calculus
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[img]http://img50.imageshack.us/img50/2246/integralrx3.jpg[/img]

Can anybody please show me how to work this step-by-step (integral by parts) in solving this integral? The "e^[x^2]" part confuses me. I understand that you have to move an "x" to "x^5" but after that I'm confused? Please explain to me in details so I can understand how to solve these type of integrals thoroughly.

Thanks.

Seeing $x^{2}$ trapped inside of the exponential function, you should think of substitution $t=x^{2}$. It's great that there is an odd power of $x$ in the integral, because you will need one copy of $x$ to form $dt=d(x^{2})=2xdx$, and the remaining term $x^{4}$ will turn into $t^{2}$. So you get \[\frac12 \int t^{2}e^{t}dt.\] This is a common example of repeated integration by parts. You should think of dividing the integrand into two factors with the purpose of replacing one of them by its derivative and the other by its antiderivative. $e^{t}$ will not change either way; $t^{2}$ will become simpler if replaced with its derivative $2t$. And so it goes: \[\int t^{2}e^{t}dt = \int t^{2}d(e^{t}) = t^{2}e^{t}-\int e^{t}d(2t)=t^{2}e^{t}-2\int te^{t}dt.\] (Here I used $\int udv=uv-\int vdu$ with $u=t^{2}$ and $v=e^{t}$.)
And now for an encore: \[\int te^{t}dt = \int t d(e^{t}) = te^{t}-\int e^{t}dt=te^{t}-e^{t}+C.\] It would be a good idea not to omit the $dx$ part of an integral. $\int x^{2}dx$ is not the same as $\int x^{2}dy$...


Let $x$, $y$, and $z$ be three positive real numbers whose sum is $1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is

(A) $\frac{1}{32}$
(B) $\frac{1}{36}$
(C) $\frac{4}{125}$
(D) $\frac{1}{127}$
(E) none of these

[hide="Solution"] Intuitively, we want to space $x, y, z$ out as far as possible to minimize their product.  (This can be made rigorous one of several ways - for example, majorization.)

[b]Case:[/b]  $x = y, z = 2x$.  Then $x = y = \frac{1}{4}, z = \frac{1}{2}, xyz = \frac{1}{32}$

[b]Case:[/b]  $y = z = 2x$.  Then $x = \frac{1}{5}, y = z = \frac{2}{5}, xyz = \frac{4}{125}= \frac{1}{30+\frac{5}{4}}> \frac{1}{32}$

Then $\boxed{A}$ is our answer.[/hide]


Alice and Bob each arrive at a party at a random time between 1:00 and 2:00. If Alice arrives after Bob, what is the probability that Bob arrived before 1:30?

Make a pretty sketch of the scenario on a version of the Cartesian plane.  Alice will be on the $ x$-axis, Bob will be on the $ y$-axis.

[asy]draw((0,6)--(0,0)--(6,0));
label("$2:00$",(0,6),W);
label("$2:00$",(6,0),S);
label("$1:00$",(0,0),SW);[/asy]

Shade in the probability that Alice arrives after Bob.

[asy]draw((0,6)--(0,0)--(6,0));
label("$2:00$",(0,6),W);
label("$2:00$",(6,0),S);
label("$1:00$",(0,0),SW);
label("$1:30$",(3,0),S);
label("$1:30$",(0,3),W);
fill((0,0)--(6,0)--(6,6)--cycle,gray);[/asy]

It becomes clear that the probability that Bob arrives after $ 1: 30$ is simply $ \boxed{\frac14}$.

Then to answer the question, we find the probability that Bob arrives before $ 1: 30$ is $ \boxed{\frac{3}{4}}$.

Challenge:  If the probability that he arrives before is 1/4, and the probability that he arrives after is 3/4, then is the proability that he arrives at exactly 1:30 0??

[hide="Answer to Challenge"]$ \boxed{0}$.  This is the same thing as picking an integer out of a pool of numbers.[/hide]

The problem is that there isn't enough information.

I assumed that Bob arrived at a random time, and then Alice arrived after that. In other words, I assumed that, though Alices's time was dependent on Bob, Bob's time was independent of Alice.

Argh I solved the problem wrong by saying Alice can before Bob. I said it was 1/4 instead of 3/4.

This was a fun question!  (yay for Alice & Bob questions haha)
[quote=minikakapo]How do I know that the ratio of the area of the shaded region to the left of the line marking 1:30 to the area of the whole shaded region is 3/4?[/quote]
We can use similar triangles.  Knowing that the side length of the smaller simiar triangle is half that of the larger shaded area, we know that the smaller triangle has $\frac14$ the area of the larger triangle...

...meaning that the trapezoid takes up $\frac34$ of the shaded region.

Oh wait!  It looks like we could also use the formula for the area of a trapezoid to solve this question.

[hide=sol]
note that if it is after 1:30, the average is 1:45, with probability of 15/60=1/4. So 1-1/4=3/4
[/hide]

Question:
Bogus solution:
since Bob arrives any time between 1 to 2 PM, and we want to figure out P(Bob arrives before 1:30), the answer is clearly 1/2.
However, using conditional probability, we get:
P(Alice AND Bob arrive before 1:30)=P(Alice arrives before 1:30|Bob Arrives before 1:30)*P(Bob arrives before 1:30).
So Alice and Bob arrives before 1:30 has a 1/4 chance, P(Alice arrives before 1:30 | Bob arrives before 1:30)=1/3.
So $1/4=1/3*p$.
$p=3/4$. So it's truly 3/4. But why isn't it 1/2?
I'm stuck.


http://mcis.jsu.edu/mathcontest/geo01.pdf

number $\#48$

OK, so copy the diagram.  We are dealing with ratios, not exact numbers, so we can make up numbers and keep a constant ratio, as long as the numbers we make up satisfy the conditions in the problem.  (We could add in x's throughout, but that takes too long!)

$\frac{AX}{AB}= \frac{1}{3}$.  So let $AX = 1$ and let $XB = 2$.
$\frac{BY}{BC}= \frac{1}{3}$.  So let $BY = 1$ and let $YC = 2$.

Now we can use Ceva's theorem:  $AX \cdot BY \cdot CZ = XB \cdot YC \cdot ZA$.

So, plugging in the values I made up:
$1 \cdot 1 \cdot CZ = 2 \cdot 2 \cdot ZA$.
$CZ = 4 \cdot ZA$.

So let $CZ = 4$ and let $ZA = 1$.  So by the diagram, $CA = 5$.  Thus, $\frac{ZA}{CA}= \boxed{\frac{1}{5}}$.

This is a nice test.  I wish I could have taken it...


The average of thirteen consecutive integers is 10. What is the smallest of the thirteen integers?

[hide]the smallest is 4. if the average is 10, then the 6th number must be 10, meaning that the 1st number is 4.[/hide]

[hide]
if you put them in a line you can have 5 numbers on each side of 10 so 10-5= $5$[/hide]

[hide]Since they are consecutive the middle number must be 10. 10-6=4 . The smallest of the thirteen numbers is 4.[/hide]


[quote="Aryabhatta"]If I am reading this correctly: there are no such $a_{i}$.

It is well known that if $a_{i}> 0$

then

$(a_{1}+a_{2}+..+a_{6})(\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{6}}) \ge 36$  (use AM>=GM on each sum)

So the product cannot be 10

Since a false statement implies any statement, we are done without any algebra, tedious or not.  :D  :P[/quote]
Not quite. $r$ could be negative, or the sequence could involve nonreal terms. :wink:

[quote="scorpius119"]
Not quite. $r$ could be negative, or the sequence could involve nonreal terms. :wink:[/quote]

Eh? The original problem statement mentions that $a_{1}, r > 0$, which implies positive real.  :)


How many positive integer values of $ x$ are there such that $ \frac{36}{x\plus{}3}$ is an integer?

Basically, $ x\plus{}3$ must be a positive integer factor of 36.  $ 36\equal{}2^2\cdot 3^2$, so it has $ (2\plus{}1)(2\plus{}1)\equal{}9$ factors.  However, if $ x\plus{}3$ is 1,2, or 3, then $ x$ is not positive, so we get $ 9\minus{}3\equal{}\boxed{6}$


The set $ S \equal{} \{\frac{1}{r} : r\equal{}1, 2, 3, ...\}$ of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, $ \frac{1}{20}, \frac{1}{8}, \frac{1}{5}$ is such a progression of length $ 3$. (and common difference $ \frac{3}{40}$). Moreover this is a maximal progression in $ S$ of length $ 3$ since it cannot be extended to the left or right within $ S$ ($ \frac{\minus{}1}{40}$ and $ \frac{11}{40}$ not being members of $ S$).

(i) Find a maximal progression in $ S$ of length $ 1996$.

(ii) Is there a maximal progression in $ S$ of length $ 1997$?

[quote="AndrewTom"]The set $ S \equal{} \{\frac {1}{r} : r \equal{} 1, 2, 3, ...\}$ of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, $ \frac {1}{20}, \frac {1}{8}, \frac {1}{5}$ is such a progression of length $ 3$. (and common difference $ \frac {3}{40}$). Moreover this is a maximal progression in $ S$ of length $ 3$ since it cannot be extended to the left or right within $ S$ ($ \frac { \minus{} 1}{40}$ and $ \frac {11}{40}$ not being members of $ S$).

(i) Find a maximal progression in $ S$ of length $ 1996$.

(ii) Is there a maximal progression in $ S$ of length $ 1997$?[/quote]

(i) : $ \left\{\frac k{1996!}\right\}_{k\equal{}1}^{1996}$ (maximal since $ 1997$ is prime)

(ii) : to be continued :)


The vapor pressure of a component of a solution is directly related to the mole fraction of that component.


Is this from alcumus?  :P

Hint:  volume is maximized in a cube, not a prism.  use surface area formulas.

Yeah, but why is it maximized in a cube?
Answer, for those who want to know:
[hide]91.125[/hide]

Say you have a prism with side lengths $ 6$, $ 7$, and $ 8$, (the average is $ 7$).  You also have a cube with side length $ 7$.

As you can clearly see, the cube has the greater volume.

I suppose it's somewhat because $ x^2>(x\minus{}1)(x\plus{}1)$.

FOr ernie's example, the surface area in the prism is 292, but for the cube, it's 294, so you can't compare those :P

Because the volume of the prism is 336, while the cube is just 343, so it's pretty close still.

[quote="Jongao"]Richard has a 9-inch by 13.5-inch sheet of cardboard. He wants to create a closed cube with the largest possible volume. There are no restrictions on the way he can cut the cardboard or tape it back together, and no cardboard is lost when he makes a cut. What is the greatest volume possible? Express your answer in cubic inches as a decimal to the nearest thousandth.[/quote]

Aww, you got this from a certain target round, but I can't remember which.

First, you want to find the total area of the cardboard, which is 121.5^2 inches. Then this is obviously the surface area of the cube. You have to divide by 6 to find the area of the face of the cube.

121.5/6 = 20.25

This is the area of the face, but the face is just x^2. So you find the square root to find the side length of the cube. The side length turns out to be 4.5. Cubing this number results in a volume of 91.125.

The answer is 91.125 as you said before.


algebra
polynomial
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algebra solved

Suppose P is a polynomial such that P(z) is real if and only if
z is real. Prove that P is linear.

We may assume the polynomial is monic. It clearly has real solutions, so it's a real polynomial. For any real $r$, the polynomial $P(x)+r$ has real roots. If $r$ is large enough (large enough for $P$ to take positive values in all the real roots of its derivative, for example), $P$ has at most one real root, so the degree of $P$ must be $1$.


limit
ceiling function
integration
floor function
logarithms
function
calculus

$\textbf{FYI-}\\ \\ \lceil \textbf{x}\rceil\;\;\;is\;\;\;the\;\;\;ceiling\;\;\;function,\\ \\ \lfloor \textbf{x}\rfloor\;\;\;is\;\;\;the\;\;\;floor\;\;\;function,\;\;\; and\\ \\ \gamma\;\;\;is\;\;\;\textbf{Euler's}\;\;\;constant$

[hide="ii"]
Partition by integers.
$\int \equiv \sum_{j=2}^{\infty}\int_{0}^{1}\frac{y}{(y+j)^{2}}dy$

$\int \frac{y}{(y+k)^{2}}dy= \int \frac{1}{y+k}-\frac{k}{(y+k)^{2}}dy = \ln (y+k)+\frac{k}{y+k}+C$

The summation is then equiv. to:
$\sum_{j=2}^{\infty}\ln \left( \frac{k+1}{k}\right)-\frac{1}{k+1}$

If we take the sum to $N$, we have:
$\sum_{j=2}^{N}\ln \left( \frac{k+1}{k}\right)-\sum_{j=2}^{N}\frac{1}{k+1}= \ln \frac{N+1}{2}-\sum_{j=3}^{N}\frac{1}{k}$ $= 1+\frac{1}{2}-\ln 2-\left( \sum_{j=1}^{N}\frac{1}{k}-\ln (N+1)\right)$ $=\frac{3}{2}-\ln 2-\left( \sum_{j=1}^{N+1}\frac{1}{k}-\ln (N+1)-\frac{1}{N+1}\right)$

As $N \to \infty$, this becomes:
$\frac{3}{2}-\ln 2-\lim_{N \to \infty}\left( \sum_{j=1}^{N+1}\frac{1}{k}-\ln (N+1) \right)=\boxed{\frac{3}{2}-\ln 2-\gamma}$

[/hide]

Here's a solution for [i]i[/i] based on the solution given above for [i]ii[/i].

Let $I = \lim_{q}\frac{1}{q}\sum_{p=1}^{q}\left( \left\lceil \frac{q}{p}\right\rceil-\frac{q}{p}\right) = \int_{0}^{1}\left( \left\lceil \frac{1}{x}\right\rceil-\frac{1}{x}\right) dx \overbrace{=}^{u=\frac{1}{x}}\int_{1}^\infty \frac{\lceil x \rceil-x}{x^{2}}dx$

Also, let $J = \int_{1}^\infty \frac{x-\lfloor x \rfloor}{x^{2}}dx$, $J_{1}$ be the integral from part [i]ii[/i] and $J_{2}= \int_{1}^{2}\frac{x-\lfloor x \rfloor}{x^{2}}dx$ so that $J = J_{1}+J_{2}$

Now, setting $u = x-1$ in $J_{2}$ we get $J_{2}= \int_{0}^{1}\frac{u-\lfloor u \rfloor}{(u+1)^{2}}du = \int_{0}^{1}\frac{u}{(u+1)^{2}}du = \ldots = \ln 2-\frac{1}{2}$ (see [b]*[/b] note below for omitting the floor function), so $J = 1-\gamma$

Note that $I+J = \int_{1}^\infty \frac{\lceil x \rceil-\lfloor x \rfloor}{x^{2}}dx = \int_{1}^\infty \frac{1}{x^{2}}= 1 \Rightarrow I = 1-J = \gamma$. (again, see [b]*[/b] note below for omitting ceiling/floor functions)

[b]*[/b] note: I'm not sure if I have to prove these two bits or how to "formally" prove them. Intuitively, the explanation is: $u-\lfloor u \rfloor = u$ for $u \in [0, 1)$ (i.e. except at $\{1\}$, where it's 0) and $\lceil x \rceil-\lfloor x \rfloor = 1$ for $x \in \mathbb{R}-\mathbb{Z}$ (it's 0 only for $x \in \mathbb{Z}$). So the transformations hold almost everywhere. How do I phrase this more formally?


Prove that if a, b, c > 0 such that 1/a + 1/b + 1/c = 1 then

:sqrt:(abc) + :sqrt:a + :sqrt:b + :sqrt:c :le: :sqrt:(a + bc) + :sqrt:(b + ca) + :sqrt:(c + ab).

This one is cute :)

A hint :

You can replace :sqrt:abc with ...

Arne wrote:Prove that if a, b, c > 0 such that 1/a + 1/b + 1/c = 1 then

:sqrt:(abc) + :sqrt:a + :sqrt:b + :sqrt:c :le: :sqrt:(a + bc) + :sqrt:(b + ca) + :sqrt:(c + ab).

This one is cute

[hide]Ok, we go ahead and replace :sqrt: (abc) with :sqrt: (ab+ac+bc). We then square both sides of the given inequality to get :sqrt:(a:^2:bc)+similar terms:le::sqrt:(ab+a:^2:c+b:^2:c+abc:^2:)+similar terms. For every term on the left, there is a term on the right that is greater than that. (for example, :sqrt:(abc:^2:) and :sqrt:(ab+a:^2:c+b:^2:c+abc:^2:)). Thus the inequality holds true (all steps are reversible).[/hide]

My original solution is a bit different :

Since 1/a + 1/b + 1/c = 1, :sqrt:(abc) = :sqrt:(ab/c) + :sqrt:(bc/a) + :sqrt:(ca/b).

So it suffices to prove that :Sigma::sqrt:(ab/c) + :Sigma::sqrt:c :le: :Sigma::sqrt:(ab + c).

Now it clearly suffices to prove that :sqrt:(ab/c) + :sqrt:c :le: :sqrt:(ab + c).
This is equivalent to ab/c + c + 2:sqrt:ab :le: ab + c.
Now ab/c = (1 - 1/a - 1/b)ab = ab - a - b since 1/a + 1/b + 1/c = 1.
So the inequality to prove is equivalent to a + b :ge: 2:sqrt:ab.
THis is obvious :)

Still two inequalities to go ... nrs. 28 & 30 :mrgreen:.

[quote="Arne"]
Since 1/a + 1/b + 1/c = 1, :sqrt:(abc) = :sqrt:(ab/c) + :sqrt:(bc/a) + :sqrt:(ca/b).
[/quote]

How on earth do you know things like that. All of these subsitutions blow me away... is there a book somewhere with all kinds of related substitutions...?

How do you mean ?
Just multiply both sides of 1/a + 1/b + 1/c = 1 with :sqrt:(abc) ...
This is natural since you can create cyclic sums like this !
Be inventive !

I don't know whether your soln is correct M.T. ...
I mean, your solution (if it were correct) implies that we have equality if ond and only if ab + ac + bc is zero (do you see it?).
But equality holds if a = b = c = 3 ... so sth must be wrong :?

I guess this is just simply above me, cause I don't even know what you mean by cyclic sums.

Hey arne, why don't you take a look at

http://www.artofproblemsolving.com/wforum/viewtopic.php?p=15764#15764

A cyclic sum :

:Sigma:f(x, y, z) = f(x, y, z) + f(y, z, x) + f(z, x, y)

Do you see why :Sigma:(ab) = ab + bc + ca ?

The concept of cyclic sums is very important in proving inequalities.
Another inequality concerning cyclic sums is nr. 28.

We also have symmetric sums :

Symsum f(x, y, z) =
f(x, y, z) + f(x, z, y) + f(y, z, x) + f(y, x, z) + f(z, x, y) + f(z, y, x)

For Symsums we have Muirhead's theorem.
Unfortunately that doesn't work for cyclic sums.

This inequality (1/a + 1/b + 1/c = 1) was proposed at APMO 2002.


Let $ n_1,n_2...n_k$ be a positive integers and $ k\ge 2$ such that :

$ n_2\mid 2^{n_1} \minus{} 1$

$ n_3\mid 2^{n_2} \minus{} 1$
$ .$
$ .$
$ .$
$ n_{k}\mid 2^{n_{k \minus{} 1}} \minus{} 1$

$ n_1 \mid 2^{n_k} \minus{} 1$

Prove that $ n_1 \equal{} n_2 \equal{} .... \equal{} n_k \equal{} 1$


A rectangular marble slab,2.1 m long,10 cm thick, and with a mass of 200kg lies on a level floor. How much workmust be done to set it vertically on end?

[hide="hint"]How high will the center of mass of the slab be compared to the initial position?? :wink: [/hide]


function
geometry
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LaTeX

As you can see on my signature, I proved that $ 1\plus{}1\equal{}1$.

(One drop of water)+(One drop of water) = One bigger drop of water.

What do you think?

Any counter examples?

It's an isolated example and doesn't hold for all 1.  1 piece of candy + 1 piece of candy = 2 pieces of candy.  Water drops just happen to be very cohesive. :D

Your wrong- 2 small drops of water just combnie and make 2, put together

your saying, 1x+1x=1x, but then x would have to be 0, so what happens is 1x+1x=2x, that looks like 1x. but it has more mass and volume so its just 2x,

[quote="DonkeyKong"]As you can see on my signature, I proved that $ 1 \plus{} 1 \equal{} 1$.

(One drop of water)+(One drop of water) = One bigger drop of water.

What do you think?

Any counter examples?[/quote]
The bigger drop of water is different from the original drops of water.

That's like saying $ 1x \plus{} 1x \equal{} 1y$. You can't divide both sides by $ x$ or $ y$ to show that $ 1\plus{}1\equal{}1$

okay... fine...

You guys made your point...

I thought this one was as smart as Einstein's theory... but I guess not..

But do you guys mind if I keep this as my signature?

what's so revolutionary about $ 1 \plus{} 1 \equal{} 1$ anyway?
true for any structure where additive identity is additively idempotent...

[quote="Temperal"][quote="DonkeyKong"]
I thought this one was as smart as Einstein's theory..[/quote]

I really, really, really, hope you're kidding.[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="i_like_pie"]Make it look a bit better (text in $ \text{\LaTeX}$ is ugly), like this:

$ \mathrm{Revolutionary\;Phenomenon}: \;1 + 1 = 1$

$ \mathrm{one\;drop\;of\;water + one\;drop\;of\;water = one\;drop\;of\;water}$

Or just plain text... Might be better.[/quote]

That is more than 3 lines.... I had to make it like that...

@Temperal: Yes, I was kidding... Don't worry, I will never be smart as Einstein.
I am sorry if that shocked you... :lol:

[quote="i_like_pie"][quote="DonkeyKong"](One [b]drop[/b] of water)+(One [b]drop[/b] of water) = One [b]bigger drop[/b] of water.[/quote]
Different variables. :wink:[/quote]

 :rotfl:

[quote="CircleSquared"]Don't laugh:
$ \sqrt { \minus{} 1} \equal{} \sqrt {\frac {1}{ \minus{} 1}}$
$ i \equal{} \frac {1}{i}$[/quote]

Oh man, that is so illegal, I cannot even describe how illegal it is.

But it's clever.

Add it to [[Fallacious proof]].

[quote="CircleSquared"]Don't laugh:
$ \sqrt { \minus{} 1} \equal{} \sqrt { \minus{} 1}$
$ \sqrt { \minus{} 1} \equal{} \sqrt {\frac {1}{ \minus{} 1}}$
$ i \equal{} \frac {1}{i}$
$ i^2 \equal{} 1$
$ \minus{} 1 \equal{} 1$
$ \frac { \minus{} 1}{4} \equal{} \frac {1}{4}$
$ \frac {1}{2} \equal{} 1$
$ 1 \plus{} 1 \equal{} \frac {1}{2} \plus{} \frac {1}{2} \equal{} 1$[/quote]

Yet another proof!!!!

Awesome!!

So $ 1\plus{}1\equal{}1$ after all!!

[quote="DonkeyKong"]So $ 1 \plus{} 1 \equal{} 1$ after all!![/quote]

You know that this isn't true.  The goal of such investigations is to strengthen your understanding of actual mathematical concepts.  Turn it into an exercise:  ask yourself, [b]why doesn't that proof work?[/b]

[hide="Answer"] $ \sqrt{ab} \equal{} \sqrt{a} \sqrt{b}$ only holds for positive $ a, b$, and does not hold generally.  The main issue is that the complex square root is a multivalued function, and for complex inputs it is very unclear which square root to take. [/hide]
[quote="CatalystOfNostalgia"]Consider the Banach-Tarski Paradox: [url]http://en.wikipedia.org/wiki/Banach–Tarski_paradox[/url], basically you can take a sphere, break it up, and put it into two spheres congruent to the first using only Euclidean motions.[/quote]

This is a paradox in the traditional sense:  it appears not to make sense at first, but further consideration deepens understanding.  It is not a contradiction.

You can perform a similar decomposition with the integers.  Suppose we take as our "sphere" the integers:

$ \{ ..., \minus{}3, \minus{}2, \minus{}1, 0, 1, 2, 3, ... \}$

Double them to produce the even integers:

$ \{ ..., \minus{}6, \minus{}4, \minus{}2, 0, 2, 4, 6, ... \}$

Add one to every element of this set to produce the odd integers:

$ \{ ..., \minus{}5, \minus{}3, \minus{}1, 1, 3, 5, ... \}$

But the integers are just the union of the even integers and the odd integers!  And this is not paradoxical at all.  :)  Banach-Tarski decomposition is the same idea, but with more complicated sets.

lol torajiru, i guess he already found out the fallacy (unless he is a real kid, even smaller than old babbit here, trying to sue pants off someone while i doubt his own pants stay in their place for long  :rotfl: )

there is a nice proof of $ \pi\equal{}1$:
take the unit line, draw $ n$ identical semicircles over it side by side (each of radius $ \frac{1}{2n}$), the sum of their perimeters is $ \pi$ regardless of $ n$, now letting $ n$ go to infinity, we get the unit line!
this was shown to us this summer in a camp, took us quite a while to figure out the fallacy, it's a nice exercise for you kids

[quote="1=2"][quote="DonkeyKong"]As you can see on my signature, I proved that $ 1 \plus{} 1 \equal{} 1$.[/quote]

...

YOU STOLE MY USERNAME! I'M GONNA SUE THE PANTS OFF OF YOU FOR COPYRIGHT ISSUES!!![/quote]

but your's is 1=2.......

[quote="bubka"]lol torajiru, i guess he already found out the fallacy (unless he is a real kid, even smaller than old babbit here, trying to sue pants off someone while i doubt his own pants stay in their place for long  :rotfl: )

there is a nice proof of $ \pi \equal{} 1$:
take the unit line, draw $ n$ identical semicircles over it side by side (each of radius $ \frac {1}{2n}$), the sum of their perimeters is $ \pi$ regardless of $ n$, now letting $ n$ go to infinity, we get the unit line!
this was shown to us this summer in a camp, took us quite a while to figure out the fallacy, it's a nice exercise for you kids[/quote]

Eh, I can't quite remember the name, but there was an axiom for something like that that applied only to non-infinite quantities...Reid Barton used it for problem number 3 on the 2001 IMO.

Anyways, that's the fallacy.

[quote="DonkeyKong"]but your's is 1=2.......[/quote]
$ 1\plus{}1\equal{}1\quad\to\quad2\equal{}1$

[quote="i_like_pie"][quote="DonkeyKong"]but your's is 1=2.......[/quote]
$ 1 \plus{} 1 \equal{} 1\quad\to\quad2 \equal{} 1$[/quote]

and equality is more often than not symmetric, so mat's claim is not entirely unfounded...

@temperal: i didn't see the fallacy you mentioned

There's this book called [i]Countdown[/i] which describes it. I can't even remember it now, but I'm fairly certain it's the same situation.

[quote="bubka"]lol torajiru, i guess he already found out the fallacy (unless he is a real kid, even smaller than old babbit here, trying to sue pants off someone while i doubt his own pants stay in their place for long  :rotfl: )

there is a nice proof of $ \pi \equal{} 1$:
take the unit line, draw $ n$ identical semicircles over it side by side (each of radius $ \frac {1}{2n}$), the sum of their perimeters is $ \pi$ regardless of $ n$, now letting $ n$ go to infinity, we get the unit line!
this was shown to us this summer in a camp, took us quite a while to figure out the fallacy, it's a nice exercise for you kids[/quote]
The sum of the perimeters is not $ \pi$
let's say $ n$ is $ 1$. the perimeter is $ \frac{\pi}{2}$
if $ n$ is $ 2$, the perimeter is still $ \frac{\pi}{2}$, etc. It's never $ \pi$

also, when $ n$ is $ \infty$, we don't get the unit line. We get infinite mini-semicircles, or basically infinite unit lines if you try to think of it another way that's straight...
Either way, the line(s) is(are) not straight and there are infinitely more points, all adding up to a perimeter of $ \frac{\pi}{2}$

[quote="dbkarp"][quote="DonkeyKong"][quote="CircleSquared"]Don't laugh:
$ \sqrt { \minus{} 1} \equal{} \sqrt { \minus{} 1}$
$ \sqrt { \minus{} 1} \equal{} \sqrt {\frac {1}{ \minus{} 1}}$
$ i \equal{} \frac {1}{i}$
$ i^2 \equal{} 1$
$ \minus{} 1 \equal{} 1$
$ \frac { \minus{} 1}{4} \equal{} \frac {1}{4}$
$ \frac {1}{2} \equal{} 1$
$ 1 \plus{} 1 \equal{} \frac {1}{2} \plus{} \frac {1}{2} \equal{} 1$[/quote]
[/quote]
how can 1=1/2 unless '1' is a variable such as 'x' where it does not equal what is shown, what u say is all well and good, but a larger drop of water has a larger weight than one, and no, u cant climb a mountain and cheat...[/quote]
He added $ \frac{3}{4}$ to both sides of the equation to make that true.

[quote="i_like_pie"][quote="DonkeyKong"]but your's is 1=2.......[/quote]
$ 1 \plus{} 1 \equal{} 1\quad\to\quad2 \equal{} 1$[/quote]
Oh....I'm sorry... :blush:

[quote="i_like_pie"][quote="DonkeyKong"]but your's is 1=2.......[/quote]
$ 1 \plus{} 1 \equal{} 1\quad\to\quad2 \equal{} 1$[/quote]
That's not true unless $ 1\plus{}1\equal{}2$ and $ 1\equal{}1$ and transitive/substitution is true.

[quote="1=2"][quote="DonkeyKong"]As you can see on my signature, I proved that $ 1 \plus{} 1 \equal{} 1$.[/quote]

...

YOU STOLE MY USERNAME! I'M GONNA SUE THE PANTS OFF OF YOU FOR COPYRIGHT ISSUES!!![/quote]

[youtube]zIllRdSzSug[/youtube]

Chill, man. Watch this, it will make you feel better.

[quote="bubka"]take the unit line, draw $ n$ identical semicircles over it side by side (each of radius $ \frac {1}{2n}$), the sum of their perimeters is $ \pi$ regardless of $ n$, now letting $ n$ go to infinity, we get the unit line![/quote]

I believe the issue here is that the result is not a line, but a fractal of some sort.  Either way, the construction is problematic.

[quote="ProtestanT"][quote="i_like_pie"][quote="DonkeyKong"]but your's is 1=2.......[/quote]
$ 1 \plus{} 1 \equal{} 1\quad\to\quad2 \equal{} 1$[/quote]
That's not true unless $ 1 \plus{} 1 \equal{} 2$ and $ 1 \equal{} 1$ and transitive/substitution is true.[/quote]

If you want to strip away the useful properties of arithmetic, you end up with a system of vacuous truths.  I'm not sure what you're trying to accomplish here.

A related result:  given a single false statement, it is possible to prove all false statements.  That is, the statement $ 1 \plus{} 1 \equal{} 1$ is equivalent to all false statements.


Find all $a$ real number such that $x_n=n\{an! \}$ is convergeant

Gabriel Dospinescu

Moubi, I have already posted this problem.

yes Moubi, http://www.mathlinks.ro/Forum/viewtopic.php?t=25241
$Q + eQ$ should work ?


calculus
integration
trigonometry
function
symmetry
calculus computations

For $f(x)=x^4+|x|,$ let $I_1=\int_0^\pi f(\cos x)\ dx,\ I_2=\int_0^\frac{\pi}{2} f(\sin x)\ dx.$

Find the value of  $\frac{I_1}{I_2}.$

Are you sure it's wrong ?
$I_1-I_2=\int_0^\pi (\cos^4(x)-\sin^4(x)+\mid \cos x \mid - \sin x)dx$
$=\int_{0}^{\frac{\pi}{2}} (\cos^2(x)-\sin^2(x)+\cos x - \sin x)dx + \int_{\frac{\pi}{2}}^{\pi} (\cos^2(x)-\sin^2(x)- \cos x - \sin x)dx$
$=\int_{0}^{\frac{\pi}{2}} (\cos(2x)+ \cos x - \sin x)dx + \int_{\frac{\pi}{2}}^{\pi} (\cos(2x) - \cos x - \sin x)dx$
$=[\frac{1}{2} \sin(2x) +\sin x+\cos x]_{0}^{\frac{\pi}{2}} + [\frac{1}{2} \sin(2x)-\sin x+\cos x]_{\frac{\pi}{2}}^{\pi}$
$=0$
which means that $I_1=I_2$ and the answer is $1$
so what is wrong ???  :?

:oops: I'm very sorry that I made a typo. :blush:  So you, felipesa and erdos, Your answers were Correct!

I will edit.

kunny

Another way to look at it, with less need for calculation. Let $f(x)$ be any continuous even function on $[-1,1].$ We have the particular case of $f(x)=x^4+|x|$ but we will not use that formula in detail - only the fact that it is even.

Let $I_1=\int_0^{\pi}f(\cos x)\,dx$ and $I_2=\int_0^{\frac{\pi}2}f(\sin x)\,dx.$

Now make the substitution $u=\cos x$ in the first integral and $u=\sin x$ in the second.

$I_1=\int_{-1}^1\frac{f(u)}{\sqrt{1-u^2}}\,du$ and $I_2=\int_0^1\frac{f(u)}{\sqrt{1-u^2}}\,du$

Since $\frac{f(u)}{\sqrt{1-u^2}}$ is an even function of $u,$ it is clear by symmetry that $I_1=2I_2.$

(Yes, you have to make sure that these improper integrals converge, but they do.)

This is closely related to erdos's second argument on this thread - I'm just trying to clear away some of the clutter.


How do you make a long division symbol in $ \text{\LaTeX}$? Thanks in advance!

Do you mean something like $ x \big)\hspace{\minus{}0.3em}\overline{\hspace{0.3em} abc\ }$? It's a bit of a fudge but even the longdiv package (for long division of integers) uses something similar.

Yes, thank you. What about for the rest of the long division problem? Like how could I write, for example, the steps to $ 17 \big)\hspace{\minus{}0.3em}\overline{\hspace{0.3em} 289\ }$?

Let the longdiv package do it all for you:
[code]\documentclass{article}
\include{longdiv}
\begin{document}
\longdiv{289}{17}
\end{document}[/code]

Oh, okay! Thanks!  :)

Can it do polynomials?

No, but the polynom package can - see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226670

Where can I download longdiv and polynom?  Do they work on OSX leopard?

longdiv and polynom are part of any complete LaTeX distro and are independent of the operating system. If you don't have them in your LaTeX then you can get them from CTAN at [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=longdiv]longdiv[/url] and [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=polynom]polynom[/url].


Let $\tau (n)$ denote the number of positive divisors of the positive integer $n$. Prove that there are infinitely many positive integers $a$ such that $\tau (an) = n$  has no positive integer solution $n$.


What can we say about the linearity of an automorphism group of a free group of rank 2? Better yet (what I'm more interested in), is the linearity of a braid group $B_4$ an equivalent result?

If by linearity you mean existence of a faithful linear representation, proofs were given recently by Stephen Bigelow and by Dirk Krammer, for all $B_n$.


I know this is a request, but is it possible to include the Answer for [url]http://www.artofproblemsolving.com/Forum/resources.php[/url] with the PDF file download? By this I mean that when you download the PDF file, the answers are included with the file. Thanks!

Not going to happen any time soon for more than one reason.

[quote="Bugi"]Ah, how helpful. Can you tell us those reasons?

And, as it isn't possible, is it possible that all (all!) problems in Resources have a S which contains a solution of that problem (1 or 2 of the best), because it was done with some problems, and I don't see a reason to do that with all problems.[/quote]

This is because nobody has posted a solution to them yet.

@Valentin Vornicu
Could you tell us the reasons? (Because it takes too long?)

Each "S" in the Resources section is simply a copy of a post in the problem's thread. To put solutions into the database, someone with write access has to go through each problem and find the solution posted in each thread. If there's none listed in the resources section, it means that either no solution has been written yet or that no one has put it into the database.


You do two events at HMMT.

HMMT - Algebra and Calc or Geo
PUMAC - Algebra


Να δειχθεί ότι το σύνολο $ A$ των τοπικών ακροτάτων μίας συνεχούς πραγματικής συνάρτησης $ f$ είναι το πολύ αριθμήσιμο σύνολο.
Με ταλαιπωρεί αρκετό καιρό το συγκεκριμένο πρόβλημα.Το μόνο που σκέφτηκα αλλά πάλι εντελώς διαισθητικά είναι ότι το $ A$ δεν είναι πλήρες επί του $ R$ αλλά και πάλι δεν μπορώ να καταλήξω κάπου. :( Αφού δειχθεί αυτό να δειχθεί ότι το ίδιο ισχύει και για το σύνολο των σημείων ασυνέχειας κάθε [b]μονότονης[/b] πραγματικής συνάρτησης.

Κατ' αρχήν για το πρώτο φαντάζομαι εννοείς τα strict extrema.

Hint και για τις δυο: Το $ \mathbb{R}$ γράφεται σαν ένωση ([i]το πολύ[/i]) πόσων pairwise disjoint ανοιχτών διαστημάτων; :)

Τι είναι τα strict extrema? :huh: Από ένα βιβλίο τη βρήκα και την παρέθεσα ακριβώς.Για το δεύτερο έχω την εντύπωση διαισθητικά ότι το $ R$ γράφεται ως ένωση το πολύ αριθμήσιμου πλήθους ξένων ανά δύο ανοιχτών διαστημάτων.Έχω δίκιο;Και πώς αποδεικνύεται κάτι τέτοιο?

Ναι ίσως μπέρδεψα τα ακρότατα με τα σημεία στα οποία η συνάρτηση λαμβάνει τα ακρότατα. Στη δεύτερη περίπτωση δεν ισχύει αναγκαστικά, γι' αυτό και νόμιζα ότι ξέχασες το strict (πχ η $ f(x) \equiv 1$ έχει υπεραριθμήσιμα σημεια στα οποία λαμβάνει ακρότατα, αλλά ένα μόνο ακρότατο)

Η διαίσθησή σου είναι σωστή. Ένα sketch είναι:

Θεώρησε, κατ' αρχήν, ότι μπορείς να γράψεις ένα δεδομένο nonempty subset του $ \mathbb{R}$ ως ένωση ενός collection από pairwise disjoint nonempty open sets (αυτό κάνει impose στο υποσύνολό σου να είναι και αυτό open). Από κάθε open set του collection σου, επιλέγεις ένα ρητό. Πόσοι είναι, το πολύ, οι ρητοί; Τι σου λέει αυτό για το μέγεθος του collection;

Το ότι μπορείς όντως να βρεις ένα collection από pairwise disjoint nonempty open sets με ένωση κάποιο δεδομένο nonempty open subset του $ \mathbb{R}$ δε σου χρειάζεται στη λύση της άσκησής σου (και χρειάζεται λίγο τοπολογία, αλλά δεν είναι κάτι το δύσκολο). Ως άσκηση δοκίμασε να βρεις ένα collection απο non-empty open sets με ένωση ολόκληρο το $ \mathbb{R}$.

(Αν δεν έγινα σαφής, το sketch ήταν για το ισχυρότερο θεώρημα ότι κάθε nonempty open subset του $ \mathbb{R}$ γράφεται σαν ένωση από το πολύ countably many nonempty open sets)

Τελικά νομίζω ότι τα κατάφερα αλλά ελπίζω να μην έκανα κάπου λάθος...
Για το πρώτο υποθέτουμε χωρίς βλάβη της γενικότητας ότι η $ f$ έχει πεδίο ορισμού όλο το $ R$ και ας είναι $ A$ το σύνολο των τοπικών μεγίστων της $ f$.Θεωρούμε τυχόν $ f(x_0) \in A$.Τότε υπάρχει ένα $ d_1>0$ ώστε για κάθε $ x \in (x_0 \minus{} d_1,x_0\plus{}d_1)$ να έχουμε $ f(x) \le f(x_0)$.Ακόμη η $ f$ είναι συνεχής στο $ x_0$ άρα $ (\forall e>0)(\exists d_2>0)(\forall x \in (x_0\minus{}d_2,x_0\plus{}d_2)) \Rightarrow |f(x)\minus{}f(x_0)|<e$.Επιλέγουμε $ d\equal{}min\{d_1,d_2\}$ οπότε τελικά $ (\forall x \in (x_0\minus{}d,x_0\plus{}d)) \Rightarrow f(x_0)\minus{}e<f(x) \le f(x_0)$.Έστω και ένα $ f(y_0) \neq f(x_0) \in A$.Τότε $ (\forall x \in (y_0\minus{}d,y_0\plus{}d)) \Rightarrow f(y_0)\minus{}e<f(x) \le f(y_0)$.Έτσι αν υποθέσουμε ότι ισχύει $ (y_0\minus{}d,y_0\plus{}d) \subseteq (x_0\minus{}d,x_0\plus{}d)$ τότε μπορούμε να διακρίνουμε δύο περιπτώσεις για το $ x_0$:
[b]1.[/b]$ x_0 \in (y_0\minus{}d,y_0\plus{}d)$.Τότε όμως αφού $ y_0$ είναι τοπικό μέγιστο της $ f$ παίρνουμε ότι $ f(x_0) \le f(y_0)$ και εντελώς ανάλογα παίρνουμε $ f(y_0) \le f(x_0)$ οπότε τελικά $ f(x_0)\equal{}f(y_0)$ άτοπο.
[b]2.[/b]$ x_0 \notin (y_0\minus{}d,y_0\plus{}d)$.Χωρίς βλάβη της γενικότητας υποθέτουμε ότι $ x_0<y_0\minus{}d$ και επιλέγουμε κατάλληλο $ d$ ώστε οι περιοχές του $ x_0$ και του $ y_0$ που επιλέξαμε να είναι ξένες μεταξύ τους.
Άρα ουσιαστικά δείξαμε ότι υπάρχει αμφιμονοσήμαντη συνάρτηση του $ A$ σε μία συλλογή $ C$ ξένων ανά δύο διαστημάτων του $ R$.Το καθένα από αυτά το απεικονίζουμε σε έναν μοναδικό ρητό του εσωτερικού του.Η τελευταία απεικόνιση είναι επίσης αμφιμονοσήμαντη γιατί τα διαστήματα είναι ξένα ανά δύο.Άρα τελικά υπάρχει αμφιμονοσήμαντη συνάρτηση του $ A$ στο $ Q$ δηλαδή $ A \lesssim Q \simeq N$ που αποδεικνύει ότι το $ A$ είναι το πολύ αριθμήσιμο.Όμοια για το σύνολο των τοπικών ελαχίστων της $ f$.Το ζητούμενο σύνολο είναι το πολύ αριθμήσιμο ως ένωση το πολύ αριθμήσιμων συνόλων. :)
Αυτό με τα σημεία ασυνέχειας πάντως φαίνεται πιο δύσκολο αλλά δεν το έχω σκεφτεί ακόμη :(

Τώρα που το ξανασκέφτομαι δεν πρέπει να είναι σωστό :blush:  αφού καταρχήν δε χρησιμοποίησα  το ότι η $ f$ είναι συνεχής.Ακόμη αυτό που λέω στη δεύτερη περίπτωση για μείωση του $ d$ ώστε τα διαστήματα να είναι ξένα δε διασφαλίζει την ύπαρξη μιας αμφιμονοσήμαντης συνάρτησης από το $ A$ σε συλλογή ξένων ανά δύο διαστημάτων του $ R$ αφού δεν αποδεικνύει ότι είναι αδύνατο οσοδήποτε κοντά στο $ x_0$ να βρούμε άλλο μέγιστο της $ f$ που είναι και το ζητούμενο ώστε να μπορέσει να υπάρξει αυτή η μονοσήμαντη απεικόνιση...καμιά ιδέα κανείς για το πώς μπορούμε να δείξουμε ότι υπάρχει περιοχή του $ x_0$ όπου η $ f$ δεν έχει άλλο μέγιστο...? :maybe:

Vasika exw tin entipwsi pws einai lathos....Ama pareis ws paradigma o sinolo tou Cantor , an se kathe komati tou sinolou ftiakseis apo ena trigonaki  tote safos mporeis naxeis mia sinexi sinartisi i opoia exei iperarithmisimou sto plithos topika akrotata(oi mites twn trigonwn) mias kai to sinolo tou ca ntor einai iperarithmisimo....

Δεν ξέρω αν είναι λάθος σε βιβλίο το βρήκα και δεν ξέρω την απάντηση...αλλά ποιο είναι το σύνολο του Cantor? :huh:

Εφόσον βασικά μιλάμε για το σύνολο των εικόνων, υπάρχει απλούστερη λύση (πάντα αναφέρομαι στο πρόβλημα με τα ακρότατα). Μην τη βλέπεις πομπώδη, απλώς την έγραψα αναλυτικά. Για οποιαδήποτε λάθη, μάλλον φταίει ο πολύς ήλιος στη Χαλκιδική :)

[hide="Solution"]
WLOG μπορείς να ασχοληθείς με τα τοπικά μέγιστα. Επειδή βλέπω ότι το έβαλες και στο αγγλικό forum, θα γράψω τη λύση στα αγγλικά και αν θες τους δίνεις το link εκεί. Εκτός και αν έχω λάθος, η συνέχεια δε χρειάζεται.

Let $ f : \mathbb{R} \to \mathbb{R}$ be our function, and let $ S^\#$ be the set of the preimages of $ f$'s local maxima. We wish to show that $ f(S^\#)$ is at most countable. We may choose $ S \subseteq S^\#$, satisfying $ f(S) \equal{} f(S^\#)$ and also the property that no two points of $ S$ have the same image under $ f$. Clearly, there exists a bijective correspondence between $ T : \equal{} f(S) \equal{} f(S^\#)$ and $ S$, so, equivalently, we wish to show that $ S$ is at most countable.

Let $ x_0 \in S$ be arbitrary. By definition, there exists a $ \delta(x_0) > 0$ such that $ f \leq f(x_0)$ on $ (x_0 \minus{} \delta(x_0), x_0 \plus{} \delta(x_0))$. Now, we may pick $ \alpha(x_0) \in \mathbb{Q} \cap (x_0 \minus{} \delta(x_0), x_0)$ and $ \beta(x_0) \in \mathbb{Q} \cap (x_0, x_0 \plus{} \delta(x_0))$.

We define $ R$ to be the set of all such pairs $ (\alpha(x_0), \beta(x_0))$, for each $ x_0 \in S$. We claim that the map $ x_0 \mapsto (\alpha(x_0), \beta(x_0))$, $ x_0 \in S$, provides a bijective correspondence between $ S$ and $ R$. Equivalently, we claim that $ (\alpha(x), \beta(x)) \equal{} (\alpha(y), \beta(y))$, for $ x, y \in S$, implies $ x \equal{} y$. Indeed, observe that $ (\alpha(x), \beta(x)) \equal{} (\alpha(y), \beta(y))$ implies that $ y \in (\alpha(x), \beta(x)) \subseteq (x \minus{} \delta(x), x \plus{} \delta(x))$, and thus $ f(y) \leq f(x)$. On the other hand, by working symmetrically we see that $ f(x) \leq f(y)$. Therefore, the points $ x, y \in S$ satisfy $ f(x) \equal{} f(y)$. By our construction of $ S$, we conclude that $ x \equal{} y$, and the claim follows.

Consequently, we see that we need to show that $ R$ (which is in a bijective correspondence with our original set $ T$ of local maxima) is at most countable. This, however, is trivial since $ R \subseteq \mathbb{Q}^2$, and the result follows.
[/hide]

[quote="cmad"]Εφόσον βασικά μιλάμε για το σύνολο των εικόνων, υπάρχει απλούστερη λύση (πάντα αναφέρομαι στο πρόβλημα με τα ακρότατα). Μην τη βλέπεις πομπώδη, απλώς την έγραψα αναλυτικά. Για οποιαδήποτε λάθη, μάλλον φταίει ο πολύς ήλιος στη Χαλκιδική :)

[hide="Solution"]
WLOG μπορείς να ασχοληθείς με τα τοπικά μέγιστα. Επειδή βλέπω ότι το έβαλες και στο αγγλικό forum, θα γράψω τη λύση στα αγγλικά και αν θες τους δίνεις το link εκεί. Εκτός και αν έχω λάθος, η συνέχεια δε χρειάζεται.

Let $ f : \mathbb{R} \to \mathbb{R}$ be our function, and let $ S^\#$ be the set of the preimages of $ f$'s local maxima. We wish to show that $ f(S^\#)$ is at most countable. We may choose $ S \subseteq S^\#$, satisfying $ f(S) \equal{} f(S^\#)$ and also the property that no two points of $ S$ have the same image under $ f$. Clearly, there exists a bijective correspondence between $ T : \equal{} f(S) \equal{} f(S^\#)$ and $ S$, so, equivalently, we wish to show that $ S$ is at most countable.

Let $ x_0 \in S$ be arbitrary. By definition, there exists a $ \delta(x_0) > 0$ such that $ f \leq f(x_0)$ on $ (x_0 \minus{} \delta(x_0), x_0 \plus{} \delta(x_0))$. Now, we may pick $ \alpha(x_0) \in \mathbb{Q} \cap (x_0 \minus{} \delta(x_0), x_0)$ and $ \beta(x_0) \in \mathbb{Q} \cap (x_0, x_0 \plus{} \delta(x_0))$.

We define $ R$ to be the set of all such pairs $ (\alpha(x_0), \beta(x_0))$, for each $ x_0 \in S$. We claim that the map $ x_0 \mapsto (\alpha(x_0), \beta(x_0))$, $ x_0 \in S$, provides a bijective correspondence between $ S$ and $ R$. Equivalently, we claim that $ (\alpha(x), \beta(x)) \equal{} (\alpha(y), \beta(y))$, for $ x, y \in S$, implies $ x \equal{} y$. Indeed, observe that $ (\alpha(x), \beta(x)) \equal{} (\alpha(y), \beta(y))$ implies that $ y \in (\alpha(x), \beta(x)) \subseteq (x \minus{} \delta(x), x \plus{} \delta(x))$, and thus $ f(y) \leq f(x)$. On the other hand, by working symmetrically we see that $ f(x) \leq f(y)$. Therefore, the points $ x, y \in S$ satisfy $ f(x) \equal{} f(y)$. By our construction of $ S$, we conclude that $ x \equal{} y$, and the claim follows.

Consequently, we see that we need to show that $ R$ (which is in a bijective correspondence with our original set $ T$ of local maxima) is at most countable. This, however, is trivial since $ R \subseteq \mathbb{Q}^2$, and the result follows
[/hide][/quote]

vasika den edosa lisi , antiparadigma edosa stin ekfonisi tis askisis , pou dixnei oti einai lathos

@Sorrowman Please μου εξηγείς αν γίνεται ποιο είναι το σύνολο του Cantor που φέρνεις ως αντιπαράδειγμα γιατί δεν το γνωρίζω;Ευχαριστώ...

@cmad η λύση σου σωστή μου φαίνεται αλλά απ'ό,τι φαίνεται δεν έχεις χρησιμοποιήσει καθόλου τη συνέχεια και αναρωτιέμαι γιατί στο βιβλίο που την βρήκα απαιτούσε η $ f$ να είναι συνεχής...  :huh:

[quote="Durandal"][quote="SorrowMan"]rikse mia matia edw http://users.auth.gr/~betsakos/Teach/IRA.htm , exei ena pdf mesa pou leei gia to sinolo tou cantor....En sintomia eiani ena ipodiastima tou [0,1] pou to vriskeis em tin diadikasia trixotomisis tou diastimatos kai petamatos tou meseou komatiou etsi sou menei [0,1/3]U[2/3,1] = I1 , meta trixotomeis ta [0,1/3] k  [2/3,1] apta opoia pali petas to meseo komati opote exeis [0,1/9]U[2/9,3/9]U[6/9,7/9]u[8/9,1]=I2 kok ...Perneis ti tomi apo 1 ews apiro tou In kai afto les pws einai to sinolo Cantor...Afto apodikniete pws einai iperarithmisimo sinolo[/quote]

Ωστόσο, το αντιπαράδειγμά σου είναι λίγο προβληματικό... το totally disconnected Cantor δεν έχει κομμάτια θετικού μέτρου για να μπορέσεις να τα χρησιμοποιήσεις σαν βάσεις τριγώνων (αν αυτό εννοείς).

Cheerio,

Durandal 1707[/quote]

san vasi twn trigwnwn perneis ta komatia pou petas

[quote="SorrowMan"]san vasi twn trigwnwn perneis ta komatia pou petas[/quote]

Έτσι όμως αποκτάς αριθμήσιμο πλήθος τριγώνων :roll:

cmad has it right :)

Cheerio,

Durandal 1707

sorry den egrapsa kala to antiparadigma....

Estw F sinartisi tetoia wste otan x anikei sto sinolo cantor kai otan x den anikei sto sinolo tou cantor(dld sta komatia pou petame) i F einai ta trigonakia pouxw anaferei kai pio prin....I f exei iperarithisima sto plithos topika akrotata , gia tin akrivia topika elaxista  kai einai kai sinexis....Ola ta topika elaxista vevaia exoun tin idia timi , to 0 , alla den iparxei kati stin ekfonisi pou na mi mou to epitrepei afto...

[quote="Durandal"][quote="giannis18"]Μας ενδιαφέρουν τα τοπικά ακρότατα και όχι οι θέσεις τοπικών ακροτάτων.Στην περίπτωση που λες είναι ένα μόνο τοπικό ελάχιστο,το 0.[/quote]

Ακριβώς, αυτό είναι και που επιτρέπει στην απόδειξη του cmad να δουλέψει. Aλλιώς, έπαιρνες την Dirichlet και είχες τελειώσει από αντιπαραδείγματα ;)

Cheerio,

Durandal 1707[/quote]

Oxi re i Dirichlet den einai sinexis :P , alla ok got it allo pragma nomiza pws ithele!!

[quote="SorrowMan"]Oxi re i Dirichlet den einai sinexis :P , alla ok got it allo pragma nomiza pws ithele!![/quote]

Καλά, ό,τι θέλω λέω, με το ένα και με το άλλο είχα ξεχάσει εντελώς το αρχικό πρόβλημα όταν έγραφα το τελευταίο post :P

Καλοκαιρινές γελοιότητες :)

Cheerio,

Durandal 1707

Μα στην απόδειξη του cmad ούτως ή άλλως δεν χρησιμοποιείται η συνέχεια που σημαίνει ότι αν είναι σωστή (και δεν βλέπω κάποιο λάθος) τότε το αρχικό πρόβλημα επεκτείνεται σε όλες τις συναρτήσεις συνεχείς και μη :maybe:


Given a natural number $ n \geq 3$. Prove that there exists a set $ A\left(n\right)$ such that:
[b]1)[/b] The set $ A\left(n\right)$ consists of $ n$ distinct natural numbers.
[b]2)[/b] For any element $ a\in A\left(n\right)$, the product of all the other elements of $ A\left(n\right)$ has remainder $ 1$ when divided by $ a$.

In other words, we want positive integers $a_1,\ldots a_n$ s.t. $\sum_{i=1}^n \frac 1{a_i}-\prod_{i=1}^n \frac 1{a_i}$ is an integer.

First construct $a_1,\ldots,a_{n-1}$ s.t. $\sum_{i=1}^n \frac 1{a_i}+\prod_{i=1}^{n-1}\frac 1{a_i}=1\ (*)$ (I'll do this in a minute). After that, take $a_n=a_1\ldots a_{n-1}-1$, and the condition above is verified.

In order to construct a system satisfying $(*)$, I'll just illustrate the inductive procedure with an example:

$1=1,\ \frac 1{1+1}+\frac 12=1,\ \frac 12+\frac 1{2+1}+\frac 12\cdot \frac 13=1$, and we continue with $\frac 12+\frac 13+\frac 1{2\cdot 3+1}+\frac 12\cdot\frac 13\cdot\frac 17=1$, and so on.


Prove that polynomial $W(x)$ with integer coefficients is reducible over rationals iff $W(x)$ is reducible over integers.

You find that in any book as Gauss's lemma, and this can be generalized to a unique factorization domain.


so today was state and obviously bobby got 1st steven someone got 2nd yury someone got 3rd (steven and yury are from austin) and lilly shen got 4th (but barely)

Dont know what my score was but sicarius got 57.

EDIT: ohh! i thought he was takling about scores, so i was like...eh?

mathblitz, don't you think this probably belongs in the texas forum?

there has already been a post concerning this here and in the texas forum.

[quote="mathblitz"]and lilly shen got 4th (but barely)
[/quote]

I know the kid that lost to her. :P


ratio
geometry
circumcircle
trigonometry
geometry unsolved

Let $O$ be the circumcenter of triangle $ABC$, and $CD$ be the altitude. Point $E$ lies on line $AB$ and $M$ is the midpoint of $CE$. The perpendicular to $OM$ at $M$ cuts $BC,AC$ at $L,K$. Show that $\frac{LM}{MK}=\frac{AD}{DB}$

[img]8496[/img]

We draw $A',B'$, the midpoints of $CA,CB$. Now obviously $M \in A'B'$.
Since $KL \perp OM$, we have $KM = OM \tan \angle MOK$. But $KOMA'$ is cyclic, then $\angle MOK = \angle MA'C = \angle BAC = \alpha$.
We conclude that $KM = OM \tan \alpha, LM = OM \tan \beta$ and $\frac{LM}{MK}= \frac{\tan \beta}{\tan \alpha}$.

Now: $\frac{AD}{DB}= \frac{CD \tan \angle ACD}{CD \tan \angle BCD}= \frac{\tan \angle ACD}{\tan \angle BCD}= \frac{\cot \alpha}{\cot \beta}= \frac{\tan \beta}{\tan \alpha}$


what are y'all bringing?

I know it's 5 months early to be discussing this, but it is best to be prepared.

Good games:
Risk
Settlers of Catan
Frisbee
Monopoly
Cards
Mafia Cards

I shall bring either risk, or settlers of catan and a six-player extension.
Unless i get more extensions for catan between now and then. Catan is pwnage with lots of extensions.

Since this is at Texas, frisbee should be played at 7, because it isn't too dark and it isn't too light out. At least one frisbee should be brought to AMSP, but don't bring lots and lots. there is only one place to play(i think). Board games are best played with more than two people, and if you are planning on not going to six flags(if we are going), then make sure that there are at least two or three other people with you(alexhhmun and i were the only ones to stay. boring day XD). Then you can play lots and lots of board games for that day.

I'll stay on six flags day, if I go to the camp again.

Hex and Set, too!

nooooooo. GO TO SIX FLAGS u guys!! its sooo muchhh fun!!!
dont stay at the dorms :'(

ITS HECKA HOT HOT HOT HOT HOT AND HUMIDDD AT SIX FLAGSSS
BUT DEF WORTH GOINGGG :]
that is unless you have phobia for heights and stufff
errr but yeah def go, its so much funn.

MAFIA CARDS! lol

someone should bring a wii :P

...altho i dont think im going back ahha

I know this isn't really a game, but I might bring my electric piano...

WHOA ELECTRIC PIANO?!?
damnnnnn

but ..how are you gonna carry everything O_O

lol. dont u need a tv to play wii tho? ... we dont have tvs.... -.- [ tho i def. didnt mind. it was hecka fun w/o it! XP]

Dang should I be like signing up for AMP right now?
If you bring electric piano it better be one with weighted action and full size 88 key keyboard.

lol i think the admission test A is out?? maybe. idk.
but u get a 20%discount or something if u apply early.

hmm is the admission test a out?
[edit:  its not out yet. it comes out on 1/30...which isnt that far aaway]
if you have the time to do the problems, then do so.
i used the money i got discounted for my plane ticket and food money there LOL [starbucks at utd...]

 i might return to the santa cruz oneeeeeeeeeeeee yeeeppeee.

hmm true that wii needs a tv...ddr pads sound a lot better to me! :PPPP
that way people dont waste money on those stupid ddr machines lol

hey suzy and everyone else...this is steve...so.........y again r u preparing so early??and admission test A doesnt come out till the 31st of jan...last year we actually never played the games...except maybe set,and the monopoly boards people brought?i still have a bruise on the side of my head from that scottish terrier that hit me on the head(we had a monopoly brawl)...adn dont bring a piano;1)people in ur room and upstairs/downstairs will start complaining...2)there are pianos there(crappy ill admit,but theres a REALLY good one in the grand hall u can use when the college students arent practicing there)...so...um...yea...............LEON,DONT STAY IN THE DORM LIKE LAST TIME!!!!!!!!!!!!!!!!!!!o,and i STILL cant guess ur age...last time u said u were -3i  :mad:

ahaha. yea u prolly dont need that much spending money. only thing is energy drinks. or STARBUCKS [duddee i spent sooo much money on that lol] . oh and SIX FLAGS. [only for souvenirs tho. the admission is in ur tuition]
wow. steve did u read everything [this thread & the other?]
oh yea and it might get damaged :// some guys violin did this past year.
 i guess 13 lol. or younger. possibly 12.
wait a sec. this is on facebook. in kevin liou's album. he has a list of people their dorms. numbers. and ages. 'cept the ages rnt very accurate cuz they had half of my roommates ages wrong lol and im pretty sure a lotta of the other ones r wrong too ://
lmfao. ddr XP
im not going back to amsp =_='' i dont think i could take it w/o everyone else. itd be too diff. -.-

whoa didnt know someone actually made a list of ppl like that...
lol i dont even have him added as a friend i think.

anyways.

  piano, yeah seriously it'll just be a burden... just make sure to bring plenty of clothes so that when you get sweaty and all stinky..you can just pop a new shirt on -______-.

[quote="gatorlover124"]...LEON,DONT STAY IN THE DORM LIKE LAST TIME!!!!!!!!!!!!!!!!!!!o,and i STILL cant guess ur age...last time u said u were -3i  :mad:[/quote]

I went to the Six Flags... what made you think I didn't?
And yes, it's been a while; I'm not $ \minus{}3i$ anymore... someone told me to get real, so I did for a while, but it was too horrible, switching back and forth between losing and gaining my rationality at every moment as I aged. It just drives one insane, you know? So I've turned into the quaternion $ (1,3,3,7)$.

Oh.

Hmm... probably not going to AMSP 09... either one...

For some reason I went to Six Flags.....

I'm probably not bringing anything if I come (which I won't)......... oh well...

Maybe I should come back as a counselor in like, 12th grade or something. Probably not...

Wait, why am I listed as "most likely to come back as a counselor" on the facebook list?

yay!
i got accepted to AMSP!

any tips on a first time comer????

I already saw that I should bring TONS of clothes...

utd or ucsc?

congrats!!! :)

swimminggg stuff!! =]]
if u forget anything, u can always order from walmart, or if u need something from home, mail comes daily xD

Texas, and thanks for the tip.

Are we going swimming everyday?

(also, anything else? thanks!!!!  :D )

you dont have to go swimming every night, but you always have the option to do so.
yes bring lots of clothes!
...KEEP CLEAN!

umm and try to get some work done :P
lol i just hung out with my roomies on nights...and bombed weekly tests LOL...[as expected]
but oh
go to dance! and go to six flags! they're fun!
and bring lots of paper&pencils and all that.
you can buy from the students tore but they're PRICEY.
i paid like 7 bucks for a binder or something...
bleghh
and bring some food if you can.
food was just ehhh...

but most of all..JUST ENJOY! and learn! :]]

lol. nice double post XP
u dont get forced to do anything there really.
oh btw, no dress code either xD

yea. agreed with sooozyy's 2nd paragraph. work. but not at the expense of fun.
[hahaha. suzy, we were all bout the fun xP but it was awesome so w/e. actually we sorta did work 1st week. thenn.... lol]
definitely go to teh dance! and the talent show!
mhmmm.

OMG I JUST SAW BASHAM'S POST.

WHOOOT.

but brian...seriously WHY a baseball bat??!?
its not like youre gonna smack anyone down...
its more likely that you'll just yell?

lol

lol so nxt time he'll brutally beat the ppl who pull the "joker" joke on him XP

Hey does anyone play Starcraft? Cuz I just quit 2 days ago, but I promised myself I would start again after USAMO/USAPhO final round (if I make that), so yea...:D
I've kinda been infected by the Korean parasite...

I got rocked by somebody in starcraft last summer haha. junggi can tell you about it haha

Would it be taylor? :) Apparently he picked it up at mop from damien jiang and brian hamrick, as did like another 1/3 of mop. Exciting stuff, exciting stuff.
I'm horrible, but it's still an awesome game.

I'm a newcomer too  :P .
I play SC and Warcraft, except I suck at both, so yeah...
From the above posts, I'm guessing playing tennis in the scorching hot sun isn't such a good idea  :maybe:
And do we need to bring any games? B/c I've got tons of board games and cards.
Finally, is there anything else we need to bring besides pencil/erasers/notebooks/binders/paper/clothes?

lol. you could try playing tennis.
i know some guys did play basketball  ^.^

u could if u wanted too. some people did last year and it got used im pretty sure. dont overload urself with board games tho. =)

money! & brain :P
swimming stuff.
oh yeah, you'll be sent an iternary just follow that. [you'll be needing your own bed sheets, pillows, etc]

Money gets you everywhere in life =P. especially at camp.

bring cards. lots of cards... 52 card pick up ftw. lol...

play capture the flag =D and pool... not like the swimming pool, but pool!


A $ 5 cm$ x $ 6 cm$ x $ 12 cm$ Box is to have an equal increase to each edge to increase its capacity by $ 1080cm^3$. Find The increase.

$ (x\plus{}5)(x\plus{}6)(x\plus{}12)\minus{}5\cdot6\cdot12\equal{}1080\implies(x\plus{}5)(x\plus{}6)(x\plus{}12)\equal{}1440$. Letting $ k\equal{}x\plus{}6$ yields $ (k\minus{}1)k(k\plus{}6)\equal{}k(k^2\plus{}5k\minus{}6)\equal{}k^3\plus{}5k^2\minus{}6k\minus{}1440\equal{}0$. Descartes' Rule of Signs tells us that there is exactly $ 1$ positive real root, which is the only possibility for $ k$ (since $ k\equal{}x\plus{}6>6$). Searching for roots, we get $ k\equal{}10\implies x\equal{}\boxed4$.


Putnam
algebra
polynomial
number theory
relatively prime
Rational Root Theorem
college contests

Let $P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0$ be a polynomial with integer coefficients. Suppose that $r$ is a rational number such that $P(r)=0$. Show that the $n$ numbers
$c_nr, c_nr^2+c_{n-1}r, c_nr^3+c_{n-1}r^2+c_{n-1}r, \dots, c_nr^n+c_{n-1}r^{n-1}+\cdots+c_1r$
are all integers.

Define $y_0=c_nr, y_1=c_nr^2+c_{n-1}r$ and so on up to $y_{n-1}=c_nr^n+c_{n-1}r^{n-1}+\cdots+c_1r^n.$ This sequence is given recursively by $y_{k+1}=(y_k+c_{n-k})r.$ We note that $y_{n-1}=-c_0$ is an integer.

Suppose $r=\frac pq$ in lowest terms. We will prove the following: if $y_k$ is not an integer, then $y_{k+1}$ is also not an integer. Then it would follow that if $y_k$ were ever not an integer, then the sequence of $y_k$'s would never be an integer again, contradicting the fact that $y_{n-1}$ is an integer.

Since the only part of the recursion for $y_k$ that could give a non-integer result is the multiplication by $r,$ it follows that if $y_k$ is not an integer, then we can write it as $\frac ab$ where $b$ divides some power of $q.$ In particular, $b$ is relatively prime to $p.$ Then $y_k+c_{n-k}$ is a fraction plus an integer and can be written as $\frac{a'}b$: a fraction with the same denominator. Then $y_{k+1}=\frac{a'}b\cdot\frac pq.$ Since both $a'$ and $p$ are relatively prime to $b$, the denominator of the resulting fraction must be divisible by $b$ and cannot be $1.$ This proves our result.

The result of this problem has algorithmic significance: if we are testing an integer-coefficient polynomial for rational roots (with the possibilities already restricted by the Rational Root Theorem), if we ever get a non-integer intermediate result in our synthetic division, we may immediately abandon that prospective root.


induction
conics
ellipse
vector
combinatorics proposed
combinatorics

Let $ n$ be an even number, and $ S$ be the set of all arrays of length $ n$ whose elements are from the set $ \left\{0,1\right\}$. Prove that $ S$ can be partitioned into disjoint three-element subsets such that for each three arrays $ \left(a_i\right)_{i \equal{} 1}^n$, $ \left(b_i\right)_{i \equal{} 1}^n$, $ \left(c_i\right)_{i \equal{} 1}^n$ which belong to the same subset and for each $ i\in\left\{1,2,...,n\right\}$, the number $ a_i \plus{} b_i \plus{} c_i$ is divisible by $ 2$.

Please show it, not just write like this  :mad:

When copying problems from other sources, please do take some care - for instance, this one is obviously wrong. It's not $ S$ but $ S\setminus\left\{\left(0,0,...,0\right)\right\}$ that can be partitioned into disjoint three-element subsets with the property you want... At least it works for $ S\setminus\left\{\left(0,0,...,0\right)\right\}$, and it obviously does not work for $ S$ because $ 3\nmid 2^n$. And I am also wondering whether, for $ 4\mid n$, we can partition $ S\setminus\left\{\left(0,0,...,0\right)\right\}$ into five-element subsets of arrays, each with sum $ \left(0,0,...,0\right)$... it's clear how to generalize from here on.

"Induction on $ n$" is probably the best hint towards the solution. Just write $ S_n$ instead of $ S$, and think about how you would partition $ S_n\setminus\left\{\left(0,0,...,0\right)\right\}$ into disjoint three-element subsets with the property you want given a "good" partition of $ S_{n \minus{} 2}\setminus\left\{\left(0,0,...,0\right)\right\}$ into three-element subsets. Notice that if $ a$, $ b$, $ c$ are three arrays with $ a \plus{} b \plus{} c \equal{} \left(0,0,...,0\right)$, then
$ \left(\left(0,1\right)\oplus a\right) \plus{} \left(\left(1,0\right)\oplus b\right) \plus{} \left(\left(1,1\right)\oplus c\right) \equal{} \left(0,0,...,0\right)$;
$ \left(\left(1,0\right)\oplus a\right) \plus{} \left(\left(1,1\right)\oplus b\right) \plus{} \left(\left(0,1\right)\oplus c\right) \equal{} \left(0,0,...,0\right)$;
$ \left(\left(1,1\right)\oplus a\right) \plus{} \left(\left(0,1\right)\oplus b\right) \plus{} \left(\left(1,0\right)\oplus c\right) \equal{} \left(0,0,...,0\right)$;
$ \left(\left(0,0\right)\oplus a\right) \plus{} \left(\left(0,0\right)\oplus b\right) \plus{} \left(\left(0,0\right)\oplus c\right) \equal{} \left(0,0,...,0\right)$,
where $ \left(a_1,a_2,...,a_k\right)\oplus\left(b_1,b_2,...,b_l\right)$ stands for $ \left(a_1,a_2,...,a_k,b_1,b_2,...,b_l\right)$.

Sorry for the brief reply, but tell me how to install a library in Eclipse and I'll have the time to write proofs on MathLinks.  :bomb:

  darij

[i]Generalization.[/i] Let $ p$ be a prime and $ n$ and $ k$ be two positive integers such that $ n\mid p^k \minus{} 1$ and $ n\neq 1$. Prove that the set $ \mathbb{F}_p^k\setminus\left\{0\right\}$ can be partitioned into disjoint $ n$-element subsets such that for each of these subsets, the sum of all vectors in this subset is $ 0$.

I am not opening a new topic for this just because it's so obvious if you look at it from the right viewpoint. But I am wondering whether this can be generalized to non-prime $ p$ - a suitable generalization of $ \mathbb{F}_p\setminus\left\{0\right\}$ being used (either $ \left(\mathbb{Z}\slash p\mathbb{Z}\right)\setminus\left\{0\right\}$ or $ \left(\mathbb{Z}\slash p\mathbb{Z}\right)^{\times}$ ?). Any ideas?

[hide="Hint to the above generalization"]
As an $ \mathbb{F}_p$-vector space, $ \mathbb{F}_p^k\cong\mathbb{F}_{p^k}$. The field $ \mathbb{F}_{p^k}$ has a primitive root $ \zeta$. Now notice that $ \sum_{i \equal{} 0}^{n \minus{} 1}\zeta^{i\left(p^k \minus{} 1\right)\slash n} \equal{} 0$...
[/hide]

  darij


Evaluate this limit I developed:

$ \lim_{x\to\infty} \csc(\frac{k}{x})\left(\sqrt{x^2 \plus{} k} \minus{} \sqrt{x^2 \minus{} k}\ \right)$

where $ k \not \equal{} 0$ is a real constant.

To me, this limit looks pretty complex at first glance, but I think it evaluates to a beautifully simple value...

Once the value of the limit is agreed upon, maybe we could suggest variations on it or other limits that look complicated at first but that actually simplify quite nicely?

[hide="Answer"]$ L \equal{} \lim_{x \to \plus{} \infty} \frac{\sqrt{x^2\plus{}k} \minus{} \sqrt{x^2\minus{}k}}{\sin(k/x)}$

$ \equal{} \lim_{x \to \plus{} \infty} \frac{k/x}{\sin(k/x)} \cdot \frac{x}{k} \cdot \frac{(\sqrt{x^2\plus{}k} \minus{} \sqrt{x^2\minus{}k})(\sqrt{x^2\plus{}k} \plus{} \sqrt{x^2\minus{}k})}{\sqrt{x^2\plus{}k} \plus{} \sqrt{x^2\minus{}k}}$

$ \equal{} \lim_{x \to \plus{} \infty} \frac{2kx}{kx(\sqrt{1 \plus{} k/x^2} \plus{} \sqrt{1 \minus{} k/x^2})} \equal{} 1$.[/hide]

Yes, that is exactly what I got!

I like how the limit evaluates to be $ 1$ and how its value does not depend upon the value of $ k$!


March 14, 2008 at 9:30 PM EST, proud AoPSers will break the record number of people online at one time.  How do I participate?  Just check the time zone conversions below and find the right time.  If it's too late or too early, leave your computer on (if this doesn't work, I apologize).  How do I make it happen?  Put a link to this thread in your signature.  We want this to work, so let's try our hardest.

Eastern: 9:30 PM
Central: 8:30 PM
Mountain: 7:30 PM (Note: Arizona does not observe daylight savings time)
Pacific: 6:30 PM
Alaskan: 5:30 PM
Hawiian: 4:30 PM

Try to be on 15 minutes before and after the turn of the hour so that we can have a better chance of us all being on.

Thanks to all who participate!

[color=darkblue]Moderator Edit: To post this thread in your signature, simply copy and paste the following code into your Signature Options in your Profile Menu.[/color]

[code]Be a part of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188613]AoPS Record-Breaking Experience[/url][/code]

YEAH!
Though it's too far away so I may forget.

Another way to do this is just keep refreshing on the page. Either get the refreshing programs for firefox, or keep an object on the f5 key for PCs. This way, even if one doesn't have time, it'll still work.

I will continue reminding everybody. :)

Put it in your signature, everybody!

how do u replace the http//....com with that

I am going to do that now.
Already a supporter!

I got a new sig too. Maybe I should state what heaven is... :roll:

I shall join as well! Must not forget... must not forget...

I probably won't forget since it's on my birthday  :roll: .
A very nice way to celebrate!

"Most users ever online was 593 on Mon Dec 10, 2007 8:19 am"
Who goes on AoPS on a Cold Monday Morning?

We'll have to have mroe than 594 people.

Please move the time to 9:30 PM ET.  The current time is just before a class starts, and a concerted effort to get on the site will make it hard for our class students to get to class.

Shoot, I was planning to come, but I had to take this practice SAT right during the time... :wallbash_red:

Heh. I spaced. Partly asleep. I'll..uh...come next time. :D

Meh there's 160 on atm

So yeah, its not that much :(

we might want to postpone it...i mean we r only getting opinions now...

There are now more people online than there were on Friday....
__________________
‧uʍop əpı̣sdn ədʎʇ uɐɔ ɪ

[quote=ObjectZ]12 YEAR /BUMP......[/quote]

how do people get here... literally

[quote=RedChameleon]When will the next event like this occur?[/quote]

yeah


modular arithmetic
calculus
induction
vector
abstract algebra
algebra
polynomial

Let $ p \ge 3$ be a prime number. The sequence $ \{a_{n}\}_{n \ge 0}$ is defined by $ a_{n}=n$ for all $ 0 \le n \le p-1$, and $ a_{n}=a_{n-1}+a_{n-p}$ for all $ n \ge p$. Compute $ a_{p^{3}}\; \pmod{p}$.

[quote="Peter"]Let $p \ge 3$ be a prime number. The sequence $\{a_{n}\}_{n \ge 0}$ is defined by $a_{n}=n$ for all $0 \le n \le p-1$, and $a_{n}=a_{n-1}+a_{n-p}$ for all $n \ge p$. Compute $a_{p^3} \; \pmod{p}$.[/quote]

I'm wondering... what makes this problem appear here and not in the L(inear Recurrences) section of PEN?

Besides: Is it really a Canada 1986 problem? I can't find it at http://www.math.ca/Competitions/CMO/examt/english86.html or on Kalva.

Now, for the solution: I claim that $a_{p^3}\equiv -1\text{ mod }p$.

The proof relies on computations with operators acting on sequences. It requires some basics of linear algebra (vector spaces, linear operators and endomorphism rings) that should be known to IMO participants. I will first introduce some notations:

A [i]bisequence[/i] in a set $K$ will mean a mapping from the set $\mathbb{Z}$ to the set $K$. For any set $K$, we denote by $K^{\mathbb{Z}}$ the set of all bisequences in $K$.

Roughly speaking, a bisequence is like a sequence, but its terms are labeled by integers and not by natural numbers.

We define a mapping $S:K^{\mathbb{Z}}\to K^{\mathbb{Z}}$ such that, if $u\in K^{\mathbb{Z}}$ is a mapping from the set $\mathbb{Z}$ to the set $K$, then $Su$ is the mapping from the set $\mathbb{Z}$ to the set $K$ defined by $\left(Su\right)\left(x\right)=u\left(x-1\right)$ for every $x\in\mathbb{Z}$. Note that we write $Su$ for $S\left(u\right)$.

Roughly speaking, the mapping $S$ shifts every bisequence one term forward. This mapping $S$ is actually called the [i]shift operator[/i]. By induction, it can be seen that $\left(S^k u\right)\left(x\right)=u\left(x-k\right)$ for every $k\in\mathbb{Z}$ and every $x\in\mathbb{Z}$.

If $K$ is a field, then $K^{\mathbb{Z}}$ can be naturally considered as an (infinite-dimensional) vector space over $K$ as follows:
- the zero vector is the "zero bisequence" $0\in K^{\mathbb{Z}}$ defined by $0\left(x\right)=0$ for every $x\in\mathbb{Z}$;
- for any two elements $a$ and $b$ of $K^{\mathbb{Z}}$, we define the element $a+b$ of $K^{\mathbb{Z}}$ by setting $\left(a+b\right)\left(x\right)=a\left(x\right)+b\left(x\right)$ for every $x\in\mathbb{Z}$;
- for any element $a$ of $K^{\mathbb{Z}}$ and any element $\lambda$ of $K$, we define the element $\lambda a$ of $K^{\mathbb{Z}}$ by setting $\left(\lambda a\right)\left(x\right)=\lambda\cdot a\left(x\right)$ for every $x\in\mathbb{Z}$.
It is easily seen that (if $K$ is a field) the transformation $S$ is linear, i. e. it is an endomorphism of this vector space $K^{\mathbb{Z}}$. Note that all endomorphisms of the vector space $K^{\mathbb{Z}}$ form a skew ring; its addition is pointwise addition, its multiplication is composition of endomorphisms; its additive unit is the zero endomorphism (this is the endomorphism sending everything to the zero vector); its multiplicative unit is the identity endomorphism.

Now we show a basic fact from algebra, but first a matter of notation: I call a [i]skew ring[/i] what most other people just call "ring with $1$" (i. e. a set with an Abelian additive group structure, an associative multiplication with a unit, and distributivity). I, personally, use the word "ring" for commutative skew rings. Two elements $a$ and $b$ of a skew ring are said to [i]commute[/i] if $ab=ba$.

[color=blue][b]Lemma 1.[/b] Let $p$ be a prime number, and let $R$ be a skew ring such that $p=0$ in $R$ (hereby, $p$ means $\underbrace{1+1+...+1}_{p\text{ ones}}$, and $0$ and $1$ mean the additive and multiplicative units of $R$). If $a$ and $b$ are two elements of $R$ which commute, then $\left(a+b\right)^p=a^p+b^p$.[/color]

[i]Proof of Lemma 1.[/i] Since the elements $a$ and $b$ commute, we can apply the binomial theorem to them, and we obtain

$\left(a+b\right)^p=\sum_{k=0}^p\binom{p}{k}a^kb^{p-k}=a^p+\sum_{k=1}^{p-1}\binom{p}{k}a^kb^{p-k}+b^p$.

Now, it is known from number theory that $\binom{p}{k}$ is divisible by $p$ for each $k\in\left\{1,2,...,p-1\right\}$ (since $p$ is prime), so that in $R$ we have $\binom{p}{k}=0$ (because $p=0$). Hence, we obtain

$\left(a+b\right)^p=a^p+\sum_{k=1}^{p-1}\binom{p}{k}a^kb^{p-k}+b^p=a^p+\sum_{k=1}^{p-1}0a^kb^{p-k}+b^p=a^p+b^p$,

and Lemma 1 is proven.

Now we can show:

[color=blue][b]Lemma 2.[/b] Let $p$ be a prime number with $p\geq 3$, and let $R$ be a skew ring such that $p=0$ in $R$. Let $x$ be an element of $R$. Then, there exists an element $t\in R$ such that $x^{p^2-1}-1=t\left(x^p+x-1\right)$.[/color]

[i]Proof of Lemma 2.[/i] First, since $p=0$ in $R$, we can apply Lemma 1 to any two commuting elements of $R$. Besides, $p$ is odd (since $p$ is a prime and is $\geq 3$), so that $\left(-x\right)^p=-x^p$.

Now, since the values of any two polynomials of $x$ with integer coefficients commute with each other, we can compute:

$xx^{p^{2}-1}=x^{p^{2}}=\left(  x^{p}\right)  ^{p}=\left(  \left( x^{p}+x-1\right)  +\left(  1+\left(  -x\right)  \right)  \right)  ^{p}$
$=\left(  x^{p}+x-1\right)  ^{p}+\left(  1+\left(  -x\right)  \right)  ^{p}$ (by Lemma 1, applied to the commuting elements $x^{p}+x-1$ and $1+\left(-x\right)$ of $R$)
$=\left(  x^{p}+x-1\right)  ^{p}+\left(  1^{p}+\left(  -x\right)  ^{p}\right) $ (since $\left(  1+\left(  -x\right)  \right)  ^{p} = 1^{p}+\left(  -x\right)  ^{p}$ by Lemma 1, applied to the commuting elements $1$ and $-x$ of $R$)
$=\left(  x^{p}+x-1\right)  ^{p}+\left(  1+\left(  -x\right)  ^{p}\right) =\left(  x^{p}+x-1\right)  ^{p}+\left(  1-x^{p}\right)  $ (since $\left( -x\right)  ^{p}=-x^{p}$)
$=\left(  x^{p}+x-1\right)  ^{p}-\left(  x^{p}+x-1\right)  +x$
$=\left(  x^{p}+x-1\right)  \left(  \left(  x^{p}+x-1\right)  ^{p-1}-1\right) +x$,

so that

$x\left(  x^{p^{2}-1}-1\right)  =xx^{p^{2}-1}-x$
$=\left(  \left(  x^{p}+x-1\right)  \left(  \left(  x^{p}+x-1\right) ^{p-1}-1\right)  +x\right)  -x$
$=\left(  x^{p}+x-1\right)  \left(  \left(  x^{p}+x-1\right)  ^{p-1}-1\right) $.

Consequently,

$x^{p^{2}-1}-1=\left( x^{p}+x\right) \left( x^{p^{2}-1}-1\right) -\left( x^{p}+x-1\right) \left( x^{p^{2}-1}-1\right) $
$=\left( x^{p-1}+1\right) x\left( x^{p^{2}-1}-1\right) -\left( x^{p}+x-1\right) \left( x^{p^{2}-1}-1\right) $
$=\left( x^{p-1}+1\right) \left( x^{p}+x-1\right) \left( \left( x^{p}+x-1\right) ^{p-1}-1\right) -\left( x^{p}+x-1\right) \left( x^{p^{2}-1}-1\right) $
$=\left( x^{p}+x-1\right) \left( \left( x^{p-1}+1\right) \left( \left( x^{p}+x-1\right) ^{p-1}-1\right) -\left( x^{p^{2}-1}-1\right) \right) $
$=\left( \left( x^{p-1}+1\right) \left( \left( x^{p}+x-1\right) ^{p-1}-1\right) -\left( x^{p^{2}-1}-1\right) \right) \left( x^{p}+x-1\right) $,

and setting $t=\left( x^{p-1}+1\right) \left( \left( x^{p}+x-1\right)^{p-1}-1\right) - \left( x^{p^{2}-1}-1\right) $, we get $x^{p^{2}-1}-1=t\left( x^{p}+x-1\right) $, so that Lemma 2 is proven.

Now, to the solution of the problem:

We define a bisequence $u\in\mathbb{Z}^{\mathbb{Z}}$ recurrently by setting $u\left(n\right)=n$ for all integers $n$ such that $0\leq n\leq p-1$ and the recursion $u\left(n\right)=u\left(n-1\right)+u\left(n-p\right)$ for all $n\in\mathbb{Z}$. Note that this recursion has to be applied "forwards" (i. e. in the form $u\left(n\right)=u\left(n-1\right)+u\left(n-p\right)$) in order to compute the terms $u\left(p\right)$, $u\left(p+1\right)$, $u\left(p+2\right)$, ..., and applied "backwards" (i. e. in the form $u\left(n-p\right)=u\left(n\right)-u\left(n-1\right)$) in order to compute the terms $u\left(-1\right)$, $u\left(-2\right)$, $u\left(-3\right)$, ....

The sequences $\left(u\left(0\right),u\left(1\right),u\left(2\right),...\right)$ and $\left(a_0,a_1,a_2,...\right)$ are equal, because they have the same $p$ starting values ($a_n=n$ and $u\left(n\right)=n$ for all integers $n$ such that $0\leq n\leq p-1$) and satisfy the same $p+1$-term recursion ($a_{n}=a_{n-1}+a_{n-p}$ and $u\left(n\right)=u\left(n-1\right)+u\left(n-p\right)$). In other words, $u\left(n\right)=a_n$ for every integer $n\geq 0$.

Next, we define a bisequence $\overline{u}\in\left(\mathbb{F}_p\right)^{\mathbb{Z}}$ as follows: For every integer $n$, let $\overline{u}\left(n\right)$ be the residue class of $u\left(n\right)$ modulo $p$.

Then, since $u\left(n\right)=u\left(n-1\right)+u\left(n-p\right)$ for every $n\in\mathbb{Z}$, we have $\overline{u}\left(n\right)=\overline{u}\left(n-1\right)+\overline{u}\left(n-p\right)$ for every $n\in\mathbb{Z}$. Since $\overline{u}\left(n-1\right)=\left(S\overline{u}\right)\left(n\right)$ and $\overline{u}\left(n-p\right)=\left(S^p\overline{u}\right)\left(n\right)$, this becomes $\overline{u}\left(n\right)=\left(S\overline{u}\right)\left(n\right)+\left(S^p\overline{u}\right)\left(n\right)$ for every $n\in\mathbb{Z}$. Hence, $\overline{u}=S\overline{u}+S^p\overline{u}$, so that $\left(S^p+S-1\right)\overline{u}=0$ (hereby, $1$ stands for the identity endomorphism of $\left(\mathbb{F}_p\right)^{\mathbb{Z}}$; in fact, we can call it $1$ because it is the multiplicative unit of the skew ring of endomorphisms of $\left(\mathbb{F}_p\right)^{\mathbb{Z}}$).

The skew ring of endomorphisms of the vector space $\left(\mathbb{F}_p\right)^{\mathbb{Z}}$ satisfies $p=0$ (since $p=0$ holds in $\mathbb{F}_p$). Thus, applying Lemma 2 to the element $x=S$ of this skew ring, we see that there exists an element $T$ of this skew ring such that $S^{p^2-1}-1=T\left(S^p+S-1\right)$. Hence,

$S^{p^3-p}-1=\left(S^{p^2-1}\right)^p-1=\sum_{k=0}^{p-1}\left(S^{p^2-1}\right)^k\cdot \left(S^{p^2-1}-1\right)$
$=\sum_{k=0}^{p-1}\left(S^{p^2-1}\right)^k\cdot T\left(S^p+S-1\right)$.

Thus,

$\left(S^{p^3-p}-1\right)\overline{u}=\left(\sum_{k=0}^{p-1}\left(S^{p^2-1}\right)^k\cdot T\left(S^p+S-1\right)\right)\overline{u}$
$=\sum_{k=0}^{p-1}\left(S^{p^2-1}\right)^k\cdot T\cdot\underbrace{\left(S^p+S-1\right)\overline{u}}_{=0}=0$,

so that $S^{p^3-p}\overline{u}=1\overline{u}$. This yields that $\left(S^{p^3-p}\overline{u}\right)\left(n\right)=\left(1\overline{u}\right)\left(n\right)$ for every $n\in\mathbb{Z}$. In other words, $\overline{u}\left(n-\left(p^3-p\right)\right)=\overline{u}\left(n\right)$ for every $n\in\mathbb{Z}$. Applying this to $n=p^3$, we get $\overline{u}\left(p^3-\left(p^3-p\right)\right)=\overline{u}\left(p^3\right)$. In other words, $\overline{u}\left(p\right)=\overline{u}\left(p^3\right)$. Since $\overline{u}\left(n\right)$ was defined as the residue class of $u\left(n\right)$ modulo $p$, this equation yields that $u\left(p\right)\equiv u\left(p^3\right)\text{ mod }p$. Since $u\left(n\right)=a_n$ for all integers $n\geq 0$, this rewrites as $a_p\equiv a_{p^3}\text{ mod }p$. Now, applying the recursion $a_{n}=a_{n-1}+a_{n-p}$ to $n=p$, we get $a_p=a_{p-1}+a_{p-p}=a_{p-1}+a_0=\left(p-1\right)+0=p-1$. Thus, $a_p\equiv a_{p^3}\text{ mod }p$ yields $p-1\equiv a_{p^3}\text{ mod }p$, so that $a_{p^3}\equiv p-1\equiv -1\text{ mod }p$. This completes the proof.

[b]PS.[/b] In the above, I showed that $a_{p^3}\equiv -1\text{ mod }p$ for every prime $p\geq 3$. By inspection you can see that this also holds for $p=2$.

  Darij

This is indeed not from Canada 1986... does anyone know the correct source?


probability
function
limit
number theory
relatively prime

What is the probability of two arbitrary positive numbers both have a given pirime $ p$ as a factor?

There is no well-defined answer to your question, because there is no canonical distribution for an "arbitrary positive number", even if you intend that to be an integer.

Assuming you mean integer when you say number:

A number is divisable by $ p$ iff its $ \equal{}0 \mod{p}$ so the chances of a number being divisable by $ p$
is $ \frac{1}{p}$ So the chances of 2 numbers both being divisable by $ p$ is $ \frac{1}{p^2}$

This actually leads to the zeta function if you ask 'what are the chances of $ n$ integers being relatively prime?'

Seamusoboyle, yes I meaned positive integer. As for me, your answer is correct. Thank you.

Saying that the probability that an arbitrary positive integer is divisible by $ p$ is $ \frac{1}{p}$ contains at least the underlying assumption that the "probability" of choosing a particular integer in a particular residue class $ \bmod p$ is $ \frac{1}{p}$, which usually implies the further assumption that the "probability" of choosing each integer is equal.

For a countably infinite set, this is clearly absurd.  If we ignore the second assumption and only use the first, we have a tautology.  We picked a probability distribution that makes the answer what we expect, and that distribution would have to change for each $ p$ to make the problem valid, which is clearly not the spirit of the original question.

fedja made some comments on this subject [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183699]here[/url].  The notion of density substitutes well for probability here.  In particular, the question is made sensible if we define $ p_n$ to be the probability that two integers selected uniformly at random in $ [1, n]$ are divisible by $ p$ and then ask for $ \lim_{n \to \infty} p_n$.


function
complex numbers
advanced fields
advanced fields unsolved

Let $\mathcal{U}$ be the set of all infinite subsets of Spm$(\mathbb{Z})$ and define the ideal in $\prod_{p\in\mathrm{Spm}(\mathbb{Z})}\overline{\mathbb{F}}_{p}$
\[\mathcal{I}_\mathcal{U}: =\left\{(f_{p})_{p\in\mathrm{Spm}(\mathbb{Z})}\in\prod_{p\in\mathrm{Spm}(\mathbb{Z})}\overline{\mathbb{F}}_{p}\text{ }\vert\text{ }\exists U\in\mathcal{U}\text{ such that }f_{p}=0\text{ for }\forall p\in U\right\}.\]

Show that
\[\prod_{p\in\mathrm{Spm}(\mathbb{Z})}\overline{\mathbb{F}}_{p}/\mathcal{I}_\mathcal{U}\cong\mathbb{C}\quad\text{as fields},\]
so that, in particular, $\prod_{p\in\mathrm{Spm}(\mathbb{Z})}\overline{\mathbb{F}}_{p}/\mathcal{I}_\mathcal{U}$ is an algebraically closed field of characteristic zero.

I am not sure, but you may want to check in Field Arithmetic by Freid and Jarden.

It is most likely easier to do this the other way round: show that it is an algebraically closed field of characteristic zero having the right cardinality, *hence* isomorphic to $\mathbb C$. owk

Can someone explain me owk *hence*:

why is an algebraically closed field of characteristic zero, whose cardinality equal the continuum, isomorphic to the complex numbers?

This is a somewhat messy exercise using transcendence bases over $\mathbb Q$. owk

I did not find a good reference, most authors either do not mention this or take it as well-known. However, [url=http://planetmath.org/encyclopedia/CardinalityOfAlgebraicClosure.html]here on PlanetMath[/url] is a sketch. owk

So, for example, $\mathbb{C}$ is isomorphic to the algebraic closure of its function field, $\overline{\mathbb{C}(T)}$.

And by this, $\mathbb{C}$ is also isomorphic to one of its subfield.

Right?

Right. (This is not uncommon, e.g. the subfield $\mathbb Q(T^{2})\subset\mathbb Q(T)$ is isomorphic to $\mathbb Q(T)$.) owk


Ο Α και ο Β βαφουν εναλλαξ με μαυρο χρωμααπο ενα τετραγωνακι μιας ολολευκης
4x4 σκακιερας. Χανει αυτος που πρωτος θα αναγκαστει να συμπληρωσει
ενα 2x2 τετραγωνο (βαφοντας δηλαδη και τεταρτο τετραγωνο). Εχει καποιος στρατηγικη νικης? :)

Edit:Διορθωσα την αρχικη μου κακη διατυπωση :oops:

[hide]Pistevw pws o paiktis B(aftos pou pezei meta ton A) exei stratigiki nikhs.
To skeptiko mou itan to eksis:
An theorisoume tin skakiera san ena pinaka $ 4x4$ $ [a_{ij}]$(etsi gia anesi to evala xexe) tote
otan o A paei na vapsei ena koutaki $ (i,j)$ o B tha pigainei kai tha vafei to koutaki $ (5 \minus{} i,5 \minus{} j)$.
Akolouthwntas aftin tin stratigiki pistevw o B mexri to telos apofeygei na symplirosei kapoio koutaki $ 2x2$.[/hide]

Προς new member
[hide]Είναι πολύ ωραίο να το πηγαίνεις συμμετρικά ως προς το κέντρο, όμως το θέμα είναι πως ο Αλέξανδρος ,οπως βλεπω την εκφωνηση, πιστευω οτι ζητά στρατηγική καθαρης νίκης και οχι μη ηττας!(Δηλαδη ισως να ειναι μερος της ασκησης να δεις την πιθανη "ισοπαλια".)
[/hide]


geometry
trapezoid
algebra
system of equations

[hide="Solution"]
To determine the individual sides of the isosceles trapezoid, we will use [url=http://www.artofproblemsolving.com/Wiki/index.php/Ptolemy%27s_Theorem]Ptolemy's Theorem[/url], which states that $ ac + bd = ef$, where $ a$,$ b$,$ c$,$ d$ are the sides of the trapezoid ABCD and $ e$ and $ f$ are the diagonals.

\begin{eqnarray*}
ac + (30)^2 &=& (82)^2 \\
ac  + 900 &=& 6724 \\
ac &=& 5824
\end{eqnarray*}

Because we also have $ a+c=160$, we can solve the system of equations

\begin{eqnarray*}
a+c &=& 160\\
ac &=& 5824
\end{eqnarray*}

Skipping ahead we find that we have the equation $ c^2 -160c + 5824$, and my calculator tells me that the $ a=56$ and $ c=104$.

Now, with all the sides, we can calculate the area with [url=http://www.artofproblemsolving.com/Wiki/index.php/Brahmagupta%27s_Formula]Brahmagupta's Formula[/url] for a cyclic quadrilateral. In other words, the area of trapezoid ABCD is $ \sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $ s$ is the semiperimeter.

$ s = \frac{56+30+104+30}{2} = 110$

\begin{eqnarray*}
[ABCD] &=& \sqrt{(s-a)(s-b)(s-c)(s-d)} \\
&=& \sqrt{(110-56)(110-30)(110-104)(110-30)} \\
&=& \sqrt{2073600}\\
&=& \boxed{1440}
\end{eqnarray*}

[/hide]

One can ascertain that the trapezoid is isosceles because the diagonals are equal. Thus:

The side  ED is
$ b \minus{} \frac{(b\minus{}a)}{2}$

and the following equation equation results:

$ (\frac{(a\plus{}b)}{2})^2 \plus{} h^2 \equal{} 82^2$
and  as $ (a\plus{}b)$ is 160:

$ (\frac{160}{2})^2 \plus{} h^2 \equal{} 82^2$
$ 80^2 \plus{} h^2 \equal{} 82^2$
$ h^2 \equal{} 82^2 \minus{} 80^2$
$ h^2 \equal{} 324$
$ h \equal{} 18$

And then use the formula for the area of a trapezoid:

$ A \equal{} \frac{1}{2}(160)(18)$
$ A \equal{} 80*18$
$ A \equal{} 1440$

And tada.

[quote="123456789"]The sum of the side lengths of the two parallel sides in an isosceles trapezoid is $ 160$. The side length of each of the remaining two sides is $ 30$. Both diagonals measure $ 82$. Find the area of the trapezoid.[/quote]

Let $ D'$ be on $ AB$ such that $ AC \equal{} CD'$; then $ \triangle ACD \cong \triangle CD'B$ and $ \overline{AD'} \equal{} AB \plus{} CD \equal{} 160$, so $ [ABCD] \equal{} [ACD'] \equal{} \frac {1}{2} \cdot 160 \cdot \sqrt {82^2 \minus{} 80^2} \equal{} 1440$.

[asy]pathpen = linewidth(0.7); pen d = linewidth(0.7)+linetype("4 4");pen f = fontsize(8);
pair A=(0,0),B=(104,0),C=(80,50),Ca=foot(C,A,B),D=(24,50),Da=(160,0);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(B--MP("D'",Da)--C--A,d); D(C--Ca,d); D(rightanglemark(C,Ca,A,150)); MP("82",(A+C)/2,SE,f); MP("160",(A+Da)/2,f);[/asy]

I think that definitely kills your's  :P


inequalities
vector
function
triangle inequality
linear algebra

Let V be a vector space with a norm $ \parallel{}.\parallel{}$

$ \left| \parallel{}u\parallel{}\minus{}\parallel{}v\parallel{} \right| \leq \parallel{}u\minus{}v\parallel{}$ for u and v in V

[quote="KBriggs"]Let V be a vector space with a norm $ \parallel{}.\parallel{}$

$ \left| \parallel{}u\parallel{} \minus{} \parallel{}v\parallel{} \right| \leq \parallel{}u \minus{} v\parallel{}$ for u and v in V[/quote]

the given inequality is equivalent to the following two inequalites:$ \parallel{}u\parallel{}\minus{}\parallel{}v\parallel{}\leq \parallel{}u\minus{}v\parallel{}$ and $ \parallel{}u\parallel{}\minus{}\parallel{}v\parallel{} \geq \minus{}\parallel{}u\minus{}v\parallel{}$

we prove the first one

 by triangle inequality of norm,we can obtain $ \parallel{}u\parallel{}\equal{}\parallel{}u\minus{}v\plus{}v\parallel{}\leq \parallel{}u\minus{}v\parallel{}\plus{}\parallel{}v\parallel{}$,that is,$ \parallel{}u\parallel{}\minus{}\parallel{}v\parallel{}\leq \parallel{}u\minus{}v\parallel{}$

the second inequality can be proved in similar way.
Proof is accomplished.

Thank you. I was missing some very simple things there. :)

This inequality implies, by the way, that the norm function is continuous.


If $0 < a_{n}< a_{2n}+a_{2n+1}$ then series $\sum_{n}a_{n}$ diverges

This is kind of cute.  [hide]Because all terms are positive, I can futz around with the order as much as I want.  Suppose $\sum_{n \geq 0}a_{n}= L$.  But then $L < \sum_{n \geq 0}a_{2n}+a_{2n+1}= \sum_{n \geq 0}a_{n}= L$, contradiction.  Thus, the sum diverges.[/hide]

[quote="JBL"] then $L < \sum_{n \geq 0}a_{2n}+a_{2n+1}$[/quote]

No, It is '' $\leq$ '' , i think so  :maybe:

[quote="JBL"]If $a_{n}< b_{n}$ for all $n$, and the sums of the two sequences converge, then $\sum a_{n}< \sum b_{n}$.[/quote]
I not think so, We can choose two sequences such that $a_{n}<b_{n}$ for all $n$ and $(a_{1}+a_{2}+...+a_{n})-(b_{1}+b_{2}+...+b_{n})=-\frac{1}{n}$ for all $n$.

No, we can't.  Your counter-example is a counter-example to a different claim.

If $(a_{1}+a_{2}+...+a_{n})-(b_{1}+b_{2}+...+b_{n})=-\frac{1}{n}$ and $(a_{1}+a_{2}+...+a_{n+1})-(b_{1}+b_{2}+...+b_{n+1})=-\frac{1}{n+1}$ then by subtraction $a_{n+1}-b_{n+1}=-\frac{1}{n+1}+\frac{1}{n}> 0$.

More generally, if the difference between the first terms of the two series is $d$, the difference between the $n$th partial sums of the two series is at least $d$ and so the difference between the limits of the two series is at least $d$.

JBL,
I don't know theorem follows, that is truth.
[quote="JBL"]If $a_{n}< b_{n}$ for all $n$, and the sums of the two sequences converge, then $\sum a_{n}< \sum b_{n}$.[/quote]
Can you post prove of it?  :)

This is the proof:
[quote="JBL"]More generally, if the difference between the first terms of the two series is $d$, the difference between the $n$th partial sums of the two series is at least $d$ and so the difference between the limits of the two series is at least $d$.[/quote]


Hey. I'm a newbie to this part of the forums. The AMC cosmos is far, but not impartically so from the MATHCOUNTS region.

I just took the tests this year, both the AMC 10A and 12A in impractical conditions because the school's "gifted" program teacher person has serious....issues. :maybe:

Anyways, I haven't prepared for any AMC tests, be it the 8, 10, or 12. This is my last year for Mathcounts, and whether I start next week, or in March, or in May, don't I need to start preparing for the AMC 10/12? I realize that you have to do lots of those types of problems, but are there certain areas to look into, to learn about or stuff like that? I'm very confused. :|

Is there any advice anyone can give to me about preparing for the AMC or just what kind of math contests there are in high school?

--Also, I heard that you can only take either the AMC 10 or the AMC 12, and then something about qualifying for something called the AIME? Could someone clarify that, too?

Thanks!
~The_Scintillator

[quote="illusiat"]However, this year I suggest you concentrate solely on Mathcounts. One of my school teammates did horrible at chapters cuz he was used to having more time on the AMC[/quote]
That's a bad thing?

It would probably be bad in preparing for a sprint, since on sprints you need to work fast and on AMC you are given plenty of time.  The first 5-10 problems or so on the AMC are almost like sprint round problems, except on sprint rounds, you would need to do them faster.  Although it is not like Sprint Rounds, I think that the AMC 10 problems #11-25 are about state target level.  I usually try and use sets of AMC (10 or 12) problems as practice for target rounds.

[quote="CatalystOfNostalgia"][quote="illusiat"]However, this year I suggest you concentrate solely on Mathcounts.[/quote]

whoa whoa whoa whoa there. Even if success at MathCounts is more likely than success at the AMC, and any type of success may be good, keep in mind that the AMC is exponentially more important than MathCounts and it's good to get an early start.[/quote]
Not only that but knowledge you can pick up from other places can be useful on everything, INCLUDING mathcounts. You seem pretty motivated, so just go work on AMC stuff. Well not necessarily studying for the amc itself, but just getting basics and all that down.

err... More time on mathcounts... But it's easier than the AMC 10... I don't see much of a difference. Since all the mathcounts questions are pretty much as easy as number 1-10 on an AMC 10. at least in chapter.

to help me, i take mathcounts state and national tests (back when i hadn't taken them like 24523 times) in 25-30 minutes so that in the actual competition it would help with the speed issue which gets most mathcounts ppl. in chap and even state i think any problem 1-20 in AMC 10 is fair game, but what do i know?

First off, thanks for all of the help and advice for me! Thanks! :D

Hmm. I'm sensing some conflicting feelings towards the amount of time spent on Mathcounts Vs. AMC 10. From what I can see, AMC seems to be fairly predominant. Sort of, anyways.

(I have heard some horror stories about when people use too much time on the AMC and fail at Mathcounts....)

I think for now--I'm still focusing on Mathcounts--I doubt I'll make it past this competition on Saturday, the Chapter competition. However, I'll try as soon as possible to start the AMC, as soon as I'm eliminated from Mathcounts.

I already have AoPS, volumes 1 & 2. I suppose I should probably do past problems--from both the AMC 10 & 12? And what's the AIME? \

Thanks, again!
~The _Scintilator

Look at the description of the AMC at the top of the forum.

Hmm. AIME.....Sounds difficult.

Does anyone else have any other reading suggestions?
Certain subjects to cover, certain things to study?
What would you consider 'the basics"?
What would you consider "advanced"?

Thanks!
~The_Scomto;;atpr


A curve has parametric equations

$ x\equal{}9t \minus{} \ln(9t), y\equal{}t^{3} \minus{} \ln(t^{3}).$

Show that there is only one value of $ t$ for which $ \frac{dy}{dx} \equal{} 3$ and state that value.

$ x^\prime_t\equal{}9\minus{}\frac{1}{t}$, $ y^\prime_t\equal{}3t^2\minus{}\frac{3}{t}$
Hence, $ y^\prime_x\equal{}\dfrac{3(t^3\minus{}1)}{9t\minus{}1}$
So $ y^\prime_x\equal{}3$ implies $ t^3\minus{}1\equal{}9t\minus{}1$ which gives $ t\equal{}0$, $ 3$ and $ \minus{}3$. But since $ t>0$, we get $ t\equal{}3$.


Let $ a,b,c$ be reals with product $ 1$.
Prove that:
$ 5 \plus{} \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \geq (1\plus{}a)(1\plus{}b)(1\plus{}c)$

[quote="epitomy01"]Let $ a,b,c$ be reals with product $ 1$.
Prove that:
$ 5 \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)$[/quote]
It's Schur for positive $ x,$ $ y$ and $ z$ such, that $ a\equal{}\frac{x}{y},$ $ b\equal{}\frac{z}{x}$ and $ c\equal{}\frac{y}{z}.$

[quote="dduclam"][quote="epitomy01"]Let $ a,b,c$ be reals with product $ 1$.
Prove that:
$ 5 \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)$[/quote]
 $ \Leftrightarrow a^2c \plus{} c^2b \plus{} b^2a \plus{} 3 \geq ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c$
Using AM-GM ineq:
       $ a^2c \plus{} ab \geq 2a$
       $ c^2b \plus{} ca \geq 2c$
       $ b^2a \plus{} bc \geq 2b$
       $ a^2c \plus{} c \geq 2ac$
       $ c^2b \plus{} b \geq 2bc$
       $ b^2a \plus{} a \geq 2ab$
 Sum 5 ineq      $ \Rightarrow$  (q.e.d) :D[/quote]
I have obtained $ 2\sum_{cyc}a^2c\geq a \plus{} b \plus{} c \plus{} ab \plus{} ac \plus{} bc.$ :maybe:


conics
parabola
projective geometry
geometry unsolved
geometry

Denote $A_{\infty},$ $B_{\infty},$ $C_{\infty},$ $D_{\infty},$ the intersections of $AB,BC,CD,DA$ with the line $\tau_{\infty}$ at infinity. By Desargues theorem, the pencil of conics through $A,B,C,D$ induce an involution on $\tau_{\infty},$ thus the parabolas $\mathcal{P}_1,\mathcal{P}_2$ through $A,B,C,D$ touch $\tau_{\infty}$ at the fixed points $X_{\infty}, Y_{\infty}$ of the involution $\mathbf{I}: \{A_{\infty} \mapsto C_{\infty}, B_{\infty}\mapsto D_{\infty} \}.$ One way two construct these directions is as follows: Let $P \equiv AB \cap CD$ and let $u,v$ be the parallels through $P$ to $AD,BC,$ resp. If $U \equiv u \cap BC,$ then $\mathbf{I}: \{B \mapsto C, U \mapsto B_{\infty} \}$ $\Longrightarrow$ $U$ is center of the involution on $BC,$ therefore the circle centered at $U$ and orthogonal to $\odot(PBC)$ (it can be any circle through B,C) cuts $BC$ at the fixed points $M,N$ on $BC$ $\Longrightarrow$ $PM,PN$ have the desired directions $X_{\infty}, Y_{\infty}.$

Now, technically the 2 parabolas $\mathcal{P}_1,\mathcal{P}_2$ are defined by 5 points $\{A,B,C,D, X_{\infty} \}$ and $\{A,B,C,D, Y_{\infty} \},$ but we can construct their foci and focal axes easily. For instance, using Pascal theorem for $AABCDX_{\infty},$ we have a construction of the tangent $\tau_A$ of $\mathcal{P}_1$ at $A$ and similarly its tangents $\tau_B,\tau_C,\tau_D.$ Miquel point $F_1$ of the quadragle $(\tau_A,\tau_B,\tau_C,\tau_D)$ is the focus of  $\mathcal{P}_1$ and the parallel from $F_1$ to $X_{\infty}$ determines its focal axis. Similarly, we construct the focus and focal axis of $\mathcal{P}_2.$


Let $ x$, $ y$, $ z$ be positive numbers. Find the minimum value of:
$ (a)\quad \frac{x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz}$

$ (b)\quad \frac{x^2 \plus{} y^2 \plus{} 2z^2}{xy \plus{} yz}$

[quote="April"]Let $ x$, $ y$, $ z$ be positive numbers. Find the minimum value of:
$ (a)\quad \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz}$

$ (b)\quad \frac {x^2 \plus{} y^2 \plus{} 2z^2}{xy \plus{} yz}$[/quote]

[hide="hint"]By AM-GM:

(a)  $ x^2 \plus{} y^2 \plus{} z^2\equal{}x^2\plus{}\frac{y^2}2\plus{}\frac{y^2}2\plus{}z^2\ge\sqrt2 (xy\plus{}yz)$

(b)  $ x^2 \plus{} y^2 \plus{} 2z^2\equal{}x^2\plus{}\frac{2y^2}3\plus{}\frac{y^2}3\plus{}2z^2\ge2\sqrt{\frac{2}3}(xy\plus{}yz)$  :) [/hide]

$(\text a)$ We have, by AM-GM:
$$x^2+y^2+z^2=x^2+\frac12y^2+\frac12y^2+z^2\ge xy\sqrt 2+yz\sqrt 2$$
so $\frac{x^2+y^2+z^2}{xy+yz}\ge\sqrt2$, with equality when $(x,y,z)=\left(\sqrt2,2,\sqrt2\right)$.
$(\text b)$ We have, by AM-GM:
$$x^2+y^2+2z^2=x^2+\frac23y^2+\frac13y^2+2z^2\ge\frac{2\sqrt6}3xy+\frac{2\sqrt6}3yz$$
so $\frac{x^2+y^2+2z^2}{xy+yz}\ge\frac{2\sqrt6}3$, with equality when $(x,y,z)=\left(2\sqrt6,6,\sqrt6\right)$.


If anyone has anything interesting that could revive this forum, please help post it.
I'm going away for winter break, so no help from me.

hmmm...this is the first time i've gone to AoPS for like 4 months...but I guess we could make an Iowa game thread?idk...


algebra polynomial trigonometry function

Here is a nice and simple problem about polynomials.

For a positive integer $n > 1$ and a fixed real number $t$, define

\[ f(x) = x^n \sin t - x \sin nt + \sin (n - 1)t,\; g(x) = x^2 - 2x \cos t + 1.\]

Prove that $f(x)$ is divisible by $g(x)$.

Quite easy as you said.
Just compute the roots of g and check that they are also the roots of f. The trigonometric identity you obtain is quite obvious. :D

yep, that's my solution as well. Surprisingly that such a complex function as f can still have roots that simple :)

[quote="Arne"]Here is a nice and simple problem about polynomials.

For a positive integer $ n > 1$ and a fixed real number $ t$, define
\[ f(x) \equal{} x^n \sin t \minus{} x \sin nt \plus{}\sin (n \minus{} 1)t,\; g(x) \equal{} x^2 \minus{} 2x \cos t \plus{} 1.\]
Prove that $ f(x)$ is divisible by $ g(x)$.[/quote]

I think is $ \sin(n\plus{}1)t$ and not $ \sin(n\minus{}1)t$.


Given triangle ABC inscribed in (O). D, E are the points respectively on AB, AC. H and K are respectively the projections of D, E onto BC. Let (O') be the cricumcircle of triangle ADE. Suppose that HK= 1/2 BC. Let O''(different from A) be the intersection of AO' with (O). Prove that:
AO'.AO''=$ AO^2$. :D

First see that $ \angle HDE \equal{} O''BA$. Hence $ sin\angle O''BA \equal{} sin\angle HDE \equal{} \frac {BC}{2DE}$. $ AO' \equal{} R_{ADE} \equal{} \frac {DE}{2sin\alpha }$ and $ AO'' \equal{} 2R_{ABC}sinO''BA$ (sine theorem for $ \triangle BO''A)$ $ \equal{} \frac {BC.R_{ABC}}{DE}$. Hence $ AO'.AO'' \equal{} \frac {DE}{2sin\alpha }.\frac {BC.R_{ABC}}{DE} \equal{} R_{ABC}^2 \equal{} AO^2$. DONE!

If $ \angle AED \equal{} \alpha$, from the given condition $ HK\equal{}\frac {BC} {2}$ we easily get $ DE \equal{} \frac {BC} {2 \cdot cos (C \minus{} \alpha)}$  and  $ AO_{1} \equal{} \frac {R} {2 \cdot cos (C \minus{} \alpha)}$ ( 1 ). It’s also easy to see that $ \angle CAO^{"}\equal{}A\plus{}\alpha \minus{}90^{o}$, which makes
$ \angle ABO^{"}\equal{}90^{o}\minus{}C\plus{} \alpha$, that is, $ AO^{"}\equal{}2 \cdot R \cdot cos(C\minus{}\alpha)$ ( 2 ). By multiplication of relations (1) and (2) we get the desired relation.

Best regards,
sunken rock


logarithms
function
inequalities
inequalities proposed

[b]1)[/b] [color=green]  [b] Let   $a_1,b_1,c_1, a_2,b_2,c_2$ be  positive numbers such that   $\begin{array}{|cc|} a_1 & b_1\\ b_1 & c_1\\ \end{array}\ge 0\; \; \; ,\; \; \; \; \begin{array}{|cc|} a_2 & b_2\\ b_2 & c_2\\ \end{array} \ge 0 \; .$ Prove  $\begin{array}{|cc|} a_1+a_2 & b_1+b_2\\ b_1+b_2 & c_1+c_2\\ \end{array}\ge 0\; .$
[/b] [/color]
[b]2)[/b] [color=blue] [b] If $f,g : (0,\infty)\to (0,\infty)$ are such that $\ln{f}$ and $\ln{g}$ are convex on $(0,\infty),$
  show that  $\ln{(f+g)}: (0,\infty)\to {\mathbb R}$  is also  a convex function on $(0,\infty) .$  [/b][/color]

for 2):
we can prove it quickly by
\[ (f(x_1)+g(x_1))^{\lambda}(f(x_2)+g(x_2))^{1-\lambda} \ge f^{\lambda}(x_1)f^{1-\lambda}(x_2)+g^{\lambda}(x_1)g^{1-\lambda}(x_2), \lambda \in (0,1) \]
which is holder's inequality

[quote="zhaobin"]for 2): .....we can prove it quickly by ..... [/quote]
Nice solution, thanks/proposer