Let $ x,y,z\in\mathbb R^{\plus{}}$ and $ x\plus{}y\plus{}z\equal{}3$. Prove that:
\[ \frac{x^3}{y^3\plus{}8}\plus{}\frac{y^3}{z^3\plus{}8}\plus{}\frac{z^3}{x^3\plus{}8}\geq\frac19\plus{}\frac2{27}(xy\plus{}xz\plus{}yz)\]

furthurmore we have:

$ \frac {x^3}{y^3 \plus{} 8} \plus{} \frac {y^3}{z^3 \plus{} 8} \plus{} \frac {z^3}{x^3 \plus{} 8}\geq\frac {1}{3}\geq\frac {1}{9} \plus{} \frac2{27}(xy \plus{} xz \plus{} yz)$

$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3) \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}$$ \geq\frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{\frac {1}{3}{(x^3 \plus{} y^3 \plus{} z^3)}^2 \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}$.
$ \frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{\frac {1}{3}{(x^3 \plus{} y^3 \plus{} z^3)}^2 \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}\geq \frac {1}{3}\longleftrightarrow$
$ (x^3 \plus{} y^3 \plus{} z^3)\geq3$

you can use couchy and also generalized jensen for convexity of $ \frac {1}{x}$ to prove first inequality

$ \frac{x^3}{(y\plus{}2)(y^2\minus{}2y\plus{}4)}\plus{}\frac{y\plus{}2}{27}\plus{}\frac{y^2\minus{}2y\plus{}4}{27} \ge x/3$

Summing up cyclically, we have

$ \frac{x^3}{y^3\plus{}8}\plus{}\frac{y^3}{z^3\plus{}8}\plus{}\frac{z^3}{x^3\plus{}8}\plus{}\frac{x^2\plus{}y^2\plus{}z^2\minus{}(x\plus{}y\plus{}z)\plus{}6*3}{27} \ge 1$

blah $ \ge 1/3\plus{}1/9\minus{}\frac{x^2\plus{}y^2\plus{}z^2}{27}$

Hence it suffices to show that $ 1/3\minus{}\frac{x^2\plus{}y^2\plus{}z^2}{27} \ge (2/27)(xy\plus{}xz\plus{}yz)$

$ 9\minus{}(x^2\plus{}y^2\plus{}z^2) \ge 2(xy\plus{}xz\plus{}yz)$

$ 9 \ge (x\plus{}y\plus{}z)^2\equal{}9$

Q.E.D.

My solution is same as you  roffler ,but we can multiple the second and third term in main inequality of
solution by  x.^3 and proving inequality by AM-GM .

Using Hölder Inequality we get :
$(\sum x +2)(\sum x^{2}-2x+4)(\sum \frac{x^{3}}{y^{3}+2}) \geq (\sum x)^{3}=27$. Hence we get :$LHS \geq \frac{3}{15-2q} \geq \frac{1}{9}+\frac{2}{27}q$ where $q=\sum xy$.
The last inequality is obviously true since it's equivalent to :$(q-3)^{2} \geq 0$
Q.E.D ;)

[quote="sumita"]furthurmore we have:
$ \frac {x^3}{y^3 \plus{} 8} \plus{} \frac {y^3}{z^3 \plus{} 8} \plus{} \frac {z^3}{x^3 \plus{} 8}\geq\frac {1}{3}$[/quote]
The following inequalities are also true.
Let $ x,y,z\in\mathbb R^{\plus{}}$ and $ x\plus{}y\plus{}z\equal{}3$. Prove that\[ \frac {y^3}{x^3 \plus{} 8} \plus{} \frac {z^3}{y^3 \plus{} 8} \plus{} \frac {x^3}{z^3 \plus{} 8}\geq\frac {1}{3}.\]
\[ \frac{y^3\plus{}8}{x^3}\plus{}\frac{z^3\plus{}8}{y^3}\plus{}\frac{x^3\plus{}8}{z^3}
\geq27.\]


inequalities
trigonometry
geometry unsolved
geometry

A chord $ AB$ and diameter $ CD$ of circle $ k$ are mutually perpendicular and intersect at $ M$. Let $ P$ be a point on the arc $ ACB$, distinct from $ A,B,C$. Line $ PM$ meets $ k$ again at $ Q$, and $ PD$ meets $ AB$ at $ R$. Prove that $ RD>MQ$.

Set $ \alpha\equal{}\angle PDC$ and $ DM\equal{}x$. Easily obtain that $ DR\equal{}\frac{x}{\cos{\alpha}}$ and $ MP^2\equal{}(1\minus{}x)^2\plus{}\sin^2{\alpha}\minus{}2(1\minus{}x)\sin^2{\alpha}$. $ RD>MQ$ is same as $ \frac{a^2}{a^2\plus{}b\minus{}2ab}<\frac{a^2}{1\minus{}b}$ where $ a\equal{}1\minus{}x$ and $ b\equal{}\sin^2{\alpha}$, which is trivial after expanding.


[quote="zvi"]Please view the attached PDF file.
Hint: use the pigeon-hole principle.[/quote]
It's an old problem.use a theorem,of which the name has been forgot by me.It's very very famous with the first letter D.I just forgot how to spell it.
And I think this problem may have been posted.

The name of the result you mention is "Siegel's lemma".  It gives a bound on the size of integer solutions to a system of linear equations with more variables than equations.

well, of course this sounds familiar!
it's an IMO 1976 [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1976]problem #2[/url]
a solution can be found [url=http://www.kalva.demon.co.uk/imo/isoln/isoln765.html]here[/url]


Prove that this equation has no natural solution:
$ x^{16}\plus{}28\equal{}y^{5}$

It is obvious through checking both ides $ mod 8$
$ LHS\equiv 4\plus{}(0\vee 1) \equal{}4\vee 5 \ (mod8)$
$ RHS\equiv 0\vee 1 \vee 3 \ (mod8)$ so no solutions :wink:

[quote="ZetaX"]RHS can be $ 0,1,3,5,7 \mod 8$ ;)[/quote]

Yes, you're right of course. After checking the 16 power I automatically checked only 0,1,2,3,4 residues, as if 5 was an even power :wink:  :oops:


The math department at a small high school has six classes with 25 students each, four classes with 20 students each, and two classes with 35 students each. No student takes more than one math class. If each student correctly fills out a questionnaire asking for the total number of students in his/her math class, what is the average of all of the numbers turned in on the questionnaires?

That was a national team round??!!
[hide]150 of the students answer 25, 80 answer 20, and 70 answer 35. 150*25+80*20+70*35 all over 300=26.[/hide]

Unless I'm missing something, I think its just $\frac{150(25)+80(20)+70(35)}{300}= 26$

Yeah, this was the easy one from last year.

i guess it could be easy to read over some key points if you're going fast.

very easy with a calculator!

jorian

4 people w/ 20 minutes at nat'ls aren't ALL going to go too fast-I'm pretty sure there were only like 1 or 2 teams that missed this one last year.


This is #7 on the Do You Need This Class test for Intro to Geometry.

If you need a picture, go to Enroll and look at the picture there.

There are two flagpoles, one of height 12 and one of height 16.  A rope is connected from the top of each flagpole to the bottom of the other.  The ropes intersect at a point x units above the ground.  Find x.

In spoiler I'm going to say how far I got, but you'll need to see the picture if you want to help me out.

[hide]

Let BD be y.  Thus, by using similar triangles, 12/y = x/(I don't know what to put here, maybe another variable, but that wouldn't be good).  Also, 16/y = x/(again I dont know what to put, I could put something like y - z but then I'd have 3 variables with 2 equations, which I can't solve)

[/hide]

I solved the problem using your exact same approach. My work in spoiler:

[hide]x=desired length

y=bottom right segment

z=bottom left segment

we have:

x/y=12/y+z

x/z=16/y+z

so 12y=16z or y=4z/3

x/z=16/(7z/3)

x=48/7[/hide]

Consult my drawing for reference.
By similar triangles, $\displaystyle\frac{A}{H}=\displaystyle\frac{X+Y}{Y}$ and $\displaystyle\frac{B}{H}=\displaystyle\frac{X+Y}{X}$.
We can conclude: $(X+Y)H=AY=BX$.
So, $X + Y = X + \displaystyle\frac{BX}{A}=X\left(\displaystyle\frac{A+B}{A}\right)$.
With this, $(X+Y)H=X\left(\displaystyle\frac{A+B}{A}\right)H=BX$. Finally, simplifying, we get $H=\displaystyle\frac{AB}{A+B}$.

Pork_chop8 has deomonstrated the crossed ladders theorem. There is a more general result which can be proven with similar triangles or mass points. According to Mathworld ([url]http://mathworld.wolfram.com/CrossedLaddersTheorem.html[/url]) it was originally shown by H. Stengel, does anyone know who this is?

Let $D$ and $E$ be on sides $\overline{BC}$ and $\overline{AB}$ of triangle $ABC$. $F$ is the intersection of $\overline{AD}$ and $\overline{EF}$. Select points $B', D', E'$, and $F'$ on $\overline{AC}$ such that $\overline{BB'} \parallel \overline{DD'} \parallel \overline{EE'} \parallel \overline{FF'}$. Then

\[ \frac{1}{BB'} + \frac{1}{FF'} = \frac{1}{DD'} + \frac{1}{EE'} \]

Of course, letting $B$ be infinitely far away from $\overline{AC}$ causes $\frac{1}{BB'} = 0$ and $\overline{AB} \parallel \overline{BC}$; hence the crossed ladders theorem is the degenerate case of Stengel's theorem.

For the record, this is mentioned in AoPS Volume 1 as a problem type frequently exploited by contest writers.  In essence, [hide]1/x +  1/y = 1/z, where x and y are the flagpole heights and z is the intersection height.[/hide]


calculus
function
derivative
calculus computations

Let $f$ be a twice differentiable function on $[a, b]$, except possibly for some interior point $x_{0}$. Suppose that $f'(x)<0<f''(x)$ for all $x<x_{0}$ and $f''(x)<0<f'(x)$ for all $x>x_{0}$. Discuss $f'$ at $x_{0}$.
[hide]Suppose that it exists, then intermediate value theorem for derivatives assures that it must be zero. Try to obtain a contradiction.[/hide]
PS This problem reminds me of another problem of mine :wink: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81869 .

Let $f$ be differentiable at $x=x_{0}$.
Then $f'(x_{0})=0$ at (intermediate mean value theorem).
Hence, for $x>x_{0}$ as $f''(x)<0$ or since $f'(x)$ decreases, $f'(x)<0$,which is a contradiction as $f'(x)>0$ for $x>x_{0}$.
hence the function is not differentiable at $x=x_{0}$.


real analysis
calculus
integration
function
vector
geometry
articles

Can somebody give me a quick explanation of what convolution operators are? (something besides the definition at mathworld :))
What exactly are they used for? Are multivariable calculus and linear algebra enough to even attempt to read and study about them, or would I just end up inevitably getting lost if I attempted to research more about them, even if I tried pretty hard to try to understand)?

One sort of convolution is

[tex]\displaystyle (f\ast g)(x)=\int f(y)g(x-y)\; dy.[/tex]

I have seen that convolution used in Fourier series and measue theory (and probably other places). You can also turn the integral into a sum and have a slightly different convolution.

[tex]\displaystyle (f\bullet g)(n)=\sum f(k)g(n-k).[/tex]

I don't know if tht's the convolution you mean though.

Yeah, that's the convolution I mean. Do you know anything more about it? =)

The reason I'm asking is that I'm hoping to do some sort of math research this summer (like for those Intel/Westinghouse projects mentioned around the board...). Since I'm also doing a math program at a university this summer and the professor that will be teaching us wrote on his faculty profile that "my main research interests concern convolution operators defined by measures on Euclidean spaces," I figured I should probably ask the professor to help me with some sort of project concerning convolution operators--if they're are a suitable project, in any case ;). So I've been looking up some stuff on the Internet about them, but figured I should probably ask on this board, too.

In a sense, both of those convolution operators that Simon mentioned are the same thing. It turns out that convolution operators are a lot like "dot products" between integrable functions. So if you think of those functions as vectors, the convolution operator gives you a sense of geometry on that space of functions.

But to really understand the importance of convolution operators and how they're used, you need to study some real analysis first. I'd start out with a book called [u]Principles of Mathematical Analysis[/u] by Walter Rudin. It's a difficult book, but if you get through that, you can start on [u]Real and Complex Analysis[/u] by Walter Rudin, and an excellent book called [u]Fourier Series and Integrals[/u] by Dym and McKean. You might not understand everything in the latter chapters of the Dym and McKean book right now, but I think you can get enough out of it to understand the basics of Fourier Analysis and how the convolution operator is used. There are lots of potential ideas for projects and applications in this book.

Argh. That sounds like an amazingly fun way to start a project, and convolution operators definitely sound interesting. The problem is that, given how hard I've heard reading through Rudin is, actually getting enough background to even start the project (by reading through the three books) sounds like it would take the whole summer.
Guess I better start now ;)

Actually the best thing to do would be to start with Dym and McKean now, and just use the Rudin books for references if you need to look something up. When doing research, you can't take the time to learn [i]everything[/i] before you get started. What happens is that you have to choose a narrow area of interest, and then learn the things necessary to advance into that area.

So scratch my advice about reading through the Rudin books first and just start some books and articles on Fourier Analysis and Convolution Operators. Use whatever Real Analysis books you need to for references.

Also, in your spare time (if you have any) read the book [u]In Code: A Mathematical Journey[/u] by Sarah Flannery. Its all about how she found and developed a prize winning mathematics project for the Intel International Science Fair.


linear algebra
matrix
linear algebra unsolved

Hi everybody

May someone help in solving this problem?

Suppose A and B are square matrices of order n and C=AB.  If [img]http://b.majidzadeh.googlepages.com/0005.gif[/img] be the ij-th cofactor of matrix [img]http://b.majidzadeh.googlepages.com/0006.gif[/img], is it correct that [img]http://b.majidzadeh.googlepages.com/0000.gif[/img]? Is there an elementary way for proving it?

I have checked it's correctness for matrices up to forth order by computer.

Thanks in advance

If $A$ and $B$ are invertible, then the formula is true because $\mathrm{cof}(A)=(A^{-1})^{T}$. And since invertible matrices are dense, it is true for all matrices by continuity.

Hi
We don't know anything about invertibility of A and B.


This is a problem I received when solving other ones.
Let $x, y, z$ be the positive real numbers satisfying $xyz\leq1$.Prove that:
$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}).(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x})\geq\frac{3}{1+\sqrt[3]{xyz}}+x+y+z$.
Try!!! :)


Let A and B be real, nonzero numbers.

The equation $5ABx^4 - 8ABx^3 + (16B-45A)x^2 + 72Ax - 144 = 0$ has four real roots: two are equal and the other two differ by 3.  Find the absolute value of the difference between the smallest root and one of the two equal roots.

[hide="What did I do wrong?"]
Let $a$, $a$, $b$ and $b+3$ be the four roots.
The sum of the roots, by Vieta's Rule, is $2a+2b+3=\frac{8AB}{5AB}=\frac{8}{5}$
$\therefore a+b=-\frac{7}{10}\Longleftrightarrow b=-a-\frac{7}{10}$

So we can rewrite the roots as $a$, $a$, $-a-\frac{7}{10}$ and $-a+\frac{23}{10}$.
The product of the roots is
$a^2(-a-\frac{7}{10})(-a+\frac{23}{10})=\frac{144}{5AB}$

$\Longleftrightarrow a^2(a^2-\frac{8}{5}a-\frac{161}{100})-\frac{144}{5AB}=0$

$\Longleftrightarrow a^4-\frac{8}{5}a^3-\frac{161}{100}a^2-\frac{144}{5AB}=0$

$\Longleftrightarrow 5ABa^4-8ABa^3-\frac{161}{20}ABa^2-144=0$

Since $a$ is a root of the original equation, we can compare the terms of the equation above and the original equation.

But... we don't have $x$ term in the equation above, which means $A=0$, which obviously makes a contradiction... :? [/hide]

Very interesting approach ... but its far different than my way.  My method is rather straight-forward and simpler.

What did you mean by "comparing the terms with the ones in the original equation"?

What I meant was, we can compare the corresponding terms of the two equations ($5ABx^4-8ABx^3-\frac{161}{20}ABx^2-144=0$ and $5ABx^4-8ABx^3+(16B-45A)x^2 +72Ax-144=0$). We know that the coefficients of the corresponding terms have to match. For instance, if we look at $x^2$ terms, we see that $\frac{161}{20}AB=16B-45A$.

I manage to factorise it into

$(Bx^2-9)(5Ax^2-8Ax+16)=0$

would that help  :P

Oh wow :o I didn't think of factoring. This obviously helps a lot :D


probability
inequalities
geometry
3D geometry
octahedron
pyramid
triangle inequality

Three numbers are chosen randomly from the interval $[0,1]$
What is the probability that they form a triangle?

......on the XY plane? define "form a triangle"

I think he means if the three sides are $a,b,c$ they satisfy the triangle inquality. I'm not sure how to solve it, though.

I'm pretty sure they can be the same (though it might not make a difference anyway, since there are still an infinite number of choices for each side).

EDIT: now that I think about it, perhaps some 3-d graphical interpretation of volumes involving the lines $a+b=c$, $b+c=a$, and $c+a=b$ might work

[quote="paladin8"]now that I think about it, perhaps some 3-d graphical interpretation of volumes involving the lines $a+b=c$, $b+c=a$, and $c+a=b$ might work[/quote]

That's how I approached it.[hide] Those are actually planes, so the solid obtained by graphing the inequalities, as well as the condition that 0<=a,b,c<=1, is an octahedron, though I'm not 100% sure. The sample space is a 1x1x1 cube with volume 1. So the volume of the octahedron would be the probability. I think it comes out to 1/2.

Edit- It's actually a hexahedron, two tetrahedra sharing a face.[/hide]

[quote="Bictor717"][quote="paladin8"]now that I think about it, perhaps some 3-d graphical interpretation of volumes involving the lines $a+b=c$, $b+c=a$, and $c+a=b$ might work[/quote]
Those are actually planes[/quote]

Heh. I knew that.

Yes, I do believe you have to use volume... It's the volume of pyramids...

You have your interval [0, 1], from which your three numbers are chosen... say x, y, and z. You also have your triangle inequalities:
x + y > z,  y + z >x, x + z > y

add them together; you get x + y + z > 0. But you also know that
0 < x <= y  <= z  <= 1. Knowing that you can obtain:

0 < x + y + z  <= 3, which is basically the total volume that you have to play around with. But you are restricted by the 0 < x <= y  <= z  <= 1. Find the volumes of the pyramids, compare them.

I got: 1/27

The way I would go about this is assuming without loss of generality that $x \ge y \ge z$ and these graphs intersecting with the possible region presented in the problem gives an area of (I think) $\frac{1}{6}$ as a feasible region.  Then because of the restrictions, the only inequality we should have to worry about is that $y+z \ge x$.  Now, I'm not good with 3d graphs, so I couldn't picture that part to finish up the problem, but I think if you find the area of the intersection of that inequality and the feasibility region, then divide by $\frac{1}{6}$, I think that's it.

[quote="Bictor717"][quote="osdes"]You have your interval [0, 1], from which your three numbers are chosen... say x, y, and z. You also have your triangle inequalities:
x + y > z,  y + z >x, x + z > y

add them together; you get x + y + z > 0.[/quote]
Well, you know that from x,y,z$\in$[0,1].

[quote="osdes"] But you also know that
0 < x <= y  <= z  <= 1. Knowing that you can obtain:

0 < x + y + z  <= 3[/quote]
Again, you know that from x,y,z$\in$[0,1]. Also, these inequalities are weaker than what was originally given.

[quote="osdes"]which is basically the total volume that you have to play around with. But you are restricted by the 0 < x <= y  <= z  <= 1. Find the volumes of the pyramids, compare them.

I got: 1/27[/quote]
I'm not exactly sure why you put that restriction on x,y,z.
Can you be more specific about which pyramids you're talking about? In other words, by which planes are they bounded?[/quote]

Well, the 0 < x + y + z  <= 3 is the equilateral triangle with vertices on (3, 0, 0), (0, 3, 0) and (0, 0, 3). The apex of the pyramid is at (0, 0, 0). That's how I "look" at the pyramid. So... that's your xy-plane, xz-plane, yz-plane, and your x + y + z plane (non-negative real numbers) with a max of value of 3. I'm not particularly sure how to write the equations for those planes, I hope what I said helps.

Oh, and I messed up on the x, y, z are in the interval (0, 1] part. I'm pretty sure that would define a unit cube, not a pyramid. The way that I think of it is... all the points which satisfy that inequality, which are all those points in the cube.

So I'd take 2/9.

But I see what you mean by saying that the inequalities are weaker.
So disregard what I said above, because what I said up there is obviously incorrect in this problem.

So instead, you'd have to take your triangle inequality theorems:
x + y > z; y + z > x; x + z > y and look at each individual one at a time.

First take the x + y > z. you know the interval of z is (0, 1]. Graph them, taking each value of z and plugging it in. But you notice that all those inequalities have a common ground: x + y > 1. You also know your restrictions on x and y. Repeat with the other inequalities; your set of possible points lie in a unit cube; you have a pyramid within that cube which takes up half the cubes' volume... so it's 1/2.

I have to admit that the last post confuses me, as it twice states two contradictory things without seeming to notice the conflict.  In any event, I would point out that there is not pyramid contained inside a cube which takes up 1/2 of its volume, so some part of your last solution is definitely wrong.  I'm inclined to think that you just mis-calculated the volume of the pyramid and that, if you calculated it correctly, you would agree with the answer given by bictor, among others.


In a land any towns are connected by roads with exactly other three towns. Last year we made a trip starting from a town coming back to the town by visiting exactly one time all towns in land. This year we are plan to make a trip in the same way as last year's trip. Note that you can't take such order completely same as last year's one or trace the order only reversely. Prove that this is possible.

You want to prove that a 3-regular graph has two Hamiltonian circuits if it has one. In fact, it has at least three.

This is an immediate consequence of [url=http://mathworld.wolfram.com/SmithsNetworkTheorem.html]Smith's Network Theorem.[/url]

Just to resolve this discussion: See
- http://math.stackexchange.com/questions/631673/hamiltonian-paths-in-cubic-graph, or
- Theorem 1 in [url=http://homepage.tudelft.nl/64a8q/papers/handshake.pdf]Dion Gijswijt, [i]The power of shaking hands[/i][/url], or
- Corollary 1.2 in [url=http://dx.doi.org/10.1016/S0022-0000(05)80063-7]A.G. Thomason, [i]Hamiltonian Cycles and Uniquely Edge Colourable Graphs[/i], Annals of Discrete Mathematics, Volume 3, 1978, Pages 259--268[/url]
- Theorem 3.2.1 in [url=http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.881.3710&rep=rep1&type=pdf]Andrew Wagner, [i]On the Existence of a Second Hamilton Cycle in Hamiltonian Graphs With Symmetry[/i][/url]
for proofs of both the original problem and of Smith's Network theorem. (Note that some of the authors discuss Hamiltonian paths connecting two given vertices $u$ and $v$ instead of Hamiltonian circuits using a given edge $e$. Of course, the two situations are equivalent: If you add an edge joining $u$ to $v$, and denote this edge by $e$ then the former paths become the latter circuits. Conversely, if you remove the edge $e$ and denote its endpoints as $u$ and $v$, then the latter circuits become the former paths.)


number theory
inequalities unsolved
inequalities

Schinzel's Theorem states that:
For every positive integer n, there exists a circle in the plane having exactly n lattice points on its circumference.
for further information:
[url=http://mathworld.wolfram.com/SchinzelsTheorem.html]mathworld.wolfram.com/SchinzelsTheorem.html[/url]

It's written that:
"Note, however, that these solutions do not necessarily have the smallest possible radius".

Can we aproximate somehow the minimal possible value of radius length of a circle passing exactly
threw $n$ lattice points ?

[Edit] Can someone move this subject to number theory ? I've mistaken sections, sorry.

I think this problem is in UPINT (Richard Guy, [i]Unsolved Problems in Number Theory[/i]) with references and the state of the art.  Basically, very little is known.


John has $ \$150$ in his piggy bank and plans to add $ \$2$ each week.
Parneshia has $ \$200$ in her piggy bank and plans to add $ \$6$ each
week. After how many weeks will Parneshia's bank have twice as
much money as John's?

After n weeks, John has 150+2n dollars, while Parneshia has 200+6n. Writing the equation 200+6n=2(150+2n), we see that 2n=100, so the answer is 50 weeks.


function
number theory unsolved
number theory

[quote="knol"]Let n, x, y, and z be positive integers. There are 666 ordered triples (x,y,z) such that x + 8y + 8z = n. Determine the maximum value of n.[/quote]

Let $ x \plus{} y \equal{} p$ such that $ 2\le p\le\left[\frac {n \minus{} 1}8\right]$: we have $ p \minus{} 1$ ordered triples : $ (n \minus{} 8p,k,p \minus{} k)$ where $ k \equal{} 1... p \minus{} 1$

And so the number of triples is $ \sum_{p \equal{} 2}^{\left[\frac {n \minus{} 1}8\right]}(p \minus{} 1)$ $ \equal{} \frac 12\left[\frac {n \minus{} 1}8\right]$ $ (\left[\frac {n \minus{} 1}8\right] \minus{} 1)$

So $ x \equal{} \left[\frac {n \minus{} 1}8\right]$ is solution of the equation $ x(x \minus{} 1) \equal{} 1332$ $ \iff$ $ x^2 \minus{} x \minus{} 1332 \equal{} 0$ and so $ x \equal{} \left[\frac {n \minus{} 1}8\right] \equal{} 37$

So $ 37\le \frac {n \minus{} 1}8 < 38$ $ \iff$ $ 297\le n < 305$ hence the answer $ \boxed{\text{greatest }n \equal{} 304}$

sorry for open a old topic. I don't know the problem is has "exactly" 666 triple (x,y,z) or not  :|  if exactly, I think the solution of pco has something wrong (I think) because if we use generating function, we have generating function of the equation is: $ (\frac {1}{1 \minus{} x}\minus{}1)*(\frac {1}{1 \minus{} x^8}\minus{}1)*(\frac {1}{1 \minus{} x^8}\minus{}1)$. so the question must be in: exist or don't exist n such that x + 8y + 8z = n has 666 triple(x,y,z) :blush:  sorry if I say wrong.

[quote="hqthao"]sorry for open a old topic. I don't know the problem is has "exactly" 666 triple (x,y,z) or not  :|  if exactly, I think the solution of pco has something wrong (I think) because if we use generating function, we have generating function of the equation is: $ (\frac {1}{1 \minus{} x} \minus{} 1)*(\frac {1}{1 \minus{} x^8} \minus{} 1)*(\frac {1}{1 \minus{} x^8} \minus{} 1)$. so the question must be in: exist or don't exist n such that x + 8y + 8z = n has 666 triple(x,y,z) :blush:  sorry if I say wrong.[/quote]

I'm sorry, I never studied "generating function" and dont know what it is.

But my proof seems to me quite elementary (except the typo $ x \plus{} y \equal{} p$ instead of $ y \plus{} z \equal{} p$ at the beginning) and I dont see any error in it.

My proof says that for any $ n\in[297,304]$, the number of triples is exactly $ 666$ (and that it is different than $ 666$ for other $ n$), so I proved existence.

Maybe you could just kindly read it and point at my error.
Thanks.


function
inequalities
real analysis
real analysis unsolved

Find all continuous functions $f: [-1,1] \longrightarrow R$ with property \[ f(x)\geq f(1-x^2),\ {\mbox{ for all }} x \in [-1,1]. \]

A partial result:
   It is sufficient to define f on [0,1]: if y is in [-1,0] let g(y)=f(1-y^2); we choose a continuous function f such that f(y)>=g(y) for y in [-1,0[ and f(0)=g(0).
  With x=0, x=1 one obtains f(0)=f(1). Let a=(sqrt(5)-1)/2.
  Let x in [0,1]:
Let the sequence U0=x,...,Un+1=1-Un^2,...;f(x)>=f(Un); if x<>a then the sequence U2n+1 decreases towards 0 and the sequence U2n increases towards 1 (or the reverse). Thus f(x)>=f(0)=f(1); by continuity this inequality is still valid for x=a.
Let the sequence V0=x,...,Vn+1=sqrt(1-Vn),...;f(x)<=f(Un); if x<>a then V2n increases towards a and V2n+1 decreases  towards a (or the reverse). Thus f(x)<=f(a).
      It seems that  f is increasing on [0,a] and decreasing on [a,1].
      It is not sufficient.


I was thinking on this "theorem" that says that if have a class on a High School with at least 23 persons, then, with probability greater than one half you'll have two persons whose birthday. This assumes that there is an uniform distribution of the birthdays around the year.

But I remember that I read somewhere that most species of mamals (specially out of the tropics) have their peaks of birthds on the Summer and early Fall (it is understandable, they are a little grown up for the winter) and I remember than when I was in High School, most of the birthdays parties were on September (Also understandable, at least in Costa Rica, the last two weeks of the year are basically for party!  :D ) But I don't know if there is an stadistic about all. I wouldn't spect to have the same pattern in the whole world, but I suppose that if you take some country, probably there would be there a pattern on the birds, and there would probably be a peak at some point during the year. Any body knows something (I don't have more than empirical data!).

Best regards,

yes, i have also noticed that in many of my classes, a lot my peers have birthdays in october (hmm... valentine's day maybe?)... so it would seem that if you had 23 people, there would be an even GREATER chance of two people having the same birthdays, would it not?

hmm....

Hmm...I'm born in October...but on the mathematical side, let me think about that. Well, for each person you select after the first one, the probability goes down that they'll have a unique birthday. (IE, first person has 366 possibilities, then 2nd has 365 that are different and so on) Soo, it does make sense. However, I'm too lazy right now to write a program right now that'll calculate the chance, sorry. :D

If birthdays really do clump, you would most assuredly see an increase in the probability of two people having the same birthday -- in the most absurd extension, imagine that [i]everyone[/i] was born on January 1.  The fact that less-extreme cases result in higher probabilities of a match than a uniform distribution can be expressed as a reasonably simple inequality with (only :-p) 365 terms.

As for whether there are in fact meaningful variations among months and so on, I have no idea -- maybe you could e-mail the census bureau and ask.

One datapoint:
[img]http://math.hope.edu/swanson/statlabs/birth_box.gif[/img]
[img]http://math.hope.edu/swanson/statlabs/birth_ts.gif[/img]
μ= 9132.2 (Average value  per day)
σ = 817.88
From http://www.dartmouth.edu/~chance/teaching_aids/data/birthday.txt.

As I thought, the pick of the birds are on september, and that is followed by July and August (Premature borns from the hollidays in december  :D )

But it would be interesting to know where is that chart about. I really don't think that, for example, in Central Africa (I believe mainly Muslim, and with no harsh winter) would have this distribution, there is no reason for an increase in the fecundations around the end of the Gregorian year.

FWIW To put the things in perspective, the distribution of birthdays is not that un-uniform. the value   (1/365) is still correct within  a few percentage..

[quote="ankur87"]I have yet to meet someone with my birthday. But there are 5 kids at my high school alone with their birhday on November 7. Valentines day Theory?[/quote]

Hmm... were you born on Feb. 29?

No, May 4th.

Oddly enough, I have met someone born on February 29.

[quote="ankur87"]Oddly enough, I have met someone born on February 29.[/quote]

Actually, I know four persons whose birthday is on February 29th (At least that I am aware of), The oldest had his 14th birthday last year!! (I am not telling that he is 14 years old  :D ).

Best,

Last year our math teacher said that  there is a knowledge like that about two football team's players. Was that really true? I cannot believe it either

[quote="mcalderbank"][quote="djimenez"]

Actually, I know four persons whose birthday is on February 29th

[/quote]

What are the odds?  :D[/quote]

Well, I don't know what the odds are. If we assume that every day the number of births is the same (that Gyan showed us a chart that disprove this belief, but the percentage is not that different) then, there is 1 in 1461 chances that an specific person was born a February 29th, and I have heard that in average each person knows around 1000 other people (I think the number should be actually a lot higher) then with this data we can say that the probability that a person knows someone who was born on that day is roughly 1/2 (according to my calculator 0.4957570), nevertheless, the chances to know at least four is roughly 1/200 ($5.3022066\cdot10^{-3}$).

Nevertheless, a cousin of my father was born on February 29th, 1948. A former High School classmate was born on February 29th, 1980. The ex-boyfriend of my sister was born on February 29th, 1984, and the son of a friend was born on February 29th, 2004!

You can say that I am, at least in this aspect, on in two hundreds (Damn, I would like to say that I am one in a million or so, but I would find something to say that).

Best,

[quote="ankur87"]Do you play the lotto often?

(Or whatever the Costa Rican equivalent of the lotto is?)[/quote]

Usually I did not, and I don't do now that I live in the US either.

By the way, all of us have some reason to be 1 in a million, or at least 1 in around 800 000. We are members of this forum  :first: (6 billion people in the world, 7334 members by this moment!)

Best,

[quote="diophantus"]Why don't you start a poll and see when the birthdays of the AoPSers are?  That would be rather interesting.[/quote]

There is already a discussion on that, even when is not exactly a poll. You can find it [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=birthday&t=24904]here[/url].

Best,


superior algebra
superior algebra unsolved

\[<A^A, \circ>\]
Is the group above a group and please explain why it is or isn't?

I guess (Im not sure) that it means Isomorphisms from $A$ to $A$

no that dos't help but i think it can't be a group since it is not bijective but i don't understand why it isn't

If $A^A$ is not the isomorphism from $A$ to $A$ but only morphism, $(A^A,o)$ can't be a group since there is $f$ in $A^A$ which is not bijective so there is no $g\in A^A$ such that $fog=gof=Id$, $f$ has no inverse.


quadratics
limit
logarithms
calculus
calculus computations

([b]4[/b]) Let $ a$, $ b$ be constants such that $ \lim_{x\rightarrow1}\frac {(\ln(2 \minus{} x))^2}{x^2 \plus{} ax \plus{} b} \equal{} 1$. Determine the pair $ (a,b)$.

[hide="Solution"]
The numerator approaches 0, but the limit does not, thus the denominator also approaches 0, and $ 1 \plus{} a \plus{} b \equal{} 0 \implies b \equal{} \minus{} 1 \minus{} a$

Now, we can use L'Hopital's rule to get that the limit equals $ \lim_{x\rightarrow1}\frac { \minus{} 2(\ln(2 \minus{} x))}{(2x \plus{} a)(2 \minus{} x)} \equal{} 1$

Again, the numerator approaches 0, so the denominator must also, and we have that $ 2 \plus{} a \equal{} 0 \implies a \equal{} \minus{} 2 \implies b \equal{} 1$

Hence, $ (a,b) \equal{} ( \minus{} 2,1)$
[/hide]

[hide="Contest Style Approach"]
The linear approximation of $ \ln (1 \plus{} x)$ at $ x \equal{} 0$ is $ x$, thus the linear approximation to $ \ln (1 \plus{} (1 \minus{} x))$ at $ 1 \minus{} x \equal{} 0$ is $ 1 \minus{} x$.

$ \ln (2 \minus{} x) \approx 1 \minus{} x \implies [ \ln (2 \minus{} x) ]^2 \approx (1 \minus{} x)^2 \equal{} x^2 \minus{} 2x \plus{} 1 \implies (a,b) \equal{} ( \minus{} 2,1)$
[/hide]

[quote="bunchhieng"][quote="TZF"]
The numerator approaches 0, but the limit does not, thus the denominator also approaches 0, and $ 1 \plus{} a \plus{} b \equal{} 0 \implies b \equal{} \minus{} 1 \minus{} a$

[/quote]
Excuse me!! How can you know exactly that the denominator equal to 0 by the following description! :D[/quote]

If the denominator is not equal to 0, then the limit is 0 over a nonzero number or just 0. As the limit is said to be equal to 1, this cannot be the case, which is only true when the denominator is equal to zero and the limit appears to be zero over zero. This is an indeterminate form, and the limit can be anything and is determined using L'Hopital's Rule.


number theory
greatest common divisor
number theory theorems

Can you tell me the various property about $gcd (a,b)$ ?

(like $gcd(a^n-1,a^m-1)=a^{gcd(n,m)}-1$, and so on...)

If $F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$ (Fibonacci numbers) then $\gcd (F_m,F_n)=F_{\gcd(m,n)}$.

Andrei

[quote]If $F_1=F_2=1$ and$ F_{n+2}=F_{n+1}+F_n$ (Fibonacci numbers) then $\gcd (F_m,F_n)=F_{\gcd(m,n)}$.[/quote]

Ajorza, could you please show a proof of this identity or at least tell it's name (if it has one)? :?

Let $m>n$. We have:
$F_m=F_2F_{m-1}+F_1F_{m-2}=F_3F_{m-2}+F_2F_{m-3}=F_4F_{m-3}+F_3F_{m-4}=\ldots=F_nF_{m-n+1}+F_{n-1}F_{m-n}$
With using that $\gcd(F_n,F_{n-1})$ we can get:
$\gcd(F_m,F_n)=\gcd(F_nF_{m-n+1}+F_{n-1}F_{m-n},F_n)=\gcd(F_{n-1}F_{m-n},F_n)=\gcd(F_{m-n},F_n)$
So we can use same steps as in Euclid algorithm.

Similar propertyies applies to Lucas numbers as well

Bomb


logarithms
limit
calculus
calculus computations

how to find $ \lim(n)^{1/n}\equal{}1$ without use of logarithm?

[hide="Answer 1"]Theorem: If $ \lim \frac{a_{n\plus{}1}}{a_n}$ exists then $ \lim (a_n)^{1/n}$ also exists, and $ \lim (a_n)^{1/n} \equal{} \lim \frac{a_{n\plus{}1}}{a_n}$.

Considering $ a_n \equal{} n$, then $ \lim n^{1/n} \equal{} \lim \frac{n\plus{}1}{n} \equal{} 1$.[/hide]

[hide="Answer 2"]We have $ \lim n^{1/n} \equal{} \lim e^{\ln(n^{1/n})} \equal{} \lim e^{\frac{\ln n}{n}} \equal{} \lim \left(1 \plus{} \frac{\ln n}{n}\right) \equal{} 1$.[/hide]

but i think taht  the first lemma folow by using ln

It is not hard to show that

$ \lim_{n \to \infty} \frac{n}{(1 \plus{} \epsilon)^n} \equal{} 0$

(Consider when $ n > \frac{1}{\epsilon}$.)  Since $ \sqrt[n]{n}$ is bounded below by $ 1$, the conclusion follows.

There's also this well-known proof, but it's rather unnatural:

Let $ \sqrt [n] n \equal{} 1 \plus{} x_n$. Then, $ n \equal{} \left( 1 \plus{} x_n \right)^n > 1 \plus{} \binom{n}{2} x_n^2$, i.e. $ 0 < x_n < \sqrt {\frac2{n}}$.

[quote="Carcul"]
We have $ \lim e^{\frac {\ln n}{n}} \equal{} \lim \left(1 \plus{} \frac {\ln n}{n}\right)$.[/quote]

How ?  :maybe:

I think $ \lim \frac {ln(n)}{n} \equal{} 0$

As lim (ln n/n) = 0, we can safelly use the power series expansion of e^x: $ e^x \approx 1 \plus{} x$. Or you can use directly lim (ln n/n) = 0 to make lim e^{ln n/n} = e^0 = 1.

We have ofc $ 1\leq\sqrt[n]{n}$, and by other side by AM-GM we have:
$ \sqrt[n]{n}\leq\frac{\sqrt{n}\plus{}\sqrt{n}\plus{}\underbrace{1\plus{}...\plus{}1}_{n\minus{}2}}{n}\equal{}\frac{2\sqrt{n}\plus{}n\minus{}1}{n}\to 1$, hence:

$ 1\leq\lim_n\sqrt[n]{n}\leq 1$, therefore $ \lim_n\sqrt[n]{n}\equal{}1$


Does anyone have a website for basic chess strategy? I know how the pieces move, but after that, I'm pretty much moving randomly.

To practice:
Software:
Fritz - Good for when you get better (practice) and want to analyze games
Chessmaster - Good for beginners

Books:
I have no idea. There are a lot of books, especially for beginners.

Sites:
ICC - Internet Chess Club, have to pay, lots of good players are there, some say the best online place to practice

I don't know good sites for learning chess, as my dad taught me.

Thanks, I think I'll give Chessmaster a try for a bit.

I tried doing the tutorials on chess.com, but felt I wasn't learning anything from them.

Most sites on the net seem to be very basic (how the pieces move) or more for intermediate/advanced players rather than the beginner that already knows how the pieces move.

Probably the best way to learn is just to play, the same way to best way to learn math is to do problems. I learned a lot by playing games and then analyzing them. Exactly how much do you know? For instance, do you know stuff like forks and skewers?

Take classes? I'm going to chess class... Or try the yahoo game Chessmaster Challenge. It's pretty good, as you can choose your level.

U can sign up for ICC:Internet Chess Club.
It's like online chess games, where u can talk, participate in tourneys, choose which rated players u want to play, etc. I use it and my USCF rating is 1700.

Hmm, it depends if you're willing to spend money or not. ICC is great if you don't mind paying the membership fee, and you could even take classes. There's this chess school near where I live called the "International Chess Academy" that I used to go to, and it helped a lot. An example of a rather successful student would be BOGTRO. :)

If you're interested in a book, I recommend [url=http://www.amazon.com/Winning-Chess-Strategies-Everyman/dp/1857443322]Winning Chess Strategies[/url] by Yasser Seirawan as a place to start (and the price is very modest).

When you learn strategies though, it's a good idea to be familiar with some chess tactics (like forks, pins and skews, using mate threats, and removing the guard, to name a few) since it greatly aids with strategizing (and is fun to use). You can quickly browse and skim through [url=http://www.chesstactics.org/]this[/url] to get some ideas.

I'd recommend a book that deals with tactics as opposed to strategy, if you're a beginner. Not only are tactics easier to understand, but at the beginner level, they're much more important. A book I used when starting out was Josh Waitzkin's Attacking Chess. It assumes that you have little to no knowledge of chess tactics, and teaches you the fundamentals, plus some more advanced tactics. I highly recommend it.


In a class of 70 people, what's the probablity that at least 2 of the people have..
A. the same birthday?
B. the same astrological sign?

[hide="A"]assume no one has the same birthday out of the first 69 people. the probability of the 70th person having the same birthday as someone else is 69/365(assuming they dont have the suckiest birthday ever: febuary 29th)[/hide]
[hide="B"]whats an astrological sign. im going to guess there are 12. therefore the answer is 1[/hide]

[quote="abacadaea"]assume no one has the same birthday out of the first 69 people.[/quote]

This is an absurd assumption given that you are trying to answer a probability question.  And you should've immediately checked the first answer you got to this problem; there's a reason it's called the birthday [i]paradox.[/i]  ;)

[hide="Hint"] What's the probability that everyone has a different birthday? [/hide]

this is a hard problem if leap years are involved...

assume they're not...as unreasonable a MC question this is already(it just would never show up), leap years push it completely over the top.

[quote="t0rajir0u"]Actually, considering the quantity in question is absurd to actually calculate by hand (presumably the OP would be satisfied with an expression), the writing of this expression isn't much harder if the standard 4-year cycle for leap years are allowed.[/quote]

how are we going to find the probabilty if we dont' do it by hand?

I mean that the calculation of the actual numerical probability would be very hard to do by hand (pages upon pages upon pages...).  To say any more would be to give away the solution ;)

abacadaea, your answer in B is right, but A was completely wrong. I think you should first find the complement of the event, that is, no 2 people have the same birthday, then subtract that from 1

oh woops. the one i was confident with i got wrong, the one i guessed, i got right.....

[url]http://www.mathlinks.ro/viewtopic.php?p=857614#857614[/url]

GO there and the first question will be easier to figure out.

In addition, please assume there are no leap years.


Solve the equation

 $p^{x}-y^{p}= 1$ for non negative integers $x,y$ and odd prime $p$.

Posted before, link would be nice...

Write it as $p^{x}= 1+y^{p}= (y+1)(y^{p-1}-y^{p-2}+-...+1)$.
Then $y=p^{m}-1$. If $m>1$ we get $y \equiv-1 \mod p^{2}$ and thus $y^{p-1}-y^{p-2}+-...+1 \equiv p \mod p^{2}$. This yields $y^{p-1}-y^{p-2}+-...+1=p \leq p^{2}-2 \leq p^{m}-2 = y-1$ or in other words $y^{p}\leq y^{2}-1 < y^{2}\iff y^{p-2}< 1$, clearly wrong.

Thus we know that $y=p-1$ and we have to solve $p^{x}-(p-1)^{p}=1$, thus $\sum_{k=1}^{p}\binom{p}{k}(-p)^{k}=-p^{x}$. We get for $2 \leq k \leq p$ that $p^{3}|\binom{p}{k}(-p)^{k}$, but $p^{3}\nmid \binom{p}{1}(-p)^{1}$.
Thus $-p^{x}\equiv p^{2}\mod p^{3}$, giving $x=2$ and $p=3$.

Well, we can handle $p=2$ even simpler (don't know why you want $p$ odd):
We have $2^{x}-1 = y^{2}$. For $x>1$ we get $y^{2}\equiv-1 \mod 4$, wrong.

no, you talked about non-negative integers, 0 isn't non-negative.


AMC
USA(J)MO
USAMO
USAJMO
AMC 10
AMC 12
AMC 10 A

http://www.unl.edu/amc/e-exams/e0-usajmo/usajmo.shtml

Mathematical Olympiad Summer Program (MOSP) Selection:
Currently we invite the approximately 25 top scoring 9th grade students on the USAMO to the MOSP.
At the MOSP, these 25 students are further divided into two groups randomly in order to have two classes, usually called Red 1 and Red 2.  Instruction and curriculum for the two groups is identical.  The two classes are approximately 12-13 students each.  Assuming continuing funding, under the new USAMO/USAJMO division  the USAMO SubCommittee and the graders will select about 12-13 [b]top scoring 10th and 9th graders[/b] from the USAMO and will subsequently select the top-scoring  approximately 12 [b] 10th and 9th graders [/b]from the USAJMO. That is, essentially the USAJMO Winners will be invited to MOSP. The first group from the USAMO will be become the Red 1, the second from the USAJMO will become Red 2. The MOSP instruction and curriculum for these two groups remains identical.  Instructors at MOSP remain free to reassign students to groups as needed or desired.

Well that appears to answer the USAJMO ---> MOP question. And makes it that much harder to get in  :(

AMCDirector,
If you see this, would you tell us:

If some one takes both AMC10 and AMC12 and qualifies for both USAJMO and USAMO, does this person have choice for taking USAJMO or USAMO ? Maybe JMO  is easier to get to red MOP now.

But anyway, looks like Red 1 will be much stronger than Red 2 since Red 2 is really a Red MOP for junior.

...red mop is now open to 10th graders? doesn't this kinda contradict the entire original purpose of red mop?

I think this is so that 10th graders can actually get something out of taking the AMC 10 and making JMO.

[quote="xpmath"]This is quite annoying.

Also, it seems to me that if you are in USAJMO, you have a higher chance of making red now. Hopefully this doesn't influence 9th and 10th graders to only take the AMC 10 in order to make USAJMO and have an easier chance of MOP.[/quote]

Well this thing is certainly going to influence me. What seemed to be a pointless USAJMO test may actually now be somewhat useful, and I have to consider taking AMC10A and AMC12B now instead of both AMC12A and B.

Or maybe I should just take both AMC12's, make USAMO, and get lucky and answer 1 and 4 or something (and I guess it's awesome I have a chance to make Red Mop?)

Tough choice.

Edit: I am in 10th grade

i dont like this at all

what will ultimately happen is that future IMO teams from USA wont be as strong as they are now.

this is simply because, when redMOP was only for freshman, after it finishes, all the freshman(even the ones not in mop) who are now 10th graders will practice for blue mop.

with this new method, we have many 10th graders now practicing for red mop which means that their standards will have dropped one notch and this would lead to a weaker group of 10th graders than before because making mop would be less competitive for them.

in the next year
they are now 11th graders aiming for blue mop.  by then the general competition is not as strong but the same goes for them as well.  so making blue mop will be just as hard as it is only everyone will not be as good.

i hope this makes sense and sort of explains why it would mean that people wouldnt be as good as before.

@shtsxc- i believe that in a previous thread someone said that if you qualify for both, you have to take the USAMO

and i find it interesting that everyone who has posted so far(except like 1 person whom i do not know) is in 9th grade :P

Hmm, well I agree that this is definitely a backward approach...For the motivated 10th grader, making blue mop is a difficult task, but when told that they have a chance to make red mop may not put as much effort into it and yes therefore potentially weaken the US team (but could it be that most black moppers are pretty motivated?)

Also, I kind of like this implementation because it will actually make me work harder (but then again, I may be biased since I am in 10th grade and this gives me a chance to be in 9th grade again)

[quote="abacadaea"](except like 1 person whom i do not know) [/quote]

hai

This is weird, like the only benefit I can think of right now is that some 10th graders have a chance to get bumped up to blue if they did worse than usual on USAMO, but...

Hmm if they're like this they might as well have let middle schoolers into red MOP last year >.<.

But what is the point of having someone in red for 2 years? And (@math154) if somebody does not do well enough to make blue MOP, and has never made blue MOP or higher before, they're not going to IMO probably anyways, so why bother?

I pretty much agree, I never said I liked this... :P

I seriously don't see why they changed it to G9/10, not G9-.

From what i'm seen, there's a ton of really smart people who are < 9th grade.

This is a good AoPSWIKI article for MOP information: http://www.artofproblemsolving.com/Wiki/index.php/MOP
Except a few changes with the introduction of USAJMO and 10th graders being eligible for Red, which can be found in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]this thread[/url]. Also,
[quote="Wikipedia"]In order to be eligible to take the USAMO, a participant must be either a U.S. citizen or a legal resident of the United States or Canada.[/quote]
So you're in good shape, but you have to get into MOP through Blue/Black this year.

[quote="mathlearner"]In reply to JSteinhardt, I understand your reputation to be one of a very bright individual, particularly in regards to math acumen, but I can’t help but wonder how someone so bright, assuming you read most of the posts both in this thread and similar ones that have been ongoing since the idea of the USAJMO was first introduced, honestly doesn’t “get” how the USAJMO system is inherently unfair to 9th graders. I personally believe I (and many others) were quite clear in our explanations, and not sure what your motives are in being what appears to be “nitpicky” with this matter. With all due respect, it seems as if you are taking it almost a bit personally, especially for someone who I believe is out of high school already.  May I ask – did you have any part in the organizing or any part of this USAJMO process?  The fact that there were a run of “posts” here, until you posted on Dec. 25, and then not a single reply, suggests, at least to me, that others may have been as exasperated as I with your comments. You are certainly welcome to your opinion, but rather than nitpick one or two poster’s comments, why not focus on the new system in place, which I will reiterate IS inherently unfair to all current (and future) 9th graders for all the reasons outlined.  I hope others who concur with me and/or who support a return to the original USAMO only system, DO continue to speak up and are not “intimidated” to do so.

  Final comment here: For the life of me, I can’t understand why an organization would put a system in place that can very possibly cause one or more of our country’s most promising 9th graders to not be welcomed to MOP (ones who may be strong enough to make the USAMO, but lose their spot to USAJMO’ers), & may be so discouraged as to just give up.  I am NOT a 9th grader, but if I were, and if I were put in such an unfair position, I think I might well be turned off permanently to “competition math”.  Worse, if you lose a few of these students every single year going forward, ultimately, the strength of the pool of students (as I believe someone else suggested) will surely be diminished, and that doesn’t just hurt a few students, - it hurts the reputation of our nation.

Btw, JSteinhardt, you wrote[quote]It seems to me that it's all more or less subjective anyways.[/quote]  Exactly WHAT is “subjective”?? The selection of Moppers??  :o[/quote]

The ninth graders this year have another opportunity to make it next year.

If a 9th grader doesn't make red one year, they can make it the next. I don't see how not making red with these rules is that much worse than not making red with the old rules. With the new rules, you get a second chance to make red though you have more competition.

Explain why not making red now is more discouraging? The only way I can see this being unfair is if there are a lot of people who go to red in both 9th and 10th, but IMO you need to be fairly lucky to make red mop.

[quote="echinodermata"]It's simple.

No matter what grade levels end up being eligible (whether it's 9 only, 9 and 10, 9-, or 10-); there are good reasons for each and I won't discuss them: but whatever the committee decides, one provision must be made. No student can be allowed to make red mop after having already made mop previously.

This solves two major problems:
1. Returning 10th graders can no longer clog everything up, and undermine the original purpose of red mop.
2. So that this year's 9th graders don't get shafted by this year's 10th graders: If a 9th grader last year made mop, well then this year they're no longer competing with you. If they didn't make it, well then they have a second shot at making red mop. Just as you will.
[/quote]

I think that this is the good compromise!

AMDDirector has not said anything yet, not sure if he has made his mind already, or they are still debating ...

[quote="niming9991"][quote="iYOA"]the new system isnt really that bad.

if its bothering you, just train harder.[/quote]

I thought that you did not like this new system at the beginning of this thread  :wink:

But anyway, try to put yourself in the shoes of a top 9th grader ...[/quote]
a top 9th grader would stop worrying about this and just make Blue/Black MOP

So I must not be a top 9th grader...

but yes many people will miss out because of the new system; however, would you rather have no red MOP (which would be the most fair) or a different red MOP?

[quote="andersonw"][quote="mathlearner"]
Final comment here: For the life of me, I can’t understand why an organization would put a system in place that can very possibly cause one or more of our country’s most promising 9th graders to not be welcomed to MOP (ones who may be strong enough to make the USAMO, but lose their spot to USAJMO’ers), & may be so discouraged as to just give up. I am NOT a 9th grader, but if I were, and if I were put in such an unfair position, I think I might well be turned off permanently to “competition math”. Worse, if you lose a few of these students every single year going forward, ultimately, the strength of the pool of students (as I believe someone else suggested) will surely be diminished, and that doesn’t just hurt a few students, - it hurts the reputation of our nation.
[/quote]

If there was a hypothetical 9th grader who would be discouraged to the point that he would give up on math if he did not make red MOP, I would think that he doesn't deserve to be in red MOP in the first place. This is because math shouldn't be done for the sake of making MOP, but for the sake of solving nice problems. Last year, as a 9th grader, I missed the red MOP cut-off by 1 point, and of course, I was quite angry (especially since I could have gotten 2 more points if I had written the answer for #2), but I'm still enjoying math now and hope to make blue MOP this year.

Also, you seem to be assuming that 10th graders are inherently smarter and will do better on math contests than 9th graders, which I don't believe is true. And besides, under this system, 9th graders this year will also have a chance to make red MOP the next year as well, as pythag already said.

I agree with echinodermata, though, that if you make red MOP in 9th grade, you shouldn't be allowed to make red MOP again in 10th grade.[/quote]

Agree with the last thing, since my main concern about this system is whether almost all of the 9th graders who make red mop will make it again in 10th grade.

[quote="pythag011"]

Agree with the last thing, since my main concern about this system is whether almost all of the 9th graders who make red mop will make it again in 10th grade.[/quote]

I don't think you need to be concerned about this. for some reason.

But this is a valid point.

[quote="niming9991"]If a 9th/10th grader takes AMC 10 and get AIME > 11, does he have a choice for taking USAMO or JMO, or he has to take USAMO ?[/quote]

The student must take USAMO.

[quote="niming9991"][quote="iYOA"]the new system isnt really that bad.

if its bothering you, just train harder.[/quote]

I thought that you did not like this new system at the beginning of this thread  :wink:

But anyway, try to put yourself in the shoes of a top 9th grader ...[/quote]

this is true,i didnt like it in the beginning and i still have some mixed feelings about it  but its not that bad once you get used to it.  this new system isnt entirely terrible or entirely perfect.

i dont see how this is bad for a top 9th grader.  returning moppers might clog up all the spots but im sure that there will be a decent amount of 9th graders in red mop from usamo.

what my main concern is that the red group from usajmo will be really far behind.

this is mainly because the highest index for a jmoer is 250.  furthermore, the best 9th+10th graders will be in usamo, leaving weaker competition in usajmo.

I already pm'd a reply to Pythag011, but decided to publicly post most of that reply. I think most students would prefer to have half the competition and a single chance to make it, and also, to only be competing for a MOP spot using the IDENTICAL TEST AND CRITERION as everyone else, but with the new system, as a few of us spelled out in the forum, what happens to those that make USAMO and then lose their space to a USAJMOer who might actually be less mathematically inclined (and the system just worked better for them)?, or another possibility, [b]what if at the end of this "experimental year" the AMC decides to revert back to the old ways, which means now not only will current 9th graders NOT get a 2nd chance, but their only chance was against twice as many and again, potentially disadvantaged by the system of a USAJMOer possibly taking their spot.[/b]

 [b][color=darkred]AMC Director: If you are reading this, can you absolutely guarantee this year's 9th graders that this will not occur, or if it does, can they be "grandfathered" to be given the same "2nd chance" as this year's 10th graders are being given?[/color][/b]
(It won't matter to people coming after since if assuming the system returns to the original one, they won't have twice the competition this year's 9th graders are having).  Some of you may think this sounds like a "premature" question as the system may [i]not [/i] change, but I think it's more important to give forethought to such very possible possibilities (sorry for the bad wording) than to see the problem or faults of the system in hindsight.


How many subsets of 6 integers taken from the numbers $1,\, 2, \, 3,\ldots ,\, 20$ are there such that there are no consecutive integers in any subset (e.g. if 5 is in the subset then 4 and 6 cannot be in it)?

How many arrangements of the integers $1,\, 2,\, 3,\ldots ,\, n$ are there such that each integer differs by one (except for the first integer) from [i]some[/i] integer to the left of it?

For the first problem:

[hide="a partial solution"]Note that the generalized problem is recursive: select $k$ exclusively nonconsecutive integers from the set of the first $n$ integers.  Suppose the first integer selected is 1 or $n$.  Then we want to find the remaining number of ways to select $k-1$ integers from a set of $n-2$ integers (since we can't select one integer which bordered on the one already selected).  If the first integer selected is not one of those two, then we want to find the remaining number of ways to select $k-1$ integers from a set of $n-3$ integers (since we can't select two integers which bordered on the one selected).  So we have

$F(n, k) = 2 F(n-2, k-1)+(n-2) F(n-3, k-1)$.

Since this function will give the ordered number of ways to select $k$ integers, we have to divide by $k!$ to get our final answer, so we let

$G(n, k) = \frac{1}{k!}F(n, k)$

and keep in mind that $G(n,k)$ is what we want to find.  Now note that $F(n,1) = n$.  Then we recur:

$F(n, 2) = 2 F(n-2, 1)+(n-2) F(n-3, 1) = 2(n-2)+(n-2)(n-3) = (n-2)(n-1).$

$F(n, 3) = 2 F(n-2, 2)+(n-2) F(n-3, 2) = 2(n-4)(n-3)+(n-2)(n-5)(n-4)$

... and I don't want to expand beyond that.  Does anybody see an easier way to solve that recurrence closed-form?

P.S. Also, after doing some experimenting plugging in specific values, I find that $G(n, k)$ is often non-integral, which is also a problem.  :P
[/hide]

[hide="1"]BALLZ N URNZ (never fails)

We have 20 balls.
Balls A B C D E F are the ones in our set. 14 remain.
The integer they represent will be indicated by the number of balls before them when they are arranged in a row.

We can insert balls into 7 urns.
- A - B - C - D - E - F -

However we must have at least one between each of the ones in the set. 14-5=9 remain.

Now these 9 balls can go into any of the 7 urns. BALLZ N URNZ formula: ${9+7-1}\choose{7-1}$ Whatever that is.[/hide]

Very interesting second problem.  One way to solve it is by induction.  Let $f(n,k)$ be the number of good permutations of length $n$ and begin with the number $k$.

The key to this problem is noting that $f(n+1,k)=f(n,k-1)+f(n,k)$, and this kills the problem.  From induction we obtain $f(n,k)={n-1\choose k-1}$, and then the number of good permutations is

$\sum_{k=1}^{n}f(n,k)=\sum_{i=0}^{n-1}{n-1\choose i}=2^{n-1}.$

I noted that as well, but can you explain why?

Which part?  The discovery of $f(n,k)$ or the inductive step?

This part: $f(n+1,k)=f(n,k-1)+f(n,k)$. Thanks.

The induction is easy to explain by thinking about Pascal's triangle.


trigonometry
inequalities
geometry solved
geometry

Prove that in every acute triangle ABC, we have:

$\sum\frac{1}{\cos A}\geq 6$.

By applying the AM-HM inequality we have that:

$\sum\frac{1}{\cos A}\geq\frac{9}{\sum\cos A}\geq\frac{9}{\left(\frac32\right)}=6$,

because $\sum\cos A\leq\frac32$.

solution:

because $\prod\cos A=\frac12\cos A\left[\cos\left(B-C\right)+\cos\left(B+C\right)\right]=\frac12\cos A\left[\cos\left(B-C\right)-\cos A\right]$, and $\cos A\left[\cos\left(B-C\right)-\cos A\right]\leq\cos A\left(1-\cos A\right)\leq\frac14$ because cos A > 0 and $x\left(1-x\right)\leq\frac14$ for any x from $\mathbb{R}$ we get that $\prod\cos A\leq\frac18$. and we know that cosA, cosB, cosC > 0 so we go AM $\geq$ GM for the initial ineq and we get that $\sum\frac{1}{\cos A}\geq 6$.

cheers! :D


geometry
3D geometry
prism
trapezoid
pyramid
LaTeX
Pythagorean Theorem

Let $ABCDEF$ a right prism which base is a right triangle, with $B\hat{A}C=90^{\text{o}}$.
Given $\overline{AB}=4,\,\overline{AC}=8$ and $\overline{AD}=3,$ where $M$ is midpoint of $EF$ and $N$ is midpoint of $DF$.
i) Find the area of the quadrilateral $ABMN$.
ii) Calculate the volume of the solid $ABEDMN$.

[hide="i)"]
It's a trapezoid. Too lazy to write everything up, but using Pythagorean, it's a trapezoid of area 15.
[/hide]

Part ii is almost directly based off part i.

[quote="Treething"]
Part ii is almost directly based off part i.[/quote]

That's what i'm trying to find out Treething. I did part (ii) using truncated prism formula, and it has nothing to do with part (i). Of course that i can be wrong.

[hide="1"]Using the Pythagorean Theorem, we get $AN=5$. Draw an auxiliary line from $M$ to segment $AB$ parallel to $AN$ and mark its intersection with $AB$ as $P$.

$[APMN]=10$ and $[PBM]=5$. Therefore, $[ABMN]=\boxed{15}$.

Or, $[ABMN]=\frac{NM+AB}{2}\cdot AN=3\cdot5=\boxed{15}$.[/hide]

[hide="2"]Using what we already know from part 1, we can easily find that the volume of $ABEDMN$ is $\boxed{36}$.[/hide]

So i must be wrong, because i got $20$ for part (ii).  :mad:
Somebody can solve part (ii)?

I used the formula of truncated prism:

$V={S}\cdot\frac{a+b+c}{3}$, where $a,b$ and $c$ are the side edges.

I got same result dividing the solid in 2 others, one prism and one piramid.

Using [b]LoboSolitario[/b] picture. Let $V$ be the volume of the prizm.
Let $BM$ intersect $CF$ at $X$. Clearly $XF=FC=h$ The volume of $V_{XCAB}=\frac{1}{3}.2h.S_{ABC}=\frac{2}{3}V$. The volume of pyramide $XFMN=\frac{1}{8}V_{XCAB}$. This makes $\frac{2}{3}V-\frac{1}{8}V=\frac{7}{12}V$. So the volume that we are looking for is $\frac{5}{12}V=20$.
Note that this way one can compute also the area of $AMNB$ as $\frac{3}{4}$ of $AXB$.

Hi everyone, this is my first post at this forum :blush: .
Delta, you are from my country.  :D
So I will tell you another way to find the searched Volume.
Connect N with E and A with E.
We get 2 piramides: ABMNE and ANDE.
Make MH perpendicular to DE(H belongs do DE).
MH=DN=AC/2=4; NM=DE/2=AB/2=2
=> S(NME)=NM.MH/2=2.4/2=4.
But BE is perpendicular to (NME)=>
V(ABMNE)=(1/3)[S(NME).EB]=(1/3).12=4.
Now let's take a look at the pyramid ANDE.
S(AND)=AD.ND/2=3.4/2=6;
DE=AB=4.
V(ANDE)=(1/3)[S(AND).DE]=(1/3)[6.4]=8.
The searched volume is V=V(ABMNE)+V(ABMNE)=4+8=12.
But Delta's idea is better.
Can someone explain me how could I type like you? With pictures? Sorry for the stupid question...

It's a typesetting language called Latex. You don't need to install anything, you just need to use certain commands. I'm pretty sure they have tutorials on this site as well.

yep. just click to Latex on the the left of the page. To use Latex in the forum, just put dollars signs around the code.

[quote="LoboSolitario"][quote="uktc"]
The searched volume is V=V(ABMNE)+V(ABMNE)=4+8=12.
[/quote]

$12$ or $20$ uktc? Looks like you're missing something.[/quote]
I mean V=$V_{ABMNE}$+$V_{ANDE}$=4+8=$12$, i think you understood.
But I cannot find my mistake, can you? And are you sure in the answer $20$? Because I checked my solution and I didn't find where my mistake is... :roll:

I'm not sure where are your mistake uktc, i spent one entire day to find that two solutions indicated above. And after delta solution i have no doubt that result is $20.$ Do you see any errors there?  :P

Try the formula $V=S\cdot\frac{a+b+c}{3}$.

$S=S_{AND}=\frac{4\cdot 3}{2}=6$ and ${a=\overline{MN}=2}, b=\overline{DE}=c=\overline{AB}=4$.

Hence $V=20$.

Hm, OK! ;). But I didn't know this formula: $V=S\cdot\frac{a+b+c}{3}$.
What is it for? For truncated prisms? Only for truncated RIGHT prisms, or for all truncated prisms? :blush:

It doesn't matter, all of us understand what you mean ;). But is it needed for the truncated prism to be right?


Given $ a, b, c > 0$ satisfy $ ab\plus{}bc\plus{}ca\equal{}3.$
Prove that: $ \frac{1}{a^2b^3} \plus{} \frac{1}{b^2c^3} \plus{} \frac{1}{c^2a^3} \plus{} \frac{1}{3a^2\minus{}2ab\plus{}b^2} \plus{} \frac{1}{3b^2\minus{}2bc\plus{}c^2} \plus{} \frac{1}{3c^2\minus{}2ca\plus{}a^2} \geq\ \frac{9}{2}$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 3.$
Prove that: $ \frac {1}{a^2b^3} \plus{} \frac {1}{b^2c^3} \plus{} \frac {1}{c^2a^3} \plus{} \frac {1}{3a^2 \minus{} 2ab \plus{} b^2} \plus{} \frac {1}{3b^2 \minus{} 2bc \plus{} c^2} \plus{} \frac {1}{3c^2 \minus{} 2ca \plus{} a^2} \geq\ \frac {9}{2}$
 :)[/quote]
The stronger one still holds:
\[ \frac {2(a \plus{} b \plus{} c)}{a^2b^2c^2} \plus{} \frac {9}{2(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3} \ge 9
\]
;) :)
This inequality implies our inequality because
$ \left( \frac {1}{a^2b^3} \plus{} \frac {1}{b^2c^3} \plus{} \frac {1}{c^2a^3}\right) \left( b \plus{} c \plus{} a\right) \ge \left( \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \right)^2 \Rightarrow \frac {1}{a^2b^3} \plus{} \frac {1}{b^2c^3} \plus{} \frac {1}{c^2a^3} \ge \frac {a \plus{} b \plus{} c}{a^2b^2c^2}$ (Cauchy Schwarz)
And
$ \frac {1}{3a^2 \minus{} 2ab \plus{} b^2} \plus{} \frac {1}{3b^2 \minus{} 2bc \plus{} c^2} \plus{} \frac {1}{3c^2 \minus{} 2ca \plus{} a^2} \ge \frac {9}{2[2(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3]}$ (Cauchy Schwarz)
:)

Njce proof, can_hang. And this ineq also trues with the condition: $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} 3$
 :)


asi como hay un topic donde ponemos problemas de combinatoria tambien seria bueno uno por cada curso (faltarian de geo y teoria de numeros) , aca voy poniendo un problema facil de algebra en el q quisiera ver diversas soluciones  :lol:

encuentre todas las funciones $F(x)$ tal q $x$ es un natural incluyendo al cero y :

$i)$ $F(0)=0$

$ii)$ $F(x+y)-F(x-y) = 4xy$

Sería bueno que en siguientes post's indiques bien el dominio y el rango de $F$.

Asumo que $F: \mathbb{N}_{0}\rightarrow \mathbb{N}_{0}$. ($\mathbb{N}_{0}=\mathbb{N}\cup{0}$).

Sea $F(x)=x^{2}+g(x)$ con este cambio el problema es equivalente a hallar todas las funciones $g$ tales que
           $g(0)=0$ y $g(x+y)=g(x-y)$ ($x\geq y$).  (*)
 Haciendo $x=y$ tenemos que $g(2y)=0$, para todo $y$; haciendo $x=y+1$ tenemos que $g(2y+1)=g(1)$, es decir que $g$ es constante en los impares. No es difícil probar que cualquier función de la forma:
\[g(x)= \begin{cases}a& \mbox{si x es impar}\\ 0& \mbox{si x es par}. \end{cases}\] (donde $a\in \mathbb{N}_{0}$ es constante) satisface (*). Luego, todas las funciones $F$ son dadas por:
\[F(x)= \begin{cases}x^{2}+a& \mbox{si x es impar}\\ x^{2}& \mbox{si x es par}. \end{cases}\] Noten que, en particular, si $a=0$ tenemos que $F(x)=x^{2}$ es solución.

$Tipe$

No veo porque poner $F(x) = x^{2}+g(x)$... si suponemos que $F(1) = k$ tenemos dos cosas que hacer, meterle $(x,x)$ a la función y meterle $(x+1, x)$
Para el primero obtenemos $F(2x) = 4x^{2}$ para todo $x$. Para el segundo otenemos $F(2x+1)-F(1)=4(x^{2}+x)$ entonces tenemos que $F(2x+1) = 4x^{2}+4x+k = (2x+1)^{2}+(k-1)$.
Como $x+y$ y $x-y$ tienen la misma pariedad tenemos que cualquier valor de $k$ entero funciona si el rango de la funcion son los enteros. Y queda lo mismo que dijo Tipe.

PS. Creo que el rango se debería poder extender a los reales y funciona cualquier valor de $k$ real.

help!!!
hallar todas las funciones.
$F : Z^{+}\to Z^{+}/ F(F(n))=1993(n^{1943})$, para todo $n$ perteneciente a los enteros positivos.

[color=blue]Toda función cuyo dominio sea los enteros positivos tiene un grado $k$, entero no negativo. Si $F(n)$ es de grado $k$, entonces $F(F(n))$ será de grado $2k$, pero surge una contradicción ya que, en el enunciado, $F(F(n))$ es de grado $1943$, impar. Por lo tanto, no existe una función que cumpla con las características requeridas.[/color]

[color=blue]Gracias por la corrección.
Lo único que podría resultar útil de lo que hallé es $F\left({1993x^{1943}}\right) = 1993F\left( x \right)^{1943}$. ¿Qué podría hacer a partir de esta deducción?
[/color]

[quote="tipe"][quote="R.G.A.M."][color=blue]Toda función cuyo dominio sea los enteros positivos tiene un grado $k$, entero no negativo. [/color][/quote]

Eso no cierto, por ejemplo las funciones(definidas en los enteros positivos):
a)  $F(n)$= Suma de divisores de n.
b)  $F(n)$= Menor número primo mayor que n
c)  $F(n)=2n$, si n es par y $F(n)=3$, si n es impar.
no tienen grado.

$Tipe$[/quote]
[color=blue]Sigo pensando que esta función tiene un grado, dada la ecuación. ¿Qué métodos existen para determinar que una función tiene o no grado?[/color]


parameterization
trigonometry
function
real analysis
real analysis unsolved

Find a particular solution using variation of parameters:

1. y'' + 9y = sin(3x)

2. y'' - 4y = xe^x

Also find the general solution.

Well first, you want to solve the homogeneous equation:
$ y"\plus{}9y\equal{}0\Rightarrow r\equal{}\pm 3i\Rightarrow y\equal{}\cos{3t}\plus{}\sin{3t}$.
Then assume the second solution takes the form $ y_{2}\equal{}v_{1}(t)\cos{3t}\plus{}v_{2}\sin{3t}$ where $ v_{i}(t)$ is some function to be determined later.  Then differentiate to get $ y'$.  Set $ v_{1}'(t)\cos{3t}\plus{}v_{2}'(t)\sin{3t}\equal{}0$ and then differentiate again to get $ y"$.  Now you can solve for $ v_{1}'$ and $ v_{2}'$ and in turn, $ v_{1}$ and $ v_{2}$ so you then have your general solution.


I want ask you about the B-1 problem because I cant evaluate the integral.I wait your answers.thanks


[quote="Dr Sonnhard Graubner"]hello, what do you mean with $ sen(2x)$?
Sonnhard.[/quote]That's $ \sin (2x)$. Sorry I don't have enough time right now to post a solution but you would do the following: 'square' the inequality, simplify, use $ \cos^2\theta\plus{}\sin^2\theta\equal{}1$ identity, simplify more, test intervals. I'll check back later.

hello, after your hint i have got
$ \minus{}2\arctan(1\minus{}\sqrt{2})\le x\le \sqrt{2}$
Sonnhard.

[quote="Chalao"]Find the values between 0 and $ 2 \pi$ for which the following equation has a solution

[img]http://i204.photobucket.com/albums/bb126/agerador/trig.jpg[/img]

Please explain it step-by-step. Thanks!!![/quote]
[color=blue]
[b]Problem:[/b]

$ 2\cos x\le\left|\sqrt {1 \plus{} \sin 2x} \minus{} \sqrt {1 \minus{} \sin 2x}\right|\le\sqrt2,\ x\in [0,\ 2\pi]$

[b]Solution:[/b]

$ \sqrt {1 \plus{} \sin 2x} \equal{} \sqrt {\sin^2 x \plus{} \cos^2 x \plus{} 2\sin x\cos x} \equal{} \sqrt {(\sin x \plus{} \cos x)^2}$

$ \implies \sqrt {1 \plus{} \sin 2x} \equal{} |\sin x \plus{} \cos x| \equal{} \begin{cases}{\sin x \plus{} \cos x\ ,} & {x\in \left[0,\ \frac {3\pi}{4}\right]\cup\left [\frac {7\pi}{4},\ 2\pi\right]} \\
{ \minus{} \sin x \minus{} \cos x\ ,} & {x\in \left(\frac {3\pi}{4},\ \frac {7\pi}{4}\right)}\end{cases}$

$ \sqrt {1 \minus{} \sin 2x} \equal{} \sqrt {\sin^2 x \plus{} \cos^2 x \minus{} 2\sin x\cos x} \equal{} \sqrt {(\sin x \minus{} \cos x)^2}$

$ \implies \sqrt {1 \minus{} \sin 2x} \equal{} |\sin x \minus{} \cos x| \equal{} \begin{cases}{\sin x \minus{} \cos x\ ,} & {x\in \left[\frac {\pi}{4},\frac {5\pi}{4}\right]} \\
{\cos x \minus{} \sin x\ ,} & {x\in \left[0,\ \frac {\pi}{4}\right)\cup\left(\frac {5\pi}{4},\ 2\pi\right]}\end{cases}$

$ \implies\left|\sqrt {1 \plus{} \sin 2x} \minus{} \sqrt {1 \minus{} \sin 2x}\right| \equal{} \begin{cases}{|2\sin x|\ ,} & {x\in\left[0,\ \frac {\pi}{4}\right)\cup\left[\frac {7\pi}{4},\ 2\pi\right]} \\
{|2\cos x|\ ,} & {x\in\left[\frac {\pi}{4},\ \frac {3\pi}{4}\right]} \\
{| \minus{} 2\sin x|\ ,} & {x\in\left(\frac {3\pi}{4},\ \frac {5\pi}{4}\right]} \\
{| \minus{} 2\cos x|\ ,} & {x\in\left(\frac {5\pi}{4},\ \frac {7\pi}{4}\right)}\end{cases}$

$ \implies\left|\sqrt {1 \plus{} \sin 2x} \minus{} \sqrt {1 \minus{} \sin 2x}\right| \equal{} \begin{cases}{2\sin x\ ,} & {x\in\left[0,\ \frac {\pi}{4}\right)\cup\left[\frac {3\pi}{4},\ \pi\right]} \\
{2\cos x \ ,} & {x\in\left[\frac {\pi}{4},\ \frac {\pi}{2}\right]\cup\left[\frac {3\pi}{2},\ \frac {7\pi}{4}\right)} \\
{ \minus{} 2\cos x \ ,} & {x\in\left(\frac {\pi}{2},\ \frac {3\pi}{4}\right]\cup\left(\frac {5\pi}{4},\ \frac {3\pi}{2}\right)} \\
{ \minus{} 2\sin x \ ,} & {x\in\left(\pi,\ \frac {5\pi}{4}\right]\cup\left[\frac {7\pi}{4},\ 2\pi\right]}\end{cases}$

Now consider the following cases:

[b]Case 1:[/b] $ \ x\in\left[0,\ \frac {\pi}{4}\right)\cup\left[\frac {3\pi}{4},\ \pi\right]$

Given inequality becomes $ \ 2\cos x\le 2\sin x\le\sqrt2$

Now $ \ \cos x\le \sin x\implies x\in\left[\frac {3\pi}{4},\ \pi\right]$ and $ \sin x\le\frac {1}{\sqrt 2}\implies x\in\left[0,\ \frac {\pi}{4}\right)\cup\left[\frac {3\pi}{4},\ \pi\right]$

$ \implies$ In case 1 we got $ \ x\in\left[\frac {3\pi}{4},\ \pi\right]$

[b]Case 2:[/b] $ \ x\in\left[\frac {\pi}{4},\ \frac {\pi}{2}\right]\cup\left[\frac {3\pi}{2},\ \frac {7\pi}{4}\right)$

Given inequality becomes $ \ 2\cos x\le 2\cos x\le\sqrt2$

Now $ \ \cos x\le \cos x\implies x\in\left[\frac {\pi}{4},\ \frac {\pi}{2}\right]\cup\left[\frac {3\pi}{2},\ \frac {7\pi}{4}\right)$ and $ \cos x\le\frac {1}{\sqrt 2}\implies x\in\left[\frac {\pi}{4},\ \frac {\pi}{2}\right]\cup\left[\frac {3\pi}{2},\ \frac {7\pi}{4}\right)$

$ \implies$ In case 2 we got $ \ x\in\left[\frac {\pi}{4},\ \frac {\pi}{2}\right]\cup\left[\frac {3\pi}{2},\ \frac {7\pi}{4}\right)$

[b]Case 3:[/b] $ \ x\in\left(\frac {\pi}{2},\ \frac {3\pi}{4}\right]\cup\left(\frac {5\pi}{4},\ \frac {3\pi}{2}\right)$

Given inequality becomes $ \ 2\cos x\le \minus{} 2\cos x\le\sqrt2$

Now $ \ \cos x\le 0\implies x\in\left(\frac {\pi}{2},\ \frac {3\pi}{4}\right]\cup\left(\frac {5\pi}{4},\ \frac {3\pi}{2}\right)$ and $ \cos x\ge \minus{} \frac {1}{\sqrt 2}\implies x\in\left(\frac {\pi}{2},\ \frac {3\pi}{4}\right]\cup\left(\frac {5\pi}{4},\ \frac {3\pi}{2}\right)$

$ \implies$ In case 3 we got $ \ x\in\left(\frac {\pi}{2},\ \frac {3\pi}{4}\right]\cup\left(\frac {5\pi}{4},\ \frac {3\pi}{2}\right)$

[b]Case 4:[/b] $ \ x\in\left(\pi,\ \frac {5\pi}{4}\right]\cup\left[\frac {7\pi}{4},\ 2\pi\right]$

Given inequality becomes $ \ 2\cos x\le \minus{} 2\sin x\le\sqrt2$

Now $ \ \cos x \plus{} \sin x\le 0\implies x\in\left(\pi,\ \frac {5\pi}{4}\right]\cup\left\{\ \frac {7\pi}{4}\right\}\ \!$ and $ \sin x\ge \minus{} \frac {1}{\sqrt 2}\implies x\in\left(\pi,\ \frac {5\pi}{4}\right]\cup\left[\frac {7\pi}{4},\ 2\pi\right]$

$ \implies$ In case 4 we got $ \ x\in\left(\pi,\ \frac {5\pi}{4}\right]\cup\left\{\ \frac {7\pi}{4}\right\}$

Hence $ \ \boxed{\boxed{x\in\left[\frac {\pi}{4},\ \frac {7\pi}{4}\right]}}$

 :first:

[/color]


vreau si eu subiectele cu rezolvari de la concursul interjudetean Gh Titeica desfasurat la Oradea 2007 :D tks

Gh. Titeica? Oradea? cine te-a pacalit? sper ca nu te-ai dus acolo:P

sorry era vorba de grigore moisil oradea 2007 :D oricum sunt bune si cele de la titeica :D ms

din nush ce motiv imi da eroare la deschiderea rezultatelor ... as vrea un link cu rezolvarile subiectelor si rezultatele concursului :D multumesc mult


Dear everyone here,
Can any of you tell me where I can find the general theory to learn in order to be able to solve  the kind of question like the Q1 in IMO2007?
Many thanks,
Tao zh

IMO2007 1 was one of those questions where knowing theory or techniques didn't give you much of an advantage. That problem was a test of your creativity with algebra (though not a particularly challenging one).

Rather than trying to learn results to solve questions like that, I would try to increase ability with algebra by doing problems.


$ x,y,z\in R^{\plus{}}$,$ xy\plus{}yz\plus{}zx\equal{}1$.
Prove that $ \frac{(xy)^{2}}{z}\plus{}\frac{(yz)^{2}}{x}\plus{}\frac{(zx)^{2}}{y}\plus{}6xyz\geq x\plus{}y\plus{}z$

[quote="Twins"]$ x,y,z\in R^{\plus{}}$,$ xy\plus{}yz\plus{}zx \equal{} 1$.
Prove that $ \frac{(xy)^{2}}{z}\plus{}\frac{(yz)^{2}}{x}\plus{}\frac{(zx)^{2}}{y}\plus{}6xyz\geq x\plus{}y\plus{}z$[/quote]
It's Schur only. :wink:

Can you explain it more? Thank you!
I know little about Schur, in fack I don't know how to use Schur

Me too. :blush:
Arqady : actually you make a deep impression to me, everytime you finished your solution just by writing :
and then by shur and muirhead ......... :)

let $ a\equal{}xy,b\equal{}yz,c\equal{}zx$ then your inequality is equivalent to:
$ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}6abc\geq ab\plus{}bc\plus{}ca$
now we know that $ a\plus{}b\plus{}c\equal{}1$ so our inequality is equivalent to:
$ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}6abc\geq (ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$
$ \iff a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}3abc\geq a^{2}b\plus{}ab^{2}\plus{}b^{2}c\plus{}bc^{2}\plus{}c^{2}a\plus{}ca^{2}$
$ \iff\sum a(a\minus{}b)(a\minus{}c)\geq 0$
the last inequality is called schur it says that:
for positive reals $ a,b,c$ and real number $ r$ the following inequality holds:
$ \sum a^{r}(a\minus{}b)(a\minus{}c)\geq 0$


This may be trivial, but how would you rigorously prove that the number of digits of any number $ n$ is always $ \lfloor \log{n} \rfloor \plus{} 1$?

[hide="hint"] Some number $ a\cdot{10^n}$ where $ 1\le{a} < 10$ has $ n \plus{} 1$ digits. So, if you can find the value of $ n$, you can find the number of digits.[/hide]

[hide="proof"] Suppose we want to find the number of digits of some number $ x$, where x is positive (otherwise just consider $ |x|$). Then, there exists some nonnegative $ n$ so that $ 10^{n}\le{x}<10^{n\plus{}1}$, implying $ n\le{\log{x}}<n\plus{}1$, so $ \lfloor{\log{x}}\rfloor\equal{}n$. However, as $ 10^n$ has $ n\plus{}1$ digits, $ x$ must have $ \lfloor{\log{x}}\rfloor\plus{}1$ digits, as desired.[/hide]


There was another problem like this somewhere else (I forgot where; somebody please remind me).  But you basically try to find the worst case scenario to get the minimum number.  (75+70)-100=45.  (45+85)-100=30. and (30+90)-100=20.  So twenty percent of the men meet all these things.  Is there a principle that can be applied to do what I just did.  I think it might be a variation of the Principle of Inclusion/Exclusion.

-interesting_move

First pick a number say 100.  There are 100 people.  There are 70 with black eyes and 30 without black eyes.  There are 75 with curly and 25 w/o.  Assume that the people with black hair don't have curly hair.  Repeat this over and over again.  This minimizes the number of people that could possibly have all of these traits.  The answer is written in binary: 10100 :wink:

Ah, I remember that problem that was "just like this."  It's at:
[url]http://www.artofproblemsolving.com/wforum/viewtopic.php?t=278[/url] and is called "2 simple problems."  Here is my response from that problem, with edited numbers and characteristics.

QUESTION: in a world, the men, 70% of them have black eyes, 75% have curly hair, 85% are over 4m and 90% have over 300pounds! what is the minimum of procent of the men to meet all these things!
35, 20, 40, 55 or 45?

In the "worst case scenario," as many people have 3, but not 4, or these characteristics (since this would cause the most to not know ALL four). In such a case, 30% don't have black eyes, 25% don't have curly hair, 15% aren't over 4m, and 10% aren't over 300lbs - for a grand total of 80% not having all four.

Hence, 20% have all three characteristics.


function
algebra
domain
real analysis
real analysis unsolved

Let $ f,g :\mathbb{R}\to\mathbb{R}$ be functions. Let $ \sigma(f) \equal{} f^{\minus{}1}(\mathcal{B}(\mathbb{R}))$, the smallest $ \sigma$-algebra with respect to which $ f$ is measurable. ( $ \mathcal{B}(\mathbb{R})$ = the Borel $ \sigma$-algebra on $ \mathbb{R}$). Prove that if $ \sigma(g)\subset\sigma(f)$ there exist a Borel-measurable function $ h :\mathbb{R}\to\mathbb{R}$ such that $ g \equal{} h\circ f$.

It is inessential that the domain of $ f,g$ is the real line. Let's say $ f,g :\Omega\to\mathbb{R}$ where $ \Omega$ is any set.
First suppose the range of $ g$ is a finite set i.e. $ g$ is simple function. If $ \{a_{i}\}_{1}^{n}$ is its range, $ g$ can be written as $ \sum_{1}^{n}a_{i}1_{E_{i}}$ where $ \{E_{i}\}_{1}^{n}$ is a partition of the domain $ \Omega$. Since $ E_{i}\in\sigma(f)$, there is Borel sets $ F_{i}$ with $ E_{i}\equal{} f^{\minus{}1}(F_{i})$. $ \{F_{i}\}_{1}^{n}$ consists of disjoint sets and without loss of generality we assume their union is $ \mathbb{R}$ (otherwise, replace $ F_{n}$ with the complement of the union of the rest). Let $ h \equal{}\sum_{1}^{n}a_{i}1_{F_{i}}$. Then $ h$ is Borel-measurable and $ g \equal{} h\circ f$.

Now to the general case : when $ g$ is not a simple function, construct a sequence $ \{g_{i}\}_{1}^{\infty}$ of simple functions such that $ \sigma(g_{i})\subset\sigma(g)$ and that $ g$ is its pointwise limit. Previous paragraph shows there is Borel-measurable $ h_{n}:\mathbb{R}\to\mathbb{R}$ with $ g_{n}\equal{} h_{n}\circ f$. Since $ g_{n}$ converges to $ g$, $ h_{n}$ converges on $ f(\Omega)$. But $ f(\Omega)$ may not be a Borel set. Let $ h(t) \equal{}\limsup h_{n}(t)$ for all $ t$ with finite $ \limsup h_{n}(t)$ and $ h(t) \equal{} 0$ for other values of $ t$. Then $ h$ is Borel-measurable and $ g \equal{} h\circ f$.


trigonometry
integration
calculus
calculus computations

Find out how much it is :

$\int_{0}^{\pi/2}(\sin(4n+2)t-\sin(4n-2)t)dt=?$

Gordmit sorry but i don't agree with you .I think that it is $\frac{-2}{4n^{2}-1}$.
How is right? Me or him?

Yes, you are right, sorry : )
This integral so much easy, and I've tried to solve it mentally, but I've done a mistake in my mental solution.


I feel like the answer is staring me right in the face, but I just cannot grasp this:

The local school is holding a raffle, and has sold  140 tickets.  They are raffling off 5 ipods and 8 stereos.  Bill bought 11 tickets for the raffle.
a) What is the probability that Bill wins exactly one ipod as well as exactly one stereo?
b) Bill just heard that the drawing for the ipods has been held and he has won exactly one of the 5 ipods. What is the probability that he will win at least one stereo in the remaining drawing?

Any help would be appreciated!

Hint part a):

[hide]
In the drawing, the tickets are divided into three subsets, one of 5 tickets (ipods), one of 8 tickets (stereos), and one of 127 tickets (no price).  Calculate the number of ways in which we may pick one ticket out of the first set, one ticket out of the second set, and 9 tickets out of the third set, that is your space of favorable cases.  Divide by the number of total cases to obtain the probability.
[/hide]

Hint part b):

[hide]
Now there are only 135 tickets (the 5 tickets of the ipods have been taken away) to be divided in two subsets of 8 and 127.  Bill holds exactly 10 of those tickets (the other one is the ipod that he has won).  Now calculate the probability that Bill will not win a stereo (where the total number of favorable cases is clearly the ways in which 10 tickets out of the 127 without price may be picked) and subtract it from 1 to find the probability that Bill will win one stereo or more.
[/hide]


Pythagorean Theorem
geometry
power of a point

1.) Five members of a basketball team are weighed and an average weight is recalculated after each member is weighed. If the average increases 3 pounds each time, how much heavier is the last player than the first?

2.) Perpendicular tangents $CA$ and $CB$ are drawn to circle $O$ with radius $4$ units. If $\overline{BE} = \text{1 unit}$, find $\overline{AD}$. Note: the line $AB$ does pass through the circle's center. A picture is attached...

We had the first one as a question in the 2nd round of the FMO :)
[hide]Suppose the first player weighs x pounds.
The second player should then weigh x+6 pounds so that the average increases with 3.
The average of the first three players should be x+6, so the third player should weigh an extra 6 pounds --> x+12.
Average of the first four should be x+9, so the fourth player has to weigh x+18.
And at last for the fifth player, you have to add an extra 6 pounds, wich makes him 24 pounds heavier than the first one.[/hide]

Ahh I see, thanks! What is the FMO by the way? Does anyone have a solution for the second problem?

Here are my ideas on the problem so far:

Since the radius of the circle is 4 units, the diameter is 8 making the length of $\overline{DB} = 9$. However, I don't know how to figure out the length of $\overline{AD}$

[hide]
If you draw in $OE$ and $OF$ with $E$ being the tangent point on $CB$ and F being the tangent point on $AC$, you get a square, $FOCE$ with side length 4.  From pythagorean theorem (or you could use power of a point) $BE=3$.  $\bigtriangleup AOF \sim \bigtriangleup ABC$.  Let $AD=x$.  $CE=4$, because it is a sidelength of a square.  Therefore, $BC=3+4=7$.   $FO=4$, it is the radius.  $\frac{CB}{FO}=\frac{AB}{AO} \rightarrow \frac{7}{4}=\frac{x+9}{x+4}$.  Solve for $x=\frac{8}{3}$.

i believe that's the right answer, if i didn't make any mistakes.[/hide]

[quote="prime_lover"]Ahh I see, thanks! What is the FMO by the way? [/quote]
Flanders Mathematical Olympiade :)
it was in the form of a multiple choice question though...


function
real analysis
real analysis theorems

Let v is a continuous in [0,1],consider map A:L2[0,1]--->L2[0,1]:Ax(t)=x(t).V(t).\\A\\=???help me!

.???[/list]

What are you asking?  I can't understand anything written there.

But it's impossible to tell what you're asking.  Also, you mean to say "An exercise in functional analysis."

Here is my best guess.

Fix $ v(t)\in C[0,1]$.

Define the operator $ A: L^2[0,1]\to L^2[0,1]$, by $ A(x(t))\equal{}x(t)\cdot v(t)$.

Find $ \|A\|$.

[quote="HilbertThm90"]Here is my best guess.

Fix $ v(t)\in C[0,1]$.

Define the operator $ A: L^2[0,1]\to L^2[0,1]$, by $ A(x(t)) \equal{} x(t)\cdot v(t)$.

Find $ \|A\|$.[/quote]

Suppose that  HilbertThm90's guess is correct :lol:
then this problem is a typical problem. Use the definition of norm $ \parallel{}A\parallel{}$ to find  $ \parallel{}A\parallel{}$, with a remark that $ x(t)\equal{}1\in L^2[0,1]$

[quote="nguyenvanteoo"][quote="HilbertThm90"]Here is my best guess.

Fix $ v(t)\in C[0,1]$.

Define the operator $ A: L^2[0,1]\to L^2[0,1]$, by $ A(x(t)) \equal{} x(t)\cdot v(t)$.

Find $ \|A\|$.[/quote]
 no! i don't think so,you are wrong.[/quote]

so what are you asking?


calculus
function
vector
derivative
limit
integration
complex analysis

There are quite a lot of Theorems in Single variable calculus but I cannot find multi-variable versions
Anyone know if we have a L'Hopspital Rule for multivariable calculus?  :?:

Let's stop to think about this: L'Hôpital's rule involves the quotient $\frac{f'(x)}{g'(x)}.$ This very much depends on $g'(x)$ being something that we can divide by.

Now suppose that $f$ and $g$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ for $n>1.$ What do we mean by $g'$ in that case? The most computational version would have it be a vector (a vector-valued function). But division by a vector is not an algebraic operation that makes any sense.

If you want to take directional derivatives, then two things happen. The first is that you are now doing one-dimensional calculus anyway, so you can use the ordinary L'Hôpital's rule anyway. The second thing? Well, consider this:

[b]Proposed Theorem:[/b]

Suppose $\lim_{t\to0^+}F(x+tu)=L$ for all unit direction vectors $u.$ Then $\lim_{y\to x}F(y)=L.$

Sounds good? Unfortunately, this proposed theorem is false. Consider this function:

$F(x,y)=\frac{xy^2}{x^2+y^4}.$

---

I have on this site consistently advocated not relying on L'Hôpital's rule as a primary method of finding limits, reserving its use for those few cases to which it is well-suited. The alternative methods for finding limits? (1) Algebra, (2) a good understanding of relative size, and (3) power series/Taylor series methods. With that in mind, note that there [b]is[/b] a multidimensional version of Taylor's Theorem.

[quote="Kent Merryfield"]
I have on this site consistently advocated not relying on L'Hôpital's rule as a primary method of finding limits, reserving its use for those few cases to which it is well-suited. The alternative methods for finding limits? (1) Algebra, (2) a good understanding of relative size, and (3) power series/Taylor series methods. With that in mind, note that there [b]is[/b] a multidimensional version of Taylor's Theorem.[/quote]

Kent,

I am curious as to why you try suggest avoiding the use of L'Hôpital's rule. I do indeed think that the other methods that you mentioned above seem more "natural" or "logical," but...why is that important?

I have several reasons. One comes from my experience as a teacher. L'Hôpital's rule is a mechanical, procedural method - a "black box" if you will. As such it is often applied uncritically, and in such a way as to obscure understanding of what is going on.

I offer as an example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49170]this topic.[/url]

The proposer there was dealing with the series $\sum_{n=1}^{\infty}\frac{(-1)^nn}{n^2+2}.$

He applied L'Hôpital's rule to the sequence $\frac{n}{n^2+2},$ getting his mechanical answer that this does indeed tend to zero - and in the process completely lost track of what it was that he was trying to do.

Of course, this belongs squarely within my choice "(2) a good understanding of relative size."

For large $n$ and in a context of multiplication or division, $n^2+2$ might as well be $n$ and thus $\frac{n}{n^2+2}$ might as well be $\frac1n.$ Saying that the terms might as well be $\frac1n$ helps keep us attentive to the question that we should be answering - and we see that our series fails to converge absolutely by limit comparison to the harmonic series.

L'Hôpital does have its place. It's a tool, and I can construct examples for which it may the best tool.

Example: $\lim_{z\to1+i}\frac{z-(1+i)}{z^4+4}.$

Example: $\lim_{x\to\infty}e^{-x^2}\int_0^xe^{t^2}dt.$

But it's only one tool among many - it's not THE WAY to do everything.

[quote="Kent Merryfield"] L'Hôpital does have its place. It's a tool, and I can construct examples for which it may the best tool.

Example: $\lim_{z\to1+i}\frac{z-(1+i)}{z^4+4}.$ [/quote]

How would one use L'Hospital rule for complex $z$?

[quote] Example: $\lim_{x\to\infty}e^{-x^2}\int_0^xe^{t^2}dt.$   [/quote]

OK.  But as always, the methods without L'H rule lead to more understanding of this problem.

L'Hôpital's rule does work for quotients of holomorphic functions - I'll defer to some complex analysis textbook for the proof. Of course, its use in the first problem I stated is merely a minor convenience, as the problem also falls to my method (1): algebra. In particular, you can factor $z^4+4$ into four linear factors, one of which cancels with the numerator. And as fleeting_guest asserts, it it quite possible to do the second problem by other means. It's just that L'Hôpital is probably quickest.

All of that just emphasizes my original point, and there are plenty of cases in which it would be just stupid to even try to use L'Hôpital, such as

$\lim_{x\to0}\frac{\sin^4(3x)}{1-\cos(2x^2)}$

Certainly L'H rule works if you are already in the context of power series.  The reason for the question was that it is not self-evident from the proof, let alone the statement, of L'H for differentiable functions of a real variable, that it works for complex $z$.

The example with $z = 1+i$  is just (the reciprocal of) the definition of $f'(z)$ at $z=1+i$ where $f(z)$ is the denominator.  It is relatively common in textbook L'Hospital problems to have a logical circularity, where one computes the derivative of the function at some point, in order to compute a limit which is... the derivative of the function at that point.  (The most famous example is the limit of $sin(x)/x$ at $0$).
Also, for the $1+i$ problem all the work lies in verifying that L'H rule actually applies, i.e. $f(1+i)=0$, which means doing sufficient algebra to already imply the solution without L'H, as you point out.

The other example can be considered as an analogue of $e^{ax}$ acquiring a factor of $1/a$ when integrated, but with variable $a$.   There are various ways of understanding this that generalize to contexts where L'Hopital is not available, and I think that some of them could probably be phrased to give a comparably short solution.

l'Hospital Rule is something that worth much discussion. It is a usual "mechanical procedure" for finiding limit of single variable function but not always work.
For example the following limit is obvious but cannot be found by l'Hospital Rule
$\lim_{x\to\infty}\frac{x+\cos x}{x} =1$
On could obtain $\lim_{x\to 0}\frac{\cos (\sin x)-\cos x}{x^4} = \frac{1}{6}$
by Taylor series Expansion but not so easy by l'Hospital Rule
I do think we should find a guideline to tell people "when to attempt l'Hospital Rule" and "when we should not" :?


If you mean $ 3\frac {6}7 \div 9$
Then
$ 3\frac {6}7 \equal{} \frac {27}7$
so it becomes  $ \frac {27}7 \div 9$ which is $ \frac {27}{63} \equal{} \boxed{\frac {3}{7}}$

Piece of cake. :)

You mean 27/7 divided by 9, which is 3/7.

[quote="gauss1181"]Piece of cake. :) [/quote]
I don't think it's nice to call things easy, because you might be offending the people who have trouble with it. But yeah, to solve problems like this in general, first put in to an improper fraction then divide the numerator with the proper number. Although should this be in Middle School section? Oh well, there's no elementary so it might as well be in here.


Would it be allowed to use a formula or fact on the USAMTS if it was, say, from Wolfram or Wikipedia?

I don't see why not.  It's not like formulas or theorems are copyrighted material.

The graders don't have access to the source that you cite, so it is pointless. If you cite something nontrivial and not well known, you will probably loose points.

Could you just copy and paste all the text into your solution?

Copying and pasting would be classified as plagiarism. That would be bad.  It's probably best just to summarize a proof for it.

What if it's not really a proof? Just a nontrivial fact that has already been discovered?

Just use common sense, thats the best I can say. You'll probably know what to do.

Guys guys guys. You can cite anything that doesn't trivialize the problem. That's definitely how it is for USAMO ergo that is probably how it is for USAMTS (note "probably"). Besides, the rules explicitly say you can research the problem.

So if something just flat out states the answer, it's not good?

I think whoever wrote the problems would have done research on it before publishing to make sure nothing made the problem too easy.

In my opinion, if you cite a fact, theorem, whatever, that trivializes the problem, it will not be good unless you can prove them. If it's not (like just one of the steps, not a specific solver), and you can't avoid that, I think it will be fine.

Enough debate.

[b]When in doubt, prove it.[/b]

Problem solved.