1) If you quote a definition from a book, how should you state the source? (Where would you put it, what would you put--author, title, etc.)

2) If you use a CAS (computer algebra systems) like Derive, should you explain what it is? What should I write? If it was Mathematica, then you wouldn't need to, but I don't think Derive is as popular as Mathematica, etc.

3) If you use a calculator (TI-83+) program, should you give a URL for where you can download it? How should you mention it?

Thanks! :)

The general rule of citation, not just on the USAMTS but in math and science in general, is to give enough information so that the reader can verify your work if he/she wishes.  Meaning if you use a program, you should include the source code and/or the URL where it can be found.  If you use a "well-known" software package (and Derive is in that category as far as I'm concerned), you should state it by name.

Citing reference books should use a standard citation: author, title, publisher, year, page numbers.


modular arithmetic
summer program
PROMYS
number theory
greatest common divisor
relatively prime
prime numbers

hello i'm new here and don't know where this might go...but anywayshere is a good problem i just solved recently:
if in a game there are only 2 possible scores $a$ and $b$, (u can choose to get either one each time) then:
i)what is the largest score that can not be made by these numbers?(proof needed)
ii)what is the number of impossible scores?(proof needed)

please don't say for part (i), based on the Chicken McNugget Theorem it has to be $ab-a-b$--it would be unacceptable :)

good luck.

It is important to note that $a$ and $b$ must be relatively prime. Else, their sum would have to be a multiple of some prime number, so there will always be a number not obtainable.

For part i (there was a similar proof in AoPS volume 2), write:\[\begin{array}[b]{ccccc}
0 & a & 2a & 3a & \ldots\\
b & b+a & b+2a & b+3a & \ldots\\
2b & 2b+a & 2b+2a & 2b+3a & \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots\\\end{array}\]
You can get all $n\equiv0\pmod{a}$ from the first row.
$n\equiv b\pmod{a}$ from the second row for $n\ge b$.
$n\equiv 2b\pmod{a}$ from the third row for $n\ge 2b$.
Eventually, all numbers can be obtainable in $\bmod{a}$ after row $a$ (that is, we can get all $n\ge(a-1)b$.
Since it is the last row, that is the smallest number of that remainder in $\bmod{a}$ that is obtainable. Since all the other remainders are already obtainable, the largest number that cannot be reached is then $ab-a-b$.

No idea yet for the second part...

Just FYI, this is very similar to a question on the PROMYS application. Though PROMYS starts tomorrow, they like to reuse problems. Not accusing or anything, but a heads up to people around here.

this is the coin exchange problem of Frobenius, and has been discussed before.

Does a closed form for the second part exist?

Yes.  One can show that if two nonnegative integers $ k,\ell$ sum to $ ab \minus{} a \minus{} b$, then exactly one of $ k$ or $ \ell$ is an impossible score (and one is possible).  From this, a closed form for ii) follows immediately.

So in that case the answer is just $ \frac{ab\minus{}a\minus{}b\plus{}1}2$, right?

I think I see the reasoning behind it: it cannot be two numbers that can both be expressible as a non-negative linear combo of $ a,b$ or otherwise $ ab\minus{}a\minus{}b$ will be able to also. So it must be at least one that is not expressible, and this number must be in the form $ N\equal{}mb\minus{}na>0$ for $ 0<m\le a\minus{}1\in\mathbb{N}$ and any natural $ n$ such that $ n>0$ (considering the complete residue system mod a $ \{0,b,2b,...(a\minus{}1)b\}$), hence we have $ (a\minus{}1)b\minus{}a\minus{}N\equal{}(a\minus{}1\minus{}m)b\plus{}(n\minus{}1)a$, which is a valid non-negative linear combo of $ a,b$.

Is this proof good enough (of course the formulation for N follows directly from the proof of the chicken mcnugget theorem)?
BTW why is it called chicken mcnugget theorem? Can't they give it a more appetizing name? :P

People were wondering the largest number of Chicken McNuggets that cannot be attained when you order small, medium, or large orders of Chicken McNuggets, which came in orders of 6, 9, and 20. Hence the name.

Also the answer for (ii) seems right because it's odd iff a and b are not both even, and if a and b are even then the GCD is more than 1.

EDIT: Wait a minute, if $ ab\minus{}a\minus{}b$ is even (which it is), then we have it over 2 plus it over 2 equals it. $ \frac{ab\minus{}a\minus{}b}{2}$ can't be both obtainable and unobtainable at the same time...

[quote="1=2"]EDIT: Wait a minute, if $ ab \minus{} a \minus{} b$ is even (which it is)...[/quote]
Correction: which it never is if $ a$ and $ b$ are relatively prime.

Just wondering, can this be generalized to more than 2 variables?

If you look at possible scores of $ a,b,c$, the problem is pretty open. But if you look at possible scores of $ ab,bc,ca$ or $ abc,bcd,cda,dab$, interesting stuff happens.

[hide="idea for three numbers"] Ask the same question using three relatively prime numbers $ a, b$ and $ c$, where $ a < b < c$. Consider all numbers now $ \pmod{a}$. Now, using x $ b$s and y $ c$s, where x and y may be as large as needed, construct a complete residue class $ \pmod{a}$, using the smallest value of $ bx \plus{} cy$ for each $ d\pmod{a}$ (note that a immediately gives the $ 0\pmod{a}$).Then take the largest $ bx \plus{} cy$, subtract a from it, and use that as your answer.

For example, consider 4, 9, and 11. We have to find the smallest sum of 4s, 9s, and 11s so we end up with numbers congruent to $ 0, 1, 2, 3 \pmod{4}$.

Smallest 0 (mod 4): 4
Smallest 1 (mod 4): 9
Smallest 2 (mod 4):18
Smallest 3 (mod 4): 11

Our answer is 18-4=14.

Unfortunately, this may take some time to calculate for larger numbers...[/hide]

[quote="MellowMelon"]But if you look at possible scores of $ ab,bc,ca$ or $ abc,bcd,cda,dab$, interesting stuff happens.[/quote] Could you please elaborate some interesting stuff about this? Thanks.

[quote="The Enigma"][quote="The Enigma"]i am very sry to express my ignorance but wat is the chicken mcnuggets theorem?[/quote]
some one help me..[/quote]

http://lmgtfy.com/?q=chicken+mcnugget+theorem


geometry
vector
analytic geometry
geometry unsolved

It might be (and I know it actually is) a simple problem but my skill in geometry is not sharp. So, what I have in my program is a line and a segment given with their two points respectively. The question is, how do I find out whether they intersect without computing the intersection point itself? In other words, I need a condition of their intersection.
I'd be very grateful for your help.

The line intersects the segment iff the two end points of the segment are on opposite sides of the line. How do we know if they are on opposite sides of the line? Denote two arbitrary points on your line A, B and the end points of the segment C, D. Compute (B-A)x(C-A), that is, a vector in the direction of the line crossed with a vector from one of the end points to the line. Then compute (B-A)x(D-A), or the same thing for the other end point. Note that both of these products will have only a z coordinate. If both z coordinates are the same sign (+ or -) then the segment does not intersect the line. Otherwise, they do.

By the way, I think this topic would be better suited for the comp sci forums but ah well.


conics
projective geometry
geometry proposed
geometry

Given three circles in the plane, none of which lies inside another (nonintersecting). Prove that the main diagonals of the convex hexagon determined by three pairs of internal common tangents of the three circles are concurrent. ( where the main diagonal means the one that exactly divides the hexagon into two convex quadrilaterals.)

My solution:

Denote these three circles as $ (O_1), (O_2), (O_3) $

By Monge's theorem we get
the intersection of two external tangents of $ \{(O2),(O3) \}, \{(O3),(O1) \}, \{(O1),(O2)\} $ are collinear ,
so by [url=http://mathworld.wolfram.com/FourConicsTheorem.html]Dual of Four Conics theorem[/url] $ \Longrightarrow $ six internal tangents of $ (O1), (O2), (O3) $ form an envelope of a conic .
By Brianchon's theorem we get the main diagonals of the convex hexagon bounded by three pairs of internal common tangents of these three circles are concurrent .

Q.E.D


algebra
polynomial
advanced fields
advanced fields unsolved

Hi all,

I have the following problem:

Let $ p \in \mathbb{C}[z]$ be a polynomial with complex coefficients, which has no roots in $ S^1$.  Show that the number of zeros of $ p$ in the interior of the unit disc is the degree of the continuous map $ \hat{p}: S^1 \rightarrow S^1$ defined by $ \hat{p}(z)\equal{}\frac{p(z)}{|p(z)|}$.

One may factor $ p(z)\equal{}f(z)g(z)$ where $ f(z)$ has no roots $ z$ with $ |z| \geq 1$ and $ g(z)$ has no roots $ z$ with $ |z| \leq 1$.  One may do this because of the Fundamental Theorem of Algebra and by reordering of the linear factors.  If I could show that $ \hat{p}$ is homotopic to $ f/|f|$, then I would be done since that would imply that $ \deg(\hat{p})\equal{}\deg(f/|f|)$, which is the number of zeros in the interior of the disc.  But that is where I'm encountering trouble.  How can I do this?

Thanks in advance.

Your notation is unclear - what does $ S_1$ mean? I thought it meant the unit circle, but clearly that's not right.

Here, $ S^1: \equal{} \{z \in \mathbb{C} | |z|\equal{}1\}$ is the boundary of the $ 2$-disc.

I was getting worried about what would happen to the map p-hat near zeros of p, but the fact that it is continuous immediately implies that all =discontinuities are removable and we fill them in as dictated.

The fact that $ p$ has no roots on $ S^1$ implies that $ \hat{p}: S^1 \rightarrow S^1$ is well-defined.  But how is it homotopic to $ f/|f|$?

argument principle.
consider $ log(\frac{p}{|p|})$, and see how many times will it go around the $ S$ when $ z$ run around $ S$.

The degree of $ p$ is the integer $ d$ such that $ p_*(x) \equal{} dx$ for every $ x \in \pi_1(S^1)$.

You can show that
$ \mathrm{deg}(pq) \equal{} \mathrm{deg}(p) \plus{} \mathrm{deg}(q)$: since $ S^1$ is a topological group, multiplication
of loops is homotopic to concatenation, from which this result follows. (This is problem 3 in May Ch. 1; do it
first to make things easier.)

Given this, you only have to show the result for linear polynomials $ (x \minus{} z)$. From here this should be much
easier. (There are many ways to proceed - since you are probably getting this problem from May's book,
reference the above argument why $ \pi_1(S^1) \equal{} \mathbb{Z}$.)


probability
geometry
3D geometry
real analysis
real analysis unsolved

Suppose that $a_1,...,a_k$ are distinct irrational numbers and $N$ is a positive integer. Is it true that the probability for $n$ to verify that $[na_1]+...+[na_k]$ is congruent to $j$ (mod $N$) is the same for all $j$? Does anyone have any idea about a proof?

I don't think it's true.  Set $a_1 = \sqrt{2}$, $a_2 = \sqrt{2} + 2$, and $N = 2$.

Sorry, I was idiot, I should have realized that. But what if $a_1,...,a_k$ are linearly independent over $Q$?

Then the answer is "Yes". All you need to do is to use the fact that the points $na_1,\dots, na_k$ taken modulo $N$ are uniformly distributed in the cube $(0,N)^k$.


function
integration
limit
calculus
derivative
real analysis
geometry

Can anyone help prove Riemann Theorem?
The definition is as following:
Assume the symbols using here are well-defined.
And $G(x)$ is a cyclic function with a cycle of $T$.
Prove:
$\displaystyle \int\limits_{a}^{b}F(x)G(px)dx=\frac{1}{T}\int\limits_{a}^{b}F(x)dx\int\limits_{0}^{T}G(x)dx$

I forget something,the left hand side should be the limit when $p$ tends to $+\infty$.
However,I don't know how to type that symbol in LaTeX. :blush:

Is this what you mean?:

$\lim_{p\rightarrow\infty}\int_{a}^{b}F\left(x\right)G\left(px\right)\,dx=\frac{1}{T}\int_{a}^{b}F\left(x\right)\,dx\int_{0}^{T}G\left(x\right)\,dx$

Yes,but I didn't know how to type in that limit when I was posting that.

[Note on language: what you called "cyclic" is usually called "periodic" in English]

You might as well start by subtracting off the average value- write $G(x)=h(x)+c$, where $h$ has integral zero over one period. $\int_a^b F(x)\cdot c\,dx=\frac1T\int_a^b F(x)\,dx\int_0^T c\,dx$, and we are left with the taxk of proving that $\int_a^b F(x)h(px)\,dx\to 0$ as $p\to\infty$ when $h$ has average value zero.

If $F$ is continuously differentiable, we can use integration by parts. Let $H$ be an antiderivative of $h$; since $\int_0^T h(x)\,dx=0$, $H(a+T)=H(a)$ and $H$ is bounded. Let $F'=f$.
We have $\int_a^b F(x)h(px)\,dx=\frac{F(b)H(pb)}{p}-\frac{F(a)H(pa)}{p}-\frac1p\int_a^b f(x)H(x)\,dx$. Each term is bounded by some constant multiple of $\frac1p$ by boundedness of $M$, and everything goes to zero.

For uglier functions $F$, we use a density result; the "nice" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).

Firstly,thanks for the advice on language.
[quote="jmerry"]
We have $\int_a^b F(x)h(px)\,dx=\frac{F(b)H(pb)}{p}-\frac{F(a)H(pa)}{p}-\frac1p\int_a^b f(x)H(x)\,dx$. Each term is bounded by some constant multiple of $\frac1p$ by boundedness of $M$, and everything goes to zero.
[/quote]
I don't really understand this,can you give a more detailed explanation?
I can prove the cases that $F(x)$ is a continuous function.But the problem is,$F(x)$ can be a lot uglier.
The original expression is that $F(x)$ and $G(x)$ are Riemann integrable...
[quote="jmerry"]
For uglier functions $F$, we use a density result; the "nice" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).[/quote]
What's the meaning of $C^1$ and $L^1$ :?:
And what is the meaning of this density result?

Firstly,thanks for the advice on language.
[quote="jmerry"]
We have $\int_a^b F(x)h(px)\,dx=\frac{F(b)H(pb)}{p}-\frac{F(a)H(pa)}{p}-\frac1p\int_a^b f(x)H(x)\,dx$. Each term is bounded by some constant multiple of $\frac1p$ by boundedness of $M$, and everything goes to zero.
[/quote]
I don't really understand this,can you give a more detailed explanation?
I can prove the cases in which $F(x)$ is a continuous function.But the problem is,$F(x)$ can be a lot uglier.
The original expression is that $F(x)$ and $G(x)$ are Riemann integrable...
[quote="jmerry"]
For uglier functions $F$, we use a density result; the "nice" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).[/quote]
What's the meaning of $C^1$ and $L^1$ :?:
And what is the meaning of this density result?Can you give an explanation?Thank you.
Sorry that I've posted the same thing twice,that's a mistake.

$C^{1}$ just means that the function's derivative is continuous.  I don't know what $L^{1}$ means...

$f \in L^1(\Omega) \Rightarrow \int_{\Omega} |f(x)|dx <\infty$
$f$ is any function.
$\Omega$ is some space (I am used to subspaces of $R^n$ but you can do other spaces if you have a measure).

And if you want $L^1$ to be a metric space, then its members are not functions but equivalence classes of functions under the relation that two functions are equivalent if they are equal almost everywhere. (Example: the function which is 1 on the rationals and 0 on the irrationals belongs to the equivalence class of the the zero function.)

Finding a nice dense subset would be saying that for any $\epsilon>0,$ we can find a "nice" function $g$ such that $\int|f-g|<\epsilon.$ If you take for granted that you can do that, it shouldn't be that hard to see how to extend jmerry's argument.

Actually proving the density would be part of the Lebesgue theory of integration.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26568]This[/url] is the best explanation I can find. Look up the Riemann-Lebesgue lemma for more information.

Thanks a lot.It seems liyi's solution,that classical one, is exactly what I'm looking for.
By the way,can any of you help give a proof to that dense result?

You prove the density step by step.

1. given a positive measurable function, one can construct a sequence of measurable simple functions the increases to the given function. (A simple function is a function with a finite range. It's a linear combination of indicator functions of sets. At this point in the construction, we have no control over the shape of the sets.)

2. Prove that if this original positive function is integrable ($L^1$), then the sequence of simple functions constructed in #1 converges to the original function not just pointwise but also in the $L^1$ norm. (This proof would use either Fatou's Lemma or the Dominated Convergence Theorem.)

3. Given a complex-valued integrable function, write it as the linear combination of four functions (positive part of real part, negative part of real part, etc.) Approximate each as in #2. We now have that simple functions are dense in $L^1.$

4. Now we need this result from measure theory: Given a measurable set $E$ of finite measure and $\epsilon>0,$ there exists a set $B$ which is a finite union of bounded intervals such that $|E\Delta B|<\epsilon.$ Here I'm using $|\cdot|$ to mean measure and $E\Delta B=(E\setminus B)\cup(B\setminus E)$ is the symmetric difference of the sets.

5. Given #4, we can approximate a characteristic function of a set arbitrarily well in $L^1$ by a step function. Now it's just an exercise in epsilon-gymnastics to say that we can approximate any simple function arbitrarily well in $L^1$ by a step function.

6. Now that we have the step functions dense in $L^1,$ we can proceed along any of several paths. One thought is to "bevel the edges" of the step functions to produce continuous functions or even $C^1$ functions. Another is this:

6'. Suppose we're working on all of $\mathbb{R}.$ (Cutting down to some interval isn't too hard.) Then translation of functions is continuous in $L^1.$ What I mean by that: Suppose $f\in L^1(\mathbb{R}).$ Define $f_t(x)=f(x-t).$ Then $\lim_{t\to0}\|f_t-f\|_1=0.$ The proof uses #5, the density of step functions.

7'. Suppose $\varphi$ is a $C^{\infty},$ compactly supportd function such that $\int_{\mathbb{R}}\varphi=1.$ Define $\varphi_{\epsilon}(x)=\frac1{\epsilon}\varphi\left(\frac{x}{\epsilon}\right).$ Then $\varphi_{\epsilon}*f\in C^{\infty}$ (that's convolution) and $\lim_{\epsilon\to0^+}\|\varphi_{\epsilon}*f-f\|_1=0.$ The proof uses #6'.

Now we have that $C^{\infty}$ is dense in $L^1.$


How many divisors does $ 2\cdot2\cdot3\cdot3\cdot3$ have?

Every factor of a number must share its prime factors and have no extra ones. For example, the number 18 has prime factorization $ 2 \cdot 3^2$. A factor of it, 9, shares those prime factors with no extras: $ 2^0 \cdot 3^2$.

This implies that for the first prime factor of the number with exponent $ n$, there are $ n\plus{}1$ ways to "choose" it. In our 18 example, the 2 could have had an exponent of 0 or 1, and the 3 could have been 0, 1, or 2. Thus, the number of factors is simply the product of the exponents added to 1.

In this problem, it's already prime factorized, so we have $ (2 \plus{} 1)(3 \plus{} 1) \equal{} 12$.


How do you prepare for State?

I use the MATHCOUNTS Drills on that one website.

The night before the competition, I eat some food with protein.
I also do the past years competition (So I would do the 2005 State Competition for the 2006 State).

The morning of the competition, I eat a banana.
I also look over the formula sheet I downloaded and printed from the Formula forum.

What do you do?

It is a good idea to get a lot of rest several nights before the competition.  You should also eat a good breakfast before the competition so you do not become hungry and have plenty of energy.

Looking over a formula sheet the morning of the competition won't do you much good.  You should find how the formulas are derived and make sure you understand that, because then you are not very likely to forget the formula, but even if you do, you could derive it.

It is a good idea to practice multiple state rounds before the competition, so you have more practice.  I do every sprint round from about 1995 to 2005 (all that I have) and do some twice to make sure I'm ready.  Practicing several different rounds will give you an advantage.  If you have done multiple rounds, you will be likely to be able to pace yourself well so you can finish the test and get the most possible correct.  If you have done only one round as practice, you won't have a good enough idea of the range the of difficulty of the test, so you probably won't do as well.

Another good idea is to do all of the warm ups and work outs provided for the current year.  The same people that write the warm ups and work outs write the actual competitions, so if you do them, you will have a better idea of what types of problems might be on the contest.

Basically, practice will help you do the best.

Our MATHCOUNTS coach is also our Algebra 2 teacher. 2 weeks before the competiton, we take time from math class and turn it into MATHCOUNTS practice time. We do all the chapter, state, and national competitions.
One day we'll do all the sprints of one year.
Another day we'll do all the targets of one year.
We take home another year of competition for homework.
We also practice countdown and team after school.
Our coach has finally been able to project the countdown problems on a board, so we can practice it the real way.

Do you think this will help us at State? (we're in Kentucky and our school has won it six years in a row)(at least I think so)

I try to win.

here are some real tips:
First, eat a good dinner the night before.
Get 9+ hours of sleep.
Eat a good breakfast.

Try to stay as relaxed as possible all that day, it really helps.

SPRINT:
Before you get the sprint round, make numbered boxes 1-30 on a few sheets of paper.
Use the boxes as scratch paper for specific problems. Circle your answer.
Read each question at least twice, and make sure you get the key information. Trust me, this is important. Many a time I have given radius instead of diameter, speed instead of distance...
Don't spend too long on one problem, if you think it will take more than a minute, skip it and come back later.
Either have a watch or timer to keep time, and with about 5 minutes left, start checking your work. You will probably get more right this way than spending the 5 minutes on the two hardest problems.
Make sure the circled answer that you have on scratch paper is what you have for an answer.
Make sure your answers are in the right form/units.
Then go back and do the problems you skipped.

TARGET:
You don't need the numbered paper thing here; you have enough scratch paper on the sheet itself.
Do whatever problem seems easier first.
Then do the harder one.
Go back to the first problem and try it a different way. Do that for the other problem.
Keep confirming your answer in different ways until time is called.

TEAM:
Most of this is based on what your team's strategy is and how strong you are in comparison of he rest of the group.
However, this is something that I think really helps.
Make boxes on paper like in the sprint round, except one sheet of paper per problem.
Write a decent solution so the other person can follow your steps.
When someone or a group is done, give to another person/group to check it over.
When there are 2 minutes left, make sure that the captain has all of the answers written on the sheet. Make sure its the captain's sheet though. At nats last year, we almost got 0 on the team round because the captain wrote on the wrong sheet (stupid girl).

COUNTDOWN:
If you make it this far congrats!
Before you go up, observe who your opponents may be.
Note their speed and whether they are prone to giving a wrong answer too quickly.
When you are up there, don't get nervous.
Read the question on the projector (or whatever else) quickly, noting the key words/numbers. If you think you can do it in your head, do so. If you need paper, use it.
Make sure your answer is in the right form, then buzz in. If it isn't, convert it and buzz in immediately before you have the right answer.
If the other guy buzzes in and gets it wrong, then do the question on paper, and take all of the time you need. Check over it several ways, and buzz in a few seconds before time is called (but don't cut it too close :) )

Just wondering, what AoPS chapters would be useful for preparing for states?

I think mainly
$1$ if you forget how to simplify radicals.
$3$
$4$
$\boxed{5}$
$7$ but its really easy
$9-18$
$25$
$26$
Thats all i think but the rest is useful a bit

[quote="Aznwithabrain"]When you start the sprint round, should you scan the first page and look for the easiest question (the one with the least words)?

I did this at chapter and I got 8th overall.[/quote]

Well, the shortest problems aren't always the easiest. I would say as a general rule, go through every question in order, and if you need to spend more than 10 seconds visualizing what to do, skip it and come back. What I did at chapter was work on the problems backward (I posted this on a thread somewhere, I just don't remember which one) and I got second on written.

I wouldn't do that. If you can't answer all of the low numbered problems, then  :oops:

at any level.
Not to be mean, but if you can't answer the easy questions, then you aren't gonna win.
For the last problems, I do that a lot, especially if one looks easy

Another question

Other than past competitions, 2006 warm-ups and workouts and etc., and the MATHCOUNTS drill page, and the AoPS books.

What do you use for practice?

mathematical circles a little. also a few amc-10 tests, but the format is so different that >.<
I don't know what else there is :?:

how often do you meet per week with your team? and what do you do during team practices?  does anyone have suggestions for team round?

and looking at these schedules, when do you do homework or extracurricular activities?

state in illinois was in a holiday inn. free jolly ranchers on all the tables. we played poker and used ours as chips. they gave us a boxed lunch, it was okay I guess. Afterwards we had an ice cream social and everyone was coming by congratulating me on the nationals thing...i guess everyone's really nice in illinois!

We practiced every monday wednesday friday morning from 7:00 to 7:50. mostly other competitions that our coach has picked up from places. around AMC 10 time she made sure we each understood all the problems, it was really random. sometimes we'd take a session and do countdown (most fun)
for team round we always set up a group answer sheet the same way. I'm not sure how widespread this is but we take a sheet of paper and make a chart with our initials at the top and rows for 1-10. when we get the answer we put it in the right box. when we get 2 or three answers the same in a row we put it down, when we get 4 different answers we make sure we're covered on all the other ones and work on that...every team round is different (duh) so that's kind of inefficient on some of them. also two people work 10-1 and two work 1-10. That way we at least get to every problem...supposedly.

We have practice wednesday mornings from 7:15 to 8:00, but it's canceled every other week, and we don't really do anything...

My team has practice Monday and Friday afternoons and Wednesday mornings from 7:00 to 7:45

we had 5 practices (literally) before states.

They're 90 minutes, but be usually get in only 2 team rounds or countdown,target,team.

So we did 5 team rounds total and 3 countdowns, 3 targets (ALL SCHOOL ROUND)

[quote="camathkid"]To prepare for state, I go over everything I've learned the night and day of State[/quote]

Actually, that's not necessarily a good idea, having to look over all that tires out your brain, and also shows you haven't prepared enough :(.

Yeah, considering that he'll get pwnd at nats.

Does the handbook with warm-ups and work-outs really help?

yeah. There are 300 or so problems per year in those, as compared to 144 in chapter, state, national. (sprint target team)

Warm-ups and Workouts are the building blocks for being good at MATHCOUNTS.


The way I save it is click Print when on the transcript, then hit command-s (control-s for windows) and click save as Webpage, complete.

The only problem with this is the latex doesnt show up.  It downloads as a folder with all the latex pictures and a html file with the transcript, and i'm wondering if there's an easier way.

The way I did it was copy-paste onto Word and save as PDF.  OR you can use something like CutePDF and print it to that.

Just copy it onto a Word document...what's so hard about doing that?

[quote="mathemagician1729"]Use opera, then save it as an MHTML file, it will retain all the images like the original. This should work in IE as well but Opera Pwns IE  :P[/quote]

I believe AIME15 has Safari.

@AIME15: Those virtual printers that print into PDF, etc. are probably the best choice.  With me, they retained formatting.  I'm not sure that CutePDF is compatible with Mac, but you can always Google and see.


This seems to easy, but if $x_{1},x_{2},x_{3},..x_{1997}=1$   it works

But there is more than one solution, right?

[quote="AstroPhys"]$-1$ does not work with the either of the statements.[/quote]
How so?

I just tried it and it worked..

A less-than-clever way to do it is to use Lagrange multipliers on the function $f = x_{1}+x_{2}+\cdots+x_{1997}$ subject to the constraint $(x_{1}^{4}-x_{1}^{3})+(x_{2}^{4}-x_{2}^{3})+\cdots+(x_{1997}^{4}-x_{1997}^{3}) = 0$ which tells us that the extrema must occur when

$1 = 4x_{i}^{3}-3x_{i}^{2}\Rightarrow (x_{i}-1)(4x_{i}^{2}+x_{i}+1) = 0$

for all $i$, i.e. $x_{i}= 1$ is the only real extremum (we can verify that this is a maximum). Thus we have $f \le 1997$ with equality iff $x_{i}= 1$ for all $i$.

If all $x_{i}\in \mathbb{R}^{+}$:

$S_{1}= \sum_{i=1}^{1997}x_{i}$
$S_{2}= \sum_{i=1}^{1997}x_{i}^{2}$
same for $S_{3},S_{4}$

By Cauchy-Schwarz we have $S_{1}S_{3}\ge S_{2}^{2}$.  By AM-GM we have $S_{2}^{2}\ge 1997S_{4}$

Hence $S_{1}S_{3}= S_{1}S_{4}\ge 1997S_{4}\Leftrightarrow S_{1}\ge 1997$ with equality iff all $x_{i}$ are equal (obviously to 1)

if one or more are negative then I dunno...maybe try splitting into negative and positive subsets...w/e.


Shubrashis Nyogi aka tittu has won the gold medal at Int Bio olympiad!!!!!! :)  :D

Congratulations Tittu  :clap2:  :first:  :10:

congratulations  :thumbup:  :clap:  :first:

and deserves muccchhhhhh  mooooooreeeee

:1:

and you are also going to make us do the same(congratulating) next year :P


A quarter weighs the same as two pennies. The quarters in one pound of quarters have a total value of $ \$25$. How many dollars would a pound of pennies be worth?

the stack of quarters is worth $ \$25$, which means that there are $ 25 \times 4$ quarters, or $ 100$ quarters

since a quarter weighs $ 2$ pennies, that means there are $ 100 \times 2$ pennies in that stack, or $ 200$ pennies

$ 200$ pennies is $ \$2$, so our answer is $ \boxed {2}$

i got this right so i'm gonna explain:the stack is $25 and 2 pennies weigh the same as 1 quarter,2500/25=100,and 100 *2=200,200/100=2 so the answer would be 2


algebra
polynomial
function
complex analysis
complex analysis unsolved

Let
\[ D \equal{} \{z\parallel{}z| \le 1\},\]

\[ A_0 \equal{} \{f: D \to \mathbb{C}| f\text{ is analytic in the interior of } D,f\text{ is continous on D}\}\]
Does the polynomials dense in $ A_0$ with the maximum norm?

The Cesaro means of the Fourier series of a continuous function converge uniformly. Given $ f(z)\equal{}\sum a_nz^n$ (on the open disk), the polynomials $ f_n(z)\equal{}\sum_{k\equal{}0}^{n\minus{}1} \frac{n\minus{}k}{n}a_kx^k$ converge uniformly to $ f$ on the circle, and thus also on the whole disk.

Another solution is to note that $ f(rz)$ is close to $ f$ if $ r<1$ is close to 1, by uniform continuity. Since $ f(rz)$ is analytic for $ {|z| < 1/r}$, the power series of $ f(rz)$ at 0 converges uniformly to $ f$ on the closed unit disc. This leads to the desired polynomial.


$\frac{\sin 10^\circ + \sin 20^\circ}{\cos 10^\circ + \cos 20^\circ}$ equals

A. $\tan 10^\circ + \tan 20^\circ$
B. $\tan 30^\circ$
C. $\frac{1}{2} (\tan 10^\circ + \tan 20^\circ)$
D. $\tan 15^\circ$
E. $\frac{1}{4} \tan 60^\circ$

[hide]we know  $\sinh +\ sin b =2sin(a+b)/2*cos(a-b)/2$
and
$\cosh + \ cosb = 2cos(a+b)/2*cos(a-b)/2$
then the ansewr is D. $\tan 15^\circ$[/hide]

[hide]$\frac{\sin 10^\circ + \sin 20^\circ}{\cos 10^\circ + \cos 20^\circ}$

$=\frac{2sin15^\circ * cos(-5)^\circ}{2cos15^\circ * cos(-5)^\circ}$

$=\frac{sin15^\circ}{cos15^\circ}$

$=tan15^\circ$

$\boxed{D}$[/hide]

wow, we have a calculator for the amc, it takes no thinking to put those into the calculator, and would be faster (maybe)

but yes, the other ways are better

I dont think you could use calculators in 80s AMCs


function
real analysis
real analysis unsolved

Is every real derivable and convex function on an open convex subset A of Rⁿ , C¹ ?

ok no one likes this kind of question : is ..? :oops:
just show that instead in R :

let f : R -> R derivable and convex . show that f' is continuous . :roll:

no one ?  :maybe:  it might be classical one ..


function
algebra
domain
inequalities
calculus
derivative
functional equation

How do you figure out if a function one-to-one algebraically instead of graphically (using the horizontal line test).

Thanks!

[quote="b555"][quote="10000th User"]Prove the implication "$ f(a) \equal{} f(a')\Rightarrow a \equal{} a'$", or its contrapositive "$ a\ne a'\Rightarrow f(a)\ne f(a')$"[/quote]

can u explain it in a better way  please[/quote]Ex: Show that $ f(x)\equal{}x^5$ is 1-1.

If you graph it, it looks like a snake, a curve similar to $ f(x)\equal{}x^3$. For any $ a<b$, $ a^5<b^5$ so it passes the horizontal line test.

To prove $ f$ is 1-1, suppose that $ f(a)\equal{}f(a')$. Then, $ a^5\equal{}(a')^5$. Take fifth roots (we can do that because the domain and range are reals) and you'll have $ a\equal{}a'$ so $ f$ is 1-1.

In general, if you can prove that "If $ f(a)\equal{}f(a')$, then $ a\equal{}a'$", then $ f$ is 1-1. I wrote the contrapositive of the statement because sometimes it is easier to prove the contrapositive.

[quote="t0rajir0u"]Really?  In my experience it's very rarely easier to prove the contrapositive.  The first way of writing the implication gets used all the time in functional equation problems to show that a function is one-to-one without knowing what it is.[/quote]Not trolling or anything. One word: "sometimes"

[quote="10000th User"]To prove $ f$ is 1-1, suppose that $ f(a) \equal{} f(a')$. Then, $ a^5 \equal{} (a')^5$. Take fifth roots (we can do that because the domain and range are reals) and you'll have $ a \equal{} a'$ so $ f$ is 1-1.[/quote]

I don't get the significance of this.  If you raise something to the fifth power and then take the fifth root, aren't you just left with what you started?

Exactly.  This means that two real numbers are distinct if and only if their fifth powers are distinct, i.e. taking fifth powers is one-to-one.

As for the other point, it is easier to turn an equality into other equalities than to turn an inequality into other inequalities.  In the former case you can apply arbitrary functions; in the latter case you must apply one-to-one functions.

[quote="t0rajir0u"]Really?  In my experience it's very rarely easier to prove the contrapositive.  The first way of writing the implication gets used all the time in functional equation problems to show that a function is one-to-one without knowing what it is.[/quote]

[quote="t0rajir0u"]As for the other point, it is easier to turn an equality into other equalities than to turn an inequality into other inequalities. In the former case you can apply arbitrary functions; in the latter case you must apply one-to-one functions.[/quote]

I'm really confused by the above two quotes.  Can someone explain what contrapositive and implication are and how they're applied to this?  Also, I don't get the second quote at all, but may that's because I don't get the first one.

Thanks!

Suppose you know that $ a \equal{} b$ and you want to prove that $ c \equal{} d$.  How would you go about doing this?

One general strategy is to find a function $ f$ such that $ f(a) \equal{} c$ and $ f(b) \equal{} d$.  This is the essence of algebraic manipulation.  If you can find any function of this form, you are done.

However, suppose you know that $ a \neq b$ and you want to prove that $ c \neq d$.  You have two options:

1.  Find a one-to-one (that is, injective) function such that $ f(a) \equal{} c, f(b) \equal{} d$.  The reason I dislike this strategy is that injective functions are harder to come by than arbitrary ones.  For example, $ x^2$ isn't injective.

2.  Suppose that $ c \equal{} d$ and use this to prove that $ a \equal{} b$; contradiction!  The first strategy applies.  This is my point.

(Another reason to philosophically dislike using inequalities to prove other inequalities is that inequality isn't [url=http://en.wikipedia.org/wiki/Transitive_relation]transitive[/url]; failure to understand this can occasionally be dangerous.)

As for your other two questions:  an implication is a statement of the form "if P, then Q" (that is, "P implies Q.").  Its contrapositive is the statement "if not Q, then not P."  The two statements are logically equivalent.  For example, "if it rains, I will not go outside tomorrow" is logically equivalent to "if I go outside tomorrow, it didn't rain."

I find the easiest way to prove whether or not a function is one-one is to assume $ x\neq y$, and then show that either $ f(x)\equal{}f(y)$ (not one-one), or $ f(x)\neq f(y)$ (one-one).  This works for more than just functions-it also works in showing uniqueness of certain elements, such as an identity element.

[quote="VP_91"]the method shown by jrav is very right.

but there is a shorter method
here we have 2 prove that the function is 1-1
in other words we have 2 prove that the function is monotic.
so if f(x) is the function
find f'(x)
then 1.if f'(x)>o for all x then it is stritcly increasing.
2.if f'(x)<o for all x then it is strictly decreasing.

so find f'(x).if it satisfies any 1 of the above 2 conditions, it is a 1-1 function.[/quote]You'll have to realize two things:

1) This is High School Basics Forum, and the material you are discussing is beyond the scope of this subforum and belongs to Calculus & Computational Forum, which is not fair to the intended audience.

2) A function can be 1-1 without it being monotonic, e.g. $ f(x)\equal{}\begin{cases}x&\text{if x is rational}\\\minus{}x&\text{if x is irrational}\end{cases}$, though the other direction is true if it is [i]strictly[/i] monotonic. But, finding derivatives is not the way to [u]prove[/u] whether a function, in general, is 1-1 or not.

@10,000thuser:i understood my mistake.i believe my logic is applicable mainly for cotinous functions.


let a,b,c, are base of triagle ABC ,S is the area of  ABC
      prove that:
(2a^2+b^2+2c^2)^2>_128S^2

you can prove that:
$ (2a^2\plus{}2b^2\plus{}2c^2)^2 \geq 198S^2$

Make it stronger.
$ 16S^2\equal{}(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\leq \frac{(a\plus{}b\plus{}c)^4}{27}$ :lol:


Here is a proof of this theorem [url=http://www.its.caltech.edu/~ari/paper-folding.html]Proof[/url].

It have an approach of the final length of the paper folded is equal to the initial length.

I tried to prove this in another way. The area of the paper folded is equal to the area of the initial paper not folded. The area here is a transverse one.

So, using the same assumption that the bend in the paper is a semi circle we need to sum all those semi-circles.

The radius of each semi circle is $ 2^{i \minus{} 1}$ where $ i$ is the number of folds. [url=http://www.pomonahistorical.org/12times.htm](See the image of this site if you have doubts about this)[/url]

So, by this conservation of area.

$ \sum _{i \equal{} 1}^n \frac {1}{2} \pi \left(2^{i \minus{} 1} h\right)^2 \equal{} L h$

Solving for $ L$ in the last equation we have:

$ L \equal{} \frac {h \pi }{6} \left(2^n \minus{} 1\right) \left(2^n \plus{} 1\right)$

But the equation founded by Britney Gallivan is:

$ L \equal{} \frac {h \pi }{6} \left(2^n \minus{} 1\right) \left(2^n \plus{} 4\right)$

So, where this difference from 1 to 4 came from?
The formula above is true, one the length is conserved but the other the area is conserved, in theory the 2 need to be equal and give the same result.

The only thing I can think that make this difference is that the length of the part bended is not actually $ j \pi h$, but less than this or the area is not $ \frac {1}{2} \pi \left(2^{i \minus{} 1} h\right)^2$ but a little bit greater (don't know how).

I would be glad if someone help me to find my mistake and correct the formula to give the correct result.

Note: If we think about volumes the result would be the same, because in this case the volume is Area times $ w$, and the factor $ w$ would cancell out

The difference between your and Britney's formula is the difference between the assumption that the area is preserved and the assumption that the length of the outer edge is preserved when making a fold. I do not know which of these two assumptions is more realistic for paper. You both have no mistakes in mathematics but your physical models of the fold are different.

hm...

Now I know the difference between. The result needed to be equal, because the paper is conserved itself.

The thing is, Britney used the radius of the paper $ j$. So it is the external part of the paper. Physically it would be more realistic to use the middle of the paper insted of the external part, (the mean of the external and internal), so the radius would be $ j \minus{} \frac{1}{2}$. After that I found this:

$ \sum _{i\equal{}1}^n \sum _{j\equal{}1}^{2^{i\minus{}1}} \left(j\minus{}\frac{1}{2}\right) \pi  h\equal{}\sum _{i\equal{}1}^n \frac{\pi  \left(h 2^{i\minus{}1}\right)^2}{2 h}$

It give the equality if we change this. The thing confusing is where you mesure the length matter, but the area not.

The 2 conservative laws are rigth and actually the same law.

Do you agree fedja?

I certainly agree that when you bend metal or plastic, your law makes more sense, so, probably, you are right about paper too. :) I also wonder if the thickness of the paper really changes a bit when folding (at least, I see no reason why it shouldn't happen :?, pehaps I should refresh my very rusty knowledge of "solid body physics" or to leave this discussion to someone who knows it better :P).

Hm... I was also thinking about this.

If we fold the paper and unfold it, the thickness in the fold will be less than in others parts of paper. I think this is just sensible for one fold, If you continuosly fold and fold, it will not make a big differente, and if the thickness of the paper is too great this differente in thickness will also be small.

So one of the consequences of folding and unfolding a paper is that the length of the paper increass (just a little)

I just made a very very approximation of the real case (don't know if this hold)

You have a paper of length $ L_0$ and thickness $ h$. After you fold it you have basically a retacgle of thickness $ 2 h$ and length $ L_1$ and a semi-circle of radius $ h$. If we use the law of conservation of area (mass) we have.

$ L_0 h\equal{} L_1 (2h) \plus{} \frac{\pi }{2} h^2$

So, If we unfold the length of the paper would be $ 2 L_1 \plus{} \delta L$

A good approximation of this $ \delta L$ would be "Cut the paper when you bend it and the put it thogeher". So we would joing 2 quartes of circle side by side.

The final lenth would be so: $ 2 L_1 \plus{} 2 h \equal{} L_0\plus{}\frac{1}{2} h (4\minus{}\pi )$

And this approximation would've be only valid if $ h$ is small.

If the paper has thickness of $ 1mm$ the difference of length would've be only $ 0.43 mm$

I don't think this is a precise result, but we could've play with it. Find the average thickness of the fold, how may chemical bonds were broken when you fold (this explain why the paper is too sensive in the fold, chemical bonds are broken when you fold a paper)

This is not a study in paper, just some interesting models to work out.

I don't know too much about physics of bending materials, so I can't say that those approximations are good or not.


calculus
integration
calculus computations

What is the sum from 1 to infinity of $ \frac{1}{n \sqrt{n^2\plus{}2}}$? Please prove it.

Numerically, it's about $ 1.1588006.$ Is there some reason why it should be a more recognizable number than that?

1.  For what class?

2.  Are you sure it wasn't an integral?

3.  Are you sure it wasn't an estimate?

Well, that's what she told me 4 calc bc, no, it's definitely a sum, but the problem was probably just to see whether it converges or not which is trivial, although she asked me 4 the exact value.

At that level, you're definitely not asking for the exact value on something like this. Most of the time, that exact value can't be written in closed form.


induction
number theory proposed
number theory

Let $ a_1\equal{}a_2\equal{}1$ and \[ a_{n\plus{}2}\equal{}\frac{n(n\plus{}1)a_{n\plus{}1}\plus{}n^2a_n\plus{}5}{n\plus{}2}\minus{}2\]for each $ n\in\mathbb N$. Find all $ n$ such that $ a_n\in\mathbb N$.

First I get $ na_n\minus{}(n\minus{}1)^2a_{n\minus{}1}\equal{}\left\{
\begin{array}{ll}
\minus{}n\plus{}2\ (n: odd) &\quad  \\
\minus{}n\plus{}3\ (n: even) &\quad
\end{array}
\right.$

Prove by induction that \[ a_n\equal{}\frac{(n\minus{}1)!\plus{}1}n\]
The rest is obvious.

[quote="anonymouslonely"]$ a_{3}=\frac{2}{3} $. your formula does not work for $ n=1,3 $. your formula works for the sequence $ a_{1}=2,a_{2}=1 $...so it was a typo in the initial problem?[/quote]
No in the initial problem $ a_{1} $ is 2 and $ a_{2} $ is 1.Right!


inequalities
calculus
trigonometry
logarithms
calculus computations

Prove that :$ (sin x)^{cos x}<(cos x)^{sinx}\  for \ 0<x<\frac{\pi}{4}$

hello
well one of much time consuming method:
[hide="hint"]
consider a func $ f(x) \equal{} (cos(x))^{sin(x)} \minus{} (sin(x))^{cos(x)}$
$ g(x) \equal{} f'(x)$which is coming$ \minus{}ve$ for the interval (0,\frac{\pi}{4})
thus$ f(\frac{\pi}{4})<f(x)$  where $ 0<x<\frac{\pi}{4}$
or$ (cos(x))^{sin(x)}>(sin(x))^{cos(x)}$
[/hide]
thank u

[quote="pankajsinha"]Prove that :$ (sin x)^{cos x} < (cos x)^{sinx}\ for \ 0 < x < \frac {\pi}{4}$[/quote]
You could do it taking the logarithm.
Let $ f(x) \equal{} (\sin x)( \log \cos x)$ and $ g(x) \equal{} (\cos x)( \log \sin x)$
Now divide both by $ \log \cos x$.
That way, you have logarithms with a base and you can compare the values accordingly for the interval $ \left[0, \frac {\pi}{4}\right]$.

For $ 0<x<\frac{\pi}{4}$, we have $ \sin x < \cos x<1$

Hence $ (\sin x)^{\cos x} < (\sin x)^{\sin x} < (\cos x)^{\sin x}$


Ring Theory
number theory
relatively prime
superior algebra
superior algebra solved

suppose R is a commutative ring and P is a prime ideal. then i know that P contains every nil potent element. is the converse true? that is, if an element belongs to every prime ideal then its nilpotent?

ok another qn, this is quite unrelated to the previous one. suppose a and b are 2 relatively prime positive integers. show that every integer greater than ab-a-b can b written as a positive linear combination of a and b. by this i mean ax+by=N,N>ab-a-b and x,y>0

For the first: when I remember correctly: yes (possibly I forgot some other condition that we need)

For the second (but please post only one problem per thread in future):
Some people call it (I don't know why) the Chicken McNugget theorem, so searching for it will help.

I remember posting a proof for the affirmative answer to your first question some time ago, but it's not worth the effort of looking for it :).

Suppose $x$ is not nilpotent. Our aim is to show that there is a prime ideal which does not contain $x$. Let $\mathcal I$ be the collection of ideals which do not contain any of the elements $x,x^2,\ldots$. Since $x$ is not niplotent, the zero ideal is an element of $\mathcal I$, so, in particular, $\mathcal I$ is non-empty. It's easy to see that this collection satisfies the hypothesis necessary for applying Zorn's Lemma: every chain (with respect to inclusion) in $\mathcal I$ has an upper bound, namely the union of all the ideals in the chain. By Zorn's Lemma, $\mathcal I$ has a maximal element $P$. We're done if we can show that $P$ is prime.

Take $a,b\not\in P$. By the maximality of $P$, we have $x^m\in P+(a),\ x^n\in P+(b)$ for some positive integers $a,b$. We then clearly have $x^{m+n}\in P+(ab)\Rightarrow ab\not\in P$, so whenever $a,b\not\in P$, we get $ab\not\in P$, which means precisely that $P$ is prime.


inequalities calculus derivative algebra function domain

show that for any x natural>=1 we have:

x=[1/a^2]; where a=arctan(1/sqrt(x)).

and [t]=t-{x}.

spoiler:

[hide]

consider the right triangle below with side lengths 1, :sqrt: x, and :sqrt: x+1.  It is clear that arctan(1/ :sqrt: x)= the angle D.  So what we want to prove is that x = [1/D^2] or equivalently x<=1/D^2<x+1.  The left hand inequality says x<=1/D^2 or D<=1/ :sqrt: x or equivalently D<=tanD, similarly with the rhs inequality, we see D>1/ :sqrt: x+1 or D>sinD, so we need to prove the equivalent identity, for all positive angles D, 0<D<90, sinD<D<=tanD.  For the lhs, the derivative of D-sinD is 1-cosD, which is positve for 0<D<90, so after they intersect at D=0, D>sinD for all D>0.  for the rhs, consider tanD-D with derivative sec^2(D)-1, now this derivative is positve for 0<D<180 since 0<cos^2(D)<1 in that domain, and D=tanD at 0, so tanD > D for 0<D<90, proving our result.[/hide]


[hide]I really am kind of lost, and I would like to know, how do you get out of math mode?

Edit: this is what happens.  $ Let n_k\,be\,the \,value \,of \,n on the kth turn, let l_k be the value of l on the kth turn, and let r_k be the value of r on the kth turn.$

[code]Let n_k\,be\,the \,value \,of \,n on the kth turn,  let l_k be the value of l on the kth turn, and let r_k be the value of r on the kth turn.[/code]

Is the only way to space it to add "\,"?[/hide]

EDIT 2:  Never mind, I got it.  Can someone please close this thread.   :oops:

:huh: Don't put text in dollar signs and you won't have this problem.   :|   There are also commands like \textrm{}, but there's no reason at all to use them on the forum.


Tom has a red marble, a green marble, a blue marble, and three identical yellow marbles. How many different groups of two marbles can Tom take?

the solution says:  There two cases here: either Tom takes two yellow marbles (1 result), or he takes two marbles of different colors ( results). The total number of distinct pairs of marbles Tom can take is 1+6=7.

but i can list
rg,by
rg,yy
rb,gy
rb,yy
ry,yy
ry,gy
ry,by
gb,yy
gy,yy
by,yy
so thats eleven different ones already.

Well, here's what I have.  I haven't checked the official solution, but...

[i]Case 1: Both marbles are yellow.[/i]

One possible way.

[i]Case 2: Both marbles are not yellow.[/i]

This will include only $ 1$ yellow marble in the set, so we have $ \binom{4}{2}$.

Thus, we have $ 1\plus{}6$ or $ \boxed{7}$.

oh i see, i just misinterpreted the question

I don't think that problem should be a level 21 problem(with 111 xp). It seemed pretty easy...

[font=TIMES NEW ROMAN]@Above I agree: [hide=Easier than it seems]If there were only one yellow marble the answer would be $\binom{4}{2}=6.$ But since we actually have 3 yellow marbles, we have to add on the $\text{Yellow, Yellow}$ combination, so our answer is $\boxed{6+1=7}.$ $$\definecolor{color}{rgb}{0.5, 0.3, 0.2} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[quote=pifinity][font=TIMES NEW ROMAN]@Above I agree: [hide=Easier than it seems]If there were only one yellow marble the answer would be $\binom{4}{2}=7.$ But we have to add on the $\text{Yellow, Yellow}$ combination, so our answer is $\boxed{6+1=7}.$ $$\definecolor{color}{rgb}{0.5, 0.3, 0.2} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$[/quote]

4c2 isnt 7......

We add the Y,Y combination: [quote]But we have to add on the  $\text{Yellow, Yellow}$ combination, so our answer is $\boxed{6+1=7}.$[/quote]

4c2 should still be $6$...….. you can add the Y,Y after but u said IF there is only one yellow

I mean that we take the 6 from 4 choose 2 and add 1: [quote]$\boxed{6+1=7}.$[/quote]

Yes but you dont add $1$ if there is only one yellow ball as u said...…

I edited it to make it clearer though. [quote][hide=Easier than it seems]If there were only one yellow marble the answer would be $\binom{4}{2}=7.$ But since we actually have 3 yellow marbles, we have to add on the $\text{Yellow, Yellow}$ combination, so our answer is $\boxed{6+1=7}.$ $$\definecolor{color}{rgb}{0.5, 0.3, 0.2} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$[/quote]

Ok, im confused now..... just if you are adding one Y,Y to the previous case of only 1 yellow then it sould be 4c2+1=7

If there is only 1 yellow ball, then there are 4c2= 6 possibilities and then u can say, Since there are 3 we need to add the case Y,Y to get 6+1=7

That's what I edited it to say. Read my post again.

You cant say there are 4c2=7 possibilities if there is only 1 yellow ball..... You are still saying 4c2 is 7.....

Oh, that was a typo. Here is a revised version: [quote=pifinity][font=TIMES NEW ROMAN][hide=Easier than it seems]If there were only one yellow marble the answer would be $\binom{4}{2}=6.$ [b]But since we actually have 3 yellow marbles,[/b] we have to add on the $\text{Yellow, Yellow}$ combination, so our answer is $\boxed{6+1=7}.$ $$\definecolor{color}{rgb}{0.5, 0.3, 0.2} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$[/quote]


Hi
cananybody tell me from where i can download article.sty files and if i want to type a paper in latex. And the style is atricle.sty then what other files along with article.sty would i be needed to download?

Tanks in advance

jigna

article.sty is obsolete and part of an old version of LaTeX.
You can't just download a few files, you need a LaTeX distribution. See [[LaTeX:Downloads]] for details assuming you are using Windows, though you don't say which operating system you are using.


function
algebra
domain
integration
real analysis
real analysis unsolved

$U \in C^2(\mathbb{R},\mathbb{R})$ satisfies: $U(x) \geq 0$, $lim_{|x|\to \infty} U(x) = +\infty$.

Consider the differential equation:
$(E)$:$X'(t)=-U'(X(t))$, $X(0)=y \in \mathbb{R}$ for $t \geq 0$
1/ show there is a unique solution defined over $\mathbb{R}^+$
2/ suppose that the set where $U'$ vanishes is discrete. Show that the solution $X(t)$ converges towards a limit $L \in \mathbb{R}$ and that $U'(L)=0$
3/ study the same pb with $U \in C^2(\mathbb{R}^d,\mathbb{R})$ ..

Are we supposed to see what happens to (2) for $u: \mathbb R^d\to\mathbb R$ as well? Because it looks a bit weird, on first sight :).

(1) works for all $d$ in pretty much the same way. First of all, $u'$ is a $C^1$ funtion, so the equation has the property of [i]local[/i] existence and uniqueness of solutions. This means that the maximal solutions are unique. All we need to do now is prove that maximal solutions are actually global solution, i.e. that every solution can be extended to the entire $\mathbb R_+$.

Let $x: [0,k),\ k<\infty$ be a solution with $x(0)=y$. We have $\frac{\mbox d}{\mbox dt}u(x(t))=-(x'(t),x'(t))\le 0$, so $u\circ x$ is decreasing. Since $u$ goes to infinity as the norm of its argument goes to infinity, this means that the image of $x$ stays within a ball conained in the domain of $u$, and this implies that $x$ can be extended strictly to the right (this is a standard result).

For (2) (we're back to the case $d=1$) we could do this:

Suppose $x(t)=x(t'),\ t<t'$. Then $u(x(t))=u(x(t'))$, so $u\circ x$ is constant on $I=(t,t')$. The assumption on the set of zeros of $u'$ now implies that $x$ must be constant on $I$, and from here we conclude that $x$ is monotonic. Its limit $L$ has to be finite, since otherwise $u\circ x$, which is decreasing, would go to $\infty$, which is absurd.

Given any $\varepsilon>0$, and any $n\in\mathbb N$, we can find $t_n>n$ such that $|x'(t_n)|<\varepsilon$: simply take $n<t<t',\ t'-t>1$, with $|x(s)-L|<\frac\varepsilon 2,\ \forall s\ge t$, and choose $t_n\in(t,t')$ such that $(t'-t)x'(t_n)=x(t')-x(t)$. We have $|u'(x(t_n))|=|x'(t_n)|<\varepsilon$, and the conclusion that $u'(L)=0$ follows.

nice  :)
and you're right, it's a bit more difficult (i think) for the case $d>1$. But it's still true.


AMC
AIME
geometry
circumcircle
3D geometry
algebra
polynomial

[For reference: the contest page: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126603[/url]]

[size=167]Mock AIME 5 for 2007[/size]

Written by Alex Anderson
with contributions from Jacob Steinhardt

[hide="Answers:"]1.	299
2.	322
3.	972
4.	997
5.	181
6.	204
7.	670
8.	601
9.	275
10.	502
11.	936
12.	003
13.	252
14.	647
15.	219[/hide]

Results:

1st place:  :winner_first:
MysticTerminator, 14/15
2,3,4,5,6,7,8,9,10,11,12,13,14,15

2nd place:  :winner_second:
paladin8, 12/15
1,2,3,4,5,6,7,8,9,10,12,14

3rd place [tie]:  :winner_third:
beta 10/15
2,3,4,5,6,7,8,11,12,13
nat mc 10/15
1,2,3,6,7,9,10,11,12,13

Comments:

Thanks a lot for jsteinhardt for helping me check the test. Also he submitted 4th and 14th problems and thought of the idea for problem 5.

Problem 12 had a small error...I decided to let the test go on, rather than trying to change anything. It was pretty clear to find the largest lower bound. Regardless, if you even looked at that problem, you should have just asked with a private message; I was quite clear about this.

I suspect that some people used a calculator for problem 5. I probably should have said that there were 7 prime factors, find the second largest one.

Many people were confused about problem #3. A geometric sequence of integers with a finite number of terms does not have to have an integer ratio.

Problem # and percentage of people who answered it correctly:
1.	75
2.	87.5
3.	71.875
4.	56.25
5.	50
6.	68.75
7.	62.5
8.	18.75
9.	46.875
10.	21.875
11.	21.875
12.	12.5
13.	12.5
14.	6.25
15.	3.125

overall average:	6.15625 of 15 correct

Difficulty:

The first two problems were significantly easier than the actual AIME. I would say that the next 7 problems [3-9] problems or so were on par with an official AIME. #10 was particularly difficult to find the actual solution. Many people found the pattern and solved that way. Perhaps I could have put this problem later. I feel that many people saw this problem, got discouraged, and stopped working on problems. I feel that more people could have solved #12. Anyway, the later geometry was pretty difficult. Also #14 was tough and interesting problem; one of my favorites on the test.

I thought that the general content was similar. I did not have any "annoying casework problems" that are prevalent on the AIME contests. I also felt that I could have included more counting problems. #6 was pretty interesting, but #8 was a standard application of recursion for expected value.

Solutions:

I don't plan to type out official solutions. Every problem has been answered by somebody. Furthermore, I have sent people solutions to some of the harder problems, like #10,14,15. Perhaps I could get some help writing out solutions...I can compile them into a one thread...maybe wiki entries too...

Finally,
I appreciate the positive feedback. I worked hard on this contest, and I think it turned out well.

--Alex Anderson

Cool. I really liked #14 as well :).

BTW, nice job to Arnav for getting them all except #1 :P.

Yes, that is rather ironic, isn't it? Problem 1 [b]was[/b] quite an easy problem.
I did 1, 2, 3, and then got too lazy and didn't finish it. I did get all three of them right, though.  :lol:

I'm willing to bet he submitted 42 or the like for #1.

Since some people were wondering...

[hide="15"]
Denote $DE=x$. Consider just $P$, $A$, $B$, $D$, $E$, and $C$. You can either use similar triangles, or some tricky polar stuff [somebody will have to fill in the pole/polar stuff] to show that $P$, $D$, $E$, and $C$ are in a harmonic progression. In particular this means [without directed lengths] that

\[\frac{PD}{DE}*\frac{EC}{CP}=1\]

Hence $CE=2x$, and using power of a point on $E$,

\[(x)(2x)=12\implies x=\sqrt{6}\]

It follows directly that $PD=3\sqrt{6}$, $PA^{2}=(3\sqrt{6})(6\sqrt{6})\implies PA=PB=6\sqrt{3}$.

Now we have to deal with $F$. Let $I$ be an inversion centered at $P$, with the circle passing through $A$ and $B$. This maps $w_{1}$ to itself [since the circles are orthogonal, or equivalently using power of a point]. Hence, $I(A)=A$ and $I(D)=C$. Furthermore, we know that the circumcircle of $I(ADFP)$ is a line because any circle through the center of inversion becomes a line, this line being $AC$.  We need to find $I(F)$. We know that this must lie on $PF$ and $AC$. Hence, $I(F)=G$. Then by definition of inversion,

\[PA^{2}=(PF)(PG)\implies 108=PF*(6\sqrt{3}+10)\implies PF=81\sqrt{3}-135\]

It follows that the answer is $81+3+135=\boxed{219}$.

Comment: Obviously this problem was quite difficult. I wanted to go all out, so to speak, on the geometry at the end.[/hide]

Woah totally forgot to even look at the questions.

5 looks way too easy. I don't want to work it out, but can't we be all "WOAH DIFFERENCE OF SQUARES!!!111lololol"?

[quote="probability1.01"]You can do #15 using only basic angle-chasing/power-of-a-point, but it's still fairly involved.[/quote]

yea...i realize...but i prefer the concise solution...the harmonic progression directly follows from triangle similarity...the inversion thing is equivalent to showing that $\triangle APF\sim\triangle GPA$...

[quote="Ravi B"]One way to factor $40001$ is to write it as $40401-400$.[/quote]

[hide]in general, that would be 4a^4+1=(2a^2-2a+1)(2a^2+2a+1)

this is the sophie-germain identity[/hide]

[quote="Altheman"][quote="PenguinIntegral"]Woah totally forgot to even look at the answers.

5 looks way too easy. I don't want to work it out, but can't we be all "WOAH DIFFERENCE OF SQUARES!!!111lololol"?[/quote]

Unless you can factor 40001 magically, you need some other techniques...I think that these problems are reasonable, and could be on an aime, but since there is nobody preventing the use of a calculator during a mock test, I don't think I will use this type of problem again.[/quote]
Why would I need magic? It only took me a few minutes to find 13, 17, and checking everything under the square root of 181 finds that it is prime. What was your intended solution, and could you give a hint for 12?

[quote="paladin8"][quote="PenguinIntegral"][quote="paladin8"][hide="hint for 12"] $x^{3}-x(y^{2}+z^{2}) = y^{3}-y(z^{2}+x^{2})$ tells you a lot more than it looks like it tells you. [/hide][/quote]
Set the difference of cubes equal to the other terms?[/quote]

Yes... and what do you get?[/quote]
Something like $(x-y)(x^{2}+xy+y^{2})=x(y^{2}+z^{2}-y(z^{2}+x^{2})$

If we expand that, $(x-y)(x^{2}+xy+y^{2})=x(y^{2}+z^{2}-y(z^{2}+x^{2})=xy^{2}+xz^{2}-yz^{2}-yx^{2}=xz^{2}-yz^{2}=z^{2}(x-y)$
$(x^{2}+xy+y^{2})=z^{2}$?

[hide="5."]1,599,999,999 can be written as $16*a^{8}-1$ if we let $a=10$, then factoring:

\[16*a^{8}-1=(4a^{4}+1)(4a^{4}-1) =(2a^{2}-2a+1)(2a^{2}+2a+1)(2a^{2}+1)(2a^{2}-1)\]

Hence, we get the factorization $(181)(221)(201)(199)$. It is easy to check these factors for divisibility by $2,3,5,7,11,13$ [we need not go higher because $15^{2}>181,221,201,199$]. We obtain the prime factorization as $3*13*17*67*181*199$. Hence the answer is $\boxed{181}$[/hide]

[hide="12."]

Subtracting equations gives:

\[(x^{3}-y^{3})-z^{2}(x-y)-(xy)(x-y)=(x-y)(x^{2}-2xy+y^{2}-z^{2})\]

The second condition gives us that $x\ne y$, so we have that $(x+y)^{2}=z^{2}$, likewise, we have $(y+z)^{2}=x^{2}$, and $(z+x)^{2}=y^{2}$. Adding these and dividing by 2 gives:

\[x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=0\]

Subtracting $(x+y)^{2}=z^{2}$ gives:

\[2yz+2zx+2z^{2}=0\implies (z)(x+y+z)=0\]

We throw out $z=0$, and have $x+y+z=0$. Plugging this into the last statement gives $x=1$. Then if $y=t$, then $z=-1-t$. We can finish by completing the square:

\[(1)^{2}+(t)^{2}+(-1-t)^{2}=2t^{2}+2t+2=2\left(t+\frac{1}{2}\right)^{2}+3\]

The minimum does not occur...but it can be approached...this was my mistake on the test. Anyway, the answer is $\boxed{3}$.
[/hide]

[quote="K81o7"]Oh hah I forgot to submit this...Well lucky for an AIME...no careless mistakes, i got 15!

I liked it: pretty much all clean problems, no major computation.  But one question I had: I got that there's only onep ossible real part for k in #10...[/quote]

o rly?

in retrospect, #10 was not so good for an aime...the point was to prove that the sum was invariant...i could not think of how to phrase the problem so that you would need to find the invariant...

#8 is pretty simple recursion, if you do it right, it becomes 1.1^n...

[quote="Altheman"]
#8 is pretty simple recursion1.1^n...[/quote]
O rly?

Can someone show me how to do this one:

9. If $p(x)=x^{3}-5x^{2}+4x-1$ has roots $a$, $b$, and $c$, then find $\left(a^{2}-\frac{1}{a^{2}}\right)\left(b^{2}-\frac{1}{b^{2}}\right)\left(c^{2}-\frac{1}{c^{2}}\right)$.

[quote="pkothari13"][quote="Altheman"][hide="9"]Note that $p(x)=(x-a)(x-b)(x-c)$ by the factor theorem. Then we factor to employ that fact. [The product is taken over a,b, and c].

\begin{eqnarray*}\left(a^{2}-\frac{1}{a^{2}}\right)\left(b^{2}-\frac{1}{b^{2}}\right)\left(c^{2}-\frac{1}{c^{2}}\right)&=&\frac{1}{(abc)^{2}}(a^{4}-1)(b^{4}-1)(c^{4}-1)\\ &=&\frac{1}{(abc)^{2}}\prod (1-a)(-1-a)(i-a)(-i-a)\\ &=&\frac{1}{(abc)}*p(1)p(-1)p(i)p(-1)\\ &=&\frac{1}{(-1)^{2}}*(-1)(-11)(4+3i)(4-3i)\\ &=&275\\ \end{eqnarray*}[/hide][/quote]

Ah! That's a great solution! Thanks man  :)[/quote]
Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?

abc is 1(from vietas), not -1, but it doesnt make a difference when you are squaring it..

I just kept on playing around with a+b+c, ab+ac+bc, and abc..until you get the answer..

Wow I would have had 11, but I missed #5 because i din't see that $3|201$.  I feel dumb, but not as dumb as arnav missing number one lol.

[quote="Altheman"][quote="PenguinIntegral"]Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?[/quote]

but that is ugly...and this technique is useful in other contexts...like this year's hmmt algebra #10, for example[/quote]
But the elegant method takes thought, and it's a nine. I really don't see the AMC putting that kind of textbook question that late in the test.

@nat mac

Arnav does not "miss" math problems. He just typed in the wrong answer.

[quote="Ravi B"]Sorry, PenguinIntegral, could you explain your (the AoPS 2) method more fully?

I solved the problem the Altheman way.[/quote]
I think he means newton's sums

Heh, I had a pretty ugly solution to 9

[hide]
Matching coefficients with the roots, we have
\begin{eqnarray*}a+b+c &=& 5\\ ab+bc+ac &=&4\\ abc &=& 1 \end{eqnarray*}

Expand out the product to eventually get
\[(abc)^{2}-\left(\frac{1}{abc}\right)^{2}+\frac{a^{4}+b^{4}+c^{4}}{(abc)^{2}}-\frac{(ab)^{4}+(bc)^{4}+(ac)^{4}}{(abc)^{2}}. \]

To find $a^{4}+b^{4}+c^{4}$, apply Newton's sums on the original polynomial. For the last term, the polynomial with roots $ab, bc, ca$ is
\begin{eqnarray*}(x-ab)(x-bc)(x-ac) &=& x^{3}-(ab+bc+ac)x^{2}+(a^{2}bc+ab^{2}c+abc^{2})x-(abc)^{2}\\ &=& x^{3}-4x^{2}+5x-1 \end{eqnarray*}
So to find $(ab)^{4}+(bc)^{4}+(ac)^{4}$, apply Newton's sum on the above polynomial.
[/hide]

Oh, and I thought question 1 was kinda hard, relative to its position in the test (I got it wrong,  :blush: )

I tried Newton's Sums but they don't like me :( . (I had to look them up in AoPS 2 and I still got it wrong :( )

[quote="Ravi B"]Sorry, PenguinIntegral, could you explain your (the AoPS 2) method more fully?

I solved the problem the Altheman way.[/quote]
Look somewhere in the polynomial section where it shows how to construct polynomials that have, for example, roots that are the reciprocals of the original polynomial.

Hm, I remember having a nice way of doing this, but I don't remember what it was, and I threw away my scratch work... :wallbash:

[quote="Ravi B"]Thanks.  I looked at AoPS 2, and see how to construct a polynomial whose roots are reciprocals of the originals, or a multiple of the originals.  But constructing a polynomial with roots $a^{2}-\frac{1}{a^{2}}$ (and same for $b$ and $c$) seems harder.  At least I don't see a nice method.  Do you have one?[/quote]
I'm sure there is a way to do it. I haven't really thought about it much...

[quote="PenguinIntegral"][quote="Altheman"][quote="PenguinIntegral"]Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?[/quote]

but that is ugly...and this technique is useful in other contexts...like this year's hmmt algebra #10, for example[/quote]
But the elegant method takes thought, and it's a nine. I really don't see the AMC putting that kind of textbook question that late in the test.[/quote]

Constructing a polynomial with such roots is complicated, cumbersome, and prone to error. Even for simple problems, it's usually easier to use an alternate approach; a problem with roots as relatively complicated as it was in Altheman's AIME is even more so unwieldy to solve with construction methods.

On the other hand, the elegant solution is quick, extremely nice, and FAR more applicable; even simple problems are slaughtered by the factor theorem approach.

[quote="yhh3002"]what are the mock AIME for?[/quote]
practice and fun


calculus
integration
trigonometry
real analysis
real analysis solved

I think that the integral $\displaystyle \int \sqrt{ \tan x} dx$ was discussed last week here. So I was thinking about computing:

$\displaystyle \int \sqrt[3]{ \tan x} dx$ and $\displaystyle \int \sqrt[3]{
ctan x} dx$ for $\displaystyle x  \in \left( 0, \frac{ \pi}{2} \right).$

Then try:

$\displaystyle \int \sqrt[n]{ \tan x} dx$ and $\displaystyle \int \sqrt[n]{ ctan x} dx$ for $\displaystyle x  \in \left( 0, \frac{ \pi}{2} \right),$ $n \geq 2,$ $n \in N.$

cheers! :D :D

Let $\sqrt[3]{\tan x} = u$, $x = \arctan u^3$, thus
$dx = \frac{3 u^2}{1 + u^6}$

$\int \sqrt[3]{\tan x} dx = \int \frac{3 u^3}{1+u^6} du$

We have reduced it to a rational form.

$\frac{u^3}{1+u^6} = \frac{A}{u^2+1} + \frac{B}{u^2-\sqrt3 u+1} + \frac{C}{u^2+\sqrt3 u+1}$

Find $A, B, C$ in this way... and we get the primitive.

For $A=\int \sqrt[n]{\tan x} dx$, $x=\arctan u^n$
$A=n\int\frac{u^n}{1+u^{2n}}du$

Partial fraction in $\mathbb C[X]$ and coming back on $\mathbb R[X]$

gives

if $m,n\geq 2$ are positive integer with $m<2n$

$\int \frac{u^{m-1}}{1+u^{2n}}du = \frac{-1}{2n}\sum_{k=1}^{n}
\cos(a_k)\ln(1-2x\cos(b_k)+x^2)+\frac{1}{n}\sum_{k=1}^{n}\sin(a_k)\arctan(\frac{x-\cos(b_k)}{sin(b_k)})$

With $a_k = \frac{m\pi(2k-1)}{2n}$, $b_k=\frac{(2k-1)\pi}{2n}$


$ a$, $ b$ are positive non-integers numbers such a+b is an integer:
a)is it possible to a^20 + b^20 be an integer?
b)and a^21 + b^21 ?

If $ a\plus{}b$ and $ ab$ integers, then $ a^n\plus{}b^n$ integer too for any natural n. For example $ a\equal{}3\minus{}3\sqrt 2, b\equal{}3\plus{}2\sqrt 2$.


Find all integers n such that
2^(n-1)/n!


I'm looking over functions and graphs.  can anyone explain to me why you move the graph $y=x^{2}$, two to the right to graph this:$f(x)=(x-2)^{2}$why is it that you move right when its$x-\textbf{3}$?  and why does the graph steches horizontally by a factor of 2 when the graph is $f(x)=(2x)^{2}$ ?  seems like everything for x is backwards...

Thank you  :P

[color=red][size=59]Preview your posts before posting  :wink:
b-flat[/color][/size]

Don't EVER think graph. Think COORDINATE SYSTEM.

If you put $x-2$ instead of $x$, you effectively move the x-axis for $-2$, that is for $2$ to the left, which appears as the graph moved to the right. Similarly, if you put $x+3$, you move x-axis for $+3$, that is $3$ to the right, which apparently moves the graph $3$ to the left.

If you put $2x$ instead of $x$, you multiply every number on x-axis by $2$, which shrinks it twice, and that looks like the graph has been stretched twice.

And so on.

Remember: Whatever happens to the coordinates, it's applicable to the COORDINATES (not the graph), therefore from the graph point of view, exactly the "opposite" happens.

but that doesnt happen for the y values.  $y=2(x)^{2}$  The numbers on the y axis aren't multiplied by 2.  and why are the numbers on the x axis doubled for $y=(2x)^{2}$ ?

thanks for your help

It DOES happen: Just as you put $2x$ instead of $x$, you put $2y$ instead of $y$, hence the y-axis is shrank twice, meaning the graph is stretched twice along the y-axis.

Therefore, if you have the graph of $y=f(x)$, then the graph of $y=Cf(ax+b)+D$ is obtained by writing $ax+b$ instead of every $x$ on x-axis and writing $Cy+D$ instead of every $y$ on y-axis. That results in the following: (1) x-axis is shrank $a$ times ("shrinking" is to be understood relatively, i.e. it IS shrinking if $a>1$ and becomes stretching if $a<1$) and moved right by $b$ (again relatively: "right" by a negative number is "left"), hence the graph is stretched $a$ times horizontally and moved left by $b$; (2) y-axis is shrank $C$ times (relatively) and moved up by $D$ (where "up" by a negative number is "down"), hence the graph is stretched $C$ times vertically and moved down by $D$


algebra
polynomial
calculus
derivative
geometry
3D geometry
calculus computations

Okay, so I had to estimate the sq. root of 7.9 with a Taylor polynomial of degree three.

I based the polynomial at 8.

First I calculated the 1st-3rd derivatives of x^(1/2):
g'(x) = 1/2(x)^(-1/2)
g''(x) = -1/4(x)^(-3/2)
g'''(x) = 3/8(x)^(-5/2)
Then there's g(x), which is just the sq. root of 8.

When plugging in 8 for x, you get for the Taylor polynomial of degree three:

T3(x) = sqrt[8] + 1/2(8)^-1/2 (x-8) + 1/2 * -1/4(8)^(-3/2) (x-8)^2 + 1/6 * 3/8(8)^(-5/2) (x-8)^3

Plugging in 7.9 for x in the polynomial above, you get 2.810693867, which is not far off from what you get when you enter the square root of 7.9 into a calculcator. When you enter it into the calculator, you get 2.810693865.

And now for the question that is confusing me:
Apply Taylor's theorem to find an upper bound on the error in this approximation.
.
If someone could explain how this is done, I would REALLY appreciate it. Thanks in advance!

(And one more thing:  I think I might have been wrong in using 8 for approximating the square root of 7.9 since 8 isn't a perfect square.  Would it be more appropriate to use 9 instead?)

[quote="SharkBites77"]I think I might have been wrong in using 8 for approximating the square root of 7.9 since 8 isn't a perfect square.  Would it be more appropriate to use 9 instead?)[/quote]

Yes.  Notice that in your polynomial you had to evaluate the square root of $8$ implicitly several times, which is not an integer.

The choice of $7.9$ is unusual for a square root, though.  Are you sure the problem didn't say [i]cube[/i] root?

Hmm.

Anyway, once you've found the correct taylor series centered around $9$, consult

http://en.wikipedia.org/wiki/Taylor's_theorem

Taylor's theorem essentially says that an upper bound for the error when you take the series to $n$ terms is given by the $(n+1)^{th}$ term, except that instead of

$\frac{ f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}$

The input for $f^{(n+1)}$ is some $\zeta$ between $a, x$ that maximizes $f^{(n+1)}$.

Thanks!  I'm a bit confused to how I would find this maximum value, though.


[color=blue][size=150]a, b and c belongs to the interval [1/3,1].
prouve that  a/(a+b)+b/(b+c)+c/(c+a)>=7/5 [/size][/color]

also you can uyse it http://www.mathlinks.ro/Forum/viewtopic.php?t=135954


If anybody has links to sites which have practice SATs, it would greatly be appreciated, thanx :)

Why not buy a book of the "10 Real SATs" sort? They're cheap at $20 and contain a lot of tests.

Yes, but I've bought three: Kaplan's and Baron's, and Master the SAT, and finished all of them.

[quote="frt"]Ok, [url=http://www.educationplanner.com/education_planner/preparing_article.asp?sponsor=2859&articleName=SAT_Practice_Tests]try this one[/url] :) It has two practice tests (in pdf files). Does it help?

You can go to google and search "SAT practice tests" etc. ;)[/quote]

Yup, good link frt  :D

Really the one and only SAT prep book you need is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874477182/learninfreed]The Official SAT Study Guide: For the New SAT[/url] by the College Board. It is very easy to waste a lot of time on test prep until test prep becomes your only EC--which makes your college application look kind of lame. Just use the REAL materials from the College Board for test prep, and use the rest of your time for worthwhile activities like reading books you enjoy on subjects you are interested in, participating in your favorite extracurricular activities, keeping up your school grades, and sports. It is DEFINITELY possible to study too much for the SAT.

Good luck on your tests, and on your college applications.

Yeah, after a certain amount of work, practicing for the (P)SAT becomes a waste.


inequalities
geometry
circumcircle
geometry proposed

Given are triangle $ABC$ and $I$ inside it. Called $R_{a},R_{b},R_{c}$ be circumradius of triangles $BIC,CIA,AIB$ and $R$ is circumradius of $ABC$ prove that
$R\ge \min\{R_{a},R_{b},R_{c}\}$  :)

Let the points $O_{A},O_{B},O_{C}$ and $O$ be the circumcentres of the triangles $BIC,AIC,AIB$ and $ABC$.Then what means $R<R_{A}$?That means that $\angle{BOC}>\angle{BO_{A}C}$!
=>$2\angle{A}>360^{0}-2\angle{BIC}$
So,if we would have that $R<min(R_{A},R_{B},R_{C})$ then
$180^{0}=A+B+C>3*180^{0}-(\angle{BIC}+\angle{AIC}+\angle{AIB})=180^{0}$ :P  :lol:

In the solution above I supossed that all the angles $AIC,BIC,AIB$ are $>90^{0}$.I think the other case is easier.

Ok that's right, thank Tiks for good solution!   :blush:


I am not sure whether I should post this in the Calculus forum or not but I feel that maybe the solution won't require calculus at all.

I proved the following results.

1. $\int^{ 1}_{0}x^{\alpha}e^{x}. dx < \frac{3}{\alpha+1}$ for $\alpha > 0$

2. There exist integers $a_{n}$, $b_{n}$ such that $\int^{1}_{0}x^{n}e^{x}. dx = a_{n}+b_{n}e$ where $n$ is a non-negative integer.

3. If $r$ is a positive rational number so that $r= \frac{p}{q}$, $(p,q)=1$ where $p,q$ are positive integers then for all integers $a,b$:

$|a+br| = 0$ or $|a+br| \geq \frac{1}{q}$.

Hence, prove $e$ is irrational.

I have a feeling that I am missing something completely trivial because it is so late.

If anyone wants to solve this who doesn't know calculus, here are the relevant facts:

There is a function $f$ with the following properties:

1) $f(\alpha) < \frac{3}{\alpha+1}$ for $\alpha > 0$
2) For $n \in \bf N$, $0 < f(n) = a_{n}+b_{n}e$ for some integers $a_{n}, b_{n}$ dependent on $n$.  (The $0 < f(n)$ was missing in the original question, but it's important.  Maybe this is where the trouble came from.  We also could have put it with given-the-first.)
3) For any rational number $r = \frac{p}{q}$ and any integers $a, b$, either $|a+br| = 0$ or $|a+br| \geq \frac{1}{q}$.

Given this knowledge, prove that $e$ is irrational.

See http://en.wikipedia.org/wiki/Euler's_continued_fraction_formula#The_exponential_function for another interesting approach.

[hide="hint"]
Use the well known series for $e$:
\[S_{n}=\sum_{k=0}^{n}\frac{1}{k!}\Rightarrow e=\lim_{n\rightarrow \infty}S_{n}\]
Now come up with bounds to show $n!(e-S_{n})$ is never an integer for each positive integer $n$, and deduce that $n!\cdot e$ is never an integer.
[/hide]


Let $ n$, $ k$, $ p$ be positive integers with $ 2 \le k \le \frac {n}{p \plus{} 1}$. Let $ n$ distinct points on a circle be given. These points are colored blue and red so that exactly $ k$ points are blue and, on each arc determined by two consecutive blue points in clockwise direction, there are at least $ p$ red points. How many such colorings are there?

This is rather easy, no? :maybe:

Well, if not then my solution is probably wrong.

[hide]Place, the blue points anywhere.  Now, place in between every $ k$ consecutive pairs of blue points the neccessary $ p$ red points.  Now, we have $ n\minus{}k\minus{}kp$ points left, all of which must be colored red.  The distinctness of the coloring will depend on how many red points are between each pair of consecutive blue points.  So, label the blue points around the circle $ b_1,...,b_k$.  Then, let $ p_i$ be the number of red points between $ b_i$ and $ b_{i\plus{}1}$ excluding the neccessary $ p$.  Thus, we have

$ \sum_{i\equal{}1}^k p_i\equal{}n\minus{}k\minus{}kp$

The number of nonnegative integer solutions for $ \{p_i\}$ is $ \binom{n\minus{}k\minus{}kp\plus{}k\minus{}1}{k\minus{}1}\equal{}\binom{n\minus{}kp\minus{}1}{k\minus{}1}$.  So, now since we overcounted because of the symmetries of the circle, we retract that by dividing by $ k$.  Thus, there are a total of

$ \frac{1}{k}\binom{n\minus{}kp\minus{}1}{k\minus{}1}$

colorings.[/hide]

[quote="cosinator"]

[hide]Now, we have $ n \minus{} k \minus{} kp$ points left[/hide][/quote]

isn't it $ n \minus{} k \minus{} (k \minus{} 1)p$ points left?

[quote="cosinator"]This is rather easy, no? :maybe:
[/quote]
I don't think so,though it's not hard.But I still haven't found a proof :blush: .
[quote="cosinator"]
Well, if not then my solution is probably wrong.
[/quote]
It's definitely incorrect,even the answer is incorrect.Just check some particular cases. :wink:


[quote="Totally Zealous"]Supposed A, B, and C are three numbers for which 1001C-2002A=4004 and 1001B+3003A=5005. The average of the numbers A, B, and C is
(A) 1
(B) 3
(C) 6
(D) 9
(E) not uniquely determined[/quote]

[hide]Add the two equations and divide by 3003 to get B[/hide]

[quote="vishalarul"]Add the two equations to get:
1001A + 1001B + 1001C = 9009
Divide by 3003:
(A+B+C)/3=3
THE ANSWER IS B.[/quote]

Cant you just divide by 1001 in the first place.

[hide]Adding the two equations, dividing by 3003 to get 3.

$3$[/hide]


calculus
integration
limit
calculus computations

Find
$ \lim_{a \to 0} \int_{ \minus{} a}^{a} \frac {1}{\sqrt {a^{2} \plus{} 2ba^{n} \minus{} x^{2} \minus{} 2bx^{n}}}dx$,$ n$ is an integer

case 1: n=1
$ I_a\equal{}\int_{ \minus{} a}^{a} \frac {1}{\sqrt {(a\plus{}b)^2\minus{}(x\plus{}b)^2}}dx$

$ \equal{}\frac{1}{a\plus{}b}\int_{ \minus{}a}^{a} \frac {1}{\sqrt {1\minus{}\left(\frac{x\plus{}b}{a\plus{}b}\right)^2}}dx$

$ \equal{}\int_{\frac{b\minus{} a}{b\plus{}a}}^{1} \frac {1}{\sqrt {1\minus{}u^2}}dx\equal{}\frac{\pi}{2}\minus{}\arcsin \left( \frac{b\minus{} a}{b\plus{}a}\right)$

so we get

$ \lim_{a\to 0} I_a \equal{}\frac{\pi}{2}\minus{}\arcsin (1) \equal{}0$

case 2: $ n \ge 2$

$ I_a\equal{}\int_{ \minus{} a}^{a} \frac {1}{\sqrt {(a^2\minus{}x^2) \plus{} 2b(a^n \minus{} x^n)}}dx$

$ \equal{}\int_{ \minus{} 1}^{1} \frac {1}{\sqrt {(1\minus{}u^2) \plus{} 2ba^{n\minus{}2}(1 \minus{} u^n)}}du$

so i would GUESS

$ \lim_{a\to 0} I_a \equal{} \int_{ \minus{} 1}^{1} \frac {1}{\sqrt {1\minus{}u^2}}du \equal{} \pi$


All these posts have fancy formal notation.  My question is two-fold, where is this formal documentation written up?  In other words, where can I learn it other than deciphering it from the posts and 2), most importantly, how do you type <ABC (angle ABC) in that formal bold lettering, is that in MS word?  What program does that?

Thanks.  HUGE newbie question.

I think you're probably alluding to latex and if you look to the very left of your screen and click on Latex , all your questions should be answered.

okay so "latex" is that formal notation program?  I clicked on Latex HElp between the Forum and Classes URL buttons on the mathlinks page, but it said it wasn't listed on the server.

That was a kind of below average response help; no idea what "top left of screen" means, but you directed me to latex.  This is the link that you should give next time some one asks about latex:

http://www.mathlinks.ro/Forum/viewtopic.php?t=123690


geometry
circumcircle
incenter
rectangle
inradius
geometry proposed

let  $ o$ is circumcircle of ABC  triangle and <B=90. $ \omega$  is tangent to AB,BC and $ o$.
prove that : Radius of $ \omega$=Diagonal of incircle of ABC.

Let $ X,Y$ denote the tangency points of $ \omega \equiv (V)$ with $ BC,BA.$ It's well-known that the polar of vertex $ B$ WRT the B-mixtilinear incircle $ (V)$ (the line XY), goes through the incenter $ I$ of $ \triangle ABC.$ If $ \angle ABC=90^{\circ}$ $\Longrightarrow$ $I$ is the center of the rectangle $BXVY.$ Thus, it follows that the radius of $ (V)$ is twice the inradius of $ \triangle ABC.$


Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $ x\in \mathbb R$ and $ y\in f(\mathbb{R})$, $ f(x\minus{}y)\equal{}f(x)\plus{}xy\plus{}f(y)$

Obviosly $ x\equal{}y$ give $ f(x)\equal{}\frac{\minus{}x^2}{2}\plus{}c, c\equal{}f(0)$, and $ x\equal{}y\equal{}0$ give $ c\equal{}0$.

$ x\equal{}y\in f(\mathbb{R})$ gives $ f(x)\equal{}{c\minus{}x^2\over 2},c: \equal{}f(0)$ for all $ x\in f(\mathbb{R})$.
If $ f\not\equiv 0$ we find some $ y\in f(\mathbb{R}),y\not\equal{}0$
and then for all $ z\in\mathbb{R}$ we find some $ x\in\mathbb{R}$ with
$ z\equal{}xy\plus{}f(y)\equal{}f(x\minus{}y)\minus{}f(x)\equal{}a\minus{}b$ for some $ a,b\in f(\mathbb{R})$.
This gives
$ f(z)\equal{}f(a\minus{}b)\equal{}f(a)\plus{}ab\plus{}f(b)\equal{}{c\minus{}a^2\over 2}\plus{}ab\plus{}{c\minus{}b^2\over 2}\equal{}c\minus{}{(a\minus{}b)^2\over 2}\equal{}c\minus{}{z^2\over 2}$
for all $ z\in\mathbb{R}$.
And now for $ x\equal{}f(0)\in f(\mathbb{R})$ we have $ f(x)\equal{}{c\minus{}x^2\over 2}\equal{}c\minus{}{x^2\over 2}\Rightarrow c\equal{}0$.
So there are $ 2$ solutions $ f\equiv 0$ and $ f(x)\equal{}\minus{}{x^2\over 2},x\in\mathbb{R}$.

[quote="olorin"]So there are $ 2$ solutions $ f\equiv 0$ and $ f(x) \equal{} \minus{} {x^2\over 2},x\in\mathbb{R}$.[/quote]
$ f\equiv 0$ is not solution, consider $ x\equal{}y\equal{}1$.

You forgot the restriction that $ y$ is in the range of $ f$, which is why your solution isn't quite right. $ x \equal{} y \equal{} 1$ is not valid because $ 1$ is not a value $ f$ can take.


calculus inequalities algorithm symmetry algebra polynomial derivative

a,b,c is real numbers, a,b,c>0 and a+b+c=3
prove that:
                    (a+b^2)(b+c^2)(c+a^2) =<13
help me!

I can prove this if $ a^2\plus{}b^2\plus{}c^2 \le 4$, can anyone prove it given $ a^2\plus{}b^2\plus{}c^2 \ge 4$ ?

The critical points lie within $ a^2 \plus{} b^2 \plus{} c^2 > 4$, so that should prove to be the more difficult part.  One proof strategy might be this:  Prove first that the maximum is achieved when one of the values is $ 0$, then the problem reduces to proving
\[ a(3\minus{}a)^2(3\minus{}a\plus{}a^2)\leq 13\]
How you would go about accomplishing either of these without calculus, however, is beyond me.  Using calculus, you can find the inequality is a bit tighter actually:
\[ (a \plus{} b^2)(b \plus{} c^2)(c \plus{} a^2)\leq 12 \plus{} \frac {9308 \plus{} 8309\cdot 2^{\frac {2}{3}} 7^{\frac {1}{3}} \minus{} 5936\cdot 2^{\frac {1}{3}}7^{\frac {2}{3}}}{9375}\approx 12.76496
\]
the inequality being strict equality for exactly 3 points.

Using calculus, how do you prove the max is when one of them is 0?

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58379

Well if we've given up on a non-calculus solution and taken the gloves off, then the proof sketch is as follows.  Let $ f(a,b) \equal{} (a \plus{} b^2)(b \plus{} (3 \minus{} a \minus{} b)^2)(3 \minus{} a \minus{} b \plus{} a^2)$.  $ f$ is of fifth-degree in both $ a$ and $ b$, so $ \frac {\partial f}{\partial a}$ and $ \frac {\partial f}{\partial b}$ are both quartic.  Thus, we can find using any algorithm for quartics exact algebraic expressions for the solutions of
\[ \frac {\partial f}{\partial a} \equal{} \frac {\partial f}{\partial b} \equal{} 0
\]
After some ridiculously tedious algebra, you find the critical points, evaluate $ f$ at those points and find that $ f(a^*,b^*) < 9$ for each of them.  Comparing this to, say, $ f(1,0) \equal{} 12$, you see that the maximum cannot be at a critical point, and therefore the maximum must be on the boundary of the region $ 0\leq a \plus{} b \leq 3$.  This boundary is composed of three line segments, one of which is $ \lbrace (a,b): b \equal{} 0,0\leq a \leq 3 \rbrace$.  You then argue by symmetry that $ f$ takes on the same range of values on each line segment (i.e. interchanging any of a,b,c doesn't change the problem), so the maximum must lie on the line $ \lbrace (a,b): b \equal{} 0,0\leq a \leq 3 \rbrace$.  This is simply the one dimensional problem of finding the maximum of
\[ f(a,0) \equal{} a(3 \minus{} a)^2(3 \minus{} a \plus{} a^2)
\]
on $ 0\leq a \leq 3$.  Again, this is a fifth degree polynomial, so its derivative with respect to $ a$ is quartic, and we can obtain exact solutions.  It has two real solutions, one is $ a \equal{} 3$, which corresponds to a minimum, the other is
\[ a \equal{} \frac {1}{15}\left(13 \plus{} 4\cdot 2^{\frac {2}{3}} 7^{\frac {1}{3}} \minus{} 2^{\frac {1}{3}} 7^{\frac {2}{3}}\right)
\]
which is where $ f$ obtains its maximum.  Substituting this back into $ f$ yields the upper bound I stated in a previous post.

[quote="Math pro"]Let $ x,y,z$ are nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 3,$ prove that

$ \left(x \plus{} y^2\right)\left(y \plus{} z^2\right)\left(z \plus{} x^2\right) < 13.$[/quote]Using the arithmetic-mean — geometric-mean inequality,

$ \frac {1}{3}\sum_{cyc}{(3x \plus{} 2y \plus{} 4z)^3\left(x \plus{} y^2\right)}\geq\prod_{cyc}{(3x \plus{} 2y \plus{} 4z)}\sqrt [3]{\prod_{cyc}{\left(x \plus{} y^2\right)}}.$

This comes down to prove that

$ 3\sqrt [3]{13}\prod_{cyc}{(3x \plus{} 2y \plus{} 4z)}>\sum_{cyc}{(3x \plus{} 2y \plus{} 4z)^3\left(x \plus{} y^2\right)}.$

Id est, it remains to the following [color=darkred]quintic[/color] homogeneous inequality:

$ 47(x \plus{} y \plus{} z)^2\prod_{cyc}{(3x \plus{} 2y \plus{} 4z)}>20\sum_{cyc}{(3x \plus{} 2y \plus{} 4z)^3\left[x(x \plus{} y \plus{} z) \plus{} 3y^2\right]},$

since $ x \plus{} y \plus{} z \equal{} 3$ and $ \sqrt [3]{13} \equal{} 2.3513\cdots > 2.35 \equal{} \frac {47}{20}.$ But

$ 47(x \plus{} y \plus{} z)^2\prod_{cyc}{(3x \plus{} 2y \plus{} 4z)} \minus{} 20\sum_{cyc}{(3x \plus{} 2y \plus{} 4z)^3\left[x(x \plus{} y \plus{} z) \plus{} 3y^2\right]}$

$ \equal{} \frac {83}{8}\sum_{cyc}{x(3x \minus{} 4y)^2\left(x^2 \plus{} 6xy \plus{} 6y^2\right)} \plus{} \frac {117}{8}\sum_{sym}{x^5} \plus{} \frac {75}{4}\sum_{cyc}{x^4y} \plus{} \frac {95}{4}\sum_{cyc}{x^3y^2}$

$ \plus{} 6441\sum_{sym}{x^3yz} \plus{} 8676\sum_{sum}{xy^2z^2} > 0.$

see also here : http://www.mathlinks.ro/viewtopic.php?t=146734


http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational

There are more proofs out there, and few can be found searching here on ML (and please try searching before you post these quite well known questions - this one, and the one with Zeta function in College Playground).

http://www.mathlinks.ro/viewtopic.php?t=162010
http://www.mathlinks.ro/portal.php?t=162051

[hide="Something to laugh about"]

http://in.answers.yahoo.com/question/index?qid=20090430032537AA7F1RP

 :rotfl: [/hide]


limit
trigonometry
function
calculus
calculus computations

Compute the ff. limits:

$ 1.)  \lim_{x\rightarrow\infty} \cos (\pi \sqrt{4x^4\plus{}3x^2\plus{}1}\minus{}2\pi x^2)$

$ 2.)  \lim_{x\rightarrow 0} \frac{| {\cos(\sin(3x)) } | \minus{}1}{x^2}$

1. $ L \equal{} \lim_{x \to \plus{} \infty} \cos [\pi (\sqrt {4x^{4} \plus{} 3x^{2} \plus{} 1} \minus{} 2 x^{2})]$

$ \equal{} \lim_{x \to \plus{} \infty} \cos \left(\pi \frac {4x^4 \plus{} 3x^2 \plus{} 1 \minus{} 4x^4}{\sqrt {4x^{4} \plus{} 3x^{2} \plus{} 1} \plus{} 2 x^{2}}\right)$

$ \equal{} \lim_{x \to \plus{} \infty} \cos \left(\pi \frac {3 \plus{} \frac {1}{x^2}}{\sqrt {4 \plus{} \frac {3}{x^2} \plus{} \frac {1}{x^4}} \plus{} 2}\right)$

$ \equal{} \cos \left(\frac {3 \pi}{4}\right)$

$ \equal{} \minus{} \frac {1}{\sqrt {2}}$.

2. For x close to zero, the function cos is positive, and the function $ \frac{\cos(\sin(3x)) \minus{} 1}{x^2}$ is even. Hence,

$ L \equal{} \lim_{x \to 0} \frac{\cos(\sin(3x)) \minus{} 1}{x^2} \equal{} \minus{} \lim_{x \to 0} \frac{1 \minus{} \cos^2(\sin(3x))}{x^2(1 \plus{} \cos(\sin(3x)))}$

$ \equal{} \minus{} \frac12 \lim_{x \to 0} \frac{\sin^2(\sin(3x))}{x^2} \equal{} \minus{} \frac92  \lim_{x \to 0} \left[\frac{\sin(\sin(3x))}{\sin(3x)}\right]^2 \cdot \left[\frac{\sin(3x)}{3x}\right]^2 \equal{} \minus{} \frac92$


limit
number theory
prime factorization
number theory unsolved

[i]Is there any elementary proofs to prove that : $ \lim_{ n \to \infty } \frac{ \pi(n)}{n} \equal{} 0$ :maybe: [/i]

$ \displaystyle S: =\lim_{n \to +\infty}{\frac{\pi(n)}{n}= \lim_{k \to +\infty}{\prod_{i=1}^k{\left(1-\frac{1}{p_i}\right)} \ge 0}}$ where $ p_i$ is the $ i$-th prime (it is true by inclusion-esclusion principle)
Now if $ y \in \mathbb{R}^+$ then $ 1-y \le e^{-y}$, with equality iff y=0. It follows that $ S \le e^{\displaystyle -\lim_{k \to +\infty}{\sum_{i=1}^k{\left(\frac{1}{p_i}\right)}}}$.
Now assume that $ \sum_{i=1}^{\infty}{\left(\frac{1}{p_i}\right)}$ converges, then there exists $ v \in \mathbb{N}$ such that $ \sum_{i=v+1}^{+\infty}{\left(\frac{1}{p_i}\right)}<\frac{1}{2}$. Define $ t: =\prod_{i=1}^v{p_i}$; because of $ p_j \nmid nt+1$ forall $ 1 \le j \le v$ we must have $ \displaystyle \sum_{n=1}^{+\infty}{\frac{1}{nt+1}} \le \prod_{i=v+1}^{\infty}{ \sum_{k=0}^{\infty}{\frac{1}{p_i^k}}}$(*), but the last rhs is $ \le \prod_{i=v+1}^{\infty}{\left(1+\frac{2}{p_i}\right)}$ that converges for assumption, contracting that lhs of (*) diverges. So such $ v \in \mathbb{N}$ does not exist. It implies that $ \sum_{i=1}^{\infty}{\left(\frac{1}{p_i}\right)}$ diverges, and so $ \displaystyle S: =\lim_{n \to +\infty}{\frac{\pi(n)}{n}}$ is exactly 0.

Your first statement isn't quite rigorous since we don't know that the limit exists; what you're computing is an upper bound on the $ \limsup$.

The second half of the argument can be made more elementary.  Equivalently, it suffices to show that

$ \lim_{k \to \infty} \prod_{i \equal{} 1}^{k} \left( \frac {1}{1 \minus{} \frac {1}{p_i} } \right) \equal{} \infty$.

But this is classic:  by unique prime factorization we have the bound

$ \prod_{i \equal{} 1}^{k} \left( \frac {1}{1 \minus{} \frac {1}{p_i} } \right) \ge \sum_{i \equal{} 1}^{p_k} \frac {1}{i}$

and it is elementary to show that the harmonic series diverges, i.e. by Cauchy condensation.  The conclusion follows.


function
limit
induction
real analysis
real analysis solved

Let the functions f_n:[e,oo)-->R, defined recurently like so f_0(x)=ln(x),
f_(n+1) (x)=ln(x*f_n(x)), n is natural. Then we have:

 \lim (x-->e) (f_n(x))^(1/(x-e)=e^(n+1)/e, n is natural.

cheers!!

I used L'Hospital for this. We are trying to prove that ln(f_(n+1)(x))/x-e tends to (n+2)/e as x->e. It's obvious by induction that f_n(x)->1 as x->e so we have an indetermination so we can apply L'Hospital.
Let's assume that we have proved P_n which states that (g_n)'(x)->(n+1)/e as x->e, where g_n(x)= ln(f_n(x)). Then, by L'Hospital we also have g_n(x)/x-e -> (n+1)/e as x->e. Now let's prove that (g_(n+1))'(x)->(n+2)/e. (g_(n+1))'(x)=(1/ f_(n+1)(x))* f_(n+1)'(x)= (1/ f_(n+1)(x))* (1/x+f_n'(x)/f_n(x)). But we can observe that f_n'(x)/f_n(x) = (g_n)'(x)->(n+1)/e as x->e, and 1/x->1/e as x->e and 1/ f_(n+1)(x)->1 as x->e, so by combining these we get that (g_(n+1))'(x) -> (n+2)/e. So by induction (g_n)'(x)->(n+1)/e as x->e for any n. Now by applying L'Hospital to these we get g_n(x)/(x-e) -> (n+1)/e as x->e. If we put e^(this limit we just got) we obtain the limit we wanted.

See ya!

 :D


geometry
power of a point
radical axis
geometry proposed

Let $w=C(O,r)$ be a fixed circle and let $A$ be a fixed point such that $OA>r$.

For a variable circle $x,\ A\in x,\ x\cap w\ne \emptyset$, I note the tangent $t$ in the point $A$ to the circle $x$

and the common chord $c$ of the circles $w,x$. Ascertain the geometrical locus of the point $L\in t\cap c$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=54479

$A$ is a circle of zero radius. The tangent $t$ is the radical axis of the circles $(A, x)$, the chord $c$ is the radical axis of the circles $(w, x)$. $L \equiv t \cap c$ is the radical center of the circles $(A, w, x)$ which lies on the radical axis of the circles $(A, w)$ perpendicular to their center line AO. Let the last radical axis intersect the center line AO at a point R.

$AR^2 = (RO + r)(RO - r) = RO^2 - r^2 = (AO - AR)^2 - r^2$

$0 = AO^2 - 2AO \cdot AR - r^2,\ \ \ AR = \frac{AO^2 - r^2}{2 AO}$