function
algebra
domain
integration
logarithms
trigonometry
complex numbers

Some text says euler's fomula canbe proved by
tailer expansion of $ e^x, cos(x), sin(x)$
but as you know, maclaurine series expansion(just power expansion)
is made by 'mean value theorem'. That's drawn on the x-y plane of real field.

$ f(x) \equal{}$some function on the real field
$ g(x) \equal{} f(0) \plus{} {{f'(0)}\over{1!}}x^{1} \plus{} {{f''(0)}\over{ 2!}} x^{2} \plus{} ... \plus{} cx^n$
$ h(x) \equal{} f(x) \minus{} g(x)$
$ h(x) \equal{} 0$ at $ 0$, $ x_1$ (f(x) and g(x) is same at these two points)

$ h(0) \equal{} 0$ and $ h(x_{1}) \equal{} 0$
$ h'(0) \equal{} 0$ and $ h'(x_{2}) \equal{} 0$
...
$ h^{(n \minus{} 1)}(0) \equal{} 0$ and $ h^{(n \minus{} 1)}(x_{n}) \equal{} 0$
$ h^{(n)}(x_{n \plus{} 1}) \equal{} 0$
So $ c \equal{} f^{(n)}(x_{n \plus{} 1}) \over n!$ and $ x_{n \plus{} 1}$ is between 0 and $ x_1$

but text simply chage the $ x$'s domain from the real field to the complex field
for catching the $ e^{ix}$'s expression.
But in point of view of above proof, that's not possible(In above proof, x's domain can't be changed).
Please give me some understandable explaination.

First of all, you've posted in the wrong forum. [color=green][It's been moved - mod.][/color]

I agree that it's not completely rigorous to just change the domain, but if we don't allow complex numbers somehow, the expression $ e^{ix}$ doesn't even make sense.  It turns out that any proofs about convergence of these series are easily transposed to the complex numbers, as long as you interpret $ |\cdot|$ to be norm (distance from zero) instead of absolute value.

There is a nice proof that uses these two formulas:
1. $ \int \frac {dx}{x^2 \minus{} a^2} \equal{} \frac {1}{2a}\ln \frac {x \minus{} a}{x \plus{} a} \plus{} C$
2. $ \int \frac {dx}{x^2 \plus{} a^2} \equal{} \frac {1}{a} \tan^{ \minus{} 1} \frac {x}{a} \plus{} K$

Setting $ a \equal{} 1$ and substituting $ zi$ for $ x$ in either 1 or 2.


What is the probability that you can find a [b]"+1" integer [/b]in a box that contains real numbers and positive integer numbers?

Hmmm...

The probability of hitting an integer if you throw a dart at the number line is 0.  In fact, the probability of getting a rational number is, in fact 0.  Don't know how to prove it, just know it's true.

Well geometrically speaking, the integers are points on the line, and points have no area.


Let $ k$ be a positive integer. Find all polynomials

\[ P(x) \equal{} a_0 \plus{} a_1 x \plus{} \cdots \plus{} a_n x^n,\]

where the $ a_i$ are real, which satisfy the equation

\[ P(P(x)) \equal{} \{ P(x) \}^k\]

[hide]
If $ n\equal{}0$, then $ P(x)\equal{}a_0$. So. $ P(P(x))\equal{}[P(x)]^k \Rightarrow a_0\equal{}a_0^k$. So $ P(x)\equal{}0,1$ for even $ k$, and $ P(x)\equal{}\minus{}1,0,1$ for odd $ k$.

If $ n \neq 0$, then $ y\equal{}P(x)$ takes on infinitely many distinct values $ y$ which must satisfy $ P(y)\equal{}y^k$. So $ P(x)\equal{}x^k$. Is the only non-constant polynomial for each $ k$.

Thus, $ P(x)\equal{}0, \ 1, \ x^k$ for even $ k$ and $ P(x)\equal{}\minus{}1, \ 0, \ 1, \ x^k$ for odd $ k$.
[/hide]

[quote="Xantos C. Guin"]So $ P(x) \equal{} x^k$. Is the only non-constant polynomial for each $ k$.
[/quote]

I get why this is correct answer but I don't understand how you eliminated other possibilities. I mean, it seems pretty trivial that there isn't any other one but I wasn't sure if you had to like "prove" this assertion.

And by the way, your explanation made more sense than the one in Art of Problem Solving book.  :lol:

[b]Silverfalcon[/b]: Let $ P(x) = y$. It is given that $ P(y) = y^{k}$, so we know that $ P(x)$ has degree $ k$. Notice that, if $ P(x) = x^{k} + G(x)$, where $ G(x)$ is a non-zero polynomial of lesser degree, then

$ \begin{align*} P(P(x)) & = P(x^{k} + G(x)) = (x^{k} + G(x))^{k} + G(x^{k} + G(x)) \\
& \neq [P(x)]^{k} = (x^{k} + G(x))^{k}$

Thus, $ P(x) = x^{k}$ is the only non-constant solution.

(It can't be $ P(x) = ax^{k}$, $ a \neq 1$, either, because then $ P(ax^{k}) = a^{k + 1}x^{k^{2}} \neq a^{k}x^{k^{2}}$.)

Also, if $ P(x)$ and $ Q(x)$ are polynomials of degree $ k$ and $ P(y) \equal{} Q(y)$ for $ k \plus{} 1$ distinct values of $ y$, then $ P(x) \equal{} Q(x)$. (For a proof, realize that $ P(x) \minus{} Q(x)$ has a degree at most $ k$, but has $ k \plus{} 1$ roots, so it must be constant, and thus $ 0$.)

Since $ P(y) \equal{} y^k$ for infinitely many values of $ k$ (which is more than $ k \plus{} 1$), $ P(x) \equal{} x^k$.

Thanks metafor, that makes a perfect sense!  :)


Let $ A\equal{}\frac{1}{21\cdot 2009},\ B\equal{}\left(1\plus{}\frac{1}{2009}\right)^{\frac{1}{21}}\minus{}1,\ C\equal{}1\minus{}\left(1\minus{}\frac{1}{2009}\right)^{\frac{1}{21}}.$

Which is the greatset or smallest among these expressions?

[hide="solution"]
By bernoulli;
$ 1 \minus{} \left(1\minus{}\frac{1}{2009}\right)^{\frac{1}{21}} \ge \frac{1}{21\cdot 2009}$
$ \Rightarrow C>A$

now we show $ B>A$
$ \left(1\plus{}\frac{1}{2009}\right)^{\frac{1}{21}} \ge \frac{1}{21\cdot 2009}$

$ \Rightarrow 1 \plus{} \frac{1}{2009} \ge 1 \ge \left(\frac{1}{21\cdot 2009}\right)^{21}$

Now we show $ B>C$
let $ a \equal{} \left(1\plus{}\frac{1}{2009}\right)^{\frac{1}{21}}$ and let $ b\equal{}\left(1\minus{}\frac{1}{2009}\right)^{\frac{1}{21}}$
We have $ \{a,b\}\ge 0$ and $ a^{21}\plus{}b^{21} \equal{} 2$

And we want $ a \plus{} b \ge 1$

$ \Rightarrow (a\plus{}b)^{21} \ge 1$

$ \Rightarrow 2 \plus{} \left(\binom{21}{1}a^{20}b \plus{} \binom{21}{2}a^19b^2...\plus{}\binom{21}{1}ab^{20}\right) \ge 1$

This is clearly true, Therefore $ \boxed{A<C<B}$

[/hide]

I also think so. The problem is difficult for Japanese high school students for university entrance examination.

(Note:   I don't use LATEX, so I was trying to fill it in based on other
people's use of it.)

If we suppose that $ B > A$ from ocha's line:

$ \left(1 + \frac {1}{2009}\right)^{\frac {1}{21}} - 1\ge \frac {1}{21\cdot 2009}$

$ \left(1 + \frac {1}{2009}\right)^{\frac {1}{21}} \ge \frac {1}{21\cdot 2009} + 1$

$ [\left(1 + \frac {1}{2009}\right)^{\frac {1}{21}}]^{\frac {21}} \ge (\frac {1}{21\cdot 2009} + 1)$^$ {\frac {21}}$

$ \left(1 + \frac {1}{2009}\right) \ge (\frac {1}{21\cdot 2009} + 1)$^$ {\frac {21}}$

1.00049776... versus 1.00049787...  (no rounding in place for the digits shown)

[/quote]

The supposition is false, then, with at least this portion.

So B < A.

Then A < C < B is not the final answer.


A yearbook sells half-page ads for $ \$100$, quarter-page ads for $ \$75$, and eighth-page ads for $ \$50$. How many different amounts of income can be obtained from various full-page arrangements?

half+half=200
half+2 quarters=250
half+4 eighths=300
4 quarters = 300
3 quarters+2 eighths=325
2 quarters+4 eighths=350
1 quarter+6 eighths=375
8 eighths=400

As you can see, there are 7 different sums.

However, the question asked for different income accounts, so the answer is $ 6$.


Simplify the following and express as a mixed number.

\[ 1\plus{}\cfrac{1}{1\plus{}\cfrac{1}{1 \plus{} \frac{1}{3}}}\]

[quote="GameBot"]Simplify the following and express as a mixed number.
\[ 1 \plus{} \cfrac{1}{1 \plus{} \cfrac{1}{1 \plus{} \frac {1}{3}}}
\]
[/quote]

$ 1 \plus{} \frac {1}{3} \equal{} \frac {4}{3} \implies 1 \plus{} \frac {4}{3} \equal{} \frac {7}{3} \implies 1 \div \frac {7}{3} \equal{} \frac {3}{7} \implies \frac {3}{7} \plus{} 1 \equal{} \boxed{1\frac {3}{7}}$

EDIT: Oh, I think I misread the problem. Sonnhard is correct.

hello, we have $ 1\plus{}\frac{1}{3}\equal{}\frac{4}{3}$, now we get $ \frac{1}{\frac{4}{3}}\equal{}\frac{3}{4}$, then we have $ 1\plus{}\frac{3}{4}\equal{}\frac{7}{4}$ and now $ 1\plus{}\frac{1}{\frac{7}{4}}\equal{}1\plus{}\frac{4}{7}\equal{}\frac{11}{7}$.
So our result is $ \boxed{1\frac{4}{7}}$.
Sonnhard.


[quote="Xixas"]Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.[/quote]

The following identity is easily verified:
\[ {{y-z}\over{x}}+{{z-x}\over{y}}+{{x-y}\over{z}}=-{{(y-z)(z-x)(x-y)}\over{xyz}}\quad\hbox{for}\quad x,y,z\neq 0.\leqno (1) \]
Set $x=b-c,y=c-a,z=a-b$ in (1). In view of $a+b+c=0$ we have $y-z=b+c-2a=-3a,z-x=-3b,x-y=-3c$, and (1) becomes
\[ {{-3a}\over{b-c}}+{{-3b}\over{c-a}}+{{-3c}\over{a-b}}=-{{(-3a)(-3b)(-3c)}\over{(b-c)(c-a)(a-b)}}, \]
i.e.,
\[ {{a}\over{b-c}}+{{b}\over{c-a}}+{{c}\over{a-b}}=-{{9abc}\over{(b-c)(c-a)(a-b)}}. \]
Now set $x=a,y=b,z=c$ in (1):
\[ {{b-c}\over{a}}+{{c-a}\over{b}}+{{a-b}\over{c}}=-{{(b-c)(c-a)(a-b)}\over{abc}}. \]
Multiplying the last two equalities we get the claimed one.

Nice work !

 [color=red]The problem comes from the Austrian- Polish competition[/color].I have recently read a very similar - almost the same -  solution in an older issue of Crux.

 Babis

Indeed  :D
http://www.kalva.demon.co.uk/aus-pol/ap85.html

The problem indeed is classic and quite well-known (at least in Greece ).
Another solution is . Let $S$ the given product

Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$

Similar for the other two .

We expand the given product and we use he above .so
$S=3+\frac{2(a^3+b^3+c^3)}{abc}$ .But $a^3+b^3+c^3=3abc$ when $a+b+c=0$

so $S=9$  and we are done

Another solution..

[hide]
By AM-HM:

$\frac{x+y+z}{3}\geq\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \implies (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9$

Equality holds only if $x=y=z$.

$(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$

$\frac{a}{b-c}=\frac{b}{c-a}=\frac{c}{a-b}$

$a(c-a)=b(b-c) \implies a^2 + b^2=ac + bc$
$b(a-b)=c(c-a) \implies b^2 + c^2=ac + ba$
$a^2 + b^2-(b^2 + c^2)=ac + bc-(ac + ba) \implies a^2-c^2=bc-ba$

Since $a \neq c$, we can divide $a^2-c^2=bc-ba$ by $(a-c)$.

$a+c=-b \implies \boxed{a+b+c=0}$

[/hide]

Ghm  :wacko:  Can we apply AM-HM to potentialy negative $x, y, z$?

Hm I have to admit I really didn't consider that restriction. Isn't it only for the inequalities with geometric mean? :? Actually it is also possible to proof this inequality using Cauchy-Schwarz: (Though it seems that this time we really have a restriction because of those d$a$mn sqrt{} things. :))

$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\sqrt{x}\cdot\sqrt{\frac{1}{x}} +\sqrt{y}\cdot\sqrt{\frac{1}{y}}+ \sqrt{z}\cdot\sqrt{\frac{1}{z}})^2=9$

[EDIT]

Ok, we just have to try $(-10, 1, 1)$ and see that my solution sucks. :oops:

That's the reason that prevented me from using Cauchy-Schwarz on the contest.

Since $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$
It follows that if $ a \plus{} b \plus{} c \equal{} 0$, then $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$
The numerator of the left factor is
$ \minus{} \sum (a^3 \minus{} a^2b \minus{} a^2c \plus{} abc) \equal{} \minus{} \sum (a^3 \minus{} a^2(b \plus{} c) \plus{} abc) \equal{} \minus{} \sum (2a^3 \plus{} abc)$
Since $ 3abc \equal{} a^3 \plus{} b^3 \plus{} c^3$, the left factor is
$ \frac { \minus{} 3 \sum a^3}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}$
The right factor is
$ \frac {b^2c \minus{} bc^2 \plus{} c^2a \minus{} ca^2 \plus{} a^2b \minus{} ab^2}{abc}$
But the denominator, $ abc \equal{} \frac {1}{3} \sum a^3$. Everything cancels, and we're left with 9.

[quote="silouan"]
Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$
[/quote]
How did you come up with that ?

The problem is very classic. In Japanese univesity entrance exam, Nara University entrance exam/Medicine has posed te same problem in early 1960 or later 1950.

I would like to know the original problem or articles regarding the problem.

Are there anyone having the information ？

Any information would be appreciated.

Thanks in advance.

kunny

I have found this problem in the following book.
В.Кречмар, "Задачник по алгебре", (1937 year, page 15, problem 18).

Since a+b+c=0, c=-a-b
Therefore, substituting this in you get the attached image.
Take out 9 from the numerator, cancel 2a^4 + 5(a^3)b - 5a(b^3) - 2b^4 from the top and bottom... and you're done!

[quote=Xixas]Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.[/quote]
[url=https://artofproblemsolving.com/community/c6h1402372p7844276]Kyrgyzstan national 2016
[/url]

Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that$$\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$$


Find the value of $n$ if
\[33 \cdot 35 \cdot 37 \cdot 39+15 = n(n+2). \]

[hide="Solution"]
$n^{2}+2n-1666680 = 0$.  Using the the quadratic formula $n = 1290,-1292$.  [/hide]

[hide="Without calculator too"]
$33 \cdot 35 \cdot 37 \cdot 39+15 = n(n+2)$

$33 \cdot 35 \cdot 37 \cdot 39 = n^{2}-2n-15$

$3^{2}\cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 39 = n^{2}-2n-15$

$3^{2}\cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 39 = (n+5)(n+3)$

At this point, we want to arrange the factors into 2 groups that multiply out to very close to each other (specifically, they differ by 8)
A bit of common sense and good judgment ($5 \cdot 7 \approx 3 \cdot 11; 3 \cdot 13 \approx 37$ ) leads to
$(5 \cdot 7 \cdot 37)(3^{2}\cdot 11 \cdot 13) = (n+5)(n+3)$

$1287 \cdot 1296 = (n+5)(n+3)$

$\boxed{n=1290}$
[/hide]


The problem is the following:
a^n-1 / b^n-1 for every postive integer n. Prove that b = a^k for some integer k.
I would find a (relatively) short soultion extremely helpful.

This is old... See http://www.mathlinks.ro/Forum/viewtopic.php?t=15361 , http://www.mathlinks.ro/Forum/viewtopic.php?t=4556 .

  Darij


inequalities trigonometry AMC AIME calculus integration logarithms

The title pretty much explains everything.  :)

If you have some good problems involving Cauchy, AM-GM, or all that other good stuff, please post them here. I'd like to practice these a bit before the AMC, in case they show up. (You'd like to practice too, right?  :D )

Here's one from AoPS. Pretty obvious which inequality to use, but that's ok. :P

Prove that $1^2+2^2+\cdots+n^2\geq \frac{(1+2+3+\cdots+n)^2}{n}$ for all integers $n\geq 1$.

Here's a problem involving trig. Yay!

Show that if $x$ and $y$ are angles in the first quadrant, then $\left( \frac{\cos^3 x}{\cos y}+\frac{\sin^3x}{\sin y} \right)\cos (x-y)\geq 1$.

Hmm, nobody's posted yet?  :(

I guess I'll just go do all the optimization problems in my calc book then. :dry:

OK, here's a few, but most are probably well-known:

1) Let $a,b,c$ be the lengths of the sides of a triangle. Prove that $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$.

2) Show that if $x,y,z>0$ and $x+y+z=1$ then $0\le xy+yz+zx-2xyz\le \frac{7}{27}$.

3) Prove that $n! < (\frac{n+1}{2})^n$ for any integer $n>1$.

4) Show that $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd>14$ if $a,b,c$ and $d$ are positive real numbers whose product is equal to $2$.

5) If $x,y,z>0$, show that $(x+y+z)\left( \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \ge \frac{9}{2}$.

6) Show that if $x,y\ge 0$ then $x^3+y^3\ge \frac{1}{4}(x+y)^3$.

This ought to keep you busy for a while :D

Here are two others. They are probably more toward AIME level though:

$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\geq 0\Rightarrow \frac{a+b+c}{3}\geq \sqrt[3]{abc}$. (Also, when is there equality?)

$n\text{ integral and }>1,\ p>0\Rightarrow \sum_{k=1}^n n^{-p}>n\left( \frac{2}{n+1} \right)^p$

Since this is very easy, and that I have little time for inequalities (at the moment), I'll just solve 3.
$\\\frac{1+\dots +n}{n}\geq\sqrt[n]{n!}\\ \frac{\frac{n(n+1)}{2}}{n}=\left(\frac{n+1}{2}\right)\geq\sqrt[n]{n!}\\\left(\frac{n+1}{2}\right)^n\geq n!\hspace{.3in}\forall n\in N$

EDIT: I have a couple too.
1. If $\triangle ABC$, prove that $\frac{aA+bB+cC}{a+b+c}\geq\frac{\pi}{3}$.

2. If $a,b,c\geq 0$, prove that $\sqrt{3(a+b+c)}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}$

[hide="Solutions"]
1. Using Chebyshev's inequality, we have
$\\\frac{aA+bB+cC}{3}\geq \frac{a+b+c}{3}\cdot\frac{A+B+C}{3}=\frac{a+b+c}{3}\frac{\pi}{3}\\ \frac{aA+bB+cC}{a+b+c}\geq\frac{\pi}{3}$
Q.E.D.

2. Using Power Means Inequality, we have
$\\\frac{a+b+c}{3}\geq\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)^2\\ 3(a+b+c)\geq(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\\\\ \sqrt{3(a+b+c)}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}$
Q.E.D.
[/hide]
Masoud Zargar

Umm I don't think inequalities are going to show up on AMC except maybe #25 or something...

[quote="4everwise"]Well, I feel like learning them anyway, as math isn't justa bout the AMC. :yup:[/quote]

Inequality problems generally show up at the Pre Olympiad level or higher.  There are some that fall under the Intermediate category but if you really want to go further than the bare basics, I'd recommend you to post in a higher forum (or look at some previous posts in those forums) :)

Also, it's not true that inequalities [i]never[/i] show up on AMC's.  They are just really unlikely to show up and if they do, it's generally an easy application of AM GM.  For example, AMC 12 2003 A #24 requires knowing that $x+\frac 1x\ge 2$ for $x>0$...which you might be able to figure out intuitively but I'll bet that 90%, at least, of the people who got that problem right knew that this was an immediate result of AM GM.

[quote="joml88"][quote="chess64"]Actually, I solved that problem without AM-GM:

[hide]
Simplifying, we get $2-\log_a b - \log_b a$. Maximizing this means minimizing $\log_a b + \log_b a = \log_a b + \frac{1}{\log_a b} = \frac{(\log_a b)^2+1}{\log_a b}$. Minimizing this means maximizing the denominator, which is when $a=b$, so we get 2. $2-2=\boxed {0}$. Unfortunately, I forgot to subtract the result from 2 when I did this problem :D
[/hide][/quote]

Actually, it appears that you used really faulty logic :P  Since when does $\log_ab$ reach a maximum...?  Also, you can only say that a fraction is at its minimum when its denominator is at its maximum when the numerator is constant...which it was not.  The only thing you got right was that $a=b$...which is the equality condition of AM GM :P :P[/quote]

$\log _a b$ is maximum when $a=b$ because $a\ge b >1$. ;) But about the numerator is constant part... I guess you're right. :)

O, whoops...good call on the $\log$...I should've looked at the problem again :P


algebra
polynomial
modular arithmetic
number theory proposed
number theory

Let $f$ be a polynomial with integer coefficeints and positive degree, irreducible over the rationals. Prove that for every positive integer $k$ we can find an integer $n$ and distinct primes $p_1,\ldots,p_k$ sucht that $p_i\|f(n),\ \forall i\in\overline{1,k}$.

($p\|a$ means that the prime number $p$ divides $a$, but $p^2$ does not)

I remember Omid Hatami posted it as Just beautiful (name of topic) and I gave a quite long solution. Do you have a short one, grobber?

I don't think mine was too long, but I solved it a long time ago, so I might be mistaken :).

Take a prime $p$ such that if $p\mid f(n)$ for some integer $n$ then $p^2 \mid f(n)$: we want to show that we can find only a finite number of such primes $p$.

Clearly $p\mid f(n+p)$, so by construction also $p^2 \mid f(n+p)$. But $f(n+p) \equiv f(n)+pf'(n) \pmod{p^2}$ , i.e. $p\mid f'(n)$. Now $f(x)$ and $f'(x)$ are coprime polynomials, so there exist polynomials $\alpha(x),\beta(x)\in \mathbb{Z}[x]$ and a constant $k\neq 0$ such that
\[ \alpha(x)f(x)+\beta(x)f'(x)=k \]
and that's a contradiction, since $\omega(k)$ is bounded, i.e. $k$ can be divisible by only a finite number of such primes $p$.

Result follows by Chinese remainder theorem.


Does anyone have any good ideas how to quickly prepare for USAMO or know any good websites?

This one is clearly a good website: the best source of information, a wealth of problems, and a good community of people to discuss them with.  Do problems here in the Olympiad/Pre-Olympiad section, or search online for olympiad problems.  There are tons of good sources, for example here, kalva, or just search online...

Another good way: if you have access to some books or some sort of Olympiad material, that's great.  I asked my teacher if he had heard of "Mathematical Olympiad Challenges".  He pulled a copy off his bookshelf and handed it to me, saying "You can return it when you're done with it,": that is, until after USAMO.  I've found it's helped my geometry skills a lot!

[quote="worthawholebean"]Rusczyk/Lehoczky->Zeitz->Engel

Edit: Also known as Art of Problem Solving -> Art and Craft of Problem Solving -> Problem-Solving Strategies[/quote]

I wholeheatedly agree; those three books are excellent. Mostly the latter two for the USAMO, though.


If I can only take 1 of these which should I take?  On the course details it has the same topics listed but the description says they are completely different.  Geometry is my strongest point and Algebra is my weakest.

If you're not looking for a focus in one subject, the problem solving classes are probably your best bet.

I don't know your skill level but I will say that I take Intermediate Algebra but I also take MATHCOUNTS which seems a little easy, so the AMC problem solving would be the safest bet.

[quote="kysuke"][quote="rrusczyk"]They cover the same essential areas of math, but there are completely different problems in the two.  So, either is fine.  If you want a thorough exploration of algebra, you might consider the Intermediate Algebra class instead of either AIME Problem Series (or, at the very least, work through the Intermediate Algebra book).[/quote]

Thanks.  I wasn't sure if one would focus more on one area than the other.  I just received the intermediate algebra book so I'm working through that now.  I'm not looking for only an algebra course though since I still need to work on all the other areas.[/quote]

do you find the intermediate algebra book to be a good book?

lol rrusczyk wrote intermediate algebra so i am pretty sure he thinks it's a good book. coming from someone who didn't right it, yes i think it is a great book!  :)

[quote="cognos599"]coming from someone who didn't right it, yes i think it is a great book!  :)[/quote]
lol *write.
Anyway yes I completely agree with cognos.
The Intermediate Algebra is great and even better with the course. (lol I could be an AoPS advertiser). However it is a little bit of a struggle for me as a 12 year old. :blush: But a great read and a great class!!!

Yes Dojo, same with me.  Being 11, there's a lot of new material, and sometimes the class goes too fast.  But you can always look at the transcripts.


geometry
circumcircle
incenter
Euler
geometric transformation
inradius
geometry proposed

Let $ ABC$ be a triangle and $ (O)$ be its circumcircle. Let $ D$, $ E$, $ F$ the tangency points of $ BC$, $ CA$, $ AB$ with the incircle $ (I)$ of triangle $ ABC$, respectively. Circle $ \Gamma_A$ is tangent to $ BC$ at $ D$, and internally tangent to $ (O)$ at $ M$. Points $ N$, $ P$ are defined similarly. Prove that lines $ MD$, $ NE$, $ PF$ are concurrent, and this concurrency point lies on the line connecting $ O$ and the orthocentre of triangle $ DEF$.

Again a nice April problem, though this one is quite easy, but I like it very much nonetheless.

As [b]pohoatza[/b] would say, we will make use of the [b]Monge-d'Alembert circle theorem[/b] three times. (I never knew it was called that way)

[u][b]Proof:[/b][/u]

Clearly, $ M$ is the external center of similitude of $ (O)$ and $ \Gamma_A$, and $ D$ is the internal center of similitude of $ (I)$ and $ \Gamma_A$. Using the Monge-d'Alembert circle theorem, we obtain that $ M,D,X_{55}$ are collinear. The point $ X_{55}$ is the internal center of similitude of $ (O)$ and $ (I)$.
Similar arguments for $ \Gamma_B$ and $ \Gamma_C$ show that the desired point of concurrency is $ X_{55}$.

The only thing we need to prove now is that $ X_{55}$ lies on the line connecting $ O$ and the orthocenter $ H'$ of $ \Delta DEF$.
This property is well-known, while the point is known as $ X_{65}$ if I am not mistaking, but we will prove the property for the sake of completeness:

Draw the orthic triangle $ UVW$ of $ DEF$, and it's circumcircle $ \Gamma_0$. Clearly, $ \Delta ABC$ and $ \Delta UVW$ are parallel. Therefore, the line connecting both incenters ($ H'$ and $ I$) is parallel to the line connecting both circumcenters (Center of $ \Gamma_0$ and $ O$). However, the center of $ \Gamma_0$ is also the center of the nine-point circle of $ \Delta DEF$, so it lies on the Euler line $ H'I$ of $ \Delta DEF$. This shows that $ H',I,O$, and hence $ H',X_{55},O$ are collinear. $ \Box$

[u][b]Figure:[/b][/u]

[img]http://img247.imageshack.us/img247/6348/aprilinternalsimilitudewx1.png[/img]

Generally, if we have three circles in general position (excuse the repetition, but the second word [i]general[/i] has different meaning from the first) $ \rho_{1}$, $ \rho_{2}$, $ \rho_{3}$ and if we denote by $ \Gamma_{e}$ and $ \Gamma_{i}$ the circles simultanously tangent to all these three circles (the two Apollonius circles) externally and internally at $ M_{1}$, $ N_{1}$, $ P_{1}$ and $ M_{2}$, $ N_{2}$, $ P_{2}$, respectively, then the lines $ M_{1}M_{2}$, $ N_{1}N_{2}$, $ P_{1}P_{2}$ concur at the insimilicenter of $ \Gamma_{e}$ and $ \Gamma_{o}$.

This follows similarly from the Monge-D'Alembert circle theorem  :D.

Here is what I'm wondering now: April's configuration is a particular case of Emelyanov's configuration from [url]http://forumgeom.fau.edu/FG2001volume1/FG200124.pdf[/url]. Let me rewrite his result:

[b]Theorem[/b]. Let $ P$ be an arbitrary point in the interior of a triangle $ ABC$ and let $ A'B'C'$ be its cevian triangle. Let $ \Gamma_{A}(\rho_{a})$ be the circle tangent to the sideline $ BC$ at $ A'$ and simultanously tangent to the circumcircle of $ ABC$. Similarly, define $ \Gamma_{B}(\rho_{b})$ and $ \Gamma_{C}(\rho_{c})$. Then, the circle tangent to all $ \Gamma_{A}$, $ \Gamma_{B}$ and $ \Gamma_{C}$ (different from the circumcircle of $ ABC$) is also tangent to the incircle.

My question is if we can write the inradius of $ ABC$ as a symmetric expression of $ \rho_{a}$, $ \rho_{b}$, $ \rho_{c}$. If not (though I doubt the answer is negative) try to do this in a few particular cases.


geometry
3D geometry
inequalities
trigonometry
inequalities unsolved

2. the real numbers $x_1, x_2, \cdots$ belong to the interval $[-1,1]$ and the sum of their cubes is zero. Prove that their sum does not exceed  $n/3$.

I have a trigonometric solution with substituting x by sin, do you have any other nicer solutions?

The best solution I know is as follows: substitute $y_i = x_i^3$ so that $y_1 + \cdots + y_n = 0$. In maximizing $\sqrt[3]{y_1} + \cdots + \sqrt[3]{y_n}$, we note that $f(y) = y^{\frac{1}{3}}$ is concave over $[0,1]$ and convex over $[-1,0]$, with $|f'(y_1)| \ge |f'(y_2)| \iff 0 < |y_1| \le |y_2|$. Hence, we may put $y_1 = \cdots = y_k = -1$; $-1 \le y_{k+1} < 0$, and $y_{k+2} = \cdots = y_n = \frac{k - y_{k+1}}{n-k-1}$.  We first show that $y_{k+1}$ leads to a maximal sum of $\sqrt[3]{y_i}$ if it is -1 or can be made positive. If $|y_{k+1}|  < |y_{k+2}|$, we set $y_{k+1}' = y_{k+2}' = \frac{y_{k+1}+y_{k+2}}{2}$, increasing the sum while making $y_{k+1}$ positive. Otherwise, set $y_{k+1}' = -1$ and $y_{k+2}' = 1 - y_{k+1} - y_{k+2}$, again increasing the sum of $\sqrt[3]{y_i}$. Now we may apply Jensen to equate all positive variables, so that we need only show

$k \sqrt[3]{-1} + (n-k) \sqrt[3]{\frac{k}{n-k}} \le \frac{n}{3}$.

But we have

$(n+3k)^3 - 27(n-k)^2k = n^3-18n^2k+81nk^2 = n(n-9k)^2 \ge 0$,

as desired. Particularly, as $k$ is an integer, equality can hold only if $9 | n$ and then if and only if one ninth of the variables $y_i$ are -1 and the rest are 1/8.

I think the trigo solution is better. You only need to use the triple angle formula and all is settled .

Wow.

Let $x_i = \sin(\alpha_i)$, and write $0 = x_1^3 + \cdots + x_n^3 =  \sin^{3}(\alpha_1) + \cdots + \sin^3(\alpha_n) = \frac{1}{4}\left((3\sin(\alpha_1)-\sin(3\alpha_1)) + \cdots + (3\sin(\alpha_n)-\sin(3\alpha_n))\right)$. It follows that $x_1 + \cdots + x_n = \sin(\alpha_1) + \cdots + \sin(\alpha_n) = \frac{\sin(3\alpha_1) + \cdots + \sin(3\alpha_n)}{3} \le \frac{n}{3}$. The only values of $\sin(\alpha)$ which lead to $\sin(3\alpha) = 1$ are $\frac{1}{2}$ and -1. The condition for equality follows.

That is nice.


Επειδή οι ανισότητες είναι το αγαπημένο μου Ολυμπιακό κεφάλαιο να μία ενδιαφέρουσα.

Δείξτε ότι:

$ 1/(a^{n}\plus{}b^{n}\plus{}1)\plus{}1/(b^{n}\plus{}c^{n}\plus{}1)\plus{}1/(c^{n}\plus{}a^{n}\plus{}1)\leq 1$ για $ abc \equal{} 1$

Θέλω να δω τις προσεγγίσεις σας και μετά θα βάλω τη δική μου.

Παραπλανητική θα την χαρακτήριζα  :D  . Αρκεί να μην σε τρομάξει το $ a^{n}$ .
Για να μην βλέπουμε το $ n$ και μας τρομάζει  :)  θέτουμε $ a^{n}\equal{}x^{3}$ ,$ b^{n}\equal{}y^{3}$ , $ c^{n}\equal{}z^{3}$
Τότε προφανώς $ xyz\equal{}1$ και έχουμε νδο
$ A\equal{}\frac{1}{x^{3}\plus{}y^{3}\plus{}1}\plus{}\frac{1}{y^{3}\plus{}z^{3}\plus{}1}\plus{}\frac{1}{z^{3}\plus{}x^{3}\plus{}1}\leq 1$

Τώρα είναι $ x^{3}\plus{}y^{3}\geq xy(x\plus{}y)\equal{}\frac{x\plus{}y}{z}$

Εφαρμόζουμε για όλα τα κλάσματα το ίδιο και έχουμε
$ A\leq\frac{z}{x\plus{}y\plus{}z}\plus{}\frac{y}{x\plus{}y\plus{}z}\plus{}\frac{x}{x\plus{}y\plus{}z}\equal{}1$

QED Are mafiozoi ti mas kanate

Αυτό ακριβώς έκανα κι εγώ και αυτή άλλωστε είναι η επίσημη λύση που βρήκα. Είναι από διαγωνισμό Baltic Way. Απλά επειδή μου φάνηκε περίεργη είπα μήπως υπήρχε κι άλλη.:wink:


Two of the edge lengths of a tetrahedron are 1 and sqrt2. If its vertices are also vertices of some cube, how many such tetrahedra exist?


1) Prove that $\log x$ cannot be expressed as $f(x)/g(x)$, with $f(x)$ and $g(x)$ polynomials over the reals.

2) Prove that $e^{x}$ cannot be expressed as $f(x)/g(x)$, with $f(x)$ and $g(x)$ polynomials over the reals.

For 1) (resp. 2)), compose by exp (resp. ln) and see what happens when you take limits.


[quote="GameBot"]The dots on the number line are equally spaced. What is the value of the common fraction $ x$?

[asy]unitsize(1inch);
for(int i = 0; i < 6; ++i){
dot((i,0));}
draw((-.5,0)--(5+1/2,0),Arrows);
label("$\frac{-5}6$",(0,0),S); label("$\frac14$",(3,0),S); label("$x$",(5,0),S);[/asy][/quote]

$ \minus{}\frac{\minus{}30}{36},\minus{}17/36,\minus{}1/9,1/4,22/36,35/36$.

$ \frac{35}{36}$.


Use the law of energy conservation and the law of momentum conversation to prove that Free Electron can not emit or absorb Light Quantum.
    ( That book gives a hint that to prove   $v_1+v_2 > 2c$ where $c$ is the velocity of light and $v_1,v_2$ are two velocities of something(??), this yields a contradition. But I don't know how to obtain it, please help!)

hi

cuon sach ma darkmaster noi la "giai toan vat li 12 le hong phong" phai khong?
minh cung co post bai nay, chua ai tra loi haaaaaaa :D

absorb case:

energy of free electron and light:(conservation of energy)

$\frac{1}{2}mv1^2+hf=\frac{1}{2}mv2^2$

conservation of momentum

$\frac{h}{lamda}n+mv1(vecto)=mv2(vecto)$
so
on the moving way:

$\frac{m(v2-v1)lamda}{h}\geq 1$

from conservation of energy equation:

$c=lamda/2\frac{(v2-v1)(v2+v1)}{h}$

from above we deduce that $v1+v2\geq 2c$

which is a papradox

phuong phap nay con chua chat lam, minh con cach khac nhung bang relativity :D

on the other hand , in Nobel price , Feynman give the idea that in order to quantuminize the electromagnetism field , we must think that two electrons impacts with each other by
electromagnetic force through the emit and absorb photon, but from the proof above this idea can't be reality. how can you support Feynman idea?(his idea is true and bring him NOBEL PRICE) :lol:

Thanks a lot, Cuong.
     Neu minh ko nham thi ban hoc lop 12A2 phai khong? Dung la ngoa ho tang long.


calculus
derivative
calculus computations

What is the derivative of:

f(x)=(1+x^2)e^(-x^2)

i got f'(x)=(-2x^3)e^(-x^2) , but im not sure if its right...

hello, using the product and the chain rule we get for your $ f(x)\equal{}(1\plus{}x^2)e^{\minus{}x^2}$: $ f'(x)\equal{}2xe^{\minus{}x^2}\plus{}(1\plus{}x^2)e^{\minus{}x^2}(\minus{}2x)$.
Sonnhard.

thanks!

can anyone confirm this answer? or show me how to do it? :P

[quote="Carcul"][f(x)g(x)]' = [f(x)]'g(x) + f(x)[g(x)]'. So, with (1+x^2) = f(x) and e^{-x^2} = g(x), we have:

$ [(1 \plus{} x^2)e^{ \minus{} x^2}]' \equal{} (1 \plus{} x^2)'e^{ \minus{} x^2} \plus{} (1 \plus{} x^2)(e^{ \minus{} x^2})'$

$ \equal{} 2xe^{ \minus{} x^2} \plus{} (1 \plus{} x^2) \cdot ( \minus{} 2x)e^{ \minus{} x^2} \equal{} e^{ \minus{} x^2}(2x \minus{} 2x \minus{} 2x^3) \equal{} \minus{} 2x^3e^{ \minus{} x^2}$.[/quote]

ah ok, so i was correct?

Thanks Carcul!


For some reason, I feel that I have become dumber since last year. I am doing terrible on Mandelbrot compared to last year and my PSAT score decreased. I also seem to have problems concentrating. Sometimes, I make silly spelling and grammatical errors. My mind may tell my hands to write "decided," but "decidid" comes out instead. It baffles me. The problem is not that I don't know proper spelling and grammar; things simply don't come out right. I think it might be caused by sleeplessness and stress, but I am apprehensive.

I have recently come to loathe sleeping. The time between when I close my eyes and when I wake up seems much too short. Thing is, why sleep when it only makes the next day come sooner? All I do in school is sit there in boredom. The only thing we do in AP chemistry and biology is take notes off PowerPoint slides. Instead of taking notes, I just read the textbook.

I also have more stress. Constant boredom has made me disillusioned and burnt-out, while parental expectations and stereotypes have made me hedonistic and cynical. My parents are nearly 50 years old and have begun to become very forgetful, which angers me endlessly. My older sisters never received the wrong type of birthday cake three years in a row. Family and social problems only add to my stress. My strong moral convictions lead me to hate many people for past wrongs, of which there is no end. There is nobody I can talk to for help. My parents are too old and opinionated to be of any help. They hold modern society in contempt. I am also loathe to talk to any of my peers, as they have wronged me and continue to do so.

[hide="Example"]This past weekend, I participated in the state science symposium with my two friends. We left home on Friday, presented on Saturday, and returned Sunday to participate in the "Who Wants To Be A Mathematician?" contest (me and one of my two friends). We were sure that both of us would win in our bracket (500 dollars) and square off (like MathCounts Countdown) for an additional 500 dollars and a shot at the 2000 dollar bonus question (Google for the contest format).

On Saturday night, my friend asked if I would like to come to a small ISEF reunion with him. He told me that we would probably be back by 11 PM. We weren't. We went downtown to a Japanese restaurant and stayed there until 2 AM Sunday morning. By the time 12 AM had passed, I had been getting pretty annoyed and sleepy. However, I could not walk back to the hotel alone (too dangerous). My two friends said that they were going to keep me awake so that I would not be able to participate well (the one participating took a nap earlier in the day). I was angry and sleepy.

Sunday came. I was sleepy and sick. I threw up before the competition and was unable to concentrate. I was going to pull out of the competition, but I stayed. As a result of being sick, I didn't even win my bracket. I made several mistakes because I was sick. It came down to a tiebreaker (also like Countdown) within my bracket. I misread and lost. None of that would have happened if I wasn't sick. My friend squished the guy I lost to and answered the bonus question, winning 3000 dollars.

I was enraged. Their dirty tactic made me lose. I had stayed to watch the rest of the competition. I would have easily beaten my friend at the square off question and got the bonus question. Instead of winning 3000 dollars, I got 0. My reputation was also destroyed. Because I won 1st at the state MathCounts back in middle school, I was previously known as the best in the state. Now, people are coming and insulting me like dirt, while my friend's reputation has ballooned past its abnormally large size. He won the state science fair in 8th and 9th grade and has won national competitions, making him famous in our state. I will never forget this. It is only one time in which I have been wronged.[/hide]

Your take?

Regardless of the reasons for these events, try to get more sleep. Sleep will definitely help you do better in everything (there's a reason we feel tired :wink: ). But mainly it helps to deal better with stress (and finding stress outlets is also very helpful--math and music work well for me).

School sucks for everyone. Just make sure that you know what you need to know, and bring more interesting things to do (like math :) ), or just random homework. When you sleep more it will be easier to learn the school material quickly so you have more time. Of course you still have to do most of what they make you do, but if not doing some busywork will only benefit you then skipping it is fine.

Also for social problems... don't just hate people, because it doesn't help at all. There will always be annoying people who won't ever care enough to understand, and personally I think apathy is the best way to deal with them. Not all opinions truly count. As for your example, I know you feel gipped, but what can you do about it now? There certainly will be more legit opportunities for you in the future. And all you can do is to build on your passions.

Listen, it seems that you just take yourself a wee to seriously. Maybe lighten up a little, take some walks ( stress reliever), read funny books, that sort of thing.

I completely agree with the sleep issue math154 brought up. It's obviously not good for you if you don't get enough sleep. On days when I have to do a lot of stuff and study like crazy, I usually end up getting like 5 to 6 hours of sleep. However, after being adequately rested, I generally work much more productively and efficiently so I need less time. I think that one of the reasons you are regressing is because of your stress levels. I've learned this from experience, try to find a way to release your stress and hopefully not build it up in the first place.

Ways I relax:

* play a musical instrument
* Work on an interesting math problem
* play tennis
* chess
* read a good book

I'd like to add one more:
Watch comedy. If you can find a good comedian that you like (in my case it's John Stewart,) then you will be much more relaxed.

Working on interesting math problems are in no way a way to relieve stress for everyone; it does the reverse of that for me.

Or, for some, write a (good?) book would be in order. As for me: I just started a writing project. The book is set on a frontier planet of the Lyran Star Empire (for those who played Starfleet Command 1/2 or Star Fleet Battles, you should be familiar with the Lyran Star Empire) whose natives award cheating charters, which are a kind of licenses to cheat in school exams. The main character has passed her cheating licensure in first grade.

about comedy: try russel peters :lol: , although some people are easily offended so he is not for the faint of heart


What tends to be an average score on the HMMT test? I took the most recent algebra one and I was only able to get 6 out of 10  :oops:

I know, it is a bad score; 60%.

Uh, they're weighted.

Go look at the results to find what the top 10 are.

i went to the website and i can't find the weighting for the individual tests.

[quote="neelnal"]Oh ok. I got 3+3+4+5+5 = 20 on the algebra. I took a bit more than 1 hour though. :blush: More like 75 minutesish.[/quote]

cool dawg

20's definitely not a bad score. Considering that pretty much all of HMMT except the general test last year was too hard, it's actually a pretty good score :-)

Really? Last year's test was harder than usual? Awesome!


$ (\forall)q \in \mathbb{Q}\cap (0,1)$ show that $ (\exists)n,a_2,a_3,...,a_n\in \mathbb{Z}$ uniques such that :

$ q\equal{}\frac{a_2}{2!}\plus{}\frac{a_3}{3!}\plus{}...\plus{}\frac{a_n}{n!}\ ,\ n>0\ ,\ 0\le a_k<k\ ,\ (k\equal{}2,3,...,n\minus{}1)$ and $ 0<a_n<n$

Exist n, suth that $ q*n!\equal{}q_n<n!$ integer. Then wecan define $ a_k\equal{}[\frac{q_k}{(n\minus{}1)!}], q_{k\minus{}1}\equal{}q_k\minus{}a_k(k\minus{}1)!, q_{k\minus{}1}<(k\minus{}1)!$ for $ k\equal{}n,n\minus{}1,...$


geometry
3D geometry
sphere
rotation
probability
trigonometry
function

Find the average area of a spherical Triangle.

under what conditions?  are you saying if we pick three points randomly on the sphere?  I believe this will require integration.

$ A \equal{} r^2 (\alpha\plus{}\beta\plus{}\gamma\minus{}\pi)$ where $ \alpha$, $ \beta$, and $ \gamma$ are the angles of the triangle.

yes you pick three points randomly.

A better approach would be to fix one point, since we are on a sphere.

wait a second... 3 arcs define two different triangles, one with three huge angles (larger than 180 degrees) and one with three small angles (less than 180 degrees). I suspect you only want me to consider the smaller triangles, but if we consider both, we can take any two triangles determined by the same lines and their areas will add to the surface area of the sphere, so our answer is $ 4\pi r^2/2 \equal{} 2\pi r^2$.

If we only consider smaller triangles...

First, fix one point.  Then we can rotate the sphere so the only freedom the second point has is how far it is from the first.  The 3rd point has the freedom to form whatever angle with the first point and whatever distance it wants.

Clearly each angle is equally likely (angle with the first point as the vertex), but we need the probability distribution for the distances.

Let the central angle between two points be $ \theta$.  Then we have a circle of circumference $ 2\pi \sin \theta$ at a distance of $ r\theta$.  Dividing this by the total surface area, we have $ \frac {\sin \theta}{2r}$ for the pdf of $ \theta$ and thus for the pdf of the distance $ r\theta$.

Now I'm a bit lost.  Not quite sure what to do from here.

i had a cool idea for a solution.

Fix one point in a spherical triangle. Have the angle vary from 0 to 180. now if the arcs continue to the south pole you get a lune. in that lune are two spherical triangles.

the average area of the triangle will then be half that lune.

if you vary the angle, the average area will be achieved at 90 degrees.

at 90 degrees you know the lune is a fourth of the surface area. and since the avg. area is half the lune.

 the average area of a spherical triangle is $ \frac{4 \pi r^2}{8}\equal{}\frac{\pi r^2}{2}$ !

nice solution fraenkel, but I don't think that the average area is achieved at 90 degrees.

The area of a lune with angle $ \theta$ is $ \theta /(2\pi) \cdot 4\pi r^2 \equal{} 2\theta r^2$.  The average value of a function $ f(x)$ on $ [a,b]$ is given by $ \frac{1}{b\minus{}a} \int_a^b f(x) dx$.  Hence we have $ \frac{1}{\pi} \int_0^\pi 2\theta r^2 d\theta \equal{} \frac{r^2}{\pi}(\pi^2\minus{}0) \equal{} \pi r^2$.

Ok so I guess you were right.  Thinking about it now, if we take a hemisphere and two opposing points on its border and divide the hemisphere with a line, then we create two lunes whose areas add to the area of the whole hemisphere, so the average lune has an area that is half a hemisphere, and hence the average lune has a 90 degree angle.

[quote="Fraenkel"]how do you derive that formula for the area of a lune?[/quote]

Simple.  A lune is bounded by two points directly opposite of eachother on the sphere and two lines segments between the two points (semicircles from great circles).   A lune with an angle $ \theta$ covers $ \theta/(2\pi)$ of the sphere because an angle of $ 2\pi$ is a full rotation.  Then we multiply this fraction by the surface area $ 4\pi r^2$.  Hence the area of a lune is $ 2\theta r^2$


[quote="madadpoor"]can we proove that a certain polynomial is reducible or not? (for any polynomial,does exsit a general way?)[/quote]

You should precise what kind of polynomial you are speaking about :

In $ \mathbb C[X]$, each polynomial with degree $ n$ may be factorized in product of $ n$ polynomial of degree $ 1$
In $ \mathbb R[X]$, each polynomial may be factorized in polynomials of degrees at most 2
In $ \mathbb Z[X]$, I dont think there is a general way (except that there are a lot of methods to detect impossibilities


number theory
least common multiple
Gauss
calculus
integration
logarithms
limit

Let's do proofs that there are an infinite number of primes.  Feel free to use whatever conjectures you'd like to use, so long as they are legit.  So, a few quick ones:
Twin Prime Conjecture.
Goldbach's Conjecture-Suppose there were only n primes.  Then there could only be 1/2*n^2 possible even numbers.
Dirilect's Theorem that there are an infinite number of primes in the arithmetic sequence a+bx, where a and b are relatively prime.

The Riemann Hypothesis clearly implies there are infinitely many primes. Anyway, here's my favorite proof:

\[\zeta(2)=\prod_{p\text{ prime}} \left(1-p^{-2}\right)^{-1}=\frac{\pi^2}{6}.\]

Suppose there are finitely many primes. Each of the terms in the product is rational, so the product of all of them must also be rational. But $\frac{\pi^2}{6}$ is irrational. Thus there must be infinitely many primes.

yeah, I think it fits your nickname well :D

no clue what is is anyway...

but I can start stuff like that too :P gauss integral says that the number of primes between a and b can be approximated by $\int{ \dfrac{1}{\ln^2 x}}$ (or something)
and well that one converges to infinity if b goes to infinite :maybe:

[quote="Peter VDD"]the most common way to prove it is that if p were the greatest prime, than lcm (1,2,...,p) is 0 mod 1,...,p so one higher is 0 mod nothing and therefor prime.[/quote]

try p1*p2*...*pn -1 where pn is the last prime...  :D

also, theres a slick proof that shows 1/pi diverges

and then there is the infamous prime number theorem...

Here's a proof that there are infinitely many composite numbers.  Suppose there were a finite number of composite numbers.  Let N be 4 times the product of those composites.  It's easy to see that N is composite, and yet it is bigger than each of the original composites.  So our list of composites must not have been complete, and we have reached a contradiction.  </joke>

One of my professors gave us a proof that there are infinitely many composites last year:

Suppose there are only finitely many composites. Take them all, multiply them together, and DON'T ADD ONE!

Yes, but there is a subtle hole in the proof: You first have to prove that there at least 2 composites.   :)

Yes, that's true. But going through that makes the proof so much less elegant.

$\frac{x}{\log{x}}$ can be used to approximate $\pi{}(x)$, the number of primes less than or equal to x.  Also, some mathematition whose name I do not know proved that

\[
\lim_{x \to \infty} \pi{}(x) = \lim_{x \to \infty} \frac{x}{\log x}
\]

By L'Hopital's (sp?) theorem,

\[
\lim_{x \to \infty} \frac{x}{\log x} = \lim_{x \to \infty} x^2
\]

As x goes to infinite, so does x^2.  So

\[
\lim_{x \to \infty} \pi{}(x) = \infty
\]

which means that there are infinitely many primes.

That's pretty sloppy notation. $\lim_{n\to\infty}\pi(x)=\lim_{n\to\infty}\frac{x}{\log x}$ is true because they're both infinite. Better notation would be something like $\lim_{n\to\infty} \frac{\pi(x)}{x/\log x}=1$.

I wonder how many of these actually involve the fact that there are infintely many primes in their proof?

Anyway, here's one (which may or may not succumb to the above):

There is always a prime between $n$ and $2n$. So $p$ can't be the last prime, because there has to be one between $p+1$ and $2p+2$.

By the way, here's another mallet-used-to-crush-a-walnut from IMO training: Find all integer solutions $m,n$ to $m^3+n^3=8^{30}$. Guess which theorem we used...

[quote="alekk"]hi complexZeta,
I've just read your proof that shows that there are infinitely many primes number. Nevertheless I'm not sure that the product of infinitely rational numbers must be a rational number .. could you prove this fact ?[/quote]

This problem doesn't look too complicated:

A rational number can be written as a/b. Therefore when you multiply any number of rational numbers, you get a fraction. A fraction must be rational.

This proof has lots of holes in it, but you should be able to convince yourself that it's true.

I haven't seen Euler's argument in this thread yet. If $P$ is the set of all primes, the sum $\sum_{p\in P} \frac1p$ diverges. (Something that can only happen if there are infinitely many primes.) Who is familiar with the argument? (It does involve both infinite products and the unique prime factorization of integers.)

alekk wondered if the product of infinitely many rationals must be rational. It does not. The proof I posted is based on the fact that the product of finitely many rationals must be rational.


Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a subring of a field $ F$ such that for each $ x$ in $ F$ we have that either $ x$ in $ R$ or $ {x}^{\minus{}1}$ in $ R$. Prove or disprove that if $ I$ and $ J$ are two ideals of $ R$ then $ I$ is a subset of $ J$ or $ J$ is a subset of $ I$.
Thanks in advance.

Suppose that $ J$ does not contain $ I$ and that $ I$ does not contain $ I$. Then there exist $ x \in I$ and $ y \in J$ such that $ x$ does not generate $ y$ and $ y$ does not generate $ x$. But we have $ \frac {x}{y} \equal{} r_1 \in R$ or $ \frac {y}{x} \equal{} r_2 \in R$. Contradiction..!
What you have proven is that ideals are linearly ordered in such a ring.


inequalities
real analysis
real analysis unsolved

If $f''$ is continuous with $|f''| \leq m$ on $[0,a]$, and $f$ assumes a maximum on $(0,a)$, show

$|f'(0)|+|f'(a)| \leq am$.

Try in following way: assume that $\zeta , \; \zeta \in (0,a) ,$ is the maximum point. Then $f^\prime (\zeta)=0 .$  On the other hand (using Lagrange mean value theorem)
\[\{\begin{array}{lcl}f^\prime(\zeta)-f^{\prime}(0)&=& \zeta f^{\prime\prime}(\xi_{1})\; \; \; ,\; \; \; \xi_{1}\in (0,\zeta) \\ f^\prime(a)-f^{\prime}(\zeta)&=&(a-\zeta)f^{\prime\prime}(\xi_{2})\; \; ,\; \xi_{2}\in (\zeta,a)\; .\\ \end{array} \]
Thus, from \[\{\begin{array}{lcl}-f^{\prime}(0)&=& \zeta f^{\prime\prime}(\xi_{1})\; \; \; ,\; \; \; \xi_{1}\in (0,\zeta) \\ f^\prime(a)&=&(a-\zeta)f^{\prime\prime}(\xi_{2})\; \; ,\; \xi_{2}\in (\zeta,a)\; , \end{array} \] we give
\[\{\begin{array}{lcl}|f^{\prime}(0)|&=& \zeta| f^{\prime\prime}(\xi_{1})|\le \zeta m \\ |f^\prime(a)|&=&(a-\zeta)|f^{\prime\prime}(\xi_{2})|\le (a-\zeta)m \end{array}\]
and by adding these inequalities you find the desired result.


In a multiple-choice test, there are 4 problems, each having 3 possible answers.
In some group of examinees, it turned out that for every 3 of them, there was a question that each of them gave a different answer to. What is the maximal number of examinees in this group?

6 is the maximum. And the 20 different groups can be fitted exactly in the test format.


geometry
geometric transformation
reflection
circumcircle
symmetry
power of a point
radical axis

Let acute triagle $ABC.$ $B', C'$are midpoint of $AC ,AB$,repective and $H$is reflection of $A$ on side $BC.$Prove that: the circumcircle of triangle$AB'C' , BC'H,B'CH$concurrent at a point and side through that point and $H$ passing through midpoint of side  $B'C'$.

[color=darkblue]The point $H$ is defined so :  $AH\perp BC$ and $H\in BC$ ??? I think that "reflection" means "symmetry" and no projection !

If it is so, than this problem is [b]very easily[/b] for [u]Treegoner[/u] or [u]Darij G.[/u] Maybe for mine !

[b]Proof.[/b] Prove easily that the circumcircles of the triangles $AB'C'$ , $BC'H$ , $CB'H$ are concurrently (in the point $P$ ). The line $B'C'$ is common tangent to the circumcircles of the triangles $BC'H$ and $CB'H\ .$ So, the its radical axis $AP$ cuts the common tangent in the its middle !

[b]Remark.[/b] In the picture is a very nice girl ! Who is she ?[/color]

I think that $H$ must be the [b]projection [/b] of $A$ on $BC$. I've made the figure for the two cases and...the plausible is that one !(projection)

On the other hand, I think that the very nice girl seems to symbolize the intersection of three  other "circles": YOUTH, BEAUTY and MELANCHOLY...
:blush:
"A Theorem is a Sentiment"- Grigore C.Moïsil
What do you think about  that [color=red]zaizai[/color] ?

[quote="mstoenescu"]On the other hand, I think that the very nice girl seems to symbolize the intersection of three  other "circles": YOUTH, BEAUTY and MELANCHOLY... . What do you think about  that [color=darkred]zaizai[/color] ?[/quote]
[color=darkblue]I think that you forgot ... the fourth circle : "INTELLIGENCE". The "Melancholy" is a consequence of the "Intelligence".
The miss (princess) Zaizai-Hoang posted some very nice and difficult (at least for my) problems. I wish as more girls as possibly be like she.[/color]

Yes, that H must be the projection  of A on BC
 PS: My English is not good and I used to get mistake in vocabulary and grammar  :(
[quote]In the picture is a very nice girl ! Who is she ? [/quote]
She is my friend  :D And I am a boy  :D


Hello.

Let ABC be an isosceles triangle with AB=AC. Supose that the angle bisector of angle B meets AC at D and that BC= BD+AD. Determine angle A.

I'll take it.  :wink:

Point $E$ is chosen on $\left[BC\right]$ so that $\left|BE\right| = \left|BD\right|$

$\frac{\left|AB\right|}{\left|BC\right|}= \frac{\left|AD\right|}{\left|DC\right|}= \frac{\left|CE\right|}{\left|CD\right|}$

And a common angle gives that $\Delta DCE$ and $\Delta BCA$ are similar. Hence, $ABED$ is cyclic.

Call $\alpha = \angle BAC$:

Hence, $\pi = \angle BED+\angle BAD = \frac{3\pi+\alpha}{8}+\alpha$
So $\alpha = \frac{5\pi}{9}(= 100^{\circ})$.

Umm how does

$\frac{\left|AD\right|}{\left|DC\right|}= \frac{\left|CE\right|}{\left|CD\right|}$

Am I missing something.. or is that a typo..cuz the denominator for both is CD???

$\left|AD\right| = \left|CE\right|=$ [b]given[/b]!!!  :maybe:

Sorry...ya it does..i missed that..thank you..


[quote="moldovan"]Prove that:
\[ (x \minus{} y)^{2009}(x \minus{} z)^{2009} \plus{} (y \minus{} x)^{2009}(y \minus{} z)^{2009} \plus{} (z \minus{} x)^{2009}(z \minus{} y)^{2009} \ge 0\]
where $ x,y,z \in \mathbb{R}$.[/quote]

Assume that $ x\ge y \ge z$  Then  For all $ n \in \mathbb{N}$ we have :

$ (x \minus{} y)^n(x \minus{} z)^n \ge (x \minus{} y)^n(y \minus{} z)^n$  and ,
$ (z \minus{} x)^n(z \minus{} y)^n \ge 0$

And the result follows  :P


Suppose we have $2n$ positive real numbers $x_1,...,x_n,y_1,...,y_n$ such that
 $x_1\leq x_2\leq ...\leq x_n\leq y_1\leq y_2\leq ...\leq y_n$.

Prove that $(x_1+x_2+...+x_n+y_1+y_2+...+y_n)^2\geq 4n(x_1y_1+x_2y_2+...+x_ny_n)$

We know that $(y_2-y_1)(y_1-x_2)+(y_3-y_1)(y_1-x_3)+...+(y_n-y_1)(y_1-x_n)\geq 0$ what rewrites as
$x_1y_1+...+x_ny_n\leq y_1(x_1+...+x_n+y_1+...+y_n) - ny_1^2$.
Denoting $A=\sum_{i=1}^{n}x_i + \sum_{i=1}^{n}y_i$, it is therefore enough to show that
$A^2\geq 4nAy_1 - 4n^2y_1^2$
which is equivalent to
$(A-2ny_1)^2\geq 0$,
QED.

I just wanted to add that my proof remains valid if we only impose the condition that $x_i\leq y_j$ for all $i,j$, but no condition between the $x_i$ and the $y_i$. It is clear that this additional assumption is useless because of the rearrangement inequality. The positivity of the numbers is not needed as well.
To make the proof valid, we just need to assume wlog that $y_1$ is the smallest out of $y_1,...,y_n$(which can be done by a cyclic shift).

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46406
Is there a connection between these inequalities ? In my opinion, the proofs are somehow similarly for both inequalities.

The connection is just that both inequalities yield a trivial inequality after the variable shift $x\to x+a$ for some $a$. In your inequality, one takes $a=-1$, in this inequality, any value of $a$ with $x_n\leq a\leq y_1$ is ok. one might invent dozens of such inequalities by writing down a trivial inequality and shift it.
(The inequality here gets after shifting $(x_1+...+x_n+y_1+...+y_n)^2\geq 4n(x_1y_1+...+x_ny_n)$ where $x_i\leq 0$, $y_i\geq 0$; of course, one may put 0 between both sides. Similarly, virgils inequality gets $c_1^2+...+c_n^2+d_1^2+...+d_n^2\geq \frac 1n(c_1d_1+...+c_nd_n)$ where $c_i\leq 0$, $d_i\geq 0$. Again, $LHS\geq 0\geq RHS$.)


Find the focus of the  rectangular hyperbola $ (x\plus{}4)(y\minus{}4)\equal{}16$

If $ xy\equal{}4^2$, foci are $ (\pm4\sqrt{2}, \pm4\sqrt{2})$.
So foci of $ (x\plus{}4)(y\minus{}4)\equal{}16$ are $ (\minus{}4\pm4\sqrt{2}, 4\pm4\sqrt{2})$.


Dear pardesi, please dont delete this thread, plz....

Which is the coolest element??

I think its hydrogen.

Please vote even if you like 2 or more elements.

my favorite is [url=http://en.wikipedia.org/wiki/Roentgenium#Proposed_name]unununium[/url] (Uuu) . sorry, but that isn't on the list   :D

gr8 mids think alike. my fav is unununium too.

I refuse to vote not because I am stupid but there is no option mentioning "NONE OF THESE". :D

My favourite element is Manganese(Mn) :) . I like the exceptional characteristics that it posseses.

My favourite element is Mercury. You can guess why.

[quote="achu_182"]my favourite element is lithium, coz theres a good song by evanescence by the name lithium![/quote]
acvtually dere's a song by nirvana by the same name too.

i know that the song by nirvana is by the same name, but i don't like it....infact, i don't like nirvana at all!!

my fav --Carbo,Boron & Fluorine
   try out speaking together their symbols!!!!!!! :rotfl:  :rotfl:  :rotfl:

Would it be prudent to say that my favourite element is antimony? Guess not.... :D I'd probably say fluorine... I can relate to that element :D

hey hav u heard the periodic table song????????

yay,i m in yang mills

search in google
i myself don't have
i heard it in a camp

i've NEVER went to an olympiad camp :huh:
and will never b able 2 reach an international olymp camp :(  :(

y u r a genius
and anythings possible for u
i read wen u said that without preparation u gave RMO and qualified so that completes the picture
u just try and the win will be on ur side
and also "FORTUNE FAVOURS THE BRAVE "
And so r u
as ur name suggests RITURAJ quite bold i believe!
bye!

:rotfl:  :rotfl:
u turned a bit senti :P

:rotfl: r u sure
did i went too senti
achha tell me that wat do u consider or wat r ur thoughts on intel isef!

[quote="rituraj007"]and will never b able 2 reach an international olymp camp[/quote]

 :rotfl:  :rotfl:
common
stop joking

hey skand u also tell me wat r ur thoughts on intel isef!

i don't have any idea
but i've tried searching and now i am thoroughly impressed
but i am a lazy one,so i can never participate :(

so you go and win there
best of luck
 :thumbup:

thats my true friend
and wat r ur thoughts on it rituraj007

best of luck
be the champion :)

but just tell wat u r making!!!!!!

oh i am so  sorry!
ill pm u!
and u reply me!


[quote="amir2"]if $x+y+z=1$ prove :

$x^3+y^3+z^3+6xyz\ge \frac 14$[/quote]

known I think.

homogenize to get

$x^3+y^3+z^3+6xyz\ge \frac{(x+y+z)^3}{4}$ $\Longleftrightarrow$

$x^3+y^3+z^3+8xyz\geq (x+y)(y+z)(z+x)$

Recall Schur's ineq $x^3+y^3+z^3+5xyz\geq (x+y)(y+z)(z+x)$

now it is obvious!

Of course, it is necessary $a, b, c >0$
(otherwise try x=-1, y=1, z=0.)

There is a stronger ineq:

$x^3+y^3+z^3+\frac{15}{4} xyz\ge \frac 14$
which ends up in Schur!

 :)


What is the sum of the "infinite square root"  :sqrt:(2 +  :sqrt:(2 +  :sqrt:(2 +  :sqrt:(2 + ...   ?

For those of you who have seen tricks like this - hush :).

an more exciting challenge that was in advanced section.
 goto a previous message that is very similar. that i sort of ruined. :(
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9954&highlight=nested+radicals[/url]

hint: if you give up. DON"T LOOK UNLESS YOU ARE DESPARATE

[hide]let x=something

and set up an equation

[/hide]


I applied and got accepted into SUMaC 06, and was wondering if anyone else here applied.

Also, if anyone has attended in the past, please let me know if you feel it was worthwhile, because there is a lot that could be accomplished in those 4 weeks.

Try the Other US Contests and Programs forum -- there's already a discussion about SUMAC going on there.


I will start sending out the Round 4 questions.  There is still one matchup from round 3 going on but I ll just work around that.  I will be giving a differerent set of questions to the loser's bracket opposed to the winner's bracket.  Saturday is the deadline.
[size=150][b]
Round 4 Matchups[/b][/size]

Winner's bracket

Dune    vs.    yif man12

Lucky707  vs.  TBD

Loser's bracket

h_s_potter2002  vs.  theone853

Singthesorrow   vs.   kool_dudy

mlee06     vs.  soccerplayer

Everyone else has a bye or is unfortunately eliminated. The updated bracket is attached.

All I can say afte that round is, I wish you well in the next rounds theone853.

Come on - everyone be a little more optimistic. I'm tired of people degrading themselves. Everyone put effort into making it this far, and everyone should be proud nevertheless.

[quote="theone853"]What makes you say that? Come on... have some faith in yourself. I'm sure you did a really good job. I mean you beat out so many other opponents to get here.[/quote]

First, I only beat one person.  Second, I only got 1 out of 5.

First,  I made the questions harder this week.  Second you were very close on all of them.  I think that if you had time to check over your work you would have gotten more.  So don't feel so bad.

Wow, I got at least close?

Anyways, I'm not saying that I feel bad.  It's just that after you try doing problems that barely know how to do, you get the feeling that a person who was a mock AIME qualifier is going to do better than you.

Crud, I'm losing computer access for the next couple days. At least that'll give some time for the jet-lag to wear off.  :lol: Even getting online's hard with a 40 KBps connection.  :maybe:

I just want to say that, if there were tiebreakers, I wouldn't be here until Sunday around 5 PM, which corresponds to Sunday 11 AM in the west of the US I think.

Hey joml

Can you post some of the scores for people that have done the round?

Round Update:

Matches:
theone853    def.    h_s_potter2002 3-1

people who have sent in answers but their oppenents haven't:

Dune-5
kool_dudy-2
mlee06-5

I need answers from the following people(by tomorrow if possible)

yif man12

Lucky707

mat_88

singthesorrow

soccerplayer

If you are on this last list and haven't received questions please inform me.  If you have sent me answers and you are on this list I might have deleted them... I doubt this has happened though.

Ugh, non-winners bracket, here I come. I have a really bad feeling about most of my answers this round.

In my opinion, this round was real hard for me.  I am surprised I was able to answer more than one question.  I'm more surprised that there are currently two people with perfect score.

Just a reminder to Singthesorrow and soccerplayer.  I need your answers TODAY.  If I don't have them you will have a loss recorded.

Have the scores and winners been posted for Round 4 yet?

The winners are the pople in the next round, losers aren't, unless they were in the winner's bracket.

Not possible

Yif man and Dune were facing off in the winners bracket in round four and neither have a match in round five

Im not having the winners participate in round 5.  If they did they wouldn't have a mach for another round or so.

Here are the results:

Dune   def   Yifman  5-4

mat_88 def  Lucky707 5-4

theone853  def  h_s_potter2002 3-1

kool_dudy defeats singthesorrow 2-0

mlee06 def soccerplayer 5-forfeit

The updated bracket is attached... if there are any errors please tell me.


In a two-dimensional coordinate system, what is the sum of the coordinates of the point $ \frac23$ of the distance from the origin to point P($ 21, \minus{}12$)?

you find 2/3 of 21 and -12 which is

                    (14,-8)

it says the sum of the coordinates, which is 6


Find all pairs of positive integers $ (n,m)$ such that $ 3^n\minus{}7^m\equal{}2$

I heard that less than 5 students solved it :(
(I also couldn't solve it.)

If $ m\equal{}0$, then $ n\equal{}1$.
If $ m>0$, then $ n\equal{}2\mod 6$ even. Consider by mod 5 give m odd.
Consider Pell's equation $ x^2\minus{}7y^2\equal{}2$. There are only one of solutions, suth that $ x\equal{}3^k$. It give $ n\equal{}2,m\equal{}1$.

Try "jumping method" :) , I learned it from [b]Ali666[/b] and it's sure way in such problems :D :
http://www.mathlinks.ro/Forum/viewtopic.php?t=211811
http://www.mathlinks.ro/Forum/viewtopic.php?t=211812

So when $ n \le 2 \vee m \le 1$ it's obvious. Unless we have $ a \equal{} n \minus{} 2 \ge 1, \ b \equal{} m \minus{} 1 \ge 1$ and the equation looks:
$ 9(3^a \minus{} 1) \equal{} 7(7^b \minus{} 1)$

And in sequence:
$ (I) \ 7|(3^a \minus{} 1) \Rightarrow 6|a \Rightarrow 13|728 \equal{} (3^6 \minus{} 1)|(3^a \minus{} 1) \\
(II) \ 13|(7^b \minus{} 1) \Rightarrow 12|b \Rightarrow 43|(7^{12} \minus{} 1)|(7^b \minus{} 1) \\
(III) \ 43|(3^a \minus{} 1) \Rightarrow 42|a \Rightarrow 7^2 \equal{} 49|(3^{42} \minus{} 1)|(3^a \minus{} 1)$

From (III) we get $ 49|9(3^a \minus{} 1)$ but it causes: $ 49|7(7^b \minus{} 1)$ - contradiction. So when $ n > 2 \vee m > 1$ that equation hasn't got any sollution.

Knowing that method it's very easy to solve [b]every[/b] equation of that kind, the one problem is factorization of numbers $ (x^y \minus{} 1)$, but I'm always doing it by taking random prime greater than recent (in that problem I took 7 -> 13 -> 43 and it's done).

P.S. Pell equations are to heavy weapon for that problem  :P
P.S.2. Not always there is: $ p|(x^y\minus{}1) \Rightarrow (p\minus{}1)|y$, it was rather luck that in this problem there is such divisibility ;)

[quote="Sylwek"]Try "jumping method" :)
[/quote]
Does anybody have any notes for the ''jumping method''
Please pm me....

Rather not, because it's my offer for name for that method  :D  and I learned how to use it from that link, where [b]Ali666[/b] used this method in solving such equation.

Essentially you go back and forth using Fermat to get restriction after restriction until you find one that works. It's an extremely effective brute forcing method.


function
algebra
functional equation
algebra unsolved

[quote="pavel kozlov"]Can anybody give example of a dense function, which satisfying to functional equation [b]f(x+y)=f(x)+f(y)[/b] ?[/quote]

Привет Павел  :)

Leepakhin, I think "dense function" should mean that the graph of the function is anywhere dense in the plane, so that f(x) = kx is not a valid answer.

What Pavel is asking for are the famous non-continuous solutions of Cauchy's functional equation f(x+y) = f(x) + f(y), whose graphs are anywhere dense in the plane (see http://www.mathlinks.ro/Forum/viewtopic.php?t=28183 ). In fact, it is well-known that, assumed the axiom of choice is true, such solutions exist, but whether you can construct them depends on what exactly you call "construct" (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=130693#p130693]http://www.mathlinks.ro/Forum/viewtopic.php?t=19470 post #4[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=376413#p376413]http://www.mathlinks.ro/Forum/viewtopic.php?t=62795 post #6[/url] for a construction using a Hamel basis of $\mathbb{R}$).

  Darij

I remember I saw the following result (don't remember where and how they proved it but it was elementary) :

$f$ is a non linear solution of Cauchy equation iff the graph of $f$ is dense in $\mathbb{R}^2$.

So just take any non linear solution of Cauchy's (which, btw, exists only if you accept Choice Axiom :P)

Thank you, Darij.
I know, that it is well-known Cauchy's functional equation, but I early can't find any non-trivial solution of this.


I have a problem here.
[url=http://imageshack.us][img]http://img80.imageshack.us/img80/9990/rootio6.jpg[/img][/url]
[url=http://g.imageshack.us/img80/rootio6.jpg/1/][img]http://img80.imageshack.us/img80/rootio6.jpg/1/w345.png[/img][/url]

This is a homework were I have to solve the roots and order them from greatest to least. But specifically, my problem is with the roots greater than square. I have no idea how to solve them. I know how to solve square roots, but roots greater than square seem to have different way of solving, and I can't solve them. No idea. Can anyone please help me with these?

this is kinda flike finding a common denominator, except here you find a common root index thingy
so for the first one, try putting everything under a common root

[hide="solution for first one"]
the root indexes are 2, 5, and 10
their LCM is 10 so we will put them all under a 10th root
you get $ \sqrt[10]{3^5}\equal{}\sqrt[10]{243},\sqrt[10]{2^2}\equal{}\sqrt[10]{4},\sqrt[10]{5}$
so we can now compare their radicands (numbers under the root at least i think that is what it is)
and order them
[/hide]

now try the second one

Similar to solving fractions (by having a common denominator), you must have a common index in order to proceed to multiplication. In number 2, you have to have these factors with index 6. Actually, the answer is $ \sqrt[6]{2^3 3^4 5^2}$

Thanks both of you. Thanks to you I could solve my problem! :D

corps, I'm sure you had the right idea, but the problem was not to multiply, but to order from least to greatest.

So it would have been
$ 2^3\equal{}8$
$ 5^2\equal{}25$
$ 3^4\equal{}81$

Giving us $ 2^3,5^2,3^4$


Dears
I work as a web developer, and sometimes when I try to read more advanced books (last one was about image processing matrix) I found a lot of algorithms and equation, but unfortunately I didn’t study computer science so I haven’t the enough math background to read such books, what do you advice? Is there a way out?

Thank you for your time.

O' reilly makes a bunch of mastering algorithims with X books, where X is your favorite language.


A proper factor of an integer $ N$ is a positive integer, not $ 1$ or $ N$, that divides $ N$.

(i) Show that $ 3^{2} \times 5^{3}$ has exactly 10 proper factors. Determine how many other integers of the form $ 3^{m} \times 5^{n}$ (where $ m$ and $ n$ are integers) have exactly $ 10$ proper factors.

(ii) Let $ N$ be the smallest positive integer that has exactly $ 426$ proper factors. Determine $ N$, giving your answer in terms of its prime factors.

[quote="AndrewTom"]A proper factor of an integer $ N$ is a positive integer, not $ 1$ or $ N$, that divides $ N$.

 Show that $ 3^{2} \times 5^{3}$ has exactly 10 proper factors. Determine how many other integers of the form $ 3^{m} \times 5^{n}$ (where $ m$ and $ n$ are integers) have exactly $ 10$ proper factors.

(ii) Let $ N$ be the smallest positive integer that has exactly $ 426$ proper factors. Determine $ N$, giving your answer in terms of its prime factors.[/quote]

[b](i)[/b] $ (2 \plus{} 1) \times (3 \plus{} 1) \minus{} 2 \equal{} 10$.  $ 3^m \times 5^n$ has $ 10 \plus{} 2 \equal{} 12 \equal{} 2^2 \times 3$ positive factors. So $ (m \plus{} 1)(n \plus{} 1) \equal{} 12$
$ \Leftrightarrow (m,n) \equal{} (0,11),(11,0),(1,5),(5,1),(3,2)$

[b](ii)[/b] $ 428 \equal{} 2^2 \times 107$, to minimize $ N$, we let $ N \equal{} 2^x3^y5^z$. Clearly $ x \ge y \ge z$ minimize $ N$.

$ 428 \equal{} 428 \times 1 \times 1 \Rightarrow (x,y,z) \equal{} (427,0,0) \Rightarrow N \equal{} 2^{427}$
$ 428 \equal{} 214 \times 2 \times 1 \Rightarrow (x,y,z) \equal{} (213,1,0) \Rightarrow N \equal{} 2^{213}3$
$ 428 \equal{} 107 \times 4 \times 1 \Rightarrow (x,y,z) \equal{} (106,3,0) \Rightarrow N \equal{} 2^{106}3^3$
$ 428 \equal{} 107 \times 2 \times 2 \Rightarrow (x,y,z) \equal{} (106,1,1) \Rightarrow N \equal{} 2^{106} \times 3 \times 5$

Among them $ N \equal{} 2^{106} \times 3 \times 5$ is the smallest.


limit
trigonometry
real analysis
real analysis unsolved

[color=darkred][b]Using $ \lim_{x\to 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}$ show [u]without the l"Lospital's rules[/u] that $ \lim_{x\searrow 0}\ \frac{1}{x}\arccos\left(\frac{\sin x}{x}\right)=\frac{\sqrt 3}{3}$.[/b][/color]

$ \arccos$ is continuous function. So $ \lim_{x\searrow 0}\arccos(\frac{\sin x}{x})=\arccos\lim_{x\searrow 0}\frac{\sin x}{x})=\arccos 1=0.$ Am I missing something?

[quote="Ognjen09"]$ \arccos$ is continuous function. So $ \lim_{x\searrow 0}\arccos(\frac{\sin x}{x}) =\arccos\lim_{x\searrow 0}\frac{\sin x}{x}) =\arccos 1 = 0.$ Am I missing something?[/quote]
[color=darkblue]You proved only that you are in the irresolute case $ \frac{0}{0}$ and you didn't solve the limit $ \lim_{x\searrow 0}\ \frac{1}{x}\arccos(\frac{\sin x}{x}) =\frac{\sqrt 3}{3}$.[/color]

[quote="Virgil Nicula"][color=darkred][b]Using $ \lim_{x\to 0}\frac{x\minus{}\sin x}{x^{3}}\equal{}\frac{1}{6}$ show [u]without the l"Lospital's rules[/u] that $ \lim_{x\searrow 0}\ \frac{1}{x}\arccos\left(\frac{\sin x}{x}\right) \equal{}\frac{\sqrt 3}{3}$.[/b][/color][/quote]

$ \lim_{x\to 0}\frac{1}{x}\cdot\arccos\left(\frac{sin(x)}{x}\right)\equal{}\lim_{x\to 0}\frac{1}{x}\cdot\arcsin\left(\sqrt{1\minus{}\left(\frac{sin(x)}{x}\right)^{2}}\right)$

$ \equal{}\lim_{x\to 0}\frac{1}{x}\cdot\arcsin\sqrt{x^{2}\frac{(x\minus{}sin(x))}{x^{3}}\cdot\frac{(x\plus{}sin(x))}{x}}\equal{}\lim_{x\to 0}\frac{1}{x}\cdot\arcsin\left(\frac{x}{\sqrt{3}}\right)\equal{}$

$ \equal{}\lim_{x\to 0}\frac{1}{\sqrt{3}}\cdot\frac{\arcsin\left(\frac{x}{\sqrt{3}}\right)}{\frac{x}{\sqrt{3}}}\equal{}\frac{\sqrt{3}}{3}$

Note: for the third step I used the well-known relation :$ \lim_{x\to 0}\frac{sin(x)}{x}\equal{}1$ and the hypothesis.
As for the last step we have $ \lim_{x\to 0}\frac{\arcsin(x)}{x}\equal{}\lim_{y\to 0}\frac{y}{sin(y)}\equal{}1$ ,by plugging $ y\equal{}\arcsin (x)$ ,and we see that it's true from the relation above.

[quote="Ognjen09"]$ \arccos$ is continuous function. So $ \lim_{x\searrow 0}\arccos(\frac{\sin x}{x}) =\arccos\lim_{x\searrow 0}\frac{\sin x}{x}) =\arccos 1 = 0.$ Am I missing something?[/quote]
We can't do it for $ arccos$in point $ 1$. look the picture
http://en.wikipedia.org/wiki/Image:Asin_acos_plot.svg
also: good work Svejk  :)


The sum of two whole numbers is 9, and their positive difference is 5. What is the product of the numbers?

$ x\plus{}y\equal{}9$, $ x\minus{}y\equal{}5$.  Add equations to find $ 2x\equal{}14 \implies x\equal{}7 \implies y\equal{}2$.  Multiply to find the answer of $ \boxed{14}$.


trigonometry
inequalities proposed
inequalities

If you will find it very easy sorry.

Find the minimum value of the expression   $A=\frac{2x^2+(cosx)^2}{x^2+1}$

Since $|x|\geq |\sin x|\Longleftrightarrow x^2\geq \sin ^2 x,$

$\frac{2x^2+\cos ^ 2 x}{x^2+1}=1+\frac{x^2-\sin ^ 2 x}{x^2+1}\geq 1.$

Thus the minimum value is $1.$ at $x=0.$


It will be really stupid to ask whether there are more irrational numbers or rational numbers because both are infinite, but it is quite sensible to ask this :-

If you pick a number between 1 and 2 randomly, what's the probability that it is rational?

$0$.

Because the set of the rationals is countable so its mesure is $0$.

Pierre.


When clicking on the ask on forum button, a new window or tab is suppose to appear.  Depending on your browser settings, your screen may not change and you will have to specifically click on the new tab or window to see the forum page.

[quote="BOGTRO"]When I click on "Ask on Forum", I get a message saying

"You are not authorized to access this page"[/quote]

Send a private message to Valentin Vornicu about this.  He's the forum god.

When I click on "Ask on Forum", a new tab or window doesn't pop up. Nothing happens at all.


The following question was asked at the NSW Mathematics Assessment. The correct answer is 45 meters, but I'm at a loss on how to work it out. Can someone please help me? (show working and reasoning please)Thanks  :P

"A car follows another car on a straight stretch of open highway, both cars are driving at 100 kmph and are 150 meters apart.

The cars approach a town. When each car reaches the town it slows down at the same rate to 30 kmph. Both cars travel through the town at 30 kmph.

How far apart are the cars when they are both travelling at 30 kmph?"

[hide]
if they are going at the same rate, and accelerate at the same rate, what about the time in between them?

think of this, say you have a mark on the ground, and the first car starts a timer, and when the second car hits the marker, it stops the timer, what can you say about this time?

[hide] the times are the same, so we have that

$\frac{d_1}{r_1}=\frac{d_2}{r_2}$

$\frac{100}{150}=\frac{30}{x}$[/hide][/hide]

[hide]When the first car reach the town, they are still 150m away. When the cars go into town, their speed slows to 3/10 their original speed. Their distance must also shorten by the same factor since they are at a constant speed, so their distance shortens to 3/10*150 = 45 m.  [/hide]


linear algebra
matrix
linear algebra unsolved

Find all endomorphismes of $M_n(C)$ such that whenever $A$ is nilpotent, $f(A)$ is nilpotent.

Is it of the form $f(A)=XAY$ with $X.Y=I_n$ ?

No, it can also be of the form $ A^{t}XA^{-1}$ and you know I don't like this answer. Too bad I'm not moderator, I would have deleted the answer.  :P . By the way, the source is wrong, as well as my initial solution. The problem seems to have been solved only recently. Sorry for posting this.

Sorry, wrong post :blush:

We consider $\mathbb{K}$ a zero characteristic field and do not limit the answer to $\mathbb{C}$.

Denote by $f$ a linear map $f~: ~\mathcal{M}_{n}(\mathbb{K}) \to \mathcal{M}_{n}(\mathbb{K})$ such that $f(XY ) = f(X)f(Y )$ for all nilpotent matrices $X$ and $Y$ .
We show that either $f=0$ or there exists $P\in \mbox{GL}_{n}(\mathbb{K})$ such that $f(M)=PMP^{-1}$ for all $M \in \mathcal{M}_{n}(\mathbb{K})$.

The case $f=0$ is obvious so let's assume that $f\not=0$.

Denote by $E_{i,j}$ the matrice of $\mathcal{M}_{n}(\mathbb{K})$ with all coefficients equal to 0 except on row $i$ and column $j$, where it equals 1 and $F_{i,j}=f(E_{i,j})$.

We proceed in two steps.

[b][u]Lemma 1:[/u][/b] $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$ such that $F_{i,j}F_{k,l}= \delta_{j,k}F_{i,l}$.

[b][u]Lemma 2:[/u][/b] If $\{M_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$ such that $M_{i,j}M_{k,l}=\delta_{j,k}M_{i,l}$, then there exists $P\in \mbox{GL}_{n}(\mathbb{K})$ such that $M_{i,j}=PE_{i,j}P^{-1}$ for all $i,j\in \{1,\ldots , n \}$.

[b][i]Proof of Lemma 1:[/i][/b]
If $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ was not a basis of $\mathcal{M}_{n}(\mathbb{K})$, we could find $\{\alpha_{i,j}\}_{i,j\in \{1,\ldots , n \}}\in \mathbb{K}^{n^{2}}$ with at least one $\alpha_{l,p}\not=0$ and $\sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}F_{i,j}=0$

For all $i,j,k,l \in \{1,\ldots , n \}$ we have $E_{i,j}E_{k,l}= \delta_{j,k}E_{i,l}$ so $F_{i,j}F_{k,l}= \delta_{j,k}F_{i,l}$.
This last relation is obvious if $i\not=j$ and $k\not=l$, because $E_{i,j}$ and $E_{k,l}$ are nilpotent. But also $f(E_{i,i}E_{k,l})= f(E_{i,j_{0}}E_{j_{0},i}E_{k,l})=f(E_{i,j_{0}}\delta_{i,k}E_{j_{0},l})=\delta_{i,k}F_{i,l}$ (where we choose $j_{0}\not=i$ and $j_{0}\not=l$).

There is at least one $F_{k,q}\not=0$ because $f\not=0$, $F_{k,l}\times \sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}F_{i,j}\times F_{p,q}= \alpha_{l,p}F_{k,q}= 0$ which is a contradiction, so $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$.

[b][i]Proof of Lemma 2:[/i][/b]
$M_{i,k}M_{k,j}=M_{i,j}$, so $\mbox{Im}M_{i,k}=\mbox{Im}M_{i,j}$, let's denote this space by $\mathcal{V}_{i}$.

There exist $\{\alpha_{i,j}\}_{i,j\in \{1,\ldots , n \}}\in \mathbb{K}^{n^{2}}$ such that $I_{n}= \sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}M_{i,j}$.
If $X \in \mathbb{K}^{n}$, $X=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i,j}M_{i,j}X$, and $\sum_{j=1}^{n}\alpha_{i,j}M_{i,j}X \in \mathcal{V}_{i}$, so $\mathbb{K}^{n}= \mathcal{V}_{1}+\mathcal{V}_{2}+\cdots+\mathcal{V}_{n}$.

If $X_{i}\in \mathcal{V}_{i}$, and $X_{1}+X_{2}+\cdots+X_{n}=0$ then $M_{i,i}(X_{1}+X_{2}+\cdots X_{n})=X_{i}=0$, so $\mathbb{K}^{n}= \mathcal{V}_{1}\oplus \mathcal{V}_{2}\oplus \cdots \oplus \mathcal{V}_{n}$.

If $k\not=j$, $\mbox{Im}M_{i,k}\subset \mbox{Ker}M_{i,j}$, so $\bigoplus_{k=1, k\not=j}^{n}\mbox{Im}M_{i,k}= \mbox{Ker}M_{i,j}$ (because they have same dimension and the previous inclusion).

Let $Y_{1}\in \mathcal{V}_{1}$, such that $Y_{1}\not=0$. For $i \not= 1, Y_{i}=M_{i,1}Y_{1}$.
We have $M_{i,j}Y_{j}=Y_{1}$.
So $\{Y_{1}, \ldots, Y_{n}\}$ is a basis of $\mathbb{K}^{n}$.

We define $P \in \mbox{GL}_{n}(\mathbb{K})$ the matrix whose columns are the $Y_{i}$. We have $PE_{i,j}=M_{i,j}P$, which concludes the proof of lemma 2.

Since $f$ is linear we have proven that $f(M)=PMP^{-1}$ for all $M \in \mathcal{M}_{n}(\mathbb{K})$.


A hexagon is inscribed in a circle. Proceeding clockwise the lengths of its edges are $1, 1, 1, 2, 2, 2$. What is the area of this hexagon?

[hide="Solution"]Let $\alpha$ be the central angle corresponding to the side of the length $1$ and $\beta$ the central angle corresponding to the side of the length $2$. If we rearrange the sides thus: $1,2,1,2,1,2$, we see that $\alpha+\beta=120^\circ$. Hence if the radius of the circle is $r$, then in the triangle $1,2,r\sqrt{3}$ there's an angle $\frac{180^\circ-\alpha}{2}+\frac{180^\circ-\beta}{2}=120^\circ$ between $1$ and $2$. Applying Cosine Law we find

$3r^{2}=1^{2}+2^{2}-2\cdot 1\cdot 2\cos 120^\circ=7\iff r=\sqrt{{7\over 3}}$

Now $S=3\cdot{1\over 2}\sqrt{{7\over 3}-{1^{2}\over 4}}+3\cdot{2\over 2}\sqrt{{7\over 3}-{2^{2}\over 4}}={3\over 2}\cdot{5\over 2\sqrt{3}}+3\cdot{2\over\sqrt{3}}={13\over 4}\sqrt{3}$[/hide]


let f bea non constant poly. with integer coeff. s.t. f(i) is a perfect p th power for i=0,1,...,p-1 and p is aprime no.what is minimum possible degree of f? :maybe:

Minimum possible? Zero. Put $f(x)\equiv 2^{p}.$

p-1 works can be shown using linear algebra but is it minimum for 3 1 works.

Sorry, I totally haven't seen "non-constant"  :maybe:

If $f(i)=i^{p}$ for i=0,1,...,p-1  it is obviosly, that $deg(f(x))\ge p-1$ (wronskian is not equal to 0).  If $f(i)=m(i)^{p}$, then we can construct sequence of differences $\Delta f(i)=f(i+1)-f(i), \Delta^{j+1}f(i)=\Delta^{j}f(i+1)-\Delta^{j}f(i)$. If f is polinom degree k, then $\Delta^{k+1}f =0$. I think, that is not hard to prove $\Delta^{K}m(i)^{p}\not =0$, when k<p.

This problem was proposed for Pascual Restrepo (he is in mathlinks) so meybe he can give a complete solution


Prove that in a triangle $ ABC$,

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} 4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

[hide]
$ \cos A \plus{} \cos B \minus{} \cos (A\plus{}B)$

=$ \cos A \plus{} 2\sin \frac{A\plus{}2B}{2} \sin \frac{A}{2}$

=$ 1\minus{}2\sin \frac{A}{2} ( \sin \frac{A}{2}\minus{}\sin \frac{A\plus{}2B}{2})$

=$ 1\plus{}4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
[/hide]


search
modular arithmetic
number theory proposed
number theory

proof that no matter what 11 natural numbers you'll coose you'll always find 6 numbers that the sum of them is divisible by 6.

There is a proof for the much more general problem: prove that one can choose $n$ numbers out of any $2n-1$ such that their sum is divisible by $n$. It has been discussed somewhere on the forum, so a search is needed.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=19496

[hide="Hint"]
For this particular case, prove that out of 3 numbers one can always choose 2 which have their sum $\vdots$ with 2 (obvious), and that out of 5 numbers one can choose 3 such that their sum is $\vdots$ with 3 (harder, but not impossible: if the 5 numbers contain a residue $\pmod 3$  which repeats 3 times, just consider those 3 numbers, otherwise there is a residue which is only once, and two which are twice. So each residue is represented, so just take one of each).

Now it's easy to see how we obtain 6 numbers divisible with 6 out of 11. [/hide]


This is an old and beautiful problem which I couldn't find here by searching - apologies if was posted before.

A dart is a simple (no crossing edges) quadrilateral which is not convex. See example here in the middle:

[img]http://www.cut-the-knot.org/Curriculum/Geometry/Quadrilaterals.gif[/img]

Can you tile a square with finitely many darts?

Every dart has some angle > 180 degrees. Suppose there are n darts. Then there are at least n interior vertices, since 2 reflex angles cannot be adjacent to a single interior vertex. Furthermore, these n interior vertices are not on the side of any dart. Note that the four corners of the square must also be vertices of some darts.

Now the sum of the interior angles of the darts must be 360*n. However, the each of the n interior vertices described above is surrounded by interior angles of darts, as well as the four corners. Thus the sum of the interior angles of the darts is > 360*(n+1). Contradiction!

Thought I'll give the new GeoGebra a spin, seeing as cut-the-knot is apparently not fond of linking to images on their website. Mods, if you can plug this into the original post that'll be really cool (I can't edit it, and that swine thing doesn't look good).

So this is a dart:

 [geogebra]05502d2043f2d9eb1df7f85870005e7e88ae2cac[/geogebra]

 :lol: that thing is awesome - a great feature for the forum and Wiki. Really cool.

@mathophile: good job!


What is the precise deadline?
I mean, is that 12:00 AM on Apr. 26? or 11:59 PM on Apr.26? or at noon? or before you off work? or before the morning of Apr.27? I really want to squeeze more time for the quiz!
Thanks!

[quote="tokenadult"]Don't you mean "precise deadline"?

  ;)

I'm advising a member of the math team I coach to finish TODAY (Monday 24 April 2006), to take into account the likely surge of log-ons the MathCamp server will have on the deadline day. There is always plenty to do each day, and tomorrow will be a day to do other math-learning projects.[/quote]

Yes, please help me with the answer.

[quote="DanZ"]Tokenadult is correct that it's best to get your quiz in early, since our server gets really busy the day applications are due.  However, we'll accept applications at any time on April 26th, so just make sure to get it in on that day.

(You might want to at least start your application earlier, so that all you need to do is upload the quiz and submit.  And, of course, you need to get the recommendations!)[/quote]

Thank you! I might submit my application on that day at 9 p.m.  :D

I sent mine in yesterday by overnight...so it should get there...

Wow, sent my in around 11:30 pm :lol: (though it was only 9:30 my time).
Very precise :D


geometry topology vector calculus algebra function domain

So, where do I begin?

I have a current interests in going into Algebraic Topology and I am currently learning about Point-Set Topology and Abstract Algebra, which are pre-requisites to Algebraic Topology.

After talking to a professor, he recommended learning some Differential Geometry, but I think this requires some knowledge of Differential Forms first (not 100% sure).

So can anyone recommend me in a good direction towards Differential Geometry?

What should I know before pursuing this topic? I'm guessing some vector calculus like Green's/Stoke's Theorem and what not.

Any suggestions or advice is greatly appreciated.

Note: I noticed that a lot of Algebraic Topologists also know a lot of Differential Geometry too, which I know is not a coincidence.

I suggest you plunge straight into Algebraic Topology and take detours for Topology or Algebra when you need them.
Topology and Algebra are huge domains and you risk getting stuck in useless (for you) areas of them.
Most of the topology problems I have seen ask to provide pathological (counter)examples, whereas in Algebraic Topology you will work most with nice spaces.
What you will need most in the begining of Algebraic Topology from algebra is Group Theory and Homological Algebra.

I think of Algebraic Topology as of a tool. It applies nicely and mostly in Differential Geometry because Geometry one of the best ways to visualise topology. With a deeper involvment from algebra, De Rham cohomology will appear. Some applications of Algebraic Topology arise in Combinatorial Commutative Algebra and also in Nonlinear Analysis.
So you see, after you have learned some Algebraic Topology you will want to find a field to apply it to.
I suggest you procure "bibles" in each collateral domain. "Bibles" are not the books from which it is easiest to learn, but from which it is most likely to extract useful information. They are like encyclopedias.

I will point out some of them:

[i]Topology[/i]
1. Munkres - Topology

[i]Homological Algebra:[/i]
1. Rotman - An introduction to homological algebra
2. Weibel-...

[i]Group Theory:[/i]
1. Rotman - An introduction to the theory of groups
1. Spivak-...

[i]De Rham Cohomology[/i]
1. Bott, Tu - ....

[i] Complex Analysis[/i] (this is just a step towards Complex Geometry)
1. Ahlfors - Complex Analysis

[i]Commutative Algebra[/i]
1. Matsumura - Commutative Algebra
2. Zariski...
3. Eisenbud - An introduction to Commutative Algebra with a view towards Algebraic Geometry(*)

[i]Combinatorial Commutative Algebra[/i]
1. Villarreal - Monomial Algebras
2. Bruns, Herzog - Cohen-Macaulay... (*)

[i]Algebraic Geometry[/i]
1. Hartshorne - Algebraic Geometry

I am not sure about the titles for the books where a "..." sign appears.
The books that have a star ("*") sign next to them require prerequisites, but certainly they are pointed out in the prefaces.

I insist on my advice to start Algebraic Topology directly and learn what you need for it when you need it.
After you have learned some Algebraic Topology start looking for applications to other fields. I am sure it applies to more than I've written.

I wish you best of luck!

I agree amfulger, just jump into algebraic topology and keep a good reference on group theory and point-set topology close by. I took an algebraic topology course without any formal study in general topology and it was fine.

Some books I like.

point-set topology:
general topology by willard - more advanced than munkres, but great preparation for analysts and has tougher problems.

algebraic topology:
I like the Massey book as an introduction.

For geometry, I like algebraic geometry by Harris. Less advanced than hartstone, and has lots of examples.

Thanks, a lot guys.

I guess the plan is to now jump right into Algebraic Topology this summer.

I was thinking of getting Massey's book too. I saw it at the library, and it looks pretty good for an introduction.

I guess I'll learn Algebra alongside as I go.

I have other questions too with regards to Complex Analysis. I may only do Algebraic Topology just for the sake of doing one thing at a time, but I'm also interested in another area. What kind of things should I know going in learning about Elliptic Functions and Transcendental Numbers? And where should I start, like book recommendations?

I'm currently in Complex Analysis right now.

Thanks.

[quote="Victor96"]If you know something about point-set topology, I strongly recommend J. P. May or A. Hatcher's book for algebraic topology.  The former would be more abstract but still filled with examples, I like the style

If you have some knowledge about basic theory on fundemental groups--which you can easily can quickly handle from any text--then Vick's book on homology theory is a fast and effective access to some interesting facts on algebraic topology.

Massey's book is good, but a little lengthy. If you decide to use that book, one suggestion is to view the images of topology in your mind instead of reading line by line.  That thick-seemed book should be finished without difficulty within at very most 2 weeks.

ps: Actually Hatcher's book is also lengthy... the only concise books on algebraic topology I have ever seen are May's Concise course and Vick's Homology.[/quote]

Two weeks?! You have to be kidding.

So, is Hatcher's text more abstract? I'm not sure about this though because after being exposed to a few Dover textbooks I'm not impressed. It's free, so I will no doubt give it a try though.

I always visualize things in my mind. I create my own images and adjust them as I learn.

Some people I know like Fulton's Algebraic Topology text (Springer paperback).  The paperback is pretty cheap from amazon; I have a copy but haven't read enough of it to comment much.

[quote="Victor96"]Hatcher's book is not abstract at all.  Examples, demonstrations, many of them.  It feels like Munkres' "Topology".  And by the way, perhaps the last part of Munkres' "Topology" is nice as an introdution to Algebraic Topology.  But I did not read it myself, I just believe this.

About May's book, I can say it is abstract, at least at the very beginning.  It uses the language of categories to describe fundemental groups...  Maybe it is not suitable for beginners.

If you capture Massey's book's idea, you can see it actually tells you nothing much useful.  A teacher in our school asked his students (graduates) to finish it within a week, so I believe 2 weeks are enough for most people.  A hint is, something like the classification of 2-dim surfaces does not deserve your time to finish at the first time of reading.  These proofs should be revisited after you finish the book, maybe some time before you sleep, it's like a novel, easy and interesting but lengthy.   I don't quite remember its contents exactly, but if there's anything about homology, I believe you can neglect it and find some better books for this subject, such as Hatcher's.[/quote]

So, you read it, but don't remember it?

I highly doubt a professor expected students to read that in a week. I'm very skeptical about this. I've read descriptions to graduate courses, and they do not even cover the whole text in one term. If it was a review, then sure yeah a week should be fine.

Read my signature. After reading one of his books (biograghy written by himself), I have an idea of what it takes to do mathematics. If it takes you week, great, but like he explains, mathematics is something you should fight and not just for one week.

Anyways, I took a look at Fulton's text and that looked reasonable for a first time read.

Note: I thought about it for a bit, and thought if I did nothing else but read that text, then sure you can be done in a week. I plan on doing other things as mentionned above, work, and have sex with my girlfriend, so I have very limited time to be spent reading this. I usually just do a bit every night, and after 3-4 months I should be done.

Maybe I should apologize because I did not read through Massey's book since when I came across it, I was not a beginner and know most things inside.  So yes, I just reviewed this book, and it took me hours.  And abouyt the professor, maybe you can assume his graduates know something about algebraic topology and not blank in this field.

But this book was used as material for those graduates with relatively weaker backgrounds in a seminar, so the professor may also assume over 5 hours per day on this book, which is also my assumption.  If you need 8 hours for sleep and 2 for meals, you still have 9 hours for sex, hoho.

[quote="Victor96"]If you need 8 hours for sleep and 2 for meals, you still have 9 hours for sex, hoho.[/quote]

Sounds good.  :D


trigonometry
irrational number
real analysis
real analysis solved

hi, i have questions about limits like those of georg

what about sin( n) and cos (n ) do they converge?

are there general techniques (i see some approximation of pi coming ) ?
are there much harder limits of this kind?

fred

Of course $\sin n,\cos n$ do not converge. Harazi has posted a few problems of this type. If I remember correctly, the latest one posted by him was to show that $|\sin n|+|\sin(n^3)|$ doesn't converge.

just an idea for its proof:
you have to show that N is dense in R/2piR (N and R are naturals and reals respectively), which means that you can get as close as you want to any irrational number with something in the form m + 2m*pi, where m and n are integers... now you have that you can find two (infinite) different subsequences converging at any number in [-1,1].
proof is nothing special, but it can be useful to prove other ones


:) i just got around this problem and found it difficult to understand:
I need to find out what number goes in the space. Here it is : 2[ ]=13 { note : [ ] means space ]}
plz help !!!!!!!!!!1

what a silly ques?
grow up and then ask ques :P

answer it man ..........
don't boast yourself? :mad:

This problem makes no sense...
[hide]6.5?[/hide]

You know that two of those spaces add up to be 13, then one of those spaces is half of 13, which is 6.5.

Not to be off topic, or anything, but I find it slightly odd that not ONE, but TWO posters above me
have the string "123456789" somewhere in their names.


According to the USAMTS Participation Rules:
[quote]We will only accept the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS.[/quote]
1) Why isn't PNG accepted? It's not lossy like GIF and JPG, and it's smaller than BMP.
2) Even if PNG is specifically disallowed, can I have PNG images in a PDF (i.e. the images are not by themselves as files)?

I've actually sent them PNG files previous years, and it was fine.

I think someone said a while ago that they sent a .zip file and it was fine, but the USAMTS staff said later that it actually was not fine, and they may or may not accept them.  So, I'm guessing that it might not be such a good idea to send a file type that they might not accept.  Plus, PDF is still the ideal file type because it's small enough, and Latex automatically makes it in PDF.  Also, you can always save any file as a PDF.

PNG is OK.

.zip files are not OK.

We [b]very strongly[/b] prefer a single PDF file with all your solutions.  Please [b]do not[/b] send each solution in a separate PDF file.  If you additionally, [b]optionally[/b], want to send .tex source and any necessary images to compile the .tex source, that's OK.

If you are having trouble getting everything into a single PDF file, then we would much rather you print out and mail your solution then for us to have to sift through multiple files.

How do you put pictures into a .tex file? For the past two rounds I put my pictures in separate pdfs and had to send several files.

Use this:

[code]\includegraphics{picture.png}[/code]

Or if your picture file is in PDF

[code]
\includegraphics[width=x in]{name of file}
[/code]

where x is a number.

That's what I've been doing, but is there any way to do it so I don't have to send 5 pdfs?

I just use includegraphics and then attach whatever files i need to. do you have a specific problem with attaching too many files (memory, upload limits, etc)?

No, but DPatrick said that it's a lot easier for them if there's only one file. And it takes my computer forever to send it  :lol: .


I think this question has been asked about a billion times, in a couple hundred threads but here goes nothing.

I placed 7th individually and we were 2nd as a team!

State is on the 11th... Blech  :(  I wish we had a longer gap between chapter and state.

For the answers to this question, go to the forum started by McCrawford... that's where everyone is posting..