How would you write 10010 base 2 in $ \text{\LaTeX}$

Is there anything wrong with $ 10010_2$?

[code]10010_{2}[/code]

gives us

$ 10010_{2}$. If the number following the underscore is a single digit, the curly braces are not necessary.

Thanks fedja, izzy i know understand  :)


What is the number of degrees in the smallest positive angle $x$ such that $8\sin x\cos^{5}x-8\sin^{5}x\cos x=1$.


calculus
derivative
geometry
rectangle
conics
parabola
analytic geometry

I have a problem about determining a maximum:

[b]Determine the dimensions of the rectangle inscribed in the region limited by the graphs of the parabolas $y=x^2-16$ and $y=16-x^2$, in a way that its area is the [color=blue]maximum[/color].[/b]

[quote="Kent Merryfield"]I assume that part of the problem specified that we are only considering rectangles whose sides are parallel to the coordinate axes. Without that restriction, I wouldn't want to touch this problem.

So let's assume that restriction. Suppose that $x$ is some number in the interval $[0,4]$ and that one of the corners of the rectangle is the point $(x,?)$

Can you now write down the precise formulas for all four corners of the rectangle?

Having identified the corners, can you tell me what the area of the rectangle is? Of course, your answer to this question will depend on $x.$

If you can do those two things, you'll be set up to use ordinary single-variable calculus to answer the question.[/quote]

I cannot identify the four corners with only a variable (ie $x$), see the figure I got below:

[quote="Kent Merryfield"]Say it's the upper right corner, which you just wrote as $(x,y).$ But it's on the upper curve! That means that $y=16-x^2$ and we can actually write that upper right corner as $(x,16-x^2).$

You should be able to get the lower right corner in a similar fashion.

Now, what about the left side of the rectangle? Have you noticed that the curves are both symmetric about the $y$-axis? Use that.[/quote]

I got it :)
Do u mean this:

$Length = \sqrt{(2x)^2} = 2x.$

$Width = \sqrt{(32-2x^2)^2} = 32-2x^2.$

So $A(x) = 2x(32-2x^2)$ is the function to be maximized.


topology
real analysis
real analysis unsolved

How to prove that $\text{Bor }\mathbb{R}=\{B\subset \mathbb{R}: \{(x,0): b\in B\}\in\text{Bor }\mathbb{R}^{2}\}$?

Let $X$ be a topological space, and let $A\subset X$. The induced topology on $A$ is defined by declaring that $U\cap A$ is open in $A$ whenever $U$ is open in $X$. Intersections, unions, and complements commute with $U\mapsto U\cap A$.


Ok that's an easy part, now the harder one. :D

Given $x,y,z>0$ and $(xy)^2+(yz)^2+(zx)^2=3$. Prove that
\[
x+y+z\leq\frac{6+3(xyz)^2}{xyz+2(xyz)^2}.
\]

but,not hard again :D ,and nice :)
first we have $(x+y+z)xyz \leq 3$
we remain to prove:$2(xyz)^2(x+y+z) \leq 3+3(xyz)^2$,
because $3+3(xyz)^2=[(xy)^2+(yz)^2+(zx)^2]+[(xy)^2+(yz)^2+(zx)^2](xyz)^2 \geq 2(xyz)^2(x+y+z)$
done :)


An integer is randomly selected from the integers 0 through 99, inclusive. What is the probability that the integer is in the range 10 through 20, inclusive? Express your answer as a common fraction.

[quote=pog][hide=solution]We see that $P_{\text{integer in the range} (10,20)} = \frac{\text{number of integers (inclusive) from }10 \text{ to } 20}{\text{number of integers (inclusive) from }0 \text{ to } 99}$

We have $99-0+1=100$ integers from $0$ to $99$ inclusive and $20-10+1=11$ integers from $10$ to $20$ (inclusive).

Thus our answer is $\boxed{\frac{11}{100}}.$

$$\definecolor{lightcoral}{rgb}{1, 0.7, 0.7}\colorbox{lightcoral}{\textcolor{white}{solution by \textbf{pog}}}$$[/hide][/quote]

why bump a 10+ year thread just to post the answer when it's already been posted?

[quote=pog]the other solution wasn't very nice[/quote]

shm

The other solution was pretty similar to yours tbh

[quote=Possible]TBH even if you have a better solution to a really easy problem. It is still best not to post it cus the problem us so easy[/quote]

lol facts

[quote=Possible]One solution is good enough for these easy problems[/quote]

agreed

You are allowed to bump a post to add a solution, especially if it is a more detailed solution. Also, it is not very nice to say a problem is easy, because some people might have trouble with it, and that makes them feel bad.


inequalities
induction
algebra proposed
algebra

Let $ x_1$ denote a positive number less than $ 1$, and consider the sequence defined by $ x_{k\plus{}1}\equal{}x_k\minus{}x_k^2$. Prove that $ x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2<1$ , no matter how large $ n$ is!

Observe that $ x_k^2\equal{}x_k\minus{}x_{k\plus{}1}$, so that $ x_1^2\plus{}\cdots \plus{}x_n^2\equal{}x_1\minus{}x_{n\plus{}1}$. But we can prove (easily by induction) that $ x_k\in (0,1)$ for all $ k$. This gives $ x_1\minus{}x_{n\plus{}1}<1$.


trigonometry
limit
function
real analysis
real analysis unsolved

Let $ a_{1}, ... , a_{n}$ and $ 0<b_{1} < ... < b_{n}\leq$ $ \pi$ be real numbers.

Let $ f(x) = \sum_{i = 1}^{n}a_i\cos (b_{i}x)$.

We know that $ \lim_{k  k\ integer\rightarrow + \infty} f(k)$ exists.

Is it true that $ f(x) = 0$ for all $ x$?

I'm not sure I follow.  If $ n \equal{} 1$ and $ a_1 \equal{} b_1 \equal{} 1$ then $ f(x) \equal{} \cos{x}$ so surely the limit cannot exist in all cases and $ f \neq 0$.

The limit exists in exactly one case: that $ b_1\equal{}0$ and $ n\equal{}1$ (or equivalently $ a_i\equal{}0$ for $ i\neq 1$). The functions available this way are all of the constants.

Actually it should be $ 0<b_{1}$.

Jmerry, what is the solution for what you wrote??


If $ \tan 15^\circ \tan 25^\circ \tan 35^\circ \equal{}\tan \theta$ and $ 0^\circ \le \theta \le 180^\circ,$ find $ \theta.$
Problems from the AoPSWiki need to be solved.

[quote="1=2"]If $ \tan 15^\circ \tan 25^\circ \tan 35^\circ \equal{} \tan \theta$ and $ 0^\circ \le \theta \le 180^\circ,$ find $ \theta.$
Problems from the AoPSWiki need to be solved.[/quote]
with my best wishes
 :play_ball:
[url=http://www.servimg.com/image_preview.php?i=249&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/image310.jpg[/img][/url]

Using the nice identity $ \tan(60\minus{}x)\tan x\tan(60\minus{}x)\equal{}\tan 3x$ we have
$ \tan 15 \tan(60\minus{}35) \tan 35 \tan (60\plus{}35) \equal{} \tan x \tan (60\plus{}35)$

$ \tan 15 \tan 105 \equal{} \tan x \tan 95 \Rightarrow \minus{}1\equal{}\minus{}\tan x \cot 5 \Rightarrow \tan x \equal{} \tan 5$

[quote="delta"]Using the nice identity $ \tan(60 \minus{} x)\tan x\tan(60 \minus{} x) \equal{} \tan 3x$ we have
$ \tan 15 \tan(60 \minus{} 35) \tan 35 \tan (60 \plus{} 35) \equal{} \tan x \tan (60 \plus{} 35)$

$ \tan 15 \tan 105 \equal{} \tan x \tan 95 \Rightarrow \minus{} 1 \equal{} \minus{} \tan x \cot 5 \Rightarrow \tan x \equal{} \tan 5$[/quote]
i think you mean : tanx . tan(60-x) . tan(60[color=red] +[/color]x) = tan3x
almonkez


find all $ f: R \to R$ such that : $ \forall x,y \in R$:
\[ f(x) \plus{} f(y) \geq f(x\plus{}y)\]
\[ x \geq f(x)\]

[quote="stvs_f"]find all $ f: R \to R$ such that : $ \forall x,y \in R$:
\[ f(x) \plus{} f(y) \geq f(x \plus{} y)
\]

\[ x \geq f(x)
\]
[/quote]

Let g(x)=x-f(x) then

g(x+y) \geq g(x)+g(y) and g(x) \geq 0

$ \forall x,y \in R$ g(x) \geq g(x-y) + g(y) \geq g(y)

So we get g(x) \geq g(y) and in similar way we get g(y) \geq g(x).

Then we get g(x)=g(y) $ \forall x,y \in R$

So g(x+y) \geq g(x) + g(y) and 0 \geq g(x)

But g(x) \geq 0 so g(x)=0 $ \forall x \in R$

And we get f(x)=x $ \forall x \in R$

p.s"\geq" means "is larger than"

Just substitute y=0 and we get f(x)≥f(x)+f(0)≥f(x), so f(0)=0
      Then substitute y=-x, 0≥f(x)+f(-x)≥f(0)=0, equality holds only when f(x)=x and f(-x)=-x
so f(x)=x is the required function


algebra
polynomial
quadratics
algebra open

[quote="abhishek1994"]if one root of a quadratic equation is 1/2-under root of 3
the the quadratic equation can be???????????[/hide][/quote]

If one root of a quadratic equation is the real $ u$, then the quadratic can be $ x^2\minus{}(u\plus{}v)x\plus{}uv\equal{}0$  :) for any real $ v$


calculus
derivative
combinatorics proposed
combinatorics

$ A\subset \{1,2,...,n\}$, $ A \neq \emptyset$ Let $ f(A)$ is difference between the biggest element of $ A$ and the smallest element of $ A$.
Find $ \sum_{\emptyset \neq A \subset \{1,2,...,n\}} {f(A)} \equal{} ?$

There are exactly $ 2^{j \minus{} i \minus{} 1}$ subsets with smallest element $ i$ and largest $ j$.
So $ S \equal{} \displaystyle\sum_{i \equal{} 1}^{n \minus{} 1} \displaystyle\sum_{j \equal{} i \plus{} 1}^n 2^{j \minus{} i \minus{} 1} (j \minus{} i)$
$ \equal{} \displaystyle\sum_{i \equal{} 1}^{n \minus{} 1} 2^{\minus{}i \minus{} 1} \displaystyle\sum_{j \equal{} i \plus{} 1}^n j \cdot 2^j \minus{} \displaystyle\sum_{i \equal{} 1}^{n \minus{} 1} i \cdot 2^{\minus{}i \minus{} 1} \displaystyle\sum_{j \equal{} i \plus{} 1}^n 2^j$
which should lead to the answer after a quarter of an hour of sweating.  :)

To simplify calculations we consider the range $ k$ with the minimum $ i$.
Given $ k$ (between $ 1$ and $ n \minus{} 1$ both inclusive), $ i$ can range from $ 1$ to $ n \minus{} k$.
Hence for each $ k$ we have $ (n \minus{} k) \cdot 2^{k \minus{} 1}$ subsets, which gives:
$ S \equal{} \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} k (n \minus{} k) 2^{k \minus{} 1}$
which you should be able to sum using the derived and twice derived geometric series.

Let a set $ K_n \equal{} \{1, 2, \cdots, n\}$
Let $ \sum_{\emptyset \neq A \subset K_n } {f(A)} \equal{} a_n$

then $ a_{n \plus{} 1} \equal{} a_{n} \plus{} \sum_{A \subset K_{n \plus{} 1}, (n \plus{} 1) \in A} {f(A)}$

$ \sum_{A \subset K_{n \plus{} 1}, (n \plus{} 1) \in A} {f(A)} \equal{} n 2^{n \minus{} 1} \plus{} (n \minus{} 1)2^{n \minus{} 2} \plus{} \cdots \plus{} (n \plus{} 1 \minus{} n) 2^0 \equal{} \sum_{k \equal{} 1}^{n}{k 2^{k \minus{} 1}}$
$ \sum_{k \equal{} 1}^{n}{k 2^{k \minus{} 1}} \equal{} \sum_{k \equal{} 1}^{n}{(k \minus{} 1)2^k \minus{} (k \minus{} 2)2^{k \minus{} 1}} \equal{} (n \minus{} 1)2^n \plus{} 1$
$ \therefore a_{k \plus{} 1} \minus{} a_k \equal{} (k \minus{} 1)2^k \plus{} 1$
$ \therefore \sum_{k \equal{} 1}^{n \minus{} 1} \big( a_{k \plus{} 1} \minus{} a_k \big) \equal{} \sum_{k \equal{} 1}^{n \minus{} 1}\big((k \minus{} 1)2^k \plus{} 1 \big) \equal{} \sum_{k \equal{} 1}^{n \minus{} 1}\big( (k \minus{} 2)2^{k \plus{} 1} \minus{} (k \minus{} 3)2^k \big) \plus{} (n \minus{} 1)$
$ \equal{} (n \minus{} 3) 2^n \plus{} n \plus{} 3$
$ \therefore a_n \equal{} a_ 1 \plus{} (n \minus{} 3) 2^n \plus{} n \plus{} 3$

$ K_1 \equal{} \{1\}$, therefore the subset of A (which is not an empty set) is only $ \{1\}$, so $ a_1 \equal{} 0$.

$ \therefore a_n \equal{} (n \minus{} 3) 2^n \plus{} n \plus{} 3$

maybe right, I checked this where $ n \equal{} 1, 2, 3$ , but I'm not sure :)

and to bubka :
You don't need no more than 10 ml of sweat. :) And also you don't have to differentiate(Of course, differentiation also give an answer). Telescopic summation is the only thing what you need. '-'

Telescopic summation would be great; let me try it out on my expression and check the result with yours.

$ \displaystyle\sum_1^m k^2 \cdot 2^k$
$ \equal{}\displaystyle\sum_1^m ((k\plus{}1)^2 \cdot 2^{k\plus{}1} \minus{} k^2 \cdot 2^k) \minus{} \displaystyle\sum_1^m (4k\plus{}2) \cdot 2^k$
$ \equal{}(m\plus{}1)^2 \cdot 2^{m\plus{}1} \minus{} 2 \minus{} 4\displaystyle\sum_1^m ((k\plus{}1) \cdot 2^{k\plus{}1} \minus{} k \cdot 2^k) \plus{} 6\displaystyle\sum_1^m 2^k$
$ \equal{}(m\plus{}1)^2 \cdot 2^{m\plus{}1} \minus{} 2 \minus{} (m\plus{}1) \cdot 2^{m\plus{}3} \plus{} 8 \plus{} 3 \cdot 2^{m\plus{}2} \minus{} 12$
$ \equal{}(m^2\plus{}2m\plus{}1\minus{}4m\minus{}4\plus{}6) \cdot 2^{m\plus{}1} \minus{} 6$
$ \equal{}(m^2\minus{}2m\plus{}3) \cdot 2^{m\plus{}1} \minus{} 6$

Similarly,
$ \displaystyle\sum_1^m k \cdot 2^k$
$ \equal{}\displaystyle\sum_1^m ((k\plus{}1) \cdot 2^{k\plus{}1} \minus{} k \cdot 2^k) \minus{} 2\displaystyle\sum_1^m 2^k$
$ \equal{}(m\plus{}1) \cdot 2^{m\plus{}1} \minus{} 2 \minus{} 2^{m\plus{}2} \plus{} 4$
$ \equal{}(m\minus{}1) \cdot 2^{m\plus{}1} \plus{} 2$

(Note: both these results can be easily proved by induction.)

Plugging in the results into
$ S \equal{} \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} k (n \minus{} k) \cdot 2^{k \minus{} 1}$
$ 2S \equal{} n \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} k \cdot 2^k \minus{} \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} k^2 \cdot 2^k$,
we have
$ 2S \equal{} n(n\minus{}2) \cdot 2^n \plus{} 2n \minus{} ((n\minus{}1)^2 \minus{} 2(n\minus{}1) \plus{} 3) \cdot 2^n \plus{} 6$
$ \equal{} (n^2 \minus{} 2n \minus{} n^2 \plus{} 2n \minus{} 1 \plus{} 2n \minus{} 2 \minus{} 3) \cdot 2^n \plus{} 2n \plus{} 6$
$ \equal{} 2[(n \minus{} 3) \cdot 2^n \plus{} n \plus{} 3]$
which gives
$ S \equal{} (n \minus{} 3) \cdot 2^n \plus{} n \plus{} 3$,
so, apparently, [i]both[/i] of us are right!   :thumbup: :D

Yes, we are right :)

By the way, you know, there is very inconsiderible problem with
$ S \equal{} \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} k (n \minus{} k) 2^{k \minus{} 1}$
because $ n$ would be 1. ^^; but, For you can ignore this trivial problem and be led to the answer, so it is not a big problem.

Huh?
For $ n \equal{} 1$, we have $ S \equal{} 0$, the emtpy sum, as it should be. :huh:  :maybe:

Oh well, I'm sorry - there's no problem.


modular arithmetic
number theory unsolved
number theory

Let $ n,m$ be positive integers such that $ n^2\equal{}3m^2\plus{}3m\plus{}1$. Prove that $ n$ is a sum of two squares.

analyse the discriminant... and use the fact that $ x^2 \not \equiv \minus{} 1 \pmod {3}$.
$ n$ is the sum of 2 consecutive square.

I think we also need the condition $ n>2$. If we add $ m^3$ to both sides and do a bit of manipulation we have:

$ n^2 \equal{} (m\plus{}1)^3 \minus{} m^3$.

This is then the 1996 Hungary-Israeli Mathematics Competition problem 2:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1299099#1299099

my solution:
for the discriminant we have :
$ \Delta \equal{} (3)^2\minus{}4.3.(1\minus{}n^2)\equal{}9\minus{}12\plus{}12n^2\equal{}\minus{}3\plus{}12n^2$
$ \Delta$ must be a square,
$ \Delta\equal{}z^2 \implies \minus{}3\plus{}12n^2\equal{}z^2 \implies 3 |z \implies 3q\equal{}z$
$ \implies \minus{}3\plus{}12n^2\equal{}9q^2 \implies \minus{}1\plus{}4n^2\equal{}3q^2  \implies (2n)^2\minus{}1\equal{}3q^2$
so we must have
$ (2n\minus{}1)(2n\plus{}1)\equal{}3q^2$
but we have $ \gcd(2n\minus{}1,2n\plus{}1)\equal{}1$
so we have two cases:
$ 2n\minus{}1\equal{}3t^2,2n\plus{}1\equal{}l^2,tl\equal{}q \implies l^2\minus{}3t^2\equal{}2 \implies l^2\equal{}3t^2\plus{}2 \implies l^2 \equiv 2 \pmod {3}$
which is a contradiction so let's check the second case;
$ 2n\minus{}1\equal{}l^2,2n\plus{}1\equal{}3t^2,tl\equal{}q \implies 2n\equal{}l^2\plus{}1$
but $ l$ must be odd and we must have $ l\equal{}2k\minus{}1$ which implies
$ 2n\equal{}(4k^2\minus{}4k\plus{}1)\plus{}1\equal{}4k^2\minus{}4k\plus{}2 \implies n\equal{}2k^2\minus{}2k\plus{}1 \implies n\equal{}(k)^2\plus{}(k\minus{}1)^2$
so we r done!
plz tell if i'm wrong :oops:


How many prime numbers are between 30 and 40?


The goal of many people on AoPS is to get a perfect score on MOEMS. How did you do?

I got a 5 on round one and 5 on round 2 as well.

Really hoping for a perfect...ICAE gives 50ish dollars to every perfect (depending on how many there are)

:(
i made a dumb mistake on the first round and now my total is a fail 9/10.

Hmm, I've never heard of MOEMS. I learned about it after I finished 8th grade  :mad: ...Anyway, what is the format? Isn't it like a really easy version of the USAMO for middle schoolers?

i was sick the day of the first round.  :(  oh well. 5 on the second

@Thunder365: They give out money for perfects?!  :wallbash_red:

[hide="@AIME15USAMO"]Not really. MOEMS gives you five rounds of five problems. They're all pretty easy.  :lol: [/hide]

I finally got my scores on round 2! I have a 10/10 now.  :)

$ MOEMS\equal{}MathOlympiadforElementaryandMiddleSchools$

You should participate. It really boosts your self-esteem.  :wink:

Do you do Moems online? or does your school have to register and etc...
Because our school only does the AMC8 (10~AIME for those who qualify) , Mathcounts(school~national), and county or invitational tourneys...

It is usually done through school, but I think sometimes you can do it as an individual (not online).

Not discussing scores on 1st one.
But, i got perfect on 2nd one  :lol:

it is also known as Math Olympiads, if that helps


Prove that there isn't f function such that f(x-f(x))=x/2 for any x∈R and f continuous.

This is pretty standard.

Set $ g(x): \equal{}x\minus{}f(x), x\in\mathbb{R}$ continuous. Then
$ (*)$ $ g(g(x))\minus{}g(x)\plus{}{x\over 2}\equal{}0$ for all $ x\in\mathbb{R}$.

$ g(x)\equal{}g(y)\Rightarrow x\equal{}2(g(x)\minus{}g(g(x)))\equal{}2(g(y)\minus{}g(g(y)))\equal{}y$ for all $ x,y\in\mathbb{R}$.
So $ g: \mathbb{R}\to\mathbb{R}$ is injective and by the intermediate value theorem strictly monotonic.
Set $ g_2(x)\equal{}g(g(x)),x\in\mathbb{R}$ and $ g_3(x)\equal{}g(g(g(x))),x\in\mathbb{R}$ and $ g_4(x)\equal{}g(g(g(g(x)))),x\in\mathbb{R}$.
Then $ g_2(x)$ and $ g_4(x)$ are strictly monotonically increasing.

But $ (*)$ gives
$ A) :$ $ g_2(x)\minus{}g(x)\plus{}{x\over 2}\equal{}0$
$ B) :$ $ g_3(x)\minus{}g_2(x)\plus{}{g(x)\over 2}\equal{}0$
$ C) :$ $ g_4(x)\minus{}g_3(x)\plus{}{g_2(x)\over 2}\equal{}0$
for all $ x\in\mathbb{R}$, and then
$ {A\over 2}\plus{}B\plus{}C):$ $ g_4(x)\plus{}{x\over 4}\equal{}0$
and $ g_4(x)\equal{}\minus{}{x\over 4}$ for all $ x\in\mathbb{R}$. Contradiction!


Please give me some examples for decreasing Lipschitz functions with the constant Lipschitz C =1 on set of real number R?
Thank you very much.

Let $ f(x)\equal{}\minus{}x$.  Then $ \left|f(x)\minus{}f(y)\right|\equal{}\left|\minus{}x\plus{}y\right|\equal{}\left|y\minus{}x\right|$, so the constant is $ K\equal{}1$.

Thanks JRav so much. It's easy example!
I've just found an other example: $ f(x)\equal{}sin(x)\minus{}x$ is also a decreasing and Lipschitz function with K=2 on R.
I really appreciate your help.
Have fun!


solve in $ R$ this equation:
$ \frac{x^2}{x\minus{}1}\plus{}\sqrt{x\minus{}1}\plus{}\frac{\sqrt{x\minus{}1}}{x^2}\equal{}\frac{x\minus{}1}{x^2}\plus{}\frac{1}{\sqrt{x\minus{}1}}\plus{}\frac{x^2}{\sqrt{x\minus{}1}}$

[hide="Solution"]Let $ a\equal{}\frac{x^2}{x\minus{}1}, b\equal{}\sqrt{x\minus{}1}\ (x>1; a,b>0)$
The equation is equivalent to:
$ a\plus{}b\plus{}\frac{1}{ab}\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}ab$
$ \Leftrightarrow (a\minus{}1)(b\minus{}1)\equal{}(\frac{1}{a}\minus{}1)(\frac{1}{b}\minus{}1)\ (*)$
It's easy to prove $ a\neq 1$.
Case 1: $ b\equal{}1$
$ \Rightarrow x\equal{}2$
Case 2: $ b\neq 1$
$ (*)\Leftrightarrow (a\minus{}1)(b\minus{}1)\equal{}\frac{a\minus{}1}{a}\cdot \frac{b\minus{}1}{b}$
$ \Leftrightarrow ab\equal{}1$
$ \Leftrightarrow x^2\equal{}\sqrt{x\minus{}1}$
We have $ \sqrt{x\minus{}1}<\sqrt{x}<x^2 ...$
Thus $ x\equal{}2$[/hide]


Given equilateral triangle $ABC$ and $d$ is an arbitrary line $A',B',C'$ are othor projection of $A,B,C$ on $d$. Prove that
\[B'C'^{4}+C'A'^{4}+A'B'^{4}=const\] for all $d$

Hi I see it isn't so ugly, any idea from everybody ? :|

[hide]
translate the line $d$ parallel to itself so it passes through a vertex of the given triangle and it does not intersect it.lets say it passes through $A$. then you have $A'B'=x$, $A'C'=y$ and $B'C'=x+y$. denote $|AB|=a$. we look at the expression
$x^{4}+y^{4}+(x+y)^{4}=2x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+2y^{4}=2(x^{2}+xy+y^{2})^{2}=$
$=2a^{4}(cos^{2}\alpha+cos\alpha cos(120-\alpha)+cos^{2}(120-\alpha))^{2}=2a^{4}(\frac{3}{4})^{2}=\frac{9a^{2}}{8}$
maybe there's a nice geometric proof that $x^{2}+xy+y^{2}$ is constant.
[/hide]


geometry
circumcircle
projective geometry
radical axis
geometry proposed

Let $ (k)$ be a circle with diameter $ AB$. Points $ C,D,E,F$ (in that order) $ \in$ semicircle $ \bar{k}$. If $ \angle AMC \equal{} \angle FMB$ and $ \angle AMD \equal{} \angle EMB$ prove that:
a)$ PM\perp AB$ where $ P \equal{} CD\cap EF$
b)$ QM\perp AB$ where $ Q \equal{} CE\cap DF$
EDIT: $ M$ is an arbitrary point on $ AB$.

$ M$ maybe center of circle?
But if it is, it's too obvious.

If $M$ is the center of the circle, the result is indeed trivial. I believe $M$ is an arbitrary point on $\overline{AB}.$

Let $R \equiv AC \cap BF$ and $S \equiv AD \cap BE.$ Since $M$ is the unique point of $AB$ such that $AB$ bisects angles $\angle CMF$ and $\angle DME$ externally, it follows that $\triangle CFM$ and $\triangle DEM$ become orthic triangles of $\triangle RAB$ and $\triangle SAB,$ respectively $\Longrightarrow$ $N \equiv CF \cap DE$ is the harmonic conjugate of $M$ WRT $(A,B),$ i.e. $RS$ is polar of $N$ WRT the circle $k$ with diameter $AB.$ But indeed the intersections $P \equiv CD \cap EF$ and $Q \equiv CE \cap DF$ lie on the polar of $N$ WRT $k$  $\Longrightarrow$ $P,Q,R,S$ are collinear.

I'm truly sorry. $ M$ is an arbitrary point on $ AB$.

I found a slighty different approach:

Let $X$ be the intersection of $DE$ with the perpendicular $\tau$ to $AB$ through $M$ and let $N \equiv DE \cap AB.$ Since $MX,MN$ bisect $\angle DME,$ then $(D,E,X,N)=-1$ $\Longrightarrow$ $\tau$ is the polar of $N$ WRT the circle with diameter $\overline{AB}.$ Similarly $CF$ passes through $N$ and then in the complete quadrilateral $CDEF,$ it follows that $P \equiv CD \cap EF$ and $Q \equiv CE \cap DF$ lie on $\tau.$

Let $O$ be the center of $(k)$. Let the circumcircle of $OMF$ hit $(k)$ at $C'$. We have $\angle FMO = \angle OC'F = \angle C'FO = \angle C'MA$, whence $C' = C$, so $CFOM$ is cyclic. Similarly, we find that $DEOM$ is cyclic. Let the tangents to $(k)$ at $C$ and $F$ meet at $U$, and let the tangents at $D$ and $E$ meet at $V$. $\angle UCO = \angle UFO = 90^{\circ}$, sp $CMOFU$ is cyclic; hence, $\angle UMO = \angle UCO = 90^{\circ}$, so $UM \perp AB$. Similarly, $VM \perp AB$. Finally, Pascal's theorem on $CCEFFD$ and $EEFDDC$ shows that $P,Q,U,V$ are collinear, so we are done.

Let make simmetry of points C D wrt AB and get points C' D' . Easy to see that M is on C'F and D'E and triangle MEF ~ MC'D' = MCD , so MEF ~ MCD . Let EF intersect (FMD) at point T , aesy to see that angle EFM = MCD , so T is on CD . So like the same T is on (MCE) . Let O is center of (k) well known that O is on (MDE) , so angle EMD = EOD = 2*TFD , so MT is angle bissector of EMD , so AB is perpendicular to MT and T = P and Q is on MT (Q = radical center of ((k) (MCP) (MDP)) . done


Prove that \[ \sum_{p\leq x} \frac{log(p)}{p} - log(x) \] converges implies the prime number theorem ($\pi(x)$ equivalent to $\frac{x}{log(x)}$).
In fact, I've found that we just need to have $\sum_{p\leq x} \frac{log(p)}{p}$ is equivalent to $\frac{x}{\pi(x)}$.

Please I can't believe no one know that. Maybe it's too easy for mathlinks ?

It's in the wrong section I think ... :)

Hello,

I can't really help you, even though the problem is interesting.

But in "Hardy and Wright : an introduction to the theory of numbers", on page 348, theorem 425 :

$\sum_{p\leq x}\frac{ln(p)}{p}=ln(x)+O(1)$
(which is a bit weaker than the convergence you speak of)

They only prove the Prime Number Theorem eventually on page 367 (to give you an idea   :( )


function
calculus
limit
topology
real analysis
real analysis solved

hi all,

here is a problem that seems quite interesting:
let f be a continuous function from the closed interval [0;1] to itself. Indeed
f has at least one fixed point xo: f(x0)=x0
Let u_n a sequence such that:
1)  u_1 is arbitrary in [0;1]
2)  u(n+1)=(1-1/n)u(n) +(1/n)f(u(n))

prove that u_n converges toward one fixed point of f ..

bye.

The recursion can be rewritten as $u_{n+1}-u_n=\frac1n(f(u_n)-u_n)$.

$(u_n)$ is a bounded sequence, so by Bolzano-Weierstrass, it has a convergent subsequence. Let $x$ be the limit of that subsequence.

First, suppose the whole sequence converges to $x$. Then $f(x)=x$. We argue as follows: Suppose $f(x)-x=a>0$. Then there is some neighborhood of $x$ in which $f(u)-u>\frac a2$. There exists $N$ such that for $n\ge N, u_n$ belongs to this neigborhood. Thus for $n\ge N, u_{n+1}-u_n\ge \frac a2\cdot\frac1n$. Thus for $m>N, u_m-u_N=\sum_{n=N}^{m-1}(u_{n+1}-u_n)\ge\frac a2\sum_{n=N}^{m-1}\frac1n$. Since this can be made arbitrarily large, $u_n$ must eventually leave the neighborhood in question, contradicting the convergence. A similar argument can be made for $f(x)-x<0$.

We're not done yet, becuase we haven't shown that the whole sequence converges to $x$. Now suppose that it's just a subsequence that converges to $x$. As the proof in the last paragraph shows, if $f(x)\ne x, (u_n)$ must be monotone whenver $u_n$ is near $x$. (Increasing if $f(x)>x$, decreasing if $f(x)<x$.) For $x$ to be the limit of a subsequence, $u_n$ must leave the neigborhood and then come back. To come back, it must at some point move in the opposite direction. But since $|u_{n+1}-u_n|\le\frac1n$, it can eventually only "crawl" back in very small steps, not "jump" back; however the forced one-way monotonicity of the sequence in a neighborhood of $x$ prevents crawling back in small increments.

That last argument shows that for $x$ the limit of a subsequence, $f(x)=x$. We're still not done, but we've narrowed it down to showing that in that case, the whole sequence converges to $x$.

nice solution ;-)
I think it's easy to conclude because since u(n+1)-u(n)->0 then the set of the limit points is an interval. So if there are two distinct limit points, that leads to a contradiction ..

Thanks, alekk, now I think I see it. Suppose the sequence has two different cluster points (that is, limits of subsequences). Let $x$ be any point in the interval in between. For both outer points to be limits of subsequences, the sequence must go back and forth between neighborhoods of those two points infinitely often. Since $|u_{n+1}-u_n|<\frac1n$, that would imply that for all $N$, there exists $n>N$ such that $|u_n-x|<\frac1N$. (Since the sequence keeps going past $x$, and since the steps of the sequence get arbitrarily short, then the some members of the sequence must get arbitrarily close to $x$.)

But that makes $x$ another cluster point of the sequence; that means there is a subsequence of $u_n$ that tends to $x$. By what I have already proved, this forces $f(x)=x$.

This holds true for every point in the interval between the two original cluster points. But the moment $u_n$ falls into an interval on which $f(x)=x$, it would "stick": that is, $u_n=u_{n+1}=u_{n+2}=\dots.$ That would keep the sequence from having those two other cluster points. And with that, we've got the whole thing.

What properties of $\frac1n$ were used in this proof? Two properties: that it goes to zero, and that is is the terms of a divergent series. Any other positive sequence with those two properties would also work. However, if we were to replace it with a sequence that doesn't go to zero, then we'd have no tools to keep $(u_n)$ from having multiple cluster points, and if we were to replace $\frac1n$ with the terms fo a convergent series, then $u_n$ could just grind to a limit without ever getting close to a fixed point for $f$.

now there is an amazing thing that a friend told: this result still holds in the complex plane. What I mean is that if f is a complex function from the unit closed ball to itself (unit ball of C) then the sequence still converges to a fixed point of f. So we can prove the brouwer theorem  :D

[quote]...this result still holds in the complex plane.[/quote]
If that's so, then there must be a better proof than the one I gave. I used too many line-specific ideas to have much hope of moving the proof to the complex plane (or to $\mathbb{R}^n$).

Can anyone give an essentially different proof?

[quote="alekk"]I think it will be quite hard to solve the pb in the complex plane because I remember it's hard to prove that such a function f has always a fixed point ..[/quote]
Does anybody know nice proof for brouwer theorem? Proof, which I know, is very ugly, in my opinion.

[quote="alekk"]now there is an amazing thing that a friend told: this result still holds in the complex plane. What I mean is that if f is a complex function from the unit closed ball to itself (unit ball of C) then the sequence still converges to a fixed point of f. So we can prove the brouwer theorem[/quote]

Unfortunately, that's wrong; the sequence does not necessarily converge, and this is not a proof of the Brouwer theorem.
Here is a function for which this iterative method fails. Expressed in polar form, let
$ f(r,\theta)=(\sqrt{r},\theta + \frac{\pi}{6}) $. This is a continuous function from the unit disk to itself, and its only fixed point is the origin.
On the other hand, if $ r<\frac34 $, the entire segment between $ (r,\theta) $ and $ f(r,\theta) $ lies outside the circle of radius $ r $, so the iteration moves you farther  away from the center. Outside that circle, the iteration brings the next point closer (if $n$ is large enough), but the nearest point of the segment is still at least $ \frac34 $ away from the origin. The sequence thus settles into a circular orbit of radius $ \frac34 $, moving about $ \frac{\pi}{6n} $ in the $ \theta $ direction at each step. This is obviously not a convergent sequence, as it goes around infinitely many times (the cluster points of the sequence are the circle of radius $ \frac34 $).

In fact, the only way for the sequence to converge for this function is to start exactly at the fixed point.

For Myth, there is one using differential calculus and the following

Theorem: Let $B$ unit ball of $R^n$ and $F:B \rightarrow B$  $C^\infty$ function
has a fixed point.

For Brouwer, construct $F_\epsilon:B\rightarrow B, C^\infty$ a sequence of functions wich converge uniformly $F:B\rightarrow B$ a continous function
for each $\epsilon >0$ we have a fixed point $x_{\epsilon}$ for $F_{\epsilon}$
and $F(x)=\lim F_{\epsilon}(x_{\epsilon})=\lim x_{\epsilon} =x$

Application of Brouwer fixed point theorem in science economy

http://cepa.newschool.edu/het/essays/math/fixedpoint.htm#brouwer

[quote="Moubinool"]Another one using algebraic topology
http://planetmath.org/encyclopedia/ProofOfBrouwerFixedPointTheorem.html
[/quote]
There are no such topic.

In the following thesis
by Tata Stuckless
Title :Brouwer's fixed point theorem methods proof generalisation

http://eprints.cecm.sfu.ca/archive/00000035/

Different proof

->Degree self - map of $S^{n-1}$ (by Brouwer 1912)page 9

->KKM theorem page 15

->Homology group (with algebraic topology) page 16

->Hairy Ball theorem page 21

->Garcia's proof with Green theorem page 31

A lot of references page 68


calculus
derivative
algebra
polynomial
search
calculus computations

Suppose that $ f\in\mathbb{R}[x]$ such that $ \deg f\equal{}n$ and $ f(x)\geq 0\ \forall\ x\in\mathbb{R}$.  Prove that $ \sum_{k\equal{}0}^nf^{(k)}(x)\geq 0\ \forall\ x\in\mathbb{R}$, where $ f^{(0)}(x)\equal{}f(x)$.


function AMC AIME USA(J)MO USAMO algebra polynomial

the nth term of (a+bx+cx^2+dx^3+ex^4)(1-x)^(-5) is n^4*x^(n-1)

find a,b,c,d,e

now i bashed it out using generating functions and compared coeffficients to get the values.  but the problem is, i only compared the first 5 coefficients.  using mathematica, it seems that this is true for all n.  any thoughts?  the value of a,b,c,d,e i've found are:

[hide]1,11,11,1,0[/hide]

It seems to me that from the problem statement, it's supposed to be true for all n.

well the problem statement was rewritten form the original problem...  there had being a few case of errors in some cases already.  also there should exist an relatively easy solution since it was #6 out of 13 questions.  more of less AIME difficulty rather than USAMO

[hide]An ugly solution (waiting an astute one)  :

n^4=n(n-1)(n-2)(n-3)+6n(n-1)(n-2)+7n(n-1)+n

f(x)=1/(1-x), d^n(f(x)) = diff (f(x), x), n times

 :Sigma:  n^4 x^(n-1) = f(x) + 7d(f(x)) + 6 d^2(f(x) + 7d^3(f(x))

d^n(f(x) = n!/(1-x)^(n+1)

Now it is easy (but very cumbersome). Since you have mathematica ....[/hide]

Why it has to be true for all n:

Essentially, this is a problem of nth differences.  In each of (1 - x)-m, m :ge: 1, the coefficients follow a polynomial of degree (m - 1).  So if you can get it to work for the first 5, you've matched a 4th degree polynomial over 5 values, so that polynomial is unique.

I don't have any idea what the nicest way to actually calculate those values is -- maybe multiply through by (1 - x)5?

Belenos, I don't think your solution is ugly.  I think it is nice.  It's basically how I solved it.

>> solution ... ugly

I think it's not nice because it requires at lot of (error prone !) calculations  :(

Another approach is to notice that multiplying a power series by 1 - x is the same as differencing the coefficients of the series.  (Joel also mentioned first differences.)  So I'll do five differences on the fourth powers:

[code]1 16 81 256 625
1 15 65 175 369
1 14 50 110 194
1 13 36  60  84
1 12 23  24  24
1 11 11   1   0
[/code]
The last row gives the values of a, b, c, d, and e.

By the way, bugzpodder, what was the original problem statement?  For example, if it just asked for a + b + c + d + e, that would be easier.

i dont have the original copy.  but the version i have asked for the individual coefficients.

alright, time for the easy way:

[hide]

(a+bx+cx^2+dx^3+ex^4)(1-x)^(-5) =1+16x+81x^3+256x^4+...

a+bx+cx^2+dx^3+ex^4=(1+16x+81x^3+256x^4+...)(1-x)^5

[/hide]

you can finish off.  By the way, this is from the Canadian Descartes contest (the one i mentioned that I liked a lot)

I claim credit for having suggested that earlier :)

Yes, Joel gets the credit.  But I am a little confused by bugzpodder's last post.  He doesn't say what is the best way to multiply those two polynomials.  Is there an easier way than the repeated differencing in my last post?

sorry, I must have missed your post, Joel


complex analysis
function
limit
complex analysis unsolved

Given a real analytic $ f(x)$ on $ \mathbb{R}$, can I find a holomorphic function $ F(z)$ on $ \mathbb{C}$ such that
$ F(x)\equal{}f(x)$  for real $ x$?  What if I assume that $ f$ is positive and $ \lim_{x\rightarrow\infty}f(x)\equal{}0$?

That doesn't suffice to archieve holomorphy. Take $ \frac {1}{x^2 \plus{} 1}$.
The only thing you can get is an open neighbourhood of the reals in which it is holomorphic.


function
analytic geometry
algebra
polynomial
real analysis
real analysis unsolved

Here is a very beautiful, but hard problem: find all continuous functions $f$ defined on $R^2$ and having values in $R$ such that $ f(A)+f(C)=f(B)+f(D)$ for any unit square $ABCD$ in the plane.

It seems like we in SPb read the same paper, because the following problem was proposed in 2004 on our tests for All-Russian: find all functions $f$ from a plane to [0,1] with the same property. Unfortunately, students did not solve it.

I believe that it is famous result. In my paper it is attributed to a bulgarian mathematician whose name I cannot remember. Wow, proposing this to a contest... even in Sankt Petersburg... I really don't wonder the students could not solve it (although they are super-smart there). :?

Is it exactly as you wrote it ($f(A)+f(D)=f(B)+f(C)$), or maybe $f(A)+f(C)=f(B)+f(D)$? :?

As usually, typos. Thanks, grobber, it is your variant.  :blush:

Just to make sure: we consider all unit squares including those with sides at an angle to the coordinate axes, right?

Yes, Fedja, we speak about any kind of unit square, not those with sides parallel to the axes.

Then it is not so difficult. The proof I found consists of 4 steps:

1) Show that $f(A)+f(C)=f(B)+f(D)$ for any square $ABCD$ with integer side length.

2) Conclude from 1) that $f$ grows at most polynomially and, therefore, can be considered as a tempered distribution.

3) Note that $\text{supp\,}\widehat f=\{0\}$ and, therefore, $f$ is a polynomial ( 2) even allows to estimate the degree of $f$ by some fairly small number).

4) Prove that the only polynomials that satisfy the property are $a+bx+cy+d(x^2+y^2)$ (this is the most boring and ugly part in my solution; I did it more or less by brute force using the estimate for the degree I got from $2$. There should be some better way of doing this part, I believe).

Hope I haven't made a stupid mistake somewhere   :) .

Let $R$ be large. Then take an integer $n$ close to $\frac R{\sqrt 2}$ and consider a square with side $n$ whose one vertex $A$ is on the circumference of radius $R$ and 3 other vertices are inside the disk of radius $\frac R{\sqrt 2}+10$ (it is exactly what will happen if the vertex $C$ lies on the line through the origin and $A$ close to the origin). Since $f(A)=f(B)+f(D)-f(C)$, we conclude that the maximum of $f$ over the circumference of radius $R$ does not exceed 3 (actually, even 2, because $C$ can be kept in a fixed disk around the origin) times the maximum over the disk of radius $\frac R{\sqrt 2}+10$. From here the power growth is not hard to derive. If one is accurate enough, one can even obtain that $f$ grows not faster than $R^2$, which makes step 4 quite easy.


Find all natural numbers $m,n\geq 2$ such that the number

$\frac{1+m^{3^n}+m^{2\cdot 3^n}}{n}$   is an integer

It is easy to prove, that $n=3^k, \ m=1 (mod \ 3)$ solution. It has not other solutios, because $p|(1+x+x^2) \Longrightarrow p=3, x=1(mod \ 3) \ or \ p=1(mod \ 6), p|(x^3-1), p\not|(x-1)$. Therefore if p|n, p>3, then $p|(m^{3^{n+1}}-1) \ and \ p\not|(m^{3^n}-1) \Longrightarrow (m,p)=1 \ ord_p(3)|(3*3^n, ord_p(3)\not|3^n.$


I'm doing a school project on (European) mathematicians who went crazy.  All I can think of are Newton, Cantor, and Godel.  Anyone I'm missing?

[quote="blahblahblah"]an addendum to my earlier post.  Using the term crazy to apply to mentally ill people is insensitive, but Hardy IIRC tried to commit suicide a few times, and I'm pretty sure Littlewood succeeded.[/quote]

Insensitive, or a nice buffer to keep one away from how horrible these things really are?

Thanks for the suggestions.  I always thought Hardy was just quirky; I didn't know he was that far.

Evidently you should take all things I say with a grain of salt - I remembered correctly, Littlewood did not commit suicide (but Ramanujan tried...)

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Littlewood.html

[quote="themonster"]Turing went crazy (but he was more of computer scientist).  I wouldn't really consider Newton to have gone crazy.[/quote]Well, Alan Turing has a sad story. First than all, he was a mathematician, or at least he studied mathematics in Cambridge. I think that he proved the central limit theorem in probability. Now, remember that in those days there was no computer science as we know it today, it was a branch of mathematics (and somehow it still does).

He died of potassium cyanide poisoning. The cyanide was found on a half eaten apple beside him. An inquest concluded that it was self-administered. Now, by that time he was in a mental institute for something nobody nowdays goes to a mental institution (at least in the civilized world): Because he was gay. Yes, if I remember well the story, one day he was in a gay bar and he "seduced" a boy. A few streets from there, probably going to Turing's place, the guy hit an robbed him. Turing went to the police, the police found the guy and the guy, in his defense said how and where he had met Turing... and they considered Turing crazy and locked him on a mental institution.

Now, I noted an interesting thing on all of the people arguing here. You consider crazy basically to every one who had enough reasons to try to kill himself. I am not that sure if I would say that a potential suicide is really crazy.

Cantor, Turing and Nash.
Magnara, I never heard Newton was crazy. He was definitely not a psychopath.
Do you simply mean he behaved different from others? Most of the scientists are crazy under this standard.

Newton was deeply into mysticism and did some crazy things with Alchemy, or so I am told in my  european history class.

First of all, I would have to object to the term "crazy." It is not even a well defined term. There are such things as mental disorders, but its hard to say for certain that someone has one just by anecdotal stories about them. That would have to be done by a professional diagnosis.

That being said, Alexander_Grothendieck was one of the greatest mathematicians of the 20th century (and when time is told, will perhaps be considered among the greatest mathematicians who ever lived).

There is a brief biography of him here on Wikipedia http://en.wikipedia.org/wiki/Alexander_Grothendieck It discusses his major accomplishments and also tells a little about his (rather eccentric) life.

All mathematicians are crazy because they are so smart.  :lol:

I thought stuff like alchemy in Newton's time was normal lol.

[quote="smbelcas"][quote="djimenez"][quote="Fedor Petrov"]As for modern mathematicians, you have to mention Alexander Grothendieck.[/quote]I know he was a Field medal long time ago, but I don't know he has been mentally ill, and I can't find references about it, could you explain?[/quote]

Check out the AMS Notices from Oct and Nov 2004; there's an excellent 2-part article on the life and mathematics of Grothendieck.  You can download it at the addresses below.

http://www.ams.org/notices/200409/200409-toc.html
http://www.ams.org/notices/200410/200410-toc.html[/quote]I would have to strongly disagree with the assertion that Grothendieck is "crazy", absent further evidence.

There is a big difference between "eccentric" and "crazy".

I have a very slight connection -- my Ph.D. advisor, Michael Artin, was one of Grothendieck's associates and is mentioned (and quoted) several times in the articles linked to above.  (Note: you have to have an AMS account to access the articles, but it's free and painless to get one.)  I've read several sections of Grothendieck's seminal works [i]EGA[/i] and [i]SGA[/i], and they're quite an astonishing leap from what came before.  Modern algebraic geometry simply would not be anywhere near where is it today without Grothendieck's ideas.

Newton wedged a blade "betwixt my eye and ye bone as near to ye back side
of my eye as I could" to see the effects of deforming the retina. I'm not sure if that's crazy or just really really curious.


If three, standard, 6-faced dice are rolled, what is the probability that the sum of the face up integers is 16?

[hide]$ a \plus{} b \plus{} c \equal{} 16\implies(6 \minus{} a) \plus{} (6 \minus{} b) \plus{} (6 \minus{} c) \equal{} 2,\quad(6 \minus{} a),(6 \minus{} b),(6 \minus{} c)\in\mathbb Z\ge0$
$ \implies\frac {\binom{2 \plus{} 3 \minus{} 1}{3 \minus{} 1}}{6^3} \equal{} \boxed{\frac1{36}}$.[/hide]

[hide=The Easiest way to solve this problem.]Total is 216. Desired is 4+6+6;5+6+5;5+5+6;6+4+6;6+6+4;6+5+5 = 6. So, our answer is 1/36[/hide]

Because we have that the sum of the face up integers is 16, each dice can be expressed as a, b, and c, respectively.

Because each dice ranges from 1-6, we have that 1<=a<=b<=c<=6. Furthermore, we also have that a + b + c = 16.

Now we can simply use casework to find the total amount of possible arrangements.

We will check for each value of a until there are no more arrangements for that value of a.

We first note that the smallest possible a is 4, so we have
a | b | c
4 | 6 | 6
5 | 5 | 6
and
6 | 5 | 5
6 | 4 | 6
but, we note that the arrangements starting with a = 6 already exist as a rearrangement of the above so we cross them out to only have two arrangements:
(4, 6, 6) and (5, 5, 6)
We have 3!/2! * 2 = 6 arrangements total, and the total amount of possibilities is 6^3 = 216 arrangements to get that the probability is 6/216 = 1/36.


limit
real analysis
calculus
calculus computations

Dear everyone.

please read the following

find the value of

${ \lim_{n\to\infty}{ (\sum_{k=1}^{2n}{\frac{1}{ \binom{2n}{k} }} })^{n} }$

thanks for reading. :)

Choose $ \epsilon > 0$ and let $ n$ be sufficiently large so that $ 8n^{-1} < \epsilon$. Then

\begin{eqnarray*}\sum_{k=1}^{2n} \frac{1}{\binom{2n}{k}}
& = & 1 + \frac{1}{n} + \frac{2}{n(2n-1)} + \sum_{k=3}^{2n-3} \frac{1}{\binom{2n}{k}} \\
& \leq & 1 + \frac{1}{n} + \frac{2}{n(2n-1)} + \sum_{k=3}^{2n-3} \frac{1}{\binom{2n}{3}} \\
& = & 1 + \frac{1}{n} + \frac{2}{n(2n-1)} + \frac{3(2n-5)}{n(2n-1)(2n-2)} \\
& = & 1 + \frac{1}{n}\left( 1 + \frac{2}{2n-1} + \frac{6n-15}{(2n-1)(2n-2)} \right) \\
& \leq & 1 + \frac{1}{n}\left( 1 + \frac{2}{n} + \frac{6n}{n^2} \right) \\
& \leq & 1 + \frac{1+\epsilon}{n}
\end{eqnarray*}

Thus we have

$ \left( 1 + \frac{1}{n} \right)^n \leq \left( \sum_{k=1}^{2n} \frac{1}{\binom{2n}{k}} \right)^n \leq \left( 1 + \frac{1+\epsilon}{n} \right)^n$.

Taking liminf and limsup, we have

$ e \leq \liminf_{n\to\infty} \left( \sum_{k=1}^{2n} \frac{1}{\binom{2n}{k}} \right)^n \leq \limsup_{n\to\infty} \left( \sum_{k=1}^{2n} \frac{1}{\binom{2n}{k}} \right)^n \leq e^{1+\epsilon}$.

Since $ \epsilon$ is arbitrary, liminf and limsup coincide. Therefore the limit exists and is equal to $ e$.

감사합니다.^^
제대하실 때까지 몸 건강하세요~
제 남동생 군대 갔을 때가 생각나서 한글로 몇자 적어봅니다.


$ (a_n)_{n \in \mathbb{N}^*}$ is a sequence and $ a_1,a_2$ are given. Now this sequence satisfies the relation: $ a_{n\plus{}2}\plus{}\alpha \cdot a_{n\plus{}1}\plus{} \beta \cdot a_n\equal{}0$. Suppose that the roots of its characteristic equation are $ \lambda _1\equal{}\lambda _2$. Now I have a question. I don't know which of the following two statements is correct for the general term. I found each one in different books:

$ a)$ $ a_n\equal{}a \cdot {\lambda_1}^n\plus{}b \cdot n \cdot{\lambda_2}^n$ (Arthur Engel's book)

$ b)$ $ a_n\equal{}a \cdot {\lambda_1}^n\plus{}b \cdot n \cdot{\lambda_2}^{n\minus{}1}$ (Vasile Pop's book)

It's the same form.  We can convert from (b) to (a) as follows:
\[ a \lambda_1^n \plus{} bn \lambda_2^{n \minus{} 1} \equal{} a \lambda_1^n \plus{} \frac{b}{\lambda_2} \cdot n \lambda_2^n.\]

See a discussion of the general solution [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833]here[/url].

Let $ a(x) \equal{} \sum_{n \equal{} 1}^{\infty}a_nx^n$
$ a(x) \equal{} a_1x \plus{} a_2x^2 \plus{} x^2\sum_{n \equal{} 1}^{\infty}( \minus{} \alpha a_{n \plus{} 1} \minus{} \beta a_n)x^n \equal{}$
$ \equal{} a_1x \plus{} a_2x^2 \minus{} \alpha xa(x) \plus{} \alpha a_1 x \minus{} \beta x^2a(x)$
$ \Leftrightarrow a(x) \equal{} \frac {(a_1 \plus{} \alpha a_1)x \plus{} a_2x^2}{\beta x^2 \plus{} \alpha x \plus{} 1}$

We donate $ y \equal{} \frac {1}{x}$. If the characteristic equation $ y^2 \plus{} \alpha y \plus{} \beta \equal{} 0$ has roots $ \lambda_1 \equal{} \lambda_2 \equal{} \lambda$ then

$ a(x) \equal{} \frac {c_1}{y \minus{} \lambda} \plus{} \frac {c_2}{(y \minus{} \lambda)^2} \equal{} c_1x\sum_{n \equal{} 0}^{\infty}(\lambda x)^n \plus{} c_2x^2\sum_{n \equal{} 0}^{\infty}(n \plus{} 1)(\lambda x)^n$ for some constants $ c_1$ and $ c_2$ which depend on $ a_1$ , $ a_2$ and $ \lambda$.
$ a(x) \equal{} c_1x \plus{} \sum_{n \equal{} 2}^{\infty}(c_1\lambda ^{n \minus{} 1} \plus{} c_2n\lambda^{n \minus{} 2})x^n \equal{} \sum_{n \equal{} 1}^{\infty}a_nx^n$

 :roll:

[quote="moldovan"]$ (a_n)_{n \in \mathbb{N}^*}$ is a sequence and $ a_1,a_2$ are given. Now this sequence satisfies the relation: $ a_{n \plus{} 2} \plus{} \alpha \cdot a_{n \plus{} 1} \plus{} \beta \cdot a_n \equal{} 0$. Suppose that the roots of its characteristic equation are $ \lambda _1 \equal{} \lambda _2$. Now I have a question. I don't know which of the following two statements is correct for the general term. I found each one in different books:

$ a)$ $ a_n \equal{} a \cdot {\lambda_1}^n \plus{} b \cdot n \cdot{\lambda_2}^n$ (Arthur Engel's book)

$ b)$ $ a_n \equal{} a \cdot {\lambda_1}^n \plus{} b \cdot n \cdot{\lambda_2}^{n \minus{} 1}$ (Vasile Pop's book)[/quote]
Hmm... Try it out! I do recognize the one from Engel's book, but Vasile's could (maybe) work too. (Maybe it's just a typo?)

[quote="t0rajir0u"]It's not a typo; the two are completely equivalent.  Read nsato's post.[/quote]
You are right, i hadn't read his post  :blush:


modular arithmetic
trigonometry
ARML
number theory
absolute value

$ \mu_n$ is a sequence in strictly positive integers such that all of the odd numbers in the sequence is defined by:

$ \mu_n \equal{} \left| \sum^{n}_{k \equal{} 0} ( \minus{} 1)^k \binom{2n \plus{} 1}{2k}\right|$

(i.e. $ \mu_1 \equal{} \left| \sum^{1}_{k \equal{} 0} ( \minus{} 1)^k \binom{3}{2k} \right| \equal{} \left|\binom{3}{0} \minus{} \binom{3}{2}\right| \equal{} 2$).

If $ \mu_0 \equal{} k$ where $ k$ is a positive integer and that:

$ \displaystyle \frac {k \plus{} \mu_1 \plus{} \mu_3 \plus{} \mu_5 \plus{} \cdots \plus{} \mu_{1001} \plus{} \mu_{1003}}{503}$ is also an integer,

then find the smallest possible value of $ k$.

[NOTE: The original answer choices were wrong..]

A. 499
B. 501
C. 503
D. 505
E. 507

[hide="Solution, probably messed up near end"]
Summing the binomial expansions of
\begin{align*} (1 + i)^{2n + 1} & = {2n + 1 \choose 0} + {2n + 1 \choose 1} i - {2n + 1 \choose 2} - {2n + 1 \choose 3} i + \cdots \\
(1 - i)^{2n + 1} & = {2n + 1 \choose 0} - {2n + 1 \choose 1} i - {2n + 1 \choose 2} + {2n + 1 \choose 3}i + \cdots \end{align*}
and then dividing by $ 2$ gives
\[ \frac {(1 + i)^{2n + 1} + (1 - i)^{2n + 1}}{2} = \sum_{k = 0}^{n} ( - 1)^k {2n + 1 \choose 2k}
\]
as desired. Thus,
\[ \mu_n = \left|\frac {(1 + i)^{2n + 1} + (1 - i)^{2n + 1}}{2}\right| = \frac {(\sqrt {2})^{(2n + 1) + 1}}{2} = 2^{n}
\]
So $ k + \sum_{m = 0}^{501} \mu_{2m + 1} = k + \sum_{m = 0}^{501} 2^{2m + 1} = k + \frac {2^{1005} - 2}{3}$. We only need to work $ \pmod{503}$, and by Fermat's Little Theorem ($ 503$ is prime) $ 2^{502} \equiv 1 \pmod{503}$, so:
\[ 2^{1005} - 2 \equiv 2^{502\cdot 2 + 1} - 2 \equiv 0 \pmod{503}.
\]
Thus $ 503|k$ which implies that I made a off-by-one mistake somewhere.
[/hide]

Blah, you're right.

I did my solution wrong at the end.

[hide="What I did WRONG.."]

For some reason, I thought $ 2^1 \plus{} 2^3 \plus{} 2^5 \equal{} 2^{3^2}$ (I somehow saw the addition as multiplication. blah...) [/hide]

So yeah, the correct answer is 503.

[hide="My approach"]

It can be easily shown that $ \mu_n \equal{} 2^n$. Here's another way:

$ \sum^{501}_{k \equal{} 0} ( \minus{} 1)^k \binom{1003}{2k} \equal{} \binom{1003}{0} \minus{} \binom{1003}{2} \plus{} \binom{1003}{4} \minus{} \binom{1003}{6} \plus{} \cdots$

Remember this. Now, note that:

$ (1 \plus{} i)^{1003} \equal{} \binom{1003}{0} \plus{} \binom{1003}{1} i \minus{} \binom{1003}{2} \minus{} \binom{1003}{3}i \plus{} \binom{1003}{4} \cdots$

See what's going on here? The real term of $ (1 \plus{} i)^{1003}$ is the desired answer for $ \mu_n$. So, using DeMoivre's Theorem (and before doing that, factoring $ \sqrt {2}$ out from 1 and $ i$ and putting the number in sine and cosine), we see that the real term is $ \minus{} 2^{501}$. But, remember, there is an absolute value so $ 2^{501}$ and we know that $ \mu_n \equal{} 2^n$ as azjps showed.

From here is basically same thing as what he did. The key part was to realize that 503 was prime so you can use Fermat's Little Theorem. [/hide]

So, (other than wrong choices), what did you guys think of this problem?  :maybe:  (Good problem, bad problem, etc..?)

[quote="t0rajir0u"]It's more of an ARML problem than an AMC problem to me.  The issue I have with is it's pretty straightforward if you're familiar with the technique and very difficult otherwise.  To me, AMC problems are more "approachable" than that.[/quote]

Thanks. I'll remember that.  :)


function
algebra
polynomial
invariant
linear algebra
matrix
search

Does there exist a function $ f : \mathcal{M}_n(F) \to F$ such that $ f(AB) \equal{} f(A) f(B)$ that does not factor through the determinant, in the sense that there exists a multiplicative function $ g$ such that $ f(A) \equal{} g(\det A)$?

What if $ F \equal{} \mathbb{C}$ and $ f$ is required to be continuous?

What if $ f$ is required to be polynomial in the entries of $ A$?

What if $ f$ is required to be invariant under conjugation by a permutation matrix?

1) If $ f: GL_n(F)\rightarrow{F}^*$ is a morphism then see:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1919526860&t=218429
   I have not reread my first post but I think that it shows that $ f(A)\equal{}\psi(det(A))$ where $ \psi$ is a morphism of $ F^*$.
2) Now let $ g: \mathcal{M}_n(F)\rightarrow{F}$ be a morphism.
Case 1: $ g(I)\equal{}0$ then $ g\equal{}0$ and we are done.
Case 2: $ g(I)\equal{}1$. If $ A$ is invertible then necessarily $ g(A)\not\equal{}0$.
Let $ B$ be a non invertible matrix.
Case 2.1: $ g(0_n)\equal{}0$. If $ N$ is nilpotent then $ g(N)\equal{}0$.
 $ B$ is product of nilpotent matrices (Sullivan,Product of nilpotent matrices,Linear and multilinear algebra,vol 56,3/2008/ p. 311-317). This result is valid for $ n\geq{3}$ and I think, for any field.
Therefore $ g(B)\equal{}0$. We are done.
Case 2.2: $ g(0_n)\equal{}1$. There exists $ B'$ s.t. $ BB'\equal{}0$;then $ g(B)\not\equal{}0$. Thus if $ U$ is idempotent then $ g(U)\equal{}1$. $ B$ is a product of idempotent (for any field): see
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1511348630&t=271857
thus $ g(B)\equal{}1$.
Let $ P$ be an invertible matrix. $ g(PB)\equal{}g(P)\equal{}1$. Therefore $ g\equal{}1$ and we are done.
Conclusion: $ g$ factors through the determinant without supplementary hypothesis.
PS: it remains to solve the case 2.1 with $ n\equal{}2$. I go to bed.

The case 2.1 and $ n\equal{}2$.
The result of Sullivan is still valid  because $ \begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\equal{}\begin{pmatrix}0&0\\0&1\end{pmatrix}$. In fact in its theorem he speaks about 4 nilpotent matrices,that explains the condition $ n\geq{3}$.


So, in 2006 we didn't have divisions at all, and we realized this was bad because some schools just don't have a strong math contest tradition and other schools have a ton of MOPpers and it's not fair to put them together.

This year, we just put teams in A and B divisions. The B division Power Test grading was more lenient than in A division, and the individual tests were less difficult for B than for A.

But it seems that people still have some issues with the way divisions were this year. Here are some problems:
1) Some teams that should have been in B somehow ended up in A.
2) Some teams may consist of very strong students and also some inexperienced students. The strong students probably want to compete in A, but that's not fair to the inexperienced students who should be taking the B tests.
3) I am sure there are others too.

So what can we do about this? In the general comments thread, diophantiet suggested that instead of putting the whole team as a group into upper and lower, we could allow each individual to choose which division to enter. Then for team and power, I suppose the team would be A if any of the students were A. (Anything else would be unfair to B teams -- what if you have one really strong dude who pretty much does the Power & team tests himself, and he could be competitive in A, but because the rest of his team is B he gets to be B and creams everyone?)

Another method would be to use the Cody Bowl (NOML) model. In that contest (in some seasons), the individual test had 9 questions and you answered 6 of them. So you select the difficulty of the test yourself by which problems you choose to do. Then we can split the tests into two groups based on the maximum possible score from questions answered, and rank by score within those groups. So for example: if there were 15 questions with values 111222333444555 and you answer 10 of them, we could have division B be anyone who did questions with maximum possible score 25 or below (the most you can get without doing the last 4 questions) and division A is anyone who did higher value questions. Teams could specify for Team and Power tests which division they were (or we could do Team the same way as individual).

Of course, these aren't the only two possibilities. We could do something based on grade level, where students can move up if they want but not down (like the Texas A&M contest). We could also have no divisions for teams at all, but still have different levels for the individual tests. But although this may be fair to all teams for the individual tests, it is harsh for weaker teams on team and power and doesn't give them much chance at sweepstakes...

I'd like to hear everyone's thoughts about this. We want to come up with a system that doesn't force people to compete below their level, but also avoids forcing people to compete too far above their level.

ARML has everyone work on the same questions, with the same grading scheme, but uses the divisions so that B Division teams who don't have as competitive a score can still get prizes for being very good in comparison to teams of their caliber.  I think this method works very well - the only thing I'd worry about is our question-writers, who have a knack for making things way too hard.  Then the median score on some tests might actually be 0 (that didn't happen at all this year, by the way, as likely as it seemed)...

Maybe you can create a general test just like the HMMT people. Only, your score on the General Test is multipled by 3/4 or something to make it worth less and the general test can be composed of the #1 and #2 from combinatorics, number thoery and #1,#2,#3 from algebra and geometry (division A questions).

I think it would be a bad idea to let individuals from the same team be in different divisions. It just makes the competition seem very individual-based. Don't get me wrong: I very much like the individual aspect. It just seems really weird to me.

Here's one suggestion: make the A test a bit easier, and then everybody can take the same test. Now have teams sign up for A or B, but smooth them out during the competition. That is, if a B team gets in, say, the top ten or two or three of its participants get in the top 10 in a subject test (or various other criteria), it gets moved to the A division. Now, instead of separating the awards into A and B, divide them into everyone (in A or B) and just B (who aren't in the first category). Of course, you would need to fix this up a lot if you used it.

[quote="Xevarion"]Another method would be to use the Cody Bowl (NOML) model. In that contest (in some seasons), the individual test had 9 questions and you answered 6 of them. So you select the difficulty of the test yourself by which problems you choose to do. Then we can split the tests into two groups based on the maximum possible score from questions answered, and rank by score within those groups. So for example: if there were 15 questions with values 111222333444555 and you answer 10 of them, we could have division B be anyone who did questions with maximum possible score 25 or below (the most you can get without doing the last 4 questions) and division A is anyone who did higher value questions. Teams could specify for Team and Power tests which division they were (or we could do Team the same way as individual). [/quote]
If I understand this correctly, you decide your division during the test. I'm going to have to disagree with this method. This emphasizes meta-testing skills over math skills, causing people to spend time deciding and strategizing about which problems to do, instead of strategizing about the problems themselves. This kind of thing should be minimized. I haven't decided between the other alternatives yet.

I think it should be somewhat similar to HMMT pre-2006. Those tests were doable for everyone to an extent. There were enough points on the table so that it was easy enough so that there were a limited number of zeros but difficult enough so that there weren't a ton of perfects. I think it should be a little bit more difficult than HMMT pre-2006, but easier than the current pumac division A problems. As in, my theory is that everyone takes the same (easier) individual test, but different team tests. I was surprised to find out that the b team had taken a different individual test than us. It led to much scoffing and what not. The team test this year wasn't that bad. Even the power test had many doable parts.

My suggestion: Same individuals, different team tests OR Same individuals, same team tests, compared relative to their divisions.

The NOML style thing is just too much strategizing, I agree.

I see one problem with the teams being split in divisions; I haven't looked at the B division problems, but I assume that they are easier, and that would help those teams in which some people are taking easier tests, which could be a problem for the sweepstakes.

I guess I'd be in favor of removing the divisions.  You could do something like HMMT with the general test, or you could just make the problems with a wider range of difficulty.

It seems like it should be possible to have maybe 15 questions instead of 10, where the first five are easier, and the last 10 are like what you have now.  Then, fewer people should be getting zeros, but the top scores won't be affected much, because the best people should be able to do the easy problems quickly and focus almost all of their time on the harder ones.

Now that I think about it, that would be exactly like combining the A and B divisions. Hmmm.

Eh c'mon now, nobody likes doing easy problems... it's not challenging! I really dislike tests where, no matter your skill, the easy problems are non-trivially valuable. So please don't combine the tests.

They don't need to be heavily weighted, the weighting could be like 111122224444888 or something like that

I think the goal should be to weight the problems such that a good competitor will get about the same number of points per unit of time, regardless of which problems he attempts (I know that's very nontrivial, but that could be the goal)

Just my $ \$0.02$

[quote="SmitSchu"]I think the goal should be to weight the problems such that a good competitor will get about the same number of points per unit of time, regardless of which problems he attempts (I know that's very nontrivial, but that could be the goal)[/quote]
But hard problems should not necessarily take such a long time - they should sometimes simply require a step that requires seeing a trick - where time is irrelevant. But I do agree that there should be some easy problems. There always need to be problems hard enough to challenge the strongest competitors, but easy problems (in moderation) can serve as a way to give less strong competitors something to work on and can provide stronger competitors with a warm-up / confidence booster (when I got the number theory test at PUMaC this year, I was like "oh crap - I'm having trouble with #1.")

If we do divisions the same way next year, we will probably aim to make B division significantly easier, and keep A division only slightly easier (if at all) while making sure that there are a couple of reasonably doable questions at the beginning of each test. (If we still have a 5 question overlap between divisions, both of these could be accomplished simultaneously.) But we are still quite open to other options. Personally, right now I like the idea of the easier General test and harder subject tests, but I'm still keeping my eyes peeled for a better method.

If the tests are otherwise the same, I don't think it's a good idea to grade a power round more "leniently" for Division B teams.  Maybe other people would disagree, but personally that seems kind of... weird.  Plus, if its anything like ARML, Division A and B teams like to be able to compare their scores to each other.  In fact, I would think it would be preferable to have teams in different divisions take the exact same tests.

I was pushing for that all along. :D

DOES ANYONE CHECK THIS ANYMORE??? I feel so disconnected from PUMaC!   :P

Hehe; hello taotao. I believe this page is asphyxiating.


How many integers from 1 to 1,000,000, inclusive, are neither perfect sqares nor perfect cubes?

[color=blue]kyyuanmathcount: 'scuse me, blue is my color - please don't use color while posting normal problems.

Changed title because "easiestproblemintheworld" is not helpful and creates spam posts.  [/color]

I wouldn't say this is the easiest problem in the world since it involves Inclusion-Exclusion Principle.

He's obviously exaggerating, but yes this is a fairly easy problem with a basic application of PIE (which I didn't know about the first time I saw it...hm).

[quote="SLX314"]What is PIE? :?:  :huh:[/quote]

Principle of Inclusion - Exclusion.

It is when you solve a counting problem by repeatedly overcorrecting your count.

[hide="huge hint"] Take the original number of numbers, subtract the number of perfect squares, subtract the number of perfect cubes, add the number of perfect sixth powers and you'll magically get the answer[/hide]

Nobody wants to, so...

[hide="Answer"]There are 1000 perfect squares, 100 perfect cubes, and 10 perfect sixth powers from 1 to 1000000, inclusive.

$1000000-(1000+100-10)=\boxed{998910}$[/hide]

Good Job

That is the right answer, and you used PIE effectively

[quote="anirudh"][hide]
You have to find how many perfect squares, add it to the number of perfect cubes, and subtract it from the perfect sixths.

So, the number of perfect squares is 1000

The number of perfect cubes is 100.

The number of perfect sixths is 10.

So, the answer is:

$1000000+1000+100-10\Longrightarrow 998910$[/hide][/quote]
You switched all minuses to pluses and vice versa, but your answer is correct.

whoops. :blush: forgot to type the brackets.

What is the reason for adding the perfect sixths?

-jorian

pie
The message is too small. Please make the message longer before submitting.

When one subtracts the perfect squares and cubes, the perfect squares and cubes, or perfect sixths, are subtracted twice.  So to make up for that you add back the perfect sixths.  That is basically what PIE is.

Oh!, I didn't know those were sixths

like $1^{6}=1$ and $2^{6}=64$

-jorian

It follows from basic exponent laws.  Like $({n^{2}})^{3}=n^{6}$.


what two-digit number belongs in the blank?

96 15 26 39 44 __

no, although i'd love to hear the explanation for that answer :P

I would assume that it is 61 because if you take the mod 100 of 96+19 you get 15 then you can add 11 to get 26, 13 to get 39, and 5 to get 44 we would then add another prime 17 to get to 61.  The reason for subtracting 10 from 15 is that it was not prime?  Either that or this is BS and we could take the systematic approach of 10, 11, 12, 13...


limit
function
logarithms
real analysis
real analysis unsolved

1) Let $ x_n $ be an unique root of the equation :
  $ (x+n^2).(2x+n^2)...(nx+n^2) = an^{2n} $
 Find $ \lim_{n\to \infty}x_n $ ?

2) Prove that : $ \forall n \in Z^+ $ , the equation :
  $ \frac{1}{x} +\frac{1}{x-1}+...+\frac{1}{x-n} = 0 $
 has an unique root : $ x_n \in (0,1) $ . Find $ \lim_{n\to \infty}x_n $ ?

Try them , MLers ! I think they are not so difficult  :(

No, they aren't difficult at all. I saw the first one given in our district olympiad in Bucharest for example.

Okay, here's a solution to the second one.

First, uniqueness.  Each term on the left-hand side is a strictly decreasing function of $x$ on $(0, 1)$.  Thus the entire left-hand side is strictly decreasing on $(0, 1)$.  As $x$ approaches 0, the left-hand side approaches infinity.  As $x$ approaches 1, the left-hand side approaches minus infinity.  Thus there is a unique root of the equation in $(0, 1)$.

Next, we will show that the limit is 0.  For the root between 0 and 1, we have
\begin{eqnarray*}
\frac{1}{x}
  & =    & \frac{1}{1-x} + \frac{1}{2-x} + \cdots + \frac{1}{n-x}  \\
  & \ge & \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} \\
  & \ge & \ln n \, .
\end{eqnarray*}
Thus $0 \le x \le \frac{1}{\ln n}$, which means the limit is 0.

Yes , the second one is easy .
  We can consider : $ F(x)=(x-1)(x-2)...(x-n) = (x-i).Q_i(x) $
    $ F'(x)= \sum{Q_i(x)} $
  On other hand : $ F(1)-F(0) = F'(c)=0 $ , where $ c \in (0,1) $
   Put $ f(x)=\frac{1}{x}+\frac{1}{x-1}+...+\frac{1}{x-n} $
   $ f'(x) <0 \forall x $
  So the equation has an unique root on $ (0,1) $ and the rest is similar to yours , Ravi B .
  I think the first one much more difficult  :)

Ok, I will suppose that $ a>1$ and $x_n>0$. The existence of such $x_n$ is easy by reasons of continuity. Next, write $ lna=\sum_{i=1}^{n}{ ln(1+\frac{i x_n}{n^2})}$. From here it easily follows that $x_n$ is bounded. Next, take $\epsilon>0$ and observe that for sufficiently large $n$ one has $ 1-\epsilon<\frac{ ln(1+\frac{i x_n}{n^2})}{\frac{i x_n}{n^2}}<1+\epsilon$. Next, just sum up and you'll find the limit.


calculus
derivative
function
analytic geometry
graphing lines
slope
floor function

This is a somewhat interesting problem I came across some time ago, and I didn't get any satisfactory answers, so I was wondering if someone here could explain it to me.

Obviously, the derivative of x squared is 2x, by the power rule.

But say you rewrite the function x squared as x times x, or the sum of x x's.

(2*x is x+x, 3*x is x+x+x, 4*x is x+x+x+x, so x*x is x+x+x+x... (x times))

In the case that x is a fraction, simply add a fractional coefficient of x to the end (so 2.5^2 would be 2.5+2.5+.5(2.5)=6.25. This makes the function continuous.

This is clearly equivalent to x^2; there are x x's being added, so when you collect like terms, you have x times x.

However, if you try differentiating the function in this form (as a sum of x's), as you differentiate each part of the sum you get a derivative of 1, and for the fractional term you get a derivative equal to the coefficent of x.

For instance, if x is 3, you get (x+x+x), which has a derivative of 3.

So you have x x's, and each x contributes 1 to the slope, so you have a derivative of X*1, which is x- but this can't be true, because the derivative of x^2 is 2x.

So why the difference if the functions are equivalent?

It's because of the chain rule. You're treating the x-variable in summing x summands as if it's a constant when differentiating.

[quote="johnnyboy"]simply add a fractional coefficient of x to the end (so 2.5^2 would be 2.5+2.5+.5(2.5)=6.25. This makes the function continuous.
...
 [b]for the fractional term you get a derivative equal to the coefficent of x.[/b] [/quote]

The bolded statement is not true.  Letting $ g(x) \equal{} x \minus{} \lfloor x \rfloor$ denote the fractional part of $ x$, what this argument does is write

$ f(x) \equal{} x^2 \equal{} x(\lfloor x \rfloor \plus{} g(x)) \equal{} x \lfloor x \rfloor \plus{} x g(x)$.

If $ x$ is not an integer, $ \lfloor x \rfloor$ is locally constant and

$ f'(x) \equal{} \lfloor x \rfloor \plus{} x g'(x) \plus{} g(x) \equal{} x \plus{} x g'(x) \equal{} x \plus{} x \equal{} 2x$;

however, the argument you give ignores the fact that $ g(x)$ is changing and therefore omits the $ x g'(x)$ term in the derivative.


1^3+2^3+...+n^3=(1+2+3+...+n)^2
Demonstrating it by 2 methods
Pleased help me and sorry I can not type in LATEX

[quote="ngan_ta2001"]$ 1^{3}\plus{}2^{3}\plus{}...\plus{}n^{3}\equal{}(1\plus{}2\plus{}3\plus{}...\plus{}n)^{2}$
Demonstrating it by 2 methods
Pleased help me and sorry I can not type in TALEX[/quote]
i edited the thingy.

umm...one is induction, once again-too lazy to do it
2, i dont know

Start with this lemma: $ \sum_{i\equal{}1}^{n}i^{2}\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$
We can prove that by induction or using the same idea we are going to use here...

First notice that:
$ \sum_{i\equal{}1}^{n}i^{4}\plus{}(n\plus{}1)^{4}\equal{}\sum_{i\equal{}0}^{n}(i\plus{}1)^{4}$
Then:
$ \sum_{i\equal{}1}^{n}i^{4}\plus{}(n\plus{}1)^{4}\equal{}\sum_{i\equal{}0}^{n}(i^{4}\plus{}4i^{3}\plus{}6i^{2}\plus{}4i\plus{}1)$
$ \sum_{i\equal{}1}^{n}i^{4}\plus{}(n\plus{}1)^{4}\equal{}\sum_{i\equal{}0}^{n}i^{4}\plus{}4\sum_{i\equal{}0}^{n}i^{3}\plus{}6\sum_{i\equal{}0}^{n}i^{2}\plus{}4\sum_{i\equal{}0}^{n}i\plus{}\sum_{i\equal{}0}^{n}1$
Its obvious that:
$ \sum_{i\equal{}1}^{n}i^{k}\equal{}\sum_{i\equal{}0}^{n}i^{k}$, $ k\in N$
Then we get:
$ (n\plus{}1)^{4}\equal{}4\sum_{i\equal{}1}^{n}i^{3}\plus{}\frac{6n(n\plus{}1)(2n\plus{}1)}{6}\plus{}\frac{4n(n\plus{}1)}{2}\plus{}(n\plus{}1)$
$ \sum_{i\equal{}1}^{n}i^{3}\equal{}\frac{(n\plus{}1)[(n\plus{}1)^{3}\minus{}1\minus{}n(2n\plus{}1)\minus{}2n]}{4}\equal{}\frac{(n\plus{}1)(n^{3}\plus{}n^{2})}{4}\equal{}\Big(\frac{n(n\plus{}1)}{2}\Big)^{2}$
Since:
$ \sum_{i\equal{}1}^{n}i\equal{}1\plus{}2\plus{}...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$, we are done.

Proof  of lemma:
Use the same idea starting from:
$ \sum_{i\equal{}1}^{n}i^{3}\plus{}(n\plus{}1)^{3}\equal{}\sum_{i\equal{}0}^{n}(i\plus{}1)^{3}$

I hope its clear!

Or look at this:
http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php
Bottom most one.


We are given convex quadrilateral ABCD. Each of its sides is divided into N line segments of equal length. The points of division of side AB are connected with the points of division of side CD by straight lines and the points of division of side BC are connected with those of side DA. This forms $N^2$ smaller quadrilaterals. From those we choose N quadrilaterals in such a way that there are no two quadrilaterals on the same "line" or "row". Prove that the sum of the areas of these is equal to the area of ABCD divided by N.

I'll just write the condensed form of the solution, then the rest (maybe tomorrow).

Let $a_{ij}$ be the area of the $ij$ "cell". I showed that the quantity $f(x,y,z,t)=\frac{(a_{xy}+a_{zt})-(a_{xt}+a_{zy})}{(x-z)(y-t)}$ is an invariant for all $x,y,z,t\in\overline{1,n},\ (x-y)(z-t)\ne 0$. We then prove the theorem for $n=2$, and this gives us $f(1,1,2,2)=0$, which then solves the problem.

One crucial observation is that if we label the points of the network (including the vertices of the quadrilateral) as $A_{ij},\ i,j\in\overline{0,n}$, then the lines $A_{i1}A_{in}$ divide the lines $A_{1j}A_{nj}$ in equal parts, and vice-versa.

Lemma I.

Given a quadrilateral XYWZ (convex or not) if P and Q are points on XY and WZ respectively such that XP/XY = WQ/WZ=u and if R and S are points on XW and YZ respectively such that XR/XW = YS/YZ=v then if  T=PQ^RS it holds that RT/RS=u and PT/PQ=v.

Lemma II.

Given a convex quadrilateral XYWZ, if X1 and X2 are points on XY such that XX1=X1X2=X2Y and if W1 and W2 are points on WZ such that WW1=W1W2=W2Z, then it holds that the area of the quadrilateral W1W2X2X1 is the arithmetic mean of the other two quadrilaterals.

Lemma III.

Given a convex quadrilateral XYWZ, if K, L, M and N are the midpoints of XY, YZ, ZW and WX respectively and if O=KM^LN, it holds that (NXKO)+(OLZM)=(KYLO)+(OMWN).

Sol.

From Lemma I we have that all lines involved in the figure are mutually divided in n equal parts.

Let's call the smallaer quadrilaterals as the element of a matrix. More precisely we call (0,0) the smaller quadrilateral  whith vertex D, (0,1) the following smaller quadrilateral in the row (towards C), (1,0) the following smaller quadrilateral in the column (towards A), and so on.

If A is the area of the smaller quadrilateral (0,0)  , let's call A+x the area of the smaller quadrilateral (0,1) and call A+y the area of the smaller quadrilateral (1,0); x and y could be, of course, positive or negative but doesn't matter.

From lemma III we have that  the area of the s.q. (1,1) is A+x+y.

Is easy to see now, applying iteratively the Lemma II, that the area of the s.q. (h,k) is equal to A+kx+hy.

Therefore choosing N quadrilaterals in such a way that there are no two quadrilaterals on the same "line" or "row", we have to add all the s.q. where k and h assume one time and only one time the valus from 0 to N-1.

Then we have a total area of N*A+x(N-1)N/2+y(N-1)N/2 given that this is indipendent from the choice on n-upla of s.q. we have that this in 1/N the total area.


I'm in eighth grade, and so far, according to my parents, there are other stuff that I should be doing instead of math. However, now, after two months of only doing the USAMTS, I can't seem to do 5 problems right on the AIME (which was my lower boundary before September).

If you're getting fives on the AIME as a 8th grader, I think you're somewhat at a good point. There's still a lot of time to get better.

Are you asking why your AIME score didn't go up after two months of USAMTS?

No, he's asking why it went down.

As a side note, is this why you're not coming to math circle anymore?

No. Whenever I DO have time to go the math circles however, my piano teacher announces a recital at her house- ON THE SAME DAY! Otherwise I don't have time because of my crushing school homework and some other stuff.

Then diminish or even stop math temporarily. Each person encounters some tight schedules.

Like worthawholebean said, don't worry about it at all. Scores go up and down [i]naturally[/i]. Since you've been busy with other tasks, it will go down more than it would. But this does not mean that you became any dumber or anything. Just relax and find a time that you can do math in regular basis. If you can't, just step back and play piano or something.

It's not that big deal. There are many people who wish they can do about 5 problems in AIME, and they are seniors in high school.

I know this sucks to deal with, but you have to trust that as you put in more effort you will get better. Generally I find the USAMTS leads you to lots of different strategies for solving the problems given, most of which don't work. Your natural intuition on how to attack some of these problems might have been thrown off because of the wide variety of choices you encountered in the USAMTS. For the scheduling commitment issue: only go to an event that is scheduled well in advance, so if your piano teacher is getting you with last-minute invitations to recitals over long-standing commitments to your math circle, use the excuse that she did not schedule far enough in advance.

Also, 5s on AIME as 8th grader? I didn't even know what the AIME [i]was[/i] as an 8th grader.

Don't worry. Just keep doing math because you enjoy it. It's okay if other things get in the way. It justs happens that these questions are two different kinds. Your scores will vary. However, the more you practice, the better you'll get.


number theory
least common multiple
floor function
logarithms
prime factorization
number theory open

Let $ m\geq 1$, $ n \geq 1$, $ m\neq n$ and let $ k\geq 2$ be natural numbers.

Find the positive integer solutions of the following equation

$ H_{n}\equal{}kH_{m}$,

where $ H_{p}\equal{}\sum\limits_{j\equal{}1}^{p}\frac{1}{j}$ denotes the $ pth$ harmonic number.

Equivalent statement of the problem:

Find the positive integers $ m\geq 1$ and $ n\geq 1$, $ m\neq n$ such that

$ \frac{H_{n}}{H_{m}}\in \mathbb{Z^{\plus{}}}$.

Assume towards a contradiction that such a solution exists.  Let $ d_n\equal{}\mbox{lcm}(1,2,\ldots,n)\equal{}\prod_{p\le n} p^{e(j)}$ where $ e(j)\equal{}\lfloor \log_p n \rfloor$.  Then,
\[ H_n\equal{}\dfrac{d_n\plus{}d_n/2\plus{}\cdots\plus{}d_n/(n\minus{}1)\plus{}d_n/n}{d_n}
\]
Let $ \nu(N)$ denote the 2-adic valuation of $ N$ (i.e. the exponent of 2 in the prime factorization of $ N$).  We let $ b$ be the largest positive integer such that $ 2^b\le n$.  With a little work I believe one can show that
\[ \nu(H_n)\equal{}\minus{}b\equal{}\minus{}\lfloor \log_2 n \rfloor.
\]
(Notice that by showing this one gets that $ H_n$ is not an integer unless $ n\equal{}1$.)  So if $ H_n\equal{}kH_m$, then we deduce that
\[ \nu(k)\equal{}\lfloor \log_2 m \rfloor \minus{} \lfloor \log_2 n \rfloor .    ~~~~~~~~~     (*)
\]
Of course we want to assume that $ n>m$.  This forces $ k$ to be odd and hence $ k\ge 3$.  It follows that $ n>3m$, indeed
\[ 1\plus{}1/2\plus{}\cdots \plus{}1/n \equal{} H_n \ge 3 H_m > H_{3m}.
\]
Altogether, from (*) we obtain the desired contradiction since
\[ 0\le \nu(k) \le \lfloor \log_2 m \rfloor \minus{} \lfloor \log_2 3m \rfloor \le \minus{}1.
\]


trigonometry
calculus
calculus computations

$ x_{n}$
$ x_{0}\equal{}a$
$ x_{n\plus{}1}\equal{}x_{n}\plus{}sin x_{n}$
Prove that $ (x_{n})$ converge.

you only mus look at the piture of your equation

I don't think that .It is depend on $ a$ but is not a .
If we call limit is $ l$ then
$ \sin{l}\equal{}0$

let's questions will be next:
for all  $ x_{0} \equal{} a \in ( \minus{} oo, \plus{} oo)$
1) prove that $ x_{n}$ converge .
2) find  the limit of $ x_{n}$ when $ n \minus{} > \plus{} oo$
3) find asymptotical expression of $ x_{n}$ when $ n \minus{} > \plus{} oo$

If no one student (only student) will not post  solution in 3 days then I will post my nice  full solution.


Hi, I just wanted to ask if any of you have any good games that are online, because I am so bored.

[hide]Its a really good thing we have AOPS[/hide]

lol, defend your castle is the craziest game. one of my favorites is snowcraft. i don't have a link but you can probably google it. its pretty fun.

I would recomend [url=http://www.freeciv.org/index.php/Freeciv]FreeCiv[/url], you can download the client and the server, and play it standalone in your computer, or play it with some friends online. It is not totally clear at the begining, but very addictive once you know how to play it :D .

Best,

[quote="MithsApprentice"]One of our favorite online games at MOP (I think at one point in time we had the entire computer lab playing it) was [url=http://www.xgenstudios.com/castle/]Defend Your Castle[/url]. It's so stupid, but it's so addictive. :-) You might be able to close the window after about 30 levels...if you're lucky.[/quote]

Youve gotten to levl 30?!?!!?! Woah....omg...a little after i get in the teens i lose...

[quote="DarkKnight"]New game i found. It's pretty fun and addictive. Heres the link:

[url=http://www.geocities.jp/lledoece/nanaca-crash_v105.swf]www.geocities.jp/lledoece/nanaca-crash_v105.swf[/url]

My records 6811.02m. and the highest i've ever gone is around 350.[/quote]

It's down.

Allow me to provide this link, then:

http://megami.starcreator.com/nanaca-crash/

It's a really addicting game. If you have ever played that penguin launching dealie, you'll get the hang of this pretty quickly.

http://www.albinoblacksheep.com/flash/metaphysik.php

Maplestory's signups are closed, so there's no point in posting the URL :P .

I don't know if this is on purpose
[hide]I hope its not[/hide]
but my computer got a parasite from defend your castle

Very simple, pointless, addicting game. Reflexes a must

[url=http://www.transience.com.au/el/elgameo.html]Immigration Game[/url]

Personally I think all computer games are gay, but try this..... http://www.slimezone.com

[quote="Treething"]www.albinoblacksheep.com/flash/metaphysik.php

Maplestory's signups are closed, so there's no point in posting the URL :P .[/quote]

[hide="Ball Revamped II Score"]
[img]http://img225.exs.cx/img225/1631/ballrevamped0tf.jpg[/img]
[/hide]

I used to play Runescape, and i have a pretty high level acct, in high scores for all but a few categories. If anyone wants it pm me. BTW, its a member.( im not gonna give you an active subscription so you'll have to subscribe.)

I kinda gave up on online games...especially runescape.  But if you go to http://www.neave.com/games you can find some fun and "old skool" games like Pac-Man and Tetris.  But for a good challenge try out hexxagon.  Quite addicting.

I have a gaming rig laptop (modified IBM thinkpad) and a ps2. I barely use them for games nowadays. TOO busy. Halo is the best game, and GTA is the best series ever. (I don't support uber-violence, BTW).

Oh, and Secret Asian, which state are you it?

I'm guessing you are either Chinese or Korean

[quote="alan"]i came across [url=http://www.zeronews-fr.com/flash/pandapang.php]this[/url]

...and was pulled in for a good half hour ;)[/quote]

i gotta admit, that game was sure addicting.  :D

[url=http://www.2flashgames.com]try this one.[/url] more than 200 flash games for free!

[quote="GoBraves"][quote="mcalderbank"]why are computer games 'gay'?[/quote]

cuz there isnt any good ones.......in plus who needs computer games when u have ps2 and x-box and gamecube???[/quote]

Hm. On the contrary, why buy ps2 and x-box and gamecube when your gonna buy a computer anyway....there are good ones, just depends on your taste...

[quote="captcha000"]Hm. On the contrary, why buy ps2 and x-box and gamecube when your gonna buy a computer anyway....there are good ones, just depends on your taste...[/quote]
cuz you can modify your xbox so that it can copy games/dvds onto a hard drive so you never have to buy another game again. (renting now seems expensive)
*this is not saying that i do this or know anyone who has done this.*

[quote="mcalderbank"]why are computer games 'gay'?[/quote]

Because they're all the same genderless, meaning they have the same gender, and since many new computer games come out, they must reproduce a lot in order to get so many games. In order to reproduce, they must - ok, I'll shut up now.

[img]http://img.photobucket.com/albums/v259/drspyder/castle36million2.jpg[/img]

I have way too much time on my hands. Stage 92.

There's a website where you can challenge other people, and if you get in the top 1000 in that game, you can see your ranking. It's

[url]http://www.gamerival.com[/url]

Halo 2 here, on xbox live, anyone got a screen name?

There's a good site I know. I think its [url=http://worldbreak.com/games]this.[/url] :wacko:  :wacko:  It's good!

you cant go wrong with this game....

http://www.miniclip.com/smashing.htm

i've got to level 39 in one go..... that was during xmas break.... when boredom = max. but i have gotten to level 50 after entering my passwords from previous levels....

[quote="Treething"]74,000 points in Panda Pang on my first try.[/quote]Thats great! :first:

I'm surprised this doesn't seem to have been mentioned...

"Orisinal: Morning Sunshine"
http://www.ferryhalim.com/orisinal/

...some simple games with a light and airy feel.

MUonline-a MMORPG. [url]http://www.muonline.com[/url]

[hide][hide][hide]
if you want to get 300X exp, try [url]mastermu.servegame.com[/url][/hide][/hide][/hide]

i keep losing @ level 12 on panda. lol. that 'blob' is HUGE!!!

yea your're supposed to not rapid fire it take it apart one part at a time


Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ at $D$. At $D$ a tangent is drawn cutting leg $CA$ at $F$. Prove each of the following:

i. $DF$ bisects $CA$.
ii. $\angle DF = FA$.
iii. $\angle A = \angle BCD$.
iv. $\angle CFD = 2\angle A$

[hide="hi"]
because $D$ is a point on the circle with $AB$ as the diameter, we can see that $m\angle CDB = 90$, hence $CD \perp AB$.
Define $m\angle CAB = \alpha$ and $m\angle CBA = \beta$.. then $\alpha+\beta = 90$ or they are complementary.
Define $R$ as the center of the circle (the only one).. we know that $RC=RB=RD$, so: $m\angle RDB = m\angle RBD=\beta$ and $m\angle RCD = m\angle RDC=\alpha$ (by def. isoceles triangle)
now since $FD$ is a tangent to the circle, we know that $FD \perp DR$, or $m\angle FDC+m\angle CDR = 90$, or $m\angle FDC = \beta$.. then $m\angle FDA=\alpha$.

ok, so:
$\alpha=m\angle FAD=m\angle FDA=m\angle RCD=m\angle RDC$
$\beta=m\angle FCD=m\angle FDC=m\angle RDB=m\angle RBD$
so its basically angle chasing i think and everything is easier after that..
[/hide]


Two circles are tangent to each other at the point P.M & Q are the centers of those circles.PA is a Chord of circle M and PB is a chord of circle Q.(PB is the extension of PA).Prove that MA and QB are parallel to each other.

--- not classroom math---
proofs do not pelong here

draw MQ
MQ passes through P since MP and QP are radii to the point of tangency (and these are part of the same line)
angles MPA and QPB are congruent by vertical angles and these angles are both congruent to angles MAP and QBP by isosceles triangles (2 sides are radii of the same circle)
since angles, MAP and QBP are congruent, lines MA and QB are parallel (converse of alternate angles theorem)

Tangent means touching at one and only one point. Use Google or wiki for these questions next time. As for the proof,

[hide]

Notice MP=AM and PQ=PB, as radii in a circle are equal. Also, <MPA=<QPB, as they are vertical angles. Since MAP and PBQ are isosceles, <AMP=<PBQ. Thu, AM||QB

[/hide]


Given $ a, b, c \geq\ 0$.
Prove that: $ \frac{1}{(a\minus{}b)^2} \plus{} \frac{(1\plus{}\sqrt[]{3})^2}{(b\minus{}c)^2} \plus{} \frac{(2\plus{}\sqrt[]{3})^2}{(c\minus{}a)^2} \geq\ \frac{3(3\plus{}\sqrt[]{3})^2}{(a\plus{}b\plus{}c)^2}$

I think this is the final problem I' ll post on forum.
I have promised with a good friend that " I'll only post my ineq on toanthpt.net". And today, toanthpt.net is come back.
I very happy but then, I feel sad.
Because althought I don't want but I must to say goodbye with mathlinks.ro, with the new friends as arquady,...who I like to discuss with.
I hope that forum 'll be improve so much  :)
And if anyone want to discuss with me, you can go to my topic:
http://toanthpt.net/toanthpt/showthread.php?t=9244
(there are many nice ineq which wait for u)
 :)

I know that it's too easy. But the thing I want to mention that how I did to creat this ineq (I don't use calculator).And because equality can occurs, so I think it's an interesting problem.

[quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac {1}{(a \minus{} b)^2} \plus{} \frac {(1 \plus{} \sqrt []{3})^2}{(b \minus{} c)^2} \plus{} \frac {(2 \plus{} \sqrt []{3})^2}{(c \minus{} a)^2} \geq\ \frac {3(3 \plus{} \sqrt []{3})^2}{(a \plus{} b \plus{} c)^2}$[/quote]
setting F(a,b,c)=$ \frac {1}{(a \minus{} b)^2} \plus{} \frac {(1 \plus{} \sqrt []{3})^2}{(b \minus{} c)^2} \plus{} \frac {(2 \plus{} \sqrt []{3})^2}{(c \minus{} a)^2} \minus{} \frac {3(3 \plus{} \sqrt []{3})^2}{(a \plus{} b \plus{} c)^2}$
assume $ c\equal{}min(a,b,c)$
$ F(a,b,c)\ge F(a\minus{}c,b\minus{}c,o)$
we need prove that $ F(a\minus{}c,b\minus{}c,o)\ge 0$ it's proven easy

Hi,and when does equality occur ?
PS: My solution is only Am-Gm :D


calculus
integration
trigonometry
calculus computations

3)

$\int \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}dx$
$= \frac{1}{a^{2}}\int \frac{\sec^{2}x}{\left(\frac{b}{a}\right)^{2}+\tan^{2}x}dx$

let $\tan x = \frac{b}{a}\tan t$, then $\sec^{2}x dx = \frac{b}{a}\sec^{2}t dt$ and

$= \frac{1}{a^{2}}\int \frac{\frac{b}{a}\sec^{2}t}{\left(\frac{b}{a}\right)^{2}+\left(\frac{b}{a}\tan t\right)^{2}}dt$
$= \frac{1}{ab}\int dt$
$= \frac{1}{ab}t+C$
$= \frac{1}{ab}\arctan \left( \frac{a \tan x}{b}\right)+C$

[quote="eshban"]1) $\int\frac{\tan^{-1}\sqrt{x}}{\sqrt{x(x+1)}}dx$[/quote]

Let $u=\tan^{-1}\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}(x+1)}$ also we have that $\tan u = \sqrt{x}\Rightarrow \sqrt{\tan^{2}u+1}=\sqrt{x+1}$ so that $\sqrt{x+1}=\sec u$ and hence the integral becomes

\[\int\frac{\tan^{-1}\sqrt{x}}{\sqrt{x(x+1)}}dx = \int\tan^{-1}\sqrt{x}\cdot\sqrt{x+1}\cdot\frac{dx}{\sqrt{x}(x+1)}=\int u\sec u\,du = \cdots\]

[quote="benorin"][quote="eshban"]1) $\int\frac{\tan^{-1}\sqrt{x}}{\sqrt{x(x+1)}}dx$[/quote]

Let $u=\tan^{-1}\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}(x+1)}$ also we have that $\tan u = \sqrt{x}\Rightarrow \sqrt{\tan^{2}u+1}=\sqrt{x+1}$ so that $\sqrt{x+1}=\sec u$ and hence the integral becomes
\[\int\frac{\tan^{-1}\sqrt{x}}{\sqrt{x(x+1)}}dx = \int\tan^{-1}\sqrt{x}\cdot\sqrt{x+1}\cdot\frac{dx}{\sqrt{x}(x+1)}=\int u\sec u\,du = \cdots \]
[/quote]

how would you continue from $\int u\sec u\,du$ ? I try integration by parts by does not seem to work ....

1) even mathematica failed to evaluate it...


Can somebody tell me the difference between an integral and a summation with the help of examples?

Sriram :maybe:

A summation just adds up a list.  An integral is the limit of summations and usually cannot be expressed as a summation itself (at least, not in a meaningful way).  Thus, for example, if we wished to find for a function f the sum $f(1)+f(2)+\ldots+f(n)$, or even $f(1)+f(2)+\ldots$, we would use a summation.  If, however, we wanted to calculate the area under the curve $y = f(x)$ between 1 and n, we find that we would have to "add up" an uncountable number of terms.  However, we can express the area as a limit of approximations, where each approximation is some summation.


geometry
geometric transformation
rotation
symmetry
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14088[/img]

 :(

let $ \Gamma_1$, $ \Gamma_2$, $ \Gamma_3$ the circle with center C and radius Ha, Ma, R. A line r from A meet $ \Gamma_1$ at K, let the perpendicolar line from K to R that meet $ \Gamma_2$ at S and T. Let the perpendicolar line to KT on T that meet $ \Gamma_3$ first on J. Let a circle with center J and radius JC that meet TK on A' and B'. Then A'B'C is equal to ABC and we can obtain ABC with a rotation or a symmetry of A'B'C.

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14094[/img]

 :thumbup:


if we have the equation $10a+7b=C,a,b\in I, a,b>=0$

find the largest C so that there's no solution to a,b

[hide]Chicken McNugget Theorem:
$C=10 \cdot 7 - 10 - 7 = 53$.[/hide] :P

wow, cool!!!
I don't really know about this theorem.

I know another solution using mod, but I am not really clear with it.

So, the equation is 10x+7y=C

then we have a table like the following: for C:

 0  7  14  21
10 17 24 31
20 27 34 41
30 37 44 51
40 47 54 61

so the first row are 0 mod7
second row 10=3 mod 7
third row 20=6 mod 7
4th 40=5 mod 7
5th 50=1 mod 7
6th 60= 4 mod 7
7th : back to 0 mod 7

numbers above 60 all have solutions.

numbers above 50 have solutions except for 4 mod7

the biggest number of 4 mod 7 smaller than 60 is 53.

Thus,that's the answer. But I do not fully understand the solution, especially the part where : the numbers above 60 all have solutions, and the numbers above 50 have solutions except for 4 mod 7.

Can somebody clarify it for me , plz

eh mathworld.com has a confusing explanation of the mcnugget theorem and of the greedy algorithm

can ne1 explain it to me better?


Given a geometric sequence with the first term $ \neq 0$ and $ r \neq 0$ and an arithmetic sequence with the first term $ \equal{}0$.  A third sequence $ 1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences.  The sum of the first ten terms of the third sequence is:

$ \textbf{(A)}\ 978 \qquad
\textbf{(B)}\ 557 \qquad
\textbf{(C)}\ 467 \qquad
\textbf{(D)}\ 1068 \\
\textbf{(E)}\ \text{not possible to determine from the information given}$

[hide]
Let the geometric sequence have first term $ a$ and the arithmetic sequence have common difference $ k$.  Then

$ a\plus{}0\equal{}1\Rightarrow a\equal{}1$.

$ ar\equal{}k\equal{}1\Rightarrow r\plus{}k\equal{}1$.

$ ar^2\plus{}2k\equal{}2\Rightarrow r^2\plus{}2k\equal{}2$.

Thus $ 2r\plus{}2k\equal{}r^2\plus{}2k\Rightarrow r^2\equal{}2r\Rightarrow r\equal{}2$.

Thus $ k\equal{}\minus{}1$.

So the sum of the first ten terms of the geometric sequence is $ 1\plus{}2\plus{}\cdots \plus{} 2^9\equal{}2^{10}\minus{}1\equal{}1023$.

The sum of the first ten terms of the arithmetic sequence is $ \minus{}(1\plus{}2\plus{}\cdots \plus{} 9)\equal{} \minus{}45$.

So the answer is $ 1023\minus{}45\equal{}978\equal{}A$.
[/hide]


inequalities
integration
real analysis
real analysis unsolved

Let $U\subset\mathbb{R}^{n}$ be open, bounded, and connected, $f\in L^{2}(U)$ and $\partial U$ is $C^{\infty}$. Consider the equation (1)
$Lu=-\Delta u=f$ in $U$
$u=0$ on $\partial U$.
Show that
$\sup_{f\in L^{2}(U)}\frac{\|u_{f}\|_{L^{2}(U)}}{\|f\|_{L^{2}(U)}}\leq\frac{1}{\lambda}$
where $u_{f}$ solves (1) and $\lambda$ is the smallest eigenvalue of $L$.

Fridrichs inequality: $\lambda_{1}\int_\Omega u^{2}dx\leqslant \int_\Omega \sum_{i}\biggl(\frac{\partial u}{\partial x_{i}}\biggr)^{2}dx$.

Besides, $\int_\Omega \sum_{i}\biggl(\frac{\partial u}{\partial x_{i}}\biggr)^{2}dx=-(\Delta u,u)=(f,u)\leqslant \|f\|\cdot\|u\|$.

From there $\lambda_{1}\|u\|^{2}\leqslant \|f\|\cdot \|u\|$.


function
geometry
geometric transformation
reflection
algebra
polynomial
floor function

A function $f(z)$ is analytic in the upper half plane (UHP) $Im z>0$ and continuous on $Im z \geq 0$.  On the real axis the boundary values $f(x+i0)$ are real for $|x|>1$ and pure imaginary for $|x|<1$.  Suppose $f$ satisfies the following growth condition:  $|f(z)|\leq A(|z|^{n}+1)$ for some integer $n\geq 0$ and some $A>0$ for $Im z \geq 0$.  Find all such functions $f$.
_______________________________________________________________

I begin by letting $g=f^{2}$ and recognize $g$ is analytic on UHP, continuous on $Im z =0$, and $g(x+i0)$ is real for all $x \in R$.  Thus, Schwarz reflection applies and I can extend $g$ to $g^{*}(z)=\overline{g(\overline{ z})}$ on LHP.  Then we have $G=g$ on UHP and $G=g^{*}$ on LHP is entire with $|G| \leq A^{2}(|z|^{2n}+2|z|^{n}+1)$ everywhere.

But $G$ entire means $G(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ where $a_{n}=\frac{1}{2 \pi i}\oint_{C_{R}}\frac{G(\xi)}{\xi^{n+1}}d\xi$.  So we can bound $|G^{(k)}(0)|\leq \frac{k!}{2 \pi}\frac{2 \pi R}{R^{k+1}}A^{2}(R^{2n}+2R^{n}+1)$ which means $G^{(k)}(0)=0$ for all $k>2n$ which just means $G(z)$ is a polynomial of order at most $2n$.

Now I want $G$ to be positive for $|x|>1$ and negative for $|x|<1$ on the real axis.  I conclude that if $n \geq 2$, then $G$ must possess roots at $\pm 1$.

I end up with the following $f$:

$n=0$, $n=1 \rightarrow$ no solution.
$n=2$, $n=3 \rightarrow f(z)=K(z^{2}-1)^{1/2}$ where $K$ is a real constant.
$\cdots$
$f(z)=K(z^{2}-1)^{1/2}\prod_{i=1}^{\lfloor{\frac{n}{2}}-1 \rfloor}(z-z_{i})$ where $K$ is as above and the $z_{i}$ are arbitrary real roots?

(I wanted them to come in pairs for $G$ so that that part would be positive or zero).

If wrong, please help me make this right!  Thanks.

I forgot about this post.  Looking at it now, it probably would have been better to post it as a proposed problem and compare solutions.

To critique my own solution, I think the answer is wrong -- though not by too much I hope.  I think the functions $f$ that I generated would be the correct ones, except that they do not satisfy the original condition of being continuous for $Im z \geq 0$ due to the branch points.  Therefore, I am modifying my conclusion to say that there can be no function $f$ for any value of $n$ which satisfies all of the conditions.

Any thoughts?


Let ABC be an acute-angled triangle. Prove that:

$\left(\frac{\sin B\sin C}{\sin A}\right)^2+\left(\frac{\sin C\sin A}{\sin B}\right)^2+\left(\frac{\sin A\sin B}{\sin C}\right)^2\geq\frac94$.

You can find the solution at

http://myhome.personaldb.net/ideahitme/sol1.pdf