2ab + 2bc + 2ca \geq 6abc
(3) 4ac + 4ba + 4cb \geq 12abc
Summing these and dividing by abc gives
\sum (a + 2ab + 4ac)/abc \geq 21.

I known that somebody write my thoughts. Thank you, Lagrangia! You have done my work!

arne, maverick has a smoother ineq!
:D :D

cheers

Arne! You are right.
But if Sum \geq 21 then Sum \geq 18! So Langrangia's proof is correct too.


Sa se arate ca ecuatzia : x ^ 2 + y ^ 3 = z ^ 4 are o infinitate de solutzii :|...

PS : Ciupane nu pune rezolvarea pe care ti-am aratat-o la nationala in camera :)

"Probleme de teoria numerelor si combinatorica pentru juniori"
Autori:Laurentiu Panaitopol,Dinu Serbanescu
Editura GIL
Pagina 14
Problema 117 :D

scriem ecuatia sub forma $(z^{2}-x)(z^{2}+x)=y^{3}$ si cautam perechi $(x,z)$ de numere naturale, astfel incat $(z^{2}-x)^{2}=z^{2}+x$. (*)Apoi luam $y=z^{2}-x$ si $(x,y,z)$ este solutie. mai ramane sa aratam ca exista o infinitate de perechi $(x,z)$ astfel incat sa aiba loc relatia (*).
(*) se scrie ca $z^{4}+x^{2}-2z^{2}x-z^{2}-x=0$, sau $x^{2}-(2z^{2}+1)x+z^{4}-z^{2}=0$, pe care o vom trata ca ecuatie de gradul doi in $x$. $\Delta=(2z^{2}+1)^{2}-4(z^{4}-z^{2})=4z^{4}+4z^{2}+1-4z^{4}+4z^{2}=8z^{2}+1$. Daca $\sqrt{\Delta}\in\mathbb{N}$, atunci vom alege $x=\frac{2z^{2}+1+\sqrt{\Delta}}{2}$, care este numar natural deoarece $\Delta$ este impar, deci numaratorul fractiei amintite anterior va fi par, deci $2$ se va simplifica. Mai ramane sa aratam ca exista o infinitate de numere $z$ astfel incat $8z^{2}+1$ este patrat perfect. Dar ecuatia $k^{2}-8z^{2}=1$ este de tip pell, iar $8$ nu este patrat perfect, deci are o infinitate de solutii. (gasim solutia minimala $3^{2}-8\cdot 1^{2}=1$). Deci exista o infinitate de $z$ pentru care $8z^{2}+1$ este patrat perfect. Conform rationamentului de mai sus, avem deci o infinitate de solutii.

Hmm...eu aveam o solutie care se baza pe identitatea : 28 ^ 2 + 8 ^ 3 = 6 ^ 4...

deci oricare ar fi n...avem (28 n ^ 6 ) ^ 2 + (8 n ^ 4) ^ 3 = ( 6n ^3) ^ 4
deci avem o infinitate de solutii pt ecuatzia x ^ 2 + y^ 3 = z^4 are o infinitate de solutii pt:
x = 28 n ^ 6
y=8n ^ 4
z=6n ^ 3

...mdea :|

io m`am uitat sa nu fie chiar rezolvarea de la raspunsuri a mea  :P , d`asta am postat`o


Find $x;y$ satisfies:
$x(x+1)+y(y+1)=3xy$

Suppose $(x,y)$ is a solution, then

\[x^2+(1-3y)x+y^2+y=0 \\
x= \frac {3y-1 \pm \sqrt {5y^2-10y+1}}{2} \]

Now, for that to be a solution the root has to be whole. Clearly $( \frac {3y-1 + \sqrt {5y^2-10y+1}}{2}, y)$ is a solution iff $( \frac {3y-1 - \sqrt {5y^2-10y+1}}{2}, y)$ is a solution. For any solution $(x, y)$, with $x \geq y, (x, y)$ is a solution iff $(y, x - \sqrt {5y^2-10y+1})$ is a solution. Now note that since $x= \frac {3y-1 + \sqrt {5y^2-10y+1}}{2}, x - \sqrt {5y^2-10y+1}= \frac {3y-1 - \sqrt {5y^2-10y+1}}{2}$. Clearly it can be verified that for all nonnegative $y, x - \sqrt {5y^2-10y+1}$ is also nonnegative, and we can also check that for $y \geq 2, y \geq x - \sqrt {5y^2-10y+1}$.

So from $(x, y)$ we can go to $(y, x - \sqrt {5y^2-10y+1})$ indefinitely, getting progressively smaller solutions. Since the numbers are nonnegative we must eventually have $y=x - \sqrt {5y^2-10y+1}$, but that means we eventually have a solution of the form $(x, x)$.

Now, if $x=y$ the only solutions are $x=0, x=2$. If the above process comes to $(0,0)$, then the previous solution was $(a,0)$, but then $a=0$, so it's impossible for it to come to $(0,0)$ unless we started with it. If it reaches $(2,2)$, then it's a solution belonging to the chain of solutions we can find using the recurrence $(x, y) \rightarrow (y, x - \sqrt {5y^2-10y+1})$ backwards to go up the chain. Then we find that the next solutions are $(3, 2), (6, 3), (14, 6), (35, 14)$.

All of them are of the form $(F_{2n+1}+1, F_{2n-1}+1)$. We will prove that all solutions are of this form. Take $(F_{2n+1}+1, F_{2n-1}+1)$, then the recursion gives $(a, F_{2n+1}+1)$, where

\[a=F_{2n-1}+1+ \sqrt {5F_{2n+1}^2-4},\]
where $\sqrt {5F_{2n+1}^2-4}=F_{2n}+F_{2n+2}$ by a well-known identity. Thus the solutions are $(0,0), (2,2)$ and $(F_{2n+1}+1, F_{2n-1}+1)$ for all $n$.


I know this is easy I just can't seem to find X  :blush:
[img]http://img148.imageshack.us/img148/3267/geop23zu8.png[/img]

Can someone show me the whole process?

Thanks

ranger, I don't think that's right... where did the last step come from.

my answer is there isn't enough information to solve the problem... I think you assumed a triangle (the one with x in it) was isosceles, when it doesn't really have to be.

I will assume all of the new lines you drew were not given. if so... how do you know that the altitude goes through the intersection of the other two lines? how do you know it's perpendicular to the given line? knowing that the entire sides are equal doesn't really help much, either. I still don't think there's enough info to solve this problem, because I think your extra ilnes are a little to lucky.

[hide][img]http://img119.imageshack.us/img119/9420/trianglecr6.png[/img]

We know that $m\angle1=60$ and $m\angle2=50,$ so $m\angle3=70.$ Vertical angles are congruent, so $m\angle4=70.$

We also know that the two top segments are congruent, so $m\angle7=m\angle8.$ This means $m\angle5=m\angle6.$

We know that $m\angle4+m\angle5+m\angle6=180,$ which is the same as $2m\angle6+70=180,$ so $m\angle6=55.$[/hide]

[quote="Imperial Effect"][quote="now a ranger"]err.. I hope this is right. I got 55 for an answer.

Let's see

60 and 50, so other angle of triangle is 70. Then with the rule vert. angles are congruent you know that one angle of the triangle with x in it is 70 degrees. Therefore, since the top has 2 sides congruent, the two angles opposite those sides are congruent. Then notice that x should be equal to 180-70-x.[/quote]

The whole line is congruent not just the top triangle :oops:
I got 55 too but this supposedly a “challenge” problem and that seemed too easy... not to mention I think I'm wrong.
[img]http://img221.imageshack.us/img221/6716/geop23xjj7.png[/img]

Thanks for helping![/quote]
You are making a mistake. The angle labeled 35 is a vertical angle of a 30 degree angle. So the answer should be 60...

[quote="bpms"][quote="Imperial Effect"]The whole line is congruent not just the top triangle :oops:
I got 55 too but this supposedly a “challenge” problem and that seemed too easy... not to mention I think I'm wrong.
[img]http://img221.imageshack.us/img221/6716/geop23xjj7.png[/img]

Thanks for helping![/quote]
You are making a mistake. The angle labeled 35 is a vertical angle of a 30 degree angle. So the answer should be 60...[/quote]
The answer is correct, but the right angle sign should not be there, because it's an $85^{\circ}$ angle.

[quote="DiscreetFourierTransform"]Something easier to try.
If the lines of the two inside triangles are parallel, and cut by a transversal, than you have Alternate Interior Angles which are always congruent. This is not a hard fact to prove, so if you are having difficulty, just look at the problem as sets of triangles..

This assumes that the base of the large triangle and the base of the small triangle are parallel, however.[/quote]
Of course, that would be easier. I immediately thought of parallel lines when I saw the problem, but the it doesn't say that they're parallel. Assuming my answer is correct, they aren't parallel.

If they were parallel then there would be alternate interior angles...

That's exactly what DiscreetFourierTransform said. And I already answered him.

[quote="i_like_pie"][quote="k8reindeer"]we don't know the line in the middle is parallel!! all these answers are [b]assuming[/b] that it is! It might not be! there's not enough given information!

[quote="i_like_pie"]We know that $m\angle4+m\angle5+m\angle6=180,$ which is the same as $2m\angle6+70=180,$ so $m\angle6=55.$[/quote]
that assumes $m\angle5=m\angle6$ which might be true, but only if you're really lucky. truth is, without more information, that angle can be whatever you want it to be.[/quote]
Read the line right before that. :wink:[/quote]

okay. it says since $m\angle7=m\angle8$ then $m\angle5=m\angle6$
hmmm...
nope. not convinced.
so I maintain: not enough info!

[quote="k8reindeer"]okay. it says since $m\angle7=m\angle8$ then $m\angle5=m\angle6$
hmmm...
nope. not convinced.
so I maintain: not enough info![/quote]
After reviewing the play, the call on the field is... Bascially, I agree. Not enough information.


function
logarithms
algebra
domain
calculus
derivative
algebra unsolved

Slove this equation: $ 6^x\equal{}2x\plus{}1\plus{}3log_{6}(5x\plus{}1)$

the function
$ 6^x \minus{} 2x \minus{} 1 \minus{} \log_6(5x \plus{} 1)$ is convex in its domain (derivative is strictly increasing)
so it may have only two solutions at most. but 0,1 are two solutions thus they are the only ones.


let f(z)=cosz and g(z) cos (z^2) then which is true

1.f and g both are bounded on C.
2 f is bounded but g is not bounded.
3.g is bounded but f is not bounded.
4.f and g both are bounded on the x-axis

on complex plane we cannot say value of cos lies betweeen -1 and 1 so we cannot say about boundedness but on real lines we can say f and g are bounded on x-axis
is it right?

Let $ D\subset\mathbb C$ and $ f: D\rightarrow\mathbb C$ be a complex function. Then $ f$ is said to be bounded iff $ \exists M\geq 0$ $ \forall z\in D: |f(z)|\leq M$.

Now, both $ f: \mathbb C\rightarrow\mathbb C, z\mapsto\cos(z)$ and $ g: \mathbb C\rightarrow\mathbb C, z\mapsto\cos(z^2)$ are non-constant entire functions, hence they cannot be bounded. However, they are clearly bounded $ \forall z\in\mathbb R\subset\mathbb C$ as we know from the real analysis ($ \forall x\in\mathbb R: |\cos(x)|\leq 1$).

Are you familiar with the notion of entire functions? Do you know what theorem I might be referring to?

Or there's the low-tech version that doesn't mention the theorem that ex falso quodlibet used but didn't name.

$ \cos(iy)\equal{}\frac{e^y\plus{}e^{\minus{}y}}2\equal{}\cosh(y).$ Hence $ f(z)\equal{}\cos(z)$ grows without bound (and exponentially fast) as you travel up the imaginary axis (set $ z\equal{}iy$ for $ y\in\mathbb{R}.$)

As for $ \cos(z^2):$ well, that is periodic and tame on the imaginary axis, but try looking at a 45 degree angle.

Let $ z\equal{}re^{i\pi/4}.$ Then $ z^2\equal{}r^2e^{i\pi/2}\equal{}ir^2.$

Hence for $ z\equal{}re^{i\pi/4},$ $ \cos(z^2)\equal{}\cos(ir^2)\equal{}\cosh(r^2)$ which also grows very rapidly as $ r\to\infty.$


$n$ people are sitting in a circle. What's the probability that three particular people sit together?

Glue the three people together to form one uber person! Six ways of doing this and $n-3$ people remaining in a circle make for $6(n-3)!$ possibilities. Divide that by $(n-1)!$ total ways and we have our desired probaiblity: $\frac{6}{(n-1)(n-2)}$ for $n\ge4$. Heh, I hope I answered the right question this time, and correctly too... :)

OK  :)
What about the probability that those none of those three people are next to each other?

For the second problem:
[hide="Hint"]This is the same as finding $diagonals + 1$ of a polygon with $n-3$ vertices, since one person (point) is surrounded by 2 empty chairs, and you must count 1 extra diagonal which normally is a side, the one formed by the two people next to the 2 empty chairs.
[/hide]
[hide="Solution"]Diagonals of a convex regular polygon is $ \frac{a(a-3)}{2} $ where $a$ is the number of sides.
Then, $ \displaystyle 1 + \frac{a(a-3)}{2} = 1 + \frac{(n-3)(n-6)}{2} = \frac{n^2-9n+20}{2} $
The probability is $ \displaystyle \frac{n^2-9n+20}{2(n-1)!} $
[/hide]

I'm wondering about the first question.

The second person in the group of three will sit on the left or right of the first person, and there $n-1$ other chairs, so $\frac2{n-1}$ chance this will work.  The third person needs to sit left or right of the group of two, so the chance is $\frac2{n-2}$. The probability both event are satisfied is then $\frac4{(n-1)(n-2)}$

TowersFreak's logic makes sense, but so does mine, so who missed what?

Using the same strategy for number two,

[hide="possible solution for the second question"]
$\frac{(n-3)(n-4)}{(n-1)(n-2)}$

[/hide]

I don't think your solution to the second problem works, because:
If there are $5$ or less people, then at least two of the $3$ people must touch, so the probability is zero. However, plugging in $5$ yields $\frac{1}{6}$

I also don't understand how you arrived at $\frac{2}{n-1}$ and $\frac{2}{n-2}$.

[quote="towersfreak2006"]Perhaps the "second" person need not sit by the first person? I don't really know if order matter, but it seems like everything would be okay if the third person sat between the first and the second.[/quote]

yeah, that makes sence because I didn't account for the possibility that they sit in the order (1)(3)(2) or (2)(3)(1).

btw, I got $\frac{2}{n-1}$ because there are two ok places for the person to sit, and $n-1$ spaces total left for the second person.  Then here is what I did wrong:  there are actually 3 places that are ok for the third person to sit: on either side of persons (1) and (2), or in between them (just move a chair), so now there are 3 "ok places" for person (3) to sit and $n-2$ open chairs, which gives the probability $\frac3{n-2}$.

Then, $\frac2{n-1} * \frac3{n-2} = \frac6{(n-1)(n-2)}$, which agrees with towersfreak's answer.

And of course for any solution we come up with for this problem there is a restriction that $n\geq 4$, because otherwise it would be guaranteed.

none of the solutions to the second problem so far are correct  ;)

[quote="behemoth571"]For the second problem, I'm assuming that none of the 3 people can be touching? i.e. any 2 of  them must have at least 1 other person between them.[/quote]

You are right

Hmm, I was assuming combinations, not perms, so the answer should be times $ 3! = 6$.
Then, the solution should be $\displaystyle \frac{3(n-3)(n-6)+6}{(n-1)!}$

I have another solution but the explaination is complicated... I'll type it up if this answer is correct.

[hide="possible solution"]
$\frac{[n-5][2+(n-6)]}{(n-3)(n-2)}$
[/hide]

[quote="calculuslover800"]
[hide="possible solution"]
$\frac{[n-5][2+(n-6)]}{(n-3)(n-2)}$
[/hide][/quote]
Well, $n-3$ should be replaced by $n-1$

[hide="the solution with explaination"]

EDIT: Remember that $n\geq6$, otherwise the chance would be zero.

Imagine you are putting people in the chairs one at a time.  Then the "first" person can sit anywhere.  Consider the two spaces adjacent to the first person to be restricted, that is, none of the other three people can sit there.  If the second person is placed in an open space adjacent to a restricted space, then the number of restricted spaces will go up by one and the number of taken spaces will go up by one.  However if the second person is placed in an open space that is not adjacent to a restricted space the number of taken spaces goes up by one and the number of restriced spaces goes up by two.  The probability that a person can randomly sit down in a space that is not taken without the space being restricted is:

$p = \frac{n - r - t}{n - t}$

where $r$ is the number of restricted spaces and $t$ is the number of taken spaces.

So the chance that the second person will sit adjacent to a restricted space, multiplied by the chance that the third person will sit in an open space given the second person's position is:

$\frac2{n-1} * \frac{n - 5}{n - 2}$

The chance that the second person will sit away from any restricted space, multiplied by the chance that the third person will sit in an open space given the second person's position is:

$\frac{n - 5}{n - 1} * \frac{n-6}{n-2}$

To get the solution you need to add together all the posibilities for the second person's position:

$\frac2{n-1} * \frac{n - 5}{n - 2} + \frac{n - 5}{n - 1} * \frac{n-6}{n-2} = \frac{2(n-5) + (n-5)(n-6)}{(n-1)(n-2)}$

Expanding and factoring:

$\frac{2(n-5) + (n-5)(n-6)}{(n-1)(n-2)} = \frac{(n-5)(n-4)}{(n-1)(n-2)}$

(the last time I accidently subtracted the number of restricted spaces from the denominator when calculating for the second person)

[/hide]


Could an instructor (or anyone else) post an answer set for the 12B? Thanks.

19 answered, all correct, got 129...i thought the B test was fairly hard, harder than A, at least...

other results??

well...it wasn't easy...but i didn't think it was necessarily harder than A

answered 19, missed the inverse one...not sure about which others I missed...should have gotten 123


Let ABC be a triangle,P on (BC) and M on (AP).
It is true that BM+MC<AB+AC?

Yes, indeed.
Denote $D$ the intersection of lines $CM$ and $AB$. From triangle's inequality $BM+MC<BD+MD+MC=BD+DC<BD+DA+AC=AB+AC$
and we are done.  :D


now that's actually the new Indian Dental Cares' Recommendation. :P
Colgate is doomed :D


$ \begin {cases} x\plus{} \frac {1}{y}\equal{}2 \\ y\plus{}\frac {1}{z} \equal{}2 \\ z\plus{}\frac{1}{x}\equal{}2 \end {cases}$

[quote="thanhnam2902"]$ \begin {cases} x \plus{} \frac {1}{y} \equal{} 2 \\
y \plus{} \frac {1}{z} \equal{} 2 \\
z \plus{} \frac {1}{x} \equal{} 2 \end {cases}$[/quote]

The first and third give $ y \equal{} \frac1{2 \minus{} x}$ and $ z \equal{} 2 \minus{} \frac {1}{x}$
Substitute in the second:

$ y \plus{} \frac {1}{z} \equal{} 2  \Rightarrow  \frac1{2 \minus{} x} \plus{} \dfrac {1}{2 \minus{} \frac {1}{x}} \equal{} 2$

$ \frac1{2 \minus{} x} \plus{} \dfrac {1}{2 \minus{} \frac {1}{x}} \equal{} 2 \Leftrightarrow (x\minus{}1)^2 \equal{} 0 \Rightarrow x\equal{}1$

Substituting x=1 back in the system, we get x=y=z=1 is the only solution.

solution by constraint $ x,y,z > 0$
by summing we obtain: $ x \plus{} y \plus{} z \plus{} \frac {1}{x} \plus{} \frac {1}{x} \plus{} \frac {1}{x} \ge x \plus{} y \plus{} z \plus{} \frac {9}{x \plus{} y \plus{} z} \ge 2*3 \equal{} 6$ with uguality iff x=y=z=1.

Hi bboypa! My solution same your post. Thank you.


geometry
3D geometry
tetrahedron
geometry solved

Given a regular tetrahedron $ABCD$ with $H$, $P$ are the midpoints of $AB$, $CD$, respectively. An arbitrary plane that contains $H$ and $P$ cuts $AD$, $BC$ at the points $F$ and $E$, respectively. Prove that $EF$ is perpendicular to $HP$.

It's not difficult:
Let $ EP \cap BD = T$. Note that $ H,F,T$ is collinear.
Use Menelaus theorem for triangles $ ABD,CBD$ with secants $ TFH,TPE$:
$ \frac{\overline{FA}}{\overline{FD}}.\frac{\overline{TD}}{\overline{TB}}.\frac{\overline{HB}}{\overline{HA}}=1$

$ \frac{\overline{EB}}{\overline{EC}}.\frac{\overline{PC}}{\overline{PD}}.\frac{\overline{TD}}{\overline{TB}}=1$
$ \longrightarrow \frac{\overline{FA}}{\overline{FD}} = \frac{\overline{EB}}{\overline{EC}}$
$ \Longrightarrow EB=FA;EC=FD$ ( because $ BC=AD$)
$ \Longrightarrow \Delta EBH = \Delta FAH; \Delta ECP=\Delta FDP$
$ \Longrightarrow HE=HF;PE=PF$
$ \Longrightarrow EF \bot HP$
Hope you like my proof :)

The proof sounds good. :)
Why are H F T collinear, once again?

$ Plane(EHP) \cap plane(ABD) = HF$
$ Plane(EHP) \cap plane(ABD) = HT$
$ \Longrightarrow H,F,T$ is collinear


calculus
derivative
real analysis
real analysis unsolved

Hey :)

im stuck on trying to work out general terms for taylor series

can someone please help me with: f(x) = 1/(1 + x)^2    i.e. (1 + x)^-2

thanks
brad

Find a series for $\frac1{1+x}$. What is its derivative?


calculus
geometry
derivative
number theory
linear algebra

Why did you take BC Calculus as a sophomore?  Do you plan to major in math?

[quote="masamune2387"]I took Calc BC my sophomore year, and as for my junior year, I wasted it in a really stupid discrete mathematics class...I did get plenty of number theory done on my own though.

Should I take multivariable or linear algebra next year?[/quote]

What part of multivariable calculus is important to economics, chemistry, and physics?  Does the geometry matter or just things like taking a partial derivative?

[quote="gauss202"]Linear Algebra is rarely taught very well in high school. The chances are nearly 100% that you will have to take it again in college. So without knowing anything specifically about your school or the classes there, I would venture to say that your best bet is to take Multivariable Calculus next year. It's a natural extension of the calculus you've already learned, and if you learn it well it will put you in a position to take more advanced chemistry, physics, and economics classes earlier on in college.[/quote]

Define what you mean to be rigorous.

[quote="rlee"]I'm taking Calc BC Junior year, and even though I'm technically two or so years ahead, I always regretted not taking higher level math while I could.

If you want to take higher level math classes, do it at a local community college or university, NOT at your high school. University classes will be more rigorous, and you might even get credit for them when you go to college.[/quote]

Rigorous as in more true to the actual material that you need to know. AP Calculus in my high school, and in most high schools I think, is watered down. We don't really explore what calculus really is, and instead we just learn how to apply the formulas to typical problems.

You won't learn that stuff until real analysis anyway, even if you take your classes at a college.

There are different levels of "rigorous". Some teachers teach Calculus in greater depth than some others - i.e. their goal for the students is more than just how to pass the AP exam.

In some cases a high school AP Calculus class is actually better than the introductory calculus class that you will get at most colleges - both because of the smaller nature of the class and the slower pace. It all really depends on the quality of the teacher though, and his or her expectations for the class.

As for higher mathematics classes (like Multivariable Calculus, Abstract Algebra, Discrete Math, or Linear Algebra), those are rarely taught to much depth in high school. If you have the preparation and opportunity to take them at a local University you probably should.

I don't like AP Calculus because the class is geared towards passing the exam, not understanding calculus. AP calculus curriculum at my school is a watered down, overly simplified version of calculus.

Simply put, the emphasis is on how instead of why.

Any other high schoolers feel like this?

It seems to me that that is largely true of high school in general, and that all classes are in some way geared toward passing some test. I don't really know if this is more so of calculus than of most other classes because I rarely paid any attention in my high school AP calculus class. But at my school, AP classes were designed to help students pass AP tests, normal classes were designed to help students pass STAR tests, and honors classes were designed to help students get into AP classes maybe.

What I think is especially annoying with my AP Calc class is the teacher is totally geared toward the AP exam and not our later college classes. We have skipped alot of topics that are no longer part of the BC Exam (like shell's method, etc.). And instead of learning everything, we get take home tests that are free-response questions (and we haven't learned how to do most of them). I think more homework covering the more difficult topics would be better than just free-response questions all the time.

church i think finding volume by cylindrical shells is still on the test. Although maybe you can just use disks, but sometimes the shells are easier. However you can very easily learn this on your own. I'm learning all of the BC parts myself right now. (Currently finishing series and sequences stuff, then review time!)


[quote="tranthanhnam"]let $a,b,c$ be positive real numbers .
 [b]Prove that [/b]:
    $\frac{1}{a^{4}+a}+\frac{1}{b^{4}+b}+\frac{1}{c^{4}+c}\geq \frac{12}{(a+1)(b+1)(c+1)}$[/quote]
There you go, at least its readable!


[hide]
i used a calculator :D
the answer is
[hide]no[/hide][/hide]

[hide][quote="236factorial"]Is 98574 divisible by 33?[/quote]
It is  divisible by 3 but not by 11 so it isn't.[/hide]

[quote="236factorial"]Is 98574 divisible by 33?[/quote]

[hide]I divided 98574 by 33 using a calculator and got: 2987.0909090909090909090909090909

That obviously is not an integer - so no.[/hide]

Use divisibility rules to show that it's divisible by 33, no computer.

[hide]its divisible by 3: 9+8+5+7+4=33, which is divisible by 3 and not 11:9-8+5-7+4=3 which is not divisible by 11 so no[/hide]

[quote="236factorial"]Why are you guys using a calculator  :roll:  what happened to divisibility rules?  :(

Is this divisible by 33:

121212121212121212121212121212121212121212121212121212121212121212?  :D[/quote]
[hide=" my answer"]well 1+2 = 3 and 3 is divisible by 3 (duh!) and there are just sets of 1, and 2 so yes its divisible by 3

it's not divisible by 11 using every other number added together then subrtacted is divisble by 11 ( icounted the amount of 1's and 2 's and then multplied) No point of doing these kind of problems with a calc - a 5 year old can use a calculator![/hide]

Divisibility by 33

since 33 = 11 X 3 and 11 and 3 are co-primes.
A number is divisible by 33 if it is divisible by 3 and 11.

98574

9+8+5+7+4 = 33 33 is divisible by 3 hence 98574 is divisible by 3.

9-8+5-7+4 = 3  98574 is not divisible by 11.

hence 98574 is not divisible by 33.

121212121212121212121212121212121212121212121212121212121212121212

this is divisible by 3.(99 is divisible by 3)
this is divisible by 11 too. (66-33 = 33 which is divisible by 11)

Hence this no is divisible by 33.