quadratics
algebra
polynomial
number theory
number theory unsolved

Why is the number of quadratic nonresidues mod p, where p is an odd prime, equal to $ \frac{p\minus{}1}{2}$?

Also note that if $ n|p\minus{}1$, the number of $ n$th powers mod $ p$ is $ \frac{p\minus{}1}{n}$; otherwise everything is an $ n$th power. (Use generators again.)

That's not completely correct for the case $ n \nmid p\minus{}1$, there are in general $ \frac {p \minus{} 1}{\gcd(n,p \minus{} 1)}$ $ n$-th powers (and $ 0$).

[quote="t0rajir0u"]Every nonzero residue $ \bmod p$ is a root of the polynomial $ x^{p \minus{} 1} \minus{} 1 \equal{} 0$, which factors as

$ (x^{\frac {p \minus{} 1}{2} } \minus{} 1)(x^{\frac {p \minus{} 1}{2} } \plus{} 1) \equal{} 0$.

The roots of the left factor are precisely the quadratic residues and the roots of the right factor are precisely the quadratic non-residues.

Alternately:  let $ g$ be a generator of the multiplicative group $ \bmod p$.  The quadratic residues are precisely the elements of the form $ g^k, 0 \le k < p \minus{} 1$ with $ k$ even, and the quadratic non-residues are precisely the elements with $ k$ odd.[/quote]
I don't understand that at all.
The roots to $ x^4 \minus{} 1 \equal{} 0$ is $ 1,i,i^2,i^3$ but what are they concurrent to $ \bmod 5$? And I surely doesn't understand the last part :).

An elementary proof: First note that $ a^2 \bmod b^2 \bmod p$ $ \iff$ $ p \mid (a \minus{} b)(a \plus{} b)$ $ \iff$ $ a \equiv \pm b \bmod p$. So $ 1^2,2^2,3^2,...,(\frac {p \minus{} 1}{2})^2$ Are all different $ \bmod p$. And they are the same as $ (\frac {p \minus{} 1}{2})^2,...,(p \minus{} 1)^2$. So there are exactly $ \frac {p \minus{} 1}{2}$ different quadratic residues.

[quote="Mathias_DK"]I don't understand that at all.
The roots to $ x^4 \minus{} 1 \equal{} 0$ is $ 1,i,i^2,i^3$ but what are they concurrent to $ \bmod 5$? [/quote]
The roots of $ x^4 \minus{} 1 \equiv 0 \bmod p$ are $ 1, 2, 3, 4 \bmod 5$ by [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url].  Because $ 2$ is a generator $ \bmod 5$, we can identify $ i$ with $ 2$; in other words, we can embed the Gaussian integers into $ \mathbb{F}_5$.  (Incidentally, it is not a large leap from here to conclude that there are $ \varphi(p \minus{} 1)$ elements of order exactly $ p \minus{} 1$ in the multiplicative group $ \bmod p$, hence that the group is cyclic.)


function
inequalities unsolved
inequalities

[quote="sylow_theory"][quote="chien than"]Prove that:$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \leq 10$
where $a;b;c$ are real numbers satisfying $1 \leq a;b;c \leq 2$[/quote]
$(2+2+2)\left(\frac{1}{1}+\frac{1}{1}+\frac{1}{1}\right) \leq 10$
 :?:[/quote]
What are $a,b,c$ in your counterexample?

Use convex function, if $f$ convex on $[a,b]$ then $f(x)\leq\max\{f(a),f(b)\}$. Problem similary

Let $0<p<q$ be positive reals. Find maximum of $(x_{1}+x_{2}+...+x_{n})(\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}})$, where $x_{i}\in [p,q](i=1,2,...,n)$.


8th grader now, next year should I take AMC 10A and AMC 12B?  Or, should I take AMC 10A and AMC 10B?  Thanks!

I'd take the 10A and see from there. If you think you got a 120+ (write down answers and compare after the test), take the 12B. If you don't think you got a 120, I'd take the 10B.

That's a rough rule of thumb, though. For making the AIME, generally the 12 is easier. For making the USAMO, though, the 10 is the better shot. If you're taking Precalc now, you probably have enough background to do the 12 (the further you are in math, the better, until calc). Also, if you took it this year, you can use that to judge how well you expect to do. And practice exams don't hurt, either. Except for the USAMO, I would say that if your average 10 score is less than 20 points above your average 12 score, take the 12 (and conversely, if it's more than 20 points above, take the 10).

For some people, there are other considerations: If your school isn't that strong, you could go for both the AMC 10 and AMC 12 medals. Even if you're good enough on the 12, you might still want to take the 10 to try to get a perfect score, if it's within reach (maybe your sophomore year).

In general, though, set a goal, like make AIME or make USAMO, or perhaps collect all the school medals you can, and then look at which tests give you the best chance of making that goal. Unless you're pretty far advanced and can do well on practice tests, though, I would start with the 10A and see how you do on that first.

definently 10.

dont take 12 unless you can get a 130+ on it, otherwise it will screw up your usamo index and youll probably complain about it later.

even if your goal is just to make aime, you should still take 10...

i dont really think the 12 is harder, yea 5 percent make it as opposed to 1 percent, but those 5 percent will include 11th/12th graders whereas the 1 percent doesnt...also there are many topics on the amc 12 where many people are not ready for (espcially if you are in <=10 grade)

Are you aiming to make USAMO? if you are, take AMC 10. Otherwise, take AMC 12.

AMC 12 is easier to pass but harder to qualify to USAMO with.

im a sophmore next year...i can do about 130 on the 10 now......i definitely want to make the aime but usamo is mayb unrealistic. I would not be happy if I didnt make aime through the 12 though.....


calculus Stanford college MIT Harvard Princeton Support

I am wondering, as a leaving high school senior, how many ap tests I should take this May.

As far as I can see, taking as 'much' as I can would be a good way to go; it would only get rid of the requirements...

But is it really the case?

I have already taken 4:

AP Statistics (4); AP Calculus BC (5); AP Chemistry (5); AP Physics C Mechanics (5)

I am thinking of taking another 10-12:

AP English LAC; AP English LRC; AP US History; AP Psychology; AP Microeconomics; AP Macroeconomics; AP Biology; AP Computer Science AB; AP Environmental Science; AP Physics C EM; AP Spanish Lit(?); AP Spanish Lan(?)

I would assume that this is do-able.

However, the question is is there any benefit other than getting score/learning that this would be helpful? Is there any reward for doing this, in other words? (such as prizes, scholarships, credits granted, etc.)

10-12 :!:  :|. That's pretty outrageous even by AoPS standards. It's probably only good for you to take ones that your colleges of choice actually accept; and even then, studying a course in two weeks and getting a 5 on the test won't necessarily put you in a good position for college classes. I would recommend cutting it down to at most 6 and even that's pushing it.

I'm not aware of many scholarships that are based solely on AP test scoring; College Board does give "AP Scholar" titles, but I'm not sure if they're worth anything.

Lastly, for reference, I have taken the 4 AP tests you have already taken as well as 4 of the ones you are thinking of taking. The 4 you already took were much, much easier (for me, at least) than the 4 in your other list. So you should keep in mind that, unless you have some kind of godlike testtaking ability, you probably won't be able to study for all of those tests.

I think I am definitely taking the following... I am pretty sure that I can pull off five on these:
AP US History; AP Psychology; AP Microeconomics; AP Macroeconomics; AP Physics C EM

Now... I might get 4 on these (but take regardless):
AP English LAC; AP English LRC; AP Biology; AP Computer Science AB; AP Environmental Science

Probably not taking these...
AP Spanish Lit; AP Spanish Lan

I am pretty sure I will do fine on them.. That was not my question at all... I was just saying what kind of benefits I would have if I were to do this... Because if it's just waste of money, I would rather not do them.

I am shooting for AP State Scholar... How many should I take, then? I am in Michigan.

As paladin8 mentioned, the value of the exams depends on what you plan to do in college.  At all but the most elite schools, you will earn college credits towards graduation or the ability to substitute courses for program requirements met through AP.  At many universities, these benefits are worthwhile.  Often, the AP-equivalent courses taught at universities are large lecture-style courses with very few opportunities for interaction with real professors.  Bypassing these courses in favor of smaller upper-level courses may enhance your college experience.

Think carefully about the benefits each of these exams.  Paladin8 is probably correct about overextending yourself.  The exams you've passed so far will be very useful in college.  After passing AP Chem and AP Physics C, will a course like AP Environmental Science be required in programs that you will be taking in college? If not, what value is it adding?

If you pass the English exams, be sure to take additional English courses in college.  I passed AP English but didn't take any additional English in college.  Later, I realized that college transcripts lacking English courses raise a red flag to graduate schools and employers. In any event, you'll need to have a good reference who can vouch for your writing skills.

I don't know if the "state scholar" designation is worthwhile - probably not in your case.  It sounds more like a marketing gimmick devised by folks who make money from students who take unnecessary AP exams.

Wow, how did you just randomly learn 10-12 AP's?

Anyway, if you can do well on all of them, I personally would suggest taking them.  You'll have more flexibility in college.  You'll have the choice of taking them again in college if you want to...but you wont have to.

Where are you planning on going to school?  If you haven't decided yet, what are your top choices?

Colleges...

I do not know yet. But, I applied to:
(private) Stanford / MIT / Harvard / Princeton
(public) UC-Berkeley|LA / UM-Ann Arbor / UW-Wisconsin

I already got accepted to Stanford and UW-Wisconsin... Currently, I am waiting replies from other colleges. Because I want to go to Princeton for a graduate school, however, the choice would probably be narrowed among Stanford, MIT, or Harvard...

Any further suggestion on where might be better? I am going to double major in math/physics...

AP Tests...

The consensus seems like "take as many AP exams as you can." Am I right on this? I thought this should be true... But taking the cost of tests into account, I was unsure about if it would be worth that much.

Ok, if you think you're going to a top university, and money is a concern, then I wouldn't just take them all.

I know MIT's policy pretty well.  They're pretty strict on the credit they give for AP exams.  You can get credit for Calc BC (4 or 5) and physics C (gotta get a 5 on both parts) and various humanities courses.  Getting credits for the humanities courses doesn't really help you much.  It doesn't fulfill your HASS requirement.  But you will be more likely able to apply for early sophomore standing in the spring of your freshman year.  They used to give AP credit for bio, but they're going to stop for the class of 2011 and future classes.

Don't be discouraged though.  It's not that they don't want people getting ahead, it's that they want to make sure people actually know their stuff.  You can take an advanced standing exam to get credit for a class without taking the class.  I never took an ASE but I imagine they're harder than your average AP exam.

I'm pretty sure that tests you take in your senior year do not count for the state scholar award. You should double check that on the College Board website.

Getting a 5 on the test is not always an indication that you have studied the material adequately. For example, I managed a 5 on the Physics C E&M section even though my teacher failed to finish half of the material. Even though I was eligible for credit, I opted to take the E&M course offered by my university, which I still found challenging.

I think you should contact the concentration advisers or department chairs of your school to find out how they would assign credits and whether you can apply them toward your degree requirements or get advanced standing.

I know many people here take 10+ APs, but will it hurt me if I end up with just the following APs?

Calc BC
Phys B
Phys C
Japanese
Chem
Bio
Econ

Seven should be plenty to make you competitive at the top schools, as far as number of AP's goes.  I took 6 and I'm here at MIT, for example.

Your AP courseload won't usually make or break you.  I see it mainly as an indication that you're able to handle courses closer to those at actual colleges and also an indication of your ambition and desire to learn.

To lend support to joml, I took only 4 AP exams during high school, as far as I can remember (Calculus BC, Physics C, Government and Microeconomics).  The way the American college application system works, there are hardly any credentials that are [i]necessary[/i] to get into top universities.  APs are one sort of credential that colleges recognize, but there are dozens of others, ranging from academic to athletic to volunteer work to just plain randomness.

A blog entry from MIT admission officer Ben Jones, "[url=http://www.mitadmissions.org/topics/before/recommended_high_school_preparation/many_ways_to_define_the_best.shtml]Many Ways to Define the Best[/url]" helps put AP courses and tests in perspective.

One thing to remember: the results of your senior year AP exams - indeed, the exams themselves - happen well after your admissions have been decided and well after you have declared what your choice is. At that point, the issue becomes: what credit can you obtain at the school you will be attending, and what impact will passing or not passing the exams have on your freshman year academic schedule? The answers to those questions will be different at different institutions, but by then you should know which institution you are talking about.


integration
function
real analysis
real analysis solved

f:R->R cont. and g:R->R g(x)=f(x)* \int {0, x} f(t)dt. Show that if g is decreasing then f(x)=0 for any real x.

Lets define g:R-->R with g(x)= (  \int {0,x} f(t)dt)^2. we see that:
g'(x)=2f(x)  \int {0,x} f(t)dt. From the hypotesys we have that g' is decreasing and because g(0)=0 we have that g'(x)>=0  for x in (-oo,0]  and g'(x)<=0 for x in [0,oo). Then function g is increasing on (-oo,0] and decreasing on [0,oo) so g(x)<=g(0)=0.

On the other hand from the definition og g we have that g(x)>=0 so g(x)=0 which gives that  \int {0,x} f(t)dt=0, for any x in R. [1]

Lets show now that we have f(x)=0 from [1].

Lets presume that there exists xo in R with f(x0)<>0, f(x0)>0 (lets consder it like that). Because f is continuous, there exists the vecinity (x1,x2) of point xo on which the function maintains the sign +.

so we have that  \int {x1,x2} f(t)dt=  \int {x1,0} f(t)dt +  \int {0,x2} f(t)dt= -  \int {0,x1} f(t)dt +  \int {0,x2} f(t)dt=0.

On the other hand from the mean theorem there exists  \varepsilon in (x1,x2) so that:

\int {x1,x2} f(t)dt=(x2-x1)f(c)>0 so we have a contradiction;

so f(x)=0 for any real x.

NOTE that \int {x,y} mean \int from x to y.

cheers! :D :D


inequalities
trigonometry
inequalities proposed

Prove that in any triangle $ABC$ the following inequality holds:

                         \[ \sqrt{3} (\sin A +\sin B +\sin C )\leq 3+\cos A +\cos B +\cos C \]

$\sqrt{3} (\sin A +\sin B +\sin C )\leq 3+\cos A +\cos B +\cos C\Leftrightarrow$
$\Leftrightarrow(2cos\frac{\alpha+\beta}{2}-cos\frac{\alpha-\beta}{2})^2+sin^2\frac{\alpha-\beta}{2}\geq0,$ where $\alpha=\frac{2\pi}{3}-A,\beta=\frac{2\pi}{3}-B.$

Hi,arqady,Your solution is nice but I don't how you come it out :(
Here is mine:
use
$\sin A+\sin B+\sin C =4\cos \frac A2 \cos \frac B2 \cos \frac C2$
Then  the inequality is equal to:
\[ 2\sqrt 3 \cos \frac A2 \cos \frac B2 \cos \frac C2 \le \cos^2 \frac A2+ \cos^2 \frac B2+cos^2 \frac C2 \]
The last one is easy by AM-GM,and $\cos \frac A2 \cos \frac B2 \cos \frac C2 \le \frac{3 \sqrt 3}8$

[quote="arqady"][quote="zhaobin"] I don't how you come it out :(
 [/quote]
$\sqrt{3} (\sin A +\sin B +\sin C )\leq 3+\cos A +\cos B +\cos C\Leftrightarrow$
$\Leftrightarrow \cos{\alpha}+\cos{\beta}-\cos(\alpha+\beta)\leq\frac{3}{2},$ when $\alpha=\frac{2\pi}{3}-A,\beta=\frac{2\pi}{3}-B.$ ;)[/quote]
Ok,I know it now.thanks very much. :)


function
trigonometry
calculus
integration
algebra
domain
calculus computations

A curve $C_{1}$ is defined by:

$\frac{dy}{dx}=e^{x}\cos{x}$ for $x\in[0,2\pi]$ and passes through the origin.

Prove that the roots of the function (other than 0) occur in the ranges $\frac{\pi}{2}<x<\pi$ and $\frac{3\pi}{2}<x<2\pi$.

using integration by parts twice we obtain

$y=\int e^{x}\cos xdx=\frac{1}{2}e^{x}(\cos x+\sin x)+C$

$y$ passes through origin yields $C=-\frac{1}{2}$

Now let $f(x)=\frac{1}{2}(e^{x}(\cos x+\sin x)-1)$ and it is easy to check that

$f(\frac{\pi}{2})>0$ , $f(\pi)<0$ , $f(\frac{3\pi}{2})<0$ , $f(2\pi)>0$ . By Intermediate Value Theorem , there exist at least one root in each of the interval $(\frac{\pi}{2},\pi)$ , $(\frac{3\pi}{2},2\pi)$

Comment removed -- I failed to read carefully.

Actually, a replacement complaint:  shyong's argument still doesn't show that there aren't roots in other places (like in the interval $(\pi, \frac{3\pi}2)$).  (Not that this is terribly hard to show.)

For $x\in(0,2\pi)$ those are the only roots, I clearly provided a domain for the function and excluded a 0 root.


symmetry
number theory unsolved
number theory

Ten mutually distinct non-zero reals are given such that for any two, either their sum or their product is rational. Prove that squares of all these numbers are rational.

I think all squares are rational, not equal.

Sorry to revive an old topic, but my solution is quite odd... it proves that the problem holds for $ n$ points, with $ n\ge 6$.  Did I make a mistake in my solution?
[hide="Solution?"]
Assume that not all of the numbers are squareroots of rational numbers.  Let the numbers be represented as ten nodes on a graph.  Connect two nodes by a white edge if they multiply to form a rational.  Connect them by a black edge if they sum to form a rational.  Then, considering hexagon $ a_1a_2a_3a_4a_5a_6$, there must exist a monochromatic triangle, say with verticies $ a_1$, $ a_2$, and $ a_3$.  If $ a_1 \plus{} a_2$, $ a_2 \plus{} a_3$, and $ a_3 \plus{} a_1$ are rational, then we see that $ 2a_1 \plus{} 2a_2 \plus{} 2a_3$ is rational, so $ a_1 \plus{} a_2 \plus{} a_3$ is rational, so $ a_1$, $ a_2$, and $ a_3$ are all rational.   Yet, for any $ a_i$, $ 1\le i\le 10$, we would have that $ a_i \plus{} a_1$ is rational, so $ a_i$ is rational, or $ a_ia_1$ is rational, so $ a_i$ is rational anyways.  This means that all of the numbers are rational, which is a contradiction.  If $ a_1a_2$, $ a_2a_3$, and $ a_1a_3$ are rational, then $ a_1a_2a_3$ is the squareroot of a rational, so $ a_1 \equal{} \frac {a_1a_2a_3}{a_3a_2}$ is the squareroot of a rational.  Similarly are $ a_2$ and $ a_3$.  Now, if an $ a_i$ was connected to $ a_j$ with a white edge when $ j \equal{} 1,2,3$, then $ a_ia_j$ is rational, so $ a_i^2a_j^2$ is rational, so $ a_i^2$ is rational, so $ a_i$ is the squareroot of a rational.  If all $ a_i$ had this property, then all $ a_i$ would be squareroots of rationals, which contradicts the assumption.  Thus, there must be at least one point, say $ a_4$, that is connected to $ a_j$ with blacks for $ j \equal{} 1,2,3$.  Yet, $ a_1 \plus{} a_4$ and $ a_4 \plus{} a_2$ being rational implies that $ a_1 \minus{} a_2$ is rational.  Yet, $ a_1^2 \minus{} a_2^2$ is rational, so $ a_1 \plus{} a_2$ is rational, so we have formed black monochromatic triangle $ a_1a_2a_4$, which is a contradiction.  Thus, all of the numbers are squareroots of rationals. [/hide]

Okay, I think this should work.

Let the numbers be $ a_1,a_2...a_{10}$. We can assume all the numbers are irrational because if one was rational, then all the others must be also. Let $ r$ denote a rational number.
Suppose WLOG $ a_1^2$ is irrational.
Suppose $ a_1$ times every other number is rational.
Suppose also there exists two number $ a_2$ and $ a_3$ such  $ a_2a_3 \equal{} r_1$. Then since $ a_1a_2 \equal{} r_2$ and $ a_1a_3 \equal{} r_3$, $ a_1^2 \equal{} \frac {r_2r_3}{r_1}$, contradiction. So the sum of any two of the numbers not including $ a_1$ must be rational. But this is obviously a contradiction because then $ (a_2 \plus{} a_3) \plus{} (a_2 \plus{} a_4) \minus{} (a_3 \plus{} a_4) \equal{} 2a_2$ would be rational.
Thus, there must exist a number $ a_2$ such that $ a_1 \plus{} a_2 \equal{} r_4$. *There also must be a number $ a_3$ such that $ a_2 \plus{} a_3 \equal{} r_5$ or we could get a contradiction, and there must even exist an $ a_4$ such that $ a_2 \plus{} a_4 \equal{} r_6$*. Then, $ a_1a_3 \equal{} r_7$ (the sum can't be rational), and $ a_1a_4 \equal{} r_8$. Also, $ a_3a_4 \equal{} r_9$ (the sum can't be rational) so $ a_1^2 \equal{} \frac {r_7r_8}{r_9}$, which is rational, contradicton.
So $ a_1^2$ must be rational, and similarly the squares of all the others must be.
I didn't fully prove *, I will finish it later.

[quote="harazi"]Suppose not, then none of them is rational.
   Pick $x_1$. There are at least 5 indices $i$ such that for these indices we have either $x_1+x_i$ are rational or $x_1 x_i$ are rational. If $x_1 x_i$ are all rational, then for i,j in this set of indices we have $x_i/x_j$ rational. Fix such i,j. If $x_i+x_j$ is rational, then either $x_i$ is rational or $x_i+x_j=0$. Thus we can find i,j in this set with $x_ix_j$ rational. But then $x_i^2$ is rational and thus $x_1^2$ also. Suppose now that $x_1+x_i$ are all rational and thus $x_i-x_j$ are rational for this set of indices. If $x_i+x_j$ for some i,j in this set, then $x_i$ is rational and we are done. If not, we have $x_i x_j$ rational for i,j differents in this set. It easily follows that if $i$ is in this set of 5 indices then $ x_i^2$ is rational. Then let $ x_1+x_i=c_i$ with $c_i$ rational. Since $x_i^2$ is rational, it follows that $ x_1^2-2c_ix_1$ is also rational and thus if $i,j$ are different in the set of indices then $ (c_i-c_j)x_1$ is rational. Of course, we can find i,j different in the set of indices such that $c_i, c_j$ are different and thus $x_1$ is rational in this case. Anyway, we have shown that $x_1^2$ is rational and by symmetry we're done. Hope it's good, with so many cases discussed...[/quote]
hi,
I think you are not right in this case`
[b]If $x_i+x_j$ is rational, then either $x_i$ is rational or $x_i+x_j=0$[/b]


[hide]
resultant vector (hypotenuse) must have magnitude of 3 m/s
the y component (y is going up and down river) must be 2 m/s northward

so hypotenuse of 3 is known and leg of 2 that is opposite the angle is known

arcsin(2/3) = angle

[/hide]

[quote="math92"]Debbie is swimming in a river.  She wishes to cross directly east to the other bank of the river.  There is a southward current of 2 m/s and Debbie can swim 3 m/s.  At what angle upstream must she swim in order to go straight across (due east)?[/quote]

[hide]She starts out at point A. The point that she is aiming for is point B, and the point that she actually ends up at is point C. The angle we're looking for is $\theta$.

It's given that BC=2 and AB=3, so $\sin \theta = \frac{2}{3}$ and $\theta=\arcsin \frac{2}{3}$ or approximately $42^\circ$.[/hide]


real analysis
integration
limit
logarithms
complex analysis
real analysis unsolved

I hope this is in the right place.

From Rudin Real and Complex Analysis:

Show that if $ \mu$ is a positive measure on $ X$, $ f: X\to[0,\infty]$ is measurable and $ \int_X f d\mu \equal{} c$ where $ 0 < c < \infty$ and $ \alpha$ is a constant then
\[ \lim_{n\to \infty} \int_X n \log [ 1 \plus{} (f/n)^\alpha]d\mu \equal{} \left\{\begin{array}{*{20}c} \infty & 0 < \alpha < 1 \\
c & \alpha \equal{} 1 \\
0 & 1 < \alpha < \infty\end{array}\right.
\]
The hint says if $ \alpha \geq 1$ the integrands are dominated by $ \alpha f$ and otherwise use Fatou's lemma.

I can do the fatou's lemma part, but does anyone have a hint on what it means with the dominated integrands?

Thanks.

It's \[ \lim_{n \to \infty}\int_X n \cdot \log\left[1 \plus{} (f/n)^{\alpha}\right]\mathrm{d}\mu,\] you missed the $ n$ term. You want to apply the dominated convergence theorem to justify bringing the limit inside the integral. For the $ \alpha \equal{} 1$ case, \[ n \log\left[1 \plus{} (f/n)^{\alpha}\right] \equal{} \log\left[\left(1 \plus{} f/n\right)^n\right],\] and the term inside the logarithm should look like something familiar :D


When dealing with a function that gives some kind of equation, e.g. $f(x) = x^{2}$, to evaluate it for various values of $x$ we simply plug into the equation.  So $f(1) = 1^{2}= 1$, etc.

When evaluating a piecewise function, we simply use a different equation depending on what $x$ is.

[b]Part a.[/b]  $x =-1$.  This value is less than $0$, so we use the first equation.  $f(-1) = (-1)^{3}=-1$.

[b]Part b.[/b]  $x = 0$.  This value is equal to $0$, so we use the second equation.  $f(0) = 3 \cdot 0+2 = 2$.

[b]Part c.[/b]  $x = 1$.  This value is greater than $0$, so we use the second equation.  $f(1) = 3 \cdot 1+2 = 5$.

[quote="Interval"]NOTE: I cannot write the function as written in the textbook because I do not have the specific keyboard.  Please, try to follow.

If

f(x) = {x^3...if x < 0
         {3x + 2...if x >or= 0

then find a - c below.

(a) f(-1)

(b) f(0)

(c) f(1)

I have bIG TROUBLE with piecewise-defined functions.

I need a clear definition of this type of function and a step-by-step method that shows me how to find the answer.[/quote]

([hide]a) $(-1)^{3}=-1$
(b) $3\times 0+2=2$
(c) $3\times 1+2=5$.[/hide]

could you explain what youre having trouble with?

also, it you want to write it the cool way, click on the "latex" tab on the left of this website.

I will go over the latex section of this website over the weekend.

Thanks!


calculus
integration
calculus computations

can you integrate :D

$ \int_0^\infty\;\frac{\sqrt[3] x }{e^x\minus{}1}\;\;\textbf dx$

$ \int_0^{ \plus{} \infty}\;\frac {\sqrt [3] x }{e^x \minus{} 1}\;\;\textbf dx \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} \int_0^{ \plus{} \infty} \; \sqrt [3]{x} e^{ \minus{} n x} \; \textbf d x \; \stackrel{t : \equal{} nx}{ \equal{} } \; \sum_{n \equal{} 1}^{ \plus{} \infty} \; \int_0^{ \plus{} \infty} \left( \frac {t}{n} \right)^{\frac {1}{3}} \cdot e^{ \minus{} t} \frac {\textbf d t}{n} \equal{}$

$ \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} \frac {\Gamma \left( \frac {4}{3} \right)}{n^{\frac {4}{3}}} \equal{} \boxed{\Gamma \left( \frac {4}{3} \right) \, \zeta \left( \frac {4}{3} \right)} .$

Observe that
$ f(x) \equal{} \frac{\sqrt[3] x }{e^x\minus{}1}\equal{}\sum_{n\equal{}0}^{\infty}\;\frac{B_{n}}{n!}x^{n\plus{}\frac{1}{2}}$

[quote="akech"]Observe that
$ f(x) \equal{} \frac {\sqrt [3] x }{e^x \minus{} 1} \equal{} \sum_{n \equal{} 0}^{\infty}\;\frac {B_{n}}{n!}x^{n \plus{} \frac {1}{2}}$[/quote]
yes, but it's  true if and only if when $ |x|<2\pi$ :(


calculus
analytic geometry
graphing lines
slope
rotation
function
conics

While reading my calculus book, it came with a question of finding the slope of a line that is tangent to a given curve f(x). f(x) in the problem was in the form of $Ax^{2}+Bxy+Cy^{2}=D$, where A,B,C, and D are the coefficiants. My calculus book says you can't graph to find the answer, because of the $y^{2}$.

My question is: What would a graph look like if the equation is in the form of $Ax^{2}+Bxy+Cy^{2}=D$?

[quote="7h3.D3m0n.117"]My question is: What would a graph look like if the equation is in the form of $Ax^{2}+Bxy+Cy^{2}=D$?[/quote]
A rotated conic.

[quote="7h3.D3m0n.117"]My calculus book says you can't graph to find the answer, because of the $y^{2}$.[/quote]

You definitely can graph to find the answer. It just won't be a function, and hence you can't use a graphing calculator.

well your textbook is a noob...or you are misunderstanding it...that is not a function, but it can easily be broken up into two functions, which can be graphed

[quote="7h3.D3m0n.117"]What would a graph look like if the equation is in the form of $Ax^{2}+Bxy+Cy^{2}=D$?[/quote]

http://mathworld.wolfram.com/ConicSection.html
http://mathworld.wolfram.com/QuadraticCurve.html

It would be a rotated conic
because
there is a xy term
which makes a slope for the axis

You can plot the equation implicitly, but that takes a lot of computing power and is slower than solving for 2 possible functions of y in terms of x.

Then again, if you know the point x,y implicit differentiation owns this problem.


Hey, this is going to seem like a stupid question, but it's driving me crazy. What is the website where you put your name in and it tells you who you are and stuff like that. For example if I put in Hikari I might get something like, "Hikari is a small cat" or Hikari is the owner of a successful firm." Please, if you know tell me. I can't seem to remember it.
Thanks.

is this wut ur talking about?
[url]http://www.googlism.com[/url]

Yes it is. Thank you so much! I thought the website name soundled kind of like google, but I wasn't sure. Thanks again ^_^


Does anyone know of any good chemistry olympiad preparatory or problem books?

Hmm...I guess not.

In that case, I'll probably just use a normal college-level text. Any recommendations that would be useful for USNCO?

I might consider going through [i]Chemistry[/i] by Silverberg.  :ninja:


It's a famous kind of limit, so I think many people know this problem. It would be new to someone, though, and I hope you to enjoy it.

$ \bold{1.}$ Let $ a_n$ be the sequence defined by

(1) $ a_1 > 0$
(2) $ a_{n\plus{}1} \equal{} 1 \minus{} \exp(\minus{}a_n)$.

Find the limit $ \lim_{n\to\infty} na_n$.

$ \bold{2.}$ What if (2) is substituted to

(2') $ a_{n\plus{}1} \equal{} \sqrt{a_n}\arctan\sqrt{a_n}$?

[hide="Answer"]
$ \bold{1.} \; 2 \quad \quad \bold{2.} \; 3$
[/hide]

We can use Stolz's theorem to solve both of problems.

first we must note that for big $ n$  $ a_{n}$ will be less than $ 1$
it's main moment  :)
after is simple:
in first we have :
$ a_{n\plus{}1}\equal{}1\minus{} (1\minus{}a_{n}\plus{}\frac{a_{n}^{2}}{2}\plus{}o())$
or
$ \frac{a_{n\plus{}1} \minus{} a_{n}   }{a_{n}^{2}}$~$ \minus{}\frac{1}{2}\plus{}o()$
or
$ a_{n}$~$ \frac{2}{n}$
and limit will be equal to $ 2$
and in seccond :
$ a_{n\plus{}1}$~$ \sqrt{a_{n}}(\sqrt{a_{n}}\minus{}\frac{a_{n}\sqrt{a_{n}}}{3})$
or
$ \frac{a_{n\plus{}1} \minus{} a_{n}   }{a_{n}^{2}}$~$ \minus{}\frac{1}{3}$
and $ a_{n}$~$ \frac{3}{n}$
and limti is equal to $ 3$


modular arithmetic
algebra unsolved
algebra

prove that the equation $ x^2\plus{}y^2\plus{}z^2\equal{}2xyz$ has no solution in natural numbers >0.

$ 2|x ,2|y,2|z$  $ x \equal{} 2a, y \equal{} 2b ,z \equal{} 2c$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4abc$
$ 2|a, 2|b, 2|c$
$ a \equal{} 2d ,b \equal{} 2e, c \equal{} 2f$  $ \implies$  $ d^2 \plus{} e^2 \plus{} f^2 \equal{} 8def$
$ p^2 \plus{} q^2 \plus{} r^2 \equal{} 2^{n}pqr$
$ 2^s|x$  $ \forall s\in\ N$  $ x \equal{} 0 ,y \equal{} 0,z \equal{} 0$ only solution
but $ x,y,z > 0$ Contradiction!

[quote="cnyd"]$ 2|x ,2|y,2|z$  $ x \equal{} 2a, y \equal{} 2b ,z \equal{} 2c$
[/quote]

Why must all a,b,c be divisible by 2?What if two of them are odd?

[quote="cnyd"]if two of them are odd
let x and y  be odd.
$ (x,2) \equal{} (y,2) \equal{} 1 ,(z,2) \equal{} 2 ,4|2xyz$
$ x^2 \plus{} y^2 \plus{} z^2\equiv 2\pmod{4}$
$ (x,2) \equal{} 1$ $ \implies$  $ x^2\equiv1\pmod{4}$
Contradiction! :wink:[/quote]

I don't understand.

[hide="if x, y, are odd and z is even"] We have $ x^2\plus{}y^2\equal{}2xyz\minus{}z^2$. Then $ x^2\equiv{y^2}\equiv1\pmod{4}$ while $ 2xyz\minus{}z^2\equiv0\pmod{4}$.

So, we have $ 2\equiv0\pmod{4}$, contradiction.[/hide]

This is a pretty classic infinite descent problem.

$ x\equiv1\pmod{2}$ , $ y\equiv1\pmod{2}$  , $ z\equiv0\pmod{2}$
$ \implies$  $ 2xyz\equiv0\pmod{4}$
$ x\equiv1\pmod{2}$ $ \implies$  $ x^{2}\equiv1\pmod{4}$
$ z\equiv0\pmod{2}$  $ \implies$  $ z^{2}\equiv0\pmod{4}$
$ x^{2} \plus{} y^{2} \plus{} z^{2}\equiv2\pmod{4}$
$ 2xyz\equiv0\pmod{4}$  $ \implies$  $ 0\equiv2\pmod{4}$  Contradiction!   :wink:


Let $p>3$ be prime and $a,\ b,\ c$ are positive integers. If all of $a+b+c,\ a^{2}+b^{2}+c^{2},\ a^{3}+b^{3}+c^{3}$ are the multiple of $p$, prove that all of $a, b,\ c$ are the multiple of $p$.

I don't understand why  :maybe: But here is 2nd solution: We have $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ , so $p|3abc$ and because $p>3$ we have $p|abc$.

We also have $2(ab+bc+ca)=(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})$ therefore $p|2(ab+bc+ca)$, so $p|ab+bc+ca$ because $p>3$. Right?  :wink:

[quote="kunny"]Cam O'n, N.T.TUAN :)[/quote]
No problem. But please use only English on this forum   :lol:

My solution:
Note that $p\mid (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc\longrightarrow p\mid 3abc\longrightarrow p\mid abc$ so $p$ must divide one of $a,b,c$ for example $p\mid a$ then $p\mid b^2+c^2,p\mid b+c \longrightarrow p\mid (b+c)^2-(b^2+c^2)=2bc\longrightarrow p\mid bc$ so one of $b,c$ is divisible by $p$.
DONE


On what adress we should send solutions for the first article of  the AMY program ?? :blush:

[quote="junggi"]and when are they due?[/quote]

they are due like november 4th (if you are emailing) like october 30th if mailing...

it said that on some sheet of paper that was included


[quote="fredbel6"]I was just wondering why so many Vietnamese are called 'Nguyen'?

When i see a list of vietnamese people, it's almost an axiom there will be a nguyen included?[/quote]
Nguyễn is just a surname.
The possible reasons are: (1) there are many Việt Nam-ese with surname Nguyễn, (2) there are not many surname(s) in Việt Nam, (3) your observation is imprecise.

[quote="fredbel6"]And second of all, is it true vietnamese always, out of respect, ask for your age?[/quote]
I think there is no problem for asking a Việt Nam-ese age. Again, there's no such thing as definite "always".

This is similar to many Russian names end with "va","za".SO I think you shouldn't think about it .I'm Vietnamese and I don't understand why ,too. :roll:

I once heard an explanation that actually it is the middle name that is usually nguyen, and that it comes first, and that's why there is the misconception that everyone is called nguyen :?

It sure makes it easy to guess what country he is from :D

Where did you hear the explanation, can I ask? Is it a reliable source?  :D

:(   Can't find it again.
It was some Australian info site for people wanting to go to Australia.

[url=http://news.pacificnews.org/news/view_article.html?article_id=d74236e9ca1d48077cf10020cecb065f]link[/url]
however this article I found coincidentally while trying to find that info site, backs up my theory.  It is about a Nguyen, apparently executed in Singapore for drug trafficking  :( , it is written by a Nguyen, interviews a travel agent called Nguyen, and a friend of the executed man,called  Nguyen, :|  replies as well

[quote="analysis"]I read the article, however, I didn't see any fact about Nguyen as a middle name. To my knowledge, Nguyen is a surname. I doubt the explanation that you heard is probably false.  :D Your theory sucks, what ever it is.

"When he asked as his last wish to chew a piece of gum it was denied to him."
What the he-ll is this?  :mad: singaporean, chinese, ... are my enemies, you know[/quote]

Analysis allow me to elaborate a bit.  When I said in my last reply that the article proofs my theory, I just meant the theory that Nguyen is so common.  I didn't mean the first/middle name theory, as I can't find the article back about that specific thing.

Yes indeed apparently they denied him the right to chew gum.  A friend whose uncle went to Singapore once said that chewing gum in public is forbidden, and that like walking a dog and uhm, looking the other way (which too many people do here  :o  sometimes) is certainly very very forbidden (apparently they want a clean city)  Dunno if what I said is really true, but guess that was reason enough to deny the man his last wish.

Why would you say people from Singapore and China are your enemies?  Because of history? death penalty?

[quote="fredbel6"]Analysis allow me to elaborate a bit. When I said in my last reply that the article proofs my theory, I just meant the theory that Nguyen is so common. I didn't mean the first/middle name theory, as I can't find the article back about that specific thing.[/quote]
[quote="analysis"] :D Your theory sucks, what ever it is.[/quote]
I mean your theory first/middle name, Nguyen is so commom, or whatever it is, it sucks.  :D
You can do something better to back up your theory right (if you want to)

[quote="fredbel6"]Yes indeed apparently they denied him the right to chew gum. A friend whose uncle went to Singapore once said that chewing gum in public is forbidden, and that like walking a dog and uhm, looking the other way (which too many people do here  sometimes) is certainly very very forbidden (apparently they want a clean city) Dunno if what I said is really true, but guess that was reason enough to deny the man his last wish.[/quote]
hmm I see why, assume what you said is true.

[quote="fredbel6"]Why would you say people from Singapore and China are your enemies? Because of history? death penalty?[/quote]
yes, because of history  :rotfl:

china i can understand (third indochina war?) but singapore?
i am not going to go at it for the name theory again, i am not certain of it myself and judging from your reaction it ain't true
but it is a common first name?

Well, I didn't say that your name theory is not true. It may be true, who knows.
The notion of enemy is my personal opinion. My list extends to many countries, not just china and singapore. Like taiwan, russia, thailand, phillipine, japan, ... You may be my enemy also. Which country are you from? (My geography knowledge approaches zero, I can't tell from your flag.) You may ask why I have too many enemies. I have the just reasons for myself, some are history, some are philosophy. Sometimes, I do thing that the question why does not exists.

I am Belgian.  We were occupied by germans twice, supported south korea in the war but were not involved in vietam war, neither are we involved in iraq war.

Your reasons interest me, unless of course they are very personal, interest me.
I should warn you that although i tend to think certain things of germans and I am scared of chineses commie system, I've had many pleasant encounters with them

For what it's worth, the single most common family name in the database of students at my university (which is located in Southern California) is "Nguyen." No other name comes close. Several other Vietnamese names, such as "Pham" and "Tran," are among our more common names. We do have to learn at least some approximation of the pronunciation of "Nguyen."

Someone I know explained to me that Nguyen was an important historical figure who resisted Chinese domination centuries ago, and that many families took his name.

The concentration of many people into a very small number of family names may be even more striking in Korea, with the most common names being "Lee,", "Kim," and "Park." But we have many more students of Vietnamese ancestry than of Korean ancestry. Also, "Lee" could be Chinese - or even (rarely) English. And I've met a Vietnamese "Kim."

fredbel6, Belgian is not on my list.  :D
If you interest, I can tell you some... china took over vietnam for a long time, which is an old long enemy. Japan, in 19??, attack vietnam and burn the grain in the north, about a million(?) of vietnamese died from hunger... Well there are more...  :rotfl:
You don't have to warn me since I have chinese, russian, ... friends

Kent Merryfield, looking for the last name in a university is a local property, so not very convincing. One can object this easily.  :D

[quote="fredbel6"]Hello

sorry if I am not allowed as Nonvietnames to ask questions here in English
If so, please delete this

I was just wondering why so many Vietnamese are called 'Nguyen'?

When i see a list of vietnamese people, it's almost an axiom there will be a nguyen included?

And second of all, is it true vietnamese always, out of respect, ask for your age?

Just some silly questions of mine..[/quote]
Hi fredbel6!
In fact, there are many Vietnamese asked the same question.
I searched the answer on the Internet and the following is what I found.
-The percentages of some surnames which have most people having these surnames in Vietnam:
 + Nguyen (Nguyễn) :38,4%
 +Tran (Trần) : 11%
 +Le (Lê) : 9,5%
 +Huynh or Hoang (Huỳnh or Hoàng) : 5,1%
 +Pham (Phạm) : 5%

-Most reasons explaining why there are a lot of people having Nguyen surname are from history.
In past, when some people established new dynasties, they were afraid of people's revenge in old dynasties (descendants or relationship of the old Kings, these people didn't have Nguyen surname). So the Kings ordered them to change their surname into Nguyen.
In other dynasties, some families helped the Kings (having Nguyen surname) establish new dynasties, so they were permitted change their surname into Nguyen.

-As I know, it is normal to ask some young people about age, but it is not police to ask old people (I don't know why, may be from traditions)

[quote="fredbel6"]Thanks for everyone's reply

that 38% is that within the male vietnamese population, or the total vietnamese population?
(it's just so weird, and the same goes for koreans, my dad met a group of at most ten koreans but there were at least three kims in there) :o[/quote]

This percentage is in total Vietnamese population.


[quote="mathematica"]Find the value of $\sum_{n=1}^{\infty}{\arctan(\frac{1}{n^2+n+1})}$.[/quote]
$\frac{\pi}{4}$ since,
$\arctan\frac{1}{n^2+n+1}=\arctan(n+1)-\arctan{n}.$ :)

[quote="arqady"]
$\frac{\pi}{4}$ since,
$\arctan\frac{1}{n^2+n+1}=\arctan(n+1)-\arctan{n}.$ :)[/quote]
the answer would be $\frac{\pi}{2}$ because $\sum_{n=1}^{\infty}{\arctan(\frac{1}{n^2+n+1})}=\lim_{n \to \infty}\arctan(n+1)=\frac{\pi}{2}$.

[quote="mathematica"]
the answer would be $\frac{\pi}{2}$ because $\sum_{n=1}^{\infty}{\arctan(\frac{1}{n^2+n+1})}=\lim_{n \to \infty}\arctan(n+1)=\frac{\pi}{2}$.[/quote]
$\sum_{n=1}^{\infty}{\arctan(\frac{1}{n^2+n+1})}=\lim_{n\rightarrow+\infty}\sum_{k=1}^n(\arctan(k+1)-\arctan k)=$
$=\lim_{n\rightarrow+\infty}(\arctan(n+1)-\arctan1)=\frac{\pi}{4}.$ :)


MATHCOUNTS
AMC
AMC 12
number theory
prime factorization
AMC 10

does anybody know any good mathcounts prep books?
besides AoPs 1 and 2 of course.

Hmm... I don't know. I guess posting in the problem marathons can be good practice though.

Try
[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=6][size=200]The All-Time Greatest MATHCOUNTS Problems[/size][/url]
Don't have it, but its probably a good book!
Also, if you need Algebra help, try the intro to algebra!  I have it and its great!
-minicon

theres a topic in high school basics called personal work over the summer. its got a ton of formulas and stuff.(nice job prime factorization :))

Frank Ho's [i]Math Contest Preparation[/i] is a book that teaches a lot of tricks and tips for use in MATHCOUNTS.

You can read a little about it at [url]http://www.mathandchess.com/catalog/item/1240083/723838.htm[/url]

Unfortunately it cannot be ordered on the internet.  You have to contact Frank Ho and request it.  The information needed to contact him is found at [url]http://www.geocities.com/danb80/ContestMathPreparation.html[/url]

Book on prep for Math Olympiads are also good, but some exceed the level of MATHCOUNTS.  One that is available from Amazon is [i]First Steps for Math Olympians: Using the American Mathematics Competitions (Problem Books)[/i] but it is expensive. :(

[quote="abacadaea"]have you tried google?[/quote]

Well, I don't think Google can give you everything that you need for the Mathcounts...

[quote="moogra"]AoPS Volume 1 is perfect for mathcounts. Also you can study the AMC problems free on this site. I have realized that the AMC12 is around the same level as state-national problems for mathcounts[/quote]

i don't really think so. easier AMC 12 questions can range from medium to the hardest state mathcounts problems.
whereas medium and beyond AMC 12 questions are easily national mathcounts level problems.

Yeah,  think that the medium AMC 10 questions are close to state mathcounts. This year, number 30 on the sprint was almost exactly the same as #20 (?) on the AMC 10.

What about the post for yasharbiz on the MathCounts resources thread.

[url]http://mathcounts.saab.org/mc.cgi/[/url][url]
has lots of problems.  In my opinion, you can't get any better than AoPS 1 + some parts of AoPS 2 + lots of practice.[/url]

[quote="mihail911"]
whereas medium and beyond AMC 12 questions are easily national mathcounts level problems.[/quote]

um
that's not true at all?

I don't think MATHCOUNTS problems require trig...

you can buy tests at
[url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=183&z=6[/url]

Ya'll might want to actually EXPEND a [i]little[/i] in finding them yourselves before asking. Like, say, reading stickys in the mathcounts forum? ;) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]such as this one on MATHCOUNTS RESOURCES.[/url]

... ;)


Prove that at most $ 4$ powers of $ 2$ can have the same number of digts.

[hide="Hint1"]$ 10 < 2^4$[/hide]
[hide="Hint2"]$ 2^n$ has $ k$ digits in base $ 10 \iff 10^{k\minus{}1} \le 2^n < 10^k$[/hide]
[hide="Hint3"]Assume that it is possible for $ 5$ and compute the quotient of the lower an upper one[/hide]

[hide="Solution"]Assume that there are five powers of $ 2$ with the same number of digits and
let the lower one be $ 2^n$ and the upper one $ 2^m$ then $ 2^n \cdot 2^4 \le 2^m$
$ 10^{k\minus{}1} \le 2^n < 10^k \Rightarrow 2^4 \cdot 10^{k\minus{}1} \le 2^4 \cdot 2^n \le 2^m$ and
$ 10 < 2^4 \Rightarrow 10^k < 2^4 \cdot 10^{k\minus{}1} \le 2^4 \cdot 2^n \le 2^m$
however $ 10^{k\minus{}1} \le 2^m < 10^k$ contradiction
[/hide]

Now show that there are no two powers of $ 2$ that are permutations of one another (no leading zeros)

The order of 2 mod 9 is 6.  In light of the previous proof, $ \boxed{\text{QED}}$.  :D

[hide="more drawn out proof"]If two powers of two are permutations of each other, they must have the same number of digits and their digital sums are equal.  As well known, the digital sum of a number is equal to it mod 9.  Now we have $ 2^n\equal{}2^m \mod 9$ means that $ 2^m(2^{n\minus{}m}\minus{}1)\equal{}0 \mod 9$.  The second factor must be the one divisible by 9, and we want the least distance between $ n$ and $ m$.  The least value possible for $ n\minus{}m$ is 6, so the powers of 2 must be 6 apart.  But we know that they can only be 4 apart to have the same number of digits. $ \boxed{\text{QED}}.$[/hide]


Can you do this inequality by pqr ?
$ a, b, c>0; a\plus{}b\plus{}c\equal{}3.$ Find min:
$ (3\plus{}2a^2)(3\plus{}2b^2)(3\plus{}2c^2)$

[quote="DangChienbn"]Can you do this inequality by pqr ?
$ a, b, c > 0; a \plus{} b \plus{} c \equal{} 3.$ Find min:
$ (3 \plus{} 2a^2)(3 \plus{} 2b^2)(3 \plus{} 2c^2)$[/quote]
we can use mixing variable
supposing$ c \ge b \ge a$
let $ (3 \plus{} 2a^2)(3 \plus{} 2b^2)(3 \plus{} 2c^2) \equal{} f(a,b,c)$
=> $ f(a,b,c) \ge f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c)$
<=>$ (a \minus{} b)^2(3 \minus{} \frac{(a \plus{} b)^2 \plus{} 4ab}{4}) \ge 0$
Find min:
$ f(c) \equal{} (3 \plus{} 2c^2)(3 \plus{} \frac {(3 \minus{} c)^2}{2})^2$     ,   ($ c \ge 1$)
$ f'(c) \equal{} \frac {3}{4}((c \minus{} 3)^2 \plus{} 6)(c^2 \minus{} 3c \plus{} 3)(c \minus{} 1) \ge 0$
=>$ f(c) \ge f(1) \equal{} 125$

[quote="fesha3t"][quote="DangChienbn"]Can you do this inequality by pqr ?
$ a, b, c > 0; a \plus{} b \plus{} c \equal{} 3.$ Find min:
$ (3 \plus{} 2a^2)(3 \plus{} 2b^2)(3 \plus{} 2c^2)$[/quote]

ặc, sao tiếng anh ko vậy nè, nói việtnamese ai hỉu ko zậy

p=a+b+c
q=ab+bc+ca
r=abc
$ (3 \plus{} 2a^2)(3 \plus{} 2b^2)(3 \plus{} 2c^2) \equal{} 27 \plus{} 8r^2 \plus{} 162 \minus{} 36q \plus{} 12q^2 \minus{} 72r\geq 125$
$ < \equal{} > 11 \plus{} 24r \plus{} 162 \minus{} 36q \plus{} 12q^2 \minus{} 72r \minus{} 125 \geq 0$
$ < \equal{} > 48 \minus{} 48r \plus{} 12q^2 \minus{} 36q\geq 0$
$ < \equal{} > 48 \minus{} 48r \plus{} 12\sum a^2.b^2 \minus{} 12(ab \plus{} bc \plus{} ca)\geq 0$
$ < \equal{} > 4 \minus{} 4r \plus{} \sum a^2.b^2 \minus{} \sum ab\geq 0$

$ \sum a^2.b^2 \plus{} 1\ge 4r$
$ 3\geq \sum ab$[/quote]

Nice;solution  :P
Không nên nói tiếng việt ở đêy nhá ông  :rotfl:

[quote="DangChienbn"]Can you do this inequality by pqr ?
$ a, b, c > 0; a \plus{} b \plus{} c \equal{} 3.$ Find min:
$ (3 \plus{} 2a^2)(3 \plus{} 2b^2)(3 \plus{} 2c^2)$[/quote]
@DangChienbn: Did you make this inequality, or do you know who originally made it? :)

$ a,b,c > 0$.Prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \left(3\sqrt [3]{4} \minus{} 2\right)\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\geqslant 3\sqrt [3]{4} \plus{} 1$

 $ a\geqslant 4$, such that $ a \equal{} b \plus{} 1$.Prove that:
$ a^b < b^a$

He is fesha3t :), Fesha3t is the author of that ineq :D


:help:
what is the harder?  apmo(geometry) or imo (geometry)  :?:
please help me ,

Definately IMO. Just think about it, problem 6 of the IMO is usually geometry and IMO as a whole is harder than APMO


Asymptote
limit
calculus
calculus computations

hi i need help with this problem

find the horizontal asymptote
sqrt((x^2)+10x+5)-x

what i tried was factoring out the x^2 from the equation, in which i got an answer of sqrt(1) or 1

so if anyone can point me in the right direction it would be great, thanks


limit
integration
logarithms
calculus
calculus computations

${\lim_{k\to \infty }\, \left(\sum_{n=0}^{k}\frac{n^{1/k}}{k+n}\right)}$

find this limit

This is Jack London Constants'

how can i proof

${\lim_{k\to \infty }\, \left(\sum_{n=0}^{k}\frac{n^{1/k}}{k+n}\right)=\lim_{k\to \infty }\, \int_{0}^{k}\frac{n^{1/k}}{k+n}\, dn=\log (2)}$

Let $x_{k}$ be your sequence. Then

$\frac{1}{k+1}+\cdots+\frac{1}{k+k}\leq x_{k}\leq \left(\max_{n=1\cdots k}\sqrt[k]{n}\right)\left(\frac{1}{k+1}+\cdots+\frac{1}{k+k}\right)$

Letting $k$ converge to $\infty$ we get that the limit is $\ln 2$.

What do you mean by Jack London's constant?

uhaha

Jack London's constant

hehe... its just a joke.... hehehe

you guys never understand good jokes


linear algebra
matrix
geometry
3D geometry
MATHCOUNTS
ARML
rich tag

An idea I've seen on other places. Everyone is free to post once a small review of their own about the movie they just saw. Include a fair mark on a 1-10 scale if you like.

I'll start: [b]StarWars: The empire strikes back[/b]. I'd give it a 6. A classical, and a "must" watch. I think I saw the revised edition. The story line is somewhat bi-linear, always working on two lines, with short intersections and crossovers. The famous "I am your father" line is always impressing. I didn't like the way Luke slides down a tube after falling something like 2 miles.

Oh well let's not forget to mention the two strangely amusing robots R2 and the other guy (I keep forgetting his name :D)

It's C3PO.  If it's that good, why'd you give it a 6?

[b]KungFu Hustle[/b]: 10! its just awesome  :lol:

Well, we just finished yesterday the Latin American Film Festival here in GATech. I would suggest a few of those, but probably those are movies you have never heard about and most likely you never will unless you are Latin American. Any way, I recomend [b]Ojos que no ven[/b] (Inexactly translated as [b]What the Eye Doesn't See[/b], there are two with that name, the one I talk about was filmed in 2003 and it is from Peru), it is interesting. Also [b]Sexo con Amor[/b], that is a Chilean comedy, very good one, but just recomended for mature audiences. As proud hispanic, I would recomend a lot of Spaniard and Latin American movies, but probably most of them would not be available in most of the places (eventhough, you would get surprised what BlockBuster may have!)

A Science fiction movie that I liked a lot is [b]Thirteenth Floor[/b], that is a must! (and I think it is very likely that many of the ideas of the Matrix were taken from that movie.)  And of course, who can forget [b]The Cube[/b]

Another that I saw recently is [b]50 First Dates[/b]. I didn't go to the movie theater to watch it, because I had the idea that Adam Sandler just do stupid movies, but the movie is actually quite good. Probably it would be one of the movies that teenager guys consider [i]girl's movies[/i], but it is a good romantic-comedy-drama.

[quote="djimenez"]

Another that I saw recently is [b]50 First Dates[/b]...........movie is actually quite good....[/quote]

That was Sandlers WORST movie. I love most of Adam Sandlers movie but that movie was pretty bad.

[b]ring 2[/b] is awesome. if you saw the first one, then you should go see this one, it's better. 8.5 out of 10(cool special effects)

[quote="GoBraves"][quote="djimenez"]Another that I saw recently is [b]50 First Dates[/b]...........movie is actually quite good....[/quote]

That was Sandlers WORST movie. I love most of Adam Sandlers movie but that movie was pretty bad.[/quote]

Well, may be it is because it is not exactly a comedy, it is more a drama, eventhough it has pretty funny scenes.

It is somehow similar to what people say about [b]Cable Guy[/b] and Jim Carrey, the problem is that it is not a comedy, even when it has a lot of funny scenes, it is a Drama, not a commedy.

i just saw [b]sin city[/b]. pretty good movie for having a storyline full of inappropriate stuff. Still give it a 5-7 becuase of that inappropriateness.

It's spring break and to start off my week, I saw [b]Godfather[/b]. Easily the greatest movie of all time. If you did not watch it, then do so! I give it a 10. :lol:

Thats a late spring break....

Anyways, I just saw [b]Good Will Hunting[/b], and this was a very good movie.

Everyone should go see it (at least sometime in their life)

Have you guys seen the movie or the book [b]I,Robot[/b]?
It's pretty much about computer--but still--I like the concepts it's introducing.
LALALA
MATH AND COMPUTING UNITED

[quote="Valentin Vornicu"][b]I, Robot[/b] pretty good actually :) I'd give it a 7 for special effects and for .... Will Smith. I just love his humour :D[/quote]

That WAS a good movie. I give it 8.5

[quote="Valentin Vornicu"]I, Robot pretty good actually  I'd give it a 7 for special effects and for .... Will Smith. I just love his humour [/quote]

Yeah, I'm a harsh harsh movie critic, and I thought this one was good. Obviously it had little to do with the actual book, which I loved, but it had a decent story, an okay twist, great special effects, and Smith was good. I'd give it a 7.5 maybe.

[quote="xxreddevilzxx"]i just saw sin city. pretty good movie for having a storyline full of inappropriate stuff. Still give it a 5-7 becuase of that inappropriateness.[/quote]

Seems a little unfair to give it a bad grade because of its content. [b]Last Tango in Paris[/b] wasn't a bad movie, now was it? But in any case, Sin City isn't really that bad. Only the violence distinguishes it from any other R-rated movie. And in terms of violence, it was a little shocking at first. Later I got bored. There's only so much bloodlust can do for ya' - and the plot wasn't exactly rewarding. Acting is also irrelevant, since the movie is mostly in black-and-white and theres much more narration that dialogue.

the movies that i just saw:Constantine,Blade Trinity,and War of the world

War of The Worlds.

Good enough to stop me from revising! :(

The little girl was somewhat annoying (for a short period) but overall I will give it $ 7 \plus{} \varepsilon$! (Where $ \varepsilon > 0$).

[quote="perfect_radio"]

[url=http://www.imdb.com/title/tt0114787/]Underground (Подземље)[/url]
It's absurd, it's funny and it's about Yugoslavia. Ceausescu has a cameo, too :lol:

....

BTW, what foreign movies do you recommend for the collection?[/quote]

Well, I've seen some more movies.

[url=http://www.imdb.com/title/tt0410730/]Taxidermia[/url] (Hungary,2006)
Minors and easily offended, please don't click
[hide]When I rented it, I thought it was just a fun hungarian movie about communism, but I soon discovered that it is one of the most twisted, sick films I have ever seen.

The story focusses on three generations,and one man plays a role in all three parts. In this movie, eating as much as possible is some sort of an olympic sport, and this man is a "huge talent".

Sex with a candle, drinking the water used by women to bathe in it, explicit masturbation scenes, close-ups of genitals of animals, excrement, excessive obesity and vomiting and even pedophilia.... this movie is just unbelievable.

But also very original :| [/hide]

[url=http://www.imdb.com/title/tt0397493/]Kako ubiv svetec[/url], or How I killed a saint (Fyrom, 2004)

This movie takes place in Skopje, Macedonia. The year is 2001 and trouble starts with Albanian separatists.  NATO troops and foreign diplomats are all over the place.  Kokan is a young Macedonian, who is very anti-NATO.  The movie starts when for some reason, his older sister comes back from the United States.
It's pretty interesting.  Kokan's character is not very nice, he is a "nationalist", but he seems mostly motivated by his own personal frustations.

And then also :
[url=http://www.imdb.com/title/tt0114787/]Bila jednom jedna zemlja[/url] , or Underground (then-Yugoslavia,1995)
Like Zivot je cudo, again a very long film (166 minutes), but this one didn't feel as streched out like the other movie.
The Kusturica-style is all over the place : characters being drunk, loud and having mood swings all the time :roll: ....
Very good though.

Hmm...

I recently saw [b]Tropic Thunder[/b] and [b]Saving Private Ryan[/b] and [b]Matrix 1, Matrix 2, Matrix 3[/b]...

OK.

So [b]Tropic Thunder[/b] deserves a 9.0; it has wonderful screenplay, and also lots of [i][u][b]VIOLENCE!!![/b][/u][/i]

[b]Saving Private Ryan[/b] gets a 9.5, for its great emotional aspect, [i][u][b]VIOLENCE!!![/b][/u][/i]
, and other [i]nice[/i] stuff...

[b]Matrix 1, 2, 3[/b] all get 6.5's. I like the deep philosophy behind the ideas, but, I can't be attracted to an actor who says like 20 words in 2 hours. Come on, we can do better than that!

Has anyone here watched the movie "Seven Pounds"?

It deserves at least 9.5 out of 10, Will Smith deserves 10 out of 10 for his outstanding play.

Gran Torino was really great. There were some gangstery looking people in the theater, and they seemed to be really deep in thought after watching it, or at least they looked like it :O
Clint Eastwood definitely captured the stereotypical grumpy old white man.

I watched twilight. great movie. so was the book! :love:

massive revival???

ok w/e

Ratatouille, 8/10. very funny
Wall-E, 6.5/10, not that funny, but very [i]realistic[/i]. :wink:

Endgame. Sorry no spoilers but if you really want to know: [hide]The avengers find thanos and kill him and they get the stones on the infinity gotlet and snap to reverse the snap that killed half the population and they succeed but then thanos from the past comes and tries to kill the avengers but since the avengers reversed the snap thanos did to kill half the population, they have every single superhero on their team including a new one (I forgot the name). Then Iron Man gets the gotlet and snaps, which kills thanos and his team but, the half population includes iron man himself so he dies which is sad. Then they have to scatter the infinity stones they gathered and they scatter them in time so it is harder for other people to find it. Also I forgot to add. They time travel with a quantum thingymajingy.[/hide]

I watch Ratatouille and so far, it is my favorite movie, I have no idea how Disney has such great movie ideas, Ratatouille is rated 999999/10
I just watch Toy Story 4, also really good, there are some sad parts in the movie but it is mostly humor. I rate 10/10

yo the last post was more than a year ago by me 0_0

This thread was literally made in 2005....

[quote=vicseb]I saw Borat, which is really funny, but I'd give it a lower grade for some VERY inappropriate parts. Probably a 6.5[/quote]

Borat has nothing to do with kazakhstan ok? there were no actual Kazakh actors, it was filmed in romania and that person was speaking hebrew and polish the whole time


limit
trigonometry
real analysis
real analysis unsolved

Let $ x \in \mathbb{R}\ \ \text{and}\ \ f_{0}(x)=\sin(x),\ f_{n}(x)=\sin(f_{n-1}(x))$. Find $ \lim_{n\to \infty}\sqrt{n}f_{n}(x).$
PS. This problem may seem easy to those familiar with numerical analysis.

[hide="solution"]
We can show that $ \lim_{n}f_{n}=0$ and there $ \exists_{m}\ \forall_{n>m}\ f^{2}_{n}$ is decreasin', thereafter:
$ \lim_{n}\sqrt{nf_{n}^{2}}=\lim_{n}\sqrt{\frac{n}{\frac{1}{f_{n}^{2}}}}$, so lets compute:
$ \lim_{n}\frac{n}{\frac{1}{f_{n}^{2}}}$, by Cesaro-Stolz lemma:
$ \lim_{n}\frac{1}{\frac{1}{f_{n+1}^{2}}-\frac{1}{f_{n}^{2}}}=\lim_{n}\frac{f_{n}^{2}f_{n+1}^{2}}{f_{n}^{2}-f_{n+1}^{2}}=\lim_{n}\frac{f_{n}^{2}\sin^{2}f_{n}}{f_{n}^{2}-\sin^{2}f_{n}}=\lim_{x\to 0^{+}}\frac{x^{2}\sin^{2}x}{x^{2}-\sin^{2}x}$ and for example by de L'Hospital rule the limit equals $ 3$, so:
$ \lim_{n}\sqrt{n}f_{n}=\sqrt{3}$[/hide]

[quote="The Captain"]Of course for $ x\neq 0$ :wink:[/quote]
Of course you're right   :)
even for $ x\neq k\pi$ for $ k\in\mathbb{Z}$ :wink:

See a extension in my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=45424


1. One plane leaves the airport at noon heading east at x mph. Another plane leaves the same
airport heading west 30 minutes later at y mph. At 3 p.m., the two planes are 1800 miles apart.
At 5 p.m., the two planes are 3100 miles apart. Find x – y.

Answer:
[hide]50[/hide]

2. Find the volume of the sphere inscribed in a regular tetrahedron with total surface area
36radical3 . Round to the nearest 0.0001 cubic unit.

Answer:
[hide]7.6953[/hide]

Hi returnofkiladi.

1.

At $ 3$pm, one plane has been travelling for $ 3$ hours and the other for $ 2.5$ hours.

$ 3x \plus{} 2.5y \equal{} 1800$

Similarly

$ 5x\plus{}4.5y \equal{} 3100$

$ 6x\plus{}5y\equal{}3600$
$ 10x\plus{}9y\equal{}6200$

$ 30x\plus{}25y\equal{}18000$
$ 30x\plus{}27y\equal{}18600$

$ 2y\equal{}600$

$ y\equal{}300$

Therefore $ 6x\equal{}3600 \minus{} 1500$
$ x\equal{}350$

$ x\minus{}y \equal{} 50$.

Problem 2:
a: side length
r: radius of the sphere

Step 1: Surface area of a tetrahedron 4 equilateral trianlges- a^2*sqrt(3)= Surface area ( given in the problem)
Step 2: radius of a inscribed spehere in a tetrahedron = a/sqrt(24).
Step 3: Volume of a spehere is 4/3*pi*r^3.

find a and r by using Step 1 and Step 2, Then you can find the volume of a sphere.


$3x-y=10$
$2x+5y=1$

Use second-order determinants.
Your answer should be an ordered pair in a solution set.

Please show your work.
AND
Explain each step please.

in $LaTex$ please!

[quote="indianamath"]what are second-order determinants?[/quote]

Nobody needs to post anymore
I think I got the correct answer already

Second-order determinants include two equations i think...if memory serves me well :ninja:

You can write it as Ax=b and then solve for x. (i.e. $x=A^{-1}b$ ) or use Gauss elimination on [A |b]. The system has a unique solution.

Cramer's rule rings a bell, but I can't recall having heard of second order determinants unless I'm computing eigenvalues.


This one is easy but quite nice  :)

In triangle $ABC$, $M$ is the midpoint of $BC$. A line passing through $M$ divides the perimeter of triangle $ABC$ into two equal parts. Show that this line is parallel to the internal bisector of $\measuredangle A$.

[hide="Proof"]Standard markings $a,b,c$. WLOG assume $b<c$. Then the line in question will intersect $AB$ and not $AC$. (Assume the opposite and denote the intersection by $N$. Then $CN={b+c\over 2}>b$, hence $N$ is outside $AC$.) If $AD$ is the bisector of $\angle A$, where $D\in BC$, then $b<c$ implies that $D$ is between $C$ and $M$. (By Angle Bisector Theorem, $CD=\frac{ab}{b+c}<\frac{ab}{b+b}={a\over 2}$.) By Angle Bisector Theorem $BD=\frac{ac}{b+c}$ and by construction $BM={a\over 2}, BN=\frac{b+c}{2}$ (since $BM+BN=\frac{a+b+c}{2}$), hence

$BD: BA=\frac{ac}{b+c}: c=\frac{a}{b+c}={a\over 2}: \frac{b+c}{2}=BM: BN$

which means that $\triangle BMN\sim\triangle BDA\implies MN\parallel AD$.

If $b=c$, then the two lines coincide and represent the symmetrial axis of the given isosceles triangle.[/hide]

[hide="Another Proof"]
We assume $AB>AC$.

Construct a point $F$ on the extension of $BA$ such that $AF=AC$. Then triangle $FAC$ is isoceles with $\angle AFC = \angle ACF = \theta$. Then $\angle BAC = 2\theta$ and the angle bisector of $A$ makes an angle of $\theta$ with the line $BF$. Hence it is parallel to $CF$.

Let the line from $M$ intersect $AB$ at $D$. Then $D$ is the midpoint of $BF$. But since $M$ is the midpoint of $BC$, triangle $DBM$ is similar to triangle $FBC$. Hence $MD$ is parallel to $CF$.

So $BF$ is parallel to $MD$.[/hide]


[b]Problem: (Truong Minh Tien)[/b]Find $ x,y,z \in Z$ such that:
$ x^5\minus{}y^3\plus{}x!\plus{}xy\minus{}3y\minus{}x\equal{}0$

Does everyone interested in this one?  :(


algebra
polynomial
calculus
integration
modular arithmetic
function
domain

Find all integers [b]a[/b] and [b]n[/b] such that

$ t^2 \minus{} t \plus{} 1$ divides $ t^n \plus{} t \plus{} a$ .

[hide="Umm..."]
By the loooong division method,

i got 2 solutions  $ ( n,a ) \equiv (4,0) \ and \ (5,1)$

Any other method?? or more solutions ??   :idea:

[/hide]

$ (t^2 \minus{} t \plus{} 1)(t \plus{} 1) \equal{} t^3 \plus{} 1$.  How does this help?

[b]Claim:[/b]  $ t^n$ leaves the same remainder as $ \minus{}t^{n\minus{}3}$ upon division by $ t^2 \minus{} t \plus{} 1$.

Prove the claim.  How does it help?

It does help.

Let $ n \equal{} m(mod 6)$ for $ 0 \le m \le 5$.

As $ t^3 \equal{} \minus{} 1(mod (t^2 \plus{} t \plus{} 1)) \Rightarrow t^6 \equal{} 1(mod (t^2 \plus{} t \plus{} 1))$
Then $ t^n \plus{} t \plus{} a \equal{} t^m \plus{} t \plus{} a \equal{} 0(mod (t^2 \plus{} t \plus{} 1))$

$ m \equal{} 0,1,2$ gives no solution for $ a$
$ m \equal{} 3$ gives $ t^3 \plus{} t \plus{} a \equal{} \minus{} 1 \plus{} t \plus{} a \equal{} 0(mod (t^2 \plus{} t \plus{} 1))$, no solution.
$ m \equal{} 4$ gives $ t^4 \plus{} t \plus{} a \equal{} \minus{} t \plus{} t \plus{} a \equal{} 0(mod (t^2 \plus{} t \plus{} 1))$, gives $ a \equal{} 0$
$ m \equal{} 5$ gives $ t^5 \plus{} t \plus{} a \equal{} \minus{} t^2 \plus{} t \plus{} a \equal{} 0(mod (t^2 \plus{} t \plus{} 1))$, gives $ a \equal{} \minus{} 1$

So $ (a,n) \equal{} (0,6k \plus{} 4), ( \minus{} 1, 6k \plus{} 5)$, where $ k$ is any non-negative integer, are the only solutions.

[hide="Alternative"]Let $ \omega^3 \equal{} 1$ with $ \omega \neq 1$. Note that the two roots of $ t^2 \minus{} t \plus{} 1$ are $ \minus{}\omega$ and $ \minus{}\omega^2$. Therefore, the first polynomial divides the second iff $ \minus{}\omega$ and $ \minus{}\omega^2$ are both roots of the second polynomial.

If $ n$ is even, then $ (\minus{}\omega)^n \minus{} \omega \plus{} a \equal{} \omega^n \minus{} \omega \plus{} a$. If $ n \equiv 0 \mbox{ mod 3}$, then $ a \equal{} \omega \minus{} 1$, yielding no solutions for $ a$. If $ n \equiv 1 \mbox{ mod 3}$, then $ a \equal{} 0$. If $ n \equiv 2 \mbox{ mod 3}$, then $ a \equal{} \omega \minus{} \omega^2$, an impossibility.

If $ n$ is odd, then $ (\minus{}\omega)^n \minus{} \omega \plus{} a \equal{} \minus{}\omega^n \minus{} \omega \plus{} a$. If $ n \equiv 0 \mbox{ mod 3}$, then $ a \equal{} \omega \plus{} 1$, which is impossible. If $ n \equiv 1 \mbox{ mod 3}$, then $ a \equal{} 2\omega$, again impossible. If $ n \equiv 2 \mbox{ mod 3}$, then $ a \equal{} \omega^2 \plus{} \omega \equal{} \minus{}1$.

Therefore, the possible integral values for $ (n, a)$ are $ (6k \plus{} 4, 0), (6k \plus{} 5, \minus{}1)$, where $ k$ is nonnegative.
[/hide]

[quote="Zhero"][hide="Alternative"]Let $ \omega^3 \equal{} 1$ with $ \omega \neq 1$. Note that the two roots of $ t^2 \minus{} t \plus{} 1$ are $ \minus{} \omega$ and $ \minus{} \omega^2$. Therefore, the first polynomial divides the second iff $ \minus{} \omega$ and $ \minus{} \omega^2$ are both roots of the second polynomial.

If $ n$ is even, then $ ( \minus{} \omega)^n \minus{} \omega \plus{} a \equal{} \omega^n \minus{} \omega \plus{} a$. If $ n \equiv 0 \mbox{ mod 3}$, then $ a \equal{} \omega \minus{} 1$, yielding no solutions for $ a$. If $ n \equiv 1 \mbox{ mod 3}$, then $ a \equal{} 0$. If $ n \equiv 2 \mbox{ mod 3}$, then $ a \equal{} \omega \minus{} \omega^2$, an impossibility.

If $ n$ is odd, then $ ( \minus{} \omega)^n \minus{} \omega \plus{} a \equal{} \minus{} \omega^n \minus{} \omega \plus{} a$. If $ n \equiv 0 \mbox{ mod 3}$, then $ a \equal{} \omega \plus{} 1$, which is impossible. If $ n \equiv 1 \mbox{ mod 3}$, then $ a \equal{} 2\omega$, again impossible. If $ n \equiv 2 \mbox{ mod 3}$, then $ a \equal{} \omega^2 \plus{} \omega \equal{} \minus{} 1$.

Therefore, the possible integral values for $ (n, a)$ are $ (6k \plus{} 4, 0), (6k \plus{} 5, \minus{} 1)$, where $ k$ is nonnegative.
[/hide][/quote]

How was $ a$ found in each case?

$ t^{6k\plus{}4} \plus{} t \equiv t^4 \plus{} t \equiv 0 \bmod t^3 \plus{} 1$ and $ t^{6k\plus{}5} \plus{} t \equiv t^5 \plus{} t \equiv \minus{}t^2 \plus{} t \equiv 1 \bmod (t^2 \minus{} t \plus{} 1)$.

I actually meant to ask how Zhero got $ a \equal{} \omega \minus{} 1$ in the first part of his solution.  And also, why is stephencheng working in $ \pmod{t^2 \plus{} t \plus{} 1}$ instead of $ \pmod{t^2 \minus{} t \plus{} 1}$?

EDIT: Also, can someone prove the claim?

[quote="sunehra"]I actually meant to ask how Zhero got $ a \equal{} \omega \minus{} 1$ in the first part of his solution.[/quote]
If $ 6 | n$, then $ \omega^n \minus{} \omega \equal{} 1 \minus{} \omega$.

[quote="sunehra"]Also, can someone prove the claim?[/quote]
$ t^n \plus{} t^{n \minus{} 3} \equal{} (t^3 \plus{} 1)t^{n \minus{} 3}$.

The point here is that $ \mathbb{Q}[t]/(t^2 \minus{} t \plus{} 1)$ is isomorphic to $ \mathbb{Q}[\omega]$ where $ \omega$ is a primitive sixth root of unity.  That's why the polynomial remainder solution and the explicit-computation solution are equivalent.  The way you're talking, it looks as if this isomorphism hasn't clicked yet; once it does, hopefully all of these computations will make sense.

Oh I understand Zhero's solution now. He was applying the fact that $ \omega^3 \equal{} 1$. If I am correct, the notation $ \mathbb{Q}[t]$ denotes the polynomial in $ t$ over the ring $ \mathbb{Q}$ (rationals?). But I don't understand what you mean by $ \mathbb{Q}[t]/(t^{2}\minus{}t\plus{}1)$ is isomorphic to $ \mathbb{Q}[w]$. An isomorphism is a bijection right?

[quote="sunehra"]Oh I understand Zhero's solution now. He was applying the fact that $ \omega^3 \equal{} 1$.[/quote]
$ \omega^3 \equal{} \minus{}1$.

[quote="sunehra"]But I don't understand what you mean by $ \mathbb{Q}[t]/(t^{2} \minus{} t \plus{} 1)$ is isomorphic to $ \mathbb{Q}[w]$. An isomorphism is a bijection right?[/quote]
An isomorphism is a bijection that respects the ring structure.  $ \mathbb{Q}[t]/(t^2 \minus{} t \plus{} 1)$ is constructed by a sort of generalization of modular arithmetic where you take remainders modulo $ t^2 \minus{} t \plus{} 1$.  Any such remainder is uniquely a polynomial $ at \plus{} b$ of degree one - this is just polynomial division - and multiplication in this ring looks like this:

$ (at \plus{} b)(ct \plus{} d) \equal{} act^2 \plus{} (ad \plus{} bc)t \plus{} bd \equiv (ad \plus{} bc \plus{} ac)t \plus{} (bd \minus{} ac) \bmod (t^2 \minus{} t \plus{} 1)$.

Addition is componentwise.  This is the [b]same[/b] as the multiplication you would get if you just repeatedly applied the relation $ t^2 \equal{} t \plus{} 1$; in other words, if you treated $ t$ as a primitive sixth root of unity.  In fact, these polynomial quotients are exactly how field extensions are constructed abstractly.

[quote="t0rajir0u"]The point here is that $ \mathbb{Q}[t]/(t^2 \minus{} t \plus{} 1)$ is isomorphic to $ \mathbb{Q}[\omega]$ where $ \omega$ is a primitive sixth root of unity.  That's why the polynomial remainder solution and the explicit-computation solution are equivalent.  The way you're talking, it looks as if this isomorphism hasn't clicked yet; once it does, hopefully all of these computations will make sense.[/quote]

I might be confused but that statement doesn't seem to say much. In $ \mathbb{Q}[\omega]$, all we know is that we can do computations using the fact that $ \omega^2 \minus{} \omega \plus{} 1 \equal{} 0$...So trivially it is an isomorphism.

As another comment, we can approach this in a slightly different way that uses a nice idea:

we have a homomorphism $ \phi: \mathbb{Q}[x]\rightarrow C$ where what we do is we "plug in" $ \omega$ for $ x$. So $ \phi(a_0 \plus{} a_1x \plus{} ... \plus{} a_n x^n) \equal{} a_0 \plus{} a_1\omega \plus{} ... \plus{} a_n (\omega \omega...\omega)$. So anyway, if we collect all $ f\in\mathbb{Q}[x]$ such that $ \phi(f) \equal{} 0$ in a set $ S$. This is an an ideal of $ \mathbb{Q}[x]$ and this is a principle ideal domain so $ S \equal{} \mathbb{Q}[x]*f_0$ where $ f_0$ is some polynomial... (that is to say this $ S$ is some fixed polynomial times any other polynomial). We can see that $ f_0 \equal{} x^2 \minus{} x \plus{} 1$ (or something that differs by a constant).

So that all is a long winded way of saying that if $ p(x)$ is a polynomial with rational coefficients, then $ p(\omega) \equal{} 0$ implies that $ x^2 \minus{} x \plus{} 1|p(x)$.

So Zhero's solution is a little sloppy. Because if we are working in $ \mathbb{C}$, he essentially showed that $ p(\omega)\equal{}0$ so of course $ (t\minus{}\omega)$ divides the polynomial. But we'd still need that the other root of $ t^2\minus{}t\plus{}1$ divides the polynomial. But this follows by the idea above. (or say conjugating everything).

[quote="Altheman"]I might be confused but that statement doesn't seem to say much. In $ \mathbb{Q}[\omega]$, all we know is that we can do computations using the fact that $ \omega^2 \minus{} \omega \plus{} 1 \equal{} 0$...So trivially it is an isomorphism.[/quote]
Well, yes.  But many people don't seem to realize that substituting roots is equivalent to working in a quotient ring.

[quote="Altheman"]we have a homomorphism $ \phi: \mathbb{Q}[x]\rightarrow C$ where what we do is we "plug in" $ \omega$ for $ x$.[/quote]
That's essentially what I meant by $ \mathbb{Q}[\omega]$; sorry if that wasn't clear.  We only talk about this structure up to isomorphism, after all.

[quote="Altheman"] if we are working in $ \mathbb{C}$, he essentially showed that $ p(\omega) \equal{} 0$ so of course $ (t \minus{} \omega)$ divides the polynomial. But we'd still need that the other root of $ t^2 \minus{} t \plus{} 1$ divides the polynomial. But this follows by the idea above. (or say conjugating everything).[/quote]
One way to think about conjugation is to realize that there are [b]two[/b] injective homomorphisms from $ \mathbb{Q}[\omega]$ to $ \mathbb{C}$, one of which takes $ \omega$ to $ e^{ \frac{i \pi}{3} }$ and one of which takes it to $ e^{\minus{} \frac{i \pi}{3} }$.  This is equivalent to the existence of a $ \mathbb{Q}$-automorphism (i.e. a conjugation, or a nontrivial element of the [url=http://en.wikipedia.org/wiki/Galois_group]Galois group[/url]).


Given any set of ten natural numbers between 1 and 99 inclusive, prove that there are two disjoint nonempty
subsets of the set with equal sums of their elements.

the maximum some of such a subset is something about 950, so it just follows from pidgeonhole.

Let the subsets be the pigeons and the sums be the holes. The smallest possible sum is $ 1$ and the largest possible sum is $ 99+98+...+90=945$. So there are $ 945-1+1 = 945$ different sums. The pigeons are the possible subsets and given a ten element set there are $ 2^{10}-1=1023$ possible proper subsets. Since $ 1023>925$ by Pigeonhole Principle  (:love:) we have two subsets having the same sum. But we still didnt finish cuz we need to show they are disjoint. If $ S$ and $ T$ are such two sets and $ S\cap T = \{ \}$ then we are done :). If $ S\nap T \not = \{ \}$ then we can remove the same elements from $ S$ and $ T$ until we have disjoint sets $ S'$ and $ T'$ with the same sum (note: we cant remove all the elements since $ S\not = T$).

This is similar to IMO 1972


for a,b,c is three sides of triangle

prove that:

$\sum\sqrt{\frac{bc}{p-a}}\geq\sqrt{12p}$

with p=(a+b+c)/2

Very Easy.
It's equal:
$\sum\frac{1}{cos\frac{A}{2}} \ge 2\sqrt{3}$

let $x=p-a,y=p-b,z=p-c$, we have to prove
$\sqrt{x+y+z+\frac{yz}{x}}+\sqrt{x+y+z+\frac{xz}{y}}+\sqrt{x+y+z+\frac{xy}{z}}\ge \sqrt{12(x+y+z)}$
Applying minkowski twice, we have
$L.H.S. \ge \sqrt{9(x+y+z)+(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xy}{z}}+\sqrt{\frac{xz}{y}})^2}$
it remains to prove
$(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xy}{z}}+\sqrt{\frac{xz}{y}})^2 \ge 3(x+y+z)$
which is equivalent to $(a+b+c)^2 \ge 3(ab+bc+ca)$ and this is easy to prove.

[quote="manlio"][quote]Applying minkowski twice, we have
$L.H.S. \ge \sqrt{9(x+y+z)+(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xy}{z}}+\sqrt{\frac{xz}{y}})^2}$[/quote]

Can you please explain it better?

Thank you.[/quote]

\[
\begin{array}{l}
 \sqrt {x + y + z + \frac{{yz}}{x}}  + \sqrt {x + y + z + \frac{{xy}}{z}}  + \sqrt {x + y + z + \frac{{xz}}{y}}  \\
  = \sqrt {\sqrt {x + y + z} ^2  + \sqrt {\frac{{yz}}{x}} ^2 }  + \sqrt {\sqrt {x + y + z} ^2  + \sqrt {\frac{{xy}}{z}} ^2 }  + \sqrt {\sqrt {x + y + z} ^2  + \sqrt {\frac{{xz}}{y}} ^2 }  \\
  \ge \sqrt {\left( {\sqrt {x + y + z}  + \sqrt {x + y + z}  + \sqrt {x + y + z} } \right)^2  + \left( {\sqrt {\frac{{yz}}{x}}  + \sqrt {\frac{{xy}}{z}}  + \sqrt {\frac{{xz}}{y}} } \right)^2 }  \\
  = \sqrt {9\left( {x + y + z} \right) + \left( {\sqrt {\frac{{yz}}{x}}  + \sqrt {\frac{{xy}}{z}}  + \sqrt {\frac{{xz}}{y}} } \right)^2 }  \\
 \end{array}
\]

[quote="manlio"]Thank you very much, but I again don't understand why you say that you apply Minkowsky two times. Sorry  :(

I think you apply it only one time  ;)

Thank you very much.[/quote]

That's because the version of Minkowski i studied consists of two components only.. i mean this,
http://planetmath.org/encyclopedia/MikowskiInequality.html
so, to me i applied it 2 times.   :P


ARML Duke college Putnam HMMT Harvard MIT

There isn't really anything much like ARML -- although I think Mu Alpha Theta has conventions and things which are more roughly comparable than anything I've done.  (Note that I haven't actually done Mu Alpha Theta, which is why I'm not sure.  Is Tom Clymer's company the company that runs MA:theta:?  Maybe he can talk about it for you.)

I don't know if they still have them, but when I was a student the NYSML (New York State Math League) and the NEAML (New England Association of Math Leagues) are both similar in format - though not in size to ARML. And NYSML even had some of the same authors for the problems as ARML.

In the south you have the Duke Math Meet (which just passed), which is also similar to the ARML format - but it has individual schools competing instead of "regions".

And in Charleston on February 28th, the College of Charleston hosts one of the largest math meets in the southeast. They have a lot of different events that schools can pick and choose to participate in for awards.

they still have NYSML, and they have NEAML, but i think it's changed, cause it's not really like arml anymore unless if im thinking of something different.  All it is is just the top 4 math teams from each state (or something like that) in new england go there and it's just like a normal meet that we have here.

I don't like the way participants have to pay for ARML. Not at all...

The AMC string of contests are comparable in prestige.

It depends where you are -- NYC students get to go free :-p

yeah, Duke math meet is modeled on ARML (same rounds: power, team, indiv., relay, tiebreaker) but easier.

CofC is not exactly like ARML, too many weird things going on. the only constant thing is the written examination and the sprints. other than that, it depends what "team" you're on: jv, varsity (both elimination tournaments), brainstorming, relay, possibly some others i can't remember. not too much like arml. but definitely fun. ugh. i can't go this year. my ******** school (i don't know what the asterisks stand for, you can fill it in) isn't going. ugh. after i did moderately well last year too. ughughughughughughugh

Right. I don't believe there's another national organized meet like ARML. Just regional ones of varying sizes and formats. I believe that ARML got it's format from a combination of the old NYSML and NEAML formats - and then some other regional contests have in turn copied that. Some of these regional contests can be quite fun though. I remember the College of Chareston Meet was also my favorite when I was in high school.

The members of our South Carolina ARML team also travel for free, because they are sponsored by the State Math Teachers Association. The downside is that we can only take a certain number of students, as opposed to some other regions where whoever wants to go, and can pay, can go.

By the way, what is the current format of the NYSML and NEAML now that they have changed it?

NYSML is still the same -- the rounds are identical to (but easier than) the corresponding ARML rounds, the times are the same, etc.  The only difference is that at NYSML, all of the rounds other than the individual round are worth fewer points, increacing the relative weight of the individual questions.  And NYC A still has yet to lose NYSML :D.

sorry, in my post it was unclear what i was talking about.  I just meant that NEAML changed, but i dont know if we're talking about the same thing.

The contest im thinking of has 5 individual rounds, each person does 3.  Each round has a certain topic like number theory or something, and then there's a team round with like 6 questions i think.

[quote="JBL"]Not so much overlap between NYSML and ARML writers any more.  NYSML is largely written by Jan Siwanowicz, a recent Putnam Fellow from CCNY.  Anyhow, I was answering a slightly different question -- there are plenty of contests which to some degree mirror ARML in format and difficulty, but very few (at least that I'm aware of) that match it in the organizational sense of having a large number of people all go to one place to do math for a while.[/quote]

JOEL, NYSML is not largely written by Jan siwanowicz, in fact, Jan doesn't even do that much with NYSML.

Jan, often work on the NYCIML.  The NYSML executive committee works on their questions.  the NYSML and ARML still have a siginficant overlap, and NYC members do get to go for free, but whether NYC has money this year to go at all remains to be seen.  However, if NYC goes, I'm sure expecting to see that bone head Joel there again ;)  And you dimwit Joel... what about the

HMMT (Harvard MIT Mathematics Competition)

Haha, always the sweet one, aren't you Jack?  Yeah, I realized later that that was wrong and didn't see any particular reason to go back and change it, although if I had known you were coming . . .

JBL mentioned me and Mu Alpha Theta, so I thought I'd respond and clarify.  By the way, do I know you, JBL, or did you gather the information from my posts?

Anyhow, Mu Alpha Theta (MAO) has a great national convention each summer which moves around (this year is Huntsville (SpaceCamp), next year is Hawaii).  While I don't think any of the contests at the convention are remarkably similar to the ARML contests, the atmosphere is very similar, with hundreds of math people doing many many math things together for a long period of time.  The MAO convention is obviously longer, which allows them to offer much more in the way of math.  There are tests that focus on specific topics (probably 15 topics a year, usually with different tests for different age/experience levels) as well as general-knowledge tests, and a number of different formats including standard 30-problem, one hour, multiple choice tests, ciphering with different points awarded depending upon how quickly you solve each problem, team tests in a variety of formats, and an interschool test which often allows an entire team (including coaches) to work for the majority of the convention to solve a number of difficult/complicated problems.

I personally have no position in the organization (other than being something of an assistant coach for a school that is a member), but have volunteered to write a lot of their tests and have been paid to lead the test writing for two of their national conventions.  In addition, my company is currently administering a mail-in contest for their organization that is free to any school with an active MAO chapter.

To the best of my knowledge we don't know each other -- that was just something that popped into my head due to earlier posts of yours.

I was going to go to the MA :theta: national convention last year, but my teacher sent in the request too late. It would have been fun, and it would have stopped my parents from yelling at me to get off the computer for a few days over the summer.

What date is ARML scheduled for this year or does anyone have a good guess on which weekend?

ARML is always on the first Saturday after Memorial Day. This year it will be on Saturday, June 5th, 2004.

I was in MA :theta: when I was in Florida.  Harold Reiter's the president of Mu Alpha Theta to clear things up.  Florida needs an ARML team...:-(.  I heard it was fun from friends who did it.

WarpedKlown1335, why don't you try and start an ARML team?  I actually know a few people down at Florida that should be fairly receptive of the idea.

What year are you now?  and how badly do you want to have a team and attend ARML? ;)

Illl speak for florida, considering that el klown is in the land of the rising sun (hehe).
Weve wanted to go to ARML for a while, even trying to get a team back a few years (as im told by a senior in my school and a soph at harvard that was at my school), but the main problem has always been funding. Basically all of florida does MAO, so thats where the funding goes. We cant afford to send a team to Pennsylvania (the closest site) for that time period.

LoL I know a lot of people in FL (inc. me and jelyman) that wouldn't mind at all.  And I haven't tried to start up an ARML team in Japan because no one here was interested.  And I'm a freshman.


quadratics
algebra
quadratic formula
irrational number
difference of squares
special factorizations

Prove that n^4-20n^2+4 is composite for all integers n.

Well [hide]n:^4:-20n:^2:+4=n:^2:n:^2:-20n:^2:+4

since n:^2: is composite, n:^2:n:^2:-20n:^2:+4:arrow:composite-20*composite+4:arrow:composite-composite*composite+composite:arrow:composite-composite+composite=composite[/hide]

Note that composite means composite number, not a variable.

That's not true. If you add composite numbers, you don't necessarily get a composite number. For example, 4 and 9 are both compositve, but 4+9=13 is prime.

Whoops, thought it was about even number...back to the drawing board.

Ok, I think I have it.

[hide]n:^4:- 20n:^2:+4 = (n:^2:-4n-2)(n:^2:+4n-2)

The only way this could be prime is if one of the factors was one or negative one, and the other was prime.

When you put this into the quadratic formula, we get an irrational number; however, n must be an integer and therefore, n:^4:- 20n:^2:+4  must always be composite.[/hide]

Correct. But how did you figure out that factorization?

[hide]I had a long tedious way of doing it...  Basically, we can express the factorization as (x:^2:+ax+c)(x:^2:+cx+d)... for b and d, the possible choices would be , (-1,-4) or (-2,-2).  After expanding the factorization, we can compare it to x^4-20x^2+4 and create some system of equations...  Then we can solve for the variables.[/hide]

Ragingg wrote:Ok, I think I have it.

[hide]n:^4:- 20n:^2:+4 = (n:^2:-4n-2)(n:^2:+4n-2)
[/hide]

wait! i didnt see the problem before. hope im not too late.

Solution:[hide]so its like n:^4:- 20n:^2:+4=n:^4:-4n:^2:-16n:^2:=(n:^2:-2):^2:-16n:^2:=(n:^2:-2):^2:-(4n):^2:, so then this is a difference of squares, so put it in the form of (a+b)(a-b), where a=(n:^2:-2) and b=(4n):. so then (n:^2:-2+4n)(n:^2:-2-4n)[/hide]

hope i helped

Sorry Syntax Error, I forgot to put on the spoiler.  :(

im uber confused. i know i put the spoiler at the end of the post, and then when i tried to edit it with spoiler at the end, it didnt work.


[quote="tranthanhnam"]$\{\begin{array}{l}a+\frac{a+8b}{a^2+b^2}=2\\b+\frac{8a-b}{a^2+b^2}=0\end{array}$[/quote]

  By putting $x=a^2+b^2$ and treat x as parametter( condition x>0)...the given system is reduced to:
$\{\begin{array}{l}xa+(a+8b)=2x \\ xb+(8a-b)=0\end{array}$
By using Cramer rule to solve this system we have the result
a=f(x) and b= g(x) with f,g are function of x only...
Give the remark that: $a^2+b^2=x$ we have :
${f(x)}^2+{g(x)}^2=x$ Eliminating common x form both 2 sides we have a reciprocal equation of variable x ...
$(x+ \frac{65}{x})^2-4(x+ \frac{65}{x})-252=0$ solve it we get: x=13 or x=5......
   From here we get: (a,b)=(-1,2) or (3,-2)..
It's a quite easy problem but i like it..... :)


What is the simplified value of the sum: $ \minus{}1^{2004} \plus{} (\minus{}1)^{2005} \plus{} 1^{2006} \minus{}1^{2007}$?

$ (\minus{}1)^{2005}\equal{}\minus{}1$ because the exponent 2005 is odd.

Therefore $ \minus{}1^{2004}\plus{}(\minus{}1)^{2005}\plus{}1^{2006}\minus{}1^{2007}$ simplifies to

$ \minus{}1\minus{}1\plus{}1\minus{}1\equal{}\boxed{\minus{}2}$.

all of them are -1 except 1^2006

so we have the equation -1-1+1-1=-2

so our answer is -2 or $\boxed{-2}$

[hide]All of the exponets are -1 except, +1^2006
-1 2004 (1) 2005 +1 2006 -1 2007 is
-1 -1 +1-1= -2  Answer is -2 [/hide]
Hello, I've provided the answer above and the solution!


Structure elucidation with stereochemistry

Citric acid (2-hydroxy-1,2,3-propanetricarboxylic acid) is the primary acid of citrus
fruits, which contributes to their sour taste. Commercial manufacturing of citric acid
involves fermentation of molasses or starch using the fungus Aspergillus niger at pH

3.5. It is widely used in food, soft drinks and as a mordant in dyeing. It is also an
important biochemical intermediate.
a. What transformation will citric acid undergo when warmed with concentrated
sulfuric acid at 45-50°C? (ive already given the answer to this) Give the structure and IUPAC name of the product
obtained. Which type of organic acids would undergo a similar reaction?
After warming citric acid with sulfuric acid, anisole (methoxybenzene) is added to the
reaction mixture and product A (C12H12O5) is obtained.

.

On heating with acetic anhydride, A forms an anhydride.

.

118 mg of A requires 20 mL of 0.05 N KOH for neutralisation.

.

Reaction with bromine indicates that the same amount of compound A requires
80 mg of bromine, to give an addition product.

b. Deduce the structure of A.
c. Identify the possible isomers of A in this reaction and give their structures,
absolute configurations and the IUPAC names.
d. In the bromination reaction, how many stereoisomers of A will be obtained?
Draw their Fischer projections.
e. Assign absolute configurations to the stereocentres in all the stereoisomers
formed in d.
Instead of anisole, if phenol and resorcinol are separately added to the reaction
mixture, compounds B and C are obtained, respectively. B does not give any
coloration with neutral FeCl3, but C does. Under identical reaction conditions, the
yield of compound C is much higher than that of B.  :ninja:  :ninja:  :showoff:

hint: ive already given the answer to the first part (a) in my other post
p.s im surprised nobody's trying this, since it was an olympiad preparatory problem

[quote="VARUNARASU"]a. What transformation will citric acid undergo when warmed with concentrated
sulfuric acid at 45-50°C? (ive already given the answer to this)[/quote]

I don not agree with the answer given. If elimination of carbon monoxide occurs to form 3-oxopentanodioic acid, then with the same reaction conditions further decarboxylation should also occur.

well, the official solutions say that such a elimination occurs only for alpha hydroxy acids :oops:

Yes, alpha hydroxy acids are known to eliminate carbon monoxide to form aldehydes or ketones, but only at around 190ºC or in the presence of oxidizing agents (LTA for example). That's why I am relutant in accepting that answer.

[hide="Anyway"]b) The structure of A is (in my opinion) $ p\minus{}CH_3O\minus{}C_6H_4\minus{}C(CH_2CO_2H)\equal{}CH\minus{}CO_2H$: at least this structure is expected to form an anhydride, it has two acidic -CO_2H groups and so consumes two moles of base, and, due to the presence of a carbon-carbon double bond, it reacts with one equivalent of bromine. Citric acid, in the presence of H2SO4 forms then $ HO_2C\minus{}CH_2\minus{}CO\minus{}CH_2\minus{}CO_2H$, which is protonated in solution in the carbonyl oxygen. Anysole, having an electron rich benzene ring, can then act as a nucleophile and attack the carbonyl carbon of protonated $ HO_2C\minus{}CH_2\minus{}CO\minus{}CH_2\minus{}CO_2H$. Subsequent dehydration then yields the product above.

c) E- and Z- diastereomers of $ p\minus{}CH_3O\minus{}C_6H_4\minus{}C(CH_2CO_2H)\equal{}CH\minus{}CO_2H$ can form.[/hide]


Select two integers withour replacement from the set of integers between 1 and 1000 inclusive. What is the probability of their sum being odd?

[hide="Solution"]
If the sum is odd, then the numbers are of opposite parities. There are $ 500$ even and $ 500$ odd numbers, and $ \binom{1000}{2}\equal{}500\cdot999$ ways to choose two numbers. Therefore, the probability is \[ \frac{500\cdot500}{500\cdot999}\equal{}\boxed{\frac{500}{999}}.\][/hide]

[hide="Another way"]
Pick a number. Now you have to pick a number of the opposite parity (even/oddness). There are 500 left that are of opposite parity out of 999 remaining.[/hide]


$f(x)$ is defined from integer to integer with the properties $f(n)=n-3$ if $n>999$ and $f(n)=f(f(n+5)$ if $n<1000$. Determine the value of $f(84)$

[hide]452?
Maybe I made a fault in my calculations.
[/hide]

[hide]

There is no way it is 452, because you can only get an integer result from the function when , otherwise  just keeps getting bigger.

Since each time the function repeats itself,  gets 5 bigger.  I took the difference between 1000 and 84, and divided it by 5 to get 183.2.  So, after 183 repetitions,  is at 999 and after 184 repetitions,  is at 1004.

[/hide]

your answer is not right because you used 82 when problem said 84.

I think a twist in this problem that those first two people did not see is the fact that f(84) is going to equal f(f(f(f(.....f(n))))))) where $n >1000$, and that once we keep subtract 3, we will have numbers that are smaller than 1000, in which we will have to use the function for $n<1000$ again.

[quote]There is no way it is 452, because you can only get an integer result from the function when , otherwise  just keeps getting bigger.

Since each time the function repeats itself,  if n gets 5 bigger. I took the difference between 1000 and 82, and divided it by 5 to get 183.2. So, after 183 repetitions,  is at 999 and after 184 repetitions,  is at 1004. $\displaystyle f(1004) = 1004 - 3 = \fbox{1001} $
[/quote]
You forgot that when you are at 1004, you still have a whole bunch of f(f(f( 's !
184 to be precise, and that explains my (wrong) answer.
I did $1004 - 184*3$ Then the solution would be 452.
We both didn't see the simple fact that once below 1000, it's going up again.
I'll try again...

errr im not sure thats right either

i got that too but i think there's more to it (check it out)

wow sorry slow post

[hide]

$f(84) = f(f(89))$
$f(89) = f(f(94))$

so plugging that in you have

$f(84) = f(f(f(94)))$ and if you keep on doing this eventually you get to $f(1004)$

but by that time you would have to take the function at least 200-300 times which would change the answer wouldn't it???

i'm not sure

[/hide]

Hmm... this is solvable. $f(84)=f(f(f(...f(1004)...))) \Rightarrow f(1004)=1001$, $f(1001)=998$, and $f(998)=1003$, so then you have $f(1003)=1000$, $f(1000)=997$, $f(997)=1002$, $f(1002)=999$, $f(999)=1004$. There are $185\; f(...)$'s in the function, and for every 8 times you repeat this pattern, you will take off $3\; f(...)$'s. So, we are left with two final ones, and I get that the answer is $998$.

Thats what i meant by the first two people not getting it. Since if the number reaches below 1000, we have to find a pattern or a sequence that could give us a generating formula in which we can just count the number of f('s we have. Good job jesusfreak

Actually I did use 84 in the computations, 82 was a typo...  but you are right I totally forgot about all of the itterations back and forth.  :blush:


Given$ a, b, c > 0$.Prove that:
$ \sqrt {3a^2 \plus{} 2ab \plus{} 4b^2} \plus{} \sqrt {3b^2 \plus{} 2bc \plus{} 4c^2} \plus{} \sqrt {3c^2 \plus{} 2ca \plus{} 4a^2}\geq\ 3(a \plus{} b \plus{} c)$
Good luck. :lol:  :)

trannguyenquyND_VNand macduchuykhtn:you can post solution?
it very easy :lol: