inequalities
algebra
polynomial
calculus
2-variable inequality

$a,b$ are non negative reals not both equal to $0$. Prove that
\[\frac{a^4+b^4}{(a+b)^4}  + \frac{\sqrt{ab}}{a+b}   \geq \frac58\]

Without loss of generality we may assume that $a+b=1$. We must prove that $a^4+b^4+\sqrt{ab} \geq \frac58$.
On the boundary $a=0$ the inequality is trivial. Than by the Lagrange principle there exists $\lambda$ such that the following equalities hold:
\[4a^3+\frac12\sqrt{\frac{b}a}=\lambda, \  \ \ \ 4b^3+\frac12\sqrt{\frac{a}b}=\lambda\]

then $8(a^3-b^3)=\frac{a-b}{\sqrt{ab}}$. So $a-b=0$ (then $a=b=\frac12$ and inequality become to equality), or $8(1-ab)\sqrt{ab}=1$. It is easy to see that
$\sqrt{ab} \in \left(0;\frac12\right)$, but there is no root of the polynomial $t^3-t+\frac18$ in that interval.

Your proof is good, Fedor.
I am thinking about a proof without calculus but I don't know if it exists.

We can prove a stronger result:
\[\frac{a^4 + b^4}{(a+b)^4} +  \frac{3ab}{(a+b)^2}  \geq \frac78 \ \ (1)\]
(by an easy computation it is equivalent to $(a-b)^4  \geq 0$.

Our desired inequality now follows from $(1)$ because
\[ \frac12 \geq \frac{\sqrt{ab}}{a+b}  \geq \frac{2ab}{(a+b)^2}\]

Congratulation Namdung, your proof is very nice and very neat.
A beautiful reasoning!!


Figure A is a single wooden cube that has edge length 3 meters.  Square holes with side length 1 meter centered in each face are cut through to the opposite face.  The edges of the holes are paralell to the edges of the cube.  Figure B is formed by connecting 27 of these cubes.  What is the number of square meters in the total surface area of Figure B?

I don't have the original diagram, but if anyone would like to post it, feel free to do so. :lol:
I've never understood this question for some reason, though the rest of the questions from that year are EASY.

this problem isn't particularly difficult...im pretty sure aops vol. 1 has a diagram. was this a target?
anyways, basically just imagine a cube with a hole in each center square on each face..the way i do it is i find the surface area of the outside faces first (which in this case is $ (9\minus{}1)\cdot 6$ then imagine what the inside of the cube looks like...that should help.  :)

But wouldn't that only apply for Figure A?  Figure B's 27 smaller squares would have some of the faces covered by other squares, right?

Im assuming that fig B is a $ 3 \times 3 \times 3$ of Fig A.
Thus you can see the interior surface from the holes is always counted.
There are 54 faces of Fig A on the cube.

interior surface area $ 8*27\equal{}216$
Exterior $ 54*8\equal{}432$
Surface Area$ \equal{}216\plus{}432\equal{}648$

abacadaea, that's what I got, but the answer's [hide]1080[/hide] :huh:

Ahh I miscounted.
The exterior surface is the same.
The interior surface of Fig A is $ 27*24\equal{}648$
Total Surface Area$ \equal{}648 \plus{}54*8\equal{}1080$

[quote="Fraenkel"][quote="abacadaea"]firgure B has 9*6=54 faces of figure A on it. each face of figure A has a surface area of 9+4=13. then do 54*13=702.

[b]frankel-[/b]where did you get you interior surface area?[/quote]

Figure A has holes in it.[/quote]
yes i realize that. the surface area of each hole is 4 more thean the side would be without a square. then there would be 5*54+8*54

EDIT: you beat me to the post i probably misinterpreted the diagram

I think this was #6 on a handout (I vaguely remember looking over some kids on my team's paper...and they were a bit confused since they got it wrong)

so just do

[hide]
6(72) for outer part + 27(24) for inner part

432+648=1080

so the 72 is just each side with 9 'blocks' of 8 squares in each so 72 (and 6 of them so 1080)

and the 24 is just the surface area for each 'small' cube (then you just multiply by 27 because all the holes are connected, so it doesn't decrease the area amount)

there's probably an easier way  :oops: [hide]

hope that helps![/hide][/hide]

[quote="Math Geek"]I don't understand.  How did you get 24?  I keep getting 30 for the surface area of the small cube.    I think that there is a 1x1x1 cube in the center of figure A, so it's faces should be included in the surface area, right? :blush:[/quote]

ok so the surface area for the inner part of a 3x3x3 cube is 24 for the following reason (imagine this...). alright, so you have 6 'tunnels' each from a different side of the cube, and in these tunnels you have open space surrounded by 4 1x1 'walls', thus the surface area of each of these is 4...multiply that by 6 and you have 24. the reason why you don't have any surface area in the middle is because it's empty space surrounded by those 'tunnels' (aka no 'walls' -> no surface area)...so 24 is it...hope that helps
(and you multiply by 27 because none of the inner surface area is blocked off by other 3x3x3 cubes)

i hope my visual (without being visual) explanation was ok

maybe an experienced mathcounts person can help on the image if you still don't understand

It's a bit hard to post any more than this...

(i'm attempting to attach it)

oh whoops. i thought there was a little indent on each side of a figure A

so, umm, the answer is 1080?  that's what i got.....i get dizzy when i see this problem.   :D

If you read some of the other entries, you would know that the answer is 1080. Please don't revive  :wink:


how do we factor this?

x^2 - 7x + 12

and

x^2 - 7x - 60

Thank you.

[quote] Knowledge is power. [/quote]

You are absolutely right, thus you'd better do your homework by yourself.  :P

Pierre.

Factors of 12 that add up to -7, (-3 and -4) (x-3)(x-4)
Factors of -60 that add up to -7, (-12 and 5) (x-12)(x+5)

i would do this by:

1.  What multiplie to 12
     and adds to -7
 well -4 and -3 do.
so it would be (x-4)(x-3)

2.what multiplies to 60
   and adds to -7
well, -12 and 5 do.
so it would be (x-12)(x+7)

[quote="nat mc"]Just like me  :lol:[/quote]
yea, i'm just trying to get better at typing.


Let the function $f: \mathbb{R}\to\mathbb{R}$ satisfy:
$f(1)=2007$ and $f(1)+f(2)+\cdots+f(n)=n^2f(n), \forall n \in \mathbb{N}$.
Find $f(2006)$.

$f(1)+f(2)+\cdots+f(n)=n^2f(n)$
$f(1)+f(2)+\cdots+f(n)+f(n+1)=(n+1)^2f(n+1)$
By subtracting first equation from the second we obtain:
$f(n+1)=(n+1)^2f(n+1)-n^2f(n)$
$f(n+1)=\frac{n}{n+2}f(n)$
By induction we have:
$\boxed{f(n)=\frac{4014}{n(n+1)}}$

Olympiad Matematik Kebangsaan. Means Malaysia National Mathematic Olympiad


we drop small balls (can considered to be points) on a concave half sphere (that is the half sphere is open upwards) of radius $ R$. We drop the balls from height $ h$ with respect to the point it hits in the half sphere. find $ h$ so that all the balls will bounce after one single fully elastic collision to the very bottom point of the sphere.

Probably i misunderstood the problem, anyway, let's try..
The balls hit the half sphere in the point P and rebounce elastically in the direction of the radius (OP). The velocity is [b]v[/b] with an angle of [b]90 - t[/b] relative to the horizontal axis (look at the picture!). Now i consider the parabolic motion of a single ball and i use the parametric equation.

[i]y = y(0) + tg(90 - t) * x - (g*x^2)/(2*(v*cos(90-t)^2)[/i]

The ball has a horizontal distance from the bottom point H equal to [b]R*sin(t)[/b], while it stands at the height [b]R*(1 - cos(t))[/b]. Setting y=0 and x=[b]R*sin(t)[/b], we calculate the velocity v

0 = R(1-cos(t)) + R sin(t)*cos(t)/sin(t) - (g*(Rsin(t))^2)/(2*(v*sin(t))^2)

1 - cos (t) + cos(t) - g*R/(2*(v^2)) = 0

v^2 = g*R/2         (1)

The ball is dropped from the height h with respect to point P, with an inizial velocity=0, so
v^2 = 2*g*h

From the (1) we have:  h= R/4

Sorry, but i haven't latex yet...let me know if it's right, even if i think as i understood the problem is quite easy..
enjoy :D

That's not quite right.

After elastic collision the angle with the normal (t) is the same. So if that is correct what you say, then t=90-t which is clearly wrong...  :wink:

so the balls don't fall in the normal direction...well, i considered the balls' direction after they hit the sphere as the direction of reflected rays...
i'm not sure of what u say, the direction should be normal to the surface, so radial!
Can you post a picture? If the direction is the same before and after they hit the sphere, we only need to substitue t to 90-t...
see u

ok here is a rough diagram  :)

so after the collision the angle of the velocity with the radius is $ \phi$ hence the angle of the velocity with the horizontal is $ 90\minus{}2\phi$.

ok!! Now it's clear..the ball rebounces with the same angle relative to the radial direction, because the collision is elastically (i didn't understood how toe express this fact)
Substituing 90 - 2t in my wrong solution, we have:

h= R/(8*cos(t)*(2cos(t)-1))
or something like that...
let's see if u can post harder problems :lol:  :P

yes thats correct :)

Well I can post harder problems, check my previous ones. some of them are really tough. Anyway I will post another hard one if you wish :)


algebra
polynomial
geometry
rhombus
national olympiad

\[\text{Day 1}\]
1. Polynomial $P(x)$ has integer coefficients. Prove, that if polynomials $P(x)$ and $P(P(P(x)))$ have common real root, they also have a common integer root.

2. $ABCDE$ is a convex pentagon and:
\[BC=CD, \;\;\; DE=EA, \;\;\; \angle BCD=\angle DEA=90^{\circ}\]
Prove, that it is possible to build a triangle from segments $AC$, $CE$, $EB$. Find the value of its angles if $\angle ACE=\alpha$ and $\angle BEC=\beta$.

3. We are given equilateral triangle with side $n$ which is divided into $n^{2}$ congruent equilateral triangles with side $1$. Let's call these small triangles plates. Each plate is either black or white. We perform the following operation:
we choose a plate which has common sides with at least two plates of diiferent colour from chosen plate and then we change the colour of our plate. For every $n\geq 2$ decide whether exists a configuration of colours that enables to perform infinite sequence of operations.
\[\text{Day 2}\]
4. $a$, $b$, $c$, $d$ are positive integers and
\[ad=b^{2}+bc+c^{2}\]
Prove that
\[a^{2}+b^{2}+c^{2}+d^{2}\]
is a composed number.

5. A convex quadrilateral $ABCD$ with $AB\neq CD$ is cyclic. Quadrilaterals $AKDL$ and $CMBN$ are rhombs with side $a$. Prove that quadrilateral $KLMN$ is cyclic.

6. $a$, $b$, $c$, $d$ are positive real numbers satisfying the following condition:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4 \]
Prove that:
\[\sqrt[3]{\frac{a^{3}+b^{3}}{2}}+\sqrt[3]{\frac{b^{3}+c^{3}}{2}}+\sqrt[3]{\frac{c^{3}+d^{3}}{2}}+\sqrt[3]{\frac{d^{3}+a^{3}}{2}}\leq 2(a+b+c+d)-4. \]
Thanks robpal and barasawala  :wink:

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=136487
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=136488
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=136490
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=136491
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=136489
6. http://www.mathlinks.ro/Forum/viewtopic.php?t=135374


What's your favorite fruit?

See Fav. fruit 1 and Fav. fruit 3 if your favorite fruit is not here. The "other" option is in Fav. fruit 3. Please only vote in one of the polls.

Fav. fruit 1 is the main thread for discussion.

Sorry if your favorite fruit is bunched up in a category and you don't like the other fruits in that category.

*bump* for those who might not know about this and number 3...


calculus
derivative
function
calculus computations

Hi, i have two problem that i cant resolve "please help me"  :P

1.-  Change in temperature along a circunference

A particle is moving in the direction of a clockwise around the circle of 1m radius centered at the origin. Rate of 2 m/s.

How fast the temperature experimented by the particle change in degrees Celsius per meter, in the point

[url=http://imageshack.us][img]http://img17.imageshack.us/img17/7596/pointuo6.jpg[/img][/url]

2.- I have to prove the wave equation:

 [url=http://img145.imageshack.us/my.php?image=waveecuationqm0.jpg][img]http://img145.imageshack.us/img145/2910/waveecuationqm0.jpg[/img][/url]

Thanks !

The first problem doesn't make any sense as written. I suspect that you just forgot half of the conditions.

As to the second, you cannot "prove the wave equation". I suspect that you were asked to prove that the given function is a solution to it. Then all you need to do is to differentiate a few times using the chain rule.

Thanks Fedja for the reply.

The first problem was taken from a book, it must be rigth :S

The only thing i could do is find the ecuation of the particle movement but i don know how to find a conection with the temperature ecuation.

The particle ecuation -> r(t) = sen (2t) i + cos (2t) j

This ecuation move the particle in the direction of a clockwise around a circle of 1m radius centered at the origin with rate of 2 m/s.

I have to find:

dt = (Directional Derivative in the point) . df

I think the direction of the derivative is the direction of the particle velocity in the point, but i dont know how to find 'df'

And in the second problem i have to 'prove that the given function is a solution',  :P but when i tried to prove that i find some complications, if someone can teach me the steps to resolve that  :oops:


[quote="digger"]What is your question? Would you like to solve this system in integers?
If I was not failed to understand the question it is extremely easy.

Subtraction gives $ x_{1}=90$ and the system becames $ 5x_{2}+x_{3}+x_{4}=-107$.
Take any $ x_{2}$ and $ x_{3}$ and calculate $ x_{4}$ from this.
Answer:
$ x_{1}=90$
$ x_{2}=u$
$ x_{3}=v$
$ x_{4}=-107-5u-v$
for arbitrary integers $ u$ and $ v$.[/quote]

Its finding the number of non-negative solutions for each equation. the two are independent from each other. I just tried number 2 if its easy to solve given a higher sum. Can't seem to find an easy way to do it.


integration
calculus
calculus computations

How can we compute $\underset{k=0}{\overset{+\infty }{\sum }}\frac{(-1)^{k}x^{2k+1}}{(2k+1)k!}$ for a certain real $x$ using fourier series ? (if possible of course  :lol: )

This is $\int_0^xe^{-t^2}\,dt.$

We're not going to find a closed form formula; we know that $e^{-t^2}$ has no elementary antiderivative.

The limit as $x\to\infty$ is well known. If we want numerical approximations, for $x$ not too large, the power series itself converges satisfactorily rapidly. There's a useful asymptotic expression we could use for large $x.$

I don't see a natural or straigtforward link to Fourier series.

Ok, I just wanted to know if there is a way to link it with fourier series, because my objectif was to compute that well known integral through it. But as it seems, it's not possible  :D
Thanks


CuSO4(aq) + Fe(s) --> FeSO4(aq) + Cu(s)
FeSO4(aq) + Cu(s) --> CuSO4(aq) + Fe(s)

[img]http://img106.imageshack.us/img106/6140/chem3fp.png[/img]

Why does the iron plate becomes brown?
I have no precise idea, but I think it might be related to the loss of electrons of the iron plate. Could someone help, please?

It is because the iron is going into solution and the copper is coming out of solution forming the brown you see on the iron plate.

CuSO4(aq) + Fe(s) --> FeSO4(aq) +  [b]Cu(s)[/b]

thats why...


[quote="mathgeniuse^ln(x)"][quote="Drunken_Math"]For what values of $x$ is this defined?

$\frac{23}{(x-1)(x+2)(x-3)(x+4)}$

Write in interval  notation if you can.[/quote]

[hide]The only values not in the domain are -4,-2,1,3

So the values in this domain are $(-\infty,-4)U(-4,-2)U(-2,1)U(1,3)U(3, \infty)$.[/hide][/quote]

Good job!  :D


trigonometry
inequalities
function
calculus
derivative
geometry
inequalities unsolved

prove that for x in (0,pi/2) we have sin(2x)<2/(3x-x^3)

can anybody tell me the proof of this inequality? and similar ones, that's sinx > a function of x. I have no idea how to prove this kind of inequality. Or are there some inequalities that are well-known to prove this kind of inequalities? Thx

would differentation help?  or one could try a geometrical approach to compare areas or side lengths.

like the proof of tan x > x > sin x and the proof of (tan x + sin x)/2 > x (i think the second's correct its an old hungarian problem)

x^3+2>=3x   and   x<pi/2<sqrt(3)
which means that  2/(3x-x^3)>=1>=sin(2x)
Cases of equality are different so equality never holds.


Suppose you could uniformly heat a liquid in a container under the normal gravitational pull of the Earth. Assume that liquid exerts a pressure due to its own weight . Will the liquid start to boil on the bottom, at the top, or uniformly throughout?

My theory is that the liquid at the bottom must bear the extra pressure of not only atmospheric but also the water above it, so it boils at a higher temperature (b/c boiling point happens when vapor presure equals atmospheric pressure. So is my logic right? Or can someone explain this in a more clear way?

[quote="Bythecliff"]Assume that liquid exerts a pressure due to its own weight[/quote]

A liquid always exerts a pressure on its container. Are you taking into account in your argument the currents of convection?

[quote="Bythecliff"]boiling point happens when vapor presure equals atmospheric pressure[/quote]

It would be more correct to say "... when vapour pressure equals the external pressure", since, for example, when cooking in a "pressure apparatus" (I apologize, I don't know how to say this in English) the boiling doesn't happen when the vapour pressure equals atmospheric pressure since the external pressure is quite greater than atmospheric pressure - because the container is closed.

The English for that is "pressure cooker."

Bythecliff's thought experiment might be difficult to do in practice, but if you could, the fluid [i]wouldn't[/i] convect. The ambient pressure would be lowest at the top, and boiling would happen first at the top. Furthermore, it wouldn't bubble. It wouldn't even look like it was boiling - but vapor would rise from it and the fluid level would drop.

Our experience of boiling, with pots of water on the stove, involves situations in which all, or almost all, of the heat is added to the lower surface of the container. So the water boils from the bottom, bubbling and convecting furiously. Even a coffee cup stuck in a microwave oven isn't that close to the heat being added uniformly - the rate of heat transfer should be greater towards the outside of the cup and lower in the interior.

[quote="Kent Merryfield"]The English for that is "pressure cooker."[/quote]

Ok, thanks.

[quote="Kent Merryfield"]the fluid wouldn't convect. The ambient pressure would be lowest at the top[/quote]

Yes, I agree with that. With uniform heating, convenction don't make sense. The external pressure is lowest at the top, that's true, since in the interior of the liquid we have to take into account the hydrostatic pressure. However, note that the hydrostatic pressure contibution is neglegible: for example, for a column of water with 50 cm tall the pressure at the bottom would be about 1.05 atm, just 0.05 atm higher than atmospheric pressure. So, I think it would be also reasonable to consider that the liquid would start to boil is a very close uniform manner throughout the liquid.


take two irrationals q1 and q2, with q1 does not equal q2. is it true that there is always a rational r1 such that q1<r1<q2? prove this if true.

new problem -
take two rationals x1 and x2, with x1 doesnt equal x2. is it true that there is always an irrational number i1 such that x1<i1<x2? prove this if true.

a) In the decimal representation of q_1 and q_2, go from right to left, and find the first time that a digit differs (this must happen, otherwise $ q_1\equal{}q_2$). Take all of the digits to the right of q_2 and change them all to 0.

e.g. 314.141285689... and 314.141285694... the first time they differ is 314.14128568 and 314.14128569. 314.14128569 is you number.

b)$ i_1\equal{}x_1\plus{}\frac{x_2\minus{}x_1}{\pi}$ it's clearly larger than x_1 and $ x_1\plus{}\frac{x_2\minus{}x_1}{\pi}<x_2$

<-->$ \frac{x_2\minus{}x_1}{\pi}<x_2\minus{}x_1$ <--> $ 0<(x_2\minus{}x_1)(\pi\minus{}1)$ which is true.

Both of these statements are true because,

[i]The set of rational numbers is dense in the reals.[/i]

Here's a proof for the first statement,

[hide="Proof"]
(i) Let $ q_1,q_2\in\mathbb{R}: q_1<q_2$ and $ r\in\mathbb{Q}$.

(ii) Therefore $ q_2\minus{}q_1>0$ and $ \frac{1}{q_2\minus{}q_1}>0$.

Lemma: Let $ x\in\mathbb{R}$. Then there exists $ n\in\mathbb{N}$ such that $ n>x$.

(iii) By the lemma, there exists $ n\in\mathbb{N}$ such that $ n>\frac{1}{q_2\minus{}q_1}>0$.

(iv) Therefore, $ 0<\frac{1}{n}<q_2\minus{}q_1$

(v) By the lemma, there exists $ m\in\mathbb{N}$ such that $ \frac{m_1}{n}>q_1$.

(Note: This is true because, by the lemma, there is an $ m_1\in\mathbb{N}: m_1>nq_1$. Dividing through by $ n$ yields $ \frac{m_1}{n}>q_1$.)

(vi) With the same logic as in (v), we know there exists $ m_2\in\mathbb{N}$ such that $ \minus{}\frac{m_2}{n}<q_1<\frac{m_1}{n}$.

(vii) Therefore, the set $ m_k\in\mathbb{Z}$ for which $ \frac{m_k}{n}>q_1$ is non-empty, and bounded below.

(viii) Thus is has a smallest element $ m$. Therefore, $ \frac{m\minus{}1}{n}<q_1$.

(ix) $ \frac{m}{n}\equal{}\frac{m\minus{}1}{n}\plus{}\frac{1}{n}<q_1\plus{}\frac{1}{n}<q_1\plus{}(q_2\minus{}q_1)\equal{}q_2$

(x) We have found a value of $ m\in\mathbb{Z}$ for which $ \frac{m}{n}>q_1$ and $ \frac{m}{n}<q_2$.

Q.E.D.
[/hide]

Actually, what we have shown is more general. We have shown that there exists a rational number between any two distinct real numbers, which statement number one is a special case of.

If you want to see the proof of the lemma used in the proof, I will be glad to post it.

What I would suggest is switch $ q$ and $ r$, so $ r_1, r_2 \in \mathbb R$ and $ q \in \mathbb Q$. It just seems more natural that way. :)

Because when I go to the middle of the proof, when I see $ q_1$ and $ q_2$, I get confused and think "Are they the rational numbers or the real numbers?"

Also:
[quote="JZambrano201"]
(v) By the lemma, there exists $ \color{red} m_1\in\mathbb{N}$ such that $ \frac {m_1}{n} > q_1$.
[/quote]

Just very minor tecnhicalities, the proof overall is great :D

Thanks for the insight. I kind of confused myself too :) . Trying to make a proof readable is almost as hard as the proof itself sometimes.


algebra
binomial theorem
complex numbers
combinatorics unsolved
combinatorics

Show that if n is a positive multiple of 6,

1) nCr(n, 1) - 3 nCr(n,3) + 3^2 nCr(n,5) - .... = 0

2) nCr(n, 1) - 1/3 nCr(n,3) + 1/3^2 nCr(n,5) - .... = 0

Combintorial argument or whatever cleverness is welcome.

Such statements can be proved using complex numbers. By binomial theorem it's quite obvious, taht the first one is equivalent to showing that the imaginary part of $(1+i\sqrt 3)^n$ is equal to 0, which is true, because $(1+i\sqrt 3)^6=2^6$, so if 6 divides n, the imginary part is 0. The second one is equivalent of showing that imaginary part of $\left(1+i\frac{\sqrt 3} 3\right)^n$ is equal to 0,
$\left(1+i\frac{\sqrt 3} 3\right)^6=\frac{2^6} {3^3}$, so the imaginary part is  0 again.


Harvard college calculus Princeton percent Stanford MIT

I am interested in majoring in computer science/mathematics.

I have reasearching the top schools and I know that Harvard has a good math department. However, I was wondering if anyone could give me any information regarding the quality of Harvard's CS department.

Also, I am currently enrolled in a Community college and will earn an AS degree and high school diploma concurrently. Does anyone know weather or not I can transfer my credits? I have read the site, but the information seems a little vague.

Any information would be greatly appreciated.

You cannot transfer community college credits to most of the highly selective elite colleges, such as Harvard. Are you planning to enter Harvard as a first-year undergraduate, or as a transfer student?

Tokenadult is 100% correct. I've talked to two Harvard representatives and they both told me the same thing. (:()

A good idea might be to take the corresponding AP test to get the credit. That's what I did for Calculus BC and I had to do about 2 hrs of self-study and got a 5, and I plan to take a few sciences to get some credit also. You might need to pay some money out of your pocket depending on your school but I'm sure that it'll cost you less than a semester or two of college.

Good luck!

Question:

It would probably be at my disadvantage to apply to Harvard as a transfer from a community college rather than a highschool applicant, correct?

Odds of admission to most of the elite colleges are much lower for transfer applicants than for freshman applicants. Princeton claims to not accept transfer applications at all. So the smart way to take college-level courses before applying to an elite college is to take them as part of your "high school" program, which is something done by about 5 percent of the high school students in the United States.

[url=http://www.ed.gov/news/pressreleases/2005/04/04062005a.html]High School Students Using Dual-Enrollment Programs[/url]

[quote="tokenadult"]Odds of admission to most of the elite colleges are much lower for transfer applicants than for freshman applicants. [/quote]

That statistic is not a reliable indicator.   Transfer students are subject to different application procedures and the standards may be higher or (sometimes dramatically) lower than for the regular population.

[quote="tokenadult"]What are successful examples of transfer applications known to readers of this thread that fit the pattern described by the original poster?[/quote]

The original poster was talking about Harvard CS.  I know several successful transfer applicants to Harvard from community college (or state institutions comparable to community college),  including at least one in CS.    Whether or not they received one-to-one transfer credits, the transfer generally did not increase the number of semesters it took them to graduate.   Transfer of credits is judged on arrival by professors from the relevant departments.

Thank you for your information regarding transferability of credits. Now how about some information regarding the quality of Harvard's computer science department

I should say, having reviewed your first message, that the issue really is that if you do your college credits as part of a high school program, that will help you be more likely to be admitted, but you won't get any transfer credits at all. For sure. As long as you are getting a high school diploma as part of the program you are doing, you are a high school student in Harvard's eyes, irrespective of the associates degree you simultaneously earn. Advanced standing at Harvard is based on test scores, not on previous courses taken.

[url=http://www.fas.harvard.edu/~fdo/publications/advancedstanding0607/]Advanced Standing at Harvard College[/url]

But you will always be placed in appropriate courses, based on your background, and Harvard takes care to make sure that it's courses are challenging even for very well prepared students.

As for whether you should apply to Harvard if you are very interested in computer science, sure, why not? The traditional list of "top five" computer science undergraduate departments in the United States includes Stanford, MIT, Carnegie Mellon, and one or two other non-Harvard colleges that I'm sure someone will fill in here, but Harvard's department is not bad.

[quote="tokenadult"] you will always be placed in appropriate courses, based on your background [/quote]

The top US universities, including Harvard,  tend to have plain vanilla placement exams to calibrate matriculants' level within the basic calculus, French, etc course sequences.   Apart from that and some perfunctory supervision by academic advisors, students can enroll in a nearly unrestricted set of classes.   Attempts to take a third-year PhD class by somebody just out of high school might require some permission, but the general philosophy, particularly at Harvard, was to "give them enough rope to hang themselves".   There has been a trend toward increased supervision in the past decade, but also increased awareness that some students enter with a specialized background in science and should be given relatively free rein.

Harvard CS in the past had some notable faculty such as Rabin and Valiant, but otherwise was not a strong department.   More recently they invested a lot of money to upgrade the department, and the quality is closer to top-level now.

Harvard faculty is strong across the board, including CS, however, the CS dept is clearly not as strong as the biggies like MIT.  Perhaps the biggest difference at Harvard is that the CS dept is much much smaller than at the elite CS schools.

From an undergrad point of view, the biggest differences would be: (a) far fewer courses to choose from (although cross-enrollement at MIT is possible, of course), and (b) smaller, more intimate atmosphere.  You'll probably see the same people in your classes and know everyone who is concentrating in CS.  The students in the Harvard CS dept are probably as good as those anywhere.  (Again, because of numbers, you probably don't hear much about them.)

Also important is how sure you are that you want to do CS.

[quote="yenlee"]Also important is how sure you are that you want to do CS.[/quote]

Yes, that is very important. Harvard offers many other majors ("concentrations," in Harvard terminology) and most of those are academically strong and well regarded. Some other schools with notable CS departments have few other strong departments, so Harvard has value as a place where you can change your mind and still end up with a good major program.


Find all positive integers $ n$ such that the following inequality holds for positive reals $ a_{1},a_{2},\cdots,a_{n}$:

$ \left(\sum^{n}_{i=1}a^{2}_{i}\right)\left(\sum^{n}_{i=1}a_{i}\right)-\left(\sum^{n}_{i=1}a^{3}_{i}\right)\geq6\;a_{1}a_{2}\cdots a_{n}$

[quote="navid"]This problem purposed in Polish Mathematical Contests And: n=3.[/quote]

well nice Navid, would you please send you solution too! :wink:

If $ n>3$, then since the degree of right side is larger than 3 and both right side and left side is positive, for big positive real, the inequality flips. In the similar way, if $ n<3$, contradiction. So, the remain case is $ n=3$, and the inequality changes to $ \sum_{sym}a_{1}^{2}a_{2}\geq 6a_{1}a_{2}a_{3}$, which is true by muirhead.


Find a closed-form formula for the following sequence: $ a_n\equal{}\frac{2}{n}(a_1\plus{}\cdots\plus{}a_{n\minus{}1})\plus{}kn$ (in terms of $ k$ and $ a_1$).

Obviosly $ na_n \minus{}(n\plus{}1)a_{n\minus{}1}\equal{}k(2n\minus{}1),n>1$.
Let $ f(x)\equal{}\sum_n a_nx^n$. Then we get $ f'(x)(1\minus{}x)\minus{}2f(x)\equal{}a_1\plus{}\frac{1\plus{}x}{(1\minus{}x)^2}.$
It give $ f(x)\equal{}\frac{a_1(x\minus{}x^2/2)\minus{}kx\minus{}2ln|1\minus{}x|}{(1\minus{}x)^2}$. It give closed form for $ a_n\equal{}\frac{f^{(n)}(0)}{n!}$.

That's not really a closed-form answer; otherwise any generating function would work as a closed-form answer
for a sequence. (Although the g.f. is often the best you can do...)

I think Rust's point is that this particular generating function has a Taylor series which is relatively straightforward to compute.

[quote="bambaman"]Find a closed-form formula for the following sequence: $ a_n \equal{} \frac {2}{n}(a_1 \plus{} \cdots \plus{} a_{n \minus{} 1}) \plus{} kn$ (in terms of $ k$ and $ a_1$).[/quote]

I dont think there is a closed form for this sequence :

$ \forall n > 1$ : $ na_n \equal{} 2\sum_{i \equal{} 1}^{n \minus{} 1}a_i \plus{} kn^2$
$ \forall n > 2$ : $ (n \minus{} 1)a_{n \minus{} 1} \equal{} 2\sum_{i \equal{} 1}^{n \minus{} 2}a_i \plus{} k(n \minus{} 1)$ and so $ (n \plus{} 1)a_{n \minus{} 1} \equal{} 2\sum_{i \equal{} 1}^{n \minus{} 1}a_i \plus{} k(n \minus{} 1)^2$

So, $ \forall n > 2$ : $ na_n \minus{} (n \plus{} 1)a_{n \minus{} 1} \equal{} k(2n \minus{} 1)$

So, $ \forall n > 2$ : $ \frac {a_n}{n \plus{} 1} \minus{} \frac {a_{n \minus{} 1}}{n} \equal{} k\frac {2n \minus{} 1}{n(n \plus{} 1)}$ $ \equal{} k(\frac {3}{n \plus{} 1} \minus{} \frac 1n)$

Adding all these equalities (for index greater than 2), we get :

$ \forall n\geq 3$ : $ \frac {a_n}{n \plus{} 1} \minus{} \frac {a_{2}}{3}$ $ \equal{} k(\sum_{i \equal{} 3}^n\frac {3}{i \plus{} 1} \minus{} \sum_{i \equal{} 3}^n\frac 1i)$

$ \forall n\geq 3$ : $ \frac {a_n}{n \plus{} 1} \equal{} \frac {a_{2}}{3} \plus{} k(\sum_{i \equal{} 3}^n\frac {3}{i \plus{} 1} \minus{} \sum_{i \equal{} 3}^n\frac 1i)$

And so, since $ a_2 \equal{} a_1 \plus{} 2k$ and simplifying the sums :

$ \forall n\geq 3$ : $ \boxed{a_n \equal{} (a_1 \minus{} 10k)\frac {n \plus{} 1}{3} \plus{} 3k \plus{} 2k(n \plus{} 1)\sum_{i \equal{} 1}^n\frac 1i}$

And finding a closed form for this expression would mean finding a closed form for $ \sum_{i \equal{} 1}^n\frac 1i$ ...

Hello Bambaman !

Since you posted in the "proposed and own problems", you have the solution to this problem.

Could you kindly give us this solution, or the hint to find the closed form since it seems most of us think there is no such closed form.
Or at least tell us where we are wrong.

Thanks a lot.

Actually, I considered $ H_n$ as a legitimate element in a closed-form formula. My solution was along the lines of Rust's solution (he has some calculation mistakes, though). I was delighted to see your elegant solution!


A jet fighter takes off at an angle of 27.0 degrees w/ the horizontal, accelerating at 2.62 m/s*s. The plane weighs 79,300 N. I need to find the thrust T of the engine on the plane and the lift force L exerted by the air perpendicular to the wings.

Should i switch the axis around, and then the Force of the thrust = (2.62cos27-gsin27)m and L  equal to (2.62sin27+gcos27)m?

I'm not sure how you came to those numbers, could you explain? According to my calculations, it should be

[hide]
$T=ma+mg\sin\phi$, $L=mg\cos\phi$[/hide]


modular arithmetic
quadratics
Vieta
number theory
relatively prime
algebra

Find all pairs of positive integers $ a,b$ such that \begin{align*} b^2 + b+ 1 & \equiv 0 \pmod a \\ a^2+a+1 &\equiv  0 \pmod b . \end{align*}

First I will show that $ a\ne b\plus{}1$. If instead $ a\equal{}b\plus{}1$, then we would have

$ b\plus{}1\equal{}a|b^2\plus{}b\plus{}1$
$ b\plus{}1|b(b\plus{}1)\equal{}b^2\plus{}b$
$ b\plus{}1|(b^2\plus{}b\plus{}1)\minus{}(b^2\plus{}b)\equal{}1$

which is impossible since $ b\ge 1$ and hence $ b\plus{}1\ge 2$.

Now, I will show the problem is equivalent to $ a^2 \plus{} a \plus{} b^2 \plus{} b \plus{} 1\equiv 0\mod ab$.

We know that $ a$ and $ b$ must be relatively prime. Otherwise, $ \gcd(a,b)|a^2 \plus{} a$ so $ \gcd(a,b)\not| a^2 \plus{} a \plus{} 1$ if $ \gcd(a,b)>1$. Now $ a^2 \plus{} a \plus{} b^2 \plus{} b \plus{} 1\equiv 0\mod a$ and $ a^2 \plus{} a \plus{} b^2 \plus{} b \plus{} 1\equiv 0\mod b$, and combining these, since $ a$ and $ b$ are relatively prime, gives $ a^2 \plus{} b^2 \plus{} a \plus{} b \plus{} 1\equiv 0\mod ab$.

This goes the other way to. If $ a^2 \plus{} b^2 \plus{} a \plus{} b \plus{} 1\equiv 0\mod ab$ then $ a^2\plus{}a\plus{}b^2\plus{}b\plus{}1\equiv 0\mod a\implies b^2\plus{}b\plus{}1\equiv 0\mod a$ and $ a^2\plus{}a\plus{}b^2\plus{}b\plus{}1\equiv 0\mod b\implies a^2 \plus{} a \plus{} 1\equiv 0\mod a$.

So now we just need to solve $ a^2\plus{}a\plus{}b^2\plus{}b\plus{}1\equiv0\mod ab$. Let

$ a^2 \plus{} a \plus{} b^2 \plus{} b \plus{} 1 \equal{} kab$

for some integer $ k$. Also, assume that $ a>b$ and that $ a,b>1$.

Then

$ a^2 \minus{} (kb\minus{}1)a \plus{} b^2 \plus{} b \plus{} 1 \equal{} 0$

This is a quadratic equation, so if $ a$ is a solution, then by Vieta's, $ a'\equal{}kb\minus{}1\minus{}a\equal{}\frac{b^2\plus{}b\plus{}1}{a}$ is also a solution. Since $ b^2\plus{}b\plus{}1$ and $ a$ are postive, $ a'$ must be positive. Now (by assumption that $ a>b$),

$ a'\equal{}\frac{b^2\plus{}b\plus{}1}{a}<\frac{b^2\plus{}b\plus{}1}{b}\equal{}b\plus{}1\plus{}\frac{1}{b}$
$ a'\le b\plus{}1$.

(because $ b>1$ by assumption, we know that $ 0<\frac{1}{b}<1$, but $ a'$ is an integer). Also, $ a'\ne b\plus{}1$ by above, so $ a'\le b$. So $ (a',b)$ is another solution if we already had $ (a,b)$. Since $ a>b$ and $ a'\le b$, our $ (a',b)$ solution is smaller. So, we know that we can always produce a smaller solution as long as $ a,b>1$ and $ a\ne b$. Now let's look at the cases where either $ a\equal{}b$ or one of $ a,b$ is $ 1$.

$ a\equal{}b$. We know that $ a$ and $ b$ are relatively prime so for $ a$ to equal $ b$ we must have $ a\equal{}b\equal{}1$.

$ a\equal{}1$ or $ b\equal{}1$. If $ a\equal{}1$ then $ b$ divides $ 1^2\plus{}1\plus{}1\equal{}3$, so the only pairs with $ 1$ in them are $ (1,3)$ and $ (1,1)$.

So we have $ (1,1)$ and $ (1,3)$. Note than $ (1,3)$ can be reduced, as $ 5ab\equal{}1^2\plus{}1\plus{}3^2\plus{}3\plus{}1$, so $ k\equal{}5$ and $ a'\equal{}kb\minus{}1\minus{}a\equal{}5(1)\minus{}1\minus{}3\equal{}1$, so $ (1,3)$ is reduced to $ (1,1)$. Thus, all pairs besides $ (1,1)$ can be reduced in this manner.

Since this reduction process does not change $ k$, and $ k\equal{}5$ for $ (1,1)$, we have $ k\equal{}5$ for all solutions. So since all solutions can be reduced to $ (1,1)$ by $ (a,b)\to (5b\minus{}a\minus{}1,b), (a>b)$ any solution can be obtained by the reverse procedure $ (a,b)\to (5b\minus{}a\minus{}1,b), (a\le b)$. Therefore if we define the sequence $ a_0\equal{}1,a_1\equal{}1$, and $ a_{n\plus{}2}\equal{}5a_{n\plus{}1}\minus{}a_{n}\minus{}1$, all solutions $ (a,b)$ are consecutive terms of this sequence.


Can you generate Golden Ratio $\phi$ using $4$ number of $4's$ .

Use any mathematical operator you can, but use exactly $4$ number of $4's$.

can you?

[quote="agolsme"][hide]\[ 4^{-1/2} + \sqrt{(4/4 + 4^-1)}  \][/hide][/quote]

[hide]your solution uses $1$ and $2$. Remember only $4$ number of $4$'s but any number of operators are allowed.

$\sqrt{4}$ can also be written as $4^{1/2}$, so this is OK.... but others are not valid.

?[/hide]


function
real analysis
real analysis unsolved

Hello,
I want this problem again suggest.Maybe we found a good solution. The problem is:

Find all $ f: \mathbb{R} \to \mathbb{R}$ continuous functin with $ F(f(x))\plus{}f(F(x))\equal{}x^2 \forall x \in \mathbb{R}$, where $ F(x)\equal{}f'(x), F(0)\equal{}0$

Perhaps you meant $ f(x) \equal{} F'(x)$, $ F(0) \equal{} 0$?

So the problem is
Find all $ f: \mathbb{R}\to \mathbb{R}$ continuous finction such that $ F(f(x))\plus{}f(F(x))\equal{}x^2$ $ (\forall) x\in \mathbb{R}$, where $ F'(x)\equal{}f(x)$, and $ F(0)\equal{}0$


It's been posted many times.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].


Show that every power of (√2 - 1) can be written in the form √(k+1) - √k.

[hide]
Let $s_n$ be the $n$th power of $\sqrt{2}-1$. We will proceed by induction. For the base case, $\sqrt{2}-1 = \sqrt{2}-\sqrt{1}$ so that works. Suppose

$s_n = a+b\sqrt{2} = \sqrt{k+1}-\sqrt{k}$.

Then $|a^2-2b^2| = 1$. So

$s_{n+1} = (\sqrt{2}-1)(a+b\sqrt{2}) = (2b-a)+\sqrt{2}(a-b)$.

But since $|(2b-a)^2-2(a-b)^2| = |2b^2-a^2| = |a^2-2b^2| = 1$, we know

$s_{n+1} = (2b-a)+\sqrt{2}(a-b)$ can be written as $\sqrt{k+1}-\sqrt{k}$ for some $k$ as well, completing the induction.[/hide]

EDIT: Boo, someone beat me to it again.


function
calculus
integration
complex analysis
complex analysis unsolved

Suppose that a function $ f$ is analytic throughout the finite plane except for a finite number of singular points $ z_1, z_2, \ldots, z_n$. Show that

$ \text{Res}_{z\equal{}z_1}f(z)\plus{}\text{Res}_{z\equal{}z_2}f(z)\plus{} \cdots \plus{} \text{Res}_{z\equal{}z_n}f(z)\plus{}\text{Res}_{z\equal{}\infty}f(z)\equal{}0$.

I don't see how to show this right now.  Any hints on where to go would be nice.  Thank you.

It's practically the definition of the residue at $ \infty$. Consider the integral of $ f$ around a very large circle, which encloses all singularities.

This is not true. A counterexample is 1/z.

Pym.

Sorry, I did not notice that the residue at infinity is also included.
The reply of jmerry is of course correct.

Pym.


let v(t) be the amount of water in a tank at time t. If the average rate of change of V between times t = 3 and t = 6 is 3, and the average rate between t = 6 and t = 9 is 4, what is the average rate of change of V between t =3 and t = 9?

[quote="blahblahblah"]Average rate = :Sigma: change /  :Sigma: time[/quote]

Are you allowed to do that, considering we are working with averages?

[quote="mikewd"]Are you allowed to do that, considering we are working with averages?[/quote]

You know the average, though, and how many numbers you averaged, so you can multiply to get the sums and add the sums together, then divide by the time.


The vertex $P$ of a rectangular block with edges $a,b,c$ is selected. Consider the plane passing through the vertices adjacent to $P$. Denote the distance of the plane from $P$ is $m$. Prove that \[\frac{1}{m^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\]

Put origin at P and x, y, z-axes along the edges a = PA, b = PB, c = PC. Equation of the plane (ABC) is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.$ Vector product of any 2 vectors parallel to (ABC) is perpendicular to (ABC).

$\vec{AB}\times \vec{AC}= (-a, b, 0) \times (-a, 0, c) = (bc, ca, ab) = abc \left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$

$\vec{PN}= k \left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right) \perp (ABC)$ for any $k \neq 0$. If $N \in (ABC),$ then $k \left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) =1,$ yielding k, and $m^{2}= \vec{PN}\cdot \vec{PN}= k^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) = k,$ $\frac{1}{m^{2}}= \frac{1}{k}= \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}.$


$n$ distinct points are on a plane.

Prove that the number of pairs of points

which have a unit distant is less than $2n\sqrt{n}$.

can the number of pairs of points which have a unit distance can be more than $2n-3$ (for $n\geq 3$)?


Question attached.

I know that it involves simple stoichiometry, but this is the first time I am applying my knowledge of stoichiometry on redox reactions (and the sort), and it is overly complicated for me to comprehend.

Please provide an outline of what to do, and why each step is done, so I may solve similar problems in the future.

Thanks

The problem gives you an equation as balanced as you need it, and asks you to do the stochiometry calculations. The fact that there is oxidation-reduction does not matter once you have a balanced equation. They are not asking for grams and making you use molecular weights. There is no change of units, just moles, liters and mililiters. It is an easy problem if you can interpret the meaning.

The problem is in the wording. I'll try to translate.
It seems like they used the same 25.00 mL sampling through the whole procedure.
The first titration was to determine the moles of $ Fe^{+2}$ ion.
The second titration was to determine the total moles of $ Fe$ (found all as $ Fe^{+2}$ after the zinc treatment).
You are only concerned with the titrations. How al the iron was reduced to $ Fe^{+2}$ does not matter. What matters is that it was all reduced to $ Fe^{+2}$ before the second titration.
In the titration each mole of permanganate oxidizes 5 moles of $ Fe^{+2}$ ion. (Each milimole of permanganate oxidizes 5 milimoles of $ Fe^{+2}$ ion).
Your permanganate titrant contains 0.0200 moles/L (0.0200 mmoles/mL).
How many milimoles of permanganate titrant were used in the first titration?
(0.0200 mmoles/mL) X (23.0 mL) = 0.46 mmoles
How many milimoles of $ Fe^{+2}$ ion were oxidized in the first titration?
(5 mmoles $ Fe^{+2}$/1 mmol permanganate) X (0.46 mmoles permanganate) = 2.3 mmoles $ Fe^{+2}$
What was the $ Fe^{+2}$ ion concentration if those milimoles were in a 25.0 mL volume?
(2.3 mmoles ) / (25.0 mL) = 0.092 mmol/mL = 0.092 M = 92 mM
The second titration would be calculated the same way to give total $ Fe$.You can find the ${ Fe^[+3}$  concentration by difference.

There's no magic and no need to memorize formulas or multistep procedures. You only need to understand the problem. It's just a "word problem." It helped to have been an analytical chemist back in the early seventies when all we could do was titrations, though.


Let $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$ prove that
\[ a\plus{}b\plus{}c\plus{}\frac{1}{abc}\ge \frac{4\sqrt{3}}{9}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\]

[quote="M.A"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that
\[ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\ge \frac {4\sqrt {3}}{9}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$ Then your inequality is equivalent to
$ uw^3\plus{}3(3u^2\minus{}2v^2)^2\geq4uv^2\sqrt{3u^2\minus{}2v^2}.$
But by Schur we obtain: $ 2uw^3\geq15u^2v^2\minus{}9u^4\minus{}4v^4.$
Thus, it remains to prove that $ \frac{1}{2}(15u^2v^2\minus{}9u^4\minus{}4v^4)\plus{}3(3u^2\minus{}2v^2)^2\geq4uv^2\sqrt{3u^2\minus{}2v^2}.$
Let $ u^2\equal{}tv^2.$ Then $ t\geq1.$
Hence, $ \frac{1}{2}(15u^2v^2\minus{}9u^4\minus{}4v^4)\plus{}3(3u^2\minus{}2v^2)^2\geq4uv^2\sqrt{3u^2\minus{}2v^2}\Leftrightarrow$
$ \Leftrightarrow45t^2\minus{}57t\plus{}20\geq8\sqrt{t(3t\minus{}2)}\Leftrightarrow(t\minus{}1)^2(2025t^2\minus{}1080t\plus{}672)\plus{}272(t\minus{}1)\geq0.$
Done!

[quote="M.A"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that
\[ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\ge \frac {4\sqrt {3}}{9}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]

I kill problem by mixing variable
$ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\ge \frac {4\sqrt {3}}{9}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$

$ \leftrightarrow a \plus{} b \plus{} c \plus{} \frac {1}{abc} \minus{} \frac {4\sqrt {3}}{9}(\frac {b \plus{} c}{a} \plus{} \frac {a \plus{} c}{b} \plus{} \frac {a \plus{} b}{c})\ge \frac {4\sqrt {3}}{3}$

setting $ f(a,b,c) \equal{} a \plus{} b \plus{} c \plus{} \frac {1}{abc} \minus{} \frac {4\sqrt {3}}{9}(\frac {b \plus{} c}{a} \plus{} \frac {a \plus{} c}{b} \plus{} \frac {a \plus{} b}{c})$

$ \rightarrow f(a,b,c) \minus{} f(a,\sqrt {\frac {b^2 \plus{} c^2}{2}},\sqrt {\frac {b^2 \plus{} c^2}{2}})\ge (b \minus{} c)^2[\frac {1}{abc(b^2 \plus{} c^2)} \plus{} \frac {4\sqrt {3}}{9a(b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)})} \minus{} \frac {1}{b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)}} \minus{} \frac {4\sqrt {6(b^2 \plus{} c^2)} \plus{} 4\sqrt {3}a}{9bc\sqrt {2(b^2 \plus{} c^2)}}]$
we needs prove that:

$ \frac {1}{abc(b^2 \plus{} c^2)} \plus{} \frac {4\sqrt {3}}{9a(b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)})} \minus{} \frac {1}{b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)}} \minus{} \frac {4\sqrt {6(b^2 \plus{} c^2)} \plus{} 4\sqrt {3}a}{9bc\sqrt {2(b^2 \plus{} c^2)}}]\ge 0$

$ A \equal{} \frac {4\sqrt {3}}{9a(b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)})} \minus{} \frac {1}{b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)}} \equal{} \frac {4\sqrt {3} \minus{} 9a}{9a(b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)})}$
assume that $ a \equal{} min{a,b,c}$ $ \rightarrow a\le \frac {1}{\sqrt {3}}$  $ \rightarrow 9a\le 3\sqrt {3} < 4\sqrt {3}$ $ \rightarrow A\ge 0$
then, we needs proof

$ B \equal{} \frac {1}{abc(b^2 \plus{} c^2)} \minus{} \frac {4\sqrt {6(b^2 \plus{} c^2)} \plus{} 4\sqrt {3}a}{9bc\sqrt {2(b^2 \plus{} c^2)}}\ge 0$

$ \leftrightarrow 4\sqrt {6}a(b^2 \plus{} c^2) \plus{} 4\sqrt {3}a^2\sqrt {b^2 \plus{} c^2} \le 9\sqrt {2}$

$ 4\sqrt {6}a(b^2 \plus{} c^2) \equal{} 4\sqrt {6}a(1 \minus{} a^2)\ge 4\sqrt {2}$ and

$ 4\sqrt {3}a^2\sqrt {b^2 \plus{} c^2} \equal{} 4\sqrt {3}a^2\sqrt {1 \minus{} a^2}\le 4\sqrt {3}a\frac {1}{2}\le 2$ (am-gm)

$ \rightarrow 4\sqrt {6}a(b^2 \plus{} c^2) \plus{} 4\sqrt {3}a^2\sqrt {b^2 \plus{} c^2} \le 2 \plus{} 4\sqrt {2} < 9\sqrt {2}$
$ \rightarrow B\ge 0$

$ \rightarrow f(a,b,c)\ge f(a,\sqrt {\frac {b^2 \plus{} c^2}{2}},\sqrt {\frac {b^2 \plus{} c^2}{2}})$

we needs prove that $ f(a,b,c)\ge \frac {4\sqrt {3}}{3}$ with $ a \equal{} \sqrt {1 \minus{} 2t^2}$ and $ b \equal{} c \equal{} t (t \equal{} \sqrt {\frac {b^2 \plus{} c^2}{2}})$
Done!

there is a really simple solution to this one.
hint:
it also holds $ a\plus{}b\plus{}c\plus{}\frac{1}{abc}\geq \frac43 (\frac1a\plus{}\frac1b\plus{}\frac1c)$.

[quote="Albanian Eagle"]there is a really simple solution to this one.
hint:
it also holds $ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\geq \frac43 (\frac1a \plus{} \frac1b \plus{} \frac1c)$.[/quote]
You are right Albanian Eagle this is an easy problem :lol:
\[ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\geq \frac43 (\frac1a \plus{} \frac1b \plus{} \frac1c)\Leftrightarrow\]
\[ (a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4\plus{}(a^4\plus{}b^4\plus{}c^4\minus{}a^2bc\plus{}b^2ca\plus{}c^2ab)\ge 0\]
so clear.


probability
expected value
probability and stats

Hi all again, here with another problem that I get:
buscar

A medical office patients arrive at a constant rate of $ 1/10$
patients per minute, following a Poisson process. If the doctor does not begin to patients unless they meet in the waiting room has three such. What is the expected time for the doctor see the first patient? What is the probability that the doctor did not see
any patient during the first hour?

I've done this, point b): $ Pr[X<\equal{}2] \equal{} Pr(X\equal{}0) \plus{} Pr(X\equal{} 1) \plus{} Pr(X\equal{}2)$ calculated with the Poisson distribution.

The a) does not come out yet, I appreciate any help.

Greetings and many thanks  :)

Just answer a simple question:

For Poisson process with intensity $ \lambda$, what is the distribution of the times between arrivals?

hello, I think I did well like this:

The question what is the expected time for the doctor to see the first patient?
May arise as follows: what is the expected time until they reach three patients?,

Since the arrival of patients is a Poisson process of rate λ = 1 / 10 patients per minute the expected time till the third patient $ E(S_3) \equal{} n/\lambda \equal{} 3/1/10 \equal{} 30min$. $ S_n$ follows a Gamma distribution.

You're right, but you put too much mathematics into this problem.

Yes, $ S_3$ is Gamma distributed.
But to solve this problem, one needs not to know what is Gamma distribution. Times between arrivals are exponentially distributed, expected value of sum = sum of expected values = 3*1/$ \lambda$ =30.

Ok. Thank you for your help.

You know anything about this?:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298329

thank you very much again and greetings

[quote="extranger"]
You know anything about this?:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298329
[/quote]
I'm not sure I know. I will be able to read your text only if you make the formulae readable.

Ok. I appreciate if suddenly I can help, especially with code R.


Is Rummy-Q the card game where you have like 13 cards and you have to try get them in order or something?

Something like that, but they are not cards, yeah u need like a combination of at least 3 things depending on color or order.

[quote="hello"]Anyone here ever play the game Rummy-Q, I think there is a lot of strategy involved there.[/quote]
Are you talking about Rummikub or is it something else?

EDIT: I guess we're talking about the same thing :)

ive played Rummy-Q, or something very like it.  Your description sounds a lot like what i played.


Prove that in any convex polygon with $4n+2$ sides ($n\geq 1$) there exist two consecutive sides which form a triangle of area at most $\frac 1{6n}$ of the area of the polygon.

Hmm... I can't prove it for $n=1$  :blush: (Kuba said me that it is the hardest thing in this problem :P )
All over cases can be derived from $n=1$.
Let's show it.
Suppose the contrary, i.e. there are no triangle of area $\leq 1/6n$.
Consider $4n+2$-gon with $n\geq 2$ and cut a 6-gon from it by mean of a diagonal ($AB$ on figure). Let's prove that the triangle of area $<1/6n$ can't adjoin the diagonal $AB$. Otherwise if $ABC$ is a such triangle then draw a line throw $A$ parallel to $BC$. We will necessary have $D$ and $E$ lies in the other halfplane than $BC$. So $4n+2$ is not convex.Therefore we can divide $4n+2$-gon into $n$ 6-gons by mean of diagonals and no triangle which adjoins these diagonals will have area $<1/6n$.

[quote="Myth"]  Hmm... I can't prove it for n=1 (Kuba said me that it is the hardest thing in this problem  :P )  [/quote] In my humble opinion, it is not harder than the part done by Myth and it uses pretty much the same idea. Here is a little [hide="hint"] Assume for a moment that the diagonals AD, BE and CF of a hexagon ABCDEF intersect at one point O. Then one of the six triangles into which they divide the hexagon has the area at most $1/6$ of the area of the hexagon. If it is, say, the triangle AOB, what can you say about the areas of ACB and AFB?  [/hide] Nice problem, by the way! :)

Nice idea, Myth! I can prove the n=1 case:

In hexagon ABCDEF, $P=AD\cap CF$, $Q=CF\cap EB$, $R=EB\cap AD$. The six triangles APB, BPC, CQD, DQE, ERF, FRA cover less than the whole hexagon (if the three diagonals are not concurrent) or the whole triangle (if they are). In either case, one of those six triangles will have area $\le \frac{1}{6}(\text{area})$, so say it is APB. Then one of AFB and ACB has area at most the area of APB, so we're done.

Sorry, I don't see it.

Indeed, the case described by Fedor is clear. But common case... I am missing something.

I really don't understand what Myth did, especially the part:
[quote="Myth"]We will necessary have $D$ and $E$ lies in the other halfplane than $BC$.[/quote]
And also, how exactly is the case $n>1$ derived from $n=1$?

If $E$ lies in the same halfplane, then $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction.

The general cases can be done in the following way: cut the polygon into $n$ hexagons by mean of diagonals. One of this hexagon have area $\leq 1/n$ and we can cut a tringle of area $\leq 1/6$ from it, therefore this triangle is at most $\frac{1}{6n}$ of the area of the initial polygon. And we prooved that this triangle can't be "inner" triangle, so it is a required triangle.

okay. now i understand the general case.

[quote="Myth"]If $E$ lies in the same halfplane, then $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction.
[/quote]

i still don't understand the first part.
Why $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$ yields a contradiction?

Because we supposed the contrary, i.e. there is no such triangle.

Case n=1 is also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27331]here[/url]


pigeonhole principle
number theory unsolved
number theory

A positive integer is [i]dapper[/i] if at least one of its multiples begins with $ 2008$. For example, $ 7$ is dapper because $ 200858$ is a multiple of $ 7$ and begins with $ 2008$. Observe that $ 200858 \equal{} 28694\times 7$.

Prove that every positive integer is dapper.

Let m be a positive integer. Let $ n \equal{} [log_{10}{m}] \plus{} 1$. Consider $ 2008 \cdot 10^n \plus{} a, 0 \le a \le m \minus{} 1 ( < 10^n)$. These are m numbers that begin with 2008 and have a different residue when divided by m, thus one of them is a multiple of m. We conclude that all positive integers are dappers.

Consider the m following numbers:

2008, 20082008, 200820082008, ..., 20082008...20082008

The i-th term have i 2008's.
If one of them is multiple of m, ok.
Suppose that there aren't multiples of m.
By the Pigeonhole Principle, there are 2 numbers with the same remainder at the divison by m.So, take the difference beetween this 2 numbers.It's obviously that this new number is multiple of m and begins with 2008. (q.e.d.)

Cya people,
Secco

let n < $ 10^{m}$

let us consider 2008 X $ 10^{m \plus{} 1}$. let it give r as remainder when divided by n.
then 2008 X $ 10^{m \plus{} 1}$ +n-r is certainly a multiple of n and hence proved


trigonometry
limit
geometric series
complex numbers
geometric sequence

$\displaystyle\sum_{n=1}^{\infty} 3^{n-1}\cdot\sin^3 \left(\frac{\pi}{3^{n+1}}\right)$

I tried doing it by changing the sin expression into the Im part of $e^{\frac{\pi}{3^{n+1}}}$, then putting it into the inf geometric series formula and then taking the imaginary part of that at the end.

However, I got $\frac{\sqrt{3}}{14}$ which is approximately .12, while Mathematica approximates this sum to 0.045293.  Anybody care to look at it?

Use $\sin 3\theta=3\sin \theta-4\sin ^ 3 \theta\Longleftrightarrow \sin ^ 3 \theta=\frac{1}{4}(3\sin \theta-\sin 3\theta),$ then telescoping method and $\lim_{\theta \longrightarrow 0} \frac{\sin \theta}{\theta}=1.$

i am pretty sure that this is not a geometric sequence

$\Im \left(\sum_{n=1}^{\infty} 3^{n-1}*e^{\frac{\pi i}{3^n}}\right)$

notice that it would if it were $e^{n}$, instead of $e^{3^{n}}$, it would be a geometric sequence

The answer is $\frac{1}{4}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right).$

can you show exactly what you did. i am not able to completely follow.

I'm able to get everything except that $\frac{\pi}{3}$ term to come up.


[hide="solution"]
formula is:
n(n+1)/2

100(100+1)/2
100(101)/2
5050
(triangle numbers formula)
[/hide]

[hide="to Treething"]yup, you got it right, but I hid it so people could still try to solve the problem...[/hide]

 :lol:  ;)  :blush:  :P

[quote="shinwoo"]This is an easy, in fact, very easy problem...

I spent 3 hours this morning to find a better formula....

1 coconut... 1st column

3 coconuts   2nd column

6 coconuts   3rd colum

10 coconuts  4th colum

now find the number of coconuts in 100th colum....

I just posted it because there weren't enough posts in this forum...[/quote]

[hide]x(x+1)/2
100*101/2=5050[/hide]

[hide]5050[/hide]

[color=darkred]<_<
>_>

Next time don't check the "Disable BBCode in this post" box.[/color]


function
calculus
derivative
trigonometry
calculus computations

I'm getting a wildly different answer from the back of the book on this one, and looking at the solutions manual, it (the book) seems to be wrong. Can someone verify? I hope this is the appropriate forum.

The function $T(x)=\frac{200}{1+x^2}$ represents the temperature in degrees Celsius percieved by a person standing $x$ meters from the fire. If the person moves away from the fire at 2 m/s, how fast is the temperature changing when the person is 5 meters away?

Thanks in advance :)

So we're talking temperature in relation to distance from fire. Then we're looking at a "change in temperature" when there's a "change in distance" at a certain distance...what does this shout to you? "DERIVATIVE!!!!! DO ME!!!!!" So we do it:

$dT=\frac{-400x}{(1+x^{2})^{2}}dx$

At $5$m away and moving at $2$m/s, we substitute $x=5$ amd $dx=2$ and we get $dT=-2.96$

My calculation might be wrong but I suck at arithmetic.

$\frac{dT}{dx} = \frac{-400x}{(1+x^2)^2} \rightarrow \frac{dT}{dt} = \frac{dT}{dx}\frac{dx}{dt} = \frac{-400x}{(1+x^2)^2} \frac{dx}{dt}$

From the problem statement, $\frac{dx}{dt} = 2 m/s$ and $x=5m$.  This should yield the same answer.  Note also that this formula is dimensionally incorrect.

[quote="FMako"]Note also that this formula is dimensionally incorrect.[/quote]

Then ... doesn't this mean the formula is incorrect...? Or am I completely misunderstanding what you meant there?  :blush:

FMako is talking about units of measurement whether stated or implied.

The biggest problem is that denominator, $1+x^2.$ We have that $x^2$ is the square of a distance; we measure it in $\text{m}^2.$ What does that make $1?$ We can only add or subtract two quantities if they have the same units.

If we assume that the given $1$ is actually meant to be $1\text{m}^2$ and the $200$ in the numerator is actually $200\text{ deg}\cdot\text{m}^2$, then we can make the whole thing make sense.

That's not very satisfying. I like to use implied units of measurement - "dimensional analysis" - as a continuing check for the accuracy of computations, perhaps especially the intermediate steps. It's hard to get that kind of help from a formula like this.

I'd be happier with $T=\frac{k}{a^2+x^2}.$

Then I read it as $a$ being a distance and $k$ being a constant of proportionality - and constants of proportionality carry whatever units they need to carry.

One little principle from this dimensional analysis outlook: suppose you see a formula like $e^{-kt}.$ Then if $t$ is a time and is measureed in units of measurement appropriate for a time, then $k$ must be the reciprocal of a time. Why? Because the argument of a "transcendental function" like the sine* or the exponential must be a pure unitless number.

--

* Footnote: an angle given in radians [i]is[/i] a "pure unitless number."


trigonometry
calculus
ceiling function
algebra
polynomial
induction
complex analysis

Let $\theta=\frac{\pi}{2n+1}\/$ where $n$ is a positive integer.
Prove that
1. $\prod_{r=1}^{n} \sin r\theta =  2^{-n}\sqrt{2n+1}\/$
2. $\prod_{r=1}^{n} \cos r\theta = 2^{-n}\/$

(I don't know if it is suitable to post this question here :?  :?  :? )

i am sure that here is a elementary way and a complex analysis way to solve it.
but for calculus ,
i tried to break sin into  a series and add them up  , but got screwed up ,

can anyone help ?

[quote="lisai"]i am sure that here is a elementary way and a complex analysis way to solve it.
but for calculus
[/quote]
Any way is good as long as it yields a solution. So feel free to post the proof you know regardless of whether you consider it "calculus" or not. :)

[quote="pipi"]Let $\theta=\frac{\pi}{2n+1}\/$ where $n$ is a positive integer. Prove that
1. $\prod_{r=1}^{n} \sin r\theta = 2^{-n}\sqrt{2n+1}\/$
2. $\prod_{r=1}^{n} \cos r\theta = 2^{-n}\/$ [/quote]
 Denote $\Theta= \frac{\pi}{2n+1}$, $P_1=\prod\limits_{k=1}^{n}\sin{\frac{k\pi}{2n+1}},\; \; Q_1=\prod\limits_{k=n+1}^{2n}\sin{\frac{k\pi}{2n+1}}$ and observe that $P_1>0 .$ Because $(*)\hspace*{0.5cm} \ds \prod\limits_{k=n+1}^{2n}T_k=\ds \prod\limits_{k=1}^{n}T_{2n+1-k}\; ,$ we have
$Q_1=\prod\limits_{k=1}^{n}\sin(2n+1-k)\Theta= \prod\limits_{k=1}^{n}\sin(\pi-k\Theta)=P_1\; .$ Further we try to find $P_1^2=P_1Q_1=\prod\limits_{k=1}^{2n}\sin{k\theta}.$
 Let $z_k=\cos{(2k\theta)}+ i\cdot\sin{(2k\theta)}.$ If we select $x=1$ in the  equalities $\ds \frac{x^{2n+1}-1}{x-1}=\sum\limits_{k=0}^{2n}x^k=\ds \prod\limits_{k=1}^{2n}\left(x-z_k\right) ,$ we find
\[ (**)\; \; \; \; \begin{array}{c}\ds \ds 2n+1=\ds \prod\limits_{k=1}^{2n}2\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\sin{k\Theta}=\\ =\ds 2^{2n}P_1^2 \prod\limits_{k=1}^{2n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)=2^{2n}P_1^2\; . \end{array}  \]
The last equality may be justified in the following way :
\[ \begin{array}{c} \prod\limits_{k=1}^{2n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)=\\ =\left( \prod\limits_{k=1}^{n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\right)\cdot \left( \prod\limits_{k=n+1}^{2n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\right)=\\ =\left( \prod\limits_{k=1}^{n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\right)\cdot \left( \prod\limits_{k=1}^{n}\left(\sin{(2n+1-k)\Theta}-i\cdot\cos{(2n+1-k)\Theta}\right)\right)=\\ =\left( \prod\limits_{k=1}^{n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\right)\cdot \left( \prod\limits_{k=1}^{n}\left(\sin{(\pi-k\Theta)}-i\cdot\cos{(\pi-k\Theta)}\right)\right)=\\ =\left( \prod\limits_{k=1}^{n}\left(\sin{k\Theta}-i\cdot\cos{k\Theta}\right)\right)\cdot \left( \prod\limits_{k=1}^{n}\left(\sin{k\Theta}+i\cdot\cos{k\Theta}\right)\right)=1 \; . \end{array}  \]
Now (**) enables us to write $\ds P_1=\ds \frac{\sqrt{2n+1}}{2^n}\; .$
Further try to find $P_2: =\prod\limits_{k=1}^{n}\cos{k\Theta}$ in  following manner: observe that $P_2>0$ and  $P_1P_2=\ds \frac{1}{2^n}\prod\limits_{k=1}^{n}\sin{2k\Theta}$ . Consider further the product $Q_2: =\ds\prod\limits_{k=n+1}^{2n}\sin{2k\Theta}$ ...
[b]Remark:[/b] In above hint/solution, equality (*) has an important  role .

[quote="zhaobin"]I have another idea.
for 2,I think you can use the conclusion posted  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40503]here[/url]
for1,you can find a solution similar way(but a little different)
sorry for my short post,if who want to know more,i will try to explain again :)[/quote]
Hi zhaobin, it is possible to share your solution (for  1)?

Here are some questions I wish to post:
Let $\theta=\frac{\pi}{2n+1}\/$ where $n$ is a positive integer. Prove that:
(i)  \[\sum_{r=1}^{n} cos \  r \theta = - \frac{1}{2} \]
(ii)  \[\sum_{r=1}^{n} cos^{2} \ r \theta =  \frac{2n-1}{4} \]
(iii) $r_1,r_2 \in \{1,2, \ldots, n\}\/$ \[\sum_{r_1 >r_2} cos \ r_1 \theta cos \ r_2 \theta=  \frac{1-n}{4} \]
(iv)  $\forall n \geq k \geq 1\/$ (more gerenal case for (i)),  \[\sum_{r=1}^{n} (cos \ r \theta)^{2k-1} = - \frac{1}{2} \]

I manage to get the solution for the above questions, and I make a conjecture (base on (i) and (ii) above), as follow:
Let $\theta = \frac{2 \pi}{2n+1}, n \in N\/$ and $r_1,r_2,\ldots ,r_k \in \{1,2, \ldots, n\}\/$.
\[\sum_{r_1 >r_2> \ldots >r_k} cos \ r_1 \theta cos \ r_2 \theta \ldots  cos \ r_k \theta = (-1)^l \frac {{n- \left\lceil \frac{k}{2} \right\rceil \choose \left\lceil \frac{k-1}{2} \right\rceil }}{2^k} \]

where $l= \left\lceil \frac{k}{2} \right\rceil\/$

Can somebody help me on this  :?  :?  :?
Thanks a lot!!

[quote="pipi"]Thanks for your solution, flip2004!
I get the idea but I think there is something not allowed if we write:
[quote="flip2004"]
 Let $ z_k=\cos{(2k\theta)}+ i\cdot\sin{(2k\theta)}. $ If we select $x=1$ in the  equalities $\ds \frac{x^{2n+1}-1}{x-1}=\sum\limits_{k=0}^{2n}x^k=\ds \prod\limits_{k=1}^{2n}\left(x-z_k\right)$,...[/quote]
thanks again.[/quote]
sure :) .
first,we can now $\frac{\sin{(2n+1)x}}{\sin x}$ is a polynomial of $\sin x$
let $P_n(2\sin x)=\frac{\sin{(2n+1)x}}{\sin x}$
we can know $P_n(x)$ have 2n roots which are $2\sin{\frac{k}{2n+1}}$ or $-2\sin{\frac{k}{2n+1}}$,$(k=1,2,\cdots,n)$
then find the relation of $P_n(x),P_{n-1}(x),P_{n-2}(x)$
with induction you will find the answer :)


function
trigonometry
symmetry
complex numbers
complex analysis

I have been taught that the roots of complex numbers are multivalued, for example $ \sqrt{3\plus{}4i}\equal{}\pm (2\plus{}i)$. What to do I have to add two complex roots, for example to compute $ \sqrt{3\plus{}4i}\plus{}\sqrt{3\plus{}4i}$? Is the correct way to do it by take any branch of the first square root and add it to any branch of the second square root, so $ \sqrt{3\plus{}4i}\plus{}\sqrt{3\plus{}4i}\equal{}4\plus{}2i, 0, \text{ or }\minus{}4\minus{}2i$? Or does such situation arise anywhere in mathematics?

$ \sqrt {3 \plus{} 4i} \plus{} \sqrt {3 \plus{} 4i} \equal{} 2\sqrt {3 \plus{} 4i} \equal{} \pm 2(2 \plus{} i)$, never $ 0$, otherwise you'd be saying $ \sqrt {3 \plus{} 4i} \neq \sqrt {3 \plus{} 4i}$. I'd recommend using the principal square root for cases like this one, though (then again, what do I know? XD).

$ \sqrt{3 \plus{} 4i}$ is not a well-defined expression.  You should never have to evaluate it unless the branch is explicitly given.

I wish to disagree:  Can it not be all of them depending on what branches are chosen?  Consider the funtion $ \displaystyle f(z)\equal{} \sqrt{z}\plus{}\sqrt{z}$.  Cannot $ f(1)\equal{}0$ if the branches are chosen appropriately?  How about $ \displaystyle g(z)\equal{}z^{1/3}\plus{}z^{1/4}$?  If $ \displaystyle g(1)\equal{}0$ then what is $ \displaystyle g(\minus{}1)$?  I think $ \displaystyle \sqrt{3\plus{}4i}\plus{}\sqrt{3\plus{}4i}\equal{}2\sqrt{3\plus{}4i}$  when the underlying square root multifunctions are being added symmetrically:  first branch of the first expression added to the first branch of second one and second branch of the first with the second of the second expression.  That's confusing I know.  Never use the principal branch unless you need to use it or the context implies or warrants it: approach roots, logs, inverse trigs, other multifunctions as multivalued quantities until specific values are needed, specific branches are specified,  or if the symmetry above is implied, that is if $ \sqrt{z}\plus{}\sqrt{z}\equal{}2\sqrt{z}$ is assumed. I think $ \sqrt{3\plus{}4i}$ is perfectly well defined.  It's $ \displaystyle \sqrt{3\plus{}4i}\equal{}r^{1/2}e^{i/2(\Theta\plus{}2k\pi)}\equal{}\pm r^{1/2}e^{i/2\Theta}$.   If this was a test question, I'd stick to my guns and hope mine is on top the next day . . .


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be an abelian finite group such that $ n$ divides $ |G|$.
Let $ H\equal{}\{ x^n, x\in G\}$. There are many ways to prove that $ |H|\equal{}|G|/n$, but I am trying to find an elementary proof that would not use the notion of quotient or factorization of homomorphism (i.e. $ G/ker (f) \simeq f(G)$), let alone Cauchy or Sylow.
Any idea?

[quote="Fleury"]Let $ G$ be an abelian finite group such that $ n$ divides $ |G|$.
Let $ H \equal{} \{ x^n, x\in G\}$. There are many ways to prove that $ |H| \equal{} |G|/n$.[/quote]
Are you sure of that? Because, if $ G \equal{} \{\minus{}1,1\} \times \{\minus{}1,1\}$ and $ n \equal{} 2$, it appears that $ H \equal{} \{(1,1)\}$ and that $ |H| \equal{} 1 \neq \frac{|G|}{2} \equal{} 2$.
Or I may have misundertood what you had written...

[quote="LaChouetteRieuse"]Are you sure of that? [/quote]
Not anymore :blush:

There are some values of $ n$ when your statement is true (although I don;t have an "elementary solution"), namely if $ |G|\equal{}\prod_{i\equal{}1}^{r}p_i^{a_i}$ ($ p_i$ primes and $ a_i>0$) and $ n\equal{}\prod_{i\equal{}1}^{r}p_i^{b_i}$ with $ b_i\equal{}a_i$  or $ b_i\equal{}0$.


If $ a,b,c,d \ge 0$  then prove
$ \frac {a}{b\plus{}c}\plus{}\frac {b}{c\plus{}d}\plus{}\frac {c}{d\plus{}a}\plus{}\frac {c}{d\plus{}a} \ge \frac {5}{2}$.

[quote="soruz"]If $ a,b,c,d \ge 0$  then prove
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {c}{d \plus{} a} \ge \frac {5}{2}$.[/quote]

I love inequalities that fail for $ a\equal{}b\equal{}c\equal{}d\equal{}1$...

See http://www.mathlinks.ro/viewtopic.php?t=140344 for the right one (I hope).

  darij

Actually, the following nice statement holds:
If $ 1 \le k \le {n\minus{}1}$ is an integer number and $ a_1,a_2,...,a_n\ge 0$
such that  $ a_1 \le a_2\le...\le a_n$   then

$ \frac {a_1}{ka{_1}\plus{}a_{k\plus{}1}}\plus{}\frac {a_2}{ka{_3}\plus{}a_{k\plus{}2}}\plus{}...\frac {a_n}{ka_n\plus{}a_k} \ge \frac{n}{k\plus{}1}$

If $ a,b,c,d,e,f \ge 0$    then prove

$ \frac {a}{b\plus{}c}\plus{}\frac {b}{c\plus{}d}\plus{}\frac {c}{d\plus{}e} \plus{}\frac{d}{e\plus{}a}\plus{}\frac{f}{a\plus{}b} \ge \frac{5}{2}$.

[quote="soruz"]If $ a,b,c,d,e,f \ge 0$    then prove

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} e} \plus{} \frac {d}{e \plus{} a} \plus{} \frac {f}{a \plus{} b} \ge \frac {5}{2}$.[/quote]
Try Cauchy - Schwarz. :wink:

[quote="soruz"]If $ a,b,c,d,e,f \ge 0$    then prove

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} e} \plus{} \frac {d}{e \plus{} a} \plus{} \frac {f}{a \plus{} b} \ge \frac {5}{2}$.[/quote]

This statement is quite as correct as the spelling of the word "correct" in your post...

  dg

I think the "correct" statement should be

$ \frac{a}{b\plus{}c} \plus{} \frac{b}{c\plus{}d} \plus{} \frac{c}{d\plus{}e} \plus{} \frac{d}{e\plus{}a} \plus{} \frac{e}{a\plus{}b} \geq \frac{5}{2}$

Is this right?

[hide="Solution"]

Assume WLOG $ a\plus{}b\plus{}c\plus{}d\plus{}e \equal{} 1$ and let $ f(x)\equal{}\frac{1}{x}$ which concaves up for $ x \in \mathbb{R}^{\plus{}}$. By Jensen's Inequality:

$ af(b\plus{}c) \plus{} bf(c\plus{}d) \plus{} cf(d\plus{}e) \plus{}df(e\plus{}a) \plus{} ef(a\plus{}b) \geq f(ab\plus{}ac\plus{}ad\plus{}ae\plus{}bc\plus{}bd\plus{}be\plus{}cd\plus{}ce\plus{}de) \equal{} \frac{(a\plus{}b\plus{}c\plus{}d\plus{}e)^{2}}{ab\plus{}ac\plus{}ad\plus{}ae\plus{}bc\plus{}bd\plus{}be\plus{}cd\plus{}ce\plus{}de}$

So we need to prove that:

$ \frac{(a\plus{}b\plus{}c\plus{}d\plus{}e)^{2}}{ab\plus{}ac\plus{}ad\plus{}ae\plus{}bc\plus{}bd\plus{}be\plus{}cd\plus{}ce\plus{}de} \geq \frac{5}{2} \iff (a\minus{}b)^{2} \plus{} (a\minus{}c)^{2} \plus{} (a\minus{}d)^{2} \plus{} (a\minus{}e)^{2} \plus{} (b\minus{}c)^{2} \plus{} (b\minus{}d)^{2} \plus{} (b\minus{}e)^{2} \plus{}(c\minus{}d)^{2} \plus{}(c\minus{}e)^{2} \plus{}(d\minus{}e)^{2} \geq 0$.[/hide]

[quote="BanishedTraitor"]

So we need to prove that:

$ \frac {(a \plus{} b \plus{} c \plus{} d \plus{} e)^{2}}{ab \plus{} ac \plus{} ad \plus{} ae \plus{} bc \plus{} bd \plus{} be \plus{} cd \plus{} ce \plus{} de} \geq \frac {5}{2}$[/quote]
This is just Cauchy. :wink:


Gauss algebra polynomial function integration calculus derivative

Let A the Gauss ring of integers and $z_1,...,z_n$ in A such that the distance from $z_i$ to $z_1$ is greater than 2 for all $i>1$. Prove that the polynomial $ 1+(x-z_1)...(x-z_n)$ is irreducible in $A[X]$.

What is Gauss ring of integers? Is it $a+bi$, $a,b\in\mathbb{Z}$?

Then it is very standart.
Consider polynomials $f(z)=(z-z_1)(z-z_2)...(z-z_n)$ and $g(z)=(z-z_1)(z-z_2)...(z-z_n)$. We see that $|f(z)-g(z)|=1<|f(z)|+|g(z)|$ for all $z$ lying on the circle $|z-z_1|=1$. So $f(z)$ has the same number of roots inside of $|z-z_1|=1$ as $g(z)$, but $g(z)$ has the only such root $z_1$.

Suppose the contrary, i.e. $f(z)=a(x)b(x)$. We may WLOG assume $z_1=0$. Then we have $a(0)b(0)=1$, but both $|a(0)|,|b(0)|\geq 1$ and at least one of these numbers is greater than 1, since one of the polynomials $a$ or $b$ has all roots are greater than 1 at absolute value. Contradiction.

Yes, Myth, this was the idea. Probably for you it is standard, but I do not think that proving Rouche's theorem in an oral examination is so easy.

And who ask to prove it in an oral examanation? Just use it!

This is also my opinion, Myth, but the problem is that sometimes the examinator does not accept high theorems during the examination if you do not know a proof. This is the problem with oral examinations! But anyway, in this case I know a nice proof of this particular case of Rouche's theorem, thus I do not have problems.

is there any elementary proof (no complex analysis) of Rouche's theorem ?
Harazi, what is the proof for this particular case you are talking about? I'm very interested. Thanx  :)

We made in a "travaux dirigé" something about "indice d'un lacet" and the teacher has shown to us that if we consider the function $ f(t)=P(re^{it})$ then the "indice" of this function, that is $\int_{0}^{2\pi}{\frac{f'(t)}{2\pi\cdot (t)}}$ is equal to the number of roots of the polynomial P that lie inside the disk $|z|<r$, assuming that P does not have roots of modulus r. This is simple to show. The hard part is to show that if a "lacet" f verifies $|f(t)-1|<1$ for all t then it has "indice" 0. It follows that if $|P(z)-Q(z)|<|Q(z)|$ for all z of modulus r, then P and Q have the same number of roots in the disk of radius r (just use the lacet $\frac{P(re^{it})}{Q(re^{it})}$). Alekk, if you come to my room (I forgot the number of your room), I can give you the TD where he proved the hard part.

Harazi, do you, by any chance, mean to say something like what I wrote in post number $16$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150&postorder=asc&highlight=curve&start=15]this topic[/url]? I think your "indice" is the degree of that function wrt the unit circle (of course, we can use any other circle instead of the unit circle, with the same results).

Yes, grobber, it is the same concept. Yet, I do not find a proof of the fact that the index of such a "lacet" is 0 of the trajectory is in a circle does does not contain the origin. With the interpretation of the "lacet" it is clear, but is it that easy to prove it rigorously?

I don't really know what "lacet" means :). I understood "indice" :).

Ok, sorry, I do not know how we call this in english:
   A "lacet" is a derivable function f from $[0,2\pi]$ into the complex numbers, such that the derivative is continuous and $f(0)=f(2\pi)$. The indice of the lacet is the number (integer ;) ) $\frac{1}{2\cdot i\pi}{\int_{0}^{2\cdot\pi}{\frac{f'(t)}{f(t)}dt}}$. The idea is to prove that if the lacet has no zeroes and $ |f(t)-1|<1$ for all t then its indice is 0. Do you have a simple solution for this?

No, not really :).

Here's how I would translate it, in the language used in that post I mentioned: given a polynomial $P$ which has no roots on or inside the unit circle, the order of $\frac{P(z)}{|P(z)|}$ is $0$. If we use the fact that the fundamental group of the circle is $\mathbb Z$, the order can be defined without resorting to integrals, and then it would become a pretty standard exercise in algebraic topology (of which I know next to nothing ;)).

First we would show that the order is invariant to homotopy, and then we would use the homotopy $H(z,t)=\frac{P(tz)}{|P(tz)|}$. This takes our path $\frac{P(z)}{|P(z)|}$ to the constant path $\frac{P(0)}{|P(0)|}$, which has order $0$ (as every constant path does).

Yes, my teacher used the same kind of made, only he translated them in the language of mere mortals that we are.  ;)  :(

ok, thank you very much harazi. I remember we did the same last year. I think "indice" is called winding number in english .


HMMT Harvard college MIT AMC AIME geometry

Hi, I was wondering how useful the Harvard-MIT math tourney problems were and if their level of difficulty corresponded to that of the AIME tests. Are there particular sections (algebra/ combo/ geo) that you found to be more useful/realistic of the AMC/AIME tests? Any years specifically that helped you or any helpful info is appreciated.

I think the HMMT problems are good practice problems for the AIME -- even though they have some differences with AIME problems. The HMMT problems are (ideally) meant to be completed in an average of 5 minutes per problem, whereas AIME problems are (ideally) meant to be completed in an average of 12 minutes per problem. The HMMT problems are more geared towards one subject at a time, whereas AIME problems are notorious for crossing subjects or topics. So a problem that starts out as a geometry problem on the AIME may require some skills from algebra or complex numbers or something.

But the HMMT problems correspond in difficulty to maybe #1-10 on the AIME, and they make good practice in solving problems that require skills or knowledge beyond your regular AMC-12 or local math contest questions. They're also good because they have solutions available to check your work.

Along with ARML problems, HMMT problems are probably the best practice problems for AIME (aside from AIME problems themselves - or problems found in the Intermediate Forum here).

What book/contest has problems for #11-15 on AIME?

The AIME would be your best source. You could also try some of the early problems on some of the easiest problems like the CMO (canada) and the JBMO.


Is it true that every singular matrix $ A \in \mathcal M_n ( \mathbb C )$ can be written as $ TN$, where $ T$ is invertible and $ N$ is nilpotent?

For a moment, I had myself convinced that $ \begin{bmatrix}1&1\\0&0\end{bmatrix}$ was trouble. But

$ \begin{bmatrix}1&1\\0&0\end{bmatrix}\equal{}\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1&1\\\minus{}1&\minus{}1\end{bmatrix},$ so we're OK there.

Yes, this is true. In the first place, there are invertible matrices $P, R$ such that $P^{-1}AR^{-1} = \begin{bmatrix} I_r & O \\ O & O \end{bmatrix}$, with $r = rank(A) \leq n-1$. Now define the invertible matrix $X = [e_{r+1} \,\, ... \,\, e_n \,\, e_1 \,\, ... \,\, e_r]$, where the $e_i$ are the column vectors of the canonical basis. Then,

$P^{-1}AR^{-1}X = \begin{bmatrix} I_r & O \\ O & O \end{bmatrix} X = \begin{bmatrix} O & I_r \\ O & O \end{bmatrix} = H$,

where $H$ is a nilpotent matrix (since it is strictly upper triangular). Making $R^{-1}X = S$, we have

$A = PHS^{-1} = PS^{-1} \cdot SHS^{-1}$.

Now let's make $PS^{-1} = T$ and $SHS^{-1} = N$: then $T$ is invertible, $N$ is nilpotent, and $A = TN$. $\diamondsuit$


linear algebra
matrix
search
linear algebra unsolved

Give A is a matrix satisfy $ a_{ij} \plus{} a_{ji} \equal{} 0, 1 \leq i,j \leq n$
Prove that  $ det(A) \geq 0$

Try to guess: how many different topics do we have with titles "det(A)", "det A", etc. ;)

This one was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=143797 , though I wouldn't mind someone posting a proof for the Pfaffian formula. ;)

  darij

Some may want to check [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1077620069&t=25572]this[/url] out.


Prove that these two sequences (called "Beatty sequences") of integers:
\[a_{n}= \left\lfloor n \alpha \right\rfloor \] and
\[b_{n}= \left\lfloor n \beta \right\rfloor \]
for $n \in \mathbb{Z^{+}}$, contain together all the positive integers, without repetition, if $\alpha+\beta = \alpha \cdot \beta$ and both $\alpha$ and $\beta$ are positive, irrational numbers.

([url=http://mathworld.wolfram.com/BeattySequence.html]See this page[/url] on Wolfram. I have basically no idea how to go about proving this. Anyone have ideas?)

I think the following argument shows that they contain all integers:

One of $\alpha, \beta$ must be in the interval $(1, 2)$.  Let's say that $\beta$ is.  Then the sequence $\{\lfloor n\beta \rfloor\}_{n \in{\bf Z}_{>0}}$ has gaps of size 1 or less.  So, suppose we have a gap somewhere, $b_{n}= m$ and $b_{n+1}= m+2$.  Thus $m < n\beta < m+1$ and $m+2 < (n+1)\beta < m+3$ and $n \leq m$.

This means that $\frac{m+2}{n+1}< \beta < \frac{m+1}{n}$, so if I write $n = m-k+1$ we have $k > 0$ and
\[\frac{m+2}{m-k+2}< \beta < \frac{m+1}{m-k+1}\]
\[\frac{k}{m-k+2}< \beta-1 < \frac{k}{m-k+1}\]
\[m-k+1 < \frac{k}{\beta-1}< m-k+2 \]
\[m+1 < k\cdot \frac{\beta}{\beta-1}< m+2 \]
But $\frac{\beta}{\beta-1}= \alpha$, so $\lfloor k \cdot \alpha \rfloor = m+1$ and the two sequences together cover all the integers.


determine k so that f(x) = 2x^2 +(k-1)x - k(k+1) is divisible by x+2 and the value of k is an exact divisor of

9
15
27
49
none of these

[hide="Hint"]
Factor theorem.

If $ f(x)$ is a polynomial then $ (x\minus{}a)$ is a factor iff $ f(a)\equal{}0$
[/hide]


function
combinatorics unsolved
combinatorics

Give two fintes set $X$ and $Y$ such that $|X|=n,|Y=m|$ where $n \geq m >0$.
Find the number of onto function mapping form $X$ to $Y$  :)

By 'onto' you mean surjective?
In that case, just look to another thread in this section.

Pierre.

Anyway ;
let
$f: X \rightarrow Y,|X|=x,|Y|=y$ define $A_{i}: =\{f: X \rightarrow Y;j\notin ImY\},j=1,2,\dots ,y$
Then $|A_{1}^{c}\cap A_{2}^{c}\cap \dots \cap A_{y}^{c}|=y^{x}-{y \choose 1}(y-1)^{x}+\dots+(-1)^{y}{y \choose y}(y-y)^{x}=\sum_{i=0}^{y}(-1)^{i}{y \choose i}(y-i)^{x}$


It isn't usually trues with all $ k \ge 0$, so you should write "Find all the calue of $ k$ for the following ineq still trues"  :maybe:
My YM: langtukhongnha1012

[quote="AngelHolder"]let a,b,c  and k are possitive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$  . Prove the inequality:
 $ \sqrt {ka^2 \plus{} b} \plus{} \sqrt {kb^2 \plus{} c} \plus{} \sqrt {kc^2 \plus{} a} \geq 3\sqrt {k \plus{} 1}$[/quote]

$ k\geq \frac{1}{2}$

The inequality  holds.


1. A car with tires having radius 33cm was driven on a trip and the odometer indicated that the distance travelled was 640km. Two weeks later, with snow tires installed, the odometer indicated that the distance for the return trip over the same route was 625 km. Find the radius of the snow tires.

I keep thinking this is $\frac{640}{625}=\frac{33}{x}$ But it does not agree with the answer given:
[hide]33.8[/hide]
2. Bob and Jim, next-door neighbours, use hoses from both houses to fill Bob's swimming pool. They know it takes eighteen hours using both hoses. They also know that Bob's hose, used alone, can fill the pool in six hours less than Jim's hose. How much time is required by each hose alone?

This one is just plain confusing, I've declared about 5 variables in the first 15 seconds of trying to solve it, the answer is:
[hide]33.25h with Bob's hose and 39.25h with Jim's hose[/hide]

3. Prove that $\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}+\sqrt{3}}=2$
I'm really confused on this one, as my calculator tells me its not 2 lol.

[b]1[/b]. $t_{1}$ - time of travel of the car with tires having radius $R_{1}= 33$cm. $S_{1}= 640$km - distance of this car .
$t_{2}$ - time of travel of the car with tires having radius $R_{2}= x$cm. $S_{2}= 625$km - distance of this car .

Then: $t_{1}\approx \frac{1}{R_{1}}$, or $t_{1}=\alpha \frac{1}{R_{1}}$ and $t_{2}\approx \frac{1}{R_{2}}$, or $t_{2}=\alpha \frac{1}{R_{2}}$, $\alpha$ - aspect ratio.

And $S_{1}\approx t_{1}$, $S_{2}\approx t_{2}$.

Then $\frac{S_{1}}{S_{2}}=\frac{\alpha \frac{1}{R_{1}}}{\alpha \frac{1}{R_{2}}}$.
$\Rightarrow \frac{S_{1}}{S_{2}}=\frac{R_{2}}{R_{1}}$
$\Rightarrow \frac{640}{625}=\frac{x}{33}$
$\Rightarrow x = 33,792$ or $x \approx 33,8$

[b]2[/b]$x$h - time, use Bob's hose to fill one swimming pool.
$(x+6)$h - time, use Jim's hose to fill one swimming pool.
For $x$h Jim's hose to fill $\frac{x}{x+6}$ part swimming pool.
Use hoses from both houses, then for $x$h  fill $(1+\frac{x}{x+6})$ pools. "We know it takes eighteen hours using both hoses".
Proportion:
$\frac{1}{1+\frac{x}{x+6}}=\frac{18}{x}$
$18(1+\frac{x}{x+6})=x$
$18\frac{x}{x+6}=x-18$
$18x=(x+6)(x-18)$
$x^{2}-30x-108=0$
$x \approx 33,25$h - Bob's hose, $(x+6) \approx 39,25$;// Excuse me for my English


An arbitrary point $B$ is chosen on the line $AC$ and on the segments $AB,BC$ and $CA$ as diameters, circles $k_1,k_2$ and $k$ are constructed. An arbitrary line through the point $B$ intersects $k$ in the points $P$ and $Q$ , and circles $k_1$ and $k_2$ (besides the point $B$ )in the points $R$ and $S$ , respectively. Prove that $PR=QS$.

Let AR and CQ meet the circle k at M and N.
Since <ARB=<CSB=$90^{o}$. So AR || CQ.
Hence arc AN = arc CM.  So arc MAN = arc AMC. Which means AC=MN. So MN is the diameter of the circle k.
Rest is easy congruents.

$RS$ is the projection of $AC$ on $PQ$, hence $R$ and $S$ are equally apart of $O$, midpoint of the diameter $AC$, but so are $P$ and $Q$, done.

Best regards,
sunken rock


search calculus algebra polynomial probability geometry ARML

3

That's it.
Nice job stretching the whole screen all the way so the text won't warp anymore.
Also, this topic has been posted before a lot.

it has? i didn't know that...

u only know 3 digits? oh...

and also i didn't know how to "unstretch" my post.  i don't know how to fix it.

You didn't have to post all those digits.
Next time use the search option.
I do not know 3 digits of pi, but only know it rounded to the nearest integer. I believe that's 3.

what's the search option?  and how do you become a mod?

3.14159265358979323846 (im not sure about the last 2 digits)

i dont kno anymore... and i bet u used the internet (unless ur lyk pifreak)

no i'm not like pifreak (who was that person) and yes i did use the internet otherwise i would've only known 3.141592653589793238462643383279 (i think) so yes.

hmm...pifreak knew way way way way more than that cardwacko.

Personally, if I were to memorize anything for a hobby, it'd probably be the Periodic Table.  It's more useful, and more practical.  Plus it's more multifaceted, since there's atomic masses, charges, oxidation numbers, and properties to remember in addition to names and symbols.

an u are A+MATH? ... i judge people by their usernames. sorry.

dont judge people by how many digits of pi they know

why $pi$ why not $pi*e$? apparantly it is not shown that it is transcendental, so you might be able to find a polynomial which ends (= and memorise that!

[quote="sunnyray"][quote="jli"]here are some pi songs to sing on March 14:

Happy Pi Day

Happy Pi day to you,
Happy Pi day to you,
Happy Pi day everybody,
Happy Pi day to you.

(to the tune of "Happy Birthday")

Oh Number PI

Oh, number Pi
Oh, number Pi
Your digits are unending,
Oh, number Pi
Oh, number Pi
No pattern are you sending.
You're three point one four one five nine,
And even more if we had time,
Oh, number Pi
Oh, number Pi
For circle lengths unbending.

Oh, number Pi
Oh, number Pi
You are a number very sweet,
Oh, number Pi
Oh, number Pi
Your uses are so very neat.
There's 2 Pi r and Pi r squared,
A half a circle and you're there,
Oh, number Pi
Oh, number Pi
We know that Pi's a tasty treat.

(to the tune of "Oh Christmas Tree")

Pi Day Song

Refrain:

Pi day songs
All day long.
Oh, what fun it is,
To sing a jolly pi day song
in a fun math class
like this. (Repeat )

Verse:
Circles in the snow,
Around and round we go.
How far did we have to run?
Diameter times pi! (Refrain )

(to the tune of "Jingle Bells")[/quote]

just wondering... did you make that up?
(I haven't been on for a while!... but now I'M BACK - why don't I hear any cheering? :D )[/quote]

Wooohoooo!!!!! ;)

You can lord all you want over pi-memorizers but memorizing pi is like exercising for some mathematicians, it keeps their brains in shape so they can think qucker.

This is not personal experience, but someone told it to me.

Why dont other numbers get memorized, like e?

[quote="mcalderbank"]Also, I heard somewhere that the probability of finding a seven digit number in pi (phone number) is 1[/quote]

haha wow... so i'm gonna randomly call 314-1592... and we'll see who picks up! :-D hahah

in geometry, on pi day, our teacher gave us one bonus point on a test for each 10 digits of pi you knew (up to 5 points)... hence i memorized the first 50 after the decimal... [hide]3.141592653589793238462643383279502884197169399375105[/hide]

oh yeah, i found this too.

[u][b]Pi vs e[/b][/u]
Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed?

[quote="sunnyray"]just wondering... did you make that up?
(I haven't been on for a while!... but now I'M BACK - why don't I hear any cheering? :D )[/quote]Hmmmm....if I remember right weren't you one of those people from Miller Middle School that spammed all over the place?

[quote="gameworld7"]the periodic table is too long...i memorized up to i think only 40 elements.  i need to get a new hobby...[/quote]

I've memorized Tom Lehrer's "The Elements." Does that count?  :lol:

I memorized Pi up to 3.141592653589793238462643383279502884 and then i think 197128944 or something back in like 3rd grade and it's stuck with me since :-)

Sorry for topping old threads but can't resist. :P I'm looking at frankthetank's previous posts.

For the 8th grade talent show, I memorized the entire periodic table. All the way to meitnerium. I still remember the tune of it. I also memorized the list of US Presidents from 1st to last and may consider memorizing all Nobel Prize Physics winners from 1900 to present.

120 digits...i got hooked when someone did a pi song at san jose arml 2004  :D

I remember pi upto 3.14159 and never get into trouble. Memorizing these things is the job of machines. My calculator has it built in.


geometry
geometric transformation
invariant
number theory proposed
number theory

Find all sequences of positive integers $a_n$ such that
$a_na_{n+3} = a_{n+1}a_{n+2} + 1 $ for all integers n

All integers n? I suppose it doesn't really matter, but is that meant to be whole numbers? Or natural numbers?

It doesn't really matter, but I meant all integers. The equation is symmetric so we can solve for either the next term or the previous term in the recursion, unless a particular term is 0. The restriction of values to the positive integers is much more important- it gets very complicated when 0 is a term.

After seeing the new site organization, I think this would fit best in number theory/proposed and own problems. It's not a perfect match, but at least the problem is based in integer arithmetic. This is also clearly an olympiad-level problem.

I am the source here- I've never seen any version of this problem elsewhere, and I first posed it to others in 1999 (at MOSP).

I'm reviving this problem, since no solutions were posted the first time.

Up to translation (which I don't care about), there are finitely many solutions. If the sequence took integer values, there would be uncountably many solutions- but almost all would take 0 as a value infinitely often.

It isn't that hard to find the solutions just by playing with the recursion; that's how I came up with this problem in the first place.

I think a crucial fact is writing $\frac{a_{n+3}}{a_{n+1}}=\frac{a_{n+2}+a_{n+4}}{a_n+a_{n+2}}$, and then multiplying these. We get $a_{2k+2}=-a_{2k}+\frac{a_0+a_2}{a_1}\cdot a_{2k+1}$ (I think; something like that, anyway). I'll complete it after I get some sleep :).

The sequence is uniquely determined by a triplet of consecutive terms, $a_0,a_1,a_2$ s.t. $a_0|a_1a_2+1,\ a_1|a_0+a_2,\ a_2|a_1+a_0$. I mean that these conditions are necessary and sufficient (all triplets of consecutive terms satisfy these).

Now take the triplet with the minimal sum. We have $(a_0,a_1)=(a_1,a_2)=(a_2,a_0)=1$, and $a_0\le \frac{a_1a_2+1}{a_0},\ a_2\le\frac{a_0a_1+1}{a_2}$. These relations show that $\max(a_0,a_1,a_2)=a_1\Rightarrow a_0+a_2\in\{a_1,2a_1\}$. If it’s $2a_1$, we get $a_0=a_1=a_2=1$.

Let’s look at the case $a_1=a_0+a_2$ now. We apply two transformations, and go to the triplet $(a_2,a_3,a_4)$. It’s easy to show that $a_3=a_2+a_4$, so the minimaltiy of the sum of the first triplet is equivalent to $a_1\le a_3\iff a_1\le \frac{a_1a_2+1}{a_0}$. Assume WLOG that $a_2<a_0$ (otherwise we use the triplet $(a_{-2},a_{-1},a_0)$ instead of $(a_2,a_3,a_4)$). We thus get $a_1\le a_1-\frac{a_1-1}{a_0}\Rightarrow a_1=1$, which is absurd. We must thus have $a_0=a_2$, and since $(a_0,a_2)=1$, we get $(a_0,a_1,a_2)=(1,2,1)$.

These are the two solutions: $(a_0,a_1,a_2)\in\{(1,1,1),(1,2,1)\}$.

Since $\frac{a_{2k}+a_{2k+2}}{a_{2k+1}}$ and $\frac{a_{2k-1}+a_{2k+1}}{a_2k}$ are invariants of the sequence and the pairs of such invariants of the sequences generated by the two triplets are different, the two sequences are, indeed, different.

I know there’s some stuff I used without proof, but I can make it clear in case someone wants me to.

Is it Ok?

It's related to: http://www.mathlinks.ro/Forum/viewtopic.php?p=15723&highlight=&sid=cb803f607f5f513e4b0956fe9f83075c#15723

More general , let $a_n$ be the numbers produced by a machine such that  :

$a_{n+3} = \frac {a_{n+1}a_{n+2} + T}{a_n}$ , where $T$ is an integer ,  $a_0,a_1,a_2$ are non zero and coprime integers with $T$ ( the common divisors with $T$ are only 1 and -1)

 If  $a_3,a_4,a_5$ are integers numbers then  the machine produces only coprime integers with $T$,  or it stopes (because division by 0 :D ) .

 :cool: Very interesting, it's a reverse machine !:cool:

Your proof looks good, grobber; maybe the algebra you started with should be expanded, including the proof that (1,2,1) and (1,1,1) actually work.

My own proof used a minimal pair (either in sum or product); it is easy to show that one of these must be a 1, giving the sequence a (1,x,1) block. If $a_{-1}=1, a_0=x,a_1=1$, then $a_3=\frac{x+2}{x}$, so either $x=1$ or $x=2$.
For the rest, I used a lemma that if six consecutive terms are nonzero integers, the next term on each side is an integer. In the positive case, this forces all terms to be integers (this lemma would also work for the problem Diogene posted, with no modifications to its proof).

A few terms of these sequences:
$\dots, 1,1,1,2,3,7,11,26,41,97,153, \dots$
$\dots, 1,2,1,3,2,7,5,18,13,47,34, \dots$

A sequence with a zero in it, just as a taste of what that allows:
$\dots, -7,3,-5,2,-3,1,-1,0,1,-1,-2,3,5,-8,-13, \dots$

In order to explain why those solutions work, it's enough to observe that if $a_n,a_{n+1},a_{n+2}$ satisfy those three relations, then so do $a_{n-1},a_n,a_{n+1}$ and $a_{n+1},a_{n+2},a_{n+3}$ (the three relations I'm talking about are: $a_n|a_{n+1}a_{n+2}+1,\ a_{n+1}|a_n+a_{n+2},\ a_{n+2}|a_na_{n+1}+1$).