Find the sum of all digits of 111....111^2  (there are a total of 1000 1's)

[quote="Suou"]The sum: $ 1^{2} \equal{} 1$, $ 11^{2} \equal{} 4$, $ 111^{2} \equal{} 9$, $ 1111^{2} \equal{} 16$
$ 4 \minus{} 1 \equal{} 3$, $ 9 \minus{} 4 \equal{} 5$, $ 16 \minus{} 9 \equal{} 7$, odd numbers.
$ 1 \plus{} 3 \plus{} 5... \plus{} 1999 \equal{} 1 \plus{} (n \minus{} 1).2$ and $ 1 \plus{} (1000 \minus{} 1).2 \equal{} 1999$[/quote]

I didn't understand the solution. Can you explain more?

What is the final answer? $ 1999$ or $ 1999^2$

Suou wrote: The sum: $ 1^{2}\equal{}1, 11^{2}\equal{}4, 111^{2}\equal{}9, 1111^{2}\equal{}16$

Pattern...$ 1^2\equal{}1$ There is [u]one[/u] 1 digit. $ 11^2\equal{}4$ there are [u]two[/u] 1 digits... $ 111^3\equal{}9$ there are [u]three[/u] 1 digits.
Based on this pattern, shouldn't it just be $ 1000^2\equal{}1,000,000$?

Note that this is also equivalent to finnding the sum of the coefficents of the polynomial
$ (x_{1}\plus{}x_{2}\plus{}x_{3}\dots\plus{}x_{999}\plus{}x_{100})^2$, for some real numbers $ x_{1}$, $ x_{2}$, $ x_{3}$ $ \dots$, $ x_{999}$, $ x_{1000}$. Thus, if we let $ x_{1}\equal{}x_{2}\equal{}x_{3}\equal{}\dots\equal{}x_{999}\equal{}x_{1000}\equal{}1$, we get the result as $ 1000^2\equal{}\boxed{1000000}$.

gaussintraining and LincolnMS nerd herd's solutions don't work where the number of 1's surpasses 9.

ex: 10*10=1*100

ex2: f(1111111111^2)=82

I see one pattern. While we do $ 11^{2}$,$ 111^{2}$...note that was forming a sequence: $ 11...1$ with $ 10$ ones has $ 12346790$ one time and $ 987654320$ one time, plus $ 1 \Rightarrow 82$; then $ 11...1$, $ 100$ ones has $ 199$ digits (in each square $ 11$, $ 111$, we add $ 2$ digits in the perfect square) and $ \frac {199}{18} \equal{} 11$ sequences of each type plus $ 1$. To $ 1000$ ones we have $ 199 \plus{} 2.900 \equal{} 1999$ digits, $ \frac {1999}{18} \equal{} 111$ sequences each type plus $ 1$.
The sum is $ 111.(1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 9 \plus{} 0 \plus{} 9 \plus{} 8 \plus{} 7 \plus{} 6 \plus{} 5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 0) \plus{} 1 \equal{} 111.(81) \plus{} 1 \equal{} 8992$.
 :?:

I've heard to use mod 9 to solve this problem. And also that this problem is way beyond the level of this forum.


inequalities
trigonometry
geometry proposed
geometry

Let $ ABC$ be an acute triangle. Prove that $ \sum (\sin 2B\plus{}\sin 2C)^2\sin A\leq 12\prod \sin A.$

I 've tried converting the problem into algebraic inequality:

$ \sum{{(a^4\minus{}(b^2\minus{}c^2)^2\plus{}(a\plus{}b\plus{}c)\prod{(a\plus{}b\minus{}c)})^2}a} \leq 48 a^3 b^3 c^3$

But it is useless and inelegant. I think there should be a geometric approach.


rotation
complex analysis
complex analysis unsolved

Let $ G\equal{}\{ z: |z\minus{}1|<\sqrt{2}, |z\plus{}1|<\sqrt{2}\}\setminus [0,i)$, find a bijective conformal map from G to the upper half-plane $ \mathbb{H}$

Any idea is appreciated. Thanks

Consider the LFT   z |->  z/(z+i)  which sends 0 to 0, -i to infinity, and i to 1/2.  ({Lines and circles} to {lines and circles}.)  The segment [0,i) is sent to [0,1/2).  Consider this new region with the new omitted segment.  Shift left by 1/2, rotate clockwise so that the upper boundary ray is along the positive (real) axis, compose with z^t (some real t) so that the segment is again along the negative (real) axis.  Now shift right and compose with z^(1/t') to bring the lower boundary ray back to the negative (real) axis.
-MrB


Mr.Franks gave a final exam worth 200 points. The distribution of scores is shown below. What was the median score on the exam?
\[ \begin{array}{c | c c c c }
 & & & & \\ \hline
11 & 0 & 3 & 9  \\
12 & 2 \\
13 & 4 & 4 & 5 \\
14 & 6 & 7 \\
15 & 1 & 1 & 2 & 4 \\
16 & 7 & 8 \\
17 & 0 & 4 \\
18 & 0 & 4 & 4 & 4 \\
19 & 5 & 7
\end{array}\]

The median score is the middle score. In this chart, we see the median is 152.


There are 2005 points given on a plane. Consider all the midpoints of line segments joining any two points among these 2005 points. Find the minimum possible number of such midpoints.

[hide]Set each point on a line equally spaced from one another.  Let's set the distance from one point to another as a unit.  Therefore, if this was a number line, each midpoint between two adjacent points is 0.5 1.5 2.5 ... from the first point (the point called 0).  However, the space between any two non adjacent points is either part of the new midpoints drawn in or of the original lines because the distances are natural numbers.  Since every two points forms a midpoint, there's 2004 new midpoints which is the minimum.[/hide]
What's the maximum number of midpoints of this configuration?


Should that (=) be a plus or a minus? And this doesn't belong here, maybe Basics or something. ;)

Chess64,

There is an = inside the radicand.  This question is from the radical equation section in my high school math book.
Can it be a typing error?  I have never seen a radical equation like this.

Enough discussing about the typo.

If it is $\sqrt{x^2+5x+5}-1=0$, then $\sqrt{(x+\frac{5}{2})^2-\frac{5}{4}}=1$. Therefore $(x+\frac{5}{2})^2-\frac{5}{4}=1$, then $(x+\frac{5}{2})^2=\frac{9}{4}$. So:
\[x+\frac{5}{2}=|\frac{3}{2}|\]
So $x=-1$ or $x=-4$. Put them in $(x+\frac{5}{2})^2-\frac{5}{4} \geq 0$, we know that they are both roots of the equation.


calculus
integration
function
Euler
Asymptote
arithmetic series
geometric series

Does anybody know how to solve this without adding lots of fractions?
Is there a pattern?
1/(1^2)+1/(2^2)+1/(3^2)+...+1/(n^2) is greater than or equal to 3/2
What is the smallest possible positive integral value of n?

[quote="Dr Sonnhard Graubner"]hello, we get
$ \sum_{i \equal{} 1}^{6}\frac {1}{i^2} \equal{} \frac {5369}{3600} < \frac {3}{2}$
and
$ \sum_{i \equal{} 1}^{7}\frac {1}{i^2} \equal{} \frac {266681}{176400} > \frac {3}{2}$
so your searched $ n$ is $ 7$.
Sonnhard.[/quote]

What formula did you use to get that fraction from the sigma if you used a formula. What is it called and what do you use it for?

Is sigma really in MC? just wondering :(

Not directly, but some problems in MC can be solved with summations/partial sums. (Its probably only for nats/hard states, but I may be wrong)

Like one question went something like:

Find: 1/(1x3)+1/(3x5)+1/(5x7)+...+1/(99x101)
Which can be solved using partial sums.

Wait, I just realized you dont even need partial sums for this. You can do this problem non-legitly pretty easily.

And there was also one nats question that asked you for the sum of an infinite arithmetic series, which doest require summation notation but can be written as a summation.

Actually, at nationals, it would probably have been a geometric series, because infinite arithmetic series tend to $ \pm\infty$, unless $ d\equal{}0$.

Infinite geometric series, on the other hand, can have a limit at a finite value, when $ |r|<1$.

Dr. Sonnhard Graubner, would you mind posting the formula for

\[ \displaystyle\sum_{i\equal{}1}^{n}\frac{1}{i^2}\]?

I just want to note that $ \displaystyle\sum_{i \equal{} 1}^{\infty}\frac {1}{i^2}$ is $ \frac {{\pi}^2}{6}$.  For more information see [url]http://en.wikipedia.org/wiki/Riemann_zeta_function[/url].

And if the upper limit is n instead of $ \infty$, the answer is the nth harmonic number of the second order, by definition.  Otherwise, I don't think there's really a general formula.  Please correct me if I am wrong.

Another argument that there is no general formula, based on the zeta function:

zeta(2), which is $ \sum_{i \equal{} 1}^{\infty}\frac {1}{i^{2}}$, or more generally, zeta(s), which is $ \sum_{i \equal{} 1}^{\infty}\frac {1}{i^{s}}$, are solved with complicated partial fractions (compliments of Euler), and I know fairly certainly that there is no general formula for which you can find horizontal asymptotes and therefore evaluate $ \sum_{i \equal{} 1}^{n}\frac {1}{i^{2}}$ or $ \sum_{i \equal{} 1}^{\infty}\frac {1}{i^{2}}$.

Well, I found the (Riemann Zeta Function) identity if the upper limit is $ \infty$, but as far as the upper limit being $ n$, I only found this:

\[ \displaystyle\sum_{i\equal{}1}^{\infty}\frac{1}{i^2}\equal{}\frac{\pi^2}{6}\] (The Riemann Zeta Function) and
\[ \displaystyle\sum_{i\equal{}1}^{n}\frac{1}{i}\equal{}H_n\]

where $ H_n$ is the $ n$th Harmonic Number.

I was just wondering, in the case that the denominator is $ i^2$, not $ i$, whether the result would be fairly simple, involving Harmonic Numbers, or whether would it be much more convoluted.

As I noted, when the denominator is $ i^2$, it is a harmonic number of the second degree.  However, that is just by definition, and does not imply a general formula.

Actually, I tried generalizing $ \displaystyle\sum_{i\equal{}1}^{n}\frac{1}{i^2}$ for all n on a computer algebra system.  It returned $ \frac{{\pi}^2\minus{}6\cdot\psi(1\plus{}p)}{6}$.  Polygamma is defined by $ \psi(z)\equal{}\frac{\Gamma'(z)}{\Gamma(z)}$.  For more information on polygamma, see[url]http://mathworld.wolfram.com/PolygammaFunction.html[/url].


WHo is the AIME problem series aimed at? PEople who can get 1-5, but not 6-15? People who can get all but the last 5?

[quote="Phelpedo"]If you can get a 10 consistently, your probably dont need to take it.[/quote]
I think I am going to take it...

what score is the minimum you should average so that you don't really need the class?

Why take AIME problem series if you can get a 12 (consistently, not just by guessing all the right answers) ? Just focus on Olympiad problems. The AIME's not really all that important.

[quote="calc rulz"]I'd say at least 11-12. You could certainly gain something from it if you're getting around 10, and possibly slightly higher.[/quote]

What?? I think if you're getting 9+, it's time to stop studying for the AIME.

[quote="kyyuanmathcount"]I guess you study until you fulfill all personal goals.  I would still take the class if I got 11's or 12's.[/quote]

If you get 11's or 12's, then the class wouldn't help you, so it would be futile.

[quote="kyyuanmathcount"][quote="nebula42"][quote="kyyuanmathcount"]Not really.  Personally, I think that even though, if you get 11's and 12's which is pretty much a dream for me right now, you might know how to do the problems in the class, the class will still help you by introducing important ideas about how to think about more complicated types of similar problems, such as those that would appear in #14's and #15's.[/quote]
That is true, but it would definitely be more advantageous for people at that level to take WOOT instead. WOOT focuses largely on preparing for USAMO, which is what people at that level would also be focusing on.[/quote]

 :)  I agree.  AIME problem series covers all topics briefly, while W00T does more in depth.[/quote]

Actually that's not what he's saying (there's not too much difference in that aspect of the classes anyway). WOOT is geared towards those who are studying for olympiads like USAMO, while the AIME problem series is geared towards those who are trying to qualify for USAMO.

You should stop taking the class when you fell confident you can, on average, qualify for USAMO in my opinion.  I believe that the class is ideal for people who are currently getting middle range scores.  People just getting started with AIME will still learn a lot and practice never hurts even when you're getting double digit scores (although you could probably practice all on your own at that point).

[quote="Karth"][quote="kyyuanmathcount"]Not really.  Personally, I think that even though, if you get 11's and 12's which is pretty much a dream for me right now, you might know how to do the problems in the class, the class will still help you by introducing important ideas about how to think about more complicated types of similar problems, such as those that would appear in #14's and #15's.[/quote]

If you're able to get 11's or 12's, I'm not really sure if AIME Problem Series would be as much of a better choice that WOOT since problems 11 to 15 tend to deal with beginning olympiad problems anyways. Take w00t. w00t pwnz.[/quote]
why not just take them both?
It's just an extra 100 bucks if you take the AIME class and no extra money if you signed up for w00t early

And a lot of time that most of us don't have  :wink:

A $100$ dollars that I don't think many of us have lying around, though I wish I did  :P


probability
analytic geometry
graphing lines
slope

1. A man leaves for work at a random time b/w 7:00AM and 8:00Am. His paper arrives at his house at a random time b/w 6:30 AM and 7:30 AM. What is the probability that he has his newspaper before he leaves for work?

I draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could you please explain this one to me?

Think about the case where the paper comes just as the man leaves the house.  This should be the very edge of the shaded region - the case where the paper time = departure time.  To figure out where the line will be, try plotting a few points:  Where is 7 on the horizontal axis? on the vertical axis?  Where is 7:30 on the horizontal axis? on the vertical axis?  You should end up with a diagonal line (slope 1) in the lower righthand section of your square.

To reason it out, you could also consider a few cases:

1.  (Probability 1/2)  The paper arrives by 7:00, and the man definitely gets it before work.
2.  (Probability 1/2)  The paper arrives after 7:00.
     a. (Probability 1/2) The man leaves after 7:30, and definitely gets his paper before work.
     b. (Probability 1/2) The man leaves by 7:30 but after 7:00, just like the paper's arrival, so he has a 1/2 probability of recieving it before work.


[quote="might"]Hi, I have to do my one test corrections. I got one problem with the table, i got it wrong. Now i gotta to explain my error. Can same body help me?
I chose the answer B). Which answer is correct and please explain my error.Thanks

sorry for three same docks.[/quote]

[hide]Wait, are you saying you chose B on the test and you got it wrong?? because choice B is correct
x=1,2,3,4,5
y=1,4,9,16,25 which is what is in the table..

so $y=x^2$[/hide]


Oklahoma because Chimmy Chong Jong is on it and he asked this not-so-repectable question

Yeah.

Maybe Individual is okay, but Team is not that good of a question.

One of the tiny states, territories, or departments.

Notice how I'm not being too specific.


geometry
geometric transformation
homothety
rectangle
geometry proposed

Squares $ BCC_1B_2; CAA_1C_2; ABB_1A_2$ are constructed from the outside at sides of triangle $ ABC$ and $ O_A;O_B;O_C$ are the centres of these squares. Let $ A_0;B_0;C_0$ be the intersection points of the straight lines $ A_1B_2$ and $ C_1A_2$; $ A_1B_2$ and $ B_1C_2$; $ C_1A_2$ and $ B_1C_2$ respectively.
Prove that the straight lines $ O_AA_0; O_BB_0$ and $ O_CC_0$ are concurrent.

It is suffice to show  triangle $ O_AO_BO_C$ and triangle $ A_0B_0C_0$ are homothety but it is obvious since $ O_{C},O_{B}$ are midpoint of $ AB_1,AC_2,O_{C}O_{B}\parallel B_{1}C_{2}.$

This problem is also true, in the general configuration of arbitrary rectangles, instead of squares, erected on the side-segments of the given triangle $ \bigtriangleup ABC,$ outwardly or inwardly to it, or some of them outwardly and the other inwardly.

Alternatively, it is easy to show that the triangles $ \bigtriangleup A_{o}B_{o}C_{o}$ and $ \bigtriangleup O_{a}O_{b}O_{c}$ are perspective, because of their sidelines, are parallel one per one.

Kostas Vittas.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 17
Sincerely
Jean-Louis


Suppose $\{x_n\}$ is a decreasing sequence that $\displaystyle\lim_{n \rightarrow\infty}x_n=0$. Prove that $\sum(-1)^nx_n$ is convergent

It is known in France as the 'theorème spécial pour les séries altérnées', and it is part of any course on series...

Pierre.

In english, it's the alternating series test.

The idea is that we can put $\sum^{n}_{i=0} (-1)^i=A_n$, and then write

$\left|\sum^q_{i=p} (-1)^ix_i\right|=\left|\sum^{q-1}_{i=p} A_i(x_i-x_{i+1})-A_qx_q-A_{p-1}x_p\right|$

$A_i\leq 1$ for each $i$, so we have

$\left|\sum^{q-1}_{i=p} A_i(x_i-x-{i+1})-A_qx_q-A_{p-1}x_p\right|\leq \left|\sum^{q-1}_{i=p} (x_i-x_{i+1})-x_q-x_p\right|=2x_p$

But as $p\to\infty$, $x_p\to 0$, so we have convergence from the Cauchy criterion (that is, Cauchy sequences converge in $\mathbb{R}$).

$A_{2n}$ decreasing and $A_{2n+1}$ increasing
 $A_{2n+1}-A_{2n}=-s_{2n+1}$ $\Rightarrow $ QED

Why do I think the only sequence is Xn=0 since Xn is decreasing?


geometry
ratio
analytic geometry
ARML
similar triangles
congruent triangles
Pythagorean Theorem

Given a square of area [i][b]A[/b][/i], each vertex is joined clockwise to the midpoint of one opposite side.  Describe the figure formed in the center and find its area. Show as many solutions as possible.

i think it is [hide]$\frac{1}{5}A$[/hide]

a simple coord bash will do this  :D  im trying it out with similar triangles atm. the shaded shape is a square because there are tonnes and tonnes of congruent triangles, and you find sides of the thing are equal and all 90 degrees.

[hide]

The answer is $\frac{1}{5} A$

Shortened soln: all four triangles, formed by three vertices; a vertex of the square, an adjacent vertex and a midpoint of an adjacent side of the line formed by the initial two points are congruent. So just look at one of them, with the two lines going through them. Now within this triangles, there are three similar triangles

ratio of the sides is $1: 2: \sqrt{5}$ by pythag.

work with this and if we assume a side length is $\sqrt{5} \implies$ side of square $=1$ hence area is one fifth[/hide]

Guile, you've come across a great proof of the pythagorean theorem. As for your question:

[hide]
All the segments in the large square intersect a right angles; if you don't believe this, set the image to coordinates and compare slopes.

The shaded figure is a square. (This can be shown a number of ways by similarity, but the basic proof is that every intersection is a right angle and every intersection is identical to every comparable segment because the intersections come from sides of a square and behave in the exact same way.)

Now let the shortest segment joining a corner of the large square and a corner of the small square be $a$, and let the side of the small square be $b$. The area of the entire figure can be expressed as the sum of the areas of the square and four congruent right triangles with legs $a$ and $a+b$.

Thus, the area = $4(\frac{1}{2})a(a+b)+b^2 = 2a^2+2ab+b^2.$
The area can also be expressed simply as the side of the large square squared. Thus the side of the large square squared is $2a^2+2ab+b^2= (a+b)^2+a^2$. Thus, we have the pythagorean theorem.

To get a more detailed description of the middle square, set the large square in the first quadrant with one corner at the origin. Call the length of a side of the large square $c$, and the four vertices are $(0,0)$, $(0,c)$, $(c,c)$, and $(c,0)$. The equations of the median lines, in clockwise order from the origin, are

$y=-\frac{1}{2}x+\frac{c}{2}$
$y=-\frac{1}{2}x+c$
$y=2x$
$y=2x-\frac{c}{2}$

By setting a couple of them equal we can get two adjacent points on the small square; for instance, two are $(\frac{2}{5}c, \frac{4}{5}c)$ and $(\frac{1}{5}c,\frac{2}{5}c)$. The distance formula gives the side $b^2=\frac{1}{5}c^2$, therefore the areas differ by a factor of $5$. [/hide]

[quote="bthings_2000"]
The distance formula gives...[/quote]

this is circular reasoning to prove the pythagorean theorem

Note that by moving around triangles, we form 5 squares. Defining the center square as $B$, we have: $[B] = \frac{[A]}{5}$

i had a problem like this for ARML homework only they gave you the area of the square and you had to find how the sides were divided but it was like an alot smaller square.

[quote="bthings_2000"]Altheman wrote:
[quote]this is circular reasoning to prove the pythagorean theorem [/quote]

Not so; my use of the distance formula took place AFTER I had proven the pythagorean theorem. I had moved on to an investigation into comparative areas. The proof of the pythagorean theorem ended before the coordinates were imposed, and thus the pythagorean theorem could be used without fear of circular reasoning.[/quote]

sorry...i suppose that i did not really read it that carefully... :blush:


choose 6 people out of 12 to sit into two columns of 3 seats each. if 3 of them only like to sit at the left hand side and 4 only like to sit at right and other 5 people willing to sit anywhere, how many ways of arrangements?


This is from an [u]old[/u] test back when they allowed calculators.

A positive integer which contains no prime factors except 2 and 3 can be divided
by exactly 24 different numbers and its square can be divided by exactly 77
different numbers. What is the integer?

I have no idea where to even begin on this problem.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273088[/url]


induction
number theory unsolved
number theory

Let $ m$ be positive integer. Prove that there are infinitely many $ n \in N^*$ which satisfy

$ n| m^n\minus{}1$

we defined the sequence $ x_k$ with: $ x_1\equal{}1$ and $ x_{k\plus{}1}\equal{}m^{x_k}\minus{}1$ ,$ k\equal{}1,2,...$
the problem can be proved by induction : $ x_k |m^{x_k}\minus{}1$      (*)
if $ k\equal{}1$ then (*) is obivious true
if $ x_k |m^{x_k}\minus{}1$ . we have to show that $ x_{k\plus{}1} |m^{x_k\plus{}1}\minus{}1$

But, $ {m^{{x_{k \plus{} 1}}}} \minus{} 1 \equal{} {m^{{m^{{x_k}}} \minus{} 1}} \minus{} 1 \equal{} {({m^{{x_k}}})^{\frac{{{m^{{x_k}}} \minus{} 1}}{{{x_k}}}}} \minus{} 1 \vdots {m^{{x_k}}} \minus{} 1 \equal{} {x_{k \plus{} 1}}$

the sequence $ x_k$ be strick increasing so that $ n\equal{}x_k$ , $ \forall k\equal{}1,2,3,...$

(Q.E.D)  :)

do not forget to mention $ m > 2$! :)
also the problem can be solved if we have the condition $ n | m^{mn\minus{}1}\minus{}1$.
use the same method,defining an appropiate sequence!

ADDED problem :
prove that for all $ m \in N,m > 2$ there exists an integer $ n$ which has exactly $ k$ distinct prime divisors and
$ n | m^{n} - 1$.
i wondered what condition(s) the set $ S = \{\alpha_1,...,\alpha_k\}^$ must have to make this statement true:
 there exists an integer $ n$ in the form $ p_1^{\alpha_1}...p_k^{\alpha_k}$ with the property that $ n|m^n - 1$ for a fixed $ m > 2$.it can be a nice problem too...

[quote="shoki"]ADDED problem :
prove that for all $ m \in N,m > 2$ there exists an integer $ n$ which has exactly $ k$ distinct prime divisors and
$ n | m^{n} \minus{} 1$.[/quote]

your problem is really nice! i am interested in seeing your full solution  :)

because of $ m>2$,$ m\minus{}1$ must have a prime divisor,let's call it $ p$ so we must have $ p^a |m^{p^a}\minus{}1$.
now we show that there exists another prime dividing $ m^{p^a}\minus{}1$ (distinct with $ p$) by increasing sufficiently the number $ a$.
let's suppose that for all $ a \in N$ the only prime divisor of $ m^{p^a}\minus{}1$ is $ p$.so we must have
$ m^{p^a}\minus{}1\equal{}p^k,m^{p^{a\plus{}1}}\minus{}1\equal{}p^{k\plus{}1}$,
(note that the exponent of $ p$ increases(+1) iff the exponent of $ m$ is mutiplied by $ p^1$).
so we must have $ \frac{m^{p^{a\plus{}1}}\minus{}1}{ m^{p^a}\minus{}1}\equal{}p$ for all $ a$ in $ N$.
now by increasing $ a$ sufficiently u can prove that there doesn't exist such equality,(prove it by using inequalities).
so there exists another prime dividing this number denote it by $ q$.so we have another number with the property
$ p^a.q|m^{p^a.q}\minus{}1$.
now for getting three distinct prime divisors do exactly the same thing as above(increasing the exponent of $ q$ and proving the existence of a third prime by supposing it doesn't exist and getting a contradiction by inequalities).
and we are done! :D

[quote="shoki"]because of $ m > 2$,$ m \minus{} 1$ must have a prime divisor,let's call it $ p$ so we must have $ p^a |m^{p^a} \minus{} 1$.
now we show that there exists another prime dividing $ m^{p^a} \minus{} 1$ (distinct with $ p$) by increasing sufficiently the number $ a$.
let's suppose that for all $ a \in N$ the only prime divisor of $ m^{p^a} \minus{} 1$ is $ p$.so we must have
$ m^{p^a} \minus{} 1 \equal{} p^k,m^{p^{a \plus{} 1}} \minus{} 1 \equal{} p^{k \plus{} 1}$,
(note that the exponent of $ p$ increases(+1) iff the exponent of $ m$ is mutiplied by $ p^1$).
so we must have $ \frac {m^{p^{a \plus{} 1}} \minus{} 1}{ m^{p^a} \minus{} 1} \equal{} p$ for all $ a$ in $ N$.
now by increasing $ a$ sufficiently u can prove that there doesn't exist such equality,(prove it by using inequalities).
so there exists another prime dividing this number denote it by $ q$.so we have another number with the property
$ p^a.q|m^{p^a.q} \minus{} 1$.
now for getting three distinct prime divisors do exactly the same thing as above(increasing the exponent of $ q$ and proving the existence of a third prime by supposing it doesn't exist and getting a contradiction by inequalities).
and we are done! :D[/quote]

it is very nice   :lol:


[quote="akai"]Let $ a,b,c\ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Find the min value of:

$ P \equal{} \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}$[/quote]
We have :

 $ \frac {a}{\sqrt {b \plus{} c}} \equal{} \frac {a}{\sqrt {3 \minus{} a}} \ge\ \frac {3a^2}{4\sqrt {2}} \plus{} \frac {1}{4\sqrt {2}}$

So $ P \ge\ \frac {3}{\sqrt {2}}$

to Onlylove_math : can you explain more about this :
$ \frac{a}{\sqrt{b\plus{}c}}\equal{}\frac{a}{\sqrt{3\minus{}a}}$
I'm very bad student  :(  :(  :(

[quote="akai"]Let $ a,b,c\ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Find the min value of:

$ P \equal{} \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}$[/quote]
By Holder Inequality, we have that
\[ P^2 \cdot \sum a(b\plus{}c) \ge (a\plus{}b\plus{}c)^3\]
Hence
\[ P^2 \ge \frac{(a\plus{}b\plus{}c)^3}{2(ab\plus{}bc\plus{}ca)} \equal{}\frac{1}{2}\sqrt{\frac{3(a\plus{}b\plus{}c)^6}{(ab\plus{}bc\plus{}ca)^2(a^2\plus{}b^2\plus{}c^2)}} \ge \frac{1}{2} \sqrt{\frac{3(a\plus{}b\plus{}c)^6}{(\frac{(ab\plus{}bc\plus{}ca)\plus{}(ab\plus{}bc\plus{}ca)\plus{}(a^2\plus{}b^2\plus{}c^2)}{3})^3}}\equal{}\frac{9}{2}\]
Hence
\[ P \ge \frac{3}{\sqrt{2}}\]
Equality holds for $ a\equal{}b\equal{}c\equal{}1$. :)

[quote="B_bnguyen"]to Onlylove_math : can you explain more about this :
$ \frac {a}{\sqrt {b \plus{} c}} \equal{} \frac {a}{\sqrt {3 \minus{} a}}$
I'm very bad student  :(  :(  :([/quote]
Oh, I must be $ \frac {a}{\sqrt {b \plus{} c}} \ge\ \frac {a}{\sqrt {3 \minus{} a}}$
But  $ \frac {a}{\sqrt {3 \minus{} a}} \ge\ \frac {3a^2}{4\sqrt {2}} \plus{} \frac {1}{4\sqrt {2}}$ is wrong
I'm sorry  :(

[quote="akai"]Let $ a,b,c\ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Find the min value of:

$ P \equal{} \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}$[/quote]
by Jensen $ P\ge (a\plus{}b\plus{}c)\frac{\sqrt{a\plus{}b\plus{}c}}{\sqrt{2(ab\plus{}bc\plus{}ac)}}\ge\frac{3}{\sqrt{2}}$

[quote="can_hang2007"]
By Holder Inequality, we have that
\[ P^2 \cdot \sum a(b \plus{} c) \ge (a \plus{} b \plus{} c)^3
\]
[/quote]
Could you write why? I don'y see it :oops:


calculus
trigonometry
function
derivative
integration
geometry
number theory

I heard some people in my school saying calculus and trigonometry are kinda same.. Is it true? If not, what's calculus anyway? And what's difference between calculus AB and calculus BA or something....

Lol whoever said that is an idiot. Yes sometimes calculus USES trigonometry, but they are far from the same...

Calculus is what makes everything else make sense.  It makes sense for where formulas come from, and why you learn certain things.  The most significant use of trigonometry is calculus.  It pops up in a lot of calc problems.

BC is harder.

Calculus is a branch of mathematics.  Roughly, what it does is tell you things about how functions behave.  In particular, it defines two operations, the derivative and integral.  The derivative is the equivalent of velocity -- it tells you how quickly a function is changing.  The integral tells you how much area is under a curve.  Alternatively, you can talk about it as the mathematics of limits -- what happens as things get very very close to something else.

Wow... Then is the calculus highest math branch before statitistics?

First of all, what is calculus? What really distinguishes it from other branches of math?

JLB had the right idea, that it has to do with limits. Heuristically, you can think of calculus as the study of "the infinitely small, and the infinitely many." Like the antithesis of Discrete Math, which is the study of "finite structures and finite things."

Although that's not an exact definition, it gets at what sets calculus apart from other branches of mathematics. Calculus is a branch of math that allows you to deal with "infinitessimal" properties of functions, like how they change from point to point; and of "uncountably large" properties like lengths and areas.

Calculus, especially as it's taught in high school, is very closely related to Geometry because it allows you to study curves and shapes. That's why it's sometimes called Analytic Geometry. There is a very deep and natural connection between Calculus and Geometry (that you'll learn in college) - but they are still not entirely the same thing. They just overlap a lot.

I don't understand what you mean by "highest math branch" though..

Math is taught (almost) chronologically in high school - from basic arithmatic, to Geometry, to topics covered in Algebra I and II, to the ideas of functions (pre-calculus), and finally to Calculus. By the time you graduate from high school, you will have been introduced to most of the math known in about the early 1700s (plus a few ideas that came later to make things more clear, like limits and complex numbers).

But statistics fits very oddly into that picture chrnologically. So I don't think you could necessarily say there's any "highest math branch" before statistics. It's sort of out on a branch by itself.

I kinda think of Calculus through the following analogy: Number Theory is to the Integers as Calculus is to the Real Numbers.

A lot of it comes from use of the Completeness Axiom, while in number theory a lot comes from the Well-Ordering Principle. And both are axioms of ordering and bounds, in some respect ... so they have seemed to me to be quite similar.

That is a good analogy Nuk. Of course you could also ask "number theoretic" questions about the real numbers, but I can see your analogy.


Is there anyway to show how many views there have been in a page?

The page view count is at the bottom of the page. For example "This page was last modified on 12 June 2009, at 08:55. This page has been accessed 8 times."

wow,:blush:,  :wallbash_red: I can't believe I didn't notice, I can see how the op could look a little dumb.


Consider a frictionless disk rotating with angular velocity w. A coin is placed on it. In ground frame, coin is stationary. Consider a man fixed to the disk. In his frame, he sees the coin going around with w. Question: How does he explain the motion?

well it depends on what frame the man chooses.if he is at the centre of the disc(and his frame is stationaryt) then it is same as earth.
if he is somewhere else and his frame rotates with him .then u can use the velocity transformation formula for a rotating frame to describe the body's motion.

But the question is how does he [u]explain[/u] the motion.
The frame of reference  in which the man is at rest is non-inertial. The forces acting on the coin are the centrifugal force $ m\omega^{2}R$ and Coriolis force $ 2m\omega v$ acting in the opposite direction. $ v$ is the velocity of the coin in the non-inertial frame, which in the beginning of the motion was equal $ \omega R$, so $ 2m\omega v=2m\omega^{2}R$. The net force on the coin is $ m\omega^{2}R = m\frac{v^{2}}{R}$, which is the correct centripedal force.

well golduck these were the formulaes i was talking about.anyway didn't u just write down those about which i was talking about.
well 'describing' motion upto which i know simply means writing down the velocity and acccelaration equation


If $ p(x)\equal{}x^{3}\minus{}3x^{2}\plus{}5x$ and $ p(h)\equal{}17$ and $ p(k)\equal{}\minus{}11$. Find $ h\plus{}k$

use the fact that $ p$ is symmetric across its inflection point.

It is more laborious but you could do [hide="this:"]
Note that $ xp(x)\ge0$ and that $ p$ has only one real root. So $ p(h)\minus{}17\equal{}0$ and $ p(k)\plus{}11\equal{}0$ have only one real root.
In $ p(h)\minus{}17\equal{}0$ substitute $ h\equal{}m\plus{}1$, in $ p(k)\plus{}11\equal{}0$ substitute $ k\equal{}n\plus{}1$ (in order to have $ 0$ as coefficient of $ h^2$ and $ k^2$), so $ h\plus{}k\equal{}m\plus{}n\plus{}2$.
We get:
$ m^3\plus{}2m\minus{}14\equal{}0$ and $ n^3\plus{}2n\plus{}14\equal{}0$
$ \Rightarrow (m\plus{}n)(m^2\minus{}nm\plus{}n^2\plus{}2)\equal{}0$ but $ m^2\minus{}nm\plus{}n^2\plus{}2\ge2$ for real $ m$ and $ n$.
$ \Rightarrow m\plus{}n\equal{}0$
So, $ h\plus{}k\equal{}2$.

 :)
[/hide]

I hope it is OK.

 :)


Just a quick question: what should I name the LaTeX file where I put my solutions?  Should it include my username and the round number?  (I'm thinking solutions_16-1_mathfanatic.tex.)

You can name the solutions file however you want.  16_1_solutions is a fine name.  You don't need your user name or id in the file name unless you want.  Your files themselves should have your name and username and userid on all pages, and when you e-mail the solutions the e-mail should indicate who you are.

Also, if you are sending LaTeX files, don't forget to send the PDF or DVI file of the solutions also.  We won't compile your LaTeX files for you.

[quote="dts"]By "username", you mean my AoPS screenname (in my case, dts), right?[/quote]
Yes


Solve the inequality

           3^(x+2)>9^(-1/x)

sorry don't know latex :(

I'll rewrite it in $ LaTeX$
 $ 3^{x \plus{} 2} > 9^{\frac { \minus{} 1}{x}}$
$ 3^{x \plus{} 2} > 3^{ \minus{} \frac {2}{x}}$

$ x \plus{} 2 > \minus{} \frac {2}{x}$
$ x^2 \plus{} 2x \plus{} 2 < 0$

Edit:So $ x\in R\plus{}$

I think you lost at the last step

x(x+2)>-2
so simply
x(x+2)+2>0

why u changed the sign at the last step....

am i not correct??

Since Obel1x messed up, the answer now is that there are infinitely many solutions.

[b]Correct answer is [/b] $ x \in R^{\plus{}}$   :?:

But why answer is positive real numbers???
every real no. satisfies the equation x(x+2)+2>0...
but if we put negative real numbers in the question then they does not satisfy why is that????

[hide]
Hello,

$ x + 2 > \frac { - 2}{x}$.  Here you can not multiply by $ x$ since you don't know if it positive or negative.

$ x + 2 + \frac{2}{x} > 0 \implies \frac {x^2 + 2x + 2}{x} > 0$ .Now multiplying by $ x^2$ we have $ (x^2 + 2x + 2)x > 0$ [/hide]

[hide="@RAGHAB GROVER"]

Yes, the $ x(x+2)+2=(x+1)^2 +1$ is always positive but we need the expresion $ (x^2 +2x+2)x$ to be positive.
It is positive when both $ x^2 +2x+2$ and $ x$ are positive ,and the solutions are  $ x\in \mathbb{R^+},$ or $ x^2 +2x+2$ and $ x$ both negative which is imposible.[/hide]


Έστω γ,δ πραγματικοί αριθμοί για τους οποίους ισχύουν οι σχέσεις |γ|<1, |δ|<1, |γ+δ+1|<2, |γ+δ-1|<2. Να αποδείξετε ότι

            2|γ+δ|+|1-γ+δ |+|1+γ-δ |<4.

Μια ξεχασμένη άσκηση.
Με μεγάλη καθυστέρηση προτείνω την παρακάτω λύση:

Θέτουμε χ=γ+δ, ψ=1-γ+δ, ζ=1+γ-δ.
Τότε 2χ+ψ+ζ=2+2γ+2δ, 2χ-ψ-ζ=2γ+2δ-2, 2χ+ψ-ζ=4γ, 2χ-ψ+ζ=4δ. Το πρόβλημα τότε επαναδιατυπώνεται:

[b]Αν |2χ+ψ-ζ|<4, |2χ-ψ+ζ|<4, |2χ+ψ+ζ|<4 και |2χ-ψ-ζ|<4 τότε 2|χ|+|ψ|+|ζ|<4.[/b]

Η συνεπαγωγή αυτή προκύπτει από την εξής παρατήρηση:
Οι αριθμοί 2χ, ψ, ζ είναι ή και οι τρεις ομόσημοι οπότε 2|χ|+|ψ|+|ζ|=|2χ+ψ+ζ| ή δύο από αυτούς είναι ομόσημοι και ο τρίτος με διαφορετικό πρόσημο
οπότε 2|χ|+|ψ|+|ζ|=|2χ-ψ+ζ| ή |2χ-ψ-ζ| ή |2χ+ψ-ζ|.


Consider the following initial value problem for a differential equation:

Assume $g$ and $\ell$ are positive constants.

$y''+g\sin\left(\frac{y}{\ell}\right)=0;\,y(0)=0,\,y'(0)=a$

There is a number $b$ such that:

* if $0<|a|<b$, then $y(t)=0$ for infinitely many positive values of $t.$

* if $|a|>b,$ then $y(t)$ is never equal to zero for $t>0.$

Find $b.$

[hide="Hint."]There's a reason this is in the Physics forum! "Reverse engineer" it to the physical problem for which it is a mathematical model.[/hide]

Imagine a massless pendulum with length $\ell$ and a point mass at the end.
Let $y(t)$ be the arclength from the position of the point mass to the lowest point in the pendulum's swing. $y(t)$ is NOT taken $\mod 2\pi$
Then $y(t)$ satisfies the differential equation.

$y(t)$ will will return to 0 infinitely often if and only if the pendulum never flips over and completes a full rotation.
Considering energy, the pendulum flips over if and only if $a \ge  (4gl)^{1/2}$

That's exactly the solution I had in mind. One of the most powerful problem solving concepts in mechanics is energy.

One very minor point: in my original statement, I intentionally neglected to mention what would happen if $a$ were to equal the critical value. In fact, if $a=(4g\ell)^{1/2}$ precisely, then the pendulum never flips over, but rather as $t\to\infty$, $y'(t)\to0$ and $y(t)$ tends in the limit to the unstable equilibrium point at $\pi\ell.$


email
abstract algebra
algebra
polynomial
function
geometry
domain

I haven't done much looking into this; I have emailed my professor though (I'm also waiting for a reply email from him). So anyways, the idea concerns the "ramification index". So basically here's my idea

Let F/E be an extension of number fields. Let $ O_F$ and $ O_E$ denote the rings of algebraic integers respectively. Let $ p$ be a prime in $ O_F$. Then $ pO_E$ is divided by a unique prime ideal $ P$ in $ O_F$. It is also true that $ O_E$ is a finitely generated $ O_F$-module. So here's my idea: localize to $ B$, then we get $ O{}_E{}_B$, a finitely generated module over the local ring $ O{}_F{}_B$. Now write down the Hilbert polynomial as follows:

$ P(n) = \sum_{i = 0}^d a_i F_i(n)$ where $ F_i(n)$ is the binomial coefficient "n pick i". The $ a_i$ are integers with $ a_d\neq 0$. My conjecture is the ramification index of $ B$ (which depends only on $ p$) is equal to $ a_d$.

Any ideas?

I am not sure whether I can follow your setting, so here are my questions on this:

- Why is $ p\mathcal O_E$ divided by a [b]unique[/b] prime? Is that an assumption or should this be implied?
- What exactly is $ B$? I think it should be some prime ideal of $ \mathcal O_F$, but I am not sure. Basically it should be or induce a multiplicative subset of $ \mathcal O_F$.
- Whose Hilbert polynomial is to be considered?

... or do you mean $ B\equal{}P$? But then the localization of $ \mathcal O_F$ w.r.t. $ B$ does not make sense, does it?

I am not sure what  I am missing here.

Ahh I'm sorry, there was a typo. I meant to write $ P$ when I wrote $ B$. I also wrote down another thing incorrectly: it is the case that if $ P$ divides $ p\mathcal{O}_E$ then it doesn't divide any other ideal of the form $ q\mathcal{O}_E$ where $ q$ is prime in $ \mathcal{O}_E$. (It shouldn't actually matter whether or not $ P$ is the unique divisor, we just want to localize to it). Now after localizing to $ P$ we get a finitely generated module over a local ring.

The general construction for the Hilbert polynomial (more correctly I should say Hilbert-Samuel function) of this finitely generated over a local ring is as follows: let $ R$ be a local ring and $ M$ a f-g. module. We pick an ideal $ q$ of finite colength (means that $ R/q$ has finite length). Then the hilbert function is $ H_{q,M}: \equal{} length(q^nM/q^{n \plus{} 1}M)$.

It is easy to see in our case that if we pick the ideal $ q \equal{} p\mathcal{O}_E$, it is of finite colength in $ {\mathcal{O}_F}_P$ because of divisibility in Dedekind rings.

I hope this clarifies everything =) sorry about the typs

Also, I'm going to clarify one last thing: when I say localize at $ P$, I mean localize by inverting everything NOT in $ P$ (this is the usual algebraic geometry terminology). Its clear also that the compliment of a prime ideal is a multiplicative system so this action is well defined.

I am sorry, but I still do not see very clear here. We have $ \mathcal O_F\subset\mathcal O_E$ and $ P$ is a prime ideal of $ \mathcal O_E$.

- What exactly is the multiplicative subset $ S\subset\mathcal O_F$ to localize $ \mathcal O_F$ to get - what you call -  $ \mathcal O{}_F{}_P$? Or do you mean $ (\mathcal O{}_F){}_p$?
- The Hilbert polynomial is calculated for $ R=\mathcal O_E$ and $ q=p\mathcal O_E$. Is that correct? And what is $ M$ in your case?

Let $ E/F$ be a finite Galois extension of number fields (sorry, I might have reversed the order in the first post >.<):

$ \mathcal{O}_E$ is in fact a finitely generated module over $ \mathcal{O}_F$. Let $ P$ be a prime ideal of $ \mathcal{O}_E$. There is a unique prime $ p$ in $ \mathcal{O}_F$ such that $ p\mathcal{O}_F$ is divided by $ P$. So localize everything to $ P$ (by inverting everything not in $ P$). We get $ (\mathcal{O}_F)_P$ is a local ring, (call this $ R$ for short) and we get that $ M: \equal{} (\mathcal{O}_E)_P$ is a finitely generated $ R$-module.

Now, we compute the hilbert polynomial for $ M$ with $ q \equal{} p\mathcal{O}_F$ being the "ideal of finite colength".  This is done by considering the "Hilbert-Samuel" function $ H(n): \equal{} length(q^n M/ q^{n \plus{} 1}M)$.  We then "approximate" $ H(n)$ with a unique polynomial $ P(n) \equal{} \sum_{i \equal{} 1}^d a_i F_i(n)$ where $ F_i(n)$ is the binomial coefficient "n pick i". This polynomial has the property that for $ n > > 0$, $ P(n) \equal{} H(n)$. Then the term $ a_d$ is the multiplicity of $ M$.

Now, my hypothesis is that $ a_d$ is equal to the ramification index of $ P$ over $ p$. I hope this clarifies, sorry if my earlier posts were messy.

[quote="LkNsngth"]We get $ (\mathcal{O}_F)_P$ is a local ring, (call this $ R$ for short)
[/quote]Well, to put things into the terminology of commutative algebra you mean $ R\equal{}(\mathcal O_F)_p$ where $ p\equal{}P\cap\mathcal O_F$.
[quote]and we get that $ M: \equal{} (\mathcal{O}_E)_P$ is a finitely generated $ R$-module.[/quote]This is not true in general. E.g. $ \mathbb Z_{(5)}$ is not a finitely generated $ \mathbb Z$-module. If you have difficulties to find a counterexample in your setting, just post it. I will see to give a counterexample for this.
[quote]
Now, we compute the hilbert polynomial for $ M$ with $ q \equal{} p\mathcal{O}_F$ [/quote]
I think we should better set $ q\equal{}p\mathcal O_E$
[quote]being the "ideal of finite colength".  This is done by considering the "Hilbert-Samuel" function $ H(n): \equal{} length(q^n M/ q^{n \plus{} 1}M)$.  We then "approximate" $ H(n)$ with a unique polynomial $ P(n) \equal{} \sum_{i \equal{} 1}^d a_i F_i(n)$ where $ F_i(n)$ is the binomial coefficient "n pick i". This polynomial has the property that for $ n > > 0$, $ P(n) \equal{} H(n)$. Then the term $ a_d$ is the multiplicity of $ M$.

Now, my hypothesis is that $ a_d$ is equal to the ramification index of $ P$ over $ p$. I hope this clarifies, sorry if my earlier posts were messy.[/quote]
To save the situation we could set $ R\equal{}M\equal{}(\mathcal O_E)_P$ and $ q\equal{}pR$. Then we get $ q\equal{}P^eR$ for some $ e>0$, the ramification index of $ P$ in $ \mathcal O_E$. Since $ PR$ is principal, we get $ q^nM/q^{n\plus{}1}M\equal{}R/P^eR$ which is independent of $ n$. Then we get $ length_{R}(R/P^eR)\equal{}e$ by standard reasoning in algebraic number theory.

I believe that this is not as surprising as you might have thought it to be. :|

I'm getting more and more certain that this proposition is true. I have reduced the problem to computing $ length(P^{e^n}(\mathcal{O}_E)_P/P^{e^{n\plus{}1}}(\mathcal{O}_E)_P)$.

My guess is that the function takes the value $ e$ for all values of $ n$ because $ P^{e^n}\equal{}P^ne$ and $ P^{e^{n\plus{}1}}\equal{}P^{n\plus{}1}e$ and the differences between the two are $ P^e$ hence leading me to think the composition series is $ e$ terms long. Is this true? If it is, then I believe my hypothesis is right.

As pointed out in my previous posting this right and your assertion becomes rather trivial since the Hilbert polynomial in question is constant. After localisation you are working in a principal ideal domain such that things can almost be read off.

The whole stuff basically needs to be written down in clear terms. Then it immediately drops out of basic results I am afraid to say. I think most of our work was to get the notation and the statement correct.

Yes you are right =) I think once I wrote it down clearly and looked at it in some greater detail, it became clear. Anyhow thanks for your time.


inequalities
logarithms
function
real analysis
real analysis unsolved

We denote by $ \pi(n)$ the number of primes not exceeding $ n$ and we denote the harmonic sequence
    $ H_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n}$, $ n\geq 1$. Show that $ \pi(n)>\frac{n}{H_{n}},\forall n\geq 1429$.

It's a very nice fact, but I think it's more about esthetic things here... The inequality is surely not optimal: a stronger one is obtained in Pierre Dusart's thesis: $ \pi(x)\geq \frac{x}{\ln(x)}\cdot (1+1/\ln(x))$ for $ x>600$. I wonder if Laurentiu Panaitopol has a not very technical solution without using advanced theorems of analytic number theory. I really doubt that a naive approach can work here.

Yes, Harazi. In solving this inequality I used the Rosser-Schoenfeld inequality, namely
\[ \pi(x)\geq\frac{x}{\log x}\left(1+\frac{1}{2\log x}\right), \forall x\geq 59.(*) \]
In fact, as noted by Professor Panaitopol himself, a mathematician (I don't remember his name now), noted that $ \pi(n)$ function behaves like $ \frac{n}{H_{n}}$. Professor Panaitopol showed that this last oen is a lower bound of $ \pi(n)$. :)

P.S. mr. Panaitopol's solution is quite technical. I remember that I told him once that it follows directly from (*). :wink:

[quote="Cezar Lupu"]a mathematician (I don't remember his name now), noted that $ \pi(n)$ function behaves like $ \frac{n}{H_{n}}$.[/quote]

You mean, the Prime Number Theorem?


Let a,b,c,d primes such that a<>b, a>2. Suppose that there exists M such that for all n>M the numbers an+b and cn+d have the same sum of digits in any base from 2 to a-1 (this means that they have the same sum of digits in base 2, which can be different from the common value of the sum of the digits in base 3, etc). Prove that a=c and b=d.
  Harazi

Is this so hard? Or is it ugly? A lot of people told me it is an ugly problem due to the hipothesys. The reason is simple: I couldn't solve the general problem (with a, b,c,d positive integers), but I had a beautiful idea that works with this one.

Very nice..

Denote with $ S_i(x)$ the sum of digit of $ x$ in its $ i$ -th base representation. For $ t$ enough large choose $ n \equal{} \frac {i^{(k \plus{} 1)t} \minus{} 1}{i^t \minus{} 1} \minus{} 1$, so $ S_i(an \plus{} b) \equal{} kS_i(a) \plus{} S_i(b) \equal{} kS_i(c) \plus{} S_i(d) \equal{} S_i(cn \plus{} d)$. But $ k \mid S_i(b) \minus{} S_i(d)$ for all $ k$, so $ S_i(b) \equal{} S_i(d)$ and $ S_i(a) \equal{} S_i(c)$ for all $ 1 < i < a$. Choosing $ i \equal{} a \minus{} 1$ we obtain $ 2 \equal{} S_{a \minus{} 1}(a) \equal{} S_{a \minus{} 1}(c) \implies c \equal{} (a \minus{} 1)^{\ell_1} \plus{} (a \minus{} 1)^{\ell_2}$ that is divisible by $ a \minus{} 1$ if $ \ell_1\ell_2 > 0$, otherwise divisible by $ a$ (the case $ \ell_1\equal{}\ell_2\equal{}0$ is trivial since $ c\equal{}2$ and $ a$ would be a power of 2), that implies $ a \equal{} c$. Now if $ b < a$ then $ 1 \equal{} S_b(b) \equal{} S_b(d) \implies b \equal{} d$ otherwise we can assume $ d \ge b > a$, so choosing enough large $ i$ and enough large $ j$ such that $ 2^i \mid aj \plus{} b$ we obtain $ S_2(aj \plus{} b) \equal{} S_2(aj \plus{} d) \equal{} S_2(aj \plus{} b) \plus{} S_2(d \minus{} b) \implies d \equal{} b$.


algebra
polynomial
Rational Root Theorem
superior algebra
superior algebra solved

Prove that $x^{3}-nx+2$ is irreducible for $n \neq-1,3,5$.

[quote="Gockel"]Irreducible over $\mathbb{Q}$ or over a finite field or something else?[/quote]

Dummit & Foote did not mention that. I assume they mean $\mathbb{Q}$.

If it were reducible over $\mathbb{Q}$, it would have a linear factor and thus a rational root. By the Rational Root Theorem, that root must be $-2,-1,1,$ or $2$. For all but the listed values. none of these are roots.


calculus
integration
logarithms
calculus computations

Find indefinite int of (x^3+1)/(x^2 +3) dx

My first step was to do long division and I got x with a remainder of -3x+1.

So I have:
indefinite int of (x^3+1)/(x^2 +3) dx
= int x - int [(3x+1)/(x^2+3)]
lt
To solve for  int [(3x+1)/(x^2+3)] do I set it equal to (Ax+B)/x^2 + C/3 and then cross multiply to find A, B, and C?  Or how would I do this?  Because the denominator is un-factorable (as far as I can see).

Help would be very much appreciated!  Thank you

There are two terms in that integral- a logarithm and an arctangent.

...I'm sorry, I'm not sure if I follow you


Let (X, d) be a connected metric space and let A be a connected subset of X. Assume that the complement of A is the union of two separated sets B and C. Prove that the union of A and B are connected.

So assume $ A \cup B$ is not connected. Then $ A \cup B \equal{} H \cup K$ for separable sets $ H$ and $ K$. Now let $ A_1 \equal{} A \cap H$ and $ A_2 \equal{} A \cap K$. Then $ A \equal{} A_1 \cup A_2$ is not connected. Contradiction. But how would you shoe that neither $ A_1$ nor $ A_2$ is empty?

[url=http://www.math.ku.dk/~moller/e02/3gt/opg/S23.pdf]Number 12[/url] is basically your problem, just with the letters changed around.  It references Lemmas 23.1, 23.2, and problem 17.6, which I've listed below.

Lemma 23.1: If $ Y$ is a subspace of $ X$, a separation of $ y$ is a pair of disjoint nonempty sets $ A$ and $ B$ whose union is $ Y$, neither of which contains a limit point of the other.  The space $ Y$ is connected if there exists no separation of $ Y$.  (I won't prove this.)

Lemma 23.2: If sets $ C$ and $ D$ form a separation of $ X$, and if $ Y$ is a connected subspace of $ X$, then $ Y$ lies entirely within either $ C$ or $ D$.

Problem 17.6: If $ A\subset B$ then $ \bar{A}\subset\bar{B}$; $ \overline{A\cup B}\equal{}\bar{A}\cup\bar{B}$; $ \overline{\bigcup_{\alpha}}\supset\bigcup\bar{A}_{\alpha}$.


function
calculus
integration
real analysis
real analysis unsolved

Suppose $f: [a,b] \to R$ is bounded and has a discontinuity only at $b$. Prove that $f$ is integrable on $[a,b]$. (Do not use the result that bounded functions on $[a,b]$ with finitely many discontinuities are integrable).

Can any help me with this integral problem? I find integrals most difficult to understand, especially when discontinuities are discussed.

Suppose $x_{0}$ is the point of discontinuity. Make a partition that traps that point inside a small interval - say the interval $[x_{0}-\delta,x_{0}+\delta]$ where we'll figure out later just how small we need $\delta$ to be.

The difference between the upper sum and the lower sum on $[a,x_{0}-\delta]$ can be made small because the function is continuous on that interval, hence Riemann integrable. The same argument shows that the difference between the upper sum and the lower sum on $[x_{0}+\delta,b]$ can also be made small.

Then if $|f(x)|\le M$ everywhere (note that you did hypothesize that it was bounded), the difference between the upper sum and the lower sum on $[x_{0}-\delta,x_{0}+\delta]$ is no larger than $4M\delta.$ We can make this small enough by choosing $\delta$ small enough.

There's still more to writing the proof up, as you should get specific and quantitive with all of my "can be made small enough" claims, but the main ideas are there.


IMO Shortlist
number theory unsolved
number theory

[quote="apollo"]Prove that for every positive integer $ n$ there exists an integer $ m$ such that $ 2^{m}\plus{}m$ is divisible by $ n$[/quote]
IMO shortlist 2006  :wink:

All problems from IMO shortlist 2006:

http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2006

Topic of this problem:
http://www.mathlinks.ro/Forum/viewtopic.php?p=867486#867486

General problem :

http://www.mathlinks.ro/Forum/viewtopic.php?p=354533#p354533


Hi I have a question involving an algebraic proof.  I tried proof by contradiction.  If someone could please check over my work, that would be greatly appreciated.  Thank you!

Question:  Prove that the equation $x^{2}=4y+3$ has no integer solutions.

Assume that there exists an integer pair $(x,y)$ such that $x^{2}=4y+3$ is true.
Square root both sides: $|x|=\sqrt{4y+3}$ [color=red][i](Do I need absolute values here?)[/i][/color]
Consider, $4y+3 = 4y+2+1$
$4y+3 = 2(2y+1)+1$, thus $4y+3$ is odd as $y$ is [u]presumed[/u] to be some integer
For $x$ to be an integer then $2(2y+1)+1$ must be an odd perfect square.
Odd perfect squares are of the form $(2k+1)^{2}=4k^{2}+4k+1=2(2k^{2}+2k)+1$ where $k$ is some integer.
Therefore $4y+3=2(2y+1)+1$ must be expressible in the form $2(2k^{2}+2k)+1$ in order for $4y+3$ to be an odd perfect square.
Hence, $4y+3=2(2y+1)+1=2(2k^{2}+2k)+1$
We have, $2y+1=2k^{2}+2k$ where $k$ is some integer and $y$ is [u]presumed[/u] to be some integer
$y=(k^{2}+k)-\frac{1}{2}$
On the right hand side, $k^{2}+k$ is an integer but the difference $(k^{2}+k)-\frac{1}{2}$ is not an integer.
From our presumption that $x$ and $y$ are integers, we arrived at a contradiction.  Thus $y$ is not an integer and consequently $x$ cannot be an integer.

That's a very long proof.  How about...

[hide="this one..."] Let $x = 4k+a$ where $a = 0, 1, 2, 3$.  Then $x^{2}= 16k^{2}+8ka+a^{2}$, where $a^{2}= 0, 1, 4, 9$.

Hence $x^{2}= 4 \cdot \text{Something}+0, 1$ which means it can never be of the form $4 \cdot \text{Something}+3$.  [/hide]

Check the quadratic residue for $x^{2}\pmod{4}$. Clearly not $3$.


Let $T: V_1\longrightarrow V_2$ be a linear transformation, $M$ be a subspace of $V_1$, which is [b]not[/b] a subset of the $\ker(T)$. Show that $\tilde{T}: V_1/M\longrightarrow V_2$, which is given as usual, namely, $\tilde{T}(M+x)=T(x)$ is not well-defined.

er. is this easy? As $M \not \subset \ker T$, $\exists m \in M$ such that $T(m) \ne 0$; then $\forall x$, $T(m+x) = T(m) + T(x) \ne T(x)$ ? :? :?

Hmm, I don't see how your statement proves that $\tilde{T}(M+x)=T(x)$ is not well-defined. Can you please elaborate?

Suppose $m\in M$ and $T(m)\neq 0$.
Then $M+(m+y)=M+y$.

$\tilde{T}(M+(m+y)) = T(m+y) = T(m) + T(y)$
$\tilde{T}(M+y) = T(y)$

thus $\tilde{T}(M+(m+y)) \neq \tilde{T}(M+y)$.

Ah thanks. Originally, I had:

Let $M+x=M+y$ and assume $T(x)=T(y)$, Then $T(x)-T(y)=T(x-y)=0$, but then $x-y\in\ker(T)$ which is (not actually) a contradiction.

I understand now though.

Yes, you can do that by considering the opposite. But I would like to construct an example here to prove it in positive-direction.  :P


Write in algebraic form:

The product of two odd numbers and one even number. (Using only one variable)

[quote="A+MATH"]Wait? Two differant odds?[/quote]

Yes, two DIFFERENT odds.

[quote="math92"][hide](x)(x+1)(x+2) where x is any odd number.[/hide][/quote]

Can you make an expression such that $x$ can be even OR odd?  :)


Welcome to [b]FTW Tourney 2[/b]! Before we begin, I would appreciate it if players (especially new ones) read the following.

[list]
[*] This tournament is a double elimination tournament. That means, if you lose one game, you're not out yet!
[*] Players will be expected to play [b]exactly[/b] three countdowns per round. Don't stop at two or one! You can look at the bracket* for your matches, but I will also post them here in the thread. The person who wins the most countdowns out of those three is the winner of that match.
[*] You will have two weeks for one match, so there are [b]no excuses[/b] for missing a match - UNLESS you post in the thread beforehand with a valid reason, such as "I will be on vacation from such-and-such date to such-and-such date" or "I've gotten sick, please extend $ x$ days".
[*] At the beginning of each round, if you don't frequently see your opponent on FTW (meaning at least once every two days), private message him or her to arrange a time. Be careful of time zones! [url=http://www.time.gov]This[/url] is a helpful website if you tend to fail with that kind of thing.
[*] At the end of a match, be sure [b]one[/b] of you and your opponent posts the results. Examples - "I won 2-1, individual games being 4-3 me, 4-2 opponent**, and 4-1 me" or "So far I'm leading 2-0 with games 4-2 me, 4-3 me, we will finish the match tomorrow".
[*] Be fair! Don't claim flashies if you don't have them, especially since I think levans said they are no more. Remember, in the end, it's only a game. Oh, and [b]memming[/b] is definitely a valid tactic, possibly the only aspect of FTW itself. [/list]

Here is the seeding***:

1. RoFlLoLcOpT
2. someonehacked
3. AIME15
4. ReseesPieces
5. manifester
6. BOGTRO
7. jonathanchou711
8. yaofan
9. AIME_is_Hard
10. saszs
11. lotsofmath
12. mousy
13. Jongao
14. rd5493
15. Mewto55555
16. louiethechewy123

Replacements:

2. Ttocs45
3. ilovepink

The bracket can be found [url=http://www.bracketmaker.com/tmenu.cfm?tid=316804&tclass=]here[/url]. Matches:

Game 1 - RoFlLoLcOpT vs. someonehacked
Game 2 - AIME15 vs. ReseesPieces
Game 3 - manifester vs. BOGTRO
Game 4 - jonathanchou711 vs. yaofan
Game 5 - AIME_is_hard vs. saszs
Game 6 - lotsofmath vs. mousy
Game 7 - Jongao vs. rd5493
Game 8 - Mewto55555 vs. louiethechewy123

Ready...set....go!  :)

* I'm not quite sure the link works. Let me know if it doesn't.
** Clearly, you would put your opponent's username, rather than just "opponent".
*** I'm a little new to this, so bear with me as I screw up the terminology.
**** What? Not a fourth footnote?

[quote="isabella2296"] Don't claim flashies if you don't have them, especially since I think levans said they are no more. [/quote]

No more flashies; however, there is the "no time bar glitch" which includes (obviously no time bar) not being able to input answers.

Can lotsofmath replace me?  I'm not very active anymore.

You are right, you are new to this and as such, made a relatively major screw-up. Actually, I count two.

1. Why need to play 3? If it's 2-0, then what's the point? It's not like we are comparing score totals or whatever.

2. Seeding is based on some measure of skill. If none applies (which is false in this case, as there is rating), do [b]not[/b] do it in order of signup. Do it randomly. If you do it in order of signup, generally the high level people with more time and motivation will wind up on top, and the good people will be gone early.

Tytia.

they need a cry emoticon

i lost first game to jonathanchou711

4-2

EH I PLAY RoFlLoLcOpT.....

SO obvious who's gonna win... sigh....

Did she like edit everything? Cause I was facing louiethechewy123 first.

[quote="AIME15"]Yes.  She made the brackets random, rather than by sign-up order.[/quote]

Well actually I meant that when I first looked at the seeding and stuff, it said that I was facing louiethechewy123. So then the next time I checked, it was different.

Yeah, I made the bracket/seeding different, like AIME15 said.

Since guests are disabled, it will be a little difficult for cowards like me to play...

Uh.... Did I just see Mewto post?!

Clarifying: The matches are best of 3 right?

Yes, you did. "Doing well at nats has its advantages." - Mewto55555  :P

Yup, best of 3.

[quote="jonathanchou711"][quote="isabella2296"]There is. But do to an excessive amount of replacements and people with Internet problems, etc.,...[/quote]

You mean "Internet problems" right? With the finger quotations?

They're prob lazy and just don't want to do anything.[/quote]

Don't accuse others of being lazy and/or ascribe various other not-so-nice attributes to them without any evidenc or justification.

If they can't get on the internet, go to the library!

Some libraries do not have the necessary ports open.  In any case, one should not sign up for a tourney unless he/she is sure that he/she can commit...

I've seen how many replacements this thread has had, so I can be a replacement if you guys run out.  Sorry if it's off-topic.

No, I mean Internet problems like Internet problems. Not like "Internet problems". I know BOGTRO in rl, and he wouldn't lie about that.  :wink: Anyway, not going on AoPS for over a week means that he does have Internet problems (not "Internet problems".)

Bleh. Maybe ask them if they still want to participate? Cause I know Internet Problems can be very frustrating and discouraging.

Yes, the party with which this is concerned has already posted.

??? Then how'd they get on? I thought they had internet probs!

Um its way past Sunday, so shouldn't they be replaced?

Patience is a virtue, young Padawan.

Seriously though, if you make a deadline, it helps if you keep it  :wink:  Otherwise no one will believe you the next time.

Hmm, I thought I posted this, but oh well. Tournament's canceled. Unless someone wishes to replace me as a mod. No one's interested and I'm kind of busy with finals, etc.


As far as I know, MikTeX comes with a lil program that makes it's own pdfs doesnt it? Why would we need the Go2PDF (besides the fact that it's a very nice tool to have around for other hw purposes)?

MikTeX can turn [tex]\LaTeX[/tex] files into pdf files. Go2pdf can turn images, for example, into pdf files.

[quote="MithsApprentice"]Why would we need the Go2PDF (besides the fact that it's a very nice tool to have around for other hw purposes)?[/quote]

If you want to take a diagram or an image or whatever and include it in a LaTeX document, you'll have to make the diagram/image into a PDF doc first, then include it as described in the LaTeX pages.

[quote="mathfanatic"]Also, Go2PDF is useful for making other things into PDFs.  It's particuarly useful for turning HTML files into PDFs, because it imitates printing.  (Basically, it pretends to print something, but instead produces a PDF.)  So you save paper and can save internet files to your hard drive.[/quote]

Does anyone know a method to complete the reverse? Turning a pdf into HTML?

[quote="MithsApprentice"]Does anyone know a method to complete the reverse? Turning a pdf into HTML?[/quote]
That seems to be a tall order - PDF files can contain lots of things that can't really be shown in HTML.

Also, PDF files are not always saved in a way that is conducive to being HTML'd -- for example, if the entire thing is one image file instead of having the "knowledge" of what is text and how it's formatted, etc.


function
inequalities
inequalities unsolved

Let $a,b,c>0,a+b+c=3$.
Prove that:
$a^k+b^k+c^k\ge a^h+b^h+c^h$
( $k,h \in R,k>h>0$).

the function $a^x$ with $a>0$ is monotonic increasing, so it's clearly that if $k>h$, then $a^k>a^h$. Sum up 3 times with $a,b,c$ and obtain the proof

No, $f(x) : = a^x$ defined on the positive reals is
increasing if $a >1$,
decreasing if $0<a<1$ and
constant if $a = 1$.

However it is not too difiicult. Let's do cases:

Case 1: $k \le 1$.
$a^h + b^h + c^h \le \frac{(a^k + b^k + c^k)^{ \frac h k -1}} {3^{\frac h k -1}} (a^k + b^k + c^k) \le a^k + b^k + c^k$
The first inequality is just the power-mean inequality comparing $k$- and $h$-powers.
The second inequality follows from the power-mean inequality comparing
$k$- and $1$-powers.

Case 2: $h \ge 1$.
$a^k + b^k + c^k \ge \frac{(a^h + b^h + c^h)^{ \frac k h -1}} {3^{\frac k h -1}} (a^h + b^h + c^h) \ge a^h + b^h + c^h$
The first inequality is just the power-mean inequality comparing $k$- and $h$-powers.
The second inequality follows from the power-mean inequality comparing
$h$- and $1$-powers.

Case 3: $k > 1 > h$.
This follows from cases 1 (applied on $(k',h') : = (1,h)$) and 2 (applied on $(k',h') : = (k,1)$).

So we are done.  :)


Given $ a, b, c \geq\ 0$. Prove that: $ \frac {\sqrt {ab}}{ ab \plus{} c^2} \plus{} \frac {\sqrt {bc}}{ bc \plus{} a^2} \plus{} \frac {\sqrt {ca}}{ ca \plus{} b^2} \geq\ \frac {9(a^3 \plus{} b^3 \plus{} c^3)}{2(a \plus{} b \plus{} c)^4}$
 :)

Equality occurs example when $ (a, b, c) \equal{} (\frac {7 \pm 3\sqrt {5}}{2}, 1, 0)$  :)

We can prove it easily by Am-Gm. Sorry for spam   :oops:


logarithms
integration
geometry
trapezoid
inequalities proposed
inequalities

1,:let n be  nature number and n>1. prove that:
$ n\ln n \minus{}n<\ln n!<(n\plus{}\frac{1}{2})\ln n\minus{}n\plus{}1$
Let:x>0, p>0.Prove that:
$ (1\plus{}\frac{x}{p})^p<e^x<(1\plus{}\frac{x}{p})^{p\plus{}\frac{x}{2}}$

By convexity, $ n\ln n \minus{} n < \ln n! \Longleftrightarrow \int_1^{n} \ln x\ dx < \sum_{k \equal{} 1}^n \ln\ k!$

For the rest, I believe the tangent line has something to do with the proof. Is trapzoid? :roll:


Four distinct circles are drawn in a plane. What is the maximum
number of points where at least two of the circles intersect?

i think the answer is 12

think of it like this (and i'm explaining because i can't draw...)

put 2 circles together like a venn diagram
put the 3rd circle on such that no place has 3 circles together
put the 4th circle on such that it is just below the 3rd circle and creates 6 intersections (it is possible)

you should get 12 intersections this way

I believe a picture is provided in the solution.

I think this diagram [hide][img]http://cdn.artofproblemsolving.com/attachments/7/9/5c24271bfbfb6ae532e26b5aea9dadb797386d.png[/img][/hide] is clearer & more logical than the one provided by Alcumus.

I had a different way of solving this. I did 4! because this reminds me of the situation where each person shakes hands with each other person. In that type of problem, you just do the total number of people factorial, and divide by 2 to eliminate repetition in order. That's the same result


It's not very interesting and it's a result valid on any ring. It's sufficient to expand the expressions and use the property: $ \sum_{k\equal{}1}^{n}{k^u}\sim\dfrac{n^{u\plus{}1}}{u\plus{}1}$ when $ n$ tends to infinity and $ u>\minus{}1$.

[quote="loup blanc"]a result valid on any ring[/quote]
This is surely not true in any ring lacking limits or where the series doesn't converge.

Obviously you are right ZetaX; it's true in a Banach algebra for example.


Let's prove this by induction.

For $n=2$ this is obvious ($\frac{5}{4}>\frac{6}{5}$).
Let's assume this is proved for some natural $n$, $n>1$. Let's then prove that it's still working for $n+1$.

$\frac{1}{1^{2}}+\cdots+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2})}>\frac{3n+3}{2n+3}$
We "know" (we assumed) that $\frac{1}{1^{2}}+\cdots+\frac{1}{n^{2}}>\frac{3n}{2n+1}$. Now, if we prove that $\frac{1}{(n+1)^{2})}\ge \frac{3n+3}{2n+3}-\frac{3n}{2n+1}$, we've solved the task. This is equivalent to
$\frac{1}{(n+1)^{2})}\ge \frac{3}{(2n+1)(2n+3)}$ which is, of course, equivalent to
$4n^{2}+8n+3\ge 3n^{2}+6n+3$, which is equivalent to $n^{2}+2n\ge 0$, and that is, of course, true for all $n$. Now we've proved that if the inequality is valid for some $n$, then it's valid for $n+1$. Since it's valid for $n=2$, it's also valid for $n=3, n=4, \cdots$, that is, it's also valid for all natural $n$, $n\ge 2$.

we proceed by induction. the case $n=2$ implies $\frac{5}{4}>\frac{6}{5}$, surely true. now assume that it's true for $n-1$. so $S=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\dots+\frac{1}{(n-1)^{2}}>\frac{3n-3}{2n-1}$. now we need to prove that $S+\frac{1}{n^{2}}>\frac{3n}{2n+1}$. we'll be done if we can show that $\frac{3n-3}{2n-1}+\frac{1}{n^{2}}>\frac{3n}{2n+1}$
$\iff (3n^{3}-3n^{2}+2n-1)(2n+1)>3n^{3}(2n-1)$
$\iff 6n^{4}-3n^{3}+n^{2}-1>6n^{4}-3n^{3}$
$\iff n^{2}-1>0$, indeed.

EDIT: sorry mornic, didn't see your post there.


Solve

$ x^{2} \plus{} y^{2} \plus{}x\plus{}y\equal{}32$

$ 12(x\plus{}y) \equal{} 7xy$

hello, solving the second equation for $ y$ we get
$ y\equal{}\frac{12x}{7x\minus{}12}$
plugging this in the frst equation and simplifying and factoring we get
$ \frac{(x\minus{}3)(x\minus{}4)(49x^2\plus{}224x\minus{}384)}{(7x\minus{}12)^2}\equal{}0$.
Now it is easy to find all solutions.
Sonnhard.


floor function
group theory
abstract algebra
number theory solved
number theory

Show that for all natural numbers $n$, $(n!)!$ is divisible by $(n!)^{(n - 1)!}$.

First note that [tex]\lfloor mx \rfloor = \lfloor x \rfloor + \lfloor x + \frac{1}{m} \rfloor + \ldots + \lfloor x + \frac{m - 1}{m} \rfloor \geqslant
\lfloor x \rfloor + \ldots + \lfloor x \rfloor = m \lfloor x \rfloor[/tex], for any real [tex]x[/tex]. Thus [tex]\alpha_p ( n!^{( n - 1 ) !} ) = ( n - 1 ) ! \sum_{j =
1}^{\infty} \lfloor \frac{n}{p^j} \rfloor \leqslant \sum^{\infty}_{j = 1} \lfloor \frac{( n - 1 ) !n}{p^j} \rfloor = \sum^{\infty}_{j = 1} \lfloor \frac{n!}{p^j} \rfloor = \alpha_p ( n!! )[/tex]. Which means that [tex]n!^{n - 1!} |n!![/tex].

[i]Combinatoric solution:[/i]

Suppose that we want to divide $n!$ person in to $(n-1)!$ group such that each group have $n$ person. :D

Write $S_n$ as a set $A \times B$ where $A$ is of size $n$ and $B$ is of size $(n-1)!$: for example, as cosets of some particular copy of $S_{n-1}$.  Consider the wreath product $W = S_n \wr S_{(n-1)!}$.  Then $W$ is a group of permutations of $S_{n!}$ and so isomorphic to a subgroup.  Hence the order of $W$, which is $n^{(n-1)!} (n-1)!$ divides $(n!)!$ and this implies the result.  ehsan2004's nice answer corresponds to the subgroup of $W$ which acts trivially on $B$.


trigonometry
trig identities
Law of Cosines

i really need help, i need a proof of the law of cosines for my class. thanks :)

Try searching [url=http://www.artofproblemsolving.com/Wiki/index.php/Law_of_cosines]here[/url] for some guidelines/hints/tips.

thanks, why didnt i think of that, lol :lol:

The theorem is called Theorem d’ Al-Kashi in French, you could understand the proof by just chasing some expressions.
See here:[url]http://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_d'Al-Kashi[/url]
[url]http://en.wikipedia.org/wiki/Al-Kashi#Law_of_cosines[/url]

This is the quickest proof I know of. Confer the diagram below, make sure you understand why the lengths of $ AD, BD, CD$ are such as they are, and then apply Pythagorean to $ \triangle BCD$:

$ (c \minus{} b\cos\alpha)^2 \plus{} (b\sin\alpha)^2 \equal{} a^2$

$ c^2 \minus{} 2bc\cos\alpha \plus{} b^2\cos^2\alpha \plus{} b^2\sin^2\alpha \equal{} a^2$

$ c^2 \minus{} 2bc\cos\alpha \plus{} b^2 \equal{} a^2$

$ \boxed{a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cos\alpha}$

Similar idea could be used for an obtuse triangle as well, where $ c \minus{} b\cos\alpha$ would become $ c \plus{} b\cos(\pi \minus{} \alpha)$ (which of course is the same).

NOTE/EDIT: The proof is of course included in the French page [b]kunny[/b] gave above.


function
algebra
domain
trigonometry
calculus
calculus computations

a)  What is the domain of $ \sqrt{x}$?  Of $ \cos \left( \frac{1}{x} \right)$?

b)  Cosines are always bounded by $ 1$.

c)  What is the actual statement of the product rule for limits?  What does it say about functions whose limits don't exist?


In 100 ml there are 0.5 Mol $ AgNO_{3}$   . In the mixture is put a Zn's tablet . Find the mass in gr. of the Ag that separates from the mixture.

How can I solve this .  Can somebody solve it.

1) What reaction is expected to occur?

2) Confirm that it will indeed occur by cheking a table of standard reduction potentials.

3) How is the limiting reagent?

4) Therefore, the quantity of silver produced is...

Assuming that there is more than enough Zinc and that you meant .5 Moles of $ AgNO_3$ and not .5 Molar $ AgNO_3$:

[hide="Solution"]
With every 0.5 moles of $ AgNO_3$ there are .5 moles of $ Ag^ \plus{}$ and .5 moles of $ {NO_3}^ \minus{}$ so:
$ (.5 Mol AgNO_3)(\dfrac {.5 Mol Ag^ \plus{} }{AgNO_3}) \equal{} .5Mol Ag^ \plus{}$
Now the $ Ag^ \plus{}$ takes the electrons given off from the $ Zn_{(s)} \to Zn^{2 \plus{} } \plus{} 2e^ \minus{}$ reaction and we have this reaction:
$ Ag^ \plus{} \plus{} e^ \minus{} \to Ag_{(s)}$
So for every .5 moles of $ Ag^ \plus{}$, .5 moles of $ Ag_{(s)}$ are formed. By the molar mass of silver, we have:
$ .5Mol Ag_{(s)} (\dfrac {107.87g}{Mol}) \equal{} 53.935g Ag_{(s)}$
[/hide]

I have 2 questions :

1)Why is given the volume 100ml in my problem? You didn't use it , did you?

2)Why in 0.5 Mol $ AgNO_{3}$  there are 5 mol $ (NO_{3})^{\minus{}}$ and 5 mol $ Ag^{\plus{}}$ ? It shuldn't be 0.5 mol $ Ag^{\plus{}}$ and 0.5 Mol $ (NO_{3})^{\minus{}}$  ?

Thanks a lot

1. only for the sake of physicality, in my opinion.

2. nobody wrote $ 5 \text{mol}$. [b]crazypianist1116[/b] wrote $ .5 \text{mol}$ which is $ 0.5 \text{mol}$ (I assume you're not familiar with the notation).


geometry
calculus
integration
angle bisector

In triangle ABC, angle C is a right angle. AD is the bisectror of A such that CD=4 and BD=5. Find the area of ABC.

Hard problem, I like it!

[hide]$AD$ is bisector of $A$, $\frac{AC}{CD}=\frac{AB}{BD}$, $AB=AC\frac{BD}{CD}$

$ABC$ is right triangle, $AC^2+CB^2=AB^2$, $AC^2+(CD+DB)^2=AC^2\frac{BD^2}{CD^2}$

$AC^2=\frac{(CD+DB)^2}{\frac{BD^2}{CD^2}-1}$

Area of $ABC=\frac{1}{2}\times AC\times (CD+DB)=\frac{(CD+DB)^2}{2\sqrt{\frac{BD^2}{CD^2}-1}}$

Put $BD=5,CD=4$, $ABC=54$[/hide]

Oh I am sorry, I forget to divide by 2  :blush:  I fix it Thx  :)

I think you forgot to divide by 2. I got 54. I may be wrong.

From angle bisector theorem,
$AB = 5a$, $AC = 4a$
Thus $BC = 3a = 9$
$a =3 $
$[\triangle ABC] = \frac{12a^2}{2} = 54$

Doesn't work all the time, but always start out by looking for triples, but it's probably a Pythagorean triple since you gave one of the legs as $9$, the other sides might be $12,15$, and so the area is $12(9)=108/2=54$.

[quote="BornForMath"]As you said, it doesn't work all  the time, since there exists another integral triple $9, 40, 41$ :)[/quote]

Yes, but using the angle bisector theorem, only 9-12-15 will satisfy the proportion.


[quote="kaydika"]For all positive reals x, y and z show that the following inequality holds
((x+y+z)^2)((xy+yz+zx)^2)<=(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)[/quote]
What if $x=y=z=1$ $?$ :wink:

i suppose you want this  :maybe:
http://www.mathlinks.ro/Forum/viewtopic.php?t=131825


pigeonhole principle
combinatorics unsolved
combinatorics

a imoboy studies at least 1 hour a day during the 35 days before a stage. Anyway he doesn't study more than 60 hours.
Show that there's a sequence of consecutive days during which he studies 13 hours totally.

enjoy!

the problem is wrong,it should be at most $ 55$ instead of $ 60$  :maybe:
(for $ 55$ it can be solved uisng Pigeonhole...)

the problem is very similar (the same!) to the chessmaster's problem (Engel)...but that technique doesn't work...

yeah youre right:
devide the set $ \{1,2,...,60\}$ into the following sets:
$ A_{1}=\{1,14\}$
$ A_{2}=\{2,15\}$
$ A_{3}=\{3,16\}$
.
.
.
$ A_{13}=\{13,26\}$
$ A_{14}=\{27,40\}$
.
.
.
$ A_{26}=\{39,52\}$
$ A_{27}=\{53\}$
$ A_{28}=\{54\}$
.
.
.
$ A_{34}=\{60\}$

now let $ S_{i}$ denote the number of hours that he studied from the first day to the $ i^{th}$ day,so we have that:
$ S_{i}\in \{1,...,60\}$
so the numbers $ S_{1},...,S_{35}$ are $ 35$ numbers from the set $ \{1,...,60\}$ so by Pigeonhole we get that there exists $ i<j$ such that $ S_{i},S_{j}\in A_{k}$ for some $ 1\leq k\leq 34$ so according to the choice of $ A_{i}$s we get that:
$ S_{j}-S_{i}=13$...
so he have studied exactly $ 13$ hours from the $ (i+1)^{th}$ day to $ j^{th}$ day...


We can suppose the origin 0 is inside of the triangle vertices (x,y,z)\inC 3

----------------------------------------------------------------------------
lemma : |x|+|y| \leq |x-z|+|y-z|
proof

the interior of the triangle  \lambda x+ \gamma y+ \varepsilon z=0
>first case
If  \varepsilon =/=0, z =-px-qy, (p;q)\in R+ 2
|x-z|+|y-z|=|(p+1)x+qy|+|px+(q+1)y| \geq |(p+1)x|-|qy|+|(q+1)y|-|px|=|x|+|y|

If y=-\alphax, \alpha \geq 0
|x-z|+|z-y| \geq |x-y|=|(1+\alpha)x|=|x|+|y|
-------------------------------------------------------------------------

Suppose 0 in inside strictely of the triangle
If t = ax+by+cz wecan suppose g=a/ \lambda  \geq b/ \gamma  \geq c/ \varepsilon
Then a-g \lambda =0,b-g \gamma  \leq 0,c-g \varepsilon  \leq 0
t=(a-g \lambda )x+(b-g \gamma )y+(c-g \varepsilon )z=-py-qz

0 is inside of the triangle vertices y,z,t

|x|+|y|+|z| \leq |x-t|+|t|+|y|+|z|
by the lemma |y|+|z| \leq |y-t|+|z-t|

we have |x|+|y|+|z| \leq |t|+|x-t|+|y-t|+|z-t|

Choose x=a+b,y=b+c,z=a+c, t=a+b+c to get the inequality


inequalities
Cauchy Inequality
inequalities proposed

[quote="CDP100"]If a, b, c, d > 0      ->

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \ge abc \plus{} abd \plus{} acd \plus{} bcd$[/quote]

True by muirhead

$ \{3,0,0,0\}\succ\{1,1,1,0\}$

Alternative solution

use Cauchy inequality cyclic

$ a^{3}\plus{}b^{3}\plus{}c^{3}\geq3abc$
$ a^{3}\plus{}b^{3}\plus{}d^{3}\geq3abd$
$ a^{3}\plus{}c^{3}\plus{}d^{3}\geq3acd$
$ c^{3}\plus{}b^{3}\plus{}d^{3}\geq3cbd$

sum up and the result follows  :)

[quote="galois01"]Alternative solution

use Cauchy inequality cyclic

$ a^{3} \plus{} b^{3} \plus{} c^{3}\geq3abc$
$ a^{3} \plus{} b^{3} \plus{} d^{3}\geq3abd$
$ a^{3} \plus{} c^{3} \plus{} d^{3}\geq3acd$
$ c^{3} \plus{} b^{3} \plus{} d^{3}\geq3cbd$

sum up and the result follows  :)[/quote]

Yes! I solved the inequality with this.

[quote="CDP100"]If a, b, c, d > 0      ->

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \ge abc \plus{} abd \plus{} acd \plus{} bcd$[/quote]
Or Holder:
$ (a^3\plus{}b^3\plus{}c^3\plus{}d^3)(b^3\plus{}c^3\plus{}d^3\plus{}a^3)(c^3\plus{}d^3\plus{}a^3\plus{}b^3) \ge (abc\plus{}bcd\plus{}cda\plus{}dab)^3$

You can also use AM-GM on it (although that followed really from muirhead)


induction
modular arithmetic
number theory
prime numbers

[quote="Rushil"]Prove that $n^4 + 4^{n}$ is composite for all values of $n$ greater than $1$.[/quote]

The case $n \ : \ even$ is trivial.

If $n \ : \ odd$ then $n+1 \ : \ even$ say $n+1=2k$

Then $n^4+4^n=(n^2+2^n+n2^k)(n^2+2^n-n2^k)$

since $(n^2+2^n+n2^k)>(n^2+2^n-n2^k)$
we shall prove that $(n^2+2^n-n2^k)\ne 1$

But i don't know how to continue..... :?

[quote="Rushil"]Prove that $n^4 + 4^{n}$ is composite for all values of $n$ greater than $1$.[/quote]

[hide="The start of the solution - I have to go now..."]

All prime numbers must end in $1, 3, 7, \text { or } 9$.

\begin{tabular}[b]{c|cccccccc} n & 1 & 2& 3&4&5&\cdots&2n&2n+1\\ 4^n &4&6&4&6&4&\cdots&4&6\\ \end{tabular}[/hide]

famous and appears in engel book

n even you are done

n odd, put n = 2k+1 and factorize $n^4 + 4^n = (2k+1)^4 + 4*2^{4k}$

so it is $= ((2k+1)^2 + 2(2^{2k}) - 2(2k+1)(2^k))((2k+1)^2 + 2(2^{2k}) + 2(2k+1)(2^k))$

To prove that the first factor doesn't equal one, notice that $a^2 + 2b^2 \ge b^2 + 2ab > 2ab$ by amgm.

yes it is on engel... it is called Sophie Germain identidy right?

[quote="Kab^2"]Try and use induction to prove it : the base case is easy.[/quote]

Try not to revive topics that are almost a year old... especially, when you don't have something REALLY pressing to post and there are already solutions.

By the way, induction can't possibly work here.

Another way to do the $n$ is odd case is to look at it in mod 5:

$n^{4}\equiv 1 \pmod 5$ (by Fermat)

$n^{4}+4^{n}\equiv 1+(-1)^{n}\equiv 0\pmod 5$

EDIT: Actually that doesn't quite work because I didn't account for the case where $n\equiv 0 \pmod 5$, but I'm pretty sure you can come up for an argument to cover that case as well.

[quote="calculuslover800"]Another way to do the $n$ is odd case is to look at it in mod 5:

$n \neq 0$ since it's odd, so $n^{4}\equiv 1 \pmod 5$ (by Fermat)

$n^{4}+4^{n}\equiv 1+(-1)^{n}\equiv 0\pmod 5$

EDIT: in fact, if $p$ is a prime greater than two, then for all $n>1$, $n^{p-1}+(p-1)^{n}$ is composite, by the same argument, except with mod p for the odd case.[/quote]

The argument is NOT valid in all the cases.

[quote]$n \neq 0$ since it's odd, so $n^{4}\equiv 1 \pmod 5$ (by Fermat)[/quote]

In the above  part   there is   a hole   :)
for example,  if n=5 (or 15) , $n^{4}$ is NOT $\equiv 1 \pmod 5$
(Just because n is odd  and is not zero, does not mean it is  not a multiple  of  5 (or  0 (mod 5))

BTW: The theme of such problems is to  know how to factor:
$x^{4}+4y^{4}$
[hide]=$x^{4}+4y^{4}+4x^{2}y^{2}-4x^{2}y^{2}= (x^{2}+2y^{2})^{2}-(2xy)^{2}$
which is easy to factor. (diff of squares)

  when n is odd,  ($n^{4}+4^{n}$) can be  reduces to the form of
$x^{4}+4y^{4}$ ($x=n, y=2^{(n-1)/2)}$ }[/hide]

Hope this helps.

Sorry... the post was already revived before I posted, and I just messed up with the n=5k case, sorry about that. so I'll edit out my post.

"but I'm pretty sure you can come up for an argument to cover that case as well."

And I am pretty sure that you can't...

trivial for n=even and for n=odd , use SOPHIE GERMAN IDENTITY :D

[b][u][i]i think that sophie german is the only way to do the problem in a short span of time . If we use modular aljebra then i dont see a neat method to solve it . if someone has a solution by a method other than factorising please share it.[color=#ff9a00] :read: [/color][/i][/u][/b]

Case 1. $n$ is even:
$$\implies n^4 \equiv 0\pmod2$$
$$\implies n^4 + 4^n \equiv 0\pmod2$$
Since $n > 1$, $2|(n^4 + 4^n)$.

Case 2. $n$ is odd:
We use Sophie Germain and then AM-GM for one of the factors to get that it isn't prime which implies it is composite

everyone knows this one, sophie germain.

$$n^4 + 4^n = n^4 + 4\cdot4^{n - 1}$$
By Sophie Germain identity,
$$n^4 + 4\cdot4^{n - 1} = ((n + 4)^2 + 4^2)((n - 4)^2 + 4^2)$$
If it is prime, then either of these factors must be $1$. $(n + 4)^2 + 4^2$ is clearly not $1$ as $4^2 > 1$. $(n - 4)^2 > 0$ by trivial inequality, $4^2 > 1$. This means that $(n - 4)^2 + 4^2 \geq 4^2 > 1$. So none of these factors are $1$, which means that it is composite. $\blacksquare$

[hide=Solution]Rewriting in Sophie-Germain form gives $$n^4+4^n=n^4+4 \cdot 4^{n-1}=((n+4)^2+16)((n-4)^2+16).$$ For the number to be prime, one of these must be $1$. Clearly $(n+4)^2+16 \neq 1$ and since $(n-4)^2 > 0$, we also have $(n-4)^2 + 16 \neq 1$. Therefore, the number is composite, as desired. $\blacksquare$[/hide]


Hello everyone! I'm new to this forum, and need help on Fourier Series. I was solving a problem where the length of the interval was given to be 2pi. The function was odd, and it's value was (x/pi +1) in the interval -2pi<=x<=0, and (x/pi -1) in the interval 0<=x<=2pi. Now, what I wanted to know is whether I would get a different answer if I integrated the function (to find the bn) over the interval -2pi to +2pi instead of the interval -pi to +pi.

Thanks in advance!  :)

Ask yourself what you mean by "over the interval ... instead of ...". One of the implicit assumptions in calculating Fourier series is that the function is defined to be periodic outside the interval we are told about. So, is your variation still the same function or not?

Yes. The given function was a periodic sawtooth waveform stretching from -infinity to +infinity. However, does this mean that the length of the interval over which we calculate the Fourier Series is irrelevant? So calculating it over -pi to +pi will give me the same result as over -2pi to +2pi or -3pi to +3pi?

Why I'm asking is because I tried calculating it over both -pi to +pi and -2pi to +2pi and got slightly different answers. In the latter, my final answer included positive fractional angles and negative integer angles, whereas in the former, I only got negative integer angles.

The allowed terms in a Fourier series for a $4\pi$-periodic function are a subset of the allowed terms for a $2\pi$-periodic function. We should get the same series for both functions, although the version with length $4\pi$ will have extra terms with zero coefficients.


Feeling charitable today, so here's a cool problem

(AHSME 1981)

If $a,b,c,d$ are the solutions of the equation $x^4-bx-3=0$, then find the polynomial with leading coefficient 3 whose roots are \[ \displaystyle\frac{a+b+c}{d^2},\frac{a+b+d}{c^2},\frac{a+c+d}{b^2},\frac{b+c+d}{a^2}. \]

[hide]
We know that $a+b+c+d=0$ since there is no $x^3$ term.

$\displaystyle \frac{a+b+c}{d^2} + \displaystyle \frac{a+b+d}{c^2} + \displaystyle \frac{a+c+d}{b^2} + \displaystyle \frac{b+c+d}{a^2} = -\displaystyle \frac{d}{d^2} -\displaystyle \frac{c}{c^2} -\displaystyle \frac{-b}{b^2} - \displaystyle \frac{a}{a^2} = -\frac{1}{a} -\frac{1}{b} -\frac{1}{c} -\frac{1}{d}$

The polynomial whose roots are the reciprocals of the original polynomial is then: $-3x^4 - bx^3 + 1$.

With roots negative:

$-3x^4 + bx^3 + 1$.

Take the negative of the polynomial so that the leading coefficient is $3$:

$3x^4 -bx^3 - 1$
[/hide]

[hide]Since the sum a+b+c+d is 0,
$\frac{a+b+c}{d^2},\frac{a+b+d}{c^2},\frac{a+c+d}{b^2},\frac{b+c+d}{a^2}=\frac{-d}{d^2},\frac{-c}{c^2},\frac{-b}{b^2},\frac{-a}{a^2}=-\frac{1}{d},-\frac{1}{c},-\frac{1}{b},-\frac{1}{a}$.
The polynomial with reciprocal roots is the original polynomial with the coefficients reversed: $-3x^4-bx^3+1=0$.
The polynomial with negative reciprocal roots is $-3x^4+bx^3+1=0$.
Transpose to the other side to get $3x^4-bx^3-1=0$.[/hide]


What is the number of the acute triangles $ABC$ for which $AB=91$, $AC=40$ and $BC=x\in N^{*}\ ?$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=108290

[hide]In an acute triangle, the sum of the square of two sides must greater than the square of the third side.[/hide]

But not all of them are [u]acute[/u] - some are obtuse, obviously.

to find the accute ones use that

$a^{2}+b^{2}>c^{2}$ for accute triangles with sides $a,b,c$ and $c\geq{b}\geq{a}$

then apply this to the solutions found my alex_ib

So use $91^{2}+40^{2}>x^{2}, 91^{2}-40^{2}>x^{2}$. So my estimate is for the number of squares between $91^{2}+40^{2}$ and $91^{2}-40^{2}$ is $2*40^{2}$ divided by $2*90$ because at the very center, the squares are spaced apart by 180...


Before I post the series I would just like to clarify I put it in this forum because I am not sure whether or not there is a combinatorial meaning. I have found most sums of this form have a meaning =)
Also this is from a friend's homework question for her stats course. Finding the actual sum is just a mark of 6...so its not a lot. =) We've been at it for two days but we can't figure it out :(
Maple:
\[ \sum_{x \equal{} 1}^n x \binom{m \plus{} n \minus{} x \minus{} 1}{m \minus{} 1} \equal{} \frac {(m \plus{} n)!}{(m \plus{} 1)!(n \minus{} 1)!}
\]
Thanks =)

I guess I should state we're looking for how to figure out the sum. The steps.

Use the Hockey-stick identity twice.  See if you can find a nice interpretation using balls-and-urns.

I got a good combinatorial meaning now. The Hockey-stick identity is something I have not heard of before. I looked it up and see why it is called that... but I am not sure how to use it here.

Write $ x {m \plus{} n \minus{} x \minus{} 1 \choose m \minus{} 1} \equal{} \sum_{i \equal{} 1}^{x} {m \plus{} n \minus{} x \minus{} 1 \choose m \minus{} 1}$ and exchange the order of summation.  (I haven't actually worked out the details.)


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Prove that the automorphism group of the algebra of quaternions is isomorphic, as an algebraic group, to the subgroup $ SO_3$ of $ O_3$ consisting of the matrices with determinant 1.

Thanks in advance.

Every automorphism of $ \mathbb{H}$ maps its center $ \mathbb{R}$ into itself, the restriction is an automorphism of $ \mathbb{R}$, i.e. the identity. Hence every automorphism of $ \mathbb{H}$ is $ \mathbb{R}$-linear.
Because it is an automorphism it maps the threedimensional space $ \{ai \plus{} bj \plus{} ck | a,b,c\in\mathbb{R}\} \equal{} \{x\in\mathbb{H} | x^2\in\mathbb{R} \wedge x^2 < 0\}$ to itself. This gives a map $ Aut(\mathbb{H})\to GL_3(\mathbb{R})$ (and this map is a morphism of algebraic groups). Now show that every automorphism is in fact norm- and orientation preserving, so the image is in $ SO_3$. Prove that every element of $ SO_3$ can be realized by a conjugation with some $ u\in\mathbb{H}\setminus\{0\}$.

Thank you very much, but could you please tell me more specific on the inverse direction. Since every orthogonal matrix can be represented for an linear orthogonal operation, and how can I do next. Thanks

Use the that the multiplication in $ \mathbb{H}\equal{}\mathbb{R}\times\mathbb{R}^3$ is $ (a,v)(b,w)\equal{}(ab\minus{}\langle v,w\rangle, av\plus{}bw\plus{}v\times w)$.

thks alot. But Is it also true for Quaternions with complex coefficients?


[b]NIVEL BASICO[/b]

[b]Problema 1[/b] [Estrategias]
Borramos el primer dígito  (comenzando de la izquierda) del número $7^{2005}$, y entonces lo
agregamos al número restante. Esta operación es repetida hasta
obtener un número de $10$ dígitos. Probar que este número tiene al
menos dos dígitos iguales.
[Sugerido por:[i]Juan Neyra[/i]]

[b]Problema 2[/b] [Geometria] [b][Reemplazado][/b]
En un triángulo acutángulo $ABC$ sean $AE$ y $BF$ dos alturas, y sea
$H$ el ortocentro. La recta simétrica de $AE$ respecto de la
bisectriz (interior del ángulo en $A$ y la recta simétrica de $BF$
respecto de la bisectriz (interior) del ángulo en B se intersecan en
un punto $O$. Las rectas $AE$ y $AO$ cortan por segunda vez a la
circunferencia circunscrita al triángulo $ABC$ en los puntos $M$ y
$N$, respectivamente. Sean: $P$, la intersección de $BC$ con $HN$;
$R$, la intersección de $BC$ con $OM$; y $S$ la intersección de $HR$
con $OP$. Indicar la naturaleza del cuadrilátero $AHSO$.
[Sugerido por:[i]Jose C. Chavez[/i]]

[b]Problema 3[/b] [Inecuaciones]
Sean $a,b,c,x,y$ reales positivos. Mostrar:
\[ \frac{a}{xb+yc} + \frac{b}{xc + ya} + \frac{c}{xa+yb} \geq \frac{3}{x+y}  \]
[b]Comentario:[/b] Tomemos $x=1, y=1$ y tendriamos la Inecuacion de Nebsitt
[Sugerido por: [i]Carlos Bravo[/i]]

Disfruten de los problemas  ;)
Los Organizadores

[b]NIVEL AVANZADO[/b]

[b]Problema 1:[/b]

Se tienen $N$ hombres y $N$ mujeres en una fiesta, cada hombre y mujer hacen una lista de preferencias respecto a con quien desean bailar del sexo opuesto. Pruebe que los puede aparear de modo que no existan dos parejas $(H_1,M_1)$, $(H_2,M_2)$ que al cambiar de parejas, las 4 personas involucradas sean mas felices.

[b]Problema 2:[/b]

Sea $P$ un primo de Fermat, pruebe que la ecuacion $\sigma(n)=P^k$ no tiene solucion en los enteros positivos $(n,k)$.

($\sigma (n)$ representa la suma de los divisores positivos de $n$)

(P.Restrepo)

[b]Problema 3:[/b]

$\triangle ABC$ es acutangulo. Su incirculo toca  $BC$ en $K$. La altura $AD$ tiene punto medio  $M$. La linea $KM$ toca al incirculo de nuevo en $N$. Muestre que el circuncirculo de $BCN$ toca al incirculo de $ABC$ en $N$.

Los Organizadores ;)

Source: Los Organizadores

[color=blue][size=150]LISTA DE PROBLEMAS: RETO #2 - 2006[/size]
SEMANA: 09-16 FEBRERO[/color]

para el problema  1 .

 cada persona  hace una lista con  todas las personas desde con  la que mas prefiere  bailar hasta con la que menos?  o como es esta  lista?

yo supongo que cada uno hace una lista del que prefiere primero al que menos le gusta....

[color=blue]Favor de enviar
sus soluciones dentro del lapso establecido:

Viernes [10] al Jueves [16]

Gracias..

Los Organizadores [/color]

Una bonita solucion de Samper al problema 1 y un intento del 2, el dos fue problema 3 de la olimpiada colombiana 2005, propuesto por mi, nadie saco 7 puntos en el, veamos si alguien supera a los colombianos, porque Samper tampoco pudo, jaja!

EL problema 2 es una generalización del problema 3 de la olimpiada colombiana del año pasado... el problema original era con p=5 que es mas manejable. Igual yo sigo intentando porque creo que con lo que llevo sale.

Aquí está mi solución al problema 2 de básico, aunque el enunciado es muy largo, y el gráfico del problema es engorroso, si se ve el problema por partes, no resulta tan complicado.

$Tipe$

De hecho no es que fuera una generalizacion, es que decri que sea un primo de Fermat ya es una pista!!!! por eso decidimos poner un caso particular y el argumento ene ste es perfectamente analogo al del caso general.

Soluciones de Federico Castillo

[color=blue][b][size=150]SOLUCIONARIO DE PROBLEMAS: RETO #2 - 2006[/size][/b][/color]
[b][color=blue]SEMANA: 09-16 FEBRERO[/color][/b]
[b][color=blue]NIVEL BASICO[/color][/b]

$\blacktriangleright$ [color=blue]Pregunta 01[/color] [Estrategias]
Borramos el primer dígito (comenzando de la izquierda) del
n\'umero $7^{2005}$, y entonces lo agregamos al número restante.
Esta operación es repetida hasta obtener un n\'umero de $10$
dígitos. Probar que este número tiene al
menos dos dígitos iguales.
[Sugerido por:[i]Juan Neyra[/i]]

[hide="Solucion"]
Notemos que : $7^3 \equiv 1 \mod 9$
Luego:
$7^{2005} = ((7^3)^{668})\cdot7 \equiv 7 \mod 9.$
Tomemos: $\sum cifras = 7 \mod 9.$ (Invariante definido)
Luego de muchas operaciones, obtendremos lo siguiente:
$\overline{a_1a_2 \cdots a_{10}}.$
Donde, supondremos que se logra tener un numero con las 10 cifras
distintas: $a_i \neq a_j, \forall i,j \in \{1,2, \cdots,10 \}.$
Sin embargo, notemos que:
$\sum cifras = 0+1+2+ \cdots + 9 = 45 = 0 \mod 9.$
Lo que nos muestra una contradiccion, es decir que por lo menos
hay dos cifras iguales.
[/hide]

$\blacktriangleright$ [color=blue]Pregunta 02[/color] [Geometria]
En un triángulo acutángulo $ABC$ sean $AE$ y $BF$ dos alturas, y
sea $H$ el ortocentro. La recta simétrica de $AE$ respecto de la
bisectriz (interior del ángulo en $A$ y la recta simétrica de $BF$
respecto de la bisectriz (interior) del ángulo en B se intersecan
en un punto $O$. Las rectas $AE$ y $AO$ cortan por segunda vez a
la circunferencia circunscrita al triángulo $ABC$ en los puntos
$M$ y $N$, respectivamente. Sean: $P$, la intersección de $BC$ con
$HN$; $R$, la intersección de $BC$ con $OM$; y $S$ la intersección
de $HR$ con $OP$. Indicar la naturaleza del cuadrilátero $AHSO$.
[Sugerido por:[i]Jose C. Chavez[/i]]

[hide="Solucion"]
Solucion presentada por Jorge Tipe.
[/hide]

$\blacktriangleright$ [color=blue]Pregunta 03[/color]  [Inecuaciones]
Sean $a,b,c,x,y$ reales positivos. Mostrar:
$\frac{a}{xb+yc} + \frac{b}{xc + ya} + \frac{c}{xa+yb} \geq \frac{3}{x+y}$
{\bfseries Comentario:} Tomemos $x=1, y=1$ y tendriamos la Inecuaci\'on de Nebsitt
[Sugerido por: [i]Carlos Bravo[/i]]

[b][color=blue][Enviada por [i]Fabycv[/i] - Lima / PERU][/color][/b]
[hide="Solucion"]
De Cauchy Schwarz :
$(y_1 z_1+y_2z_2+\ldots+y_nz_n)^2\leq(y_1^2+y_2^2+\ldots+y_n^2)(z_1^2+z_2^2+\ldots+z_n^2)$
Haciendo un cambio: $y_i=\frac{a_i}{\sqrt{x_i}}$ y $z_i = \sqrt{x_i}.$
Tendriamos:
$\frac{a_1^2}{x_1} + \frac{a_2^2}{x_2} + ... +\frac{a_n^2}{x_n} \geq \frac{(a_1+a_2+ ...+a_n)^2}{(x_1+x_2+ ... + x_n)}.$
Llamado \emph{El Lema de Titu Andresscu.}
En el problema:
$\frac{a(a)}{a(xb+yc)} + \frac{b(b)}{b(xc + ya)} + \frac{c(c)}{c(xa+yb)} \geq \frac{(a+b+c)^2}{(x+y)(ab+bc+ca)}.$
Sin embargo, conocemos que: $a^2 + b^2 + c^2 \geq ab+bc+ca; \qquad a,b,c \in \mathbb{R}$
o tambien: $(a+b+c)^2 \geq 3(ab+bc+ca).$
De aqui, tenemos:
$\frac{a}{xb+yc} + \frac{b}{xc + ya} + \frac{c}{xa+yb} \geq \frac{(a+b+c)^2}{(x+y)(ab+bc+ca)} \geq \frac{3}{x+y}.$
[/hide]

Los Organizadores ;)

me gustaria ver la solucion  del problema 3 de este examen.  yo lo  intente  con inversion  pero sin suerte.  no le encontraba otra manera de atacarlo.  porfavor posteen  la solucion gracias
                                       saludos marko avila

 p.d: muchas felicidades  a  los organizadores han  hecho u  muy  buen trabajo.

[quote="marko avila"] me gustaria ver la solucion  del problema 3 de este examen.   .[/quote]

Este problema es de la shortlist de la IMO 2002  puedes ver una solución en esta dirección:  [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2002&p=118682]www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2002&p=118682[/url]

$Tipe$


if gcd(a;b)=1 then prove that a^2+b^2 has not divisor of the form 4k+3

for $ p$ prime such that $ p|a^2\plus{}b^2$ we have $ (ab^{\minus{}1})^2\equiv \minus{}1[p]$ then $ p\equiv 1[4]$

If $ a^2\plus{}b^2$ has a divisor of the form $ 4k\plus{}3$
then there is a prime number $ p\equal{}4l\plus{}3$, that $ a^2\plus{}b^2 \vdots p$
From this we can say that $ a \vdots p$ and $ b \vdots p$, or $ (a,b)\neq1$.

[quote="kimnimalar"][quote="aviateurpilot"]for $ p$ prime such that $ p|a^2 \plus{} b^2$ we have $ (ab^{ \minus{} 1})^2\equiv \minus{} 1[p]$ then $ p\equiv 1[4]$[/quote]

what this mean (ab^(-1))^2=(-1) [p]  ?

 I can't understand it?[/quote]
$ b^{\minus{}1}$ it's just notation,
we now that if $ p\not|b$, $ \exists u$ such that $ ub\equiv 1[mod\ p]$ (bezout) we note $ u\equal{}b^{\minus{}1}$.
if $ p|a^2\plus{}b^2$ then it's evident that $ p\not|b$ and $ a^{2}\equiv \minus{}b^2[mod\ p]$
we take $ x\equal{}au$ : $ x^{2}\equiv a^2u^{2}\equiv \minus{}b^2u^2\equiv \minus{}1 (mod\ p)$ (here $ u\equal{}b^{\minus{}1}$ !! ok)
so if $ p\equal{}4h\plus{}3$ then $ x^{p\minus{}1}\equal{}(x^{2})^{2h\plus{}1}\equiv (\minus{}1)^{2h\plus{}1}\equiv \minus{}1[p]$ (-impossible, by th.ferma)
then $ p\equiv 1[4]$


vector
linear algebra
matrix
integration
probability
real analysis
real analysis unsolved

Let $A$ and $B$ be nonsingular matrices of order $p$ ,and let $\alpha$,$\beta$be independent random vectors of dimension $p$,show that if $\alpha,\beta$ and $\alpha.A+\beta.B$ have the same distribution,if their first and second moments exists ,and if their covariance matrix is the identity matrix ,then these random vectors are normally distributed.

Which covariance matrix is the identity? $\alpha$ with $\alpha$? $\alpha$ with $\beta$? $\alpha$ with $A\alpha+B\beta$? $A\alpha+B\beta$ with itself? (Oh - I habitually write vectors as column vectors, hence the notation.)

all of them are identity (excuse me I have edited the old one). :blush:

"all of them" can't possibly be right; it contradicts independence.

The place to start is the "identically distributed" fact. We get $\frac1{\det(AB)}\int r(A^{-1}u)r(B^{-1}(x-u))\,dx=r(x)$, where $r$ is the density function.

Now, what's the right transform method for vector-valued probability distributions?

by "all of them" I did not mean all combination each rv with itself


[quote="pascal_1623"]Bobby is having a party at his pool. He needs to see how big the swimming pool.
(He has a way of building things very fast.) :D
He invites 10 people.
The average amount of space a person wanted was 729 cubic yards.
He makes the pool so that everyone had that average space.
If Bobby intends to swim, how big should the pool be?
Express your answer as the surface area of the water.[/quote]
[hide="my answer  :D "]
Ok so you must do 11(# of people) times the cube root of 729 (which is 9) , squared.

$11*81= 891.$

[size=200]891
[/size][/hide]

[hide]If he and 10 other people swim, there are 11 people in the pool.  $\sqrt[3]{729}=9$  Then, square 9 to convert to the surface area for each person.  $9^2=81$  Multiply by 11 because there are 11 people.  $81\cdot11=\boxed{891}$[/hide]