There are no solutions.  $2^{x}-3^{y}-4^{z}\equiv 1\pmod{2}$.

Unless $x,y,z\in\mathbb{R}^{+}$, which leads to infinitely many solutions.

Can they be negative numbers as well?

Sorry for lack of clarity
Ok. That is...
Let $x ,y ,z \in \mathbb{R}$

If $x, y, z \in \mathbb{R}$ it seems to me there is not a unique answer.

I think they have to be negative integers.

How will that help?
Clearly $x>9$, otherwise the expression will be less than 512, and if either $y$ or $z$ are negative, then both are (otherwise the sum is clearly not integer), but then $3^{y}+4^{z}<2$, and so we would have to have $2^{x}$ about 578, which is clearly untrue  :huh:

Oops. :oops: I didn't really think about it.  I just said they can't be positive integers, can't be reals or else we have an infinite number of solutions, so the closest we can get to have a unique solution is negative integers.

In conclusion, we have an infinite number of solutions.

While we may have an nifinite number of solutions to the first eqn, maybe the second is invarient wrt them?

I don't think so (though given the number of crass errors in my posts recently, I could be wrong  :) ), because we could make all three of $x,y,z$ really big, and so the value $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ would get really close to $0$, though we can vary $x$ and $y$ to make $z \rightarrow 0$ and hence $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ can get really big.


My proof is the following:

A "first result" is: [i] Let $ a_k\ge0.$ $ \sum a_k$ converges $ \iff$ $ \sum {a_k\over 1 \plus{} a_k}$ converges[/i]

The proof at the end.

Now the term $ {a_k\over 1 \plus{} k a_k}$ has the same monotonicity of $ a_k.$ By the Cauchy-condensation test,
the series converges or diverges with $ \sum 2^k{a_k\over 1 \plus{} 2^ka_k} \equal{} \sum {2^ka_k\over 1 \plus{} 2^ka_k}.$ By the "first result" this series converges or diverges with $ \sum 2^k a_k,$ which in turn, also by monotonicity, converges or diverges with $ \sum a_k$

The proof of "first result". If $ \sum a_k$ converges, the convergence of $ \sum{a_k\over 1 \plus{} a_k}$ is assured by
$ 0 \le {a_k\over 1 \plus{} a_k} \le a_k.$  Let's suppose that $ \sum a_k$ diverges. We can bound
$ \sum {a_k\over 1 \plus{} a_k} \ge \sum_{k: a_k\le 1} {a_k\over 2} \plus{} \sum_{k: a_k > 1} 1/2 \doteq S_1 \plus{} S_2.$ $ S_1$ and/or $ S_2$ contains inifinite terms determining the divergence.

[quote="yzymat"]$ a_n > 0$,    $ a_n \geq a_{n \plus{} 1}$,  and  $ \sum a_n$  divergence,  prove  that

$ \sum \frac {a_n}{1 \plus{} na_n}$  also  divergence[/quote]

Should  it   be  ?

$ \sum 2^{k}{a_{2^k}\over 1\plus{}2^{k}a_{2^k}}\equal{}\sum{2^{k}a_{2^k}\over 1\plus{}2^{k}a_{2^k}}.$  By the "first result" this series converges or diverges with  $ \sum 2^{k}a_{2^k},$


1989 equal circles are arbitrarily placed on the table without overlap. What is the least number of colors are needed such that all the circles can be painted with any two tangential circles colored differently.

the basic idea of this problem is that at most, only three circles can be pairwisely tangent at a time, so the answer is always three no matter how many circles there are [geogebra]c423a75149a4f31b237c9e6bfad7adc79c9403e9[/geogebra]


find all positives integers $ a$ ,$ b$, and  $ c$ $ >\equal{}2$  such that :
$ a^b\plus{}1\equal{}(a\plus{}1)^c$

This has been done before.  Actually, look below this post on the previous page--it is there too.
Use Zsigmondy's Theorem and see
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196016

I solve this problem.
If you want to know solution, said me!

There is a solution without Zsigmondy, but it technically uses similar ideas (but is less difficult).

@Tigr: Either post a solution or not, but do not post something like "I solved it"only.

I wright solution in two words.If you have quetions, please, said me.

a^n+1=(a+1)^m,a,m,n∈N,if n:2 →(a^n+1):not (a+1)→n is odd!then a^n=ma+m(m-1)/2*a^2+⋯+a^m→m:2.then proof that a:2.then let m=2m_1,b=(a+1)^(m¬_1 )¬:not 2,then a^n=(b-1)(b+1)→gcd(b-1,b+1)=2,b-1=2b_1,b+1=2b_2,proof that b_1:not 2,and b_2 ∶2→b_1=a_1^n,b_2=2^(n-2) a_2^n,a_1:not 2,then (b-1):a=2a_1 a_2,→a_2=1,then we have that 2^(n-2)=a_1^n+1,→a_1+1=2^α,then proof that α=n-2,after this proof that n=3,then a_1=1,b_1=1,b=3=(a+1)^(m_1 ),then m_1=1→m=2,a=2.We get the solution a=2,m=2,n=3.


Salam everybody.

I have noticed that this forum is one of the most inactive sub-forums in the National Communities. Now I don't think that Pakistani people do not like math and problem solving, but I think the problem is unawareness.

So, to all of you Pakistani people out there, I say spread the word about AoPS, and bring in at least 1 new member. Post there usernames in this thread. Let's make this a competition. The person who can bring the most Pakistani people to join AoPS in 1 month wins!!! And the prize is going to be our respect and admiration of your hard work that resulted in people coming to know about AoPS, and how there is math after the school textbook. So let's go people, the competition starts today on August 29, 2006.
You have to bring in new users to AoPS, and ask them to post in this thread confirming that they were introduced to AoPS by you. The competition ends on September 29, 2006. The person who brings the most people wins. :lol:


For x,y,z positive real numbers prove that

(x/y)^3 + (y/x)^3 + (y/z)^3 + (z/y)^3 + (x/z)^3 + (z/x)^3  \geq x^2/yz +yz/x^2 + y^2/xz + xz/y^2  +z^2/xy + xy/z^2

Obviously,the problem is equivalent to the following one: for any positive a,b,c with product 1, we have the inequality: a^3+b^3+c^3+a^3b^3+b^3c^3+c^3a^3>=ab(a+b)+bc(b+c)+ca(c+a). All I did was to put a=x/y,b=y/z and c=z/x.But the above inequality is an immediate consequence of Schur inequality, because
  ab(a+b)+bc(b+c)+ca(c+a)<=a^3+b^3+c^3+3abc<=a^3+b^3+c^3+a^3b^3+b^3c^3+c^3a^3.


geometry
trapezoid
geometric transformation
reflection
circumcircle
ratio
angle bisector

Let $ ABCD$ be a isosceles trapezoid ($ AD\parallel BC$ , $ AB \equal{} CD$) for which $ A > 90^{\circ}$ and

$ AB^2 \equal{} AD\cdot BC$ . Denote the reflection $ R$ of the vertex $ A$ w.r.t. the diagonal $ BD$

and the point $ S\in BD$ so that $ \widehat {SCA}\equiv\widehat {SCR}$ . Prove that $ \frac {SB}{SD} \equal{} \sqrt {\frac {BC}{AD}}$ .

To: Virgil Nicula

Attached, the solution for subject question.

Best regards,
sunken rock

A blind man sees better the details

We know that in every triangle, the angle bisector and the perpendicular bisector of the opposite side intersect on the circumcircle of the triangle. It thus follows that $ ACRS$ is cyclic.

We know that $ \frac {BR}{RD} \equal{} \frac {BA}{AD} \equal{} \frac {BC}{CD} \equal{} : \lambda$, so the circumcircle of $ \triangle ACR$ is the Circle of Apollonius of $ B$ and $ D$ w.r.t. to the ratio $ \lambda$. But $ S$ lies on that circle, so $ \frac {BS}{SD} \equal{} \lambda$. Hence,
\[ \frac {BS}{SD} \equal{} \sqrt {\frac {BA}{AD}\cdot\frac {BC}{CD}} \equal{} \sqrt {\frac {BC}{AD}}.\qquad\Box
\]


This is a wellknown ineq. Here's proof.

by Cauchy-Swartz ineq.

$(x+y+z) (1- \frac{1}{x} + 1- \frac{1}{y} +1- \frac{1}{z} ) \geq ( \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} )^2$

cuz $1- \frac{1}{x} + 1- \frac{1}{y} +1- \frac{1}{z} =1$

We get the problem.

Q.E.D.

yes cauchy is most classic way of solving the problem...
although there are at least two more ways that one can approach the problem.


Let $ \frac{\sin x\plus{}\sin y\plus{}\sin z}{\sin(x\plus{}y\plus{}z)}\equal{}\frac{\cos x\plus{}\cos y\plus{}\cos z}{\cos(x\plus{}y\plus{}z)}\equal{}2\sqrt{2}$. Find $ \cos x\cos y\plus{}\cos y\cos z\plus{}\cos z\cos x$

Goodness  :!:
[hide="maybe...a hint"]
I think that the only way this will work is if two of $ x,y,\text{ or }z$ are complimentary (in effect) and the final is $ \frac{x\pi}{4}$.

But I'm not sure.
[/hide]

Well, actually the answer does not depend on $ x,y$ or $ z$.

[hide="?"]
Let $ a \equal{}\cos x\plus{}i\sin x,\ b \equal{}\cos y\plus{}i\sin y,\ c \equal{}\cos z\plus{}i\sin z$. Then $ a,b,c$ have absolute value 1 and satisfy the equation
\[ a\plus{}b\plus{}c\equal{}2\sqrt{2}abc\]
Conjugating both sides of the above (note that $ |a| \equal{} 1\Rightarrow\overline{a}\equal{}\frac{1}{a}$, etc.),
\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{2\sqrt{2}}{abc}\Rightarrow ab\plus{}bc\plus{}ca\equal{}2\sqrt{2}\]
Now we compute: $ \cos x\cos y\plus{}\cos y\cos z\plus{}\cos z\cos x$ is half of $ \sum_{\text{cyc}}\left( a\plus{}\frac{1}{a}\right)\left( b\plus{}\frac{1}{b}\right)$, which is
\[ \sum_{\text{cyc}}\frac{c(a^{2}\plus{}1)(b^{2}\plus{}1)}{abc}\]

\[ \equal{}\frac{abc(ab\plus{}bc\plus{}ca)\plus{}(a\plus{}b\plus{}c)\plus{}(a^{2}b\plus{}a^{2}c\plus{}b^{2}a\plus{}b^{2}c\plus{}c^{2}a\plus{}c^{2}b)}{abc}\]

\[ \equal{}\frac{abc(ab\plus{}bc\plus{}ca)\plus{}(a\plus{}b\plus{}c)\plus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}3abc}{abc}\]

\[ \equal{}2\sqrt{2}\plus{}2\sqrt{2}\plus{}8\minus{}3\equal{}4\sqrt{2}\plus{}5\]

\[ \Rightarrow\cos x\cos y\plus{}\cos y\cos z\plus{}\cos z\cos x\equal{}\boxed{\frac{4\sqrt{2}\plus{}5}{2}}\]
Of course, having the sum much greater than 3 is ridiculous, so either I messed up horribly or there's no solution.
[/hide]


ratio
geometry
rectangle
integration
calculus
calculus computations

Can i get some help with this problem? i don't really understand what to do, it looks related to area between curves but it's kinda confusing to me :oops:

The figure below shows part of the curve $y=x^3$ and a rectangle with two vertices at $(0,0)$ and $(c,0)$. what is the ratio of the area of the rectangle to the shated part of it above the cubic?

a) 4:3
b) 4:1
c) 3:2
d) 3:1
e) 5:4

[img]http://img91.imageshack.us/img91/2659/pic1pe.jpg[/img]

One way to solve the problem will be:
The are of the graph y = x^3 (from 0 to c) is the area of its inverse on the interval(0 to c^3)
(If u flip the ghraph 90 degrees u get the graph of the inverse). Hope this helps!

Looking at the inverse is overkill. You only need to compute one integral.

Here's the question you should be asking: what, exactly is the point at the upper right corner of the rectangle? And what are the dimensions, hence the area, of the rectangle?

[hide]The upper right point is $(c,c^3),$ so the area of the whole rectangle is $c^4.$

The white area is $\int_0^cx^3\,dx=\frac{c^4}4.$

That is, the area under the curve is $\frac14$ of the area of the rectangle. And that means that the shaded area is $\frac34$ of the area of the rectangle. Reading carefully what the probem is asking, we get the answer $4: 3.$

If we were to replace the curve $y=x^3$ by $y=x^m$ for any positive $m$ (positive $m$ so we can draw essentially the same picture), then the answer would be $(m+1): m.$[/hide]


articles
summer program
Mathcamp
calculus
derivative
geometry
Ross Mathematics Program

[quote="masamune2387"]Haha.  From what I hear, the average quant has a gold medal at the IMO.[/quote]

That's not funny, because it's a total waste of talent.

I think I'd disagree with both statements - the average quant at DE Shaw is not an IMO gold medalist (I know, I was there, and I'm not an IMO gold medalist).  Also, I'll note that there are a lot of great quants in a lot of trading houses.

Nor is it true that it's a waste of talent.  What would you have these people do?  Become PhD mathematicians?  The world has way more than enough of these - every year financial firms turn away thousands of PhD mathematicians who have first failed to find employment in academia.  Moreover, I'll point out that the financial world is no more inherently evil than the academic world is inherently good.  In fact, I'd argue that there're just as many people wasting their time in academia as there are in finance (and just as many of each pursuing important, fruitful careers in each).  There are many jobs in the financial world that are instrumental into helping people secure their financial futures, which is extremely important.  Even arbitrageurs help normalize markets (though of course some hedge funds do the opposite - I’m not arguing that all trading jobs are inherently good).

Besides, it's more important to do what you enjoy than to do what others might deem suitable for your skills.  I'm sure people think I'm wasting my talent, having walked away from a very promising and lucrative career trading to ultimately build AoPS.  I'd strongly disagree with that comment - I can't imagine a better way for me personally to spend my time, as this neatly fits my skills and interests.  But it would have been a complete waste of time for me to continue in academia, as it simply didn’t fit my temperament, so I would have been very poor at it.

And if it’s the accumulation of wealth that you find a waste of time, I’ll note that it’s hard to deem that a waste of time until you see what people do with it.  This site, for example.

For anyone wondering about "quants," [url=http://www1.excite.com/home/careers/company_profile/0,15623,811,00.html]here[/url] is an article you should look at.

I'll note that much of what that article says about Shaw in terms of culture is not unique to Shaw.  There are now many places with that sort of culture.

(It's also not true that as a rule people there didn't work that hard, though I'll note that they didn't work as much as people starting in investment banking say they work.)

And I'll note that that "vague sense" of losing money in the summer of 1998 wasn't vague to everyone there ;).

What was the 1998 collapse due to? I mean, there are some details in the article, but what led to those vents?

Wow that is cool.  I really hope I get into Mathcamp now...

[quote="rrusczyk"]Depends on who you ask.  Some will say the 'Asian flu'.  Others will say Russia, others will say the tech market hiccuped.  They're all pretty much wrong.

I was there.  It was caused in part by greed on the part of a great many people in the financial world.  A lot of people chasing worse and worse trades with more and more money, and bankers too willing to loan those people money to do the chasing.  It's a fairly long story, and perhaps I'll write about it some day.

If you're coming to Mathcamp again this year, I'll talk about in one of the classes - I'll be doing a math in finance class in which I'll teach some finance basics and tell some stories (which will be mostly true :) ).[/quote]

I've heard about the Asian Flu. It all began around the Southeastern Asian Countries, such as Malaysia, Indonesia, Philipines, Myanmar, and most notably, Korea. I did not expect Korea to collapse in such a quick manner. Anyways, it was caused by people who wanted to earn more money through trading and mostly buying. It became so big that people began buying investments out of their respective countries, and they began borrowing money.

It was bad. I heard Korea had to request a lot of money from IMF (international Monetary federation???). They have all paid now, but back then the country ran campaigns to gain money in selling or donating jewelrys or recycling things to save up money. About 70% of the Seoul Residence lost their job. My dad's computer programming corporation was big enough to stand the hit, but many companies, big ones in korea such as Daewoo fell. It was tragic.

[quote="ffdbzathf"]*note: IMF stands for International Monetary Fund

you were close =)[/quote]

lol. my bad :lol:

[quote="smbelcas"]Well, this seems like the most appropriate thread for this request... I'm on the editorial board of Math Horizons (see [url]http://www.maa.org/mathhorizons/[/url]), which is doing a series of profiles on mathematical employers.  We would like to do a piece on DE Shaw... does anyone on AoPS have a contact at DE Shaw who might be willing to assist in this endeavor?[/quote]

well... Mr. Ruscyzk did work for DE Shaw for a while... right??? so then you could ask him.

During the summer of 1998, besides the "Asian flu" there was also a Russian default, and besides (and bigger than) D. E. Shaw's difficulties was the implosion of LTCM.  Two general books about the big picture of "what happened" are "When Genius Failed : The Rise and Fall of Long-Term Capital Management" by Roger Lowenstein, a reporter for the Wall Street Journal, and (much less good, but including some stuff Lowenstein left out) "Inventing Money : The Story of Long-Term Capital Management and the Legends Behind It" by Nicholas Dunbar.  Both also shed some light on the general role of quants in finance and the machinery of relative value hedge funds in particular.

You might also check out Emmanuel Derman's recent memoir, "My Life as a Quant."  For those who get interested and have the basic math (I don't think anything more than standard 1st-year calculus is really required, though the rigorous underpinnings require some knowledge of basic measure theory, linear algebra, and stochastic calculus) you might want to read Hull's "Options, Futures, and Other Derivatives" which is a basic first text to acquaint you with the jargon, and then a lot of the finance-specific books become a lot more accessible.  Taleb has written a couple of books, both popular and specialized, though he has a unique view on things.

I would echo that few quants are mathematically talented at the IMO level. However, the job requires (in my opinion) an ability to work at a fast pace, more facility with computers, sometimes a more practical bent and often, a level of patience with others, beyond what work as an academic usually does.

I'd be interested in hearing more comments about Nicholas Taleb's writings here. I really like his book [url=http://www.amazon.com/exec/obidos/tg/detail/-/158799190X/learninfreed]Fooled by Randomness[/url] and wonder how many of you have read it.

[quote="maokid7"]what are some classes that you would recommend for someone to take as an undergrad if they plan to pursue this line of employment.

I plan to double major in mathematics and economics.
Also would there be something better to pair with a major in mathematic other than economics.[/quote]

If you are aiming at being a quant, learn how to program and get some mathematical modeling (applied math) experience, too.  That is probably more important than economics.

[quote="rrusczyk"][quote="NoYoo-hooForMe"][quote="rrusczyk"]and such a candidate is also unlikely to be happy at Shaw - if pure theoretical research is the only thing you love, then you'll be happier elsewhere...[/quote]

This is true but there is nothing about this that is unique to Shaw.  There is nowhere in finance, AFAIK, where someone who prefers pure theoretical research will be happy.[/quote]

Good point; I agree, and should have made this clear.[/quote]

Jim Simons' fund is located on an academic campus where he used to be head of department, employ mainly academics (I remember e.g. a top-level string theorist giving up a professorship to work there) and has a degree of freedom not found at most other funds including control-freak-owned DE Shaw.     If magazines like Fortune, Forbes, Trader, etc are to be believed, Simons consistently out-earns Shaw both absolutely and proportionally.

Other than that an a few other exotic funds, the previous posters' statements are accurate as far as I know.

the academic campus you are referring to is Stony Brook University...but is that where the actual fund is based?  i know he runs it on long island right near where he lives, but as far as i know it is not actually run AT Stony Brook.  by the way, does anyone know how he bridged the gap between theoretical work and running a hedge fund?  his research as a professor was in theoretical mathematics (as well as his PhD), so did he just learn about finance and some programming and applied stuff from textbooks?  it just seems strange that he was a professor of math and then suddenly started his own hedge fund (yet he appears to be insanely good at it).

[quote="fleeting_guest"]
Jim Simons' fund is located on an academic campus where he used to be head of department, employ mainly academics (I remember e.g. a top-level string theorist giving up a professorship to work there) and has a degree of freedom not found at most other funds including control-freak-owned DE Shaw.    [/quote]

This is a pretty unfair assessment of David Shaw.  I worked there for more than 4 years, and had enormous freedom, and David, as far as I could tell, virtually never was involved in our strategies and certainly was not controlling.  Perhaps that's because he had complete faith in the two people between me and him on the food chain, but nonetheless, 'control freak' doesn't describe my working relationship with anyone at Shaw.

(As for the earnings, I don't have the data to discuss it.  As I understand it, Shaw is significantly larger than Renaissance, which would likely lead to a percentage return differential.  I'd bet a smaller place like Jane Street that focuses on a different style of trading beats the pants off both of them.  But I doubt Jane Street (or anyone else) is yet in a position to manage 20 billion dollars of capital and still pull off that sort of return.  (Disclaimer: Though I know people at Shaw and Jane Street still, I have no data on their recent performance.)

I will give Simons points for his recent interest in education through Math for America and his being on Bush's panel for reforming math education, though it's not clear what the purpose of the latter is (Vern Williams is on the panel, too, so I have some hope that at least someone will be fighting to have math included in the math curriculum.).

Jim Simons may not be a “control freak” at the Renaissance Technologies investment firm (his firm), but he does like to direct how to spend his donated money. The following quoted story would tell.

The family's charitable foundation has committed $38 million to find the causes related to autism in recent years, and plans to spend another$100 million in what is becoming the largest private investment in the field of autism research, while Simons personally exerts extraordinary control over where and how his money is spent. Simons has provided DNA from his family for study, and has given assistance in helping solve research problems. But Simons has stirred debate among scientists, by directing research funding toward genetically based explanation for the disease, and controversy among parents, many of whom would prefer to have researchers investigate possible environmental triggers. When MIT asked for brain research funding, he stipulated that the project focus on autism and include scientists of his choosing.

rrusczky, in a different thread, we discussed about Simons 2005’s earning (1.5 Billion). I was ignorant about the investors’ money, not their own money at risk. I was quite wrong. The Renaissance Technologies has been returning investors’ money all the time when profit made. They have been trying to keep the fund size around 5 billion (it is reported that the fund size is currently at 12 B). A 2005 article from Wall Street Journal said that they actually planned to return all investors’ money and would only manage the fund with their own money (the firm has 60 people according to the article), therefore, the fund would not be too big. It was also reported their fund has been earning 34% after fees (5% asset plus 44% earning), and has been 10% higher since 1988 than the funds managed by such star traders Bruce Kovner, George Soros, Paul Tudor Jones. I wonder how they did at the end of 1998 when JM (John Meriwether) went under and DE Shaw group also lost a billion at that time (I read the book “when genius failed”). I assume you joined DE Shaw after that time.

[quote="rrusczyk"][quote="fleeting_guest"]
Jim Simons' fund is located on an academic campus where he used to be head of department, employ mainly academics (I remember e.g. a top-level string theorist giving up a professorship to work there) and has a degree of freedom not found at most other funds including control-freak-owned DE Shaw.    [/quote]

This is a pretty unfair assessment of David Shaw. [/quote]

It is an accurate paraphrase of Shaw's reputation among various people in finance.  In any event, Renaissance is closer to an academic working environment than Shaw, Shaw is closer to a corporate working environment than Renaissance, and both have good conditions compared to the industry at large and certainly compared to the US economy in general.

[quote] (As for the earnings, I don't have the data to discuss it.  As I understand it, Shaw is significantly larger than Renaissance, which would likely lead to a percentage return differential.  I'd bet a smaller place like Jane Street that focuses on a different style of trading beats the pants off both of them.  But I doubt Jane Street (or anyone else) is yet in a position to manage 20 billion dollars of capital and still pull off that sort of return.  (Disclaimer: Though I know people at Shaw and Jane Street still, I have no data on their recent performance.)  [/quote]

Simons, who started his funds years before Shaw, had returns that outpaced Shaw both absolutely and proportionally in most or all years of their funds' existence, including the years when Simons' fund(s) were much larger.    Simons' fund is now smaller because for years he has been returning money to the investors and progressively turning the company into a private investment vehicle for himself and his associates.   Simons is perennially in the top 3 of hedge fund managers with earnings several multiples that of Shaw, who may or may not crack the top 10 in any given year.

[quote]  I will give Simons points for his recent interest in education through Math for America and his being on Bush's panel for reforming math education, though it's not clear what the purpose of the latter is (Vern Williams is on the panel, too, so I have some hope that at least someone will be fighting to have math included in the math curriculum.).[/quote]

One sign that this board is becoming a DE Shaw fan club is when rival fund managers' philanthropic activities are up for criticism.  Look, rich people at that level tend to be a little crazy, and have more than a little vanity.   Shaw has his vanity projects as well, such as Juno email or I-am-still-a-scientist funding of "R & D" work, and I have never heard of any sensible or non-sensible large-scale philanthropy that he funds.

[quote="fleeting_guest"][quote="rrusczyk"][quote="fleeting_guest"]
Jim Simons' fund is located on an academic campus where he used to be head of department, employ mainly academics (I remember e.g. a top-level string theorist giving up a professorship to work there) and has a degree of freedom not found at most other funds including control-freak-owned DE Shaw.    [/quote]

This is a pretty unfair assessment of David Shaw. [/quote]

It is an accurate paraphrase of Shaw's reputation among various people in finance.  [/quote]

Perhaps that's true -- I can only state my personal experience, which for me trumps the rumor mill.

As for Simons' philanthropy, I'm a fan - we were one of the first sites his Math for America project linked to, and that project seems like a really good idea.  I can't speak to Shaw's philanthropy, as I really don't know anything about it.

As for the board becoming a Shaw fan club, I'll note that if true, it is somewhat earned by Shaw (the firm, not the person), as it actively funds various activities of interest to people on this site.  This isn't philanthropic - it's a recruiting investment as they are actively courting these students as future employees.   (I will add the note that other fund managers being up for criticism is hardly a sign of the board being a Shaw fan club - look at the early threads discussing math in finance, for example.)  And as for rich people being a little crazy, I tend to think my ex-co-worker from Shaw was right when he claimed that rich people are no nuttier than you or me - it's just that they can freely be who they are because they don't have to worry as much about the consequences (and they have the money to back up their idiosyncracies).

I'm not too surprised to hear the return statistics you suggest - Renaissance always had a very good reputation when I was trading.

As for my timing at Shaw - my group was the cause of that 1998 blowout, so I know more about it than I care to.  I don't think Renaissance was active in those markets, so I would guess they fared very well through that time period.  (And just to keep the record straight - Shaw didn't lose anywhere near a billion dollars.  That probably would have been the end of the firm.  It was my group responsible for the losses, so I'm pretty familiar with the numbers :) )

[quote="fleeting_guest"]One sign that this board is becoming a DE Shaw fan club is when rival fund managers' philanthropic activities are up for criticism.[/quote]

If Richard had been a McDonalds manager I imagine that students on these forums would be sending in their applications to the golden arches en masse.  I think it's just hero worship, no more, no less (and this certainly isn't intended as an insult to Richard).

[quote="blahblahblah"]
If Richard had been a McDonalds manager I imagine that students on these forums would be sending in their applications to the golden arches en masse.  I think it's just hero worship, no more, no less (and this certainly isn't intended as an insult to Richard).[/quote]

I don't think that's it -- I don't see the students lining up to become educators, which is what you would see if the primary influence were hero-worship.  I was a trader by circumstance, but am an educator by choice (or by genetics, take your pick).

I think the interest in Shaw (and Jane Street, which certainly has more name recognition in this crowd than in any other group of sub-22 year old peopel) is semi-educated self-interest.  Students hear that they have a skill that draws very high pay, and that several other people have pursued careers with that skill, and naturally they are interested.  To those of us who are older, it seems only obvious that strong math/CS students are in demand.  Think back to when you were in 8th grade or 9th grade - it wasn't so obvious then, unless you went to a very special school.  So, I think it has a lot less to do with me personally than it does with what my 'career' indicates is possible: people with strong problem solving skills can find high-paying jobs surrounded by intelligent people.  This is not as apparent to a 14-year-old in an average school as it is to a college student or someone who has been in the workforce for a while.

That the attention focuses on Shaw and Jane Street is a by-product of their having a presence here, and that they have a track record of hiring people like those who frequent this site.

Besides, for the most part, I think the crowd on this board is more resistant to hero worship than your average middle or high schooler, and most teens tend to pick heroes a little closer to their peer group.  (Everytime I see a username with something like 071393 in it, I feel very old - students born after I graduated college...)

I forgot to mention the high paying bit.

I disagree strongly with the assertion that the members of this board are resistant to hero worship - one needs only read one of the MathCounts result threads, or the 'WHO IS THE BEST IMO CONTESTANT EVAR!!!11!!!!1??!?!', or any of the threads on Reid Barton or Gabriel Carroll to see this.  The heroes are just different.  For what it's worth, I think that you make a very fine role model for younger students, especially since there don't seem to be a lot of mathematical role models outside of academia.

[quote="rrusczyk"][(Everytime I see a username with something like 071393 in it, I feel very old - students born after I graduated college...)[/quote]

At least your username doesn't contain something like "59."  Just imagine how old they feel. ;)

[quote="ffdbzathf"]the academic campus you are referring to is Stony Brook University...but is that where the actual fund is based? [/quote]

It started off-campus, then moved on campus.  They may be off-campus by now due to the expansion in their workforce, but I don't know for certain as it has been more than a few years since I last dealt with people at Renaissance.

[quote]  by the way, does anyone know how he bridged the gap between theoretical work and running a hedge fund?  his research as a professor was in theoretical mathematics (as well as his PhD), so did he just learn about finance and some programming and applied stuff from textbooks?  it just seems strange that he was a professor of math and then suddenly started his own hedge fund (yet he appears to be insanely good at it).[/quote]

He had some practical interests such as cryptography.  I think he started trading in the 1970's.    To the extent I'm familiar with it, the type of math Simons did was theoretical but not super-abstract by today's standards  (Chern-Simons secondary invariants in differential topology, Cheeger-Simons differential characters, differential geometry/topology more generally).   I don't know what he got the Veblen prize for, but that prize is one of the topical awards that is a notch below the Fields medal in prestige.    Due to the later work of Witten (for which he got a Fields prize) on knot invariants using the Chern-Simons form, there are now hundreds and possibly thousands of papers on what is called "Chern-Simons theory".   Simons was accomplished before becoming a trader, to be sure.

[quote]I think if there is a wastage of talent anywhere in this picture, it is in the numbers of people who pursue PhDs--especially in very theoretical areas like pure mathematics--without realistic hope of being able to pursue academic careers.
[/quote]

I don't think this is a waste really.  I'm doing a phd in pure math now, and having the best time of my life.  The past three years have been the first since I was fifteen where I haven't had to do "real" work, because I'm getting a scholarship to just study.  I think it's really changed the way I think about careers - rather than always thinking "what can I do that people will pay me for?", I find myself thinking "what do I think needs to be done in the world?" and so have started doing a lot of volunteer work.  I think it would be lovely if everyone could have this few years to just think really hard about something and spend the rest of the time reflecting on things. It doesn't matter if I get an academic job afterwards.  Doing the phd has been a great job just in itself.


Any help with a "formal" proof for this problem?

Let a1,a2,...,a44 be 44 integers such that a1<a2<...<a44<125. Show that at least one of the 43 differences dj=aj+1-aj occurs at least 10 times.


Find the acceleration of the masses, no friction and massless pulleys.
This is the problem, but hasn't the pulley on the left a net force and since it is massless, it has a infinite acceleration?

If the pulley is massless the net force acting on it is zero.

if the surfaces are absolutely frictionless then the tension in the rope remains constant so the left pulley should have a net force downwards .....i see no reason why the net force ought to be 0 on it

[quote="Astronum"]That's the point.

If it is massless, the net force on it should be 0, otherwise it would get infinite acceleration.

But, as pardesi says due to the fact that the tension is constant along the string, it has a downward net force.[/quote]
Yes, the net force should be 0, and we have got a downward net force $ 2T-T=T$. However this two conditions can be fulfilled at the same time - when $ T=...?$ :)

The pulley is massless so the net force on it should be 0, otherwise it would get infinite acceleration. The net force acting on the pulley is $ 2T-T=T$ ($ T$ is the tension of the rope) so we must have $ T=0$. Then the only forces acting on the masses are gravity forces so both of them have got downward $ g$ acceleration.

OK, I see. But suppose that we replace one of the masses with 2m. Is it true that in such a case mass 2m will be moving downward with acceleration of g and mass m with acceleration of g but in upward direction? (just like 2 masses on a very single pulley)

And the last question:
If we lower left mass (m) by x, will the right mass (m) raise also by x?

no, it will rise by x/2. let parts of the string be named L1, L2, L3, L4 starting from left mass. L4+L2 is always L3. If right mass is lowered by x, L2+L3+L4 will increase by 2x and left mass will be raised by 2x.

[b]shreyas_patankar[/b] but in such a case will the energy be conserved? If both masses are m, than right mass will lose energy of mgx and right will gain 2mgx  :?:

[quote="qoqosz"]OK, I see. But suppose that we replace one of the masses with 2m. Is it true that in such a case mass 2m will be moving downward with acceleration of g and mass m with acceleration of g but in upward direction? (just like 2 masses on a very single pulley)[/quote]
No, in that case both masses are moving downwards with constant acceleration $ g$ and the pulley is moving upward with constant acceleration $ 3g$ (this can be obtained from the fact, that the rope has got constant lenght).

[quote="qoqosz"]but in such a case will the energy be conserved? If both masses are m, than right mass will lose energy of mgx and right will gain 2mgx[/quote]
you forgot about the kinetic energy...

[quote="golduck"]No, in that case both masses are moving downwards with constant acceleration $ g$ and the pulley is moving upward with constant acceleration $ 3g$ (this can be obtained from the fact, that the rope has got constant lenght).[/quote]
But pulley is massless and it is moving with constant acceleration so there should be a force which makes that pulley move, but it is massles :| I must admit that I don't understand that "thing" quite good.

I don't know whether I think right, but if masses hanged like it is presented on the picture are "standstill", moving one of them with [u]constant[/u] speed upward (or downward) for distance x should cause the other mass to move in opposite direction also for distance x. Am I right?

No. It would cause the string either to go slack or to break. Remember there is only one string in the picture.

I'll try to explian more precisely how I see this problem and I will be very delighted if you would like to show imperfections in my thinking.
Let's consider generalized problem with masses m and M as it is shown on the picture:
[img]http://img518.imageshack.us/img518/6020/flowroot18041zf7.png[/img]
Of course T1 = T2 = ...
Because pulleys are massles $ T_{i}= 0$ and $ N=0$.
The only force which acts on each mass is gravitational force so both masses are moving downwards with acceleration of $ g$, right?

Now, what with pulleys - if masses are falling, two pulleys must move (or just one pulley?). If pulley in such a case is moving than it must moves with constant acceleration, but they area massless! So...?

BTW. Thanks for being patient with me  :wink:

The pulley connected with mass $ M$ moves in the same way as that mass - downward with acceleration $ g$.
Now, because the rope has got constant lenght we have got the following equation to find the acceleration of the second pulley:
\[ \Delta x_{m}+2\Delta x_{M}=\Delta x_{p}\]
$ \Delta x_{m}$ and $ \Delta x_{M}$ are the distances covered by the masses and $ \Delta x_{p}$ is the distance covered by the pulley (the pulley mowes upward). Integrating this equation twice we get the acceleraton of the pulley - $ 3g$.

[quote="qoqosz"]If pulley in such a case is moving than it must moves with constant acceleration, but they area massless! So...? [/quote]
If the pulley is massles and the force acting on it is $ 0$ then the second Newton's law $ F=ma$ is satisfied ($ 0=0$ :) ) no matter what the acceleration is (we can found the acceleration as above).

Thank you very, very much :) Now everything seems much clearer  :D


Let be $ m,n \in \mathbb{N}$ such that $ \sqrt{7}\minus{}\frac{m}{n}>0$ . Prove that $ \sqrt{7}\minus{}\frac{m}{n}>\frac{1}{mn}$ .

[hide][quote="alex2008"]Let be $ m,n \in \mathbb{N}$ such that $ \sqrt {7} \minus{} \frac {m}{n} > 0$ . Prove that $ \sqrt {7} \minus{} \frac {m}{n} > \frac {1}{mn}$ .[/quote]
$ \sqrt{7}\minus{}\frac{m}{n} > 0$
$ \leftrightarrow \sqrt{7}n\minus{}m > 0$
$ \leftrightarrow 7n^2 > m^2$
$ \leftrightarrow 7n^2 \geq m^2 \plus{}1$
we can easily check that $ m^2 \plus{} 1$ can never be divisible by 7 . Therefore the equality of the inequality above never hold .
$ \leftrightarrow 7n^2 > m^2 \plus{}1$
$ \leftrightarrow 7n^2 \geq m^2 \plus{}2$
with the same idea $ m^2 \plus{}2$ can never be divisible by 7
$ \leftrightarrow 7n^2 \geq m^2 \plus{}3$
$ \leftrightarrow 7n^2m^2 \geq m^4 \plus{} 3m^2 \geq  m^4 \plus{}2m^2 \plus{}1$
$ \leftrightarrow \sqrt{7}mn \geq m^2 \plus{}1$
$ \leftrightarrow \sqrt{7} \minus{} \frac{m}{n} \geq \frac{1}{mn}$

note that the equality here doesnt occur since we must have both $ m \equal{} 1$ and $ m^2 \plus{} 3$ is divisible by 7

so $ \sqrt{7} \minus{} \frac{m}{n} > \frac{1}{mn}$[/hide]


function calculus integration trigonometry algebra polynomial geometry

I have the following question to answer:

Let S_n be the nth Simpson approximation for a function f.
Let M_n be the nth Midpoint Rule approximation.
Let T_n be the nth Trapezoidal Rule approximation.
Let B = sup{abs(f''(x)}
Show that one has the estimate,

absolute value( (S_n) - integral f(x)dx ) <= [B((b-a)^2)]/18n^2 where B >= abs(f''(x)) for all x in [a,b].

The hint in my book said to use the identity,
(S_2n) = (2/3)(M_n) + (1/3)(T_n).
Therefore, (S_n) = (2/3)(M_n/2) + (1/3)(T_n/2).

I take the error bounds for M_n and T_n as follows:
absolute value ( (2/3)(M_n/2) - (2/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2
absolute value ( (1/3)(T_n/2) - (1/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2

If I add them together I get,
absolute value ( (2/3)(M_n/2) + (1/3)(T_n/2) - integral f(x)dx ) <= [2B((b-a)^3)]/9n^2, but this isn't what I want. Why won't this work?

Thanks for any of your help.

There is some inconsistency with $ n$ vs $ 2n$ in Simpson's Rule in the literature. Perhaps the authors of your book meant that the error for $ S_{2n}$ is at most $ \frac {B(b \minus{} a)^3}{18n^2}$.

Note that it's $ (b \minus{} a)^3$, not $ (b \minus{} a)^2$. There can't be any error bound of the form $ C\frac {(b \minus{} a)^2}{n^2}$, for any rule.

In any case, this is a strange way to estimate the error for $ S_n$. The whole point of choosing $ 2/3$ and $ 1/3$ as coefficients is that the trapezoidal and midpoint errors mostly cancel each other. Which book are you using?

It's Introduction to Real Analysis by Bartle.

Yeah, I checked it a couple times and it is [B((b-a)^2)]/18n^2 (I even checked two separate editions).  It just doesn't make any sense. Thanks for answering.

In case someone is interested, I'll explain why the error estimate cannot be of the form $ C\frac {(b \minus{} a)^k}{n^k}$. This quantity depends only on $ \frac {b \minus{} a}{n}$ (the size of subintervals, sometimes called the norm of partition). As long as we use partitions with subintervals of length $ \pi/10$, the putative error estimate is the same for, say, $ \int_0^{\pi}\sin^2 x\,dx$ and for $ \int_0^{10^6\pi}\sin^2 x\,dx$. However, the actual error for the second integral is $ 10^6$ times greater than for the first one.

Here's a table I've prepared (made an overhead projector transparency out of it) for a lecture in a second semester calculus class about numerical integration. The integral in question is
\[ \int_0^1\frac{2}{(x\plus{}1)^2}\,dx \]
This example was chosen because its exact value is $ 1,$ making it possible to easily see the approximation error without needing to do arithmetic, because the function isn't polynomial, and because the function is monotone and has the same sense of convexity (making its graph easy to visualize).

In doing the lecture, I have the transparency covered by a piece of paper, and I only reveal one column at a time, when we're ready for it.

$ \begin{tabular}{cccccc}n&Left Endpt.&Right Endpt.&Midpoint&Trapezoid&Simpson\\
1&2.00000000000&.50000000000&.88888888889&1.25000000000& 1.00925925926\\   2&1.44444444444&.69444444444&.96653061224 & 1.06944444444 & 1.00083522298\\
  4&  1.20548752834 & .83048752834  & .99109587296 & 1.01798752834  &1.00005975809\\
  8 & 1.09829170065  &.91079170065  & .99773497986 & 1.00454170065  &1.00000388679\\
 16 & 1.04801334025 & .95426334025 &  .99943119811 & 1.00113834025 & 1.00000024549\\
 32 & 1.02372226918 & .97684726918 &  .99985763849 & 1.00028476918  &1.00000001538\\
 64 & 1.01178995383 & .98835245383 &  .99996439953 & 1.00007120383  &1.00000000096\\
128  &1.00587717668 & .99415842668  & .99999109925 & 1.00001780168  &1.00000000006\\
256 & 1.00293413797 & .99707476297 &  .99999777477 & 1.00000445047  &1.00000000000\end{tabular}$

I find that doubling the number of intervals between rows of this chart makes it easy to see what is happening to the error.

First, the left endpoint Riemann sums and the right endpoint Riemann sums. The left sums are high and the right sums low. Why? Because it's a decreasing function. Hence the error seems to depend on the first derivative. But how does the error depend on $ n?$ Watch what happens as you go from row to row. Doubling $ n$ seems to approximately cut the error in half. Hence it seems that the error in each of these cases is proportional to $ n^{\minus{}1}.$ (And for the two, opposite in sign and approximately equal in size.)

At this point, I ask the class for an idea (other than increasing $ n$) for improving our accuracy. Nearly always, the first suggestion I get is to take the midpoint Riemann sums (that's why that's the next column). I usually have to draw out of them the idea of averaging together the right and left hand sums (hence, the trapezoid rule).

Now we have the midpoint and trapezoid columns. Much more accurate than the the left and right endpoint columns. The midpoint column is low and the trapezoid column is high. Why? Because the graph of the function is concave upwards. (Hence, both of these seem to depend on the second derivative.) Now, what happens when you double $ n?$ I can usually get people to see that doubling $ n$ cuts the error by about a factor of $ 4.$ And that gives us error proportional to $ n^{\minus{}2}.$

Oh, and the error for the trapezoid rule is just about twice as big as the error for the midpoint rule, and oppositely directed. Again I throw this open for suggestion, and I usually get the idea of the weighted average with the $ 2: 1$ ratio of weights. I express the expectation that we have a right to expect that this error will be proportional to $ n^{\minus{}3}$, improving by a factor of $ 8$ for each doubling of $ n.$

And then I show the last column. (One note: the "$ n\equal{}1$" row shows the weighted average of the midpoint and trapezoid rules on the same row, and hence involves $ f(x)$ computed at three points; many would call that Simpson's rule with $ n\equal{}2.$ The same comment applies to all rows.) Now, is the error improving from row to row by a factor of $ 8?$ What? It's better than that? Maybe even (lower down in the chart) a factor of $ 16?$ Could we have an error proportional to $ n^{\minus{}4}?$

-----

I will add that because this all uses the same interval, just different methods and different $ n,$ then I'm not getting at the issue that mlok raised at all. I'm also not addressing what the constants of proportionality are, or how to prove the estimates.


How many different six digits number can be formed by using 1,2,3,4,5 and 7.how many of them is divisible by 11

I assume we can only use each digit once.

[hide="Solution"]There are $ 6!$ or $ 720$ numbers that can be made.

The criterion for divisibility by 11 is that the alternating sum is $ 0$.

ie. $ a\minus{}b\plus{}c\minus{}d\plus{}e\minus{}f\equal{}0$

or rather $ a\plus{}c\plus{}e\equal{}b\plus{}d\plus{}f$

The sum of the numbers $ 1,2,3,4,5,7$ is $ 22$. Thus we need to pick 3 numbers which sum to $ 11$.These will become digits $ (a,c,e)$ and the other 3 remaining digits automatically will have the same sum. There are $ 2$ ways to do this. $ (1,3,7),(2,4,5)$ For each of these $ 2$ ways there are $ 3!$ ways of arranging $ (a,c,e)$, and there are $ 3!$ ways of arranging $ (b,d,f)$. Thus the number of these numbers divisible by 11 is $ 2 \cdot 3! \cdot 3!\equal{}72$[/hide]


Prove that, if $b\not | a$, then $(\frac{a}{b})^{n}$, or, which is the same, $\frac{a^{n}}{b^{n}}$ is not a positive integrer number, for $a, b$ and $n$ positive integrer numbers

[hide]Well if b doesn't divide a, then \frac{a}{b} is not an integer. We then reduce the fraction until it is in simplest form. Then the two resulting numbers, call them $a_{1}$ and $b_{1}$, have no common factors, and never will no matter how high of a power we raise it to. Thus $\frac{a^{n}}{b^{n}}$ can't be an integer.[/hide]

[hide]
We know that a/b cannot be reduced to x/1 or x because a cannot be divided equally into b groups. This means that a/b is a fraction, and no fully reduced fraction in which the denominator doesn't equal 1 can be raised to any power that would make it an integer.
[/hide]

[hide]
Let $k = gcf(a,b)$

$\frac{a}{k}= a'$

$\frac{b}{k}= b'$

Note that $gcf(a',b') = 1$

Note that $b' \neq 1$ since b does not divide a.

$\frac{a}{b}= \frac{a'}{b'}$

$\left(\frac{a'}{b'}\right)^{n}= \frac{a'^{n}}{b'^{n}}$, which cannot be an integer
[/hide]

This is pretty trivial in base $b$.  The size of the "decimal" part increases when we increase $n$, so it can't be an integer.

[hide="Or alternatively:"]
assume $\frac{a^{n}}{b^{n}}\in \mathbb{Z}$, reduce until $(a,b)=1$
$a^{n}=kb^{n}, k\in mathbb{Z}, (k,a)=1 \implies k | a^{n}\implies k | a$
$a=ka_{1}\implies k^{n}a_{1}^{n}=kb^{n}\implies k | b \implies k| a_{1}$ etc. ad infinitum, but this can't be ad infitum because these are integers so there is no solution[/hide]


analytic geometry
graphing lines
slope
slope formula

What is the slope of a line perpendicular to the line containing the points $ (\minus{}1,2)$ and $ (1,\minus{}2)$? Express your answer as a common fraction.

[hide=Easy $ Easy Life!!]Slope= Δy coordinate/Δx coordinate
[hide=Delta meaning (Δ)]Δ=delta or change in the coordinates[/hide]
In this case, it is Δx=2 and Δy=4 so 4/2=2 (change in the coordinate x and y)
Perpendicular to slope 2, so it is 1/2 (or reciprocal) (correct me if I'm wrong)
Ans: 1/2[/hide]


limit
calculus
integration
calculus computations

Let \[ x_n = \frac{1}{2} + \frac{1}{3 \sqrt[3] {2}} + \ldots + \frac{1}{(n+1) \sqrt[3] {n}} ~ , ~~~ n \geq 1 . \]

Find $\displaystyle \lim_{n \to \infty} x_n$.

Merry Christmas to all on ML.         :lol:

Given a combination of direct computation of partial sums, estimates of the error in the partial sums, and Richardson extrapolation, I put the limit as approximately $2.79001829.$

I see no reason to expect this to be a number with a closed-form name.

This was given on a local olympiad in Romania. (asking whether is the sequence convergent)

I figured that either it is too complicated for 11th grade to find the limit, or it has no closed form. ( :( )

Anyway, how exactly did you compute the partial sums?

By "partial sums" I mean $x_n.$ This is a series, after all, and $x_n$ is the $n$th partial sum of that series. The series converges by limit comparison to $\sum_nn^{-4/3}.$ For small enough $n,$ you  can find $x_n$ just by adding up the terms.

The integral test then gives an estimate of the remainder: $3n^{-1/3}.$

Using a spreadsheet, I directly calculated $x_n$ for $n$ up to 256. I then entered $x_n+3n^{-1/3}$ for $n=1,2,4,8,\dots,256$ into a Richardson extrapolation table assuming the exponents are $\frac43,\frac73,\frac{10}3,\dots.$

For a description of Richardson extrapolation, see a numerical analysis textbook.

I'm not saying that it doesn't have a closed-form expression - just that I did not see a way to find one, and that decimal approximation doesn't suggest anything to me.

Hello !

We look for $x>1$ (to use the limit comparison) so that

 $\frac{1}{(n+1)\sqrt[3]{n}}<\frac{1}{n^x}$

Solving this inequation we find $x<\frac{1}{3}+\frac{ln(n+1)}{ln(n)}$

But  $\frac{ln(n+1)}{ln(n)}>1$ for every $n$,  so I can choose $x=\frac{4}{3}$

It is wellknown that for any $t>1$ the sequence $(*)\ A_n (t)=\sum\limits_{k=1}^n \frac{1}{k^t}\nearrow L_t\in R$ (is convergent). But
$\left(\forall\right)k\in \overline {1,n}\ \frac{1}{(k+1)\sqrt [3] {k+1}}<\frac{1}{(k+1)\sqrt [3] k}<\frac{1}{k\sqrt [3] k}$ $\Longrightarrow A_{n+1} \left(\frac 43\right)-1< x_n< A_n \left(\frac 43\right)$. Therefore, the sequence $(x_n)$ is convergent, i.e. $x_n\nearrow l\in R$, where $l\in \left(L_{\frac 43} -1,L_{\frac 43}\right].$

[b]Remark.[/b] In the our school the remarkable sequence (series)  $(*)$ (in the general case $t>0$) is only enunciated in the beginning of the year and is proved against the end with the Lagrange's theorem.


algebra
polynomial
group theory
abstract algebra
inequalities
superior algebra
superior algebra unsolved

Let p q be primes such that p divides q-1. If t is an interger which satisfies the conditions t=/ 1 (mod q) and t^p=1 (mod q) prove: any integer which satisfies these conditions must be congruent (mod q) to one of t,t^2,...,t^(p-1).

Try to show that $ t, t^{2},\ldots, t^{p\minus{}1}$ are distinct modulo $ q$.  Also recall that over a field, a polynomial of degree $ p$ has at most $ p$ roots.

this also confused me. here the right version:

Let $ p,q$ be primes such that $ p$ divides $ q\minus{}1$. Assume $ t$ is an interger which satisfies the conditions $ t\not\equiv 1 (\text{mod~}q)$ and $ t^{p}\equiv 1 (\text{mod~}q)$. Prove: Then any integer which satisfies these conditions must be congruent $ (\text{mod~}q)$ to one of $ t,t^{2},...,t^{p\minus{}1}$.

[quote="Ravi B"][quote="pengyijie"]Let p q be primes such that p divides q-1. Assume the following fact:If t is an interger which satisfies the conditions t=/ 1 (mod q) and t^p=1 (mod q) then any integer which satisfies these conditions must be congruent (mod q) to one of t,t^2,...,t^(p-1).[/quote]
I'm confused by the question.  The second sentence says to "[a]ssume the following fact".  But then what are we trying to prove?[/quote]
Sorry! As English is not my native language,I am not quite familiar with the formula to express my thoughts.

[quote="-oo-"]the $ t$ satisfying the condition constitute a subgroup of the cyclic $ (\mathbb{Z}/q)^{*}$, so it's also cyclic. obviously, a generator has order $ p$. thus, the group has order $ p$, and [i]every[/i] nontrivial element is a generator. this is exactly the claim.[/quote]
Thanks a lot!But how to prove $ (\mathbb{Z}/q)^{*}$ is cyclic.

[quote="-oo-"]http://www.math.umass.edu/~bush/Teach/proof_multgrpofFFiscyclic.pdf
http://www.mathlinks.ro/Forum/viewtopic.php?t=75806
etc.[/quote]
I think it is not necessary that every element of the group be the root of x^m-1,but every element of the group just need to be the root of x^m-1 (mod q) which can be more than m.

[quote="Beginner"]Let $ G$ be finite abelian group in which the number of solutions in $ G$ of equation $ x^{n}\equal{} e$ is at most $ n$ for every positive integer $ n$. Prove that $ G$ must be cyclic group.[/quote]
[quote="ZetaX"]Yes you are right, it can be done more directly:

Let $ \psi(d)$ be the number of elements of order $ d$. Every single $ x$ with order $ d$ gives already $ d$ different $ y$ with $ y^{d}\equal{} e$, namely the powers of $ x$, and exactly $ \phi(d)$ of these powers have order $ d$ again. As a result, $ \psi(d)\leq\phi(d)$.
Now $ n \equal{}\sum_{d|n}\psi(d)\leq\sum_{d|n}\psi(d) \equal{} n$, thus we always have equality in  $ \psi(d)\leq\phi(d)$, especially $ \psi(|G|) \equal{}\phi(|G|) > 0$, which means that the group is cyclic.[/quote]
I am wondering why should $ phi(|G|) > 0$ hold?

[quote="pengyijie"]Thanks a lot! But can you explain it more clearly? What is the field you mentioned? And this is just the exercise behind the chapter of Sylow Theorem. I think it is none of the business of field.[/quote]

The field I mention is $ \mathbb{Z}/q\mathbb{Z}$.  And if you can show that $ t, t^{2},\ldots, t^{p\minus{}1}$ are distinct in $ \mathbb{Z}/q\mathbb{Z}$, then together $ 1, t, t^{2},\ldots, t^{p\minus{}1}$ are $ p$ distinct roots of the polynomial $ x^{p}\minus{}1$ in $ \mathbb{Z}/q\mathbb{Z}[x]$.  Over a field, polynomial has at most $ p$ roots so the $ p$ distinct $ 1, t, t^{2},\ldots, t^{p\minus{}1}$ must be all of them, i.e. any $ y$ with $ y^{p}\equal{} 1$ is one of $ 1, t, t^{2},\ldots, t^{p\minus{}1}$.  Here you don't need to know that $ (\mathbb{Z}/q\mathbb{Z})^{*}$ is cyclic.

[quote="pengyijie"]I think it is not necessary that every element of the group be the root of x^m-1,but every element of the group just need to be the root of x^m-1 (mod q) which can be more than m.[/quote]
We already work  in the field $ \mod q$.

[quote="pengyijie"]I am wondering why should $ phi(|G|) > 0$ hold?[/quote]
$ \phi(n)$ is the number of residues $ \mod n$ that are coprime to $ n$. As $ 1$ is coprime to $ n$, we get $ \phi(n)\geq 1$.

Thanks for everyone!
I still have one question for ZetaX.If $ \phi (n)$ is defined as that,why should $ \psi (d)\leq\phi (d)$ hold? $ \phi (d)$ is only the number of the powers of a fixed x which have order d .

If such $ x$ exists, then it's powers are solutions of $ x^{d}\equal{} e$. As there are $ d$ different ones, and there can be at most $ d$ solutions, we get that the powers of $ x$ are exactly the solutions of $ x^{d}\equal{}e$. Thus we get that $ \phi(d)$ elements of order $ d$ and no more.
If such $ d$ doesn't exist,then $ \psi(d)\equal{}0$ by definition.
In both cases the inequality is true.


inequalities
real analysis
real analysis unsolved

Len be a positive integer.Solve the equation
$\sum_{k=1}^{n}cos^kkx=\frac{n(n+1)}{2}$

Because cos(x)<=1 for every x so  use the inequality we have n=1.
The new equation is
cos(x)=1.It's very easy to solve this:D

Sorry,I don't understand your solution?Can you clarify it?

Your solution doesn't make any sense nor does the wording you use.  I'm pretty sure you didn't solve anything.  Could you explain a lot more?

he just said that:
$\frac{n(n+1)}{2}=\sum_1^k .. \leq n$


[quote="Temperal"]Show that if $ abc \equal{} 1$, then
\[ \sum_{a,b,c}a^2c \plus{} a^3 \plus{} a^2c^3 \plus{} a^4c^2 \le \frac {4}{3}
\]
This was taken from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171990 .[/quote]
I think it wrong because if $ a;b;c>0$ we have $ a^3\plus{}b^3\plus{}c^3 \geq 3abc \equal{}3$
 :blush:

Hmm, I didn't notice that.

I seem to have applied Jensen's incorrectly somehow... can someone show my mistake?


MATHCOUNTS
AMC
AIME
AMC 12
AMC 12 A
AMC 10
AMC 10 B

so heres a post for all texans to introduce yourself, and also if you go to states, please tell

example:

i am ding zhou.
i am 13
i like potatoes
i'm in 8th grade
going to states

Yo soy ding zhou.
Tengo trece anos.
Quieros papas.
Estoy en octavo grado.
estado? si.

bonjour
je m'appelle kevin tian
j'aime les pommes de terre aussi
j'ai treize ans
je vais aller a (with that weird accent thingy on the a) la "state mathcounts"
blah
...
aka
hi guys, i'm kevin tian and pretty much everything ding said can be said for me too
=)

Okay...

I'm Alan Zhao.
Same stuff, but I turned 14 two months ago.

Maybe we could post records?

2007:
1st place Chapter (didn't give us score)
6th place State (38)

2008:
1st place Chapter (45)
awaiting States

24 AMC 8 (I hate lemmings :wallbash: )
132 AMC 10A
awaiting AIME I scores

fine
2007: first place chapter (no scores)
7th place state (38)
amc12a: 114 (nooooo!!! 3 careless)
amc10b: 136.5
aime: 7 (ouch ouch ouch)

2008:
1st place chapter (46)
25 amc8
144 amc10a (noooooo!!! 1 careless)
118.5 amc12b (noooooooo!!!!! 3 careless)

bonjour je m'appelle cognos599
j'habite a indiana
j'adore les maths.

7th grade
AMC 8: 21
4th place chapter (? score)
not telling state  :P
AMC 10A: 117
AIME II: 2
ARML: 1  :rotfl:

8th grade:
AMC 8: 23 :huh:
3rd place chapter (45)
not telling state cuz you guys haven't state yet.
AMC 10A/B = 121.5(3 careless)/126(dang 2 careless).
AIME: lol not telling 'til tomorrow.
ARML: we'll see.  :P

[quote="distracted523"]Apparently cognos is a Texan at heart :lol:

Wow, Kevin, you really attribute a lot to careless mistakes. If I don't make any careless mistakes, I anticipate a <fill in the blank> on AIME.[/quote]

uhhh... lol
yeah, i do make a lot of careless mistakes T.T
but seriously, what would you call mistakes like "60/3=11" and "volume of pyramid is bh/6"?

[quote="cognos599"]that's nothing on [size=150]AIME [/size](i am trying to stay vague), i had the right numbers for something but i added them wrong. still no, i am not a Texan at heart. i am a hoosier and hoosiers don't roll well with texans. cuz we're like all time mc rivals. the hoosiers would hurt me if they found me posting on the texas forum.  :P[/quote]

that happened twice last year on the aime for me, did something random like 2000+500 = 2300 or something like that on problem 1 and screwed up division on problem 10-ish

on number 9 this year, i made a really dumb mistake.

[quote="cognos599"]this quote has been edited thanks to uldivad[/quote]

ouch
but edit your post, it gives information about the content of the test

[quote="uldivad9"]Well, Ra, now you have to edit your post. :D[/quote]
lol, good point
dang, so the texas forum has been invaded by michigan and indiana
*runs*

Let's take personal information to the private forum.

http://www.artofproblemsolving.com/Forum/index.php?f=178

[quote=countdown1234]
 Also, I live in Mars. Does that count as Texas?[/quote]

No.

[quote=countdown1234]Hello
 I am Classified User. Here are my scores:
 10th: 0 on AMC 8 ( no AMC 10 for me...)
 11th: -1 on AMC 8 (better, but still...) Yay made AMC 10 but fail got a 2 on it :oops_sign:  :mad:
 12th: -30 on AMC 8 (Yay!) and -12 on AMC 10 (yay!)
  A lot of significant improvement, if I do say so myself. All that hard work has helped.
 Also, I live in Mars. Does that count as Texas?[/quote]

Hello shankarmath!

 Fine here is me:
Hello I am classified user
 I can get a 2 on a math counts sprint with luck
 I am really really good at math (2 on mc is op)
 I am really really bad at math (2 on mc is bad)
 I am really really tired of typing
 so I will just stop now

[quote=countdown1234]Why do you guys have to go telling everyone...
 wait ancy you are naveen?[/quote]

[quote=Ancy]Naveen Mukkatt, 7th grade, 12 years.
Made states only to get 72nd ;-;
AMC 10A: 97.5
Why do I fail on actual tests[/quote]

[quote=Ancy]Naveen Mukkatt, 7th grade, 12 years.
Made states only to get 72nd ;-;
AMC 10A: 97.5
Why do I fail on actual tests[/quote]

Who is Naveen?
ewwwwww lol :P :D

Boy child I saw you today and you're just like who's Naveen
You wanna fight
Fite me

[quote=ac_math]andrew cai
10 yrs old
4th grade
Sugar Land(near Houston)

3rd Grade:
amc8:22[/quote]

Finally! Another person who lives in Sugar Land!

[quote=Ancy]A lot of people that posted in this thread are in Sugar Land...
Probably more than every other place in Texas XD[/quote]

sugar land too OP with quail valley, fort settlement, and sartarita ms.


can you change the look of a weblink like [url]http://www.artofproblemsolving.com/[/url] to a smiley or at least some other text that doesn't look like the actual URL address? because i've seen other people do it and i've been wondering how.

not to rickroll anybody or anything, just to put a more interesting link to my blog from my signature.

Yes, you just have to put the following code:
[code][url=THE LINK URL]CLICK TEXT[/url][/code]

You have to put an actual URL where it says THE LINK URL, so I will just put the one for this page:

[code][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=285659]CLICK TEXT[/url][/code]

And it comes out like this:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=285659]CLICK TEXT[/url]

You can put smileys in the CLICK TEXT too (I think).


Let the sum of digits of  $ \overline{999...9991}^{2} $, where $9$ occurs $222$ times be $A$. Prove that $A = 2008$.

[quote="gouthamphilomath"]Let the sum of digits of  (9999…99991)2  (9999....999 square) where 9 occurs 223 times be A
Prove that A = 2008[/quote]

Are you speaking about $ \overline{999...999}^2$ or $ \overline{999...9991}^2$ ? :

If the required number is $ N \equal{} \overline{999...999}$ with $ 223$ digits $ 9$, then :
$ N \equal{} 10^{223} \minus{} 1$ and $ N^2 \equal{} 10^{446} \minus{} 2\cdot 10^{223} \plus{} 1$ $ \equal{} \overline{999...9998000...0001}$ with $ 222$ digits $ 9$ and $ 222$ digits $ 0$. And sum of digits of $ N^2$ is $ 9\cdot 222 \plus{} 9 \equal{} 2007$

If the required number is $ N \equal{} \overline{999...9991}$ with $ 223$ digits $ 9$, then :
$ N \equal{} 10^{224} \minus{} 9$ and $ N^2 \equal{} 10^{448} \minus{} 18\cdot 10^{224} \plus{} 81$ $ \equal{} \overline{999...99982000...00081}$ with $ 222$ digits $ 9$ and $ 222$ digits $ 0$. And sum of digits of $ N^2$ is $ 9\cdot 222 \plus{} 8 \plus{} 2 \plus{} 8 \plus{} 1 \equal{} 2017$

:(

So none is $ 2008$, IMHO

In order to obtain $ 2008$, maybe you thought to $ N \equal{} \overline{999...9991}$ with $ 222$ digits $ 9$, then :
$ N \equal{} 10^{223} \minus{} 9$ and $ N^2 \equal{} 10^{446} \minus{} 18\cdot 10^{223} \plus{} 81$ $ \equal{} \overline{999...99982000...00081}$ with $ 221$ digits $ 9$ and $ 221$ digits $ 0$. And sum of digits of $ N^2$ is $ 9\cdot 221 \plus{} 8 \plus{} 2 \plus{} 8 \plus{} 1 \equal{} 2008$

Yeah you are right. Sorry for the wrong question. It should have been 222 nines followed by a 1 the whole square. Is it right then? If it consists of only nines as you told and I did a typing mistake, the sum of digits is a multiple of nine and can't be 2008. Thanks for correcting.


Euler
function
number theory
totient function

Please solve the following problem of Combinatorics:
[b]Find the number of positive integers less than 29106 and prime to it.[/b]

This is a typical Inclusion-Exclusion problem.

Counting the numbers not prime with $29106$ : add the numbers multiple of $2,3,7$ or $11$, sustract the numbers multiple of $2*3,2*7,2*11,3*7,7*11$, add the number multiple of $2*3*7, 2*3*11,3*7*11$, substract .....

[hide]Final answer may be $7560$ [/hide]

tµtµ, I got the same answer :D . Just checking my answer.

I always wonder how you write that 'µ'

[quote="amirhtlusa"]why don't you just find $\phi (29106)$?? ;)[/quote]

Ah what is it? I think it is Euler's function. Can you tell me about it?

if $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_i^{\alpha_i}$ then the number of integers relativly prime to $n$ are found by the Euler Phi function:

$\phi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})...(1-\frac{1}{p_i})$

OK, the answer comes the same this way too.
I'll find how this formula came. No need to tell.
(Obviously it came before my birth :rotfl: )

[quote="amirhtlusa"]why don't you just find $\phi (29106)$?? ;)[/quote]

Yes of course, but I assumed that someone asking this question didn't know about totient function (and it proved correct  ;) ).


geometry
incenter
geometric transformation
reflection
trigonometry
circumcircle
inradius

I was solving a problem and I reduced it to prove the following

Given a triangle $\triangle ABC$, let $I$ be the incenter and let $U,V,W$ be the reflections of $I$ trhough $BC, CA$ and $AB$ respectively. Show that $AU, BV, CW$ conccur.

I posted this on the spanish forums, and Grobber told me the following generalization:

Given a triangle $\triangle ABC$, let $I$ be the incenter and consider the lines trough $I$, perpendicular to $BC,CA,AB$. We choose $X, Y, Z$ respectively on the lines, such that the segments $IX, IY, IZ$ are congruent and equally oriented (were positive is to be considered as going from $I$ to the respective side of the triangle). Show tha $AX, BY $and $CZ$ conccur.

The thing is that the proof that grobber gave me used some projective geometry I couldnt understand (he said something about proving that the mapping $BY\rightarrow CZ$ is a projective map from the pencils of lines trhough $B$ to the pencils of lines trhough $C$, but actually I dont know what a projective mapping is  :( ) Well, however, i would like to see if some of you could help me finding a sintetic proof, or helping understand grobbers proof. Thanks :)

Ouch - I now remember having been asked this question by you via private messaging. I completely forgot the message in my mailbox and never came to answering it. Sorry!

To solve your problem, I first will start with a proof of the Jacobi theorem.

We will use directed angles throughout the following post.

[color=blue][b]Jacobi theorem.[/b] Let ABC be a triangle, and let X, Y, Z be points in its plane such that < YAC = < BAZ, < ZBA = < CBX and < XCB = < ACY. Then, the lines AX, BY, CZ concur at one point.[/color]

[i]Proof.[/i] Call

< CAB = A; < ABC = B; < BCA = C;
< YAC = < BAZ = x;
< ZBA = < CBX = y;
< XCB = < ACY = z.

Since the lines AX, BX, CX concur at one point (namely, the point X), the trigonometric version of the Ceva theorem yields

$\frac{\sin\measuredangle CAX}{\sin\measuredangle XAB}\cdot\frac{\sin\measuredangle ABX}{\sin\measuredangle XBC}\cdot\frac{\sin\measuredangle BCX}{\sin\measuredangle XCA}=1$.

But

< ABX = < ABC + < CBX = B + y;
< XBC = - < CBX = -y;
< BCX = - < XCB = -z;
< XCA = < XCB + < BCA = z + C = C + z.

Thus, the equation above becomes

$\frac{\sin\measuredangle CAX}{\sin\measuredangle XAB}\cdot\frac{\sin\left(B+y\right)}{\sin\left(-y\right)}\cdot\frac{\sin\left(-z\right)}{\sin\left(C+z\right)}=1$.

Similarly, we can find

$\frac{\sin\measuredangle ABY}{\sin\measuredangle YBC}\cdot\frac{\sin\left(C+z\right)}{\sin\left(-z\right)}\cdot\frac{\sin\left(-x\right)}{\sin\left(A+x\right)}=1$;
$\frac{\sin\measuredangle BCZ}{\sin\measuredangle ZCA}\cdot\frac{\sin\left(A+x\right)}{\sin\left(-x\right)}\cdot\frac{\sin\left(-y\right)}{\sin\left(B+y\right)}=1$.

Multiplying all these three equations together, we get

$\left(\frac{\sin\measuredangle CAX}{\sin\measuredangle XAB}\cdot\frac{\sin\left(B+y\right)}{\sin\left(-y\right)}\cdot\frac{\sin\left(-z\right)}{\sin\left(C+z\right)}\right)\cdot\left(\frac{\sin\measuredangle ABY}{\sin\measuredangle YBC}\cdot\frac{\sin\left(C+z\right)}{\sin\left(-z\right)}\cdot\frac{\sin\left(-x\right)}{\sin\left(A+x\right)}\right)$
$\cdot\left(\frac{\sin\measuredangle BCZ}{\sin\measuredangle ZCA}\cdot\frac{\sin\left(A+x\right)}{\sin\left(-x\right)}\cdot\frac{\sin\left(-y\right)}{\sin\left(B+y\right)}\right)=1$,

what immediately simplifies to

$\frac{\sin\measuredangle CAX}{\sin\measuredangle XAB}\cdot\frac{\sin\measuredangle ABY}{\sin\measuredangle YBC}\cdot\frac{\sin\measuredangle BCZ}{\sin\measuredangle ZCA}=1$.

But this equation implies, using the trigonometric version of the Ceva theorem, that the lines AX, BY, CZ concur at one point. Thus, the Jacobi theorem is proven.

[color=blue][b]Kariya theorem.[/b] Let I be the incenter of a triangle ABC, and let D, E, F be the points where the incircle of triangle ABC touches the sides BC, CA, AB. We will use directed segments on the lines ID, IE, IF; hereby, these lines will be directed in such a way that the segments ID, IE, IF will all be positive.

Now, let X, Y, Z be three points on the lines ID, IE, IF such that IX = IY = IZ. Then, the lines AX, BY, CZ concur at one point.[/color]

[i]Proof.[/i] Being the points of tangency of the incircle of triangle ABC with the sides AB and BC, the points F and D are symmetric to each other with respect to the angle bisector of the angle ABC, i. e. with respect to the line BI. Thus, the triangles BFI and BDI are inversely congruent. Now, the points Z and X are corresponding points in these two inversely congruent triangles, since they lie on the (corresponding) sides IF and ID of these two triangles and satisfy IZ = IX. Corresponding points in inversely congruent triangles form oppositely equal angles; thus, < ZBF = - < XBD. In other words, < ZBA = < CBX. Similarly, < XCB = < ACY and < YAC = < BAZ. Thus, by the Jacobi theorem, it follows that the lines AX, BY, CZ concur at one point. This proves the Kariya theorem.

[b]PS.[/b] There is much more to say about the point of concurrence of the lines AX, BY, CZ. For instance, one can show that the isogonal conjugate T of this point with respect to triangle ABC lies on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. More precisely, the point T lies on the line OI and satisfies the relation $\frac{IT}{TO}=t\frac{r}{R}$, where r is the inradius and R is the circumradius of triangle ABC, and where $t=\frac{IX}{ID}=\frac{IY}{IE}=\frac{IZ}{IF}$. This all is a nice exercise for proving (although way more difficult than the concurrence of the lines AX, BY, CZ). There is a nice synthetic proof, as always. ;)

[b]PPS.[/b] Another consequence of the Jacobi theorem is the Kiepert theorem:

[color=blue][b]Kiepert theorem.[/b] Let ABC be a triangle, and let BXC, CYA, AZB be three directly similar isosceles triangles erected on its sides BC, CA, AB (with bases BC, CA, AB, respectively). Then, the lines AX, BY, CZ concur at one point.[/color]

In fact, this trivially follows from the Jacobi theorem, since, as the isosceles triangles BXC, CYA, AZB are directly similar, they all have the same base angle, i. e. we have < YAC = < BAZ = < ZBA = < CBX = < XCB = < ACY, so that, particularly, < YAC = < BAZ, < ZBA = < CBX and < XCB = < ACY, and we can apply the Jacobi theorem.

  Darij

Thanks a lot!, all that can be quiet usefull :). I'll try solving the problem you wrote on the ps, (actually, to finish the problem I was originally trying, i must prove it  :D). However, thanks again.

Darij, can we define the isogonal conjugate of a line (with respect to the triangle [i]ABC[/i] ) ? Or, if not , what is the geometric loci of the isogonal conjugates of the points on a segment (with respect to [i]ABC[/i] ) ?

[quote="Andrei"]Darij, can we define the isogonal conjugate of a line (with respect to the triangle [i]ABC[/i] ) ?[/quote]

Yes, of course. Given a line g which doesn't pass through any of the vertices of a triangle ABC, the locus of the isogonal conjugates of all points on the line g with respect to triangle ABC is a circumconic of triangle ABC (i. e., a conic which passes through the vertices A, B, C of triangle ABC). This circumconic is called the [b]isogonal conjugate of the line g with respect to triangle ABC[/b]. In particular, the isogonal conjugate of the line OI with respect to triangle ABC is the so-called [url=http://mathworld.wolfram.com/FeuerbachHyperbola.html][b]Feuerbach hyperbola[/b] of triangle ABC[/url].

Of course, the fact that the isogonal conjugate of the point of concurrence of the lines AX, BY, CZ lies on the line OI is therefore equivalent to the fact that the point of concurrence of the lines AX, BY, CZ itself lies on the Feuerbach hyperbola of the triangle ABC. But my proof of the first fact doesn't use conics. ;)

  Darij

[quote="Leonhard Euler"]What is definition of sin$ \measuredangle ABC$?  $ \measuredangle ABC \equal{}\measuredangle ABC\plus{}180$ but sin$ x$$ \equal{}\minus{}$sin$ (x\plus{}180)$[/quote]

Yes, the proof above is arrangement-dependent. There are better ones anyway, e. g. at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=870445#870445]http://www.mathlinks.ro/Forum/viewtopic.php?t=154396 post #9[/url].

As for the other topic where you posted lately: I think your proof can be improved to make it arrangement-independent, I will answer in a few days (I have unbelievably bad internet access these days).

  Darij


Thanks you. Can you reply me soon.

$\left\{ \begin{array}{l} xy^2-2y +3x^2= 0 \\ yx^2 + y^2 + 2x =0 \end{array} \right.$

You can do it by many way?


Let triangle ABC. Prove that
$ m_a\plus{}m_b\plus{}m_c\leq\sqrt{3p^2\plus{}\frac{1}{2}[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2]}$
 :roll:

[quote="mufc"]Let triangle ABC. Prove that
$ m_a + m_b + m_c\leq\sqrt {3p^2 + \frac {1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]}$
 :roll:[/quote]

I think you should use the identity ${ \frac {1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]}= \frac{\sum a^{3}-3abc}{a+b+c}$

What does $ m_a$ denotes ?

My result is   $ m_a\plus{}m_b\plus{}m_c\leq\sqrt{3p^2\plus{}\frac{3}{4}\left[\left(a\minus{}b\right)^2\plus{}\left(b\minus{}c\right)^2\plus{}\left(c\minus{}a\right)^2\right]}$.
Is it right?

Using Schwartz inequality, we have:
$ \left(m_a\plus{}m_b\plus{}m_c\right)^2\leq3\left(m_a^2\plus{}m_b^2\plus{}m_c^2\right)$
$ \equal{}3\left(\frac{b^2\plus{}c^2}{2}\minus{}\frac{a^2}{4}\plus{}\frac{c^2\plus{}a^2}{2}\minus{}\frac{b^2}{4}\plus{}\frac{a^2\plus{}b^2}{2}\minus{}\frac{c^2}{4}\right)\equal{}\frac{9(a^2\plus{}b^2\plus{}c^2)}{4}$
$ \equal{}\frac{3a^2\plus{}3b^2\plus{}3c^2\plus{}6ab\plus{}6bc\plus{}6ca\plus{}6a^2\plus{}6b^2\plus{}6c^2\minus{}6ab\minus{}6bc\minus{}6ca}{4}$
$ \equal{}3\left(\frac{a\plus{}b\plus{}c}{2}\right)^2\plus{}\frac{3}{4}\left[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\right]$
$ \equal{}3p^2\plus{}\frac{3}{4}\left[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\right]$
$ \equal{}>m_a\plus{}m_b\plus{}m_c\leq\sqrt{3p^2\plus{}\frac{3}{4}\left[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\right]}$

Please show me my mistakes, thanks!

It's true, but the estimate $ \left(m_a\plus{}m_b\plus{}m_c\right)^2\leq3\left(m_a^2\plus{}m_b^2\plus{}m_c^2\right)$ is not strong. :wink:

[quote="arqady"]It's true, but the estimate $ \left(m_a \plus{} m_b \plus{} m_c\right)^2\leq3\left(m_a^2 \plus{} m_b^2 \plus{} m_c^2\right)$ is not strong. :wink:[/quote]

Thank you, argady.
Merry Christmas!


topology
advanced fields
advanced fields unsolved

If $ O \in \mathbb{R}$ is an open set, show that $ O$ is union of a countable collection of disjoint open segments (or form $ (a,b)$) or open rays (of form $ (\minus{}\infty, a)$ or $ (a, \infty)$

Here's an idea.

Define a relation $ \sim $ on $ O$ such that $ a\sim b$ iff there is an open interval $ I \subset O$ which contains both $ a$ and $ b$.  Prove that $ \sim $ is an equivalence relation on $ O$.  Prove that each of the equivalence classes induced by $ \sim $ are open intervals (possibly unbounded).  Countability comes from the fact that each open interval contains a rational number.


Dear all, I had difficulties in solving the below 3 questions with limit. Can anyone shed me some light on how to solve this? The Chapter on Limit seem a bit hard to me.

Here are some of my lecture notes on calculus.  This section should have what you're looking for:

Thank You, JRav for your lecture notes. That is very helpful indeed.


Let $0 < a,b,c,d \le \frac{1}{2}$. Show that

\[
\frac{abcd}{(1-a)(1-b)(1-c)(1-d)} \le \frac{a^4 + b^4 + c^4 + d^4}{(1-a)^4 + (1-b)^4 + (1-c)^4 + (1-d)^4}
\]

This one appeared in Titu's legendary MOP 2004 lecture (legendary for being so hard) for which I had been searching for over a year until I found it today as I was cleaning my room. :D

This one appears here not the first time... http://www.mathlinks.ro/Forum/viewtopic.php?t=1834 , http://www.mathlinks.ro/Forum/viewtopic.php?t=41543 , http://www.mathlinks.ro/Forum/viewtopic.php?t=27254 . Haven't checked whether any of the solutions presented is true...

  darij

Why is this called Sweden 2002 ??
Could someone please post a link or some information about on
contest it was given?


quadratics
number theory proposed
number theory

Let o(k) denote the number of positive integer n satisfying gcd(n,k)=1 and n <= k. Suppose that o(5^m-1)=5^n-1 for some positive integer m,n. Prove that gcd(m,n)>1. :D

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=quadratic+reciprocity&t=4983

Pierre.

Sorry for posting it again :blush: You gave a very nice solution :D

I am almost sure it comes from Taiwan 2001


algebra
polynomial
invariant
blogs
linear algebra

Let $ \mathbf{M} \in \mathcal{M}_n(F)$ where $ F$ is a perfect field.  The characteristic polynomial of $ \mathbf{M}$ clearly has coefficients in $ F$.  Is this also true of the minimal polynomial of $ \mathbf{M}$ (in particular, when the eigenvalues do not belong to $ F$)?

(My understanding of matrices whose minimal and characteristic polynomials don't coincide is quite poor.)

Yes. [url=http://en.wikipedia.org/wiki/Frobenius_normal_form]Rational canonical form[/url] exists over arbitrary fields.

Ah.  My question was motivated by rational canonical form, but I didn't notice that the last invariant factor was in fact the minimal polynomial.  That clears a lot up (it'll help me immensely in my next blog post).  Thanks.


what's the maximum number of non-overlaping regions formed by 100 intersecting lines?

[color=blue][size=75]kyyuanmathcount: Could you be a little more specific in the title in the future? Most problems here are mathcounts problems.[/size][/color]

[quote="M4RI0"]$5051$ if i'm not wrong. Just use induction :D[/quote]

That's not wrong. I got that too.

[hide]For n lines the answer is $\frac{n(n+1)}{2}+1$.

So $n=100$, then the answer is $5051$

$5051$[/hide]

[hide]5050.

for x number of lines, m.l.=x(x+1)/2.
where m.l.=maximum lines

[/hide]

EDIT: I realized that i m wrong. I see my mistake[/b]


algebra
polynomial
function
geometry unsolved
geometry

hi,

how to prove that maximum number of pieces obtained from dissecting the plane by k n-sided regular polygons can be expressed as a polynomial W(k) of degree 2? how about higher dimensions and any convex objects?
generalization of this problem is true for any convex object in any dimension d. the only problem i have is to prove that function returning max. num. of pieces is a polynomial.

check out http://mathworld.wolfram.com/PlaneCutting.html so you'll know what i mean.

A simple question: prove that two equal convex figure on the plane have the finite number of border intersections.


Let a,b,c  are postive real numbers, proof
$ \frac {a \plus{} b}{c^2} \plus{} \frac {b \plus{} c}{a^2} \plus{} \frac {a \plus{} c}{b^2}\geq{\frac {2}{a} \plus{} \frac {2}{b} \plus{} \frac {2}{c}}$

Thank you ...

It is because:

$ \frac{b}{a^2}\plus{}\frac{a}{b^2}\geq \frac{1}{a}\plus{}\frac{1}{b}$

$ \Leftrightarrow \frac{(a\minus{}b)^2(a\plus{}b)}{a^2b^2} \geq 0$

Adding cyclic permutations yield the result.

nice inequality,
suppose that, $ a\geq b\geq c$
then By Chebychev,
$ S\equal{}\sum \frac{a\plus{}b}{c^2} \geq \frac{2}{3}(a\plus{}b\plus{}c)\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)$
By Cauchy-Schwarz,
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\geq \frac{1}{3}\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)^2\geq \frac{1}{3}\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\left(\frac{9}{a\plus{}b\plus{}c}\right)$
then
$ S \geq \frac{2}{a}\plus{}\frac{2}{b}\plus{}\frac{2}{c}$

[quote="peine"]nice inequality,
suppose that, $ a\geq b\geq c$
then By Chebychev,
$ S \equal{} \sum \frac {a \plus{} b}{c^2} \geq \frac {2}{3}(a \plus{} b \plus{} c)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)$
By Cauchy-Schwarz,
$ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\geq \frac {1}{3}\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2\geq \frac {1}{3}\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\left(\frac {9}{a \plus{} b \plus{} c}\right)$
then
$ S \geq \frac {2}{a} \plus{} \frac {2}{b} \plus{} \frac {2}{c}$[/quote]
I got it ,thanks very much

hello,
multiply both sides bye $ \frac{1}{a} \plus{}\frac{1}{b} \plus{}\frac{1}{c}$
we have to prove,
$ (\frac{a}{c^2} \plus{}\frac{b}{a^2} \plus{}\frac{c}{b^2})(\frac{1}{a} \plus{} \frac{1}{b}\plus{} \frac{1}{c})
   \plus{}  (\frac{b}{c^2} \plus{} \frac{c}{a^2} \plus{} \frac{a}{b^2})(\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c})
                       \ge 2 \cdot (\frac{1}{a} \plus{}\frac{1}{b} \plus{} \frac{1}{c})^2$
which is true cuz by cauchy,
$ LHS \ge (\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}) \plus{}  (\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}) \equal{}  2 \cdot (\frac{1}{a} \plus{}\frac{1}{b} \plus{} \frac{1}{c})^2$ :)

peine i have a doubt for chev in your proof the second term is $ \frac{b\plus{}c}{a^2}$ which will have the orders of terms in LHS as $ \frac{1}{c^2} , \frac{1}{a^2} ,\frac{1}{b^2}$ which is not similarly sorted by assumption


analytic geometry
graphing lines
slope
calculus
calculus computations

Line $L$ is tangent to curve $y = \frac{1}{x-3}$ and cuts the $x$-axis at $x=5$. Find the slope of $L$.

The slope of the tangent line to $y$ is $-\frac{1}{(x-3)^{2}}$ and it is tangent to the point

$\left(x, \frac{1}{x-3}\right)$.

If we want to move down $\frac{1}{x-3}$ units to the $x$-axis and our slope is $-\frac{1}{(x-3)^{2}}$, we will need to move

$\frac{-\frac{1}{x-3}}{-\frac{1}{(x-3)^{2}}}= x-3$

units horizontally. Then our $x$-intercept will be $x+(x-3) = 2x-3 = 5$ so our $x$ value must be $4$ and the slope $-\frac{1}{(4-3)^{2}}=-1$.


The date for ARML next year is June 4 and 5.

Please note: this does conflict with an SAT date. Plan accordingly.

I am just wondering how sure you are about that?  I know this reply is late, but I just saw this post now.

Sure about the ARML date or the SAT date?
I can tell you for sure that it does conflict with the SAT, but I don't know about ARML.
SAT dates can be found [url=http://www.collegeboard.com]here.[/url]

same!  :P   don't have to worry about SATs anymore!

I'm still in grade 8... think ill make it to arml?
if i do, meet u there!!!!!!!!
(im on waterloo, black hair, fat) :P  :wink:  :rotfl:

yeah...my first ARML...hope it gotta be great


Dear Mathlinkers
I want many exercises about symedians. Can you help me ?

[geogebra]f7a25a2f24721b794aff0d6f3d8d6d67a93d781f[/geogebra]

[color=orange][b]Theorem[/b][/color] [Steiner's theorem]

If $ ABC$ is a triangle and $ \angle BAM\equal{}\angle NAC$ then prove that: $ \boxed{\frac{BM}{MC} \cdot \frac{BN}{NC}\equal{}\left( \frac{AB}{AC} \right) ^2}$

Note: The median and the symendian are isogonal.

Using this theorem, prove that the symedians of a triangle are concurrent.  :)

[quote="moldovan"][geogebra]f7a25a2f24721b794aff0d6f3d8d6d67a93d781f[/geogebra]

[color=orange][b]Theorem[/b][/color] [Steiner's theorem]

If $ ABC$ is a triangle and $ \angle BAM \equal{} \angle NAC$ then prove that: $ \boxed{\frac {BM}{MC} \cdot \frac {BN}{NC} \equal{} \left( \frac {AB}{AC} \right) ^2}$

Note: The median and the symendian are isogonal.

Using this theorem, prove that the symedians of a triangle are concurrent.  :)[/quote]
Oh, I know it. But it is very simple.I want some Problems about it. Who can help me, please ?


function
topology
Functional Analysis
real analysis
real analysis unsolved

Let $ 1\leq p < \infty$. For which values of $ p$ is the following property true :
for every continuous linear function $ f$ on $ l^p$  there exists $ x \in l^p$ such that $ \parallel{}x\parallel{}\equal{}1$ and $ f(x)\equal{}1$ ?

The same question for $ L^p$.

* I think this is true whenever the space is reflexive, i.e. $ p>1$.

In reflexive spaces, use weak* - compactness (Banach-Alaoglu). Indeed, continuous linear functionals are also continuous in the weak topology, which coincides with the weak* topology when the space is reflexive.

In the opposite direction, consider the sequence $ (1 \minus{} 1/n)\in\ell^{\infty}$ as a functional on $ \ell^1$.

More general result (James's reflexivity theorem): a Banach space $ X$ is reflexive iff every continuous linear functional attains its norm on the unit ball of $ X$.


A finite sequence $a_{1}, a_{2}, \ldots, a_{n}$ is called $p$-balanced if any sum of the form $a_{k}+a_{k+p}+a_{k+2p}+\cdots$ is the same for $k = 1, 2, \ldots, p$. Prove that if a sequence with $50$ members is $p$-balanced for $p = 3, 5, 7, 11, 13, 17$, then all its members are equal to zero.

[hide="hint 1"]
Generating functions.[/hide]
[hide="hint 2"]
$3+5+7+11+13+17 = 56$[/hide]


Without any multiplication, write down the prime factorization of $67!+68!+69!$.


"There are x real numbers, the sum of any 10 of them is greater than the sum of others. Prove that all the numbers are positive." Now we know x is a integer and not equal to 20. Find x.

this is not true. take ten 1's and one -1.

The problem is to find a value of $ x$ such that the statement holds.

well that just makes the problem ever so deeper  :)

19 easily works. If $ a$ is any of those numbers, split the other 18 into two groups of 9 (it doesn't matter how). Let the sum of the first group be $ S$ and the sum of the second group be $ T$. Then $ a \plus{} S > T$ and $ a \plus{} T > S$. That gives $ a>0$.

And a counterexample for 18 is $ \{ \minus{} 1, 65, 66, 67, \ldots, 81\}$.  The sum of the smallest ten of these is $ \minus{} 1 \plus{} 65 \plus{} \ldots \plus{} 73 \equal{} 620$ and the sum of the largest eight is $ 74 \plus{} \ldots \plus{} 81 \equal{} 620$.  (Or, for that matter, take 17 $ 2$s and one $ \minus{}1$.)


[quote]Determine $(x,y)$ that satisfy the following system $\{\begin{array}{l}y^{6}+y^{3}+2x^{2}=\sqrt{xy-x^{2}y^{2}}\\x^{3}+y^{3}+\frac{1}{2}\ge 2x^{2}+\sqrt{1+(2x-y)^{2}}\end{array}$[/quote]

Plz show me a way :)

I presume $x$ and $y$ are real numbers. We want to determine the pairs ($x$,$y$) (if they exists) which satisfies the equation/inequality

\begin{eqnarray}y^{6}+y^{3}+2x^{2}& = & \sqrt{z-z^{2}}, \\ x^{3}+y^{3}+{\textstyle \frac{1}{2}}& \geq & 2x^{2}+\sqrt{1+(2x-y)^{2}}, \end{eqnarray}

where $z = xy.$ From equation (1) we obtain $z-z^{2}\geq 0$, i.e. $z \in [0,1]$. This fact combined with equation (1) implies that

$(3) \;\; y^{6}+y^{3}+2x^{2}\; \leq \;{\textstyle \frac{1}{2}}.$

Futhermore inequality (2) is equivalent to

$x^{3}+y^{3}-2x^{2}\; \geq \;-{\textstyle \frac{1}{2}}+\sqrt{1+(2x-y)^{2}},$

hence

$(4) \;\; x^{3}+y^{3}-2x^{2}\; \geq \;{\textstyle \frac{1}{2}}.$

Combining (3) and (4) we obtain

$y^{6}+y^{3}+2x^{2}\; \leq \; x^{3}+y^{3}-2x^{2},$

In other words

$x^{3}-4x^{2}\; \geq \; y^{6}\; \geq \; 0,$

i.e.

$x^{2}(x-4) \; \geq \; 0.$

Therefore $x=0$ or $x \geq 4$ (which is impossible according to inequality (3)). So $x = 0$, giving $z = 0$ and

$y^{3}(y^{3}+1) \; = \; 0$

by equation (1). This means that $y = 0$ or $y =-1$, giving ($x$,$y$) = (0,0) and ($x$,$y$) = (0,-1) as the only possible solutions. But neither of these two solutions satisfies inequality (2).

[u][b]Conclusion:[/b] [/u]The system equation/inequality (1)-(2) has no solution in real numbers $x$ and $y$.


There are 16 playing cards on the table:

[b]Heart:       A, 4, Q
Spade:      2, 3, 4, 7, 8, J
Club:         4, 5, 6, Q, K
Diamond:  A, 5[/b]

One card is chosen and two perfect logicians A and B with A given the number, B given the suit, respectively, then have the following conversation:

[b]A: I cannot determine the card.
B: I knew that.
A: Now I can determine it.
B: so can I.[/b]

If the above statements are true, which one is the card?

[hide] Since $A$ cannot determine the card , it means that the number given to him must have more than one suit . Cancelling all those cards with only one type of suit , theres only

Heart : A , 4 , Q
Spade : 4
Club:4 , 5 , Q
Diamond :A, 5

Then $B$ says he knew it because all the number in the suit of card given to him must have more than than one suit .So cancel again the remainding we will have

Heart : A , 4 , Q
Diamond :A , 5

After that $A$ says he can determine it and this can only happen if the remaining cards above have exactly one suit . Which means both diamond A and Heart A are gone . Immediately , $B$ says he knew the answer and this can only happen if the suit he get have only one choice left to guess which is $\text{Diamond 5}$  :D [/hide]

[quote="Jan"]That's exactly how I did it. Nice explanation, it's not so easy to explain a logic question[/quote]

You are such a perfect logician!!


geometry
circumcircle
combinatorics solved
combinatorics

Let S be a set of n distinct points in the plane, where n \geq 2. Show that there are two distinct points P_i and P_j of S such that the circle on diameter P_iP_j contains in its interior and on its circumference a total of at least li(n/3) distinct points of S, where li(m) denotes the greatest integer \leq m.

Consider circle C with smallest possible radius and Ps \in C (s=1..n) (russian name is Young's circle).
We have two alternatives:
1) there exists segment PiPj and PiPj is a diameter of C ==> Pi and Pj are points we looking for.
2) there exists acute-angled triangle PiPjPk and C is a circumcircle of PiPjPk. It is easy to show three circles with diameters PiPj, PjPk, PkPi cover C with its interior ==> they cover all Ps (s=1..n).


I have a problem with developing equations with a type of pattern (I hope this is the correct word to use) that would be given.

An example would be on number 8 of AMC 12 2000

Basically, the question would give you some information and ask you to find the 100th term of this pattern.
For example. one equation for a pattern would be 2n+3 or something similar to that.

My question is how to uncover the equations that help follows these patterns.

One trick you can use is finding the difference between the rows.  Basically, what you do is subtract each number from the one next to it.  For example:

Row 0:  2   6   12   20   30 ...

Row 1:  [hide]-[/hide]4[hide]-[/hide]6[hide]-[/hide] 8[hide]--[/hide]10    ...

Row 2:  [hide]--[/hide]2[hide]-[/hide]  2 [hide]-[/hide]   2

Because they are all the same on the 2nd row, this pattern is a polynomial of the 2nd degree.  You could then extend the 2's to find the next in the pattern and work backwards, up.  Or you could solve for the polynomial this way:  Since you know it is a polynomial of degree 2, it is of the form ax^2+bx+c=f(x).  Then, letting the first one be f(1), the second f(2), etc:

a(1^2)+b(1)+c=2     -> a+b+c=2

a(2^2)+b(2)+c=6     -> 4a+2b+c=6

a(3^2)+b(3)+c=12   -> 9a+3b+c=12

Now you have a system of equations, with 3 variables and 3 equations.

Solving for a, b, and c gives you a=1, b=1, c=0.  Therefore you're equation is f(x)=x^2+x.

However, the fastest way is usually looking at the numbers themselves and finding the pattern, without using a system of equations.  Some good questions to ask about the pattern are: Is it linear?  Is it a polynomial (use the test above)?  Is it geometric?  Does it repeat?  If I add 1, or 2, or something to each number in the pattern, does it become obvious?

Aah...thank you Wumbate.

That was really helpful.  I never realized that by looking at the number differences could tell me that if the equation would be a 2nd degree or not.

[quote="Wumbate"]It can tell you other things, too.  For example, if you get something that repeats, it could be geometric or periodic.
The sequence 1 2 4 8 16 32 has differences of 1 2 4 8 16, for example.
Or 3 5 7 5 3 5 7 5 ... has 2 2 -2 -2 2 2 ...[/quote]

And, even better, examining the first differences can allow us to easily sum

1^n + 2^n + 3^n .... + k^n, n an integer.

Wow...I often used differences to understand basic patterns but I never knew that the differences could be this helpful in so many ways.


Prove that for any x,y,z >0 we have:

sum_sym x^{2}yz  >=  sum_sym x^{2}y^{2}.

by AM-GM
[tex]x^2y^2+y^2z^2 \geq 2y^2xz [/tex]
[tex]y^2z^2+x^2z^2 \geq 2z^2xy[/tex]
[tex]y^2x^2+x^2z^2 \geq 2x^2zy[/tex]
add all up and divide by 2
[tex]x^2y^2+x^2z^2+z^2y^2 \geq x^2zy+y^2zx+z^2xy[/tex]
which is the opposite of what we are supposd to prove.

yes, the opposite inequality is a direct consequence of Muirhead's inequality.

Pierre.

this ineq is an example in high school textbook in China ...


group theory
abstract algebra
superior algebra
superior algebra unsolved

How would I go about showing, If K is a subgroup of a finite group G, prove that the number of left cosets of K is the same as the number of right cosets of K.  Thanks.

Show that $ x \mapsto x^{ \minus{} 1}$ induces a bijection of cosets of type $ xK \mapsto Kx^{ \minus{} 1}$ (the inverse map is the same map again, just sides swapped).


Hello there. I recently came across the following(imo interesting) problem(given at the entrance exam of the University of Sofia in 2003).:
Prove that if m,n,p are non-negative numbers and $ m\plus{}n\plus{}p\equal{}1$, then $ \frac{1}{1\minus{}m^2}\plus{}\frac{1}{1\minus{}n^2}\plus{}\frac{1}{1\minus{}p^2} > \frac{189}{62}$

[quote="NoThanks"]Hello there. I recently came across the following(imo interesting) problem(given at the entrance exam of the University of Sofia in 2003).:
Prove that if m,n,p are non-negative numbers and $ m \plus{} n \plus{} p \equal{} 1$, then $ \frac {1}{1 \minus{} m^2} \plus{} \frac {1}{1 \minus{} n^2} \plus{} \frac {1}{1 \minus{} p^2} > \frac {189}{62}$[/quote]
$ \sum {\frac{1}{{1 \minus{} {m^2}}} \ge \frac{9}{{3 \minus{} {m^2} \minus{} {n^2} \minus{} {p^2}}}}  \ge \frac{9}{{3 \minus{} \frac{{{{(m \plus{} n \plus{} p)}^2}}}{3}}} \equal{} \frac{{27}}{8} > \frac{{189}}{{62}}$
 :lol:

nice solution. The first inequality follows from average-mean >= harmonic mean?


geometry
trapezoid
inequalities
trigonometry
triangle inequality

In trapezoid $ ABCD$, $ AB \parallel CD$, $ \angle CAB < 90^\circ$, $ AB \equal{} 5$, $ CD \equal{} 3$, $ AC \equal{} 15$. What are the sum of different integer values of possible $ BD$?

$\textbf{(A)}\ 101  \qquad\textbf{(B)}\ 108  \qquad\textbf{(C)}\ 115  \qquad\textbf{(D)}\ 125  \qquad\textbf{(E)}\ \text{None}$

[hide]Let $ T$ be the intersection of the diagonals. Then from $ \triangle TAB\sim\triangle TCD$ we get $ TA \equal{} {5\over 8}\cdot 15$. Denote $ x: \equal{} BD$. Then $ TB \equal{} {5\over 8}x$

By the triangle inequality, $ {5\over 8}\cdot 15 \minus{} 5 < {5\over 8}x < 5 \plus{} {5\over 8}\cdot 15\iff 7 < x < 23$. However, $ \angle CAB$ must be acute, hence $ \left({5\over 8}x\right)^2 < 5^2 \plus{} \left({5\over 8}\cdot 15\right)^2\iff x^2 < 8^2 \plus{} 15^2 \equal{} 289\iff x < 17$

Therefore $ 8\leqslant x\leqslant 16$, hence the desired sum is $ {8 \plus{} 16\over 2}\cdot 9 \equal{} 108$, so the answer is $ B$.[/hide]

It's conventional to label ABCD clockwise, and it wouldn't be "none".

[img]http://i44.tinypic.com/2lcvjtg.jpg[/img]

Length of BD, $ l$, is equal to:

$ \sqrt {(15\cos \theta \plus{} 2)^2 \plus{} (15\sin \theta)^2}, \theta \in \left(\frac {\pi}{2}, \pi \right]$

The range is $ [7, 17)$.

So it's just $ \sum^{16}_{n \equal{} 7}n \equal{} 115$

EDIT2> Okay, I see the difference. I added in the 7 from the degenerate case when <CAB measured zero degrees. I suppose it is no longer a "trapezoid", so 108 should be the correct answer... But this is a bad question in that it doesn't specify what should be considered a trapezoid.

Draw parallel line from $ A$ to $ BD$. Let this line intersect $ DC$ at $ E$. $ \triangle ACE$ is a triangle with $ 15,8$ and acute angle. So $ AE$ must be smaller than $ 17$.
$ 15\minus{}8\equal{}7<BD<17$.


What is the greatest common factor of $ (a^2 \minus{} 9)$ and $ (3a^2 \minus{} 3a \minus{} 36)$?

We factor $ a^2\minus{}9$ as $ (a\minus{}3)(a\plus{}3)$ and $ 3a^2\minus{}3a\minus{}36$ as $ 3(a\minus{}4)(a\plus{}3)$, so their greatest common factor is $ \boxed{a\plus{}3}$

Alternatively, note that any factor of both must be a factor of $ 3a^2\minus{}3a\minus{}36\minus{}3(a^2\minus{}9)\equal{}\minus{}3a\minus{}9\equal{}\minus{}3(a\plus{}3)$, but since $ \minus{}3$ is not a factor of the first, it must be $ \boxed{a\plus{}3}$. This is an extension of the Euclidean Algorithm.