The symbolism $ \lfloor x\rfloor$ denotes the largest integer not exceeding $ x$. For example. $ \lfloor3\rfloor\equal{}3$, and $ \lfloor 9/2\rfloor\equal{}4$. Compute
\[ \lfloor\sqrt1\rfloor\plus{}\lfloor\sqrt2\rfloor\plus{}\lfloor\sqrt3\rfloor\plus{}\cdots\plus{}\lfloor\sqrt{16}\rfloor.
\]$ \textbf{(A)}\ 35 \qquad
\textbf{(B)}\ 38 \qquad
\textbf{(C)}\ 40 \qquad
\textbf{(D)}\ 42 \qquad
\textbf{(E)}\ 136$

[hide]
For $ x\in[n^2, n^2 \plus{} 2n]\longrightarrow \lfloor \sqrt {x} \rfloor \equal{} n$

$ \Rightarrow \sum_{k \equal{} 1}^{16} \lfloor \sqrt {k} \rfloor \equal{} 3(1) \plus{} 5(2) \plus{} 7(3) \plus{} 4 \equal{} 38 \longrightarrow \boxed{B}$[/hide]

[hide="solution"]

1 to 3-->1
4 to 8-->2
9 to 15-->3
16-->4

$ 3(1) \plus{} 5(2) \plus{} 7(3) \plus{} 1(4) \equal{} \boxed {38},$ or choice EDIT: $ (B)$ (had E before, with the right ans of 38)

[/hide]

If you dont no summations, another way to see it is just to always round down the integers... this would look like $ 1 \plus{} 1 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 4 \equal{} 3 \plus{} 10 \plus{} 21 \plus{} 4 \equal{} 38$ $ \longrightarrow \boxed{B}$

...thats exactly what i did, i just didnt write them out, i categorized them into groups of floor values of 1,2,3,and 4.

same thing, less tedious and less room for error. your way works too, though.

Sry, nik, totally skipped over your post. Kinda happy to find one where I could post a solution to  :blush:

[quote="nikeballa96"]...thats exactly what i did, .[/quote]

How ironic, i was just about to post the same thing :wink:


Give $ a, b, c$ are positive real numbers prove that:
$ (a^5\minus{}a^2\plus{}3)(b^5\minus{}b^2\plus{}3)(c^5\minus{}c^2\plus{}3) \ge (a\plus{}b\plus{}c)^3$

Should be locked.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5383]See here[/url]


inequalities
trigonometry
inequalities unsolved

Let a acute triangle ABC be given,Prove
(1)${{ \sin{\frac {A}{2}} + \sin{\frac {B}{2}}} + \sin{\frac {C}{2}}} > \sqrt {2}$
(2)$ \frac{h_{a}}{a}+\frac{h_{b}}{b}+\frac{h_{c}}{c}>\frac{5}{2}$
Thank you!

I think this should work for 1..

Assume $ \frac {\pi}{2}\ge A\ge B\ge C\ge 0$

Differentiate the expression wrt A we get a negative value.

So put A = $ \frac {\pi}{2}$

Then using $ \cos(\frac {B}{2} \minus{} \frac {\pi}{8})\ge \cos\frac {\pi}{8}$

We get the required result...

Clearly the equality condition is $ A \equal{} B \equal{} \frac{\pi}{2}$

We have:$ \frac{1}{tanB\plus{}tanC}\equal{}\frac{h_a}{a}$
So: $ \sum\frac{h_a}{a}\equal{}\sum\frac{1}{tanB\plus{}tanC}$
Let $ x\equal{}tanA;y\equal{}tanB;z\equal{}tanC$ $ \Rightarrow xy\plus{}yz\plus{}zx\equal{}1$
We need prove: $ \sum\frac{1}{x\plus{}y} \geq \frac{5}{2}$
It is a familiar problem.

[quote="math10"]Let $ x \equal{} tanA;y \equal{} tanB;z \equal{} tanC$ $ \Rightarrow xy \plus{} yz \plus{} zx \equal{} 1$[/quote]
Are you sure? :wink:

[quote="math10"]We have:$ \frac {1}{tanB \plus{} tanC} \equal{} \frac {h_a}{a}$
So: $ \sum\frac {h_a}{a} \equal{} \sum\frac {1}{tanB \plus{} tanC}$
Let $ x \equal{} tanA;y \equal{} tanB;z \equal{} tanC$ $ \Rightarrow xy \plus{} yz \plus{} zx \equal{} 1$
We need prove: $ \sum\frac {1}{x \plus{} y} \geq \frac {5}{2}$
It is a familiar problem.[/quote]
You kind of confused $ \tan$ with $ \cot$ :) (But the idea is the right)

Sory,I post without thinking too much :blush: .
I will try to fix the mistakes of their :oops: .

I really stupit :blush:
I mistakenly damaging.
We have:$ \frac{1}{cotB\plus{}cotC}\equal{}\frac{h_{a}}{a}$ not $ \frac{1}{tanB\plus{}tanC}\equal{}\frac{h_{a}}{a}$
We need prove:$ \sum\frac{1}{x\plus{}y}\geq\frac{5}{2}$ where $ x,y,z$ such that:$ xy\plus{}yz\plus{}zx\equal{}1$
This is a familiar problem :lol:

a familiar problem ?
but i can't prove it. :(
can you express carefully?thank you.


conics
parabola
geometry
geometric transformation
reflection
geometry proposed

i want the constrution of the tangents to a parabola from a point

Reflect $ F$ over $ V$ to obtain point $ D$
Through $ D$ bring a perpendicular to $ FV$ to obtain the director $ d$
Draw the circle $ (P,PF)$
Let $ A,B$ be the points where the circle intersects $ d \ \ (*)$  There are two points because $ PF> d(P,d)$
Bring perpendicular lines to $ d$ through $ A,B$
Call $ A',B'$ the respective points of intersection between those lines and the parabola
$ PA'$ and $ PB'$ are the tangent lines

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=13839[/img]

 :lol:


Asymptote
analytic geometry
AoPSwiki
geometry

Currently, I can't draw ANYTHING with Asymptote but a straight line, a unit circle, and a unit square. I can't provide adequate solutions to problems in other forums because of this (neither can I post problems myself)!

How do you inscribe circles in squares in circles, draw funky shapes, draw bisectors, angles, lines from certain points to another, circumscribe triangles, draw scalene triangles, and a WHOLE BUNCH OF OTHER STUFF.

I thought you could basically do anything in Asymptote. Do I have to find the coordinates for each and everything thing???? And even that's the only way, aside from lengthy periods of work, THATS IMPOSSIBLE. I don't know the mathematical formula's to find all that. If I did, I WOULD BASICALLY KNOW THE ANSWER TO THE PROBLEM!

PLEASE PLEASE HELP!!

(excuse my screaming)

---EDIT---
The AoPSWiki tutorial provides no additional information; I have gone through it already.

Very nice.

The way I learned asymptote is just asking on the forum:

How do you draw this?
How do you draw that?

Also, there is a great guide at:

[url=http://asymptote.sourceforge.net/asymptote.pdf]Manual[/url]

(written by Mr. Ivaldi himself.)

Hi,
[quote="Dojo"]Also, there is a great guide at:
[url=http://asymptote.sourceforge.net/asymptote.pdf]Manual[/url]
(written by Mr. Ivaldi himself.)[/quote]
Not at all, [b]this manual is written essentially by John Bowman[/b].
I have written only very short parts.

@akhil0422: You can take a look [url=http://piprim.tuxfamily.org/asymptote/index.html]here[/url] and, for geometry purpose, [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html]here[/url].

Regards.

You don't need to find the coordinates of everything. I'm still a newb at Asymptote, but there is the [b]dir()[/b] command that allows you to draw in the direction of.

[asy]pair A=(0,0);
pair B=6+A*dir(30);
dot(B);[/asy]

[code]pair A=(0,0);
pair B=6+A*dir(30);[/code]

The [b]6+[/b] stands for "6 units," the [b]A*[/b] reads "from A" and [b]dir(30)[/b] means "with an angle of 30 degrees." (0 degrees is horizontal going to the right).

There are also other stuff, but I'm feeling too lazy to post them

[quote="Dojo"]
The way I learned asymptote is just asking on the forum:
How do you draw this?
How do you draw that?
[/quote]
Yes, that's the easiest way to go. Just make sure you understand the answer before asking the next question, read the manual about all commands that were new to you, and try to make a similar but different picture yourself from scratch every time somebody does it for you :). So, what do you want to draw?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that there does not exist a groups $G$ with the following conditions :
a) $G$ is finite and simple
b) $G$ has at least two maximal subgroups
c) for two distinct maximal subgroups $G_{1},G_{2}$ of $G$, $G_{1}\cap G_{2}=\{1\}$

[quote="Omid Hatami"]
c) for two maximal subgroups $G_{1},G_{2}$ of $G$, $G_{1}\cap G_{2}=\emptyset$[/quote]

First of all, this can never happen: all subgroups of $G$ contain $1$, the identity element of $G$. But this is not important (I assume you meant $G_{1}\cap G_{2}=\{1\}$). What I really wanted to ask: is this supposed to hold for [i]any[/i] pair of distinct maximal subgroups, or just for some specific pair?

I corrected the typo, and answered the question, Excuse me for the bad typo.


inequalities
trigonometry
function
inequalities proposed

prove this inequality  $sinx.siny.sinz\leq\sin^3\frac{1}{3}\ (x+y+z)$

It's Jensen for the concave function $lnsinx$ ;)

I suppose $x,y,z\in (0,\frac{\pi}{2})$ ;)


This is what happens on the bottom of some topics when I view:

Rules_post_can
Rules_reply_can
Rules_edit_can
Rules_delete_can
Rules_vote_can
You [b]can[/b] attach files in this forum
You [b]can[/b] download files in this forum
Rules_calendar_cannot

Yeah.  It says that at the bottom of the Support, for one.

[quote="herefishyfishy1"]It doesn't happen in any forum until I view a specific thread.

I think that this is happening for ALL the threads...[/quote]

Yeah, we must have lost a few language file entries in our last update.  We'll fix it this week.


Is this true? (Feel free to add conditions on $X$ if useful.)

Suppose that $X$ and $Y$ are independent, identically distributed random variables and $a$ is a fixed positive number. Then
\[P(|X-Y|\le 2a)<3P(|X-Y|\le a)\]

I can prove this if $X$ is integer-valued and $\frac12\le a<1.$ That is, for integer-valued random variables, $P(|X-Y|\le 1)<3P(X=Y).$

I now know that this is both highly nontrivial and published.

See

      N. Alon and R. Yuster
      [i]The 123 Theorem and its extensions[/i]
      Journal of Combinatorial Theory, Series A 72 (1995), 322-331

You can find a downloadable copy [url=http://research.haifa.ac.il/~raphy/papers.htm]here[/url].

When I saw your first post, it reminded me of some work by Noga Alon.  But I didn't realize it was an exact match.


I was trying to make a small program that make a Latin square.

So the algorithm is basically this:

Make a List of the elements from $ 1$ to $ n$ and call it $ T$

Make a matrix $ n \times n$. Make the element $ M_{1,1}$ randomically.

To compute $ M_{a,b}$:
Make a list from $ M_{a,1}$ to $ M_{a,b\minus{}1}$
Make a list from $ M_{1,b}$ to $ M_{a\minus{}1,b}$
Make the complement: $ (T \setminus Lx ) \setminus Ly$

Assign to $ M_{a,b}$ a random element from the last list.

[hide="code in Python"][code]
import random

n = 5

def RandomL(L):
    """Take on random element of the list L"""
    random.shuffle(L)
    return L[0]

def Dif2(A, B):
    """The Difference between A and B"""
    return [i for i in A if not i in B]

def Dif3(A, B, C):
    """The Difference between A and B and C"""
    return Dif2( Dif2(A, B), C)

# Make a Zero Square Matriz
M = [[0 for i in range(1,n+1)] for i in range(1,n+1)]
# Make a list from 1 to n
T = [i for i in range(1,n+1)]

def P(a,b):
    """Elements from the matrix"""
    if a == 1 and b == 1:
        M[0][0] = random.randint(1, n)
    else:
        Lx, Ly = [],[]
        i, j = 0, 0
        while j < b - 1:
            Lx.append( M[a - 1][j] )
            j = j + 1
        while i < a - 1:
            Ly.append( M[i][b - 1] )
            i = i + 1

        M[a-1][b-1] = RandomL( Dif3(T, Lx, Ly) )

    return M[a - 1][b - 1]

for i in range(1, n + 1):
    for j in range(1, n + 1):
        print P(i,j),
    print
[/code][/hide]

There is only a small problem. Some times it leads to matrix like:
[code]
5 2 1 3 4
1 5 3 4 2
3 1 2 5
[/code]

I wanted to correct this bug. I could use brute force, but I made this code because I did not want to use Brute Force.

Some hints and ideas?

Thanks.

I know nothing about Python so I didn't really read your source code, but the problem is obvious - your algorithm doesn't work :). Look at the matrix you posted - if your program now tries to calculate the next element (last element of the third row), it finds out that all numbers from 1 to 5 were already used! In other words, if you choose the numbers randomly according to your algorithm, you may end up with a matrix that is impossible to finish.

I've never tried generating random latin squares so I'm not sure if there is any efficient algorithm for this. But I've just tried googling a little bit and there seem to be quite a lot of pages that deal with this problem so you could check them out.

Oh yes and also I don't think there's any better approach to solving sudoku than brute force.... but on the other hand, it works reasonably fast.


modular arithmetic
floor function
algebra
binomial theorem
number theory unsolved
number theory

Find all $ n\in N^*$ which satisfy

$ n^n\mid (n \minus{} 1)^{n^{n \plus{} 1}} \plus{} (n \plus{} 1)^{n^{n \minus{} 1}}$

First we notice that $ n$ must be odd. Indeed, suppose that $ n$ is even, we have
\[ (n \minus{} 1)^{n^{n \plus{} 1}} \plus{} (n \plus{} 1)^{n^{n \minus{} 1}} \equiv ( \minus{} 1)^{n^{n \plus{} 1}} \plus{} (1)^{n^{n \minus{} 1}} \equiv 2 \pmod{n}\]
so $ n|2$ and hence $ n \equal{} 2$, but this does not satisfy the problem.

Now consider $ n$ odd. We will prove the two following facts:

[b]Claim 1.[/b] $ (n \plus{} 1)^{n^{n \minus{} 1}} \equiv 1 \pmod{n^n}$.
Indeed, from the binomial theorem we have:
\[ (n \plus{} 1)^{n^{n \minus{} 1}} \equal{} \sum_{k \equal{} 0}^{n^{n \minus{} 1}} \binom{n^{n \minus{} 1}}{k} n^k\]
Let $ p$ be a prime divisor of $ n$ and let $ p^{\alpha} \parallel n$ for $ \alpha > 0$.
We have
\[ v_p (k!) \equal{} \sum_{s \equal{} 1}^{\infty} \lfloor \dfrac{k}{p^s} \rfloor < k \cdot \sum_{s \equal{} 1}^{\infty} \dfrac{1}{p^s} \equal{} \dfrac{k}{p \minus{} 1} \leq \dfrac{k}{2}\]
(since $ n$ is odd then $ p \geq 3$). Thus for $ k \geq 2$ we have
\[ v_p \left( \binom{n^{n \minus{} 1}}{k} n^k \right) \geq \alpha(n \plus{} k \minus{} 1) \minus{} v_p(k!) \geq \alpha(n \plus{} k \minus{} 1) \minus{} \dfrac{k}{2} \equal{} \alpha n \plus{} k \minus{} 1 \minus{} \dfrac{k}{2} \geq \alpha n\]
Hence
\[ (n \plus{} 1)^{n^{n \minus{} 1}} \minus{} 1 \equal{} \minus{} 1 \plus{} \sum_{k \equal{} 0}^{n^{n \minus{} 1}} \binom{n^{n \minus{} 1}}{k} n^k \equiv n \cdot n^{n \minus{} 1} \plus{} 1 \minus{} 1 \equiv 0 \pmod{p^{\alpha n}}\]
Therefore, Chinese Remainder Theorem yields what desired.

[b]Claim 2.[/b] $ (n \minus{} 1)^{n^{n \plus{} 1}} \equiv \minus{} 1 \pmod{n^n}$.
This is easy to prove using the same argument as used in the proof of [b]Claim 1.[/b] Notice that the zero coefficient is $ \minus{} 1$ since $ n$ is odd.

Base on these two claims, we conclude that ever odd $ n$ satisfies the problem.

mashimaru, can you tell me what does $ v_{p}$ and $ \alpha$ mean and are they useful in number theory? please
thanks.  :blush:

$ p^ {\alpha} \parallel{} n$ means that $ p^ {\alpha}$ is the highest power of $ p$ dividing $ n$. $ v_{p}(n)\equal{} \alpha$ means the same thing.


$ x_1, x_2, \ldots, x_n \; \in (0,1) \qquad \sum_{k\equal{}1}^n x_k \equal{}1$

Prove, that

$ \sum_{k\equal{}1}^{n} \frac{x_k}{\sqrt{1\minus{}x_k}} \ge \frac{\sum_{k\equal{}1}^n \sqrt{x_k}}{\sqrt{n\minus{}1}}$

[quote="micf"]$ x_1, x_2, \ldots, x_n \; \in (0,1) \qquad \sum_{k \equal{} 1}^n x_k \equal{} 1$

Prove, that

$ \sum_{k \equal{} 1}^{n} \frac {x_k}{\sqrt {1 \minus{} x_k}} \ge \frac {\sum_{k \equal{} 1}^n \sqrt {x_k}}{\sqrt {n \minus{} 1}}$[/quote]
Use Chebyshev or Cauchy-Schwarz inequalities :)

I tried applying CS, but I don't know how to end the proof  :(

I thought about

$ LHS \ge \frac{\left(\sum_{k\equal{}1}^{n} \sqrt{x_k}\right)^2}{\sum_{k\equal{}1}^n \sqrt{1\minus{}x_k}} \ge \frac{\sum_{k\equal{}1}^{n} \sqrt{x_k}}{\sqrt{n\minus{}1}}$

Is it good? If yes, how can I continue it?

I know it's easy for all of you, but unfortunately not for me  :(
Could someone post here a nice proof?


Find all polynomials $ f(x) \in \mathbb Z[x]$ such that $f(x) + f(y) + f(z)$ is divisible by $ x + y + z$ for all integers $ x,y$, and $z$ such that $x+y+z\neq 0$.

[quote="shortlist"]find all polynomicals $ f(x)$ in Z[x] satisfy:
$ f(x) \plus{} f(y) \plus{} f(z)$ is divisible by $ (x \plus{} y \plus{} z)$ for all every$ x,y,z$ in Z such that $ x\plus{}y\plus{}z\ne 0$[/quote]

So $ x\plus{}y\plus{}z|f(x)\plus{}f(y)\plus{}f(z)$ and $ x\plus{}y\plus{}z|f(x)\plus{}f(y\plus{}t)\plus{}f(z\minus{}t)$ and so $ x\plus{}y\plus{}z|f(y\plus{}t)\minus{}f(y)\plus{}f(z\minus{}t)\minus{}f(z)$ $ \forall x\plus{}y\plus{}z\ne 0$

So $ f(y\plus{}t)\minus{}f(y)\plus{}f(z\minus{}t)\minus{}f(z)\equal{}0$ $ \forall y,z,t$

Choosing $ t\equal{}z$ in this last line, we get $ f(y\plus{}z)\equal{}f(y)\plus{}f(z)\minus{}f(0)$ which is a classical Cauchy's equation for $ f(x)\minus{}f(0)$ whose solution (since $ f(x)$ is continuous) is $ f(x)\equal{}ax\plus{}b$

Plugging this in original requirement, we get $ b\equal{}0$ and solutions are $ \boxed{f(x)\equal{}ax}$ for any $ a\in\mathbb Z$

[quote="r3d30m3j"]
This solution uses a theorem which might be 'overkill' for this problem but easy to prove directly.
[/quote]

I'm sorry for my so complex solution. In my mind, Cauchy equation is solved in 5 lines and, in my humble opinion, is not an overkiller solution.
[hide="5 lines Cauchy solution"]
============ beginning of hidden part ===============
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$ and then an easy induction gives $ f(nx)\equal{}nf(x)$

So $ f(\frac pqx)\equal{}\frac pqf(x)$ and so $ f(x)\equal{}xf(1)$ $ \forall x\in\mathbb Q$

Then continuity gives $ f(x)\equal{}xf(1)$ $ \forall x$
=========== end of hidden part ===============
[/hide]
And I like my 5 lines solution (+5 lines for Cauchy).

Congrats anyway for yours.

[quote="r3d30m3j"]Dear pco,
I meant the theorem I used in my long proof might be overkill for this problem, obviously your solution is way more elegant.[/quote]

Oooops!  :oops:  Sorry for my misunderstanding

[quote="irantst"]hi
I didn't understand your soloution?
can you help me?
 :?:[/quote]

Could you indicate where is the first statement you dont understand ?


Prove that all Tournament has one hamiltonian path.

*one tournament is one complete orientad graph.  :)

It's not true, unless the graph is strongly connected. It is true, however, that every tournament has a hamiltonian path (i.e. a cycle missing an arc).

thanks grobber.

sorry for my mistake.

could you prove that all tournamets have one hamiltonian path?

[quote="ZetaX"]It's clearly equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=60233 , so you are done.[/quote]

thanks  :)


An unfinished attempt...

$ f(x) = \sqrt {\frac 1 x - x}$ is convex on interval $ \bigg[0, \frac 1 2\bigg]$. Because of that, for all $ x, y$ in that interval we have ${ f(x) + f(y) \ge 2f\bigg(\frac {x + y}2}\bigg)$ by Jensen. That allows us to replace $ (a_i, a_j)$ with $ \bigg(\frac {a_i + a_j}2, \frac {a_i + a_j}2\bigg)$ if $ a_i, a_j$ are less or equal $ \frac 1 2$. We do it untill all such $ a_i$ are all equal. Now, there is at most one $ a_i$ which is greater than $ \frac 1 2$. In case there is no such element, we came up with all $ a_i$ equal and easily check the inequality.

In other case, we have the sequence in form $ (a, b, b, ..., b)$ in which $ a$ is the element greater than $ \frac 1 2$ and $ a + b(n - 1) = 1$. This leads to an ugly 2-variable inequality which I think isn't hard, but I still can't break it.  :?:


The digits 1-5 are each used exactly once to form three prime numbers. Exactly one of the numbers is a one-digit prime. What is the greatest possible sum of the three primes?

2, 51, 43 would make the largest sum. (Minimize the one digit number)

Sum is $ 51\plus{}43\plus{}2\equal{}\boxed{96}$.

Right answer, except it is 53 and 41 (51 is not prime)


logarithms
limit
calculus
calculus computations

How to show that $ \frac{\ln n}{n^2}$ converges or diverges?

We can use Cauchy Condensation Test: if $ a_n \equal{} \frac{\ln n}{n^2}$ then $ 2^n a_{2^n} \equal{} \frac{2^n \cdot \ln(2^n)}{(2^n)^2} \equal{} \frac{n \ln 2}{2^n}$. As the series with this last general term converges, then also our series converges.

The original question doesn't mention a sum. Taken at face value, it's simply a question about the sequence. Since the logarithm grows slower than any positive power of $ n,$ yes, we have $ \lim_{n\to\infty}\frac{\ln n}{n^2}\equal{}0,$ which is convergence.

---

Carcul seemed to be answering the question of whether $ \sum_{n\equal{}1}^{\infty}\frac{\ln n}{n^2}$ converges, which isn't what the original poster asked. (As to whether it's what he meant - I'm not a mind reader.)

Faced with that question, I wouldn't have bothered with the condensation test. Instead, since the logarithm grows slower than any positive power of $ n,$ we have that $ \frac{\ln n}{\sqrt{n}}\to0$ and is hence bounded. That means that

$ \sum\frac{\ln n}{n^2}\le C\sum\frac{1}{n^{3/2}},$ which is known to be a convergent "$ p$-series."


Solve the eqn $ z^{2}+2\bar{z}+6=0,$ giving the soln. in the form $ z=a+ib$ where a and b are real

-------------

The point P representing the complex number z moves in an  Argand diagram so that $ |z|=|z-4+2i|$

(i) Show that the locus of P has eqn $ 2x-y-5=0$

Answer

If $ |z-a| = |z-b|$ then the locus representing $ z$ is the perpendicular bisector of $ a$ and $ b$.

In this case: $ a = (0,\,0)$ and $ b = (4,\,-2)$

$ m_{ab}= \frac{-2}{4}=-\frac{1}{2}$

$ m_{\perp ab}= \frac{-1}{m_{ab}}= 2$

$ mid_{ab}= (2,\,-1)$

Hence locus is:

$ y-b = m(x-a)$

$ \Rightarrow y+1 = 2(x-2)$

$ \Rightarrow y+1 = 2x-4$

$ \Rightarrow 2x-y-5 = 0$

hence find the minimum value of $ |z|$ ??

Answer

$ |z|$ gives the distance of a point on the line $ 2x-y-5 = 0$ from the origin. Hence $ |z|_{min}$ is the shortest distance between the line $ 2x-y-5 = 0$ and the origin. If we "draw" a line representing the minimum distance, it will pass through the origin and be perpendicular to $ 2x-y-5 = 0$. Hence it will have equation: $ y =-\frac{1}{2}x$.

Finding the intersection of the two lines by substitution:

$ 2x+\frac{1}{2}x-5 = 0$

$ \Rightarrow \frac{5}{2}x = 5$

$ \Rightarrow x = 2$

$ \Rightarrow y =-1$

$ \Rightarrow |z|_{min}$ is given by the length of the line joining $ (0,\,0)$ and $ (2,\,-1)$.

$ \Rightarrow{|z|_{min}}^{2}= 2^{2}+1^{2}= 9$

$ \Rightarrow |z|_{min}= 3$


Let $a, b, c$ be real numbers.  Suppose that $|ax^2 + bx + c| \le 1$ when $|x| \le 1$.  Prove that $|cx^2 + bx + a| \le 2$ when $|x| \le 1$.

(Good exercise.  I want to see if anyone comes up with a proof more elegant than mine  :) )

A more general result is true: if \[ \left|a_nx^n + \cdots + a_2x^2 + a_1x + a_0\right| \leq 1 \] when $|x| \leq 1$, then \[ \left|a_0x^n + a_1x^{n - 1} + \cdots + a_{n - 1}x + a_n\right| \leq 2^{n - 1}\textrm{ if } -1 \leq x \leq 1. \]

Don't try to prove this :D


Let $ a$, $ b$ and $ c$ be positive reals such that $ ab$, $ bc$ and $ ca$ be sides of triangle.

Prove that
\[ \sum{\dfrac{a}{ca\plus{}ab\minus{}bc}}\ge\dfrac{a^2\plus{}b^2\plus{}c^2}{abc}\]

:)

i think the inequality is not true, and can't be fixed (trivially)... the condition on $ab, bc$ and $ca$
being sides of a triangle is equivalent to $x=1/a, y=1/b$ and $z=1/c$ being
sides of a triangle. The inequality then rewrites as

\[
\sum \frac{yz}{y+z-x}\geq \sum \frac{yz}{x}.
\]

Taking $y=z$ we get that the following inequality must hold:

\[
\frac{y^2}{2y-x}+2y\geq \frac{y^2}{x}+2x,
\]

given that $2y>x$. Taking $x\rightarrow 0$ we get a contradiction. The inequality
is not in the reverse order because we would obtain also a contradiction taking $x\rightarrow 2y$.


[i][b]Problem:[/b] Let $ABCD$ be a quadrilateral with squares $ABMM'(O_{1})$, $BCNN'(O_{2})$, $CDPP'(O_{3})$, and $DAQQ'(O_{4})$ constructed on sides $AB$, $BC$, $CD$, and $DA$, respectively. The $O_{x}$'s are the centres of the squares they correspond to.  Prove that $O_{1}O_{3}\perp O_{2}O_{4}$ and $O_{1}O_{3}=O_{2}O_{4}$.[/i]

I have already proved this, but I am looking for different solutions. :)

Masoud Zargar

Complex numbers? If the vertices correspond to the numbers $a,b,c,d$, then (assuming $A,B,C,D$ are positively oriented) $o_{1}: =m_{AB}+i(a-m_{AB})$ where $m_{AB}={a+b\over 2}$ is the complex number corresponding to the midpoint of $AB$. And so on. After that it's easy and uneventful algebra :)

That's the exact solution I had.  Can you come up with a euclidean proof?


2. Find all natural $n>1$ for which value of the sum $2^2+3^2+...+n^2$ equals $p^k$ where p
is prime and k is natural

First of all we rewrite:

$\sum_{i=2}^{n}i^2=\frac{n(n+1)(2n+1)}{6}-1=\frac{(n-1)(2n^2+5n+6)}{6}$.

We easily can check the cases $n\leq 7$ so that we can assume $n>7$. Then both factors $n-1$ and $2n^2+5n+6$ must have the same prime factor after the division with 6. Because of the Euclidean Algorithm $\gcd(n-1,2n^2+5n+6)=\gcd(n-1,13)$. So we become $p=13$ and the problem reduces to:

$(n-1)(2n^2+5n+6)=6\cdot 13^k$.

The highest power of 13 which can divide n-1 is 1 because if $n\equiv1\;mod\;169$ then $2n^2+5n+6\equiv\;13\;mod\;169$ and $\frac{2n^2+5n+6}{13}$ is greater then $6$ because of our assumption (it would have a prime divisor $\neq$ 13). So $n-1$ can only be equal

to $13$, $13\cdot2$, $13\cdot3$, $\13\cdot6$.

We only have to check $n\leq7$ and $n=14,27,40,79$ and become then all possible solutions.

Misha

[quote=Dr_Vex]/bummpp....[/quote]

But the $FS $ is right, indeed after notice $\gcd (n-1,2n^2+5n+6)=\gcd (n-1,13) $ we can see that if $\gcd=1$ this numbers are relatively primes but with
$$(n-1)(2n^2+5n+6)=6p^k $$
This implies $n\in\{2,3,4,7,p^k+1\} $. After check $2,3,4,7$ and get solutions we must check $n=p^k+1$. Which implies $2n^2+5n+6=6$ contradiction.

If $\gcd=13\implies 13\mid 6p^k $ which yields $p=13$. Now we have
$$(n-1)(2n^2+5n+6)=6\cdot 13^k $$
But since $\gcd(n-1,2n^2+5n+6)=13$ this implies $\nu_{13}(6\cdot 13^k)=1\implies k=1$ so we have
$$(n-1)(2n^2+5n+6)=6\cdot 13$$ which give us no solutions.

So the unique solution is $\boxed{n=2} $.

I had a different solution from everyone here and what I got matches up with post #3

[hide = Solution]

Note $$\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6} \implies \sum_{k=2}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}-1 = \dfrac{n(n+1)(2n+1)-6}{6}.$$

Note that $$n(n+1)(2n+1) - 6 = 2n^3+3n^2+n-6  = (n-1)(2n^2+5n+6) \implies p^k =  \dfrac{n(n+1)(2n+1)-6}{6} = \dfrac{(n-1)(2n^2+5n+6)}{6}.$$

Now we have two cases. Either $p \nmid n-1$ or $p \mid n-1.$

[b]Case [/b]$1: p \nmid n-1.$ Note that we must have $n-1 = 1,2,3,6.$ Otherwise, we would have "leftovers" and the resulting sum would not be a perfect prime power.  Solving these equations we see that all $\boxed{n=2,3,4,7}$ all work (all the possible values of $n$ work).

[b]Case[/b] $2: p \mid n-1.$ Let $n-1  = px$ for some integer $x.$ Now we rewrite the whole thing in terms of $px.$

Note $$p^k = \dfrac{(n-1)(2n^2+5n+6)}{6} = \dfrac{(n-1)(2(n-1)^2+9(n-1)+13)}{6} = \dfrac{px(2(px)^2+5px+13)}{6}.$$

We know that $2(px)^2+5px+13$ must be a multiple of $p$ or containing factors of $6.$ Clearly if we want $p \mid 2(px)^2+5px+13$ then $p \mid 13$ which through testing doesn't work. Futhermore, $2(px)^2+5px+13$ is too big to be a factor of $6$ so there are no solutions for this case.

[quote=Megus]your solution looks good but possible values of n are 2,3,4 and 7[/quote]

he forgot to mention them in the end but[quote=Misha123]
We easily can check the cases $n\leq 7$ so that we can assume $n>7$.
[/quote]

[quote=Al3jandro0000]since $\gcd(n-1,2n^2+5n+6)=13$ this implies $\nu_{13}(6\cdot 13^k)=1\implies k=1$[/quote]
Why?


trigonometry
AMC
AIME
analytic geometry
USAMTS
function
LaTeX

Find the positive integer $ n$ such that \[\arctan\frac{1}{3}\plus{}\arctan\frac{1}{4}\plus{}\arctan\frac{1}{5}\plus{}\arctan\frac{1}{n}\equal{}\frac{\pi}{4}.\]

[hide] Since we are dealing with acute angles, $ \tan(\arctan{a})\equal{}a$.

Note that $ \tan(\arctan{a}\plus{}\arctan{b})\equal{}\dfrac{a\plus{}b}{1\minus{}ab}$, by tangent addition. Thus, $ \arctan{a}\plus{}\arctan{b}\equal{}\arctan{\dfrac{a\plus{}b}{1\minus{}ab}}$.

Applying this to the first two terms, we get $ \arctan{\dfrac{1}{3}}\plus{}\arctan{\dfrac{1}{4}}\equal{}\arctan{\dfrac{7}{11}}$.

Now, $ \arctan{\dfrac{7}{11}}\plus{}\arctan{\dfrac{1}{5}}\equal{}\dfrac{23}{27}\equal{}\arctan{\dfrac{23}{24}}$.

We now have $ \arctan{\dfrac{23}{24}}\plus{}\arctan{\dfrac{1}{n}}\equal{}\dfrac{\pi}{4}\equal{}\arctan{1}$.Thus,

$ \dfrac{\dfrac{23}{24}\plus{}\dfrac{1}{n}}{1\minus{}\dfrac{23}{24n}}\equal{}1$;
$ 23n\plus{}24\equal{}24n\minus{}23$
$ n\equal{}\boxed{047}$.[/hide]

[quote="AdorableAddict"]haha i got the right answer by using a geometric approach with coordinates and stuff. anyone else did that?[/quote]

I'd actually be really interested in seeing that. I did it the "take tangent of both sides" way.

So did I, but I decided to group $ \arctan\dfrac{1}{5}$ with $ \arctan \dfrac{1}{n}$ first. It worked anyways. (although I messed up in computation at first)

From the expansion of $ e^{iA}e^{iB}e^{iC}e^{iD}$, you can see that
\[ \cos(A\plus{}B\plus{}C\plus{}D) \equal{} \cos A \cos B \cos C \cos D \minus{} \tfrac{1}{4} \sum_{\rm sym} \sin A \sin B \cos C \cos D \plus{} \sin A \sin B \sin C \sin D, \]
and
\[ \sin(A\plus{}B\plus{}C\plus{}D) \equal{} \sum_{\rm cyc}\sin A \cos B \cos C \cos D \minus{} \sum_{\rm cyc} \sin A \sin B \sin C \cos D .\]
If we divide both of these by $ \cos A \cos B \cos C \cos D$, then we have
\[ \tan(A\plus{}B\plus{}C\plus{}D) \equal{} \frac{1 \minus{} \sum \tan A \tan B \plus{} \tan A \tan B \tan C \tan D}{\sum \tan A \minus{} \sum \tan A \tan B \tan C}, \]
which makes for more direct, less error-prone computations.

This problem is also almost exactly the same as USAMTS Round 1, number 3.

[quote="the future"]This problem is very similar to a cotangent problem from 20 years ago...

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=392631#p392631

who says doing old AIME prlobelms doesnt help?[/quote]

I know, seriously! I saw that problem the night before AIME as well, I was practicing... :)

It was easy, just brute force with tangent sums identities twice.

[hide="Or..."]Interpret the arctangents as arguments and rewrite as

$ (3 \plus{} i)(4 \plus{} i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (11 \plus{} 7i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (48 \plus{} 46i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ 48n \minus{} 46 \equal{} 46n \plus{} 48 \implies$
$ n \equal{} \boxed{47}$.
[/hide]

[quote="t0rajir0u"][hide="Or..."] Interpret the arctangents as arguments and rewrite as

$ (3 \plus{} i)(4 \plus{} i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (11 \plus{} 7i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (48 \plus{} 46i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ 48n \minus{} 46 \equal{} 46n \plus{} 48 \implies$
$ n \equal{} \boxed{47}$.

:) [/hide][/quote]

lol, you win this problem.

[quote="t0rajir0u"][hide="Or..."] Interpret the arctangents as arguments and rewrite as

$ (3 \plus{} i)(4 \plus{} i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (11 \plus{} 7i)(5 \plus{} i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ (48 \plus{} 46i)(n \plus{} i) \equal{} m \plus{} mi \implies$
$ 48n \minus{} 46 \equal{} 46n \plus{} 48 \implies$
$ n \equal{} \boxed{47}$.

:) [/hide][/quote]What do you mean by arguments?

[b]Definition:[/b]  Given a complex number $ z \equal{} a \plus{} bi$, let $ r \equal{} |z| \equal{} \sqrt {a^2 \plus{} b^2}$.  Then $ \theta \equal{} \arg(z)$, the [b]argument[/b] of $ z$, is the unique angle $ \bmod 2\pi$ such that $ z \equal{} re^{i \theta}$.

[b]Lemma:[/b]  $ \arg(zw) \equiv \arg(z) \plus{} \arg(w) \bmod 2 \pi$.

Proof:  This is actually tangent summation in disguise.  It's just much easier to work with complex numbers than with tangent summation, which is why my solution is short.  :)

[b]Lemma:[/b]  With the appropriate restraints on quadrants, $ \arg(z) \equal{} \arctan \frac {b}{a}$.

We can therefore rewrite the problem as $ \arg(3 \plus{} i) \plus{} \arg(4 \plus{} i) \plus{} \arg(5 \plus{} i) \plus{} \arg(n \plus{} i) \equal{} \arg(m \plus{} mi)$ for some real number $ m$ and use the first lemma.  :)

[hide="Motivation"] This solution is much easier to see if you're familiar with the proof of [url=http://mathworld.wolfram.com/MachinsFormula.html]Machin's formula[/url] via the identity

$ \frac{(5 \plus{} i)^4}{239 \plus{} i} \equal{} 2 \plus{} 2i$.

See also [url=http://mathworld.wolfram.com/Machin-LikeFormulas.html]Machin-like formulas[/url].  [/hide]

Ha this was my solution too. (I have another nice problem on one of my mock aimes that uses a similar idea). Not to be picky, but the arguments should be taken modulo $ \pi$ because the $ \arg$ function is analogous to the inverse tangent function, which has that period.

(wow I was going to edit this but anyway)

I suppose that it is possible to have the argument function with a range of $ \minus{}\pi$ to $ \pi$, but it makes the function have a much more ugly form. Instead of $ \arg (a\plus{}bi)\equal{}\tan^{\minus{}1}\frac{b}{a}$, you have to deal with the fact that if $ b<0$, then you have to add $ \pi$. I suppose this is doable, but I prefer the other definition.

[quote="Altheman"]Ha this was my solution too. (I have another nice problem on one of my mock aimes that uses a similar idea). Not to be picky, but the arguments should be taken modulo $ \pi$ because the $ \arg$ function is analogous to the inverse tangent function, which has that period.[/quote]Do you do AMC/AIME/USAMO?

[quote="Altheman"]I suppose that it is possible to have the argument function with a range of $ \minus{} \pi$ to $ \pi$, but it makes the function have a much more ugly form. Instead of $ \arg (a \plus{} bi) \equal{} \tan^{ \minus{} 1}\frac {b}{a}$, you have to deal with the fact that if $ b < 0$, then you have to add $ \pi$. I suppose this is doable, but I prefer the other definition.[/quote]

The argument function outputs angles.  I consider its range to be $ \mathbb{R} / 2 \pi \mathbb{Z}$.  :)

i just did tangent on both sides, and split the 4 values into 2 groups of 2, using the tangent addition formula.  pretty self explanatory from there.

[quote="Math Geek"][quote="t0rajir0u"][quote="Altheman"]the arguments should be taken modulo $ \pi$ because the $ \arg$ function is analogous to the inverse tangent function, which has that period.[/quote]

Not the way I've defined it.  The condition that $ r \ge 0$ puts $ \arg(z)$ in the correct quadrant, hence the conditions on the second lemma.[/quote]

OMG, t0raji0u, you're too smart for me.  Cuz I'm a middle schooler who has never even touched upon trig like this, I just guessed.  My friend on the other hand, added $ \frac {1} {3}, \frac {1} {4}, and \frac {1} {5}$ to get $ \frac {47} {60}$, so he just put $ \boxed {047}$.  It was pretty stupid...  I got so angry, cuz he guessed two questions right last year on the AIME as well...[/quote]
Okay wow that is really random...47 just happened to be the right answer too.
Wow the complex numbers thing is nice :D great job.

I just did it by taking $ tan$ of both sides, but at the end........

"hm so the answer is 23 + 24 = 45. yay! i got another problem!"

the incorrect answer was never corrected.  :(

:maybe: sorry, but what is "arctan"???????

$ \arctan$ is the inverse function of $ \tan$, so if $ \tan \theta \equal{} r$, then $ \arctan r \equal{} \theta$.
Another example: Since $ \tan 45^\circ\equal{}1$, then $ \arctan 1 \equal{} 45^\circ$.
http://en.wikipedia.org/wiki/Arctangent

If you're referring to Math Geek's friend's method, that was almost surely a numerical coincidence.

[quote="123456789"][quote="mchoi815"]Is there a correlation between the numerator of the sum of the fractions and the actual answer?  If so, can one show the proof of it?[/quote]What do you mean?[/quote]

He's referring to this post:

[quote="Math Geek"]My friend on the other hand, added $ \frac {1} {3}, \frac {1} {4}, and \frac {1} {5}$ to get $ \frac {47} {60}$, so he just put $ \boxed {047}$.[/quote]

And I'm pretty sure that's just a coincidence.

[quote="123456789"]Wow, his friend must be extremely lucky...one correct guess on this year's AIME, and two correct guesses on last year's AIME.  :o  I never get any of my guesses correct...[/quote]

Wow. Now I believe in magic.

I sometimes can pick up a crazy random recursion pattern that seems to work, and guess that the pattern defines the sequence, like Fibonacci on 1990 #9. But it didn't work for me on #11 this year. :( I thought the sequence was sum up two terms & add 2 to get every other term, then sum up two terms & subtract 1 for the other parity of the terms.

But I think for random guesses (no computation involved), to guess a small number, guess 005 or 012. For trig problems that ask you to find the sum of all possible arguments, guess 840. For other large numbers, ... oh well.

[quote="Wikipedian1337"]For trig problems that ask you to find the sum of all possible arguments, guess 840.[/quote]Why 840?

Yea, usually when I see questions that don't involve much computation, I guess a low number.

i got 48 by drawing it on graph paper...

I didn't take this aime, but did do this problem later and i found a geo/coordinates solution that was intriguing.
So first, draw the triangle with coordinates (0,0) (15,3) (12,12).  Notice that the angle between the lines from (0,0) to (12,12) and the x axis is pi/4 and the angle formed by the lines from (0,0) to (12,12) and (0,0) to (15,3) is essentially subtracting arctan1/5 from the pi/4.  Now, draw the line from (0,0) to (15,9).  The angle fomred by the lines (0,0) to (12,12) and (0,0) to (15,9) is arctan1/4.  Next, draw the point (18,4) and notice that the angle from (0,0) to (15,9) and from (0,0) to (18,4) is arctan 1/3.  Finally, then u realize that the angle betweeen the lines from (0,0) to (18,4) and from (0,0) to (15,3) is equal to arctan1/n.
Next, when you solve (9x+2)/(2x-9)=5, you get x=47.  This is the x coordinate of the point that, when you draw a perpendicular from, you intersect the line y=5x.  But then realize that then tan of the angle we're looking for is merely 47/1=47.
This is more intuitive if you draw it out.

too lazy to use latex :)

That's the geometric interpretation of the complex-number solution.

[hide=Solution]We have that $\arctan x + \arctan y = \arctan \left(\frac{x+y}{1-xy}\right)$. This means $\arctan \frac{1}{3} + \arctan \frac{1}{4} = \arctan \frac{7}{11}$. Next, we have $\arctan \frac{7}{11} + \arctan \frac{1}{5} = \arctan \frac{23}{24}$. This means $\arctan \frac{23}{24} + \arctan \frac{1}{n} = \arctan 1$, or $\arctan \frac{23n+24}{24n-23}=1$. This means $23n+24=24n-23$, or $n=\boxed{47}$.[/hide]

Is it a coincidence that $n=3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5$?


My deepest Ragards to Leigh family ,May Robert rest in peace.

Its not an appropriate time to post this ,just as a reminder:
The canadian IOI [hide="results"]
Competition was over in $\text{AUG},25^{th}$

.........Name .....................Medal..........Total score
CAN-03 Peng Shi................Silver..........350
CAN-01 Andy Kong........... Silver..........334
CAN-02 Richard Peng.........Silver..........328
CAN-04 Qiyu Zhu...............Bronze........235

The Maximum score was 480 achieved by a poland guy.

http://www.ioi2006.org/index.php?page=Official_Results&menu=0110[/hide]

Congratulations to the Canadian IOI participants! Great job, guys! :)

Congratulations guys! (Especially Andy! Yay Vincent Massey!)

Very nice.  Anyone know if Qiyu is coming to Waterloo?  Not that our ACM needs any more help... then again, ACM-style programming is completely different than IOI-style.

[quote="weileaf"]I could be wrong, but I think richard told me Qiyu is not graduating this year.

And yeah, even if our ACM team doesn't need help, we could use some help in topcoder, Waterloo's ranking on topcoder is getting lower and lower[/quote]

Hmm... i see...

Topcoder is a contest that I just can't do.  I'm definitely not a speed-hacker.  I wonder if Richard will have the same problem, cuz he is more of a "knows every algorithm in the universe" coder rather than a "hacks through ad hoc problems like a lawnmower" coder.  Simon started a few months ago though, and he seems to be doing pretty well, so maybe Waterloo has a chance to reclaim its topcoder glory.

[quote="weileaf"]I could be wrong, but I think richard told me Qiyu is not graduating this year.

And yeah, even if our ACM team doesn't need help, we could use some help in topcoder, Waterloo's ranking on topcoder is getting lower and lower[/quote]

Really? Andy told me the other three on the team are all graduating this year.


I think this problem is about the level of a Chapter Target question:

An equilateral triangle is inscribed in a circle of diameter 5.  A small circle is then inscribed in the equilateral triangle.  What is the ratio of the area of the smaller circle to the area of the larger circle?  Express your answer as a common fraction.

[quote="Treething"][hide="solution"]
If you noticed, the length of the line segment from the center and the smaller circle is 2/3 of that length on the larger circle. Thus, the ratio of the areas is 4/9.
[/hide][/quote]

Close .. but may be some error ? check again...

[hide]The radius of the circumcircle is just the longer part of the median, not the whole median. And the radius of the smaller circle is the smaller part. It is well known that the medians intersect at a trisection point, so the ratio of radii is 1:2. Ratio of area of similar figures is the square of the ratio of corresponding parts, so the answer is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$.[/hide]

[hide]
Let ABC be an equilateral triangle, R be the radius of larger circle and r be the radius of smaller circle,A' be  the area of the smaller circle,A be the area of the larger circle
 |AB|=|AC|=|BC|=2*R*SIN(60)=5*3^(1/2)/2
 Since A(ABC)= |AB|*|AC|*sin(BAC)/2=|AB|*|AC|*|BC|/(4*R)=|AB|*|AC|*BC|/10=
r*( |AB|+|AC|+|BC| )/2 then r=5/4
A'/A= (r/R)^2=1/4
 [/hide]


You enter a draw by picking a number between $ 1$ and $ 100$. You know that the winning number is two-thirds of the average of the numbers picked by eveyone. Which number should you pick?

If the game began a long time ago then I enter $ 1$.


percent calculus geometry trigonometry MIT college integration

I hear about them every now and then on the board, so I decided to create a topic about it after searching and seeing that nothing has been devoted to it. So, to those who have taken it or have heard about it - how good is it? Do they actually teach you the stuff, or give you assignments and say 'good luck', or... Also, do you have to apply to take EPGY courses, cause I was browsing the site and it said it wanted a transcript of a PSAT/SAT? If so, is it hard to get in...?

Just anything about the courses offered would be a great help. Thanks!

Lol was it me? I just mentioned it around an hour ago...

But i'm also interested. I'm worrying that my SAT score is too low to get in...

The EPGY courses have online "tutors" who can be very good. They sometimes use fairly standard college textbooks that are generally very readable and easy to understand. I like EPGY courses a lot; my son has taken four different EPGY math courses, and is now taking the EPGY computer science (C programming) sequence.

EPGY admissions policies are to admit about the top 15 percent of the younger (elementary-age) population--that's the age at which both of my two older sons were admitted. I'm not sure about high-school-age admissions policies, but I would apply if you are interested--the worst that can happen is that they decline your application. EPGY seems to give applicants the benefit of the doubt, because some people thrive better in EPGY courses than others. Your persistence in working on the course will make a difference, and will help you become smarter.

I took algebra 1 through half of calculus 1 from EPGY. The courses are great. i would definetly reccomend them to anyone.

not having the opportunity to talk to your tutor in person takes some getting used to.. (fortunatley, my mom could work me through most of the problems i had).. but my tutor was helpful when i really needed help.

overall the high school courses were difficult. they were pretty nice with grades though. roughly an 70% on my geometry final was the equivelant of an A in the class. (it was a hard final)

on admission.. i dont know. i started taking the courses before i had SAT or PSAT scores to speak of.. once you take one class, assuming you do alright, you basically dont have to worry about taking the next one.

good luck.

PM me if you have any specific questions you think i might be able to answer

They only require SAT/PSAT if u are taking a class that has age limits or some kind of limits.... for example, AP English or high humanities require one. Math courses and physics, computer science courses only require a class of its own. For example, if you want to take Calculus, you need a transcript that verifies a completion of a pre-calculus or trigonometry class, whether in EPGY or elsewhere.

if you just want to sign up without going through an application process,  and you don't care much about interacting with the tutor, there are cheaper courses for calc (thru multi), linear alg, and number theory here:
http://www.utexas.edu/cee/dec/
(menu for college courses is on the right side of screen)
Not much instruction. You have assignments and exams, but basically you have to be able to work through the material on your own. It's self paced, so you could take anywhere from a few weeks to a year to finish a course. If you compare prices, keep in mind that these are based on semesters, and EPGY courses are based on quarters. So, for instance, you need 4 EPGY courses to cover the calc sequence, but only 2 UT courses.

if you just want to cover the course material, and you don't care about an official grade on a transcript, check out MIT Opencourseware. Some courses have the entire course, including online lectures in streaming video. Other courses might just have a syllabus and exams. The price is right (free)!
http://ocw.mit.edu/index.html

You do have to apply but I belive that it is not competitive, i.e. there are set scores that you must get over in order to become eligible to take courses.  I took a long distance course with CTY, but they actually get their materials such as CDS and tests from EPGY.  The course was great I learned a lot and got a 5 on the AP physics exam.

Will they look at combined SAT score (<2005) cause I totally guessed the English section (my languages are not that good)? ie. If you can get into CTY, does it mean you can also get into EPGY? Like mna851, I took Honours Algebra II with CTY simultaneously with Algebra I in school, and considering that i'm not as smart as everyone else here, I actually understood most stuff (wow applause), though I used Academic Systems which I don't know is a part of EPGY.

I don't know the specifics of the cutoffs, all I know is that when I was in 7th grade anything greater than 550 on the math on the SAT qualified you for a CTY math course.  I dont know if its necessarily the same for EPGY.

If you deal directly with EPGY, the cut-off scores should be no higher, and may be substantially lower, than the cut-off scores for CTY. If you are already in CTY distance learning, you should have that channel for taking EPGY courses. I would expect that most regular participants on AoPS can get into EPGY distance learning courses in math without any question.

[quote="chess64"]How do they compare with CTY courses? I like CTY distance education a lot!  :)[/quote]

Which CTY distance education courses have you taken? My son has taken two, and if I know which ones you have taken, I'll know better how to compare them to EPGY courses.

They might even be identical.  Last year when I took AP Physics B through CTY the CDS and tests and all materials came from EPGY.

Ok I see that CTY is also mentioned - what is the difference between that and EPGY'?

And I saw mentioned that the prices for EPGY are not for the full course, but only like per 9 weeks or something along those lines. If I were to take a full AP Calc AB course what would that cost? And AP Calc BC?

I took Honors Algebra II from CTY (College Algebra from Thinkwell).

That's strange... I took Honours Algebra II from CTY also and I used College Algebra (the green book) with Academic Systems.

I think they just started using Thinkwell. (http://www.thinkwell.com)

the way EPGY does calculus is in three parts, Calculus A, Calculus B, and Calculus C.  A is the differential calc, B is the integral calc, and C is the rest of the topics on the BC exam such as series.  For the AP Calc AB exam you would need both Calc A and Calc B, and for the Calc BC exam you would need Calc A, B, and C.

EPGY offers quite generous financial aid. To apply for financial aid, your family has to be willing to reveal financial information, but the discounts from list price even for clearly middle-class (definitely not poor) families can be quite substantial.

Don't let list price scare you off from applying. By the way, the same rule applies to colleges. Apply where you think would get a good education, and see what the financial aid offer (um, and the admissions decision) is before deciding where to enroll.

[quote="Tunio"]These EPGY courses can count as a real credit in college? or high school? both? i'm confused :P[/quote]

Yes, I believe so. After all, if you take Multivariable Calculus or Linear Algebra I would assume you would want credit for it...

credit is not inherent in any course or exam. It is totally up to the whim of the institution that grants you a diploma. If it is important to you to get credit for EPGY courses from either your high school or from colleges you will be applying to, you need to contact them about their policies.

If you take a college level EPGY class in high school, and then go to Stanford they will grant you credit for the EGPY class.

My parents were able (after a lot of effort, I'm told) to get our school district to pay for EPGY courses when I was in middle school, since there wasn't really any other way I could take math courses at the appropriate level (Algebra 2 through Calc A and B).  Unfortunately, at the time I didn't really have the self-discipline to do well in a course that required that much independent work.  Now, though, I took the Number Theory and Modern Algebra courese this year and did very well (and loved the coureses, too).

Yeah, this year I took multivariable calc through EPGY.  I thought the courses were very well done.  I especially liked the multimedia lectures.  The diagrams were better than any teacher could ever draw.  The school district didn't pay for mine, though.

Next year, however, I am going to do WOOT + subject courses rather than EPGY.  We'll see what happens after that.

I have only three possible options: I can take Multivariable, Topology, or Number Theory. Which do you think would be the best choice?

first of all, point-set topology requires the prerequesites of real/complex analysis which requires the prerequesites of multivariable calculus.  so i don't understand how you could be choosing between multivariable calculus and point-set topology.  that brings us to 2 choices, multivariable or number theory.  that depends entirely on what you're interested in.  multivariable is the next step in a college calculus sequence, single variable being the first (equivalent to AP Calculus BC).   if your goal is to work your way into college mathematics, this is the right way to do it.  number theory is something more abstract, so to speak, and it's not in the sequence of college mathematics.  if the theory of numbers and arithmetic really interests you, then that's where you should be.  one thing to mention though is that introductory number theory and soon intermediate number theory are offered right here at AoPS for a much lower price.  the courses i'm sure are not entirely the same, but that's definitely something to look into.

yeah i just saw the prereq's so you're right about that topology.
anyways i guess i'll take multivariable, thanks for the clarification!

Here is my two cents...

I am taking a Computer Science class from them right now.  You have lectures (power point presentations with voice overs) and then exercises which involve answering questions and programming small applications.  I personally am not satisfied with the course so far.  The instructor seems fairly uninterested in giving any serious help (I've been having problems with the EPGY software) and seems to only humor me because it's necessary.  I don't think she has any genuine interest in making sure that I am learning anything from the course.  I've received no constructive feedback or any really for that matter.  I would say that AoPS is 1000 times better, even with the independent study which is generally removed from instructors altogether.

[quote="SnowStorm"]The instructor seems fairly uninterested in giving any serious help (I've been having problems with the EPGY software) and seems to only humor me because it's necessary.  I don't think she has any genuine interest in making sure that I am learning anything from the course.  I've received no constructive feedback or any really for that matter.  I would say that AoPS is 1000 times better, even with the independent study which is generally removed from instructors altogether.[/quote]

I'd have to agree with you... I took physics classes, and Gary Oas was a horrible tutor. Nelia Mann was pretty nice, she answered all my questions and helped me but whenever I asked Mr. Oas a question, he'd say some random two-word thing that didn't even make sense... lol.

I think the lectures are fine, as long as you are willing to spend extra time learning on your own. If you want to take a course where a tutor explains everything and you don't have to do a lot of self learning, epgy probably isn't for you.

Well i don't think you can judge EPGY as a whole by one course, particularly because that course only represents the computer science department of EPGY.  I have never taken a computer science course on EPGY, but I have taken a physics course and it was extremely well done.  I have not yet, but plan to at one point, take a math course on EPGY as my friend has done so and gave the same compliments to it as he did with the physics course that I took there as well (he did too).  It might just be that the computer science courses are poorly made and the instructors are not fit for it.  I really don't know enough about that area of EPGY to judge.  All I can say is that some things don't work for everyone, and I have had good experiences with EPGY.

Honestly, I think most people would derive far more enjoyment from number theory than from multivariable calculus. Multivariable calculus really isn't very different from single variable calculus -- many things generalize to more dimensions pretty easily, and so there is a lot of painful repetition. Number theory will be a refreshing change from the normal calculus sequence, and I think it would be more beneficial to take that.

I am also considering taking a Physics course at EPGY, and I was wondering if anyone knew how strict the prerequisite requirements are. I want to take "Light and Heat," but I see that the prerequisite is AP Physics C. I signed up for AP Physics C next year at my school, so I would be taking it in conjunction with the EPGY course. Does anyone have any thoughts on this? Thanks.

well, i took ap physics c this year in school but i have never taken light and heat.  i'd imagine that they might apply some basic physics formulas such as F = ma that rely on the understanding of forces, momentum, energy, etc.  i'm sure it's possible to do both at the same time, but you should definitely make sure you have a good understanding of ap physics c first.  i'm assuming you have calculus as a prerequesite.  have you taken a physics course in school or online ever before (such as a non-calculus course)?  if you have taken for example ap physics b, and you have calculus as a prerequesite, you should be more than fine in light and heat.

I'm currently taking AP Physics B (general physics) through CTY, which uses the EPGY course material, so I can skip it next year in school.  Although I've already gotten a 5 on BC Calc and could have skipped straight to Physics C E&M, I had to take Physics B because my school requires a general physics class for graduation.

I also took Precalc at Duke through TIP at the summer program, and I have to say that I prefer the 3 weeks on campus experience a lot more than distance education.  Since I'm taking the "intensive summer session" during summer vacation, I feel like there's not enough time for me to wait for the tutor to answer any questions because I need to get moving through the material.  I have to say that I definitely prefer having a real teacher face to face, even if the the teacher I had at Duke wasn't very good, and I basically ended up figuring everything out myself and then helping the people with whom I was working.

Given a choice between the two, I'd go for the summer camp option any day.  The opportunity to work with classmates of a similarly intellectual caliber to me was very edifying.

I took 2 courses through EPGY (Linear Algebra and Diff. Eq.), and I didn't like them so much. Basically you have to buy a book and you get a bunch of shockwave-based lectures, and someone whom you can call/email for help. In terms of credit, I doubt it goes for that much at most prestigious colleges, and in terms of placement, you can probably get your way by talking to the math department. The only thing you might get out of these courses is instruction, and I definitely felt like it was a big waste of money in that area. The amount of money you spend versus what you get is absolutely ridiculous. The course material was very ho-hum in terms of presentation and pedagogy. I didn't really need the instructor's help, and if you're taking higher-level classes at this point you are probably a pretty independent learner as well. You can doubtless get more help with stuff online through forums and such than you can get from whatever tutor they set you up with. What I suggest doing, if you really want to learn, is pick up a good mathematics text, and do ALL of the exercises. You will get a better view of the material than you will through their (not very good IMHO) math texts. I suggest Loomis and Sternberg's "Advanced Calculus", which you can get for free online: http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf . It is an EXTREMELY lucid, clear, and well-developed presentation of material, and the viewpoint you will develop through it will help you a lot in your future mathematical studies. You should also try to get a feel for various proof techniques, such as mathematical induction. Wikipedia is good for any mathematical concepts you are not familiar with.

Doed epgy offer clac B or C over the summer, and if so, do i need to take both b and c for the calc bc ap or just c if im taking calc ab right now?

I looked on the CTY website and the AP courses cost more than 1000 dollars!  :?

Regarding whether to take multivariable or number theory:

I'm pretty sure EPGY's number theory course covers more than the AoPS course.  Although I've never taken AoPS classes, it looks like the two number theory ones cover the basic ideas of congruence, divisibility, prime factorization, etc. in the intro, and things like cool theorems (Fermat's Little, etc.) and Diophantine equations in intermediate.  EPGY's course covers all of this, plus a little bit on magic squares, continued fractions, and a whole lot of stuff on quadratic fields--classes of numbers of the form $a+b \sqrt{d}$ for a, b, and d integers.  It also approaches the material from a very rigorous perspective--which AoPS might do as well; I'm not the one to ask about that.

And on the subject of multivariable, I didn't take it from EPGY.  Be aware that your college may want you to retake it eventually; this is the case for me, but my multi course was much less rigorous.  You also may be able to take an honors-level course in multivariable calc when you get to college, and this course may be better and/or more rigorous than EPGY.  If you plan on coming to Stanford, for example, you will definitely want to wait to do multivariable, linear algebra, or diffeq and take the honors course here instead.  But all this really depends on where you're looking at going to college.

EPGY is in fact a very independent way of learning.  You should be prepared to not have a great deal of interaction with your tutor.  My tutors have always been very willing and able to help; it's just that the fact that you're communicating by e-mail and phone tag makes it difficult to get the kind of help you'd get by talking in person.


$ \left(1\minus{}\frac{1}{2^2}\right)\left(1\minus{}\frac{1}{3^2}\right)\dots\left(1\minus{}\frac{1}{n^2}\right)$ $ \equal{}\left(1\minus{}\frac 12\right)\left(1\plus{}\frac 12\right)\left(1\minus{}\frac 13\right)\left(1\plus{}\frac 13\right)\dots \left(1\minus{}\frac 1n\right)\left(1\plus{}\frac 1n\right)$

$ \equal{}\frac 12\cdot\frac 32\cdot\frac 23\cdot\frac 43\dots\frac{n\minus{}1}{n}\cdot\frac{n\plus{}1}{n}\equal{}\frac{n\plus{}1}{2n}$


Find the equation of a line containing (2,-2,3) and (-5,2,-3).  :( I forgot how to solve this.

The directional vector is $\langle-7,4,-6\rangle$, so the equation for the line would just be:

$\langle x,y,z\rangle = \langle2,-2,3\rangle+\langle-7,4,-6\rangle t$

[quote="blahblahblah"]Jimmy, you may want to try $\langle$ (\langle) and $\rangle$ (\rangle).[/quote]

Thanks, I changed it (took me 3 edits to get it right, I must be really tired). :)


calculus
logarithms
trigonometry
geometric series
calculus computations

Can anyone tell me where I can find a list of differential calculus summations? I can't find them on google or yahoo.

Example would be...

1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ... = e^x

"Taylor series" or "power series" is what you are looking for -- try using those words in your searches.  Why do you need a list of power series?

For my own second semester calculus courses:

[b]Seven power series to memorize:[/b]

1. The geometric series

$\frac1{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty}x^{n}$

Series valid for $|x|<1.$

2. The logarithm

$\ln(1+x)=x-\frac{x^{2}}2+\frac{x^{3}}3-\frac{x^{4}}4+\cdots =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n}}n$

Series valid for $-1<x\le1.$

3. The arctangent

$\arctan x=x-\frac{x^{3}}3+\frac{x^{5}}5-\frac{x^{7}}7+\cdots =\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$

Series valid for $|x|\le 1.$

4. The exponential

$e^{x}=1+x+\frac{x^{2}}2+\frac{x^{3}}6+\frac{x^{4}}{4!}+\cdots =\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

Series valid for all $x.$

5. The cosine

$\cos x = 1-\frac{x^{2}}2+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}$

Series valid for all $x.$

6. The sine

$\sin x=x-\frac{x^{3}}6+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$

Series valid for all $x.$

7. The binomial series: for any real number $\alpha,$

$(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)x^{2}}2+\frac{\alpha(\alpha-1)(\alpha-2)x^{3}}6+\cdots+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)x^{n}}{n!}+\cdots$

This series is valid for (at least) $|x|<1.$ It converges at one or both endpoints for certain values of $\alpha.$ When $\alpha$ is a nonnegative integer, this becomes a finite sum and converges everywhere.

Thanks a lot you guys. I need this because right now, we're learning about solutions about a singular point using Frobenius' method.  :)

In my mind, I group the above list into families. The binomial series is one of a kind. We get the logarithm and arctangent series by integrating geometric series, so those #1, #2, and #3 are grouped together. The exponential, cosine, and sine series have an obvious relationship, one made explicit by Euler's identity for the complex exponential.

I didn't include such things as $\cosh,\sinh,$ and $\tanh^{-1}$ (there's only so much room on an "_ things to memorize" list), but they can be found fairly readily, and assigned to one or the other of the families.


Well, I have a tabular. How can I define height of one row? Default height doesn't suits to me.  :lol:

[code]
\begin{tabular}{|c|c|}
\hline
30 & \hspace{0.8cm}\hpedeset \hspace{0.8cm} \\
\hline
\end{tabular}
[/code]

P.S. \hpedeset is one hieroglyph

Try \renewcommand{\arraystretch}{0.5} before the \begin{tabular} command.

The 0.5 is a scaling factor -- this will cut the row spacing in half.  You can change it to whatever suits you.


A circular lamina of radius a has a density which varies as the distance frrom a point P of its circumference. Find the moment of inertia:

(1) about the perpendicular at P to the plane of the lamina

(2) about the tangent at P.


I'm not sure how to fully factor 3(x+2w)^3 - 3p^3r^3.

take out the coefficient and then substitute a variable into the $x+2w$ and $(pr)^3$ and use those simplifications to solve it.  Then substitue back

To expand on that,

[hide]
let $a=x+2w$, $b=pr$

Then by substitution we have $3a^3-3b^3= 3(a^3-b^3)$

$=(a-b)(a^2+ab+b^2)$

Now we substitute back, and it's really pretty nasty, but this has got to be it since a and b share no common variable:

$=3(x+2w-pr)(x^2+4xw+4w^2+xpr+2wpr+p^2r^2)$

...is as far as it goes. [/hide]

you made a mistake in the solution. You said that $3(a^3-b^3)=(a-b)(a^2+ab+b^2)$. It was really $3(a-b)(a^2+ab+b^2)$

oh my bad... but I did throw the 3 in the final answer


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines
AMC

Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad
\textbf{(C)}\ \sqrt{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \dfrac{4}{\sqrt{3}}$

[hide]$\frac{1}{\sin 30^o} = \frac{OB}{\sin \angle A}$. $\Longrightarrow$ $OB = 2\sin \angle A$.

Hence the maximum occurs when $\sin \angle A$ is the maximum, that is $1$.

So the maximum of $OB$ is $2$.[/hide]

[quote="Silverfalcon"]Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is

A. 1
B. $\displaystyle \frac{1 + \sqrt{3}}{\sqrt 2}$
C. $\sqrt 3$
D. 2
E. $\frac{4}{\sqrt 3}$[/quote]

[hide] Let $OA=y$ and $OB=x$ (to simplify!).The law of cosinus gives
\[ 1=x^2+y^2-xy\sqrt{3} \]
so \[ y^2-xy\sqrt{3}+x^2-1=0 \]
Because this equation has real solution (for $y$)  that means $\Delta >0$ so
\[ 3x^2-4(x^2-1)>0 \] that means \[ x^2<4 \] so \[ x<2 \].
the answer is D.[/hide]

I was sitting here for 20+ minutes...and then I realized that Andreas used the law of sines >.<.

[hide]Well if you think of a 30 60 90 you already have D, eliminating a, b, and c. Now if you draw the diagram, the right triangle that you have created places A at the farthest point from O, as once the 90 degree angle OBA either decreases or increases, the position of A will move towards O. So with my algebra 2 knowledge and laziness in not using the law of cosines, I got D[/hide]

[quote="Andreas"][hide]$ \frac {1}{\sin 30^o} \equal{} \frac {OB}{\sin \angle A}$. $ \Longrightarrow$ $ OB \equal{} 2\sin \angle A$.

Hence the maximum occurs when $ \sin \angle A$ is the maximum, that is $ 1$.

So the maximum of $ OB$ is $ 2$.[/hide][/quote]

Why does the Law of Sines give us the maximum value?

Thanks in advance!

[quote="314.math"][quote="MellowMelon"]It doesn't "give" us the maximum. It's a way to write $ OB$ in terms of other quantities with less constraints (like the measure of angle $ A$).[/quote]

Then how does the Law of Sines help us?  Doesn't it just return a side length - how would we know if it is the longest?

Thanks![/quote]

it writes OB in another way. the max of sin is 1

By Law of Sines we have $ sin(30) \equal{} \frac {sin(OAB)}{OB} \implies OB \equal{} 2sin(OAB).$ The maximum value of $ sin \equal{} 1$ so the maximum value of $ OB \equal{} 2(1) \equal{} 2 \implies \boxed{D}$

@ [i]Adihockey[/i] - are u adi rajhoobian?


It is easy to see that $x=-3$ is a solution.

Now since $f(x)=\sqrt[3]{x}$ is monotonically increasing in $R$, and $x+2$, $x+3$, $x+4$ are all monotonically increasing in $R$, so $g(x)= \sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is always increasing of $R$, therefore $x=-3$ is the only solution.

[quote="shobber"]Solve the equation
\[ \sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0.  \][/quote]
I have another way: using the well-known formula:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
 :D  :)  :lol:
Happy Asian new year !!! :coolspeak:  :winner_first:


algorithm
geometry
real analysis
real analysis theorems

hai friends

      a image will be drawn in the screen ( computer) as show below .   I need to pick only the outer contour from the drawn image. How to do that ..? is there any algorithms ...???

    provided  :  I know each point on the closed curve .( including the red line ).

One possibility is to use some [url=http://groups.csail.mit.edu/graphics/classes/6.837/F01/Lecture03/Slide05.html]area fill algorithm[/url] starting from a point outside of the image (say, above it). Outer contour will be easy to identify either after or during the run of the algorithm.


In the figure shown, side $ \overline{CD}$ of the rectangle is also the diameter of circle $ O$. What is the area of the shaded region?

[asy]size(5cm,5cm);

pair A,B,C,D,O,a,b;

A=(-12,-4);
B=(-12,4);
C=(0,4);
D=(0,-4);
O=(0,0);

fill((A--B--C--D--cycle),gray);
draw(A--B--C--D--A);

a=(-4,0);
b=(4,0);

fill((a..C..b..D..cycle),white);
draw(a..C..b..D..a);

draw(C--D);

label("$A$",A,S);
label("$B$",B,N);
label("$C$",C,N);
label("$D$",D,S);
label("$O$",O,E);
dot(O);

label("$8$",midpoint(A--B),W);
label("$12$",midpoint(B--C),N);[/asy]

The shaded region is the the area of the semicircle subtracted from the area of the rectangle.

Area semicrcle: $ \frac{4^2 \cdot \pi}{2} \equal{} 8 \pi$
Area rectangle: $ 12 \cdot 8 \equal{} 96$

Shaded region: $ 96 \minus{} 8 \pi$


For the questions in MathCounts, I was wondering, is there an easy way of divisibility for numbers in the form 101, 1001, 10001, 100001, etc. ?
Well, not 101 because it's prime.
$1001=7\times11\times13$
But how to do other numbers like that?

you can see that if there is an even number of zeroes then it can be divided by $11$.

[hide="hm"]
Lemma: $a^{n}+b^{n}$ is divisible by $a+b$ when n is odd.
$a^{2n+1}+b^{2n+1}=0$
$a=-b$
$a+b=0$
So $10^{2n+1}+1$ is divisible by 11.[/hide]
That helps a litttle.

[quote="xscapezaer"]every number of the form $10000...((4n+3) 0's)1$ is divisible by 101. And 101 is prime[/quote]
That formula doesn't work.

10001=73*137
100000001=17*5882353 (which is prime)

Is there any easy way that you could find these factors?

I noticed that the primes all end in 7, 3, or 1.. :ninja:


function
trigonometry
calculus
calculus computations

finding zeros of functions

many forms of finding solutions to equations of form f(x)=0. one mwthod requires the equation to be expressed in the form x=g(x) so that asolution can be interpreted as the value of x where the line x=y intersect the curve y=g(x). this is described as the fixed point of the function g(x).
let g(x)=cosx. show graphically that g(x) has one fixed point between x=0 and x=2.

We need to show that there exists $c$ such that $\cos c=c$.  Define $g(x)=\cos x - x$.  Clearly $g(0)=1$, $g(\pi/2)=-\pi/2$.  Hence by continuity there exists $c\in (0,\pi/2)$ such that $g(c)=0$.  This $c$ is a fixed point of $f(x)=\cos x$


ratio
inequalities
real analysis
real analysis unsolved

Let $ (x_n)$ be sequence of real numbers, such that $ x_1 > 0$ and $ x_2 > 0$ and
\[ x_{n \plus{} 2} \equal{} 1 \plus{} \frac {x_{n \plus{} 1}}{x_{n}} \qquad \text{for } n \geq 1
\]
Prove that $ (x_n)$ converges.

We start from the inequality $ x_n\geq 1$ for $ n\geq 3$. Note that
\[ x_{n \plus{} 3} \equal{} 1 \plus{} \frac1x_n \plus{} \frac1x_{n \plus{} 1}.
\]
Hence $ x_n\leq 3$ for all sufficiently large $ n$. Suppose we proved that $ a\leq x_n\leq b$ for all sufficiently large $ n$. Then using the formulae above we obtain $ 1 \plus{} \frac2b \leq x_n\leq 1 \plus{} \frac2a$ for all sufficiently large $ n$ and furthermore $ \frac {3a \plus{} 2}{a \plus{} 2} \equal{} 1 \plus{} \frac {2}{1 \plus{} \frac {2}{a}}\leq x_n \leq 1 \plus{} \frac {2}{1 \plus{} \frac {2}{b}} \equal{} \frac {3b \plus{} 2}{b \plus{} 2}$. Consider the sequences $ a_0 \equal{} 1$, $ a_{k \plus{} 1} \equal{} \frac {3a_k \plus{} 2}{a_k \plus{} 2}$; $ b_0 \equal{} 3$, $ b_{k \plus{} 1} \equal{} \frac {3b_k \plus{} 2}{b_k \plus{} 2}$. Now it is a piece of cake to show $ a_k$ is increasing and $ a_k\to 2$; $ b_k$ is decreasing and $ b_k\to 2$.

thanks. should it  be $ b_0\equal{}3$? but it is not a big deal of course  :lol:

Copy-paste from $ a_0\equal{}1$  :blush: Edited


search
real analysis
real analysis unsolved

Find the cool sum  :D

$ \displaystyle R \equal{} \sum\limits_{m \equal{} 1}^{n} tan^{\minus{}1}({\frac {1}{{(m \plus{} 1)^2} \minus{} m}})$

waiting your nice replay.


geometry
circumcircle
inradius
incenter
trigonometry
projective geometry
geometry proposed

Given $\triangle ABC$ with incircle and $\mathcal{C}(I,r)$ circumcircle $\mathfrak{C}(O,R)$. Let
$D=\mathcal{C}\cap \ovl{BC}$, $E=\mathcal{C}\cap \ovl{CA}$,
$F=\mathcal{C}\cap \ovl{AB}$. Let $\mathfrak{T}_{a}$ be a circle through $A$ tangent to $\mathcal{C}$ at $D$. Let
$B_{a}=\mathfrak{T}_{a}\cap \ovl{AB}$, $C_{a}=\mathfrak{T}\cap \ovl{CA}$
and $\rho_{a}$ the measure of the inradius of $\triangle AB_{a}C_{a}$. Similary we define
$\rho_{b}$ y $\rho_{c}$. Show that
(i) $\frac{bc}{\rho_{b}\rho_{c}}+\frac{ca}{\rho_{c}\rho_{a}}+\frac{ab}{\rho_{a}\rho_{b}}=\frac{16R^{2}}{r^{2}}$
(ii) $\frac{\rho_{a}}{p-a}+\frac{\rho_{b}}{p-b}+\frac{\rho_{c}}{p-c}=2(\frac{\sqrt[]{2}}{\sqrt[]{r}})^{3}\sqrt[]{\rho_{a}\rho_{b}\rho_{c}}$

[quote="hucht"]Given $\triangle ABC$ with incircle and $\mathcal{C}(I,r)$ circumcircle $\mathfrak{C}(O,R)$. Let
$D=\mathcal{C}\cap \ovl{BC}$, $E=\mathcal{C}\cap \ovl{CA}$,
$F=\mathcal{C}\cap \ovl{AB}$. Let $\mathfrak{T}_{a}$ be a circle through $A$ tangent to $\mathcal{C}$ at $D$. Let
$B_{a}=\mathfrak{T}_{a}\cap \ovl{AB}$, $C_{a}=\mathfrak{T}\cap \ovl{CA}$
and $\rho_{a}$ the measure of the inradius of $\triangle AB_{a}C_{a}$. Similary we define
$\rho_{b}$ y $\rho_{c}$. Show that
(i) $\frac{bc}{\rho_{b}\rho_{c}}+\frac{ca}{\rho_{c}\rho_{a}}+\frac{ab}{\rho_{a}\rho_{b}}=\frac{64R^{2}}{r^{2}}$
(ii) $\frac{\rho_{a}}{p-a}+\frac{\rho_{b}}{p-b}+\frac{\rho_{c}}{p-c}=(\frac{\sqrt[]{2}}{\sqrt[]{r}})^{3}\sqrt[]{\rho_{a}\rho_{b}\rho_{c}}$[/quote]

no solutions??  :maybe:

[color=darkblue]Denote : the circle $w$ through the point $A$ and tangent to the line $BC$ in the point $D$ ; the second intersections $M$, $N$ of the circle $w$ with the lines $DE$, $DF$ respectively ; the incenter $I_{a}$ of the triangle $AB_{a}C_{a}$. From a well-known property, the rays $[DE$, $[DF$ are the bisectors of the angles $\widehat{ADC_{a}}$ and $\widehat{ADB_{a}}$ respectively. Thus, the rays $[B_{a}M$, $[C_{a}N$ are the bisectors of the angles $\widehat{AB_{a}C_{a}}$, $\widehat{AC_{a}B_{a}}$ respectively, i.e. $I_{a}\in B_{a}M\cap C_{a}N\ .$ From the Pascal's theorem applied to the hexagon $NDMB_{a}AC_{a}$ inscribed in the circle $w$ results that the points $F\in ND\cap B_{a}A$, $E\in DM\cap AC_{a}$, $I_{a}\in MB_{a}\cap C_{a}N$ are collinearly. But the incenter $I_{a}$ belongs to the bisector of the angle $\widehat{BAC}\ .$ Thus, $I_{a}\in EF$ and it is the middlepoint of the segment $EF$. Therefore, $\boxed{\ \rho_{a}=\frac{1}{2}(p-a)\sin A\ }\ .$
$1.\blacktriangleright \sum\frac{bc}{\rho_{b}\rho_{c}}=$ $\sum\frac{4bc}{(p-b)(p-c)\sin B\sin C}=$ $\Longrightarrow$ $\boxed{\ \sum\frac{bc}{\rho_{b}\rho_{c}}=\frac{16R^{2}}{r^{2}}\ }\ .$
$2.\blacktriangleright$ $\sum\frac{\rho_{a}}{p-a}=\frac{p}{2R}\ ;\ \sqrt{\rho_{a}\rho_{b}\rho_{c}}=\frac{pr\sqrt r}{4R}$ $\Longrightarrow$ $\boxed{\ \sum\frac{\rho_{a}}{p-a}=\frac{2}{r}\cdot\sqrt{\frac{\rho_{a}\rho_{b}\rho_{c}}{r}}\ }\ .$[/color]


Solve in real: $ x^{4} \minus{} 4x^{3} \plus{} 5x^{2} \minus{} 4x \plus{} 1 \equal{} 0$.


Solve $ x^{2007}\equal{}y^{2007}\minus{}y^{1338}\minus{}y^{669}\plus{}2$ in integers. I've arrived to the fact that $ (y^{669}\minus{}x^{669})$ must divide $ (y^{669}\minus{}1)(y^{669}\plus{}2)$. From here, is it correct to just put $ x^669\equal{}1$ and $ x^669\equal{}\minus{}2$ to find that $ x\equal{}y\equal{}1$ are the only solutions?

The problem with that logic is that $ y^{669} \minus{} x^{669}$ is not necessarily prime; in other words, some of its factors might divide $ y^{669} \minus{} 1$ while others might divide $ y^{669} \plus{} 2$, rather than it being equal to one factor or the other.

You also missed the solution $ x \equal{} 1, y \equal{} \minus{}1$.

[hide="Hint"] The LHS is a cube.  Bound the RHS between two cubes. [/hide]

Mmh, I don't how can I bound the RHS between two cubes :blush:.

Do you mean to see that the LHS is greater than $ (y^{669} \minus{} 1)^3$ and less then $ (y^{669} \plus{} 1)^3$?

Careful - you need to do casework for small values of $ y$ (both positive and negative), but that's the general idea.  For large values of $ y$ the upper bound can be $ y^3$.


rotation
function
integration
calculus
analytic geometry
geometry
calculus computations

Well, I think it is anyway. This is from my homework:

The curve $ y\equal{}x^5$, for $ 0\leq{x}\leq3$, is rotated around the $ y$-axis to form a tank. Torricelli’s Law states that water will empty from a hole in the bottom of the tank at a rate directly proportional to the square root of the height of the water in the tank. If the tank is full to begin and is still 3/4 full after 1 hour, find a function $ V(t)$ that gives the volume of the water left in the tank at any time $ t$, where  $ t$ is in hours. Be sure to resolve any constants and round them to three decimal places.

I started by deciding:

$ \cfrac{dV}{dt}\equal{}k\sqrt{h}$,

where $ k$ is a constant. I knew that the volume of water as a function of height was:

$ V(h)\equal{}\int_0^h\pi\sqrt[5]{y^2}\;dy$

The tank is 3/4 full when $ h\approx197.863$, with $ V\approx3680.712$.

However, I don't know how to put the pieces together. I've dealt with the easy problems asking for a definitive value, but not ones like these that ask for a general function. Help is appreciated! :)

Height is a function of time.  That is, $ h\equal{}h(t)$.  Just determine how height is affected by time and you've got your function.

aghh I emailed my teacher for some hints, but she just said to solve the integral for $ h$ and plug into the proportion, which was my first step anyway..

What I don't get is how I can incorporate time into this problem without deriving $ h(t)$ myself (instead of manipulating the given equations.. Is this what you meant? It's confusing, though)

When I solve for $ h$,

\begin{align*} V(h(t)) & = \pi\cdot\frac {5}{7}\;[h(t)]^{7/5} \\
h(t) & = \left(V(h(t))\cdot\frac {7}{5\pi}\right)^{5/7} \end{align*}

but that's not valid (is it?) since we have $ h$ on both sides.. anyway:

\begin{align*} \frac {dV}{dt} & = k\sqrt {\left(V(h(t))\cdot\frac {7}{5\pi}\right)^{5/7}} \\
V(t) & = \int{k\sqrt {\left(V(h(t))\cdot\frac {7}{5\pi}\right)^{5/7}}\;dt} + C \end{align*}

Whatever that may be.. I must've forgotten something.

Hi.
I could be completely wrong, but I think this is how to go at it.

[hide]
Using cylindric coordinates, the volume of you vessel is given by:
$ V(h(t)) \equal{} \int_0^{2\pi}d\theta\int_0^{h(t)}dy\int_0^{\sqrt [5]{y}}dr$
$ \equal{} 2\pi\int_0^{h(t)}\sqrt [5]{y}\,dy$
$ \equal{} 2\pi\frac {5}{6}h(t)^{\frac {6}{5}} \equal{} \frac {5\pi}{3}h(t)^{\frac {6}{5}}$

You know that $ h(0) \equal{} 3^5$, so $ V(h(0)) \equal{} \frac {5\pi}{3}3^6 \equal{} 5\pi3^5$
You also know that the vessel is at $ 3/4$ after one hour, so $ V(h(1)) \equal{} \frac {3}{4}5\pi3^5$
Assuming that the flow of liquid through the hole is constant, then, after another hour, it will be half full and so on... a linear function!
So, the total volume should be given by $ V(t) \equal{} V(0) \plus{} k*t$
where $ k \equal{} V(1) \minus{} V(0) \equal{} \minus{} \frac {5\pi}{4}3^5$
[/hide]

I was just looking at the problem statement and I noticed that I did something wrong. I assumed linearity when it is clearly stated that the flow is proportional to the square root of the height.
Well then,
$ V(h(t)) \equal{} V(0) \plus{} k \sqrt{h(t)} t$
$ V(0) \equal{} V(h(0)) \equal{} \frac{5\pi}{3}h(0)^\frac{6}{5} \equal{} 5\pi3^5$
Now we have two different expressions for $ V$. They should give the same result at all $ t$
$ V(0) \plus{} k t \sqrt{h(t)} \equal{} \frac{5\pi}{3}h(t)^\frac{6}{5}$ (1)

$ V(1) \equal{} \frac{3}{4}V(0)$ helps to get at $ k \equal{} \minus{}4^{\minus{}\frac{7}{12}}5\pi3^2$

All you need to do is solve this weird equation (1) in order of $ h(t)$. This doesn't look trivial at all and I can't do it. sorry.
When the solution to some problem becomes too ugly, it's because the solution is wrong... :(

I'm not familiar with cylindrical coordinates, so I'm going to stay in rectangular coordinates for now, but I took some hints from your posts. I still don't get how to find $ h(t)$, though. Here's what I've done:

Let $ V_h(h)$ be the volume $ V$ as a function of height $ h$, and $ V_t(t)$ be the volume as a function of time $ t$. Then
\[ V_h(h(t)) = \int_0^{h(t)}\pi\sqrt [5]{y^2}\;dy = \frac {5}{7}\pi[h(t)]^\frac {7}{5}
\]

\[ h(t) = \left[\frac {7}{5\pi}\cdot{}V_h(h(t))\right]^\frac {5}{7}\qquad\qquad\qquad(1)
\]
And:
\[ V_t(t) = V_t(0) - k\sqrt {h(t)}\cdot{t}\qquad\qquad\qquad(2)
\]
We need to find $ k$ and $ h(t)$.

Using given information,
\[ V_t(0) = V_h(h(0)) = V_h(3^5) = \frac {5}{7}\pi3^7.
\]
We can plug this into equation (2).

Let's use $ V_t(1) = \frac {5}{7}\pi3^7 - k\sqrt {h(1)}\cdot1$ to find $ k$.
\begin{align*} & V_t(1) = \frac {3}{4}[V_t(0)] \\
& = V_h(h(1)) = \frac {3}{4}[V_h(h(0))] \\
& = \frac {3}{4}\cdot\frac {5}{7}\pi3^7\end{align*}
And
\[ h(1) = \left[\frac {7}{5\pi}\cdot{}V_h(h(1))\right]^\frac {5}{7} = \left(\frac {3}{4}\right)^\frac {5}{7}\cdot3^5.
\]
Plugging in the two numbers produced above,
\begin{align*} \frac {3}{4}\cdot\frac {5}{7}\pi3^7 & = \frac {5}{7}\pi3^7 - k\sqrt {\left(\frac {3}{4}\right)^\frac {5}{7}3^5}\cdot1 \\
k^2\left[\left(\frac {3}{4}\right)^\frac {5}{7}3^5\right] & = \frac {1}{4}\left(\frac {5}{7}\right)^2\pi^23^{14} \\
k & \approx128.701. \end{align*}
Now that we have $ k$, we can plug it into equation (2) again:
\[ V_t(t) = \frac {5}{7}\pi3^7 - 128.701\sqrt {h(t)}\cdot{t}
\]
..how would I obtain $ h(t)$?

The cylindrical coordinate system is the best for this problem. Look it up in wikipedia!
I don't understand how you get the y-squared in
$ V_h(h(t)) \equal{} \int_0^{h(t)}\pi\sqrt [5]{y^2}\;dy \equal{} \frac {5}{7}\pi[h(t)]^\frac {7}{5}$

Maybe your teacher will let you keep the $ h(t)$ without solving it?...  :roll:

heh maybe.. I turned it in anyway. :P

The y-squared was because I was in rectangular coordinates, as $ A_\text{circle} \equal{} \pi{r^2}$ and $ r \equal{} \sqrt [5]{y}$. You used cylindrical coordinates, so the area was $ \int_0^{2\pi}\;d\theta\int_0^{\sqrt [5]{y}}\;dr$, right? (Though it doesn't make intuitive sense to me yet.)

Hmm I knew it had something to do with polar coordinates extended to the third dimension.. :D It does look useful (as well as spherical coordinates, maybe?) - I'll look into it. Thanks for the pointer and your help on this problem!

$ V(h) = \int_0^h\pi y^{2/5} dy = \frac {5\pi}{7} h^{7/5}$ (constant of integration is 0 because 0 height -> 0 volume).  Now differentiate both sides with respect to time (gets back to the original integrand)

$ \frac {dV}{dt} = \pi h^{2/5} \frac {dh}{dt}$

But we are given $ \frac {dV}{dt} = k\sqrt {h}$ for some $ k$...

$ k\sqrt {h} = \pi h^{2/5} \frac {dh}{dt}$
$ \frac {k}{\pi} h^{1/10} = \frac {dh}{dt}$

Ugh.. differential equation...

Mathematica tells me that the solution is $ h(t) = \frac {9 (3^{2/9} 10^{8/9} kt (kt + \pi C)^{1/9} + 3^{2/9} 10^{8/9} \pi C (kt + \pi C)^{1/9})}{100 \pi^{10/9}}$

or simpler: ${ h(t) = \frac {9 \cdot 3^{2/9} \cdot 10^{8/9}}{100}}(\frac {k}{\pi}t + C)^{10/9} \approx 1.4495 (\frac {k}{\pi}t + C)^{10/9}$

Now we can go back and determine $ C$:  Let $ t = 0$.  $ h(0) = 3^5$ and plugging this in we find that $ C = 100.4498$.

Now all we need is $ k$.  To find this we can go back to find $ V(t)$ from the formula based on $ h$ we found at the beginning:

$ V(t) = \frac {5\pi}{7} \bigg(1.4495 (\frac {k}{\pi}t + 100.4498)^{10/9}\bigg)^{7/5} = \frac {5\pi}{7} \cdot 1.6815 \bigg(\frac {k}{\pi}t + 100.4498\bigg)^{14/9}$
$ V(0) = 5\pi/7 \cdot 3^7$
$ V(1) = 3/4 V(0) = 5\pi/28 \cdot 3^8$
but from our formula $ V(1) = \frac {5\pi}{7} \cdot 1.6815 \bigg(\frac {k}{\pi} + 100.4498\bigg)^{14/9}$
Setting these two equal and solving, we find $ k = -53.2792$.

So our solution is $ V(t) = 3.773 \bigg(-16.959t + 100.4498\bigg)^{14/9}$

As a side note, I have no clue how to solve the differential equation without mathematica.

jeez now I feel stupid... I just figured out how to do that differential equation.

$ \frac{k}{\pi} h^{1/10} \equal{} \frac{dh}{dt}$
$ \frac{k}{\pi} \equal{} \frac{dh}{dt}h^{\minus{}1/10}$
$ \frac{k}{\pi}t \plus{} C \equal{} \frac{10}{9}h^{9/10}$
$ \bigg(\frac{9k}{10\pi}t \plus{} 9/10C\bigg)^{10/9} \equal{} h$


Prove that the measure of the segment passing through the point of intersection of the diagonals of a trapezoid and parallel to the bases with its endpoints on the legs, is the harmonic mean between the measures of the parallel sides.

[hide="hint, I think"]similar triangles[/hide]


hello,

Prove that :   $ (\forall (a,b)\in \mathbb{Z}^{2}):    3 | ab(a^{2}\minus{}b^{2})$

Note that $ ab(a^2 \minus{} b^2) \equal{} ab(a \plus{} b)(a \minus{} b)$. If any of $ a$ and $ b$ are multiples of 3, then we are trivially done.

If $ a$ and $ b$ are congruent to the same number mod 3, then $ a \minus{} b$ is congruent to 0 mod 3; again, we are done.

In the last case, the two are nonzero and congruent to different numbers mod 3; they must be 1 and 2. Thus, $ a \plus{} b$ is congruent to 0 mod 3, finishing the proof.


logarithms
limit
integration
real analysis
real analysis unsolved

for every n>0 natural number we consider the equation 1/(1+x)+1/(2+x)+...+1/(n+x)=ln2.
prove that
a) the equation has a unique solition(pozitive); and we denote it with x_n
b) the sequence (x_n)_n is convergent and fin its limit.

$ f(x) \equal{} \frac {1}{x \plus{} 1} \plus{} ... \plus{} \frac {1}{x \plus{} n} \minus{} \ln{2}$
1)$ f'(x) < 0$ and $ f(n) < 0,f(0) > 0$
so $ f(x)$ has unique solution on $ R^{ \plus{} }$
2)$ \lim_{n\to\infty}\frac{x_n}{n} \equal{} 1$

[quote="TTsphn"]
$ \lim_{n\to\infty}x_n \equal{} 1$[/quote]

check your proof, please

Think Riemann sums .
$ \sum_{k\equal{}1}^n\frac 1{n\plus{}k} < \int_0^1\frac {dt}{1\plus{}t} < \sum_{k\equal{}0}^{n\minus{}1}\frac 1{n\plus{}k}\equal{}\sum_{k\equal{}1}^n\frac 1{n\minus{}1\plus{}k}$ , so $ n\minus{}1 < x_n < n$ and finally $ \frac{x_n}n \rightarrow 1$
 :cool:


function
calculus
derivative
real analysis
real analysis theorems

could you please tell me why

 \[ \sum_{n\equal{}\minus{}\infty}^{\plus{}\infty}\exp(\minus{}\pi n^2x)\in C^{\infty}((0,\infty))\]

??

thanks :)

Perhaps I am not looking at this the right way or I am misunderstanding the question, but $ C^{\infty}$ implies differentiable of all orders and obviously, the exponential function is differentiable infinitely many times.

Theorem: If $ \sum f_n$ converges uniformly to $ f$ and $ \sum f_n'$ converges uniformly, $ f$ is differentiable and $ \sum f_n'$ converges to $ f'$. Repeat for each derivative to get that if $ \sum f_n^{(k)}$ converges uniformly for each $ k$, $ f\equal{}\sum f_n$ is $ C^{\infty}$ and its derivatives are the $ \sum f_n^{(k)}$.
This principle is what lets us differentiate power series term by term.

Now, show that $ \sum \frac{d^k}{dx^k}\exp(\minus{}\pi n^2 x)$ converges uniformly on $ [c,\infty)$ for each $ c>0$ and each $ k$.

it will be $ \sum\frac{n^{2k}}{\exp(\pi n^2 x)}$

if $ c>0$, for $ x\ge c$ we will have estimation from above by $ \sum\frac{n^{2k}}{\exp(\pi n^2 c)}$ - convergent, and Weierstrass works fine :)

thanks very much !


algebra
polynomial
function
algebra unsolved

Let $P(x)$ be a polynomial of degree $n$ with real zeros $a_{1},\cdots,a_{n}$, and let $P^{\prime}(x)$ have zeros $b_{1},\cdots,b_{n-1}$. Let $f$ be a convex function defined on the real line. Prove that
\[\frac{f(a_{1})+\cdots+f(a_{n})}{n}\geq \frac{f(b_{1})+\cdots+f(b_{n-1})}{n-1}. \]


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $E=F(\alpha)$ where $\alpha$ is algebraic over F of degree m, and F contains all roots of the polynomial $t^n-1$ for some $n \in \mathbb{N}$. Show that $[E:F(\alpha ^n)]$ divides $n$.

I believe I can prove this in the case when $x^n-1$ has distinct roots in $F$ (or, in other words, $F$ either has characteristic $0$ or, if it has characteristic $p$, $(n,p)=1$). In this case, the extension $K(\alpha):K$ is Galois, where $K=F(\alpha^n)$. The degree of $K(\alpha):K$ is equal to the order of the Galois group $G=G(K(\alpha),K)$.

An element of $G$ is determined by the image of $\alpha$, and takes $\alpha$ to $\alpha z^k$ for some $k\in\overline{0,n-1}$, where $z$ is a primitive root of $x^n-1$. To each element of $G$ we can associate the $k$ defined above, and this defines an isomorphism between $G$ and some subgroup of $(\mathbb Z_n,+)$, which means that $|G|\ |\ n$, as desired.

I'm rather new to this topic, so sorry for any mistakes.

[quote="you could have chosen a better name"]Let $E=F(\alpha)$ where $\alpha$ is algebraic over F of degree m, and F contains all roots of the polynomial $t^{n}-1$ for some $n \in \mathbb{N}$. Show that $[E: F(\alpha^{n})]$ divides $n$.[/quote]

Is the following wrong?

We know that $F$ contains all roots of the polynomial $t^{n}-1$; in other words, $F$ contains all $n$-th roots of unity. Thus, any extension of $F$ which contains one $n$-th root of $\alpha^{n}$ must contain all $n$-th roots of $\alpha^{n}$ (because from one such root we can obtain all such roots by multiplication with $n$-th roots of unity, which lie in $F$ and thus in our extension of $F$ as well). In particular, since the field extension $F(\alpha)$ of $F$ contains one $n$-th root of $\alpha^{n}$ - namely $\alpha$ -, it must contain all $n$-th roots of $\alpha^{n}$, and thus the polynomial $t^{n}-\alpha^{n}$ splits into linear factors over $F(\alpha)$.

Now consider the polynomial $t^{n}-\alpha^{n}$ over the (possibly) smaller field $F(\alpha^{n})$. Over this field, this polynomial splits into irreducible factors: $t^{n}-\alpha^{n}=p_{1}p_{2}...p_{k}$, where $p_{1}$, $p_{2}$, ..., $p_{k}$ are irreducible polynomials over $F(\alpha^{n})$. Since $t^{n}-\alpha^{n}$ splits into linear factors over $F(\alpha)$, its divisor polynomials $p_{1}$, $p_{2}$, ..., $p_{k}$ also split into linear factors over $F(\alpha)$, and hence have zeroes in $F(\alpha)$. Let $i\in\{1,2,...,k\}$ be an integer, and let $s$ be some zero of $p_{i}$ in $F(\alpha)$. Since $t^{n}-\alpha^{n}=p_{1}p_{2}...p_{k}$, this yields that $s$ is a zero of $t^{n}-\alpha^{n}$, so that $s$ is an $n$-th root of $\alpha^{n}$. Thus, the field $F(\alpha^{n},s)$, being a field extension of $F$ that contains one $n$-th root of $\alpha^{n}$ (namely $s$), must contain all $n$-th roots of $\alpha^{n}$, and particularly it must contain $\alpha$ (since $\alpha$ is an $n$-th root of $\alpha^{n}$), and since it is a subfield of $F(\alpha)$, it thus satisfies $F(\alpha^{n},s)=F(\alpha)$. On the other hand, $F(\alpha^{n},s)$ is the extension of the field $F(\alpha^{n})$ with a root of the irreducible over $F(\alpha^{n})$ polynomial $p_{i}$ (this root is $s$), and thus its degree is $[F(\alpha^{n},s): F(\alpha^{n})]=\deg p_{i}$. Hence, $[F(\alpha): F(\alpha^{n})]=\deg p_{i}$. Note that we have proven this for every $i\in\{1,2,...,k\}$. Thus,

$n=\deg(t^{n}-\alpha^{n})=\deg(p_{1}p_{2}...p_{k})=\sum_{i=1}^{k}\underbrace{\deg p_{i}}_{=[F(\alpha): F(\alpha^{n})]}$
$=\sum_{i=1}^{k}[F(\alpha): F(\alpha^{n})]=k[F(\alpha): F(\alpha^{n})]$,

and thus $[F(\alpha): F(\alpha^{n})]\mid n$. And since $F(\alpha)=E$, this solves the problem.

Hope that I was not tooo boring and maybe even correct.

  darij


find all the linear polynomials $ f(x)$ so that
$ (fof)(\displaystyle\frac{1}{x})\equal{} \displaystyle\frac{4\minus{}x}{x}, x\neq0$
and give the sum of all independent terms of the obtained functions


Iranian summer olympiad program ended and Here are the questions of final exams.

1-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16862
2-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16682
3-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16815
4-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16776
5-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16597
6-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16445
7-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16864
8-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16777
9-   http://www.mathlinks.ro/Forum/viewtopic.php?t=16567
10- http://www.mathlinks.ro/Forum/viewtopic.php?t=16683
11- http://www.mathlinks.ro/Forum/viewtopic.php?t=16566
12- http://www.mathlinks.ro/Forum/viewtopic.php?t=16865
13- http://www.mathlinks.ro/Forum/viewtopic.php?t=16684
14- http://www.mathlinks.ro/Forum/viewtopic.php?t=16599
15- http://www.mathlinks.ro/Forum/viewtopic.php?t=17354

Problems 3 and 12 were the most difficult problems.

I thought it will be nice to send all questions in PDF format


[quote="nguyenvuthanhha"][i][u]  Problem :[/u]

If $ f$ is a polynomial with integer coefficients, and $ f(n)$ is a perfect

square for every natural number $ n$, then there is a polynomial $ g$ with integer

coefficients such that $ f \ \equal{} \ g^{2}$[/i]

[i]
   Drij , Nayel , Help me Please :lol: [/i][/quote]

Nice result!