AMC
AIME
analytic geometry
geometry
trigonometry
modular arithmetic
algebra

This is a continuation of Poincare's marathon here (it is locked) [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1624540#1624540[/url]
and I think this time we'd better continue that thread in the Intermediate topics forum here... so the problems would range from difficult AMC to hard AIME (Pre-Olympiad level also acceptable) problems.
(As a suggestion, if a problem is unsolved for days, perhaps a hint would help in continuing the marathon)
Here is where we left off:

[hide="problem 265"]
Five pirates and a monkey are shipwrecked on an island. The pirates have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the others, one pirate wakes up during the night and divides the coconuts into five equal parts with one left over, which he gives to the monkey. The pirate then hides his portion of the pile. During the night, each of the other pirates does exactly the same thing by dividing the pile he finds into five equal parts leaving one coconut for the monkey and hiding his portion. In the morning, the pirates gather and split the remaining pile of coconuts into five equal parts and again one is left over for the monkey. What is the smallest number of coconuts the pirates could have collected for their original pile?[/hide]

[hide="solution"]Let the answer be $ N \equal{} k_0$ ($ k_i\in\mathbb{N}$).

$ k_0 \equal{} 1 \plus{} 5k_1$
$ 1 \plus{} 4k_1 \equal{} 1 \plus{} 5k_2$
$ 1 \plus{} 4k_2 \equal{} 1 \plus{} 5k_3$
$ ...$
$ 1 \plus{} 4k_5 \equal{} 1 \plus{} 5k_6$

from here $ k_5 \equal{} 5k_6/4$, $ k_4 \equal{} 5k_5/4 \equal{} 5^2k_6/4^2$ and so on to get $ k_1 \equal{} 5^5k_6/4^5\implies N \equal{} k_0 \equal{} 1 \plus{} 5^6k_6/4^5$ and the minimum occurs when $ k_6 \equal{} 4^5\implies N \equal{} 1 \plus{} 5^6$ [/hide]

[hide="Problem 266"]

Let $ f(x) \equal{} x^2 \plus{} 6x \plus{} 1$, and let $ R$ denote the set of points $ (x,y)$ in the coordinate plane such that
$ f(x) \plus{} f(y) < 0$ and $ f(x) \minus{} f(y) < 0$,

The area of $ R$ is closest to

            (a)   21         (b)  22         (c)   23        (d)   24         (e)   25   [/hide]

I may be interpreting this wrong, but in #226, $ f(x)-f(y)$ cannot be less than 0.
I will assume you mean $ f(x)-f(y)\le0.$

[hide="Solution #226"]
Complete the square in \[ f(x)+f(y)=(x^2+6x+1)+(y^2+6x+1)<0\] to get \[ (x+3)^2+(y+3)^2<16.\] The graph of this is the interior of a circle with center $ (-3,-3)$ and radius $ 4.$
Completing the square in \[ f(x)-f(y)=(x^2+6x+1)-(y^2+6y+1)\le0,\] we get \[ (x+3)^2+(y+3)^2\le0.\] This is a degenerate hyperbola. In other words, it's just two straight lines that intersect at $ (-3,-3),$ splitting the circle into $ 4$ congruent pieces, and $ 2$ of them are "successful" areas. Since it has radius $ 4,$ half of its area is \begin{align*}\frac{4^2\pi}{2}&=\frac{16\pi}{2}\\&=8\pi\\&\approx\boxed{25}.\end{align*}[/hide]

[hide="Problem #267"]
Given that $ \sum_{k=1}^{35}\sin5k=\tan\frac{m}{n},$ where angles are measured in degrees, and $ m$ and $ n$ are relatively prime positive integers that satisfy $ \frac{m}{n}<90.$ Find $ m+n.$[/hide]

[hide="Solution to 267"]Using this $ \frac{\sin(\alpha\plus{}\frac{(n\minus{}1)\beta}{2})\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}$.The expression reduces to $ \frac{\sin \frac{175}{2}}{\sin \frac{5}{2}}$.
Which equals $ \cot\frac{5}{2}$.From here we get $ m\plus{}n\equal{}\boxed{177}$[/hide]

[b]Problem 268[/b]
Express $ (x^{2}\plus{}a^{2})(x^{2}\plus{}b^{2})(x^{2}\plus{}c^{2})$ as the sum of two squares.

[hide="solution"]C'mon, nobody? Actually this problem is straightforward if one notices the identity $ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \plus{} bd)^2 \plus{} (ad \minus{} bc)^2$ So $ (x^2 \plus{} a^2)(x^2 \plus{} b^2) \equal{} (x^2 \plus{} ab)^2 \plus{} (ax \minus{} bx)^2$, which is again in the form $ a_1^2 \plus{} b_1^2$, and then applying the identity again to get $ \left(x(x^2 \plus{} ab) \plus{} c(ax \minus{} bx)\right)^2 \plus{} \left(c(x^2 \plus{} ab) \minus{} x(ax \minus{} bx)\right)^2$ There are other ways to write the sum of squares as well.[/hide]
[b]Problem 269[/b] Evaluate $ \sum_{r\equal{}1}^{25}(x^{r}\plus{}\frac{1}{x^{r}})^{2}$ where $ x$ satisfies $ x^{2}\plus{}x\plus{}1\equal{}0$

[hide="Solution to 269"]
The roots of the equation are $ \omega$ and $ \omega^{2}$
So $ \left(\omega\plus{}\frac{1}{\omega}\right)^{2}\equal{}1$
$ \left(\omega^{2}\plus{}\frac{1}{\omega^{2}}\right)^{2}\equal{}1$
$ \left(\omega^{3}\plus{}\frac{1}{\omega^{3}}\right)^{2}\equal{}4$

 so continuing like this we get $ 49$ as the answer..
Dont forget to use $ \omega^{3}\equal{}1$ and $ 1\plus{}\omega\plus{}\omega^{2}\equal{}0$
[/hide]

[b]Problem 270[/b]

Evaluate $ (1\plus{}i)^{40}\plus{}(1\minus{}i)^{40}$ where $ i^{2}\equal{}\minus{}1$

[hide="Solution to 270"]
$ 1\plus{}i\equal{}\sqrt 2\,\text{cis}\,45^\circ$ and $ 1\minus{}i\equal{}\sqrt 2\,\text{cis}\,(\minus{}45^\circ)$, and since $ 40\cdot \pm 45\equal{}\pm 1800\equiv 0\pmod{360}$, it follows that $ (1\plus{}i)^{40}\plus{}(1\minus{}i)^{40}\equal{}2^{20}\,\text{cis}\,0^\circ\plus{}2^{20}\,\text{cis}\,0^\circ\equal{}\boxed{2^{21}}$.[/hide]

[b]Problem 271[/b]

A certain polynomial leaves a remainder of 7 when divided by $ x\minus{}3$ and a remainder of -6 when divided by $ x\plus{}3$. Calculate the remainder when the polynomial is divided by $ x^2\minus{}9$.

[hide]$ P(x) \equal{} (x\minus{}3)(x\plus{}3)Q(x) \plus{} ax \plus{} b$. $ P(3) \equal{} 7$, $ P(\minus{}3) \equal{} \minus{}6$ implies $ 7 \equal{} 3a \plus{} b$ and $ \minus{}6 \equal{} \minus{}3a \plus{} b$. Adding, we get $ b \equal{} \frac{1}{2}$ and $ a \equal{} \frac{13}{6}$. Thus, we leave a remainder of $ \frac{13}{6}x \plus{} \frac{1}{2}$. [/hide]

[b]Problem 272[/b] Evaluate $ \sum_{k\equal{}0}^{670} \binom{2010}{3k}$.

[quote="Zhero"][b]Problem 272[/b] Evaluate $ \sum_{k \equal{} 0}^{670} \binom{2010}{3k}$.[/quote]

It's not exactly pretty, but it works.

[hide]
Let's look at the more general form of this Evaluate $ \sum_{k \equal{} 0}^{n} \binom{3n}{3k}$

Let's look at the row of Pascal's triangle that is used when you do "6 choose x".

1 6 15 20 15 6 1

The numbers we are going to add are 1, 20, and 1. Let's move up a row and see what numbers added to create those.

(1)1, (0)5, 1(10), 1(10), 0(5), (1)1

Move up another row.

1(1), 1(4), 2(6), 1(4), 1(1).         Note that the 6 is the only number used twice. What created that?

0(1), 1(3), 1(3), 0(1)
1(1), 2(2), 1(1)                       Note that the 2 is used twice.

1(1), 1(1)                                We now have three complete rows that made up the answer. Using the property that the nth row of pascals triangle is equal to $ 2^{n \minus{} 1}$ We get $ 2^4 \plus{} 2^2 \plus{} 2^1$

If you start with n=12, you get $ 2^8 \plus{} 2^6 \plus{} 2^4 \plus{} 2^2 \plus{} 2^1$ and in general you get ($ 2^{n \minus{} 2} \plus{} 2^{n \minus{} 4}... \plus{} 2^4 \plus{} 2^2) \plus{} 2^1$ (This only works if n is a multiple of 6)

From here you can sum a geometric sequence.

$ \frac {(2^{3n \minus{} 2})(1 \minus{} (\frac {1}{4})^\frac {3n \minus{} 2}{2})}{1 \minus{} \frac {1}{4}} \plus{} 2$

Which reduces to

$ \frac {2^{3n} \plus{} 2}{3}$

So the answer is $ \frac {2^{2010} \plus{} 2}{3}$[/hide]

I don't have a problem, so anyone who wants to can post one.

[b]Problem 273.
In the sequence $ \frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$.
The denominator of the $ {nth}$ term $ {n>1}$ is the sum of the numerator and the denominator of the $ {(n-1){}^t{}^h}$ term.
The numerator of the $ {n{}^t{}^h}$ term is the sum of the denominators of the $ {n{}^t{}^h}$ and $ {(n-1){}^t{}^h}$ term.
Find the LIMIT of the sequence.

[quote="zserf"][hide]
Let's look at the more general form of this Evaluate $ \sum_{k \equal{} 0}^{n} \binom{3n}{3k}$

Let's look at the row of Pascal's triangle that is used when you do "6 choose x".

1 6 15 20 15 6 1

The numbers we are going to add are 1, 20, and 1. Let's move up a row and see what numbers added to create those.

(1)1, (0)5, 1(10), 1(10), 0(5), (1)1

Move up another row.

1(1), 1(4), 2(6), 1(4), 1(1).         Note that the 6 is the only number used twice. What created that?

0(1), 1(3), 1(3), 0(1)
1(1), 2(2), 1(1)                       Note that the 2 is used twice.

1(1), 1(1)                                We now have three complete rows that made up the answer. Using the property that the nth row of pascals triangle is equal to $ 2^{n \minus{} 1}$ We get $ 2^4 \plus{} 2^2 \plus{} 2^1$

If you start with n=12, you get $ 2^8 \plus{} 2^6 \plus{} 2^4 \plus{} 2^2 \plus{} 2^1$ and in general you get ($ 2^{n \minus{} 2} \plus{} 2^{n \minus{} 4}... \plus{} 2^4 \plus{} 2^2) \plus{} 2^1$ (This only works if n is a multiple of 6)

From here you can sum a geometric sequence.

$ \frac {(2^{n \minus{} 2})(1 \minus{} (\frac {1}{4})^\frac {n \minus{} 2}{2})}{1 \minus{} \frac {1}{4}} \plus{} 2$

Which reduces to

$ \frac {2^{n} \plus{} 2}{3}$

So the answer is $ \frac {2^{2010} \plus{} 2}{3}$[/hide][/quote]
I think you meant $ n\equal{}670$ instead for this specific problem. By the way can you prove the general formula for $ n$? It seems quite reasonable from the example but I'm wondering how to actually prove the formula formally in an olympiad...and [b]Zhero[/b], where did you get this interesting problem?

[quote="mathwizarddude"]
I think you meant $ n \equal{} 670$ instead for this specific problem. By the way can you prove the general formula for $ n$? It seems quite reasonable from the example but I'm wondering how to actually prove the formula formally in an olympiad...and [b]Zhero[/b], where did you get this interesting problem?[/quote]

Method for proving.

[hide]I think that using the equation $ \binom{n}{r}\equal{}\binom{n\minus{}1}{r\minus{}1}\plus{}\binom{n\minus{}1}{r}$ would make it simpler to prove it.

In our case, we have

$ \binom{2010}{0}\plus{}\binom{2010}{3}\plus{}\binom{2010}{6}\plus{}...$

$ \binom{2009}{0}\plus{}\binom{2009}{2}\plus{}\binom{2009}{3}\plus{}\binom{2009}{5}\plus{}\binom{2009}{6}\plus{}...$

$ \binom{2008}{0}\plus{}\binom{2008}{1}\plus{}2\binom{2008}{2}\plus{}\binom{2008}{3}\plus{}\binom{2008}{4}\plus{}2\binom{2008}{5}\plus{}\binom{2008}{6}\plus{}...$

using the fact that the sum of all binomials in the form $ \binom{n}{r}$ is equal to $ 2^n$, you can change this to

$ 2^{n\minus{}2}\plus{}\binom{2008}{2}\plus{}\binom{2008}{5}\plus{}....$ and continue the proof.[/hide]

[hide="problem 273"]
let $ n$ and $ d$ denote the numerator and denominator
$ d_i \equal{} n_{i \minus{} 1} \plus{} d_{i \minus{} 1} \qquad(1)$
$ n_i \equal{} d_i \plus{} d_{i \minus{} 1} \qquad(2)$

$ (1) \plus{} (2): n_i \equal{} n_{i \minus{} 1} \plus{} 2d_{i \minus{} 1}$

$ \therefore \frac {n_i}{d_i} \equal{} 1 \plus{} \frac {d_{i \minus{} 1}}{n_{i \minus{} 1} \plus{} d_{i \minus{} 1}} \equal{} 1 \plus{} \frac {1}{\frac {n_{i \minus{} 1}}{d_{i \minus{} 1}} \plus{} 1}$

So let $ a_i \equal{} \frac {n_i}{d_i}$

$ |a_{i \plus{} 1} \minus{} a_i| \equal{} \frac {|a_i \minus{} a_{i \minus{} 1}|}{(a_i \plus{} 1)(a_{i \minus{} 1} \plus{} 1)} < \frac {|a_i \minus{} a_{i \minus{} 1}|}{3}$

Where the inequality comes from AM-GM
$ (a_i \plus{} 1)(a_{i \minus{} 1} \plus{} 1) \equal{} a_ia_{i \minus{} 1} \plus{} a_i \plus{} a_{i \minus{} 1} \plus{} 1 > \frac {1}{a_{i \minus{} 1} \plus{} 1} \plus{} 1 \plus{} a_{i \minus{} 1} \plus{} 1 \ge 3$

So the sequence is contractive, therefore it converges. Let the limit be $ a$
$ a \equal{} 1 \plus{} \frac {1}{1 \plus{} a} \longrightarrow a \equal{} \boxed{\sqrt {2}}$[/hide]

[b]Problem 274[/b]
Each of 30 students in a class has the same number of friends among their class mates. What is the highest possible numbers of students that can learn better than the majority of their friends?
(given that you can distinguish between everyone's level of learning)

[hide="What I had in mind for my problem"]I'll prove this for general $ 3n$. Let $ P(x) = (1 + x)^{3n} = \sum_{k=0}^{3n} \binom{3n}{k} x^k$. Basically, we do a roots of unity filter on this. Let $ \omega$ be a primitive third root of unity. Then for any $ k$ with $ (3,k) = 1$, we have $ 1^k + \omega^k + \omega^{2k} = 0$. Thus, $ \frac{P(1) + P(\omega) + P(\omega^2)}{3}$ is our desired sum (since all $ k$ not multiples of 3 vanish.) $ 1 + \omega = \cis(\frac{\pi}{3})$, and $ 1 + \omega^2 = \cis(-\frac{\pi}{3})$. It's easy from here.

For our specific example, a matter of simple computation tells us that our answer is just $ \frac{2^{2010} + 2}{3}$[/hide]

By the way, I made it up myself, but I've seen problems similar to this one before (like one on one of Mildorf's mock AIMEs, though I do not recall solving it on the mock AIME.)

[quote="ocha"][b]Problem 274[/b]
Each of 30 students in a class has the same number of friends among their class mates. What is the highest possible numbers of students that can learn better than the majority of their friends?
(given that you can distinguish between everyone's level of learning)[/quote]
Could you elaborate on "majority"?


geometry
parallelogram
rectangle
analytic geometry
vector
absolute value

Given the points A(4,2), B(0,5), C(-6,2), and D(-2,-1): Find the area of the parallelogram.

AB and CD = 5
BC and AD= (3)sqrt(5)
Diagonal AC= 10
Diagonal BD= (2)sqrt(10)

I drew an altitude but I dont have another angle so I dont know how to find the height...

[hide="clicky"]
i'm not sure  :oops:
isn't the area the product of the diagonals then divided by two??
sooo you would get:
(10)sqrt(10)

according to my method....
please tell me if this is wrong, because this seems like a really simple solution..[/hide]

[hide="Hints"]What can we say about the position of AC relative to the axes? Can we find the length of AC and the length of altitude BM of triangle ABC? What will be the altitude of triangle ADC?[/hide]

[b]@kimmy:[/b] Your method doesn't work  :stink: It works only for convex quadrilaterals with perpendicular diagonals - namely, squares, rhombuses, and kites.

What Farenhajt posted is probably the easiest way of doing this:
[hide="Fast"]Draw a rectangle around the parallelogram with sides parallel to the axes, such that there is one vertex of the parallelogram on each side of the rectangle. Draw a line from A to C. Now draw perpendicular segments from B to AC and from D to AC. What do you see? In particular, [b]how is the area of the parallelogram related to the area of this rectangle?[/b][/hide]

The following shows an application of matrices to this problem, and it works for any problem in general.

Let's pull one of the vertexes into the origin. To do so, we add 2 to each x coordinate and 1 to each y coordinate:

A(6,3), B(2,6), C(-4,3), D(0,0)

Note that A+C = B. So, we can say that this parallelogram is spanned by the vectors $ (6,3)$ and $ ( \minus{} 4,3)$.

Our answer is the absolute value of $ \left| \begin{array}{cc} 6 & 3 \\
\minus{} 4 & 3 \end{array} \right|$, which equals $ | (6)(3) \minus{} ( 3)( \minus{} 4) | \equal{} |30| \equal{} 30$

One of the random "Proofs without words" shown in the left navigation column nicely illustrates why this determinant works.

[hide="Generalization to parallelepipeds"]This in fact generalizes to higher dimensions very cleanly - suppose I tell you A=(1,1,2), B=(1,2,1), C=(2,1,1), and give you a parallelepiped with vertexes O, A, B, C, A+B, B+C, A+C, and A+B+C. This parallelepiped is spanned by vectors A, B, and C, so it's volume is given by the absolute value of $ \left| \begin{array}{ccc} 1 & 1 & 2 \\
1 & 2 & 1 \\
2 & 1 & 1 \end{array} \right|$, or $ | \minus{} 4 | \equal{} 4$[/hide]


The quadratic $ 3x^2\plus{}4x\minus{}9$ has two real roots. What is the sum of the squares of these roots? Express your answer as a common fraction in lowest terms.

Notice that $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$, so all we have to do is find the sum and product of the roots. By Vieta's, the sum is $ \minus{}\frac{4}{3}$ and the product $ \frac{\minus{}9}{3} \equal{} \minus{}3$. Thus, we have $ \frac{16}{9} \minus{} 2(\minus{}3) \equal{} \frac{70}{9}$.


integration
calculus
absolute value
calculus computations

Let $ c\in(a,b)$ and let $ k$ be defined on $ [a,b]$ by $ k(c)\equal{}1$ and $ k(x)\equal{}0$ for $ x\in[a,b],x\neq c$. If $ f: [a,b]\rightarrow \mathbb{R}$ is continuous at $ c$ show directly that $ f$ is $ k$-integrable , that $ k$ is $ f$-integrable and that

                   $ \int^{b}_{a} f dk \equal{} \int^{b}_{a} k df \equal{} 0$

Some background on the [url=http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral]Riemann-Stieltjes integral[/url].
Consider a Riemann-Stieltjes sum for $ \int_a^b f\,dk$ with partition points $ x_i$ and sample points $ x_i^*\in [x_{i\minus{}1,x_i}$. If none of $ x_i$ in the sum are equal to $ c$, the sum is zero. If $ x_j\equal{}c$, then the sum is $ f(x_j^*)\minus{}f(x_{j\plus{}1}^*)$. As the mesh of partition approaches zero, we have $ x_{j}^*\to c$ and $ x_{j\plus{}1}^*\to c$, hence $ f(x_j^*)\minus{}f(x_{j\plus{}1}^*)\to f(c)\minus{}f(c)\to 0$. You may want to use $ \epsilon$ and $ \delta$ to make this more precise.

The second integral is treated similarly.

just to be clear,

by def of Riemann Stieltjes integral of with respect to g over [a,b]

  $ \int^{b}_{a} f dk \equal{}0 \rightarrow \forall \epsilon > 0, \exists [x_{i\minus{}1}, x_{i}]$-partition of $ [a,b]$,
then $ | \sum_{i\equal{}1}^{n} f(x_{i}^{*})[k(x_{i}\minus{}k(x_{i\minus{}1})]\minus{}0|< \epsilon$.

Can I say that where $ x_{i}^{*} \neq c$ since $ k(x) \equal{} 0$, then $ [k(x_{i})\minus{}k(x_{i\minus{}1})] \rightarrow 0$ thus showing the stuff inside the absolute value is $ < \epsilon$?


Prove that for $x$ and $y$ integers \[17\ |\ 2x+3y \iff 17\ |\ 9x+5y\]


I took alternate AIME at Apr.1st and I think I passed... :D
How far can you go?, Can you be USAMO winner and go to MOSP??

Yes you can be a winner, but you can't go to MOP or be on the US team.


number theory
greatest common divisor
function
algebra unsolved
algebra

There is none, but using $pgcd$ is like using $rsin(x)$ for $\sin(x)$.

To the probelm: as I see this, we can choose the value at any prime power $\neq 2$ arbitrary and in contrary, this gives all solutions.
But this can't be the true problem...

Let
(1) ${q_{p,n}=f(p^{n}})$, were p prime, n - natural. If these numbers defined we know f(x) by ii).
Therefore all numbers $q_{p,n}, q(2,1)=2$ defined one function f satisfyed i),ii) and (1).

[quote="Rust"]Let
(1) ${ q_{p,n} = f(p^{n}})$, were p prime, n - natural. If these numbers defined we know f(x) by ii).
Therefore all numbers $ q_{p,n}, q(2,1) = 2$ defined one function f satisfyed i),ii) and (1).[/quote]
yes ,we choose $ g: N^*\to N^*$ arbitrary such that $ g(2)=2,g(1)=1$
$ \forall n\in\mathbb{N}^*\ such\ that\ n = 2^a\prod_{i = 1}^{m}p_i^{a_i}$ where $ (p_i)$ are primes $ \neq 2$,$ (a_i) > 0$ and $ a\ge 0$
$ f(n) = g(2^a)\prod_{i = 1}^{m}g(p_i^{a_i})$


trigonometry
inequalities
calculus
derivative
inequalities solved

prove this seemingly simple inequality(i cant find the trick to do this one)
 (sinx)/x >  \sqrt cosx
and x belongs to the open interval (0, \pi/2)
thanx

The inequality can also be proved using the following lemas:

lema 1: sin(x)>x-x^3/4 ,  0<x<pi/2
lema 2: cos(x)<1-x^2/2+x^4/24, 0<x<pi/2
use derivates to prove them.. or not...
for the first lema I found a prove without derivates:

sin(x)=2*tan(x/2)*cos^2 (x/2)=2*tan(x/2)*(1-sin^2 (x/2)) > 2*x/2*(1-(x/2)^2)=x-x^3/4
I used the fact that for 0<x<pi/2 we have
0<sin(x/2)<x/2<tan(x/2)

so the inequality is equivalent to :
(sin(x)/x)^2>cos(x) ; we denote f(x)= sin^2 (x)/x^2-cos(x) and we want to show that f(x)>0 for 0<x<pi/2  and using the lemas you find that (if my calculus was correct )  we get that f(x)>x^4/48>0 so the inequality was proved...

hope that it is correct!  :D

Louis.

I hope you can find a prove without derivates for this lema or ...a solution with derivates... I would be glad to see a full prove .please.:D

lema 2: cos(x)<1-x^2/2+x^4/24,
 0< x< pi/2.

Louis.

Looks to me as a taylor series :D ... but then again that's advanced calculus too ... I sincerely doubt that anyone can find a pure elementary proof of that fact ... it's mostly a waste of time trying :D

hey
thanx for the proof valentin. i also found out a proof(using the series approx of the sin x and cos x dint occur to me first).so as for the proof of the  inequality involving cos x,i think it's just the matter of seeing the series of cos x
which is well known and can be easily proved by writing
e^(ix) in two ways.one as cos x + isinx and the other as
1 + (ix)/1! + (ix)^2/2! + (ix)^3/3! .... and comparing the real and imaginary parts which yields the desired inequality (i guess)
regards

There is also an elementary solution to this one:

the ineq is equivalent to sqrt(sin(x)*tan(x))>x.

We also have that sqrt(sin(x)*tan(x))>=(2sinx*tan(x))/(sinx+tanx) and it's easy to prove that (2sinx*tan(x))/(sinx+tanx)>x which is equivalent to x<2sinx/(1+cosx), which gives that tan(x/2)>x/2 which is always true!

problem solved!!! :D :D


calculus
integration
function
trigonometry
algebra
rational function
calculus computations

$ \int_0^\infty\;\dfrac{\cos\,(a\,x)\cdot x^{b\minus{}1}}{1\plus{}x^2}\;\textbf dx$

where $ a>0$, and $ 0<\mathbb Re\,(b) <3$

$ \int_{0}^{\infty}{x^{\alpha \minus{} 1}f(x)dx} \equal{} \minus{} \frac {\pi}{\sin \pi\alpha}e^{ \minus{} \pi i \alpha}\sum_{k \equal{} 1}^{q}\mathop\text{Res}_{z \equal{} z_k}\left(z^{\alpha \minus{} 1}f(z)\right)$

Now : $ \mathop\text{I} \equal{} \int_{0}^{\infty}{x^{b \minus{} 1}\cdot \frac {\cos (ax)}{x^2 \plus{} 1}dx} \equal{} \mathop\text{Re}\int_{0}^{\infty}{x^{b \minus{} 1}\cdot \frac {e^{axi}}{x^2 \plus{} 1}dx}$

$ \mathop\text{I} \equal{} \minus{} \mathop\text{Re}\left\{\frac {\pi}{\sin \pi b}e^{ \minus{} \pi i b}\left(z^{b \minus{} 1}\cdot\frac {e^{azi}}{z \plus{} i}\right)_{z \equal{} i}\right\}$
$ \mathop\text{I} \equal{} \minus{} \mathop\text{Re}\left\{\frac {\pi}{\sin \pi b}e^{ \minus{} \pi i b}\left(i^{b \minus{} 1}\cdot\frac {e^{ \minus{} a}}{2i}\right)\right\}$

$ \mathop\text{I} \equal{} \frac {\pi}{\sin \pi b}\cdot \frac {e^{ \minus{} a}}{2}\mathop\text{Re}\left\{i\cdot e^{ \minus{} \pi i b} i^{b \minus{} 1}\right\} \equal{} \frac {\pi}{\sin \pi b}\cdot \frac {e^{ \minus{} a}}{2}\mathop\text{Re}\left\{e^{ \minus{} \pi i b} i^{b}\right\}$

$ \mathop\text{I} \equal{} \frac {\pi}{\sin \pi b}\cdot \frac {e^{ \minus{} a}}{2}\mathop\text{Re}\left\{(\cos(\pi b) \minus{} i\sin (\pi b))\cdot (\cos(\frac {\pi}{2} b) \plus{} i\sin (\frac {\pi}{2} b))\right\}$

 :D
[hide="Answer"]
In your problem should be : $ 0<\mathop\text{Re}(b)<2$ and

$ \int_{0}^{\infty}{x^{b \minus{} 1}\cdot \frac {\cos (ax)}{x^2 \plus{} 1}dx} \equal{} \frac {\pi}{4\sin\left(\frac {\pi b}{2}\right)}\cdot e^{ \minus{} a}$[/hide]

The formula $ \int_0^\infty x^{\alpha \minus{} 1} f(x)\, dx \equal{} \minus{} \frac {\pi\, e^{ \minus{} i\alpha\pi}}{\sin \pi \alpha}\sum_{z_k\neq 0} \text{res}(z^{\alpha \minus{} 1}\, f(z),z_k)$

is valid for rational functions $ f(z) \equal{} \frac {P(z)}{Q(z)}$ with $ \deg(Q)\ge \deg(P) \plus{} 2$ ,

whereat $ Q$ has no zeros on the positive real axis and at most a single zero at $ z \equal{} 0$.

In the proof of this formula it is essential that $ f$ tends to $ 0$ for $ |z|\to\infty$ .

But unfortunately $ f(z) \equal{} \frac {e^{iaz}}{1 \plus{} z^2}$ is not a rational function.

Hence the formula isn't useable.

If we choose $ a \equal{} 1$ and $ b \equal{} \frac32$ for example then

$ \int_0^\infty x^{b \minus{} 1} \, \frac {\cos ax}{1 \plus{} x^2}\, dx \equal{} 0.209...$ and $ \frac {\pi}{4\sin\left(\frac {b\pi}{2}\right)} e^{ \minus{} a} \equal{} 0.4086...$

For me it is hard to believe that $ \int_0^\infty x^{b \minus{} 1} \, \frac {\cos ax}{1 \plus{} x^2}\, dx$ can be expressed with common analytic functions.

Do you know such an expression, misan?

If so, please tell it use first and afterwards we can try to prove it.

[quote="Doe John"]
For me it is hard to believe that $ \int_0^\infty x^{b \minus{} 1} \, \frac {\cos ax}{1 \plus{} x^2}\, dx$ can be expressed with common analytic functions.

.[/quote]
it can be calculated exact.
consider :$ \cos az \equal{} \frac{e^{iaz}}{2}\plus{}\frac{e^{\minus{}iaz}}{2}$ Ok.
Lets' try to find such itegral as
1) $ \frac{1}{2}\int_{0}^{\infty} \frac{x^{b\minus{}1}e^{iaz}}{1\plus{}x^{2}}$
and
2) $ \frac{1}{2}\int_{0}^{\infty} \frac{x^{b\minus{}1}e^{\minus{}iaz}}{1\plus{}x^{2}}$
 (also it can be generalized if I take instead $ \frac{1}{x^{2}\plus{}1}$ some rational function $ R(x)$ which have no pole on real axis )
for first and 2-nd integral I take contuor such as on picture.
it's easy to note that for above integral when we integrate on big and small arc then it's tend to zero ( because $ a >0$) after is easy, if you want you can finish yourself and get exact answer.

For $ f(z)\equal{}\frac{e^{i\alpha z}}{1\plus{}z^2}$ we can write $ \text{Re}\int_0^\infty f \,dz\equal{}\frac12\, \text{Re}\int_{\minus{}\infty}^\infty f \, dz$ .

But we can't do this for $ f(z)\equal{}\frac{z^{b\minus{}1}\, e^{i\alpha z}}{1\plus{}z^2}$, because $ z^{b\minus{}1}$ is real on the positive real axis but not on the negative real axis.

I was thinking that after $ \plus{}$ above contour our $ Im\plus{}Im$ will be equal to $ 0$ but it's false.  :(


linear algebra
matrix
function
linear algebra unsolved

lLet $M$ be a matrix in $Mn(R)$. We can define a function $f$ on $Mn(R)$ with $f(M)=\prod_{i=1}^{n}{M_{i,i}}$.
Let $S$ be a symetric positive definite matrix in $Mn(R)$.
consider $E=\{OSO^{-1} | O \in O_n\}$, where $O_n$ is the orthogonal group.
Find $max \{ f(M) | M \in E \}$ and $min \{ f(M) | M \in E \}$

[quote="alekk"] $f(M)=\prod_{i=1}^{n}{M_{i,i}}$.
[/quote]

alekk what is the notation $M_{i,i}$ ?

Moubi, I think it is the element of M that is in the intersection of line and column i.  ;)

[quote="Moubinool"][quote="alekk"] $f(M)=\prod_{i=1}^{n}{M_{i,i}}$.[/quote]alekk what is the notation $M_{i,i}$?[/quote]

I am not alekk, but I think that is kind of semi-standar notation (I have seen it in some books, actually not many times :P). Just note that if the matrix is $M$ sometimes you name the entry in the position $(i,j)$ as $M_{ij}$, but as sometimes either $i$ or $j$ may be greater than 9, and then there may be some confusion if, for example $M_{111}$ means the entry in the position $(1,11)$ or $(11,1)$, then a few books include a comma to separate the entries $M_{i,j}$, then, I suppose alekk means  that $f$ is the product of the entries in the diagonal of the matrix.

Does anyone have a solution for this problem?


counting
Combinatorial Identity
combinatorics
binomial coefficients
bijection
IMO Shortlist

In town $ A,$ there are $ n$ girls and $ n$ boys, and each girl knows each boy. In town $ B,$ there are $ n$ girls $ g_1, g_2, \ldots, g_n$ and $ 2n \minus{} 1$ boys $ b_1, b_2, \ldots, b_{2n\minus{}1}.$ The girl $ g_i,$ $ i \equal{} 1, 2, \ldots, n,$ knows the boys $ b_1, b_2, \ldots, b_{2i\minus{}1},$ and no others. For all $ r \equal{} 1, 2, \ldots, n,$ denote by $ A(r),B(r)$ the number of different ways in which $ r$ girls from town $ A,$ respectively town $ B,$ can dance with $ r$ boys from their own town, forming $ r$ pairs, each girl with a boy she knows. Prove that $ A(r) \equal{} B(r)$ for each $ r \equal{} 1, 2, \ldots, n.$

Let $ A(n,r)$ and $ B(n,r)$ be the number of ways of possible dance pairs from town A and Town B out of total $ n$ girls respectively.

Enumerating $ A(n,r)$ we get,
\[ A(n,r)\equal{}{n\choose r}^2 r!\]

Let us construct a recurrence relation for $ B(n,r)$ for $ 2\le r \le n$,

Case1: When $ g_n$ is dancing.

Let the rest of the $ r\minus{}1$ girls be paired in $ B(n\minus{}1,r\minus{}1)$ ways.The $ g_n$th girl can then be paired with any of the remaining $ 2n\minus{}r$ boys.

Case2: When $ g_n$ is not dancing.

We simply have $ B(n\minus{}1,r)$.

Therefore,
\[ B(n,r)\equal{}(2n\minus{}r)B(n\minus{}1,r\minus{}1) \plus{} B(n\minus{}1,r)\quad \forall \quad 2 \le r \le n\]

and $ B(n,1)\equal{}1\plus{}3\plus{}\dots\plus{}2n\minus{}1\equal{}n^2$.

It may be observed that $ A(n,r)$ and $ B(n,r)$ satisfy the same recurrence relation with same initial values and hence the conclusion.

This problem either says something about matchings in certain graphs or equivalently about rook polynomials for boards of certain shapes.  Anyone have a bijective proof?


number theory
relatively prime
number theory solved

a, b be relatively prime positive integers
Prove that the equation ax+by=c has an obvious solution

x_0 = c.a^(phi(b)-1)

y_0= (-c/b)(a^phi(b) -1)

Bunyakovskii 1831

I think everyone understand (x0,y0) is an obvious solution for ax+by=c because b | aphi(b)-1.


Who's planning on participating in the first round of the AMATYC competition? What school?

http://www.amatyc.org/SML/

Hopefully, I will (Oakland Community College).


geometry
circumcircle
geometric transformation
reflection
trapezoid
geometry unsolved

What do you mean by the "pedal circle" of P?? Is it a circle circumscribed on a pedal triangle of point P?? Or maybe something else...??

Yes, pedal circle is circumcircle of pedal triangle!
If $ D,E,F$ are perpendicular feet from $ P$ to $ BC,CA,AB$ then pedal circle is circumcircle of triangle $ DEF$.

[quote="tdl"]When a point $ P$ moves along a line through the circumcenter of a given triangle $ \Delta$, the pedal circle of $ P$ with respect to $ \Delta$ passes through a fixed point (the Griffiths point) on the nine-point circle of $ \Delta$.[/quote]
That's funny, I know this as the [url=http://mathworld.wolfram.com/FonteneTheorems.html]2nd Fontene theorem[/url], but the same theorem is also listed as [url=http://mathworld.wolfram.com/GriffithsTheorem.html]Griffiths' theorem[/url],  the posted problem being an exact quotation from Mathworld.

[color=#FF0000][b]Lemma[/b] (Clifford's circle theorem)[b]:[/b][/color] Let $ABCD$ be arbitrary quadrilateral (convex, reflex, concave) with diagonal intersections $ E \equiv AC \cap BD,$ $ X \equal{} AB \cap CD,$ $ Y \equiv BC \cap AD$ ($Y$ is not used). Let $F$ be the intersection of circumcircles of the $ \triangle ABE,\ \triangle CDE$ other than $E$. Then the quadrilaterals $AFCX, BFDX$ are both cyclic.

[color=#FF0000][i]1st Proof:[/i][/color]  Let $(O), (P)$ be circumcircles of the $ \triangle ABE,\ \triangle CDE$ intersecting at $E, F$. $ \triangle AFC \sim \triangle BFD$ with $AC, BD$ passing through $E$ are spirally similar with center $F$, which means that the $ \triangle AFB \sim \triangle CFD$ are also spirally similar with center $F$, $ \measuredangle XBF \mod \pi \equal{} \measuredangle ABF \mod \pi \equal{} \measuredangle CDF \mod \pi \equal{} \measuredangle XDF \mod \pi,$ making $BFDX$ cyclic and similarly for $AFCX$.

[color=#FF0000][i]2nd Proof:[/i][/color]  The Simson line of the $ \triangle ABE$ with the pole $ F \in (O)$ is identical with the Simson line of the $ \triangle CDE$ with the pole $ F \in (P).$  Thus the feet $K, L, M, N$ of normals from $F$ to $ AE \equiv AC,$ $ BE \equiv BD,$ $ AB \equiv XA \equiv XB,$ $ CD \equiv XC \equiv XD$ are all collinear, which means that $F$ is also pole of the identical Simson line $KLMN$ of the $ \triangle AXC,\ \triangle BXD,$ hence $F$ lies on their respective circumcircles.
________________________________________________________

[color=#0000FF][i]Proof of 2nd Fontene theorem:[/i][/color] Let $(O)$ be circumcircle of $ \triangle ABC$, $A', B', C'$ midpoints of $BC, CA, AC,$ and $X, Y, Z$ feet of normals from a point P to $BC, CA, AB.$ Medial $ \triangle A'B'C'$ is pedal triangle of $O$ and $ \triangle XYZ$ is pedal triangle of $P$. Let $ \mathcal C_O,\ \mathcal C_P$ be circles with diameters $AO, AP,$ intersecting at $A$ and one other point $R$. Since $ \measuredangle ARO \equal{} \measuredangle ARP \equal{} \tfrac{\pi}{2}$, $ R \in OP$. Because of the right angles at $B', C'$ and $Y, Z,$ $ \mathcal C_O$ is circumcircle of $AB'OC'$, while $ \mathcal C_P$ is circumcircle of $AYPZ$. Let $C'B'$ meet $YZ$ at $U$. By the above lemma, $C'RZU$ and $B'RYU$ are both cyclic with circumcircles $ \mathcal C_Z,\ \mathcal C_Y.$

Let $ X' \in XP$ be reflection of $X$ in $B'C'$. Then $ X'A \parallel BC \parallel B'C' \perp XX' \equiv XP \equiv PX',$ $ \measuredangle PX'A \equal{} \tfrac{\pi}{2}$, $ X' \in \mathcal C_P$. From the circle $ \mathcal C_Z,$ we have $ \measuredangle URZ \equal{} \measuredangle UC'Z \mod \pi,$ but $ \measuredangle UC'Z \equal{} \measuredangle B'C'A \equal{} \measuredangle CBA \mod \pi.$ From the circle $ \mathcal C_P,$ we have $ \measuredangle ZRX' \equal{} \measuredangle ZAX' \mod \pi,$ but $ \measuredangle ZAX' \equal{} \measuredangle BAX' \equal{} \measuredangle ABC \mod \pi.$ Combining,

$ \measuredangle URX' \equal{} \measuredangle URZ \plus{} \measuredangle ZRX' \equal{} \measuredangle CBA \plus{} \measuredangle ABC \equal{}$ $\measuredangle CBA \minus{} \measuredangle CBA \equal{} 0 \mod \pi,$

which makes the points $U, R, X'$ collinear. Since the line $XU$ is reflection of the line $X'U$ in $B'C'$ and $ R \in X'U,$ reflection $L$ of $R$ in $B'C'$ lies on the line $XU$. If the line $OP$ is fixed, the foot $R$ of normal to $OP$ from the vertex $A$ is fixed, and the reflection $L$ of $R$ in $B'C'$ is also fixed. From the circle $ \mathcal C_P$, $ UY \cdot UZ \equal{} UX' \cdot UR$. Since the line $XUL$ is reflection of the line $X'UR$ in $B'C'$, $UX = UX'$ and $UL = UR$, yielding $ UY \cdot UZ \equal{} UX \cdot UL,$ making $XYZL$ cyclic.
_______________________________________________________

Let $(N)$ be 9-point circle of the $ \triangle ABC,$ identical with the circumcircle of the medial $ \triangle A'B'C'.$ The circumcenter $O$ of the $ \triangle ABC$ is the orthocenter of the medial $ \triangle A'B'C'$. Reflection $O'$ of the orthocenter $O$ in $B'C'$ is on $(N)$ and reflection $D$ of $A$ in $B'C'$ (foot of the A-altitude of the $ \triangle ABC$) is also on $(N)$. Thus $AR$ is reflection of $DL$ and $O'L$ is reflection of $OR$ in $B'C'$. Since $ AR \perp OR,$ it follows that $ DL \perp O'L$ and $ \angle DLO' \equal{} \tfrac{\pi}{2}$. Because $ \angle DA'O' \equal{} \tfrac{\pi}{2}$ is right, $DO'$ is diameter of $(N)$, hence $ L \in (N)$. $OO'$ meets $(N)$ again at $A'$ and let $ LR \perp B'C'$ meet $(N)$ again at $K$. $OO'LR$ is isosceles trapezoid, $A'O'LK$ is cyclic trapezoid, isosceles, $ A'K \parallel OR \equiv OP.$ Thus by construction, $ L \in (N)$ is pole of the Simson line of the $ \triangle A'B'C'$ parallel to $OP$. Since $ AR \perp OP,$ $ RL \perp B'C' \parallel BC,$ $ L \in (N),$ and since similar argumens can be repeated for the remaining vertices $B, C$ of the $ \triangle ABC,$ $L$ is orthopole of the line $OP$ with respect to the $ \triangle ABC.$

I have posted a synthetic proof of this theorem here:-
http://www.artofproblemsolving.com/blog/68754


A chord ST of constant length slides around a semicircle with diameter AB.
M is the midpoint of ST and P is the foot of the perpendicular from S to AB.
Prove that angle SPM is constant for all positions of the chord ST.

[i]Canadian Mathematical Olympiad 1986[/i]

Let O be the midpoint of AB, i.e, the center of the semi-circle. Thus <SMO=90=<SPO, so S,M,O,P are concyclic. <SPM=<SOM.
sin <SOM = SM/SO= ST/2R which is constant.
Hence <SOM=<SPM is constant.


I'm practicing problems for an upcoming quiz, so if anyone could help me with this particular one, I would appreciate it!

Show that x^4 + a3x^3 + a2x^2 + a1x + a0 is the determinant of the following matrix:

(When I say a3, a2, a1, a0, that means a sub 3, a sub 2, a sub 1, a sub 0)

In row 1 of the matrix, we have x, -1, 0, 0
In row 2 of the matrix, we have 0, x, -1, 0
In row 3 of the matrix, we have 0, 0, x, -1
In row 4 of the matrix, we have a0, a1, a2, a3+x

Never mind!  I just figured it out.  It's a lot easier than I thought.


what is the limit as h->0 of $ \frac{e^h\minus{}1}{2h}$?
 there must be a formula or a trick somewhere that I've forgotten...  The answer is not does not exist.

[hide="Solution"]$ \lim_{h\to 0}\frac{e^h\minus{}1}{2h}\equal{}\lim_{h\to 0}\frac{e^h}{2}\equal{}\boxed{\frac{1}{2}}$.[/hide]

It does exist.

Note that $ \lim_{h\to 0}\frac{e^h\minus{}1}{h}$ is the definition  of the derivative of $ e^x$ in $ 0$.

hello, setting $ e^h\minus{}1\equal{}\frac{1}{m}$ we get $ \lim_{h \to 0}\frac{e^h\minus{}1}{h}\equal{}\lim_{m \to \infty}\frac{1}{\ln\left(1\plus{}\frac{1}{m}\right)^m}\equal{}\frac{1}{\ln(e)}\equal{}1$
Sonnhard.


function
integration
limit
calculus
calculus computations

Help!!

Suppose f:[a,b]->R is an integrable function,
and define g:[0,1]->R by g(x)= f$($a+(b-a)x)

1. Show that g is integrable
2. Show that (b-a) $\int_{0}^{1}$ g =$\int_{a}^{b}$ f

Riemann-integrable? Well, I assume it.

Given a partition $P_1$ of $[0,1]$, say, $x_1=0<x_2<x_3<\cdots<x_n=1$, it corresponds to a partition $P_2$ of $[a,b]$ as $y_i = a+(b-a)x_i$. And it is clear that $\max \Delta y_i$ in $P_2$ goes to zero as $\max \Delta x_i$ tends to zero.

$\sum g(\xi_i)\Delta x_i = \sum f(a+(b-a)\xi_i)\Delta x_i = \frac{1}{b-a} \sum f(\eta_i) \cdot (b-a)\Delta x_i = \frac{1}{b-a} \sum f(\eta_i) \Delta y_i$

Since $f$ is Riemann-integrable, taking limit $\max \Delta x_i\to 0$ on both sides,
$\lim_{\max \Delta x_i\to 0} \sum g(\xi_i)\Delta x_i = \frac{1}{b-a} \lim_{\max \Delta y_i\to 0} \sum f(\eta_i)\Delta y_i = \frac{1}{b-a} \int_a^b f(x)dx$.

Hence $g$ is Riemann-integrable, and the limit above has shown that $\int_0^1 g(x)dx = \int_a^b f(x)dx$


prove that $ \forall n \in N$
$ \sin(\frac{\pi n}{3})$  $ \leq$ $ \frac{\sqrt3} {2})$

[quote="greatestmaths"]prove that $ \forall n \in N$
$ \sin(\frac {\pi n}{3})$  $ \leq$ $ \frac {\sqrt3} {2})$[/quote]
Just consider 3 case:
1.$ n\equal{}3k$.
2.$ n\equal{}3k\plus{}1$.
3.$ n\equal{}3k\plus{}2$.

It 's easy since $ n\equal{}6k\plus{}r$ then $ \sin(\frac{n \pi}{3})$ can only be  $ 0, \pm \frac{\sqrt{3}}{2}$


calculus
integration
function
inequalities
real analysis
real analysis unsolved

The first part of the question is: find min (over a,b,c) of the integral from -1 to 1 of |x^3-a-bx-cx^2|^2. I think I have this part.

But the second part is find max of the integral of (x^3)(g(x)), where g(x) is orthogonal to constant, x, and x^2 using the L^2 bilinear product, and g(x) has L^2 (-1,1) norm of 1.

I can't figure it out, and it's driving me nuts. Someone please help!

Thanks

Let $f$ be your maximizing function from part 1, and write $x^{3}=af+h$, where $h$ is in the span of $1,x,x^{2}$. $\langle x^{3},g(x)\rangle=a\langle f,g\rangle+\langle h,g\rangle=a\int_{-1}^{1}f(x)g(x)\,dx$. Apply Cauchy-Schwarz.

I know this is old, but I just came across it searching for something. How do you do the first part?

I'm assuming that since we are working with $ L^2[\minus{}1,1]$ we can let $ f\equal{}x^3\minus{}a\minus{}bx\minus{}cx^2\in L^2$, and then we are looking to minimize $ (f,f)$.

There doesn't seem to be any obvious inequality to get a min, though. They seem to be going the wrong way.

All of the squares have positive coefficients. The inequalities point in the right direction.

Alternately, use Gram-Schmidt on $ 1,x,x^2,x^3$.


topology
real analysis
real analysis unsolved

In a metric space E, a subset A is connected.B is an other subset contains in E-A ;B is both open an close in E-A
Show that A UNION B is connected.

It's false. Can you see how to build a counterexample?

No ,can't see.It's dark for me.
I would like to have an intuitive idea about an open and closed set at the same time.

working on some examples , we need to have E CONNECTED , otherwise its false.


we are currently studying random processes
Im having a bit of difficulty with this question

[img]http://i39.tinypic.com/2hpt4ie.jpg[/img]

I think that the reason why you might be puzzled is because in this exercise there is o assumption about when A is known.
however this should not be a problem

$ E[X_t]\equal{}cos (t). E[A)]\equal{}cos (t). (1/3*\minus{}2\plus{}2/3*1)\equal{}0$

$ E(X(t)X(s))\equal{}cos(t).cos(s)E[a^2]\equal{}cos(t).cos(s)(1/3*4\plus{}2/3*1)\equal{}cos(t).cos(s)*5/3$

one way to see that it is not stationary is to show that for some h>0
$ E(X(t)X(s))\not\equal{}E(X(t\plus{}h)X(s\plus{}h))$ or equivalently that
$ cos(t).cos(s)\not\equal{}cos(t\plus{}h).cos(s\plus{}h)$

For the last one I have no idea at the moment cause I can't remember the definition

but it should be easy

bye

tthanks for the reply

d/w about the last one i got it


show that $ \cos {\pi\over 7} \minus{} \cos {2\pi\over 7} \plus{} \cos {3\pi\over 7} \equal{} {1\over2}$

This is IMO 1963.

See here. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55965[/url]


function
linear algebra
matrix
algorithm
graph theory
abstract algebra
combinatorics proposed

Suppose that you are flip a coin $ n$ times, writing an $ H$ for a head and a $ T$ for a tail.  Let $ P$ be a word of length k from the alphabet $ \{H,T\}$.  Let $ f_n$ be the number of such $ n$ flips that do not have $ P$ as a subword.  Write a system of recurrence relations for that compeltely describes this problem.  Can you find a generating function for $ (f_n)$?

I've been thinking about this exact problem.  Here's my conjecture:  I think it is possible to construct, for every $ P$, a (possibly directed) graph $ G$ such that the number of closed walks from a particular vertex to itself is precisely $ f_n$; see the example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195724&start=20]here[/url].  The adjacency matrix of this graph then describes the problem completely.

Edit:  Here's the solution in the case that $ P$ consists of $ l$ consecutive $ H$s.  Construct a graph on $ l$ vertices as follows:  $ l \minus{} 1$ of them are to be labeled $ H_1, H_2, ... H_{l \minus{} 1}$ and the last one is to be labeled $ T$.  Now construct the directed edges

$ T \to H_1 \to H_2 \to ... \to H_{l \minus{} 1}$
$ H_1 \to T, H_2 \to T, ... H_{l \minus{} 1} \to T$
$ T \to T$.

$ f_n$ is then equal to the number of walks that begin from either $ H_1$ or $ T$ of length $ n$.  A similar construction should be possible for all $ P$, but I haven't worked out the details.  (For example, if $ P \equal{} HT$, only two vertices are necessary and the directed edges are $ H \to H, T \to T, T \to H$.)

I think I have the construction, but I can't quite prove that it works.  In any case, the natural generalization of this problem (beyond two letters) is as follows:  given a directed graph $ G$ with no multiple edges (but possibly with loops) and a collection $ C$ of "forbidden" walks between vertices of $ G$, is it possible to add vertices and add or delete edges to $ G$ to obtain a graph $ G'$ such that there exists a function $ \phi : V(G') \to V(G)$ which is the identity on the vertices of $ G$ such that $ \phi$ is a bijection between the set of walks on $ G'$ starting on a vertex of $ G$ and the set of walks not containing a subwalk in $ C$ on $ G$?  (The special case of words on $ v$ letters occurs when $ G$ is the complete graph (with loops) on $ v$ vertices.)

Edit:  Anyway, here's the construction.  Start with two vertices $ H, T$ and from whichever vertex is the beginning letter of $ P \equal{} P_1 P_2 ... P_l$ (letting $ P_0$ denote whichever of $ H, T$ is not $ P_1$), add vertices corresponding to the letters of $ P$ with directed edges $ P_0 \to P_1 \to P_2 \to ... \to P_{l \minus{} 1}$.  Now construct additional edges according to the following algorithm:

[i]For each $ P_i, i < l$, construct an edge from $ P_i$ to whichever of $ H, T$ is not $ P_{i \plus{} 1}$.[/i]

This graph $ G$ has the property that $ P$ is a forbidden walk but has the additional property that at any point during an "attempt" to walk $ P$ it is possible to "start over," which is precisely the desired behavior.  Note that $ P_0, P_1$ also behave correctly:  if $ P_2 \neq P_1$ then we construct the edge $ P_1 \to P_1$ and so forth.  The general case of words on $ v$ letters is similar.

@tor: I think your conjecture follows from the [url=http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm]Knuth-Morris-Pratt algorithm[/url] for string matching.  Basically, for every word $ P$, the KMP algorithm constructs a finite-state machine.  And a finite-state machine is basically a directed graph.

I might be wrong.

This seems related: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228842

(I searched for "overlap*" to find that.)

[quote="t0rajir0u"][i]For each $ P_i, i < l$, construct an edge from $ P_i$ to whichever of $ H, T$ is not $ P_{i \plus{} 1}$.[/i][/quote]
Whoops, this doesn't work.  We need something like the auto-correlation computation:  we want an edge from $ P_i$ to the last vertex $ P_j$ such that $ P_k P_{k\plus{}1} ... P_i P_j$ is a substring of $ P$ obtained by deleting letters from the end (if such a vertex exists) for some $ k$, or the previous construction otherwise.  (I hope I phrased that correctly.)

[quote="t0rajir0u"]$ P_k P_{k \plus{} 1} ... P_i P_j$ is a substring of $ P$ obtained by deleting letters from the end
...
 (I hope I phrased that correctly.)[/quote]  It sounds right to me, although I would have replaced the first line I've quoted with "$ P_k P_{k \plus{} 1} ... P_i P_j$ is a prefix of $ P$."   :)

BTW, are you taking Prof. Stanley's class this semester?  (I don't know what you look like, so I sort of figure I'll run into you sometime, but if I do I don't know how I'll know it's you  :wink:  Also, I don't know how I'd check if it's you, since I don't know how to pronounce your username.   :lol: )

What's Stanley teaching this semester?  I'm taking Musiker's class, but I took 18.100B last semester - you were a TA, right?

Anyway, I think one way to solve the multiple-forbidden-walks problem is to introduce a sort of restricted [url=http://en.wikipedia.org/wiki/Tensor_product_of_graphs]tensor product[/url] on the graphs $ G_1, ... G_k$ that solve each individual forbidden walk problem.  The vertices consist of all ordered tuplets $ (p_1, ... p_k)$ such that either every $ p_i$ is identified with heads or with tails (i.e. their image under $ \phi$ is identical) and the edges are defined as in the tensor product.  I'm not sure if this approach overcounts, though.

Again, your generalized problem sounds like the multiple-string matching problem.  The [url=http://en.wikipedia.org/wiki/Aho-Corasick_algorithm]Aho-Corasick algorithm[/url] starts with the given forbidden strings, and constructs a finite-state machine that checks another string for the presence of one of the forbidden strings.  By ignoring the labels on the state machine, you get a directed graph.

It's possible I didn't understand your problem.

Ah, you're right; it looks like the directed edges generated by Aho-Corasick are pretty much the ones we need for this problem, except for some corrections because the forbidden strings need to, in fact, be forbidden (and some other cosmetic changes).  It looks like the crux of the algorithm is that it provides a way to record how close you are to completing any of the forbidden strings, which is also what the algorithm I gave does (the index of $ P_i$) for the one forbidden string case.

It's probable that the fully general problem (where words are replaced by walks on an arbitrary graph) can also be solved in this manner; I suppose one could characterize walks on an arbitrary graph as words on the set of vertices with the removed edges standing in for forbidden strings.

[quote="t0rajir0u"]What's Stanley teaching this semester?  I'm taking Musiker's class, but I took 18.100B last semester - you were a TA, right?[/quote]  He's teaching 18.31(something) -- it's symmetric function theory (ch. 7 of EC2).  I was a TA for 18.100 last semester, but I didn't grade homeworks -- I was working with the "C" part of the class.  So I didn't realize you were in it  :(


The residents of Buckahmberg are not that interested in movies, so much that they have made the following custom: 1 person may see the 1 o'clock movie, 2 people may see the 2 o'clock movie, and so forth. The process repeats every 12 hours. If this custom is followed on all days of the week excepting Sundays when nobody goes to the movies, how many sales of movies will there be after two weeks?

[hide="Assuming all the movie showings are booked fully"]$ 2\,(1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\plus{}10\plus{}11\plus{}12)\equal{}156$. Multiplying by $ 6$ and then by $ 2$, the answer is $ 1872$[/hide]

the key is to be able to recognize how it repeats. (and to count correctly i guess.)  :D


instead of asking for tips on approaching problems, i'm asking what you guys do to keep yourself from choking.

what do you guys do to keep the distractions aside? i feel nervous, and i'm afraid i'll go in and just blow something i'm capable of doing well on. any suggestions?

i'm sure i'm not the only one who feels this way..

I feel very nervous too, especially after I know that I didn't make the AIME.

I don't have any specific tips, except don't slow down through the middle problems (#11-18 or so).  That's what hurt me (and prevented me from checking) last time.

As the AMC10B and AMC12B are tomorrow, I'd like to wish everyone who is taking the exams the best of luck.

Since this year the cut off seems low just double-check everything once you get enough problems if you just want to make AIME.

I believe that the cutoff for the A and the B exams are different because it is determined by the top 1% of scores.

Yes, but the cutoffs will definitely be equal to or below 100 (for the AMC 12) and 120 (for the AMC 10), meaning that if you assure yourself of a 100 (11 right and none wrong or 12 right and up to 1 wrong or 13 correct and up to 3 wrong) or a 120 (17 right and 0 wrong or 18 right and up to 2 wrong or 20 right and up to 5 wrong) and you'll be in.  So if you can make sure that out of 20 questions you do, you're absolutely positive of 18, you don't even have to spend time on the other 5.


Let $n$ be a positive integer with $n \ge 3$. Show that \[n^{n^{n^{n}}}-n^{n^{n}}\] is divisible by $1989$.

We consider $n^{n^{n^n}}=n^{n^n}$ in $\mathbb{Z}_m$ for coprime integers m=9, 13, 17, and the result will follow from the Chinese Remainder Theorem.

Lemma: If $n$ is coprime with $m$, then, dividing by $n^{n^n}$ (which I may do since m and n are coprime), I must prove that $n^{n^{n^n}-n^n}\equiv 1\mod m$, which one can reduce, by Euler-Fermat, to showing that $\phi(m)|n^{n^n}-n^n$

For 9, if 3|n it is trivial, so we consider when 3 and n are coprime. Using the lemma, we must show $6|n^{n^n}-n^n$. Modulo 2 this is either 0-0 or 1-1, both of which are clearly zero. Modulo 3, since $n^2\equiv 1$ by FLT, if $n$ is even we get 1-1, and if $n$ is odd we get n-n, so in both cases we get a zero, so $2*3|n^{n^n}-n^n$ and we are done.

For 13, again, if 13|n then it is trivial, so we assume 13 and n are coprime. Again we can apply the lemma and must show $12|n^{n^n}-n^n$, but we showed 3 works already, so it remains to show $4|n^{n^n}-n^n$. If n is even this is obvious. If n is odd, then $n^2 \equiv 1 \mod 4$ so $n^{n^n}-n^n \equiv n-n \equiv 0$ and we are done.

For 17, clearly if 17 divides n then we are done, so let us suppose it is not, so n must be coprime to 17.
If $n$ is even, $n\geq 4$, so $16|n^n$, since $2^4=16$. But then by Fermat's Little Theorem the above equation reduces to $1=1$ which is clearly true.
If $n$ is odd, then I use the lemma. $n^{n^n}-n^n=n^n(n^{n^n-n}-1)$. Now, since $n^2\equiv 1 \mod 8$ for all odd n, $n^n\equiv n \mod 8$ for all odd n, so $8|n^n-n$, and $\phi(16)=8$, so since 2|n, n being coprime to 16, Euler's Theorem states that $(n^{n^n-n}-1) \equiv 0 \mod 16$, which is $\phi(17)$, so we are done.

$ 1989 \equal{} 3^2 \cdot 13 \cdot 17$, hence $ 1989 \mid n^{n^{n^n}} \minus{} n^{n^n} \iff 3^2,13,17 \mid n^{n^{n^n}} \minus{} n^{n^n}$
First $ n^{n^{n^n}} \equiv n^{n^n} \bmod 3^2$. If $ 3 \mid n$ we are done, so wlog assume $ (3,n) \equal{} 1$. By euler-fermat it is enough to prove that $ n^{n^n} \equiv n^n \bmod \varphi(3^2) \equal{} 6$. That is:
$ n^{n^n} \equiv n^n \bmod 2 (*)$ and $ n^{n^n} \equiv n^n \bmod 3 (**)$
$ (*)$ is obvious since it suffices to consider $ n$ odd and even. For proving $ (**)$ we again note that $ 3 \mid n$ makes it obvious, so it is enough to prove $ n^n \equiv n \bmod \varphi(3) \equal{} 2$. And $ n^n \equiv n \bmod 2 (***)$ is obvious by parity too.

$ n^{n^{n^n}} \equiv n^{n^n} \bmod 13$. It is again enoughh to prove $ n^{n^n} \equiv n^n \bmod \varphi(13) \equal{} 12$ that is:
$ n^{n^n} \equiv n^n \bmod 4 (****)$ and $ n^{n^n} \equiv n^n \bmod 3 (**)$. We proved $ (**)$ beforehand and $ (****)$ follows by considering $ 2 \mid n, n \equiv \pm 1 \bmod 4$.

$ n^{n^{n^n}} \equiv n^{n^n} \bmod 17$. It is enough to prove $ n^{n^n} \equiv n^n \bmod \varphi(17) \equal{} 16$. From there it is enough to prove $ n^n \equiv n \bmod 8$ when $ 2 \nmid n$. This follows again by parity, and the fact that $ x^2 \equiv_8 1$ when $ (x,2) \equal{} 1$.


Given a,b,c are prositive real numbers ;  $ a^2\plus{}b^2\plus{}c^2\equal{}1$
Find max of P:

$ P\equal{}\frac{ab}{1\plus{}c^{2}}\plus{}\frac{bc}{1\plus{}a^{2}}\plus{}\frac{ca}{1\plus{}b^{2}}$

hello, after some calculations i have got
$ \frac{ab}{1\plus{}c^2}\plus{}\frac{bc}{1\plus{}a^2}\plus{}\frac{ca}{1\plus{}b^2}\le\frac{3}{4}$ the equal sign holds if $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{3}}$
Sonnhard.


maybe you don't mean to have 2 negatives affecting the $(x^3)^{\frac{1}{3}}$

nm on the above

EDIT: again...

i don't think that this has a limit because was we approach infintiy, we only need to concern ourselves with the greatest degree terms, so

we get $2x+3x$ or $5x$ or infinity

The first term is increasing, the second term is decreasing, but since we subtract the second, the whole thing is increasing, so it's $\infty$

[quote="tµtµ"]I'm sure he meant : $\lim_{x\rightarrow\infty}{\sqrt{4x^2+x}+\sqrt[3]{1-27x^3}}$  ;)[/quote]
no it's $\lim_{x\rightarrow\infty}{\sqrt{4x^2+x}-\sqrt[3]{1-27x^3}}$  ;)

I have both evaluated the limit algebraically and plugged it into my calculator, and I can assure you that it is $\boxed{\infty} \neq 11 \neq 5$.

Like Altheman already said, if you consider only the leading terms of both parts of the limit, they are, respectively, $2x$ and $3x$.  It is fairly clear that $\displaystyle\lim_{x\to\infty} 2x + 3x = \infty$


Find all the integer solutions to

$ k!\equal{}6n^2\plus{}2n\minus{}2$

Please explain it step-by-step

Thanks! :lol:

Since $ k!\equal{}6n^2\plus{}2n\minus{}2\equal{} 2(n(3n\plus{}1) \minus{} 1)$, and $ n(3n\plus{}1)$ is even, hence $ n(3n\plus{}1) \minus{}1$ is odd, it follows that RHS is only divisble by $ 2$, and never by $ 4$.
But any factorial with $ k\geq 4$ is divisible by $ 4$, so $ k\leq 3$.
For $ k\equal{}0$ or $ 1$, $ k! \equal{} 1$, which is odd, so there is no solution;
For $ k\equal{}2$ we have $ 2!\equal{}2\equal{}6n^2\plus{}2n\minus{}2$, so $ 3n^2\plus{}n\minus{}2\equal{}0$, with solutions $ n\equal{}\minus{}1$ and $ 2/3$, and only $ n\equal{}\minus{}1$ is integer;
For $ k\equal{}3$ we have $ 3!\equal{}6\equal{}6n^2\plus{}2n\minus{}2$, so $ 3n^2\plus{}n\minus{}4\equal{}0$, with solutions $ n\equal{}\minus{}4/3$ and $ 1$, and only $ n\equal{}1$ is integer.

[hide="Solution"]

\[ k!\equal{}2(3n^2\plus{}n\minus{}1)\]

We have the 2, now we need to find 3 and higher. First, note that $ 3n^2\plus{}n\minus{}1$ must be divisible by 3. 3n^2 will always be divisible by 3. $ n$ thus must be 1 mod 3 so that $ (n\minus{}1)$ is divisible by 3. We can obtain the 3 using $ n\equal{}1$.

Now, we shall demonstrate that k can be no greater than 3 using a parity argument.
If n is even, then $ 3n^2$ is even, but $ n\minus{}1$ is odd. Likewise, if n is odd, then $ 3n^2$ is odd, but $ n\minus{}1$ is even.
Thus, $ 3n^2\plus{}n\minus{}1$ is never divisible by 2, let alone 4.
The only solution in $ (k,n)\equal{}(3,1)$.

[/hide]

Very clean solutions!!

Thank you Caelestor and mavropnevma! :lol:


In this case you can't do much, since the inner sum depends on the value of the outer index.

And there's a typo: The upper limit of the inner sum should've been $n-i$.

Well then, if you denote the expression under the sums by $f(i,j)$, then you have this:

[code]
f(1,0)+f(1,1)+...+f(1,n-3)+f(1,n-2)+f(1,n-1)
f(2,0)+f(2,1)+...+f(2,n-3)+f(2,n-2)
f(3,0)+f(3,1)+...+f(3,n-3)
.
.
.
f(n,0)
[/code]

Then just sum the terms vertically instead of horizontally and you get the latter form of the sum. But as for finding the value, I don't see that as a big improvement.

Actually, it's an improvement, because we can calculate
$\sum_{i=1}^{n-j}\binom{n-j}{i}(-1)^{i+1}$
using (carefully) Newton's Binomial theorem :)

We could do that in the initial form too :)

[quote="Phelpedo"]When you have, for example, double summations, is there a systematic way of interchanging the order? i.e., what's the best way to switch the order of summation of

$\sum_{i=1}^{n}(-1)^{i+1}\sum_{j=0}^{n-i}\binom{n-j}{i}\frac{1}{n-j}$ [/quote]
[quote="Phelpedo"]Actually, my book does interchange them for this sum (although it's not exactly the same, I suppose). I'm just wondering if there is a generic method. In case you're wondering, it's

$\sum_{j=0}^{n-1}\sum_{i=1}^{n-j}(-1)^{i+1}\binom{n-j}{i}\frac{1}{n-j}$.[/quote]
Sometimes it is useful to use Iverson bracket: if $p$ is a logical expression, we have $[p]=\begin{cases}{1\text{ if }p\text{ is true}}\\{0\text{ if }p\text{ is false}}\end{cases}$. Then \[\sum_{i=1}^{n}(-1)^{i+1}\sum_{j=0}^{n-i}\binom{n-j}{i}\frac{1}{n-j}= \sum_{i=1}^{n}(-1)^{i+1}\sum_{j=0}^{n-1}[j\leqslant n-i]\binom{n-j}{i}\frac{1}{n-j}=\]  \[=\sum_{i=1}^{n}\sum_{j=0}^{n-1}[i+j \leqslant n](-1)^{i+1}\binom{n-j}{i}\frac{1}{n-j}\] and now you can change the order: \[\sum_{i=1}^{n}\sum_{j=0}^{n-1}[i+j \leqslant n](-1)^{i+1}\binom{n-j}{i}\frac{1}{n-j}= \sum_{j=0}^{n-1}\sum_{i=1}^{n}[i\leqslant n-j](-1)^{i+1}\binom{n-j}{i}\frac{1}{n-j}=\]  \[=\sum_{j=0}^{n-1}\sum_{i=1}^{n-j}(-1)^{i+1}\binom{n-j}{i}\frac{1}{n-j}= \sum_{j=0}^{n-1}\frac{1}{n-j}\sum_{i=1}^{n-j}(-1)^{i+1}\binom{n-j}{i}.\] I don't say that it is the best thing you can the do with your sum but sometimes it is useful technique to change the sum order. :)
PS. There's a typo. It should be [i]nested[/i] instead of [i]nestled[/i].


I'm sorry if this was answered before somewhere but where can I find in-depth descriptions of all the classes offered?

Also, I have some minor questions that I just wanted to clarify. Equipment like frisbees and basballs will be provided? There will be access to computers? Do we get homework problems each day?

There was a pretty good ordering form system that they had last year. If anything not outrageously expensive (I'd think like, <$\$$50) is needed, you can give them the cash and they'll go out to buy it for you. I got my blanket that way :D

[quote="13375P34K43V312"]Hmmm, I'm planning to bring a frisbee. I imagine that on campus somewhere there must be a computer lab or a library where we can get internet access.[/quote]

i dont think there is.  couldnt find one last year but it wasnt as bad as i thought it would be.

[quote="sapphyre571"]does anyone wanna volunteer to bring a laptop? I'm pretty sure the whole campus has wireless internet access[/quote]

i have a laptop that i will probably bring

Stephen Worlow

i dont think they would like it because i asked at the Q&A thing for last year's awesomemath and they said no

[b]Please do not bring laptops to the camp unless it is absolutely necessary.[/b]

If people bring laptops, then, chances are, they'll be playing games. If you wanted to play games, you could have just stayed at home and saved $\$$3000. You will be very busy anyway, as there is tons of things to do.

Every apartment will have a counsellor who will mostly likely bring a laptop. If you need to check your email or something, you can just ask your residential counsellor and I'm sure that he/she won't have a problem with it.

Oh, and speaking from last year's experience, you probably won't be able to use the campus wireless network unless you're a counsellor.


come vi è sembrato?

personalmente l'ho trovato facile.....anche se per soliti errori dovuti alla mia stupidità. :lol: ..caratteristica degli esseri umani....ho fatto un misero 98 :blush:

Nettamente più facili di quelli dell'anno scorso... Ma sono riuscito a fare più  meno lo stesso ridicolo punteggio..

Eppure ho provato a risolvere un problema delle olimpiadi internazionali e l'ho fatto in un'ora :wacko:


problem 1

\[ 4\left( {x\minus{}\minus{}{\rm{ }}\frac{1}{x}} \right)^2  \plus{} 8\left( {x{\rm{ }} \plus{} {\rm{ }}\frac{1}{x}} \right)\minus{}\minus{}29{\rm{ }} \equal{} {\rm{ }}0,{\rm{ }}x{\rm{ }} \ne {\rm{ }}0
\]

members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com

Shift this to MSM.
Also, don’t advertise here. Most AoPS users don’t face difficulty in Math at school !

@२ नीचे (2 Below) !
Lol I didn’t see that  :rotfl:

[hide=hint (basically solves the problem)]Note that $(x-\frac{1}{x})^2=(x+\frac{1}{x})^2-4$[/hide]

[quote=Prabh1512]Shift this to MSM.
Also, don’t advertise here. Most AoPS users don’t face difficulty in Math at school ![/quote]

its 13 years old lol


[b]Given $ a,b,c \ge 0$. Prove that:
$ \frac {a}{a \plus{} \sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{b \plus{} \sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{c \plus{} \sqrt {a^2 \plus{} b^2}}\ge1$
When do we have a quality?[/b]

$ \sqrt {a^2 \plus{} b^2} \le a \plus{} b$

So simply

$ \sum \frac {a}{a \plus{} \sqrt {b^{2} \plus{} c^{2}}} \ge \frac {a}{a \plus{} b \plus{} c} \equal{} 1$

Equality when $ a \equal{} b \equal{} 0, c > 0$ and permutations

[quote="centry57"]$ \frac {a}{a \plus{} \sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{b \plus{} \sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{c \plus{} \sqrt {a^2 \plus{} b^2}}\ge1$[/quote]

 I'm not sure but perhaps
ineq <=> $ ab\plus{}bc\plus{}ca \ge a\plus{}b\plus{}c$


calculus
integration
trigonometry
function
logarithms
algebra
rational function

Solve  $\int (x^{2}+1)^\frac{1}{2}dx$
Let  $x=tan\theta$
Draw a right triangle with $(x^{2}+1)^\frac{1}{2}$ for the hypotenue and x opposite the angle $\theta$ and 1 for the remaining side
By reverse substitution $dx=\frac{1}{cos^{2}\theta}d\theta$
but we want $\frac{d\theta}{dx}$ not $\frac{dx}{d\theta}$

so $d\theta$=$cos^{2}dx$
since $(x^{2}+1)^\frac{1}{2}$ = $\frac{1}{cos\theta}$

then $\frac{1}{cos^{3}\theta}d\theta$=$\frac{cos^{2}\theta}{cos^{3}\theta}dx$=$\frac{1}{cos\theta}dx$

Now write the original integral with $\theta$ as the new variable and solve

$\int\frac{1}{cos^{3}}d\theta$

and this is where I got stuck. Any ideas?

It's ugly. Have patience.

The next step is an integration by parts.

well... maybe you can try again with the substitution $x = \sinh t$...

I have the full solution for $\int\frac{1}{cos^{3}\theta}d\theta$

As usual I'll come back in two days and post the solution. Hint: Take jmerry's advice!

[quote="fedja"]Or, if you want to finish it from where you got with your trigonometric substitution, write $\frac 1{\cos^{3}x}=\frac{\cos x}{(1-\sin^{2}x)^{2}}$ and make the change of variable $u=\sin x$. This will reduce the problem to integrating a rational function $\frac 1{(u-1)^{2}(u+1)^{2}}$, which is a routine exercise in elementary fractions.[/quote]

Just find $a,\ b,\ c,\ d$ such that $\frac 1{(u-1)^{2}(u+1)^{2}}=\frac{a}{(u-1)^{2}}+\frac{b}{u-1}+\frac{c}{u+1}+\frac{d}{(u+1)^{2}},$ however we can find these values directly. :lol:

Anyone want to see the full solution for $\int(x^{2}+1)^\frac{1}{2}dx$

Well for the record the complete solution is

Here is my rather long post If there are any errors please let me know!

Solve $\int (x^{2}+1)^\frac{1}{2}dx$

Let $x=tan\theta$

Draw a right triangle with $(x^{2}+1)^\frac{1}{2}$ as the hypotenue and $x$ opposite the angle $\theta$ and $1$ as the remaining side

By reverse substituion $dx=\frac{1}{cos^{2}\theta}d\theta$

But  we want $\frac{d\theta}{dx}$ not$\frac{dx}{d\theta}$

so $d\theta=cos^{2}dx$

According to the triangle $(x^{2}+1)^\frac{1}{2}= \frac{1}{cos\theta}$

Then $\frac{1}{cos^{3}}d\theta= \frac{cos^{2}}{cos^{3}}dx = \frac{1}{cos\theta}dx$

Write the original integral with $\theta$ as the new variable

Solve$\int\frac{1}{cos^{3}}d\theta$  (Here is where I got stuck. jmerry suggested integration by parts but he did not say that I would have to
integrate by parts twice. Of course jmerry knew that!)

Write integral like this $\int \frac{1}{cos\theta}\cdot\frac{1}{cos^{2}\theta}d\theta$

Integrate by parts and let the 2 choices be

$\frac{1}{cos\theta}\prime = \frac{sin\theta}{cos^{2}\theta}$

and

$\int\frac{1}{cos^{2}\theta}d\theta=tan\theta$

so $\int\frac{1}{cos\theta}\cdot\frac{1}{cos^{2}\theta}d\theta = tan\theta\cdot\frac{1}{cos\theta}-\int \frac{sin\theta}{cos^{2}\theta}\cdot tan\theta d\theta$

=$\frac{tan\theta}{cos\theta}-\int sin\theta\cdot\frac{sin\theta}{cos^{3}\theta}d\theta$ (Kunny comes to the rescue just in time, lucky for vector1)

Solve $\int sin\theta\cdot\frac{sin\theta}{cos^{3}\theta}d\theta$

integrate by parts and let the 2 choices be

$sin\theta\prime=cos\theta$

and

$\int\frac{sin\theta}{cos^{3}\theta}d\theta=\frac{1}{2cos^{2}\theta}$

so$\int{sin\theta}\cdot\frac{sin\theta}{cos^{3}\theta}d\theta = sin\theta\cdot\frac{1}{2cos^{2}\theta}-\int cos\theta\cdot\frac{1}{2cos^{2}\theta}d\theta$

$=\frac{tan\theta}{2cos\theta}-\frac{1}{2}\int\frac{1}{cos\theta}d\theta =\frac{tan\theta}{2cos\theta}-\frac{1}{2}ln(tan\theta+\frac{1}{cos\theta})$

then $\int\frac{1}{cos^{3}\theta}d\theta = \frac{tan\theta}{cos\theta}-(\frac{tan\theta}{2cos\theta}-\frac{1}{2}ln(tan\theta+\frac{1}{cos\theta}))$

$=\frac{tan\theta}{2cos\theta}+\frac{1}{2}ln(tan\theta+\frac{1}{cos\theta})=\frac{1}{2}(\frac{tan}{cos\theta}+ln(tan\theta+\frac{1}{cos\theta}))$

According to the triangle $x=tan\theta$ and $(x^{2}+1)^\frac{1}{2}=\frac{1}{cos\theta}$

then finally $\int(x^{2}+1)^\frac{1}{2}dx=\frac{1}{2}(x(x^{2}+1)^\frac{1}{2}+ln(x+(x^{2}+1)^\frac{!}{2}))$

This result agrees with the CRC table of integrals

He pauses for a moment before pressing the send button knowing that at least 20 hours of work went into this post.

$\int{\sec^{3}xdx}= \int{\sec x\sec^{2}xdx}$

Set $u = \sec x \Longleftrightarrow du = \sec x\tan xdx\ \wedge\ dv = \sec^{2}xdx \Longleftrightarrow v = \tan x$, this yields

\begin{eqnarray*}\int{\sec^{3}xdx}&=& \sec x\tan x-\int{\sec x\tan^{2}xdx}\\ &=& \sec x\tan x-\int{\sec x(\sec^{2}x-1)dx}\\ &=& \sec x\tan x-\int{\sec^{3}xdx}+\int{\sec xdx}\\ \implies 2\int{\sec^{3}xdx}&=& \sec x\tan x+\ln \left|{\sec x+\tan x}\right|\\ \ \therefore\ \int{\sec^{3}xdx}&=& \frac{1}{2}\left({\sec x\tan x+\ln \left|{\sec x+\tan x}\right|}\right)+c \end{eqnarray*}

[b]Solution 1.[/b] Substitute $x=\cosh t$ and simplify the integral to $\int\sinh^{2}t dt=\frac{1}{2}(\sinh t\cosh t-t)+C$. You can then write the expression in terms of x.$\Box$

[b]Solution 2.[/b]  Substitute $x=\tan\theta$:
\[I=\int \sqrt{x^{2}+1}dx = \int \sec^{3}\theta d\theta. \]
Let $u=sec\theta,\ dv=\sec^{2}\theta d\theta,\ du=\sec\theta\tan\theta d\theta,\ v=\tan\theta$.  Hence,
\[\\ I= \int \sec^{3}\theta d\theta\\ = \sec\theta\tan\theta-\int sec\theta\tan^{2}\theta d\theta\\ = \sec\theta\tan\theta-\int \sec^{3}\theta d\theta+\int\sec\theta d\theta\\ \&\implies\& I=\frac{1}{2}\left(\sec\theta\tan\theta+\int\sec\theta\frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}d\theta\right)\\ = \frac{1}{2}\left(\sec\theta\tan\theta+\int\frac{\sec^{2}\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta}d\theta\right). \]
Let $G=\int\frac{\sec^{2}\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta}d\theta$.  Substitute $u=\sec\theta+\tan\theta,\ du=\sec\theta\tan\theta+\sec^{2}\theta$.  Hence,
\[\\ G= \int\frac{\sec^{2}\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta}d\theta\\ = \int \frac{1}{u}du\\ =\ln |\sec\theta+\tan\theta|+c. \]
As a result,
\[I=\frac{1}{2}\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta| \right)+C. \]
Since $x=\tan\theta$, $\sec\theta=\sqrt{x^{2}+1}$.  As a result,
\[I= \frac{1}{2}\left(x\sqrt{x^{2}+1}+\ln(\sqrt{x^{2}+1}+x) \right)+C. \]
$\Box$

Vector1, I have not looked at your solution, but our answers agree.

Thanks all. I printed this thread for later reference

I think the best way is (similar to MathTeX method):

$I = \int \sec^{3}x \, dx = \int \sec x \, (1+\tan^{2}x) \, dx = \int \sec x \, dx+\int \sec x \, \tan^{2}x \, dx =$

$= \ln |\sec x+\tan x|+\int \tan x \, d (\sec x) =$

$= \ln |\sec x+\tan x|+\sec x \, \tan x-\int \sec^{3}x \, dx =$

$= \ln |\sec x+\tan x|+\sec x \, \tan x-I \dots$

Note: (reduction formula) For $n\ge 2,$

$\int\sec^{n}x\,dx=\frac1{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int sec^{n-2}x\,dx.$

To derive this formula requires an integration by parts and some algebraic manipulation involving the identity $\sec^{2}x=\tan^{2}x+1.$ Several of the posts in this topic show the method; MathTex's is probably the cleanest exposition.

Similar reduction formulas, similarly proved using integration by parts and trigonometric identities, can be found for the integrals of arbitrary powers of each of the trig. functions.

[quote="Kent Merryfield"]Note: (reduction formula) For $n\ge 2,$

$\int\sec^{n}x\,dx=\frac1{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int \sec^{n-2}x\,dx.$
[/quote]
[color=darkblue][u]Proof[/u][/color]

$\int{\sec^{n}x}= \int{\sec^{2}x\sec^{n-2}xdx}$

Now, let's set

\[\begin{array}{ccccccccccccccc}u &=& \sec^{n-2}x &\Longleftrightarrow& du &=& (n-2)\sec^{n-2}x\tan xdx\\ dv &=& \sec^{2}xdx &\Longleftrightarrow& v &=& \tan x \end{array}\]

Therefore, integrating this by parts yields

\begin{eqnarray*}\int{\sec^{n}x}&=& \sec^{n-2}x\tan x-(n-2)\int{\sec^{n-2}x\tan^{2}xdx}\\ &=& \sec^{n-2}x\tan x-(n-2)\int{\sec^{n-2}x(\sec^{2}x-1)dx\\ &=& \sec^{n-2}x\tan x-(n-2)\int{(\sec^{n}x-\sec^{n-2}x )dx}\\ &=& \sec^{n-2}x\tan x-(n-2)\int{\sec^{n}x}+(n-2)\int{\sec^{n-2}x dx}\\ \int{\sec^{n}x}+(n-2)\int{\sec^{n}x}&=& \sec^{n-2}x\tan x+(n-2)\int{\sec^{n-2}x dx}\\ \implies (n-1)\int{\sec^{n}x}&=& \sec^{n-2}x\tan x+(n-2)\int{\sec^{n-2}x dx}\\ \ \therefore\ \int{\sec^{n}x}&=& \frac{{\sec^{n-2}x\tan x}}{{n-1}}+\frac{{n-2}}{{n-1}}\int{\sec^{n-2}x dx}}\end{eqnarray*}


Hi!

$x,y,z$ are positive real numbers, prove the following:
\[{{x^{2}(y+z-x)}\over{y+z}}+{{y^{2}(z+x-y)}\over{z+x}}+{{z^{2}(x+y-z)}\over{x+y}}\leq{{xy+yz+zx}\over{2}}.\]

[quote="yuyang"]
$x,y,z$ are positive real numbers, prove the following:
\[{{x^{2}(y+z-x)}\over{y+z}}+{{y^{2}(z+x-y)}\over{z+x}}+{{z^{2}(x+y-z)}\over{x+y}}\leq{{xy+yz+zx}\over{2}}. \]
[/quote]
Easy problem  :D
We have: \[\sum{{x^{2}(y+z-x)}\over{y+z}}=\sum\frac{x}{y+z}\cdot x(y+z-x)\le\sum\frac{x}{y+z}\cdot\left(\frac{x+(y+z-x)}{2}\right)^{2}=\sum\frac{x(y+z)}{4}=\frac{\sum yz}{2}\]

The above is not hard enough but very nice because the solution is based on even a stronger inequality

\[{{x^{2}(y+z-x)}\over{y+z}}+{{y^{2}(z+x-y)}\over{z+x}}+{{z^{2}(x+y-z)}\over{x+y}}\leq\frac{9xyz}{2(x+y+z)}.\]


[quote="GameBot"]Tess has a fish tank in the shape of a right rectangular prism, with inside dimensions of 24 feet x 12 feet x 16 feet. One cubic foot holds 7.5 gallons of water. To the nearest gallon, how many gallons of water does Tess need to fill her fish tank?[/quote]

Her fish tank has $ 24 \times 12 \times 16 \equal{} 4608$ cubic feet. Since each cubic foot holds 7.5 gallons of water, Tess needs $ 4608 \times 7.5 \equal{} 34560$ gallons of water to fill her fish tank.

That does sound like a mighty big fish tank.

If it were 24 inches by 12 inches by 16 inches, the volume would be 20 gallons, which is what FTW wants for a correct answer.  Someone needs to correct the question or the answer.

I think it means inches... REPORT ERRORS


1)find all nature(m,n) such that:
mn-1|(n^3)+1


integration
calculus
probability
expected value

[quote="aspiringchemlete"]A die is rolled repeatedly until a "1" appears.

Show that the expected number of rolls until a "1" appears is 6.

Hint: Prove that $ \sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^{2}}$.[/quote]

You need the condition that $ |x|<1$.

[hide="Proof hint"]This can be proved fairly easily by differentiating the equation
$ \sum_{n=0}^{+\infty}x^{n}= \frac{1}{1-x}$,

but I'll offer a different proof.

Define $ S$ as:

$ S = \sum_{n=1}^{\infty}nx^{n-1}= 1+\sum_{n=2}^{\infty}nx^{n-1}$

Multiply both sides by $ x$:
$ Sx = \sum_{n=1}^{\infty}nx^{n}= \sum_{n=2}^{\infty}(n-1)x^{n-1}$

Subtract:
$ S-Sx = 1+\sum_{n=2}^{\infty}x^{n-1}= \sum_{n=0}^{\infty}x^{n}= \frac{1}{1-x}$

Dividing both sides by $ 1-x$ proves the desired result.[/hide]

Right I didn't mention that because we are dealing with $ x=\frac{1}{6}$. But that's pretty much it.

[quote="AstroPhys"]vishalarul , $ \frac{1}{(1-x)^{2}\ dx}= \frac{2}{(1-x)^{3}}\not = \frac{1}{(1-x)}= \int \frac{1}{(1-x)^{2}}\ dx$.[/quote]
 :huh: What are you talking about?

I said that you could differentiate that equation, not integrate, or whatever that you are doing...
Here is an outline
[hide="Warning! Calculus content ahead"]$ \sum_{n=0}^{+\infty}x^{n}= \frac{1}{1-x}\quad \text{for }|x|<1$

On the left is a power series with radius of convergence equal to $ 1$. So we can differentiate both sides:
$ \sum_{n=1}^{+\infty}n x^{n-1}= \frac{1}{(1-x)^{2}}\quad \text{for }|x|<1$

It's as simple as that...[/hide]

:oops: Sorry about that I understood your original post as if you had ALREADY differentiated getting $ \frac{1}{1-x}$.


abstract algebra
vector
superior algebra
superior algebra theorems

Hello, this problem has been bugging me for some time, for some reason I can't get any results :

Consider two exact sequence of modules :

$0\rightarrow N_1\buildrel{i_1}\over{\rightarrow} P_1\buildrel{p_1}\over{\rightarrow}M\rightarrow 0$
$0\rightarrow N_2\buildrel{i_2}\over{\rightarrow} P_2\buildrel{p_2}\over{\rightarrow}M\rightarrow 0$

where $P_1,P_2$ are projective modules

How can I show that $P_1\oplus N_2 \cong P_2\oplus N_1$?

I am not experienced in projective modules at all.  When working over a field and with finite dimensions, it is very clear, the dimension are equal.  But of course I can't do that here.  For a brief moment I thought I could do it in this way

My isomorphism u will consist of a map $u_1: P_1 \rightarrow P_2\oplus N_1$
and $u_2: N_2 \rightarrow P_2\oplus N_1$

now as i have a surjective map $p_2 : P_2\rightarrow M$ and a morphism $p_1: P_1\rightarrow M$, projectivity of $P_1$ gives me a map from $P_1$ to $P_2$ , but I don't think this helps at all for finding $u_1$

All thoughts on this are welcome :?

I am really beginning to get frustrated about this one.

It's so nasty!

I had this idea though : what if I could prove that $P_1\cong N_2 \oplus M$ and $P_2\cong N_1\oplus M$
...then the result would be obvious

But is this even true?  It is definitely true for just vector spaces of finite dimension over a field, that is where I got the idea, but it seems I would have to use projectivity, ...of M, not the $P_i$  :ninja:

Seriously , is this even true?


inequalities
trigonometry
inequalities unsolved

I can only hope that I guessed your problem correctly:

http://www.mathlinks.ro/Forum/viewtopic.php?t=21405

  Darij

but there i dont see my inegality!!!!!!
mine is another

The way you have posted it, your inequality is :spam: . Correct the typos, please.

[quote="akhrif"]Let a,b,c positive integers then prove that       $\sqrt{a^2+b^2-ab}+\sqrt{a^2+b^2-ab}\geq\sqrt{a^2+c^2-ac}$[/quote]

  Darij

a, b, c > 0 so a^2 + c^2 - ac < a^2 + c^2 + ac. LOL
This seems stupid BUT it will solve the problem ;)
Pick a pt P in the plane. Draw PB = b, PA =a, PC = c, so that angle APB = angle CPB = 60 degrees and angle APC = 120 degrees.
Then AB = sqroot(a^2 + b^2 - ab) by Cosine Law and in the same way, BC = sqroot(c^2 + b^2 - bc).
AC = sqroot(a^2 + c^2 + ac).
AB + BC >= AC because ABC is a triangle or a straight line, and the result follows.
I am pretty sure the equality case doesn't exist.

Then it follows from $\sqrt{a^2+b^2-ab}+\sqrt{b^2+c^2-bc} \geq \sqrt{c^2+a^2+ac}$, and this was proven in the link I gave (just read the complete thread).

Bye ML for a few days (not that I won't be online, but probably I will post even less than I do now what is already not much because of the exam weeks).

  Darij


Bob and Meena play a two-person game which is won by the first person to accumulate 10 points. At each turn Bob gains a point with probability of 1/3. If he doesn't get a point, then Meena gets a point. Meena is now ahead 9 to 8. What is the probability that Meena will win? Express your answer as a common fraction.

There are two cases to consider.

Case 1: Meena gets the next point and wins.

There is a $ \frac{1}{3}$ chance that Bob gets the point, so our answer is $ 1 \minus{} \frac{1}{3} \equal{} \frac{2}{3}$ for this case.

Case 2: Bob gets the next point, and then Meena gets the second and wins.

Our answer for this case is $ \frac{1}{3} \times \frac{2}{3} \equal{} \frac{2}{9}$.

Now we just add these two numbers to get $ \frac{8}{9}$.

I just calculated the probability that Bob wins, and subtracted by 1.  I think it's faster, especially in a time-pressured situation.

There's $ \frac {1}{3}$ chance that Bob gets the first point, and $ \frac {1}{3}$ chance that he gets the next point.  Multiplying, we get $ \frac {1}{9}$.  Subtracting from $ 1$, we get $ \boxed{\frac {8}{9}}$.

That's a stupid game, isn't it.


Let $a \ge b \ge c$, then
$\sum (a-b)^{2}(a+b)+b(a-b)(a-c)+(b-c)(a^{2}+bc-2ac) \ge 0$
which is true since $a^{2}+bc \ge a^{2}+c^{2}\ge 2ac$

Sorry! I made a mistake in computations. I will try again.

It is interesting to find a positive number $p$ as great as possible such that
$\frac{\sum a^{4}(a-b)}{\sum a(a-b)}\geq pabc$.
Certainly, the inequality holds for $p=2$.

[quote="zhaobin"]Did you have some other solution,
I can only prove $p=2$ using SOS.[/quote]
My proof is SOS too:
Our inequality equivalent to follows
$\sum_{cyc}(a-b)^{2}S_{c}\geq0,$ where $S_{c}=4a^{3}+3a^{2}b+2ab^{2}+b^{3}-5abc.$
If $a=\max\{a,b,c\}$ then $S_{b}\geq0,$ $S_{c}\geq0$ and $S_{a}+S_{b}\geq0$ ( or $S_{a}+S_{c}\geq0$ ).
The last inequality is very nice.


Hey, I was trying to solve this question but I didn't manage to solve it, although it seems to be simple:

Proove that, for any $ n$ non-negative integer, the number $ 1^n\plus{}2^n\plus{}3^n\plus{}4^n$ is divisible by 5 if, and only if, $ n$ is not divisible by 4.

[quote="Fabiomr1"]Hey, I was trying to solve this question but I didn't manage to solve it, although it seems to be simple:

Proove that, for any $ n$ non-negative integer, the number $ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n$ is divisible by 5 if, and only if, $ n$ is not divisible by 4.[/quote]
$ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n \equiv 1^{4\plus{}n}\plus{} 2^{4\plus{}n} \plus{} 3^{4\plus{}n} \plus{} 4^{4\plus{}n} \bmod 5$ So you just have to check the cases $ n \equal{} 1,2,3,4$ :)

Alternative solution:
$ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n \equiv 1^n \plus{} 2^n \plus{} (\minus{}2)^n \plus{} (\minus{}1)^n \bmod 5$ so for odd $ n$ it is divisible with $ 5$. Assume $ n \equal{} 2k$:
$ 1^{2k} \plus{} 2^{2k} \plus{} 3^{2k} \plus{} 4^{2k} \equal{} 2 * 1^k \plus{} 2 ^(\minus{}1)^k \bmod 5$. So for odd $ k$ we know that it is divisible by $ 5$. For even $ k$ (ie $ 4 \mid n$) we it is not divisible by $ 5$. QED

[quote="Fabiomr1"]Hey, I was trying to solve this question but I didn't manage to solve it, although it seems to be simple:

Proove that, for any $ n$ non-negative integer, the number $ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n$ is divisible by 5 if, and only if, $ n$ is not divisible by 4.[/quote]
assume that $ n$ is some odd integer,therefore both $ 1^n\plus{}4^n$ and $ 2^n\plus{}3^n$ are divisible by $ 5$.
Now let $ n$ be some even number that is not divisible by $ 4$, $ 1^n\plus{}2^n\plus{}3^n\plus{}4^n\equiv$ it follows that $ 2\plus{}2^{n\plus{}1}\equiv 1\plus{}2^{n}\equiv 0$ modulo $ 5$.
this is true,simply because $ n$ is some integer number of the form $ 4k\plus{}2$ and $ 2^{4k}\equiv 16^{k}\equiv 1$ modulo $ 5$.
on the other hand if $ n\equal{}4k$,then $ 1\plus{}2^{4k}\equiv 2$,done...

Hey, thanks. Indeed it was kind of simple but... well   :lol:

See http://www.mathlinks.ro/viewtopic.php?t=40171 or the lemma for the proof of Chevalley-Warning in http://www.mathlinks.ro/viewtopic.php?t=150886 .


I'm writing a paper such that the sections are written by the command \section and the refences are written by the command \begin{thebibliography}{9}. But now I have two chapters whose number is 17. And furthermore command section inserts more spaces between the number and the letter that thebibliography. How to write references like sections as thebibliography inserts section title automatically?

[code]\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{stmaryrd}
\usepackage{euscript}
\begin{document}
\section{Beginning}
I will prove the main theorem from the article~\cite{ax1}.
\begin{thebibliography}{9}
\bibitem{ax1} Ax, J: Solving Diophantine problems modulo every prime, The Annals of Mathematics, 2nd Ser., Vol. 85, No. 2 (Mar., 1967), 161-183.
\end{thebibliography}
\end{document}[/code]
I would like to see the chapter as
2      References

The article class only has sections not chapters. You can add the section number to References using
[code]\renewcommand{\refname}{\stepcounter{section}\thesection \quad References}[/code]

Thanks, but I still won't get the correct numbering in the table of contents. It shows just 1 Beginning.
[code]
\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{stmaryrd}
\usepackage{euscript}
\begin{document}
\renewcommand{\refname}{\stepcounter{section}\thesection \quad References}
\tableofcontents
\newpage
\section{Beginning}
I will prove the main theorem from the article~\cite{ax1}.
\begin{thebibliography}{9}
\bibitem{ax1} Ax, J: Solving Diophantine problems modulo every prime, The Annals of Mathematics, 2nd Ser., Vol. 85, No. 2 (Mar., 1967), 161-183.
\end{thebibliography}
\end{document}
[/code]

You didn't ask for that. You must tell LaTeX that you want to include references in the table of contents. You do this by adding the line
[code]\addcontentsline{toc}{section}{\thesection \hspace{0.4em} References}[/code]after the bibliography (you may need to adjust the spacing yourself).


calculus
integration
geometry
3D geometry
sphere
analytic geometry
trigonometry

I am trying to find the volume of a sphere with equation
x^2 + y^2 + z^2  = a^2

I know it can be done with spherical coordinates but I need to do it with rectangular coordinates.

Since I can't seem to figure out right now how to get integral signs into my post.  I will attach a word doc with the integral I am trying to do.

I did the first integral fine cause that is an easy one.  How in the world would you integrate with respect to y^2 while there is a Z^2 under the sqrt also?  I would do trig sub if the z^2 wasn't there.

Any help would be greatly appreciated.  Let me know if you this is confusing in any way.

[hide]u can find the volume by revolving the semicircle $x^2+y^2=a^2$, $y\geq 0$, about the $y$-axis.

Since $y\geq 0$, this becomes $y=\sqrt{a^2-x^2}$.

Use the disc method (split into little cylinders instead of rectangles). The volume becomes $\pi \int_{-a}^a \left( \sqrt{a^2-x^2} \right)^2 dx=\pi \int_{-a}^a a^2-x^2 dx=2\pi \int_0^a a^2-x^2 dx$

$=2\pi \left[a^2x-\frac{1}{3}x^3\right]_0^a=2\pi\left(a^3-\frac{1}{3}a^3\right)$

$=\boxed{\frac{4}{3} \pi a^3}$.[/hide]

I appreciate the response and I realize I wasn't clear enough.  I need to use a triple integral to prefrom this.  So I am pretty sure the intregral I have is correct (feel free to tell me otherwise) but just don't know how to evaluate it.  Thanks for the help.


geometry
3D geometry
sphere
conics
trapezoid

I found 2 problems I couldn't solve :(

$1)$ We have a conoidal with it's latheral area $36\pi$ and with volume of $56\pi$. Knowing that the sum of it's bases radii is equal to 3 times of their difference, find:
$a)$ $R,r$ and $h$.
$b)$ The volume of the entire cone.

$2)$ In a cube we inscribe a sphere, in that sphere we inscriebe another cube and so on. Proof that the volumes of the cubes creates a geometric progresion.

Thanks!

2
[hide]Let the edge length of the first cube be $e$. $e$ is then easily seen as the space diagonal of the next cube. Then the $\text {n}th$ cube then has an edgelength of $\frac {e}{(\sqrt 3)^n}$[/hide]

I wrote the second one because I wanted that this post to be longer :P. The First one gave me a headache  :help:

Is that a truncated cone?
If it is
[hide]$R + r = 3R - 3r \Rightarrow r = \frac R2$

The volume of the cone is then

$\frac 13 \pi h (R^2 + \frac 14 R^2 + \frac 12R^2) = 56 \pi \Rightarrow \frac 7{12} R^2 h = 56$

The lateral surface area is

$R \sqrt {R^2 + h^2} - \frac 12 R \sqrt {\frac 14 R^2 + \frac 14 h^2} = 36 \Rightarrow \frac 34 R \sqrt {R^2 + h^2} = 36$

Then solve the messy system.[/hide]

[quote="Slizzel"]You got an ugly system.  :read: ... I'll try to solve it[/quote]

Do not spend too much time on it, it leads to a cubic equation with no rational roots. Let $H$ be the height of the full cone, $V$ its volume and $v$ the volume of conic frustrum. The frustrum volume is the difference of volumes of the full and cut-off cones:

$v = \frac H 3 \pi R^2 - \frac{H - h}{3} \pi r^2$

$\frac H R = \frac{h}{R - r},\ \ H = h \frac{R}{R - r}$

$v = \frac{\pi h}{3} \left (\frac{R^3}{R - r} - \frac{R - (R - r)}{R - r} r^2 \right) = \frac{\pi h}{3} \cdot \frac{R^3 - r^3}{R - r} = \frac{\pi h}{3} (R^2 + Rr + r^2) = 56 \pi$

The lateral surface of the frustrum is calculated like the area of a trapezoid with bases $2 \pi R,\ 2 \pi r$ and altitude $\sqrt{h^2 + (R - r)^2}$:

$s_L = \frac 1 2 (2 \pi R + 2 \pi r) \sqrt{h^2 + (R - r)^2} = \pi(R + r) \sqrt{h^2 + (R - r)^2} = 36 \pi$

$R + r = 3(R - r),\ \ 4r = 2R,\ \ r = \frac R 2$

$\frac h 3 (R^2 + Rr + r^2) = \frac h 3 \left(R^2 + \frac{R^2}{2} + \frac{R^2}{4}\right) = \frac{h}{3} \cdot \frac 7 4 R^2 = 56$

$hR^2 = 96$

$\left(R + \frac R 2\right) \sqrt{h^2 + \left(R - \frac R 2 \right)^2} = \frac 3 2 R \sqrt{h^2 + \frac{R^2}{4}} = 36$

$R \sqrt{h^2 + \frac{R^2}{4}} = 24,\ \ R^2(4h^2 + R^2) = 48^2$

Substituting $R^2 = \frac{96}{h}$ into the last equation leads to the following cubic equation for the frustrum height $h$:

$\frac{96}{h} \left(4h^2 + \frac{96}{h}\right) = 48^2$

$4h + \frac{96}{h^2} = 24,\ \ h^3 - 6h^2 + 24 = 0,\ \ (h - 2)^3 - 12(h - 2) + 8 = 0$

The only rational roots $h - 2$ the last cubic equation can have are integers $\pm 1,\ \pm 2,\ \pm 4,\ \pm 8$. None of them is a root. Hence, you have to solve numerically. The equation has 3 real roots $h_1 \doteq 5.0642,\ h_2 \doteq 2.6956,\ h_3 \doteq -1.7578$, the negative one not acceptable. Then

$R_1 = \sqrt{\frac{96}{h_1}} \doteq 4.3539,\ \ R_2 = \sqrt{\frac{96}{h_2}} \doteq 5.9688$

On the other hand, the full cone volume can be easily calculated:

$V = \frac H 3 \pi R^2 = \frac{\pi h}{3} \cdot \frac {R^3}{R - r} = \frac{\pi h}{3} \cdot 2R^2 = \frac{2 \pi}{3} hR^2 = \frac{2 \pi}{3} \cdot 96 = 64 \pi$


inequalities
induction
Cauchy Inequality
inequalities proposed

For me, the following is one of the most beautiful inequalities:
   Prove that if the geometric mean of the positive numbers $ a_1,...,a_n$ is at least $n-1$ then the following holds:
  $ \frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n}\geq\frac{n}{1+G} $ where G is the geometric mean of the n numbers.
  An extraordinary problem and I just found a beautiful solution. I would like to see others. Believe me, this problem worths the effort!

I see no movement on this problem. Too bad, since anyone would be proud to find a solution for this problem. It is absolutely fantastic!

In my paper "Extension of some cyclic inequalities" from G.M. 10, 1991,
I proved this inequality by the induction way.
Now I have a new idea. Based on the substitutions $a_1=\frac{tx_2}{x_1}$,   ,
$a_n=\frac{tx_1}{x_n}$, the inequality becomes

$\frac{x_1-x_2}{x_1+tx_2}+...+ \frac{x_n-x_1}{x_n+tx_1}\geq0$,

where $t\geq n-1$. This one can be proved beautifully by induction, in the hypothesis $x_{n+1}=max(x_1,...,x_{n+1})$.

You are absolutely incredible with these solutions! Thank you very much, I learn a lot from you and your problems and solutions. I have another approach, based on Huygens inequality and some AM-GM. I will post it some other time.

1) Harazi, I'm still waiting to see your beautiful solution  :) .

2) Is there a simple way to prove that $\frac{n-1}{1+x^n}+\frac{1}{1+y^n} \geq \frac{n}{1+x^{n-1}y}$ where $x^{n-1} y \geq n-1$?  If there is, then I will also have a nice solution because I have a neat way of forcing $n-1$ of the $a_i$ to be equal.

Just suppose $a_i=e^{x_i}$ the inequality is just Jen-sen inequality

Your talks have ignited me,Harazi.Thanks,because I am really happy now. :D
Dear Michael Lipnowski,
It seems that you are going to do by induction.So I will help you for the rest:
[tex] \frac{n-1}{1+x^n}+\frac{1}{1+y^n} \geq  \frac{n}{1+x^{n-1}y} [/tex]
[tex] \Leftrightarrow nx^{n-1}y-(n-1)x^n-y^n + x^{n-1}y[(n-1)y^n+x^n - nxy^{n-1}] \geq 0 [/tex]
Replace $x^{n-1}y=(n-1)$ then we must prove that:
[tex] nx^{n-1}y-(n-1)x^n-y^n +(n-1)[[(n-1)y^n+x^n - nxy^{n-1}]\geq 0[/tex] [tex] \Leftrightarrow x^{n-1}y+(n-2)y^n - (n-1)xy^{n-1}\geq 0 [/tex] (The last one is Cauchy inequality)
Now we have done,right  :)  ?

Omid Hatami,how can you prove that:
[tex] \frac{1}{1+e^{x_i}} [/tex] is convex.
It is not true  for $x_i+x_j<0$ :? .

Omid Hamati: $f(x) = \frac{1}{1+e^x}$ is not convex everywhere, only for $x \geq 0$.

Anh Cuong: I had in mind a revised version of the approach Omid is using (definitely not induction).  Make the substitution $a_i = e^x_i$.
$f(x) = \frac{1}{1+e^x}$ is convex for $x \geq 0$ and concave for $x \leq 0$.  Obviously, if the $x_i$ are all non-negative, then direct Jensen works, just as Omid suggested.  If not, then fix $G$, and so the sum of the $x_i$.  Applying the Majorization inequality, because of $f$'s shape, the right hand side is minimized when $n-1$ of the $x_i$ are equal.  That's where the last question I posed comes in handy, the case where n-1 of the x_i are equal.  With that, we've just proven the result very cleanly, without induction (though I must admit that Vasc's induction is beautiful).

So,I will post my induction solution.
If $a_i \geq 1 ,\forall i$ then it is trivial.Unless,assume that:$a_n\leq1$ then:
$a_1.a_2...a_{n-1} \geq a_1a_2...a_n \geq (n-1)^n \geq (n-2)^{n-1}$
$\Leftrightarrow ^{n-1}\sqrt{a_1...a_{n-1}}\geq n-2$
So we can have:
$f(a_1,...,a_{n-1},a_{n}) \geq f(^{n-1}\sqrt{a_1...a_{n-1}},...,^{n-1}\sqrt{a_1...a_{n-1}},a_{n})$
Hence,let $x=^{n-1}\sqrt{a_1...a_{n-1}},y=a_n$
For the rest,I have solved it :)
And what is Majorization inequality,could you please tell me Michael Lipnowski?

I am very interested in the proof by induction of Vasc

Has anyone a solution for

$ \frac{x_n-x_{n+1}}{x_n+tx_{n+1}} +\frac{x_{n+1}-x_1}{x_{n+1}+tx_1} \geq  \frac{x_n-x_1}{x_n+tx_1} $

I think this is a formidable challenging !!!!!!!!!

Thank you very much Anh Cuong to have indicated my misprint.

Dear Manilo,maybe I am wrong but I think you mean that:
[tex] \frac{x_n-x_{n+1}}{x_n+tx_{n+1}}+\frac{x_{n+1}-x_1}{x_{n+1}+tx_1} \geq \frac{x_n-x_1}{x_n+tx_1}[/tex]
where [tex] t \geq n [/tex]  and [tex] x_{n+1}=max \lbrace{x_1,...,x_{n+1}}\rbrace [/tex],right? :?

We get
$(x_{n+1}-x_1)(x_{n+1}-x_n)(x_1t^2-x_n)\geq 0$.
It remains the case
$x_n>t^2x_1$.
Who find a nice end ?

I hope I won't dissapoint you with the following solution and I apologise for posting it so late. Let us take a number $ N\geq 0$ such that $ 1+\sqrt[n]{a_1\cdot...\cdot a_n}=n(N+1) $. Suppose we don't have $\displaystyle \frac{1}{1+a_1}+...+\frac{1}{1+a_n}\geq\frac{1}{1+N} $ then we can find a number k<1 and $ x_i$ positive numbers which add up to 1 such that $\displaystyle\frac{1}{1+a_i}=\frac{k\cdot x_i}{1+N} $. We find that $\displaystyle (n(N+1)-1)^n=\prod_{i=1}^n {(\frac{1+N}{k\cdot x_i}-1)}>\prod_{i=1}^n {(\frac{1+N}{x_i}-1)} $. Using Huygens inequality, we infer that $\displaystyle\prod_{i=1}^n {(\frac{1+N}{x_i}-1)}\geq (\frac{N}{\sqrt[n]{x_1\cdot...\cdot x_n}}+\sqrt[n]{\prod_{i=1}^n {\frac{1-x_i}{x_i}}})^n$. I hope that you agree that now the problem is trivial: $\displaystyle\sqrt[n]{x_1\cdot...\cdot x_n}\leq\frac{1}{n}$ and $\displaystyle\sqrt [n]{\prod_{i=1}^n {\frac{1-x_i}{x_i}}}\geq n-1 $.

Ha ha, that's so amusing. I forgot to give logout gil and to write login harazi and so I posted the solution as gil. Well, it's me again, harazi, speaking from behind the courtains.

harazi...what is huygens inequality?
nice solution anyway :)

Peter

It says that $\sqrt[n]{(1+x_1)\cdot (1+x_2)\cdot...\cdot (1+x_n)}\geq 1+\sqrt[n]{x_1\cdot...\cdot x_n} $. Of course, this can be written as:
  $\sqrt[n]{(a_1+b_1)\cdot...\cdot (a_n+b_n)}\geq\sqrt[n]{a_1\cdot...\cdot a_n}+\sqrt[n]{b_1\cdot...\cdot b_n} $.

So it's just a special case of Hlder ... why then give it a special name?

I don't know if it is right to say or not Huygens or Holder inequality and sincerely I don't care. In fact, so many inequalities with different names can be proved using inequalities with other names... But I don't find this very important. Important is the fact that this inequality is very productive and has beautiful applications.

Huygens Inequality has a short beautiful proof with AM-GM.
That's why it's a difference between it and Holder

I think Holder have a short proof by AM-GM too. :)

Anh Cuong says it is trivial for a_i >1  (i=1,..,n).

I don't understand that.

Please, can you help me.

Thank you very much.

hlder(one writes him with the german vocal  but i think that letter is only used on german keyboards) actually can be proven easily with AM-GM: let a, b be arbritarily and $p,q>0$ such that $\frac{1}{p}+\frac{1}{q}=1$. then $\frac{a}{p}+\frac{b}{q}\geq a^{\frac{1}{p}}b^{\frac{1}{q}}$ by AM-GM. now set $a=\frac{a_i}{(\sum a_k^p)^{\frac{1}{p}}}$, $b=\frac{b_i}{(\sum b_k^q)^{\frac{1}{q}}}$ and sum up over all $i=1,...,n$. the left side gets one, multiply with the denominator of the right and *tada* you have the hlder inequality.

Peter

Huygens Inequality has a different proof by AM-GM and much nicer


algebra
polynomial
superior algebra
superior algebra unsolved

Why does $ x^{n\minus{}1} \plus{} x^{n\minus{}2} \plus{} \cdots \plus{} x \plus{} 1$ irreducible over $ \mathbb{Z}$ imply that $ n$ is prime?

If $ n\equal{}j\cdot k$, it factors as $ (x^{j\minus{}1}\plus{}x^{j\minus{}2}\plus{}\cdots\plus{}x\plus{}1)(x^{j(k\minus{}1)}\plus{}x^{j(k\minus{}2)}\plus{}\cdots\plus{}x^j\plus{}1)$.


induction
algebra
polynomial
function
Putnam
Rational Root Theorem

Note: A[n] will denote A sub n, or A[size=75]n[/size]

Let F(x) = A[n]x^n+A[n-1]x^(n-1)+....+A[1]x+A[0] be a polynomial function with integer coefficients. Let r be a rational number such that F(r)=0. Prove that: A[n]x, A[n]x^2+A[n-1]x, A[n]x^3+A[n-1]x^2+A[n-1]x, ... , A[n]x^n+A[n-1]x^(n-1)+...+A[2]x^2+A[1]x are all integers.

*I'm pretty sure that this can be done by induction. However, I don't know where to start.

PS. Happy New Years =]

I saw it some days ago in the Advanced Section. Try to search.

It's Putnam 2004 B1.  If it looks, tastes, and smells like the rational root theorem, then you might want to approach it in the same manner as you would the proof of the rational root theorem.


I have never heard of AwesomeMath before, so could somebody post some questions from it, so I know more about it?

Okay, I get it google is my friend. I am just always too busy *cough*lazy*cough* and I am never having time. I have a lot of things to do lately... ( :sleeping: )

it doesnt take more than 30 seconds to type "www.google.com" then type "awesomemath"

jst go to

http://www.awesomemath.org

wow im so nice.
not.

doesn't it take longer to get here and read these posts than to google it?   :wink:

ummmm... awesomemath is awesome (duhhh)
and go to the website & google (also duhhh)

Alright, you guys don't need to say it 5 times on one topic.

[quote="Altheman"]DBR posted the problems from the team contest so you can look at those. Also there are some problems in the geometry take-home test (which nobody has done BTW).[/quote]

Alright, I will take a look at it. Thanks for helping.


I was trying to post a picture on the AoPSWiki, but I can't figure out how. Can someone please explain how? Thanks! :D


calculus
derivative
integration
real analysis
real analysis unsolved

f(x) is twice-differentiable, f(0) = 0, f(1) = 1, f'(0) = 2, |f''(x)| is bounded by 2 for all x. Is it possible to find f''(1/2)?

When you say "for all $ x$," do you mean for all $ x\in [0,1]$ or for all $ x\in \mathbb{R}$? If nothing else I would guess that using $ f(x)\equal{}x\cdot (2\minus{}x)\Rightarrow f''\left( \frac{1}{2}\right)\equal{}2$ is what they were going for, but it might be a bit of a challenging proving that $ f''\left( \frac{1}{2}\right)\equal{}2$ is the only possible value of $ f''\left( \frac{1}{2}\right)$.

After a little bit more thought, $ f''\left( \frac {1}{2}\right)\equal{}\minus{}2$ is the only possible value of the second derivative at $ x\equal{}\frac{1}{2}$, and in fact $ f''(x)\equal{}\minus{}2$ for $ x\in [0,1]$. To see this, we proceed as follows: by assumption, $ f''(x)\geq \minus{}2$ for $ x\in [0,1]$. Assume $ f''(x)>\minus{}2$. Then by integrating we see that:

$ f'(x)\equal{}2\plus{}\displaystyle\int _0^x f''(x)\cdot dx>2\minus{}\displaystyle\int _0^x 2\cdot dx\equal{}2\minus{}2x$

$ f(x)\equal{}0\plus{}\displaystyle\int_0^x f'(x)\cdot dx>\displaystyle\int_0^x(2\minus{}2x)\cdot dx\equal{}2x\minus{}x^2$

Thus, $ f(x)>2x\minus{}x^2\rightarrow f(1)>1$, a contradiction. Thus, $ f''(x)\equal{}\minus{}2$ for $ x\in [0,1]$, and $ f''\left( \frac {1}{2}\right)\equal{}\minus{}2$ follows trivially.


When I try to post an image from a gallery, it doesn't show up:

[img]http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=12881[/img]

[quote="Valentin Vornicu"]That's not the correct syntax. It's [code][img]12881[/img][/code]

 [img]12881[/img][/quote]

Ahhh, thanks, Valentin!