What is the ordered 4-tuple of positive integers $(x,y,z,w)$, each as small as possible, that satisfies the simultaneous inequalities:
$2x\prec x+y\prec x+z\prec2y\prec x+w\prec y+z\prec2z\prec y+w\prec z+w\prec 2w$?

This was a CAML question, if I remember correctly. A number 6, also....

I don't remember the answer though...I got it wrong on the test...

This was a problem which I brute forced during the competition.  :)

Are $\prec$'s $<$'s?

[hide="If so..."][hide="By Brute Force"]The answer is (1,4,5,8)
[/hide][/hide]

how do u brute force this one? and how do u solve it w/o brute force? isnt the test only 30 mins too, so...

I plugged in random numbers, starting with w=1.

The CAML is only 30 minutes, but the first 4-5 questions only take ten minutes max.

This is a problem from the PAML, so it is probably a little different than the CAML. None of you have the correct answer. :(

[hide][tex](1,5,7,10)[/tex][/hide]

Doesn't take that much time even with brute force.

[quote="ankur87"]By brute force, I got the following

I don't know if it is correct

[hide="Here"]

w=1
z=4
y=6
x=10

[/hide][/quote]
That has it backwards =P .

That just goes to show

No matter how much math I try, English will always be my downfall.

What does:
w=1
z=4
y=6
x=10
have to do with English? :)
By the way: Elemennop has the correct answer. Are you from PA?

I mixed up the letters

It's an attempt on my part to blame my being wrong on something other than math...to save grace.


The gasoline gauge on a van initially read $ \frac 18$ full. When
15 gallons of gasoline were added to the tank, the
gauge then read $ \frac 34$ full. How many more gallons
would be needed to fill the tank?

Let x be number of gallons when full:
(3/4-1/8)x=15, so x=24, and the answer is 6


I have this combinatorics book that does not contain solutions( :wallbash_red: ), I'm just wondering if anyone here can tell me if my solution is correct or wrong.

Problem:
Prove that
$ \binom{2n}{n} \equal{} \binom{n}{0}^2 \plus{} \binom{n}{1}^2 \plus{} \binom{n}{2}^2 \plus{} \cdots \plus{} \binom{n}{n}^2$

The hint that the book gave,
[hide="hint"]
Consider the coefficients of $ x^n$ in the following equation:
$ (1 \plus{} x)^{2n} \equal{} (1 \plus{} x)^n \cdot  (1 \plus{} x)^n$
[/hide]

My solution:
[hide="solution"]
The coefficient in the left side of the equation of $ x^n$ is $ \binom{2n}{n}$, which means that the coefficient of $ x^n$ in the right hand of the equation is also $ \binom{2n}{n}$.

Now I start by expanding each $ (1 \plus{} x)^n$:
$ (1 \plus{} x)^n \cdot  (1 \plus{} x)^n \equal{} \left(\binom{n}{0}1^n \plus{} \binom{n}{1}x \plus{} \cdots \plus{} \binom{n}{n}x^n\right) \cdot  \left(\binom{n}{0}1^n \plus{} \binom{n}{1}x \plus{} \cdots \plus{} \binom{n}{n}x^n\right)$

Substituting $ \binom{n}{j}$ for $ \binom{n}{n \minus{} j}$, for $ 0\le j\le n$ in the second expansion:

$ \left(\binom{n}{0}1^n \plus{} \binom{n}{1}x^1 \plus{} \cdots \plus{} \binom{n}{n}x^n\right) \cdot \left(\binom{n}{n}1^n \plus{} \binom{n}{n \minus{} 1}x^1 \plus{} \cdots \plus{} \binom{n}{0}x^n\right)$

Now multiplying all the terms $ x^j \cdot x^{n \minus{} j}$ to equal $ x^n$ we find that the coefficients of $ x^j$ and $ x^{n \minus{} j}$ are the same, $ \binom{n}{j}$, adding all those up we get $ \binom{2n}{n}x^n \equal{} \binom{n}{0}^2\cdot x^n \plus{} \binom{n}{1}^2\cdot x^n \plus{} \cdots \plus{} \binom{n}{n}^2\cdot x^n$

Dividing by $ x^n$, we find that $ \binom{2n}{n} \equal{} \binom{n}{0}^2 \plus{} \binom{n}{1}^2 \plus{} \cdots \plus{} \binom{n}{n \minus{} 1}^2 \plus{} \binom{n}{n}^2$

Any feedback on my solution will be great, thanks in advance.

Edit: Fixed where JBL indicated for a 0, and changed the multiplication sign to a dot.
[/hide]

That solution works.

Alternatively, one can use [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's Identity[/url]: $ \binom n0^2\plus{}\binom n1^2\plus{}\cdots\equal{}\binom n0\binom nn\plus{}\binom n1\binom n{n\minus{}1}\plus{}\cdots\equal{}\binom{2n}n$, where $ \binom nk\equal{}\binom n{n\minus{}k}$ has been used.

[quote="tonypr"]Substituting $ \binom{n}{j}$ for $ \binom{n}{n \minus{} j}$, for $ 0\le j\le n$ in the second expansion:

$ \left(\binom{n}{0}1^n \plus{} \binom{n}{1}x \plus{} \cdots \plus{} \binom{n}{n}x^n\right) * \left(\binom{n}{n}1^n \plus{} \binom{n}{n \minus{} 1}x \plus{} \cdots \plus{} \binom{n}{1}x^n\right)$[/quote]  The last binomial coefficient should be $ \binom{n}{0}$, but otherwise it's okay.  A Latex suggestion: use \cdot instead of * for a nice multiplication dot.

math154, suggesting someone use Vandermonde to prove this is circularity at its worst ;)

[quote="JBL"]math154, suggesting someone use Vandermonde to prove this is circularity at its worst ;)[/quote]

Eh, just apply the combinatorial idea behind it, then.

The combinatorial idea and the binomial coefficients proof that the book is hinting at are largely the same.

The combinatorial proof is to try to choose some $ n$ objects from a pile of $ 2n$ objects. We can obviously do this in $ \binom{2n}{n}$ ways. Alternatively, we can break the $ 2n$ pile into two piles of $ n$ objects, and then choose, say, $ k$ objects from the first pile in $ \binom{n}{k}$ ways. Then we need to choose $ n\minus{}k$ more objects from the second pile, which can happen in $ \binom{n}{n\minus{}k}$ ways. Of course, $ \binom{n}{k}\equal{}\binom{n}{n\minus{}k}$, so this is equivalent to $ \binom{n}{k}^2$. $ k$ can range from $ 0$ to $ n$, of course ($ 0$ objects from one pile, $ n$ from the other, and vice versa), so we have $ \binom{2n}{n}\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}^2,$ QED.


vector
calculus
integration
function
algebra
domain
real analysis

This problem is really getting at me. It's a very conceptual problem rather than a "math-math" problem. I don't even know where to start.

It's problem 129 in the pdf packet
http://www.math.ufl.edu/%7Ehuang/calc3/fall2007.pdf

This is the most famous (I think) example of an irrotational vector field in $ \mathbb{R}^2$ that is not conservative. You can easily show that the field is irrotational, by taking derivatives. Let the unit circle (traversed counterclockwise) be
$ \left \{ \begin{array}{l}
             x\equal{}\cos(t)\\
             y\equal{}\sin(t)
         \end{array} \right.$
with $ t \in [0,2\pi]$
If you evaluate your field along the unit circle, you find that there the field is $ \minus{}\sin(t)\vec{i} \plus{} \cos(t)\vec{j}$, that is exactly the tangent unit vector to the circle. So when you have to take the product between the tangent unit vector and the field along the circle you find 1, so your integral has to be >0. So this field is not conservative, cause its integral over the unit circle, that is a closed simple curve, is not equal to 0. The fact that the function $ \arctan \left( \frac{y}{x} \right)$ is a potential for the field in the region of the plane $ x>0$ is not a contraddition, cause a smooth irrotational field is conservative only on a simply connected domain. $ \mathbb{R}^2 \setminus \{(0,0)\}$ is not a simply connected domain, but the half-plane $ x>0$ is. So in this region a potential can exists.


limit
logarithms
calculus
calculus computations

consider $ f : \mathbb{R^ \plus{} } \to \mathbb{R^ \plus{} }$, $ a > 1$ nondecreasing such that:
$ \lim_{x \to \plus{} \infty} f(ax) \minus{} f(x) \equal{} 0$
then prove:
$ \lim_{x \to \plus{} \infty} \frac {f(x)}{ln(x)} \equal{} 0$

Let $ f(e^u) \equal{} g(u)$.  We are given that $ \lim_{x \to \infty} g(u \plus{} \ln a) \minus{} g(u) \equal{} 0$ and asked to prove that $ \lim_{x \to \infty} \frac{g(u)}{u} \equal{} 0$.  The result then follows by [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz-Cesaro[/url].

how? could you please explain

Do you mean using the fact that $ \lim_{\infty} \frac {g(u)}{u} \equal{} 0 \Leftrightarrow \forall u_n$ such that $ u_n$ diverges to $ \infty$ we have $ \frac {g(u_n)}{u_n}$ converges to $ 0$?


[quote="arqady"][quote="nthd"]Prove that $\forall a,b,c>0$ we have :
$(a^2+b^2+c^2)(\frac{a^2}{(2a^2+bc)^2}+\frac{b^2}{(2b^2+ca)^2}+\frac{c^2}{(2c^2+ab)^2})\ge 1$
 :)  :)[/quote]
I think this is wrong. Try $a=1,$ $b=c=2.$ ;)[/quote]
Oh,soory.A month ago, I proved a similarly problem  :)  so I thought this was true ,too. :( .Thank you for pointing out my mistake.


geometry
ratio
search
vector
geometry solved

An old problem. You can use Menelaus or Area Ratios, or search the site.

[quote="Lovasz"]Let $ABC$ be a triangle. $I$ is the midpoint of $BC$. An arbitrary line cut $AB, AC, AI$ at $P, Q, J$. Prove that:
$\frac{AB}{AP}+\frac{AC}{AQ}=2\frac{AI}{AJ}$.[/quote]

Let's use directed segments. The transversal theorem (Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=134343#p134343]http://www.mathlinks.ro/Forum/viewtopic.php?t=20289 post #1[/url]), applied to the points B, C, I on one line, to the point A not on this line, and to the collinear points P, Q, J on the lines BA, CA, IA, respectively, yields

$CI\cdot\frac{BA}{PA}+IB\cdot\frac{CA}{QA}+BC\cdot\frac{IA}{JA}=0$.

But since the point I is the midpoint of the segment BC, we have $CI=IB=\frac12\cdot CB$ and BC = - CB, so this becomes

$\frac12\cdot CB\cdot\frac{BA}{PA}+\frac12\cdot CB\cdot\frac{CA}{QA}+\left(-CB\right)\cdot\frac{IA}{JA}=0$.

Division by $\frac12\cdot CB$ transforms this into $\frac{BA}{PA}+\frac{CA}{QA}-2\frac{IA}{JA}=0$, what is equivalent to $\frac{BA}{PA}+\frac{CA}{QA}=2\frac{IA}{JA}$. But this is exactly your equation.

  Darij


Why is it that one has interchange rows in order to find the LU decomposition of the following matrix?

\[ \left( {\begin{array}{*{20}{c}}
   { \minus{} 2} & 1 & 2  \\
   4 & 1 & { \minus{} 2}  \\
   { \minus{} 6} & { \minus{} 3} & 4  \\
\end{array}} \right)\]

I used a calculator to find LU but when I found its product I got

\[ \left( {\begin{array}{*{20}{c}}
   { \minus{} 6} & { \minus{} 3} & 4  \\
   { \minus{} 2} & 1 & 2  \\
   4 & 1 & { \minus{} 2}  \\
\end{array}} \right)\]

Thanks

We need a row interchange any time we get a zero in a would-be pivot position, starting from the upper left. We need a row interchange anytime an upper left square submatrix fails to be invertible - as happens here. Look at that $ 2\times 2$ upper left submatrix. If you have to do a row interchange, then it isn't an LU decomposition any more; rather it is $ A\equal{}PLU$ where $ P$ is a permutation, $ L$ is lower triangular, and $ U$ is upper triangular. That's exactly what happens here.

We determine the required permutation just by keeping track of the row swaps we perform.


How come the centering tag doesn't work in blogs?

[hide="Example"]
\[ \frac {1}{2}
\]

\[ \vdots
\]
[/hide]

[code]
\[
\frac {1}{2}
\]

\[
\vdots
\]
[/code]

This question is best asked in Support so I am transferring it to that forum.

You can use mimetex. Search on google 'latex for blogger'.

Ugh, Mimetex is so ugly, and plus can't fix the centering issue. Why not just fix the built-in latexrender support for LaTeX?


trigonometry
real analysis
real analysis unsolved

We define the sequence $ (a_n)$ by:
$ a_0\equal{}0, a_1\equal{}1$ and:
$ a_{n\plus{}1}\equal{}a_n\plus{}sin$ $ a_{n\minus{}1}$ if $ a_n>a_{n\minus{}1}$
$ a_{n\plus{}1}\equal{}a_n\plus{}cos$ $ a_{n\minus{}1}$ if $ a_n \leq a_{n\minus{}1}$
Find $ lim$ $ a_n$.

Are you sure you typed everything correctly? The sequence, as it is defined now, seems to have no limit at all :?

Well, I tried it as written, experimentally, and this seems to converge rapidly to $ \frac{\pi}2.$ By tinkering around with the initial conditions, I can get it to converge to $ \left(2k\plus{}\frac12\right)\pi$ for various integers $ k.$

Well, to converge to $ \frac\pi 2$, it would have to be eventually decreasing, so let $ b_n\equal{}a_n\minus{}\frac \pi 2$. We have $ b_{n\plus{}1}\equal{}b_n\minus{}\sin b_{n\minus{}1}\approx b_n\minus{}b_{n\minus{}1}$, when $ b_n$ are small and positive. But then you rather clearly overshoot $ \frac\pi 2$ in 3 steps.  :maybe: Kent, are you sure that you are doing $ a_n\plus{}\cos a_{\color{red}{n\minus{}1}}$ in your spreadsheet and not $ a_n\plus{}\cos a_{\color{red}{n}}$?

Ah ... fedja found my mistake.

Does not appear to converge.


limit
factorial
function
real analysis
real analysis unsolved

Could someone explain me why

 $ \lim_{n\to  \infty} \sum_{k \equal{} 1}^n(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1!)})$ equals $ \lim_{n\to  \infty} (1 \minus{} \frac {1}{(n \plus{} 1!)})$ ?

Found it in a book and I'm a little confused.:blush:
Thanks.

First of all, I believe it was meant to be $ n\to\infty$, since factorials (and Gamma function itself) are not defined for negative integers.

Second: notice that $ \sum_{k \equal{} 1}^{n}(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}) \equal{} \frac {1}{1!} \minus{} \frac {1}{2!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \cdots \plus{} \frac {1}{(n \minus{} 1)!} \minus{} \frac {1}{n!} \plus{} \frac {1}{n!} \minus{} \frac {1}{(n \plus{} 1)!}$ so all members of the sum except for the first and the last cancel out, hence the result.

What book was it in? I don't know what to make of:

$ \lim_{n\to\minus{}\infty}\sum_{k \equal{} 1}^{n}(\frac{1}{k!}\minus{}\frac{1}{(k\plus{}1!)})$

Because if we look at the partial sum... n=-3, for example:

$ \sum_{k \equal{} 1}^{\minus{}3}(\frac{1}{k!}\minus{}\frac{1}{(k\plus{}1!)})$

We need to find -2!... and that's not defined, not even with a gamma function.

So, since this makes so little sense, give us some context-- what book is it?

$ \infty$ indeed, sorry it's a typo.
Thanks for explaining, was really simple  :D \o/


If $ a,b,c>0$ and $ abc\equal{}1$, then
$ \sum(a\plus{}\sqrt{b}\plus{}\sqrt[3]{c})\geq\frac{9}{2}$


limit
integration
calculus
trigonometry
calculus computations

calculate :  $ \lim_{n\to\infty}\sum_{k\equal{}0}^{n}\frac{2^k}{C_n^{k}}$

[quote="mr.inequalities"]calculate :  $ \lim_{n\to\infty}\sum_{k \equal{} 0}^{n}\frac {2^k}{C_n^{k}}$[/quote]
What is $ C_n^k$?

:D

$ C_n^k\equal{}\dbinom{n}{k}$

http://en.wikipedia.org/wiki/Binomial_coefficient

Well in that case the limit is certainly $ \infty$.  The contribution of just the last term alone is $ {2^n} \to \infty$.  Surely the problem must be misstated?

sorry the problem is $ \lim_{n\to\infty}\sum_{k\equal{}0}^n\frac{2^k}{C_{2k}^k}$

The series does converge, as $ \binom{2k}{k}$ is asymptotic to $ \frac{2^{2k}}{\sqrt{\pi k}}.$

The series is comparable to $ \sum_k\frac{\sqrt{\pi k}}{2^k}.$

Now I'm looking for a probabilistic interpretation of the sum.

Numerical estimate: the sum appears likely to be exactly $ 2\plus{}\frac{\pi}2.$

$ \frac1{\dbinom{2k}k}\equal{}\int_0^\infty\frac{kx^k}{(1\plus{}x)^{2k\plus{}1}}\,dx,$ it's enough to transform the sum into an integral, remaining calculations are straightforward.

Note that $ k\equal{}0$ must be treated separately in the formula given by |-.-|.

If we interchange sum and integral, and use the fact that $ \sum_{k\equal{}1}^{\infty}kr^k\equal{}\frac{r}{(1\minus{}r)^2},$ we get

$ \sum_{k\equal{}0}^{\infty}\frac{2^k}{\binom{2k}{k}}\equal{} 1\plus{}\int_0^{\infty} \sum_{k\equal{}1}^{\infty}\frac{k2^kx^k}{(1\plus{}x)^{2k\plus{}1}}\,dx$

$ \equal{}1\plus{}\int_0^{\infty}\frac{2x(1\plus{}x)}{(1\plus{}x^2)^2}\,dx$

$ \equal{}1\plus{}\int_0^{\infty}\frac{2x}{(1\plus{}x^2)^2}\,dx\plus{} \int_0^{\infty}\frac{2x^2}{(1\plus{}x^2)^2}\,dx\equal{}1\plus{}1\plus{}\frac{\pi}2$

This verifies the numerical suggestion.

Maybe we can find the value by Telescoping Method.

Since the O.P. wants to know, how do we obtain the identity that [-.-] claims, namely

$ \frac1{\dbinom{2k}k}\equal{}\int_0^\infty\frac{kx^k}{(1\plus{}x)^{2k\plus{}1}}\,dx\ ?$

Try making the substitution $ u\equal{}\frac1{1\plus{}x},$ so that $ x\equal{}\frac1u\minus{}1.$ The integral becomes

$ k\int_0^1\left(\frac1u\minus{}1\right)^ku^{2k\plus{}1}u^{\minus{}2}\,du$

$ \equal{}k\int_0^1(1\minus{}u)^ku^{k\minus{}1}\,du$

This is an instance of the well-known [b]Beta integral.[/b] Searching for that term should probably turn up plenty of hits. The general result is that for $ \alpha,\beta>0,$

$ \int_0^1(1\minus{}u)^{\alpha\minus{}1}u^{\beta\minus{}1}\,du\equal{} \frac{\Gamma(\alpha) \Gamma(\beta)} {\Gamma(\alpha\plus{}\beta)}.$

If we particularize this to $ \alpha,\beta$ integers, we get

$ \int_0^1(1\minus{}u)^mu^n\,du\equal{}\frac{m!n!}{(m\plus{}n\plus{}1)!}$

So $ k\int_0^1(1\minus{}u)^ku^{k\minus{}1}\,du\equal{}\frac{k\cdot k!(k\minus{}1)!}{(k\plus{}k\minus{}1\plus{}1)!}\equal{} \frac{(k!)^2}{(2k)!}\equal{}\frac1{\binom{2k}{k}}.$

A re-casting of the same computation, starting more directly from the Beta integral:

$ \frac{1}{\binom{2k}{k}}\equal{}k\int_0^1x^k(1\minus{}x)^{k\minus{}1}\,dx$

$ \sum_{k\equal{}1}^{\infty}\frac{2^k}{\binom{2k}{k}}\equal{}\sum_{k\equal{}1}^{\infty} k2^k\int_0^1x^k(1\minus{}x)^{k\minus{}1}\,dx$

$ \equal{}\int_0^1\frac1{1\minus{}x}\sum_{k\equal{}1}^{\infty}2^kx^k(1\minus{}x)^k\,dx$

$ \equal{}\int_0^1\frac1{1\minus{}x}\cdot\frac{2x(1\minus{}x)}{(1\minus{}2x(1\minus{}x))^2}\,dx$

$ \equal{}\int_0^1\frac{2x}{(2x^2\minus{}2x\plus{}1)^2}\,dx$

For this integral, let $ u\equal{}2x\minus{}1,$ so that $ x\equal{}\frac12(u\plus{}1).$

$ \equal{}\int_{\minus{}1}^1\frac{2(u\plus{}1)}{(u^2\plus{}1)^2}\,du$

Split into even and odd parts; the odd part integrates to zero.

$ \equal{}4\int_{0}^1\frac1{(u^2\plus{}1)^2}\,du$

Now make the trigonometric substitution $ u\equal{}\tan\theta$ to obtain

$ \equal{}4\int_0^{\frac{\pi}4}\cos^2\theta\,d\theta \equal{} 2\int_0^{\frac{\pi}4}1\plus{}\cos 2\theta\,d\theta \equal{}\frac{\pi}2\plus{}1$

(Note that I started the series at $ k\equal{}1,$ so this agrees with the previous calculation.)

This is really the same argument that [-.-] gave; I just rearranged the details.


Jasper feels extreme gladness because today is the month of February and  later on he'll become sad again for the rest of the months unless another February comes, Will he ever be happy again, $ 6^{333333333}$ months from now???

I feel like this question doesn't make much sense.

First solution:

[hide]Yes.

$ 12 \equal{} 2 \times 6$. $ 6^{333333333} \equal{} 6^{333333331} \times 6^2 \equal{} 6^{333333331} \times 12 \times 3$.

Therefore, $ 6^{333333333}$ months from now, he will be at a February since $ 6^{333333333} \rm{mod} 12 \equiv 0$, which makes him happy.[/hide]

Second solution:

[hide]$ 6^{333333333}$ is greater than $ 12$, so there must be a February some time between now and $ 6^{333333333}$ months from now, so he will be happy sometime in the middle.[/hide]

Of course $ \boxed{YES}$, because he will have a lot of February(s) in the rest of his life and after he died by years as well  :rotfl: .

[quote="maximos"]Honeslty, I made up the question cause I find it  soooooo funny (It's still educational though)!!  wahahahha!!

By the way... "Jasper" might not enjoy $ 6^{333333333}$ months from now because he might be already dead when that $ 6^{333333333}th$ will come...

 :rotfl:[/quote]
No problem maybe he still can't be happy when is died until February come  :rotfl:


[hide]
First let's reduce it to the case $a,b,c,d > 0$. If two are negative, they cancel each other out so we may assume that at most one is negative. But in this case $4abcd$ is negative so the sum $a^2+b^2+c^2+d^2$ is greater than it is in the case in which $4abcd$ is positive. So proving it for positive $a,b,c,d$ is sufficient.

Let $x = \sqrt{abcd}$.

We have $4x \le a^2+b^2+c^2+d^2$ by AM-GM, so $4x+4x^2 \le 1$.

This simplifies to $(2x+1)^2 \le 2 \Rightarrow x \le \frac{\sqrt{2}-1}{2} \Rightarrow x^2 \le \frac{3-2\sqrt{2}}{4}$.

Hence $a^2+b^2+c^2+d^2 = 1-4x^2 \ge 1-4\left(\frac{3-2\sqrt{2}}{4}\right) = 2\sqrt{2}-2$.[/hide]


I have two stat question (don't know where to post it) :?

All bags entering a research facility are screened.  97% of the bags that contain forbidden material trigger an alarm.  15% of the bags that do not contain forbidden material also trigger the alarm.  If 1 out of every 1000 bags entering the building contains forbidden material, what is the probability that a bag that triggers the alarm will actually contain forbidden material.
a) .00097
b) .00640
c) .03000
d) .14550
e) .97000

A summer resort rents rowboats to customers but does not allow more than four people to a boat.  Each boat is designed to hold no more than 800 pounds.  Suppose the distribuiton of adult males who rent boats is normal with a mean of 190 pounds and standard deviation of 10 pounds.  if the weights of individual passengers are independent, what is the probaility that a group of four adult male passengers will exceed the acceptable weight limit of 800 pounds?
A .023
b .046
c .159
d.317
e.997

Thanks  :)

Question 1:

Let $F$ be a bag containing forbidden material, and $A$ be a bag that trips an alarm.  Then

$P(F | A) = \frac{P(FA)}{P(A)} = \frac{(0.97)(0.001)}{(0.97)(0.001) + (0.15)(0.999)} = 0.00643$,

which is answer B.

Question 2:

$z = \frac{x - \mu}{\sigma} = \frac{800 - (4)(190)}{\sqrt{(4)(10)^2}} = 2 \Rightarrow P(z \geq 2) \approx 0.0228$,

which is answer A.  If my answers aren't clear to you, post more specific questions and I'll try to address them.

[quote="quantum leap"]On this note, how is AP stat, the class?  It's the only math i haven't taken at my (loser) school (with no ap calc bc), and i was wondering what kind of challenge it'd be.  Taking Calc III concurrently with pre-cal based stat is... well, there's a gap.[/quote]

AP Stats is just easy. It's really not heavy on math; most of it in my opinion was just learning what things meant and what your calculations meant. I self-studied it about a week before the test and got a 5; admittedly, I didn't really learn much, but it tells you how hard the AP test is.

it's pretty easy, my teacher is a new one, his first time teaching ap stat, so i learned NOTHING in the class. i self studied everything.  The only hard part is trying not to get confused on what formula to use. :huh:

I've heard it's really easy, but i'm wondering if it'd just be a waste of a class.  Right now, I'm leaning towards taking it just because it's a least a math and and not an ap class which i don't like but am just taking for the credit.

[quote="paladin8"]
AP Stats is just easy.[/quote]
Oh, :oops: ouch, now I feel really dumb......

and...

[quote]Yeah and you can use your calculator on the test so just put all the formulas in and you're set[/quote]

"put the formulas in" when there are tens of long formulas..... :oops:  :roll:

[quote="chess64"][quote="zanttrang"][quote="chess64"]Yeah and you can use your calculator on the test so just put all the formulas in and you're set :lol:

I'm about to take it in like an hour.[/quote]
Except the AP booklet has all the formulae anyway, so you don't even need to do that  :P[/quote]

It doesn't have all of them. I have a beastly program that gives you the assumptions and formula for every single test/interval that you learn. And I can't remember assumptions, so it's cool :P[/quote]
Pshh, that takes all the fun out of doing hypothesis tests and confidence intervals - assumptions are the most interesting part of one of those problems!

[quote="chess64"]
It doesn't have all of them. I have a beastly program that gives you the assumptions and formula for every single test/interval that you learn. And I can't remember assumptions, so it's cool :P[/quote]

That's just cheap!!! :o  :roll:

it's considered cheating to put notes on your calculator but no one really cares.


function
algebra
polynomial
calculus
integration
combinatorics unsolved
combinatorics

Alice has two urns. Each urn contains four balls and on each ball a natural number is written. She draws one ball from each urn at random, notes the sum of the numbers written on them, and replaces the balls in the urns from which she took them. This she repeats a large number of times. Bill, on examining the numbers recorded, notices that the frequency with which each sum occurs is the same as if it were the sum of two natural numbers drawn at random from the range 1 to 4. What can he deduce about the numbers on the balls?

after some lengthy case by case split up we can say
the no's are following

URN1--------- 1,2,3,4
URN2----------1,2,3,4

or

URN1----------1,2,2,3
URN2----------1,3,3,5

or

URN1----------1,3,3,5
URN2----------1,2,2,3

and these three are only possibilities.

Translating to generating functions, the question asks for factorizations of $(1 + x + x^2 + x^3)^2$ into two polynomials with nonnegative integral coefficients and coefficient sum equal to 4.


Ivan rents a car for $ \$$25 a day and $ \$$0.20 a mile. If he rents it for 4 days and drives it 400 miles, how many dollars does he pay?

$ 25 x4 days= 100 dollars.
400 miles x$0.2 per mile= 80 dollars.

100+80=180 dollars.


Prove if $n$ is a perfect pair number then:
$\sum\limits_{\left. d \right|n} {\frac{1}{d}} = 2$

[hide="Solution"]$n$ is a number such that $f_1 + f_2 + ... + f_k = n$, where $f_k$ denotes the kth proper divisor of $n$.  If we include all integer divisors of $n$ (which includes $f_{k+1} = n$), we get $f_1 + f_2 + ... + f_k + f_{k+1} = 2n$.  Dividing this equation by $n$ gives $\frac{f_1}{n} + \frac{f_2}{n} + ... = 2$.  Notice that since $f_i | n$ for all $1 \le i \le k$, we have $\frac{f_i}{n} = \frac{1}{f_q}$ for some integer $q$ (we also note that all integer divisors of a number $n$ must occur in pairs because perfect numbers cannot be square), so we have $\sum_{i=1}^{k} \frac{1}{f_i} = 2$, the desired result.  Q.E.D.[/hide]


floor function
number theory unsolved
number theory

Let $ \{x\} \equal{} x \minus{} \lfloor x \rfloor$
Prove that for every natural number $ n$

$ \sum_{k \equal{} 1}^{n^2} \{ \sqrt {k}\} \leq \frac {n^2 \minus{} 1}{2}$

$ \sum_{k \equal{} 1}^{n^{2}}\{\sqrt {k}\} \equal{} \sum_{i \equal{} 1}^{n^2}{(\sqrt {k} \minus{} \lfloor \sqrt {k}\rfloor)}$.
$ \sum_{i \equal{} 1}^{n^2}{\lfloor \sqrt {k}\rfloor} \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}{(2i \plus{} 1)i} \plus{} n$.


Fie $M$ mijlocul laturii $(AB)$ a triunghiului $ABC$ si $N$ punctul din plan pentru care $MN\perp{AC}$ si $NB\perp{BC}$. Fie $P$ punctul de pe latura $(BC)$ astfel incat lungimea segmentului $(NP)$ este egala cu raza cercului circumscris triunghiului $ABC$. Sa se arate ca $P$ este mijlocul laturii $(BC)$.

[b][color=blue]
Sa da triunghiul $ABC$ ce este inscris in cercul cu centrul in $O$ iar $M \in \left( {AB} \right)$ este
mijlocul laturii $AB$. Fie $R \in AC$ astfel incat $MR \bot AC$ si $\left\{ Q \right\} = MR \cap BC$.
Fie $P \in \left( {BC} \right)$ astfel  incat $BP = PC$ si $N \in MR$ astfel incat $NB \bot BC$.
Sa se arate ca $NP$ este egal cu raza cercului circumscris triunghilui $ABC$
[/color][/b]
[img]http://img.photobucket.com/albums/v218/hudrea/mathlink50895.jpg[/img]
[b][u]Solutie:[/u][/b] Evident $m\left( {\widehat{RQC}} \right) = 90^0  - m\left( {\widehat{ACB}} \right)$
Se poate demonstra usor ca $B,P,O,M$ sunt patru puncte conciclice. (vezi $OM \bot AB,OP \bot BP$)
Din $MP||AC \Rightarrow m\left( {\widehat{MPB}} \right) = m\left( {\widehat{ACB}} \right)$ iar din $PM \bot QR$ (vezi $MP||AC$) si din $NB \bot BC$
rezulta ca $N,B,P,M$ sunt patru puncte conciclice si deci avem:
$m\left( {\widehat{MNO}} \right) = m\left( {\widehat{MPO}} \right) = 90^0  - m\left( {\widehat{ACB}} \right) \Rightarrow $
$ \Rightarrow m\left( {\widehat{RQC}} \right) = m\left( {\widehat{RNO}} \right) \Rightarrow NO||BC \Rightarrow NBPO$ este dreptunghi $ \Rightarrow NP = OB$


[quote="DaReaper"]If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?

is there a certain formula for this? if so, what is the solution please. :huh:[/quote]
[hide]
The maximum number of points of intersection should be 6 choose 2, because you are looking for the number of ways for 2 lines out of 6 lines to be chosen.

Also, the formula for $ nCr$ is $ \frac {n!}{r!(n \minus{} r)!}$

The formula for maximum number of points with $ n$ lines would be $ \frac {n!}{2(n \minus{} 2)!}$
[/hide]
edit-oops forgot to hide

not inner section, its intersection-crossing

[quote="i_like_pie"][quote="DaReaper"]If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?

is there a certain formula for this? if so, what is the solution please. :huh:[/quote]
What do you mean by inner section?[/quote]
it's in[b]t[/b]er section (intersection) (in-ter-sec-tion by sound)

so does anyone know how to find the number of intersections?

Oh... I took "inter section" to be a typo for "inner section" while it should have been "intersection".

[hide="Solution"]For each pair of lines, there is one intersection. The maximum number is therefore $ \binom{6}{2} \equal{} \boxed{15}$.

To guarantee that pairs always intersect in distinct points, the points can be placed in a circle. [/hide]

how does (6  2) = 15? what does (6  2)  mean?

It is combinations. Look it up in the Wiki.
Edit:[url]http://www.artofproblemsolving.com/Wiki/index.php/Combinations[/url]


hello.  kinda easy algebra problem i guess.

Find the remainder when $ P(x)\equal{}x^{2009}\minus{}2x^{1004}\plus{}3$ is divided by $ (x\minus{}1)^2(x^4\plus{}1)$

[quote="maths486"]hello.  kinda easy algebra problem i guess.

[/quote]
STOP with these subjective comments of yours, AND
they're insincere.  JUST post the problems straight.

hello, the remainder is $ 4x^5\minus{}6x^4\plus{}5x\minus{}1$
Sonnhard.

maths 486 is right, everyone has an opinion.
if u look at other posts in forums their opinionated as well
btw, how does it matter to u tan? :rotfl: :roll:

[quote="maths486"]@ Arrange your tan:

what's your problem?  people can always post their opinions in forums.  There are tons and tons of topics here where people post their opinions on things.  if you don't like my opinion, then just don't mind it.   :mad:[/quote]Although you can post your opinions, it is best that the opinion should stick with the topic in question. A remark like "kinda easy algebra" is, in my opinion (note that I'm referring to the subject), pointless. And yes, Arrange your tan has no right to claim whether they were insincere or not just because you have posted a pointless opinion. You can post a lot of pointless opinions but I doubt anyone would care except annoying AoPSers.


Let $ a,b,c,d$ real numbers, no all of them are zero. Prove that

$ \left.\frac{a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}bd\minus{}cd}{2a^2\plus{}3b^2\plus{}3c^2\plus{}d^2}\plus{}\frac{a^2\plus{}b^2\plus{}d^2\minus{}ac\minus{}ad\minus{}cd}{3a^2\plus{}2b^2\plus{}c^2\plus{}3d^2}\plus{}\frac{a^2\plus{}c^2\plus{}d^2\minus{}ab\minus{}ad\minus{}bd}{3a^2\plus{}b^2\plus{}2c^2\plus{}3d^2}\plus{}\frac{b^2\plus{}c^2\plus{}d^2\minus{}ab\minus{}ac\minus{}bc}{a^2\plus{}3b^2\plus{}3c^2\plus{}2d^2} \geq 0\right.$

When does equality hold?

REMARK
Maybe this inequality seems to be "ugly" and "unusual", but after solving it you'll see that the idea is really nice!

[quote="marin.bancos"]Let $ a,b,c,d$ real numbers, no all of them are zero. Prove that

$ \left.\frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} bd \minus{} cd}{2a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} d^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} d^2 \minus{} ac \minus{} ad \minus{} cd}{3a^2 \plus{} 2b^2 \plus{} c^2 \plus{} 3d^2} \plus{} \frac {a^2 \plus{} c^2 \plus{} d^2 \minus{} ab \minus{} ad \minus{} bd}{3a^2 \plus{} b^2 \plus{} 2c^2 \plus{} 3d^2} \plus{} \frac {b^2 \plus{} c^2 \plus{} d^2 \minus{} ab \minus{} ac \minus{} bc}{a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 2d^2} \geq 0\right.$

When does equality hold?
[/quote]
wlog $ a,b,c,d\ge0$, as replacing $ a$ by $ |a|$ etc. can only decrease the LHS.

Multiplying all by 2, we see it is equivalent to

$ \frac {(b\plus{}c\plus{}d)^2 }{2a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} d^2} \plus{} \frac { (a\plus{}c\plus{}d)^2}{3a^2 \plus{} 2b^2 \plus{} c^2 \plus{} 3d^2}$ $ \plus{} \frac {(a\plus{}b\plus{}d)^2 }{3a^2 \plus{} b^2 \plus{} 2c^2 \plus{} 3d^2} \plus{} \frac { (a\plus{}b\plus{}c)^2}{a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 2d^2} \le4$.

Now it's just about cutting correctly. By C-S,

$ LHS\le \frac {b ^2}{a ^2\plus{}b ^2\plus{}c^2}\plus{} \frac {c ^2}{a ^2\plus{}b ^2\plus{} c^2}\plus{} \frac { d^2}{b^2\plus{} c^2\plus{}d ^2}$ $ \plus{} \frac {a ^2}{a^2\plus{}b^2\plus{}d ^2}\plus{} \frac {c ^2}{ a^2\plus{}c^2 \plus{}d^2}\plus{} \frac {d ^2}{a^2\plus{}b^2\plus{}d ^2}$

$ \plus{} \frac {a ^2}{a^2\plus{}c^2\plus{}d ^2}\plus{} \frac {b ^2}{ a^2\plus{}b^2 \plus{}d^2}\plus{} \frac {d ^2}{a^2\plus{}c^2\plus{}d ^2}$  $ \plus{} \frac { a^2}{a ^2\plus{}b ^2\plus{}c^2}\plus{}\frac {b ^2}{b^2\plus{} c^2\plus{}d ^2}\plus{} \frac {c ^2}{b^2\plus{} c^2\plus{}d ^2}\equal{}4.$  :lol: Did it.

Equality holds iff $ a\equal{}b\equal{}c\equal{}d\neq0$.

Thank you for your solutions to this "ugly" inequality. Both proofs follow the same idea of solving. I made this problem using this idea!
Dear [b]bakerbakura[/b], you are a little ironical! You know very well that the content of that lemma is something else. Of course, you've used Titu's lemma (or Cauchy-Engel, or Cauchy-Schwarz in a particular form), and this result could be considered as an application of it (see at http://www.mathlinks.ro/viewtopic.php?t=310550 ). But this doesn't mean that I claim you to admit that this result is my result.

So I guess I took a bit too long to type out my solution... :blush: [b]spanferkel[/b] got there before me...

When I used the result above I actually took $ n \equal{} 1$ in your lemma, so [b]technically[/b] it could be considered an application of your lemma. :P


[quote="phatltv"]a,b,c>=0
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \ge 4\sqrt [3]{{abc\left( {a^2 \plus{} bc} \right)\left( {b^2 \plus{} ca} \right)\left( {c^2 \plus{} ab} \right)}}
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262971


We at Art of Problem Solving are pleased to offer two books from the creator of MOEMS, Dr. George Lenchner.  These books are [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=4]Creative Problem Solving in School Mathematics[/url] and [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=5]Math Olympiad Contest Problems for Elementary and Middle Schools[/url].

Is AoPS going to also offer some previous tests?

I'm a teacher from Romania. How can I order these books?

Im ordering them once I finish my book!!

How does [i]Creative Problem Solving[/i] by Lenchner compare to AoPS's [i]Prealgebra[/i]? I am interested in the Lenchner book but if it is easier than [i]Prealgebra[/i] or covers the same material then maybe I don't need it since I'm already working through [i]Prealgebra[/i].

What are other books, can you please kindly mention


This question is perhaps not so hard, but I'm putting it in this forum because it requires calculus.  I'd like to see as many different solutions as possible.

Consider any everywhere-differentiable function, $f: \mathbb{R} \to \mathbb{R}$.  Prove that there exists a number $a$ such that $-a = f(a)\cdot f'(a)$.

Let g(x) = x 2  + f(x) 2.g is differentiable, continuous, so since g(x) --> infinity as x ---> +/-  infinity (since g(x)  >= x^2 for all x),  we can find c, d distinct such that g(c) = g(d), then by Rolle's Theorem, g'(x) = 0 for some a between c and d, that is,

2 f'(a)•f(a) + 2a = 0 for some a. 2

I found the same solution.  Does anyone have a substantially different one?

Well, I know of at least one other solution, but it's considerably less nice -- roughly, though, the idea is that f*f' cannot always be < -x when x > 0 and it cannot always be > -x when x < 0, so it must be equal to -x at some point.  The solution you both found was essentially the way I came across the problem, but I'm curious if there are other possible solutions.

*Random thought: if we assume that f has a convergent Taylor expansion, could that be used to prove it?  (Not that it would be a [i]nice[/i] proof, but I'm just curious about other ways to solve the question -- not necessarily other nice ways.)

That solution resembles very much a Putnam problem from not so far back.  I'll try to find it.

EDIT: It was 1999 B4, and they're actually not that similar.


[quote="Myth"]I noted that some special chars (like leq or x symbols) replaced with ? in old topics.[/quote]There should be no special chars to represent mathematic symbols. All should be translated into $\LaTeX$ :)

[quote="Myth"]Indeed, but I said about old topics (before April 2004).[/quote]Where possible and in their time limits moderators should edit those old topics to make them $\LaTeX$ compatible.

If you are not sure what the old character was, just change the encoding of the page from UTF-8 to Western (ISO-8859-1) and you will see what they are.

I am solving April problems now, so I think I will meet "broken" characters  not too often :)


[color=blue]A solid, homogeneous ball with radius $R$. Before falling to the floor its center of mass is at rest, but the ball is spinning with angular velocity $\omega_0$ about a horizontal axis through its centre. The lowest point of the ball is at a height $h$ above the floor.
When released, the ball falls under gravity, and rebounds to a new height such that its lowest point is now $ah$ above the floor. The deformation of the ball and the floor on impact may be considered negligible. Ignore the presence of the air. The impact time, although small, is finite.
The mass of the ball is $m$, the acceleration due the gravity is $g$ the dynamic coefficient of friction between the ball and the floor is $\mu$, and the moment of inertia of the ball about the axis which goes through its centre is $\large I=\frac{2}{5}mR^2$
Find the minimum value of $\omega_0$ for the situation that the ball slips during the entire impact time.[/color]

N = vertical force applied by the floor
t = impact time

Limit condition:
$\frac{2mR^2 \omega_0}{5}=fRt$
$\omega_0=\frac{5\mu}{2R} \cdot Nt$

From momentum conservation (i think momentum is the right word; exscuse me if i'm wrong):
$Nt=m(\sqrt{2gh}+\sqrt{2gha})$

so i get $\omega_0 \geq \frac{5\mu}{2R} \sqrt{2gh} (1+\sqrt{a})$

Is it correct?

bacco , sorry your solution is wrong

answer:

$\frac{7muy\sqrt{2gh}(1+\sqrt{a})}{2R}$

hard and nice one

phucnv87: have you had medal in IPHO? :D

sincerely

Cuong,
could you write out how you derived that?   :)

cuong, you went to IPhO? If you did, I guess I'll see you there this year since I'm gonna be on the US Team. ha ha, jk.

yeah, bacco, you forgot to factor in that that no slip conditions occur if v > Wr. IE, if f/m(t) > WoR - (alpha)RT. In your equation, i think you assumed that W would be 0 after time t.

[color=blue]Here is the solution
The velocity of the ball's centre right before the collision
$v_0=\sqrt{2gh}$
Let $v_{2x}$ and $v_{2y}$ are the horizontal and vertical components of the velocity of the ball's centre right after the collision
$v_{2y}=-\sqrt{2gah}$
The reduction of the momentum in vertical direction
$mv_0-mv_{2y}=m(1+\sqrt{a})\sqrt{2gh}=\int_{t_1}^{t_2}N(t)dt$
The augmentation of the mometum in horizontal direction
$mv_{2x}=\int_{t_1}^{t_2}f(t)dt$
The reduction of the momentum moment
$I(\omega_0-\omega_2)=R\int_{t_1}^{t_2}f(t)dt$
where $\omega_2$ is the angular velocity after the collision
And we also know that
$f=\mu N$
From the above equations, we can infer that
$\omega_2=\omega_0-\frac{1}{I}R\mu m(1+\sqrt{a})\sqrt{2gh}$
In order to satisfy the condition of the problem, we must have
$\omega_2R\geq v_{2x}$
$v_{2x}=\mu(1+\sqrt{a})\sqrt{2gh}$
$\Longrightarrow \omega_o\geq\frac{7\mu\sqrt{2gh}(1+\sqrt{a})}{2R}$[/color]
[color=green][b]Cuong[/b] I haven't taken part in any IPhO [/color]

Ok, I hadn't correctly understood the text. I thought that before and after the collision the horizontal speed should have been 0.

can we consider the normal force to be constant if there is no deformation of the surfaces?


Let $d(n)$ denote the number of positive divisors of a positive integer $n$. Prove that the sequence defined by $u_n = d(n^2 + 1)$ does not become strictly montonic from any point onwards.

oh! it reminds me the problem from vojtech jarnik
http://prf.osu.cz/kma/dokumenty/vjaimc/j13problems1.html

It is Leningrad TST'98 (11 grade) for All-Russian MO.

Let it \[ {f(n)=d(n^2-1)}. \]
Suppose that exists $n_0$ such that
\[{n_0 \geq n \Rightarrow f(n) < f(n+1)}\]
There is a serie of lemmas which will be very useful in this proof (prove it!):

LEMMA 1: \[ f(n+k) \geq f(n)+2k\] for all $n_0 \geq n$ and k.
LEMMA 2: if n is even, then $f(n) \leq n$.

Now take k such that $k > n_0 - f(n_0)$ and $n_0+ k$ is even.
Then \[{ n_0+k \geq f(n_0+k) \geq f(n_0)+2k  }\]
or $k \leq n_0 - f(n_0)$, contradiction!
And this is the end!


A two-pan balance scale comes with a collection of weights.
Each weight weighs a whole number of grams. Weights can be
put in either or both pans during a weighing. To ensure any
whole number of grams up to 100 grams can be measured, what
is the minimum number of weights needed in the collection?

Thank You!

I am not that sure if this is right, but powers of 2 accomplish everything. So I would say from 2^0 to 2^6 which 7 weights. Not sure if theirs anything else, cause powers of 3 do not work until 40.

Sorry, but the answer is 5. I kept getting 6; I just did weights of 50, 25, 12, 6, 3, 2

Thanks anyways! :lol:

[hide="Perhaps..."]
Powers of 3?  Not sure...
[/hide]

Yes. Powers of Three works. Answer is thus $ \boxed{5}$ as FullMetalNaruto said.

So do you just guess and check, or is there some kind of formula you have to memorize? :huh: [/i]

Sorry to revive, buy you set up the equation $3^x>100$. Now we want the greatest possible value of $x$ so $x=\boxed{5}$. Why we use powers of $3$, I do not know.

This is in another thread already that is pretty much right below your thread. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=363659[/url]
It's called "Least Number of Weights Needed to Weigh up to 100 Grams".


function
induction
logarithms
algebra
polynomial
calculus
calculus computations

Consider the function $ f(x) \equal{} x\ln(x)$, apply the Leibniz formula to the identity $ xf'(x) \minus{} f(x) \equal{} x$ on $ x\in(0,\infty)$ in order to compute the Taylor series of f(x) about the point $ 1$.

What will happen to the $ x$ in the identity? :maybe:

Can someone please help. Thank you very much. :P

[quote="bookworm_vn"]how do you mean?[/quote]

I would want to know how to compute the Taylor series of $ f(x)$ about point $ 1$ with the Leibniz formula ([url]http://en.wikipedia.org/wiki/Leibniz_rule_%28generalized_product_rule%29[/url]).  :)

Can someone please help. Thank you. :P

You're assuming we can write $ f(x)\equal{}\sum_{n\equal{}0}^{\infty}a_n(x\minus{}1)^n.$

Let's rewrite $ xf'(x)\minus{}f(x)\equal{}x$ as $ [1\plus{}(x\minus{}1)]f'(x)\minus{}f(x)\equal{}1\plus{}(x\minus{}1)$

$ f'(x)\equal{}a_1\plus{}2a_2(x\minus{}1)\plus{}3a_3(x\minus{}1)^2\plus{}4a_4(x\minus{}1)^3\plus{}\cdots\plus{}
(n\plus{}1)a_{n\plus{}1}(x\minus{}1)^n\plus{}\cdots$
$ (x\minus{}1)f'(x)\equal{}0\plus{}a_1(x\minus{}1)\plus{}2a_2(x\minus{}1)^2\plus{}3a_3(x\minus{}1)^3\plus{}\cdots\plus{}
na_n(x\minus{}1)^n\plus{}\cdots$
$ \minus{}f(x)\equal{}\minus{}a_0\minus{}a_1(x\minus{}1)\minus{}a_2(x\minus{}1)^2\minus{}a_3(x\minus{}1)^3\minus{}\cdots\minus{}a_n(x\minus{}1)^n\minus{}\cdots$

Adding up the columns and equating this series to $ 1\plus{}(x\minus{}1),$ we get a chain of equations:

$ a_1\minus{}a_0\equal{}1$
$ 2a_2\equal{}1$
$ 3a_3\plus{}a_2\equal{}0$
$ 4a_4\plus{}2a_3\equal{}0$
$ \dots$
$ (n\plus{}1)a_{n\plus{}1}\plus{}(n\minus{}1)a_n\equal{}0.$

This chain of equations does not determine $ a_0,$ but we know that from other sources. Since $ f(1)\equal{}0,$ we have $ a_0\equal{}0.$ Then:

$ a_1\equal{}1$
$ a_2\equal{}\frac12$
$ a_3\equal{}\minus{}\frac13a_2\equal{}\minus{}\frac16$
$ a_4\equal{}\minus{}\frac24a_3\equal{}\frac1{12}\equal{}\frac{2!}{4!}\equal{}\frac1{3\cdot4}$
$ a_5\equal{}\minus{}\frac35a_4\equal{}\minus{}\frac1{20}\equal{}\minus{}\frac{3!}{5!}\equal{}\frac{(\minus{}1)^5}{4\cdot 5}$

The recursion $ a_{n\plus{}1}\equal{}\minus{}\frac{n\minus{}1}{n\plus{}1}\,a_n$ and an induction can be used to find the general formula

$ a_n\equal{}\frac{(\minus{}1)^n}{(n\minus{}1)n}$ for $ n\ge2.$

$ x\ln x\equal{}(x\minus{}1)\plus{}\sum_{n\equal{}2}^{\infty}\frac{(\minus{}1)^n}{(n\minus{}1)n}\,(x\minus{}1)^n.$

This series can be seen to converge on $ [0,2],$ and thus includes the continuous extension of $ f(x)$ to having a value at $ 0.$ Evaluation at the other end produces this formula for $ \ln 2:$

$ \ln 2\equal{}\frac12\left(1\plus{}\frac12\minus{}\frac1{2\cdot3}\plus{}\frac1{3\cdot4}\minus{}
\frac1{4\cdot5}\plus{}\cdots\right).$

[There were easier ways to get this particular series; this is really the general technique for finding series solutions to a differential equation (preferably a polynomial-coefficient linear differential equation) about an "ordinary point."]


I've always wondered this.  How do you become a mod?  Is there a way to qualify, or did the admins just choose people who they thought were skilled at problem-solving?

[quote="minicon"]You don't qualify.  When a forum needs a moderator, the admins will decide who is a good, active user of that forum.  Then they go into the Administrator Control Panel and add them to the Moderator group.  For example, the Mathcounts forum has a lot of posts per day, so they must have someone watching it.  Basically the only things you get as a mod are ability to lock, move, delete and split topics.  They might be able to ban users, but I am not sure.[/quote]

Most importantly they can post more than 3 smilies per post and have no 10 minute double posting restrictions.  :rotfl:  :rotfl:

[quote="ZetaX"]By using the search function.[/quote]

 :rotfl:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186114]Here[/url] is where you can get more information.

[quote="madness"]Yes and some mods like 'x' can resemble hitler n can irritate you a lot :D :rotfl: and do whatever cyber shreyas said :D[/quote]
Please explain...

that was a friendly jibe at my friend who's a mod :D
Sorry but i didnt want to post his name so that it doesnt affect him if people take that seriously :)


function
Euler
integration
algebra
partial fractions
complex analysis
calculus

How do you do sums like $ \sum_{k\equal{}0}^{\infty}\frac{1}{4k^2\plus{}1}$ or $ \sum_{k\equal{}0}^{\infty}\frac{1}{4k^2\plus{}4k\plus{}1}$?

First: Does the denominator factor, so the sum telescopes (by partial fractions, even)?
That's a good sign that other trickery can be used.

Telelscopic sum will apply for summation $ \sum_{k \equal{} 0}^\infty \frac {4k}{4k^4 \plus{} 1}$

$ 4k^4 \plus{} 1 \equal{} 4k^4 \plus{} 4k^2 \plus{} 1 \minus{} 4k^2 \equal{} (2k^2 \plus{} 1)^2 \minus{} (2k)^2 \equal{} (2k^2 \plus{} 2k \plus{} 1) (2k^2 \minus{} 2k \plus{} 1)$

So $ \frac {4k}{4k^4 \plus{} 1} \equal{} \left [ \frac {1}{2k^2 \minus{} 2k \plus{} 1} \minus{} \frac {1}{2k^2 \plus{} 2k \plus{} 1} \right ]$

Note that $ 2k^2\plus{}2k\plus{}1 \equal{} 2(k\plus{}1)^2 \minus{} 2(k\plus{}1) \plus{} 1$ and so a telescopic sum has been set up.

Sometimes it's also helpful to just look at the expansion and play with the numbers. Comparing it with other different series, even those better understood, may help.

[hide="Answer for second sum"] $ \sum_{k \equal{} 0}^{\infty}\frac {1}{4k^2 \plus{} 4k \plus{} 1} \equal{} \sum_{k \equal{} 0}^{\infty}\frac {1}{(2k \plus{} 1)^2} \equal{} 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} \frac {1}{9^2} \plus{} ...$

It's clear the above is the sum of the reciprocals of the squares of the odd natural numbers.

This looks familiar to the Reimann Zeta Function: $ \zeta(2) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n^2} \equal{} 1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \equal{} \prod_{p}\left(\frac {1}{1 \minus{} \frac {1}{p^2}}\right) \equal{} \frac {\pi^2}{6}$

We can also do this (inspired from [url=http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function]a proof of Eulers Identity[/url]):
\[ \frac {1}{2^2} \plus{} \frac {1}{4^2} \plus{} \frac {1}{6^2} \plus{} ... \equal{} \frac {1}{2^2}\zeta(2)
\]

\[ \left(1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \right) \minus{} \left(\frac {1}{2^2} \plus{} \frac {1}{4^2} \plus{} \frac {1}{6^2} \plus{} ...\right) \equal{} \zeta(2) \minus{} \frac {1}{2^2}\zeta(2) \equal{} \left(1 \minus{} \frac {1}{2^2}\right)\zeta(2)
\]

\[ 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} ... \equal{} \left(1 \minus{} \frac {1}{2^2}\right)\zeta(2)
\]
Therefore
\[ \sum_{k \equal{} 0}^{\infty}\frac {1}{4k^2 \plus{} 4k \plus{} 1} \equal{} 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{}\frac{1}{9^2}\plus{} ... \equal{} \boxed{\frac {\pi^2}{8}}
\]
[/hide]

[hide="First sum"]$ \sum_{k\equal{}1}^{\infty}\frac{1}{4k^2\plus{}1} \equal{} \frac{1}{4}\sum_{k\equal{}1}^{\infty}\frac{1}{k^2\plus{}\left(\frac{1}{2}\right)^2}$.

Now, if I am in no mistake, $ \sum_{k\equal{}1}^{\infty}\frac{1}{k^2\plus{}a^2} \equal{} \frac{\pi a \cosh(\pi a)\minus{}1}{2a^2}$

which can be proved by using Complex Analysis or Fourier Series, for example. Now just consider a = 1/2.[/hide]

[quote="Patterns_34"]Sometimes it's also helpful to just look at the expansion and play with the numbers. Comparing it with other different series, even those better understood, may help.

[hide="Answer for second sum"] $ \sum_{k \equal{} 0}^{\infty}\frac {1}{4k^2 \plus{} 4k \plus{} 1} \equal{} \sum_{k \equal{} 0}^{\infty}\frac {1}{(2k \plus{} 1)^2} \equal{} 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} \frac {1}{9^2} \plus{} ...$

It's clear the above is the sum of the reciprocals of the squares of the odd natural numbers.

This looks familiar to the Reimann Zeta Function: $ \zeta(2) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n^2} \equal{} 1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \equal{} \prod_{p}\left(\frac {1}{1 \minus{} \frac {1}{p^2}}\right) \equal{} \frac {\pi^2}{6}$

We can also do this (inspired from [url=http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function]a proof of Eulers Identity[/url]):
\[ \frac {1}{2^2} \plus{} \frac {1}{4^2} \plus{} \frac {1}{6^2} \plus{} ... \equal{} \frac {1}{2^2}\zeta(2)
\]

\[ \left(1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \right) \minus{} \left(\frac {1}{2^2} \plus{} \frac {1}{4^2} \plus{} \frac {1}{6^2} \plus{} ...\right) \equal{} \zeta(2) \minus{} \frac {1}{2^2}\zeta(2) \equal{} \left(1 \minus{} \frac {1}{2^2}\right)\zeta(2)
\]

\[ 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} ... \equal{} \left(1 \minus{} \frac {1}{2^2}\right)\zeta(2)
\]
Therefore
\[ \sum_{k \equal{} 0}^{\infty}\frac {1}{4k^2 \plus{} 4k \plus{} 1} \equal{} 1 \plus{} \frac {1}{3^2} \plus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} \frac {1}{9^2} \plus{} ... \equal{} \boxed{\frac {\pi^2}{8}}
\]
[/hide][/quote]
I can't believe I missed that

By the way, is it true that the alternating version can't be done?

Try testing it with the alternating series test.. Does it converge or diverge?

[quote="0714446459923"]Well obviously it converges. What I'm asking is whether it can be put in closed form or not.[/quote]

it is called catalan's constant, no closed form is known as of yet.  as a matter of fact,  you should know that all of

$ 1\plus{}\dfrac{1}{2^3} \plus{}\dfrac{1}{3^3} \plus{}\dfrac{1}{4^3} \plus{}\dfrac{1}{5^3} \plus{} \ldots$

$ 1\plus{}\dfrac{1}{2^5} \plus{}\dfrac{1}{3^5} \plus{}\dfrac{1}{4^5} \plus{}\dfrac{1}{5^5} \plus{} \ldots$

$ 1\plus{}\dfrac{1}{2^7} \plus{}\dfrac{1}{3^7} \plus{}\dfrac{1}{4^7} \plus{}\dfrac{1}{5^7} \plus{} \ldots$
etc.

and

$ 1\minus{}\dfrac{1}{3^2} \plus{}\dfrac{1}{5^2} \minus{}\dfrac{1}{7^2} \plus{}\dfrac{1}{9^2} \minus{} \ldots \plus{}\;\minus{} \ldots$

$ 1\minus{}\dfrac{1}{3^4} \plus{}\dfrac{1}{5^4} \minus{}\dfrac{1}{7^4} \plus{}\dfrac{1}{9^4} \minus{} \ldots \plus{}\;\minus{} \ldots$

$ 1\minus{}\dfrac{1}{3^6} \plus{}\dfrac{1}{5^6} \minus{}\dfrac{1}{7^6} \plus{}\dfrac{1}{9^6} \minus{} \ldots \plus{}\;\minus{} \ldots$
etc.

do not have any known closed forms.

[hide="First sum"]Actually we have $ S \equal{} \sum_{k\equal{}0}^{\infty}\frac{1}{4k^2\plus{}1} \equal{} 1 \plus{} \frac{1}{4} \sum_{k\equal{}1}^{\infty}\frac{1}{k^2 \plus{} \left(\frac{1}{2}\right)^2}$

and $ \sum_{k\equal{}1}^{\infty}\frac{1}{k^2 \plus{} a^2} \equal{} \frac{\pi a \coth(\pi a) \minus{} 1}{2a^2}$. Therefore, by considering a = 1/2 we get

$ S \equal{} \frac{\pi \coth\left(\frac{\pi}{2}\right) \plus{} 2}{4}$.[/hide]

[quote="Carcul"]$ \sum_{k \equal{} 1}^{\infty}\frac {1}{k^2 \plus{} a^2} \equal{} \frac {\pi a \coth(\pi a) \minus{} 1}{2a^2}$[/quote]

How do you prove this?

Here is one wave to prove this identity

Recall $ Sin(z) = z\prod_{n=1}^{\infty}\left({1 - \frac {z^2}{n^2\pi^2}}\right)$ .
We then take natural logs of both sides to find $ Log(Sin(z)) = Log(z) + \sum_{n=1}^{\infty}Log\left({1 - \frac {z^2}{n^2\pi^2}}\right)$ .
Differentiating both sides with respect to z gives us   $ Cot(z) = 1/z + \sum_{n=1}^{\infty}\frac{-\frac{2z}{n^2\pi^2}}{1 - \frac {z^2}{n^2\pi^2}}$
Rearranging we find $ Cot(z) = 1/z - \frac{2}{\pi^2}\sum_{n=1}^{\infty}\frac{z}{n^2 - \frac {z^2}{\pi^2}}$ Letting $ z = i\pi\phi$
we find that ${ Cot(i\pi\phi)= -iCoth(\pi\phi) = \frac{1}{i\pi\phi} - \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{i\phi}{n^2 + \phi^2}}$
Simplifying, one finds ${ \frac{\pi Coth(\pi\phi)}{2\phi} = \frac{1}{2\phi^2} + \sum_{n=1}^{\infty}\frac{1}{n^2 + \phi^2}}$ or
${ \frac{\pi\phi Coth(\pi\phi)- 1} {2\phi^2} = \sum_{n=1}^{\infty}\frac{1}{n^2 + \phi^2}}$ as desired

There are also Fourier series/Fourier transform methods based on the Poisson summation formula. Here's a possible starting point:
\[ \frac{1}{k^2\plus{}a^2}\equal{}\frac1{2a}\int_{\minus{}\infty}^{\infty}e^{\minus{}a|x|}e^{ikx}\,dx\]

Or you can consider the Fourier series for the function f(x) = cosh(ax).


After seeing [b]Altheman[/b]'s mock ARML individual round problems, I felt like making my own; but ARML is fast approaching, and I both do not have a full 8 problems and am not sure if these would be fair questions (and I'm not sure, as well, of the quality of these problems). :P So, I am just going to present them here for any who would like some practice (try to stay within the 10 or so minute time restriction, if you'd like :) )

[hide="Sheet 1"]
$(1)$ The equation $\sqrt{ 2x^{2}+12-2 x \sqrt{x^{2}+12}}=2x+\sqrt{x^{2}+12}$ has at most two real roots. Find all of its real roots, or write "$\text{No Real Roots}$" if there are none.

$(2)$ Let $\phi^{2}-\phi-1=0$ and $\phi > 0$. Evaluate $\phi^{12}-\frac{72}{\phi^{3}}$.[/hide]

[hide="Sheet 2"]
$(3)$ Find the sum of all $x$ such that $0 \leq x < 100$ and $23$ divdes $P(x) = x^{4}-49x^{3}+72x^{2}-24x+299$.

$(4)$ Evaluate $\dbinom{90}{0}+\dbinom{90}{6}+\dbinom{90}{12}+\ldots+\dbinom{90}{90}$.[/hide]

[hide="Sheet 3"]
$(5)$ $N$ is the largest integer less than $5000$ that has exactly $12$ positive divisors, including $1$ and itself. Find $N$.

$(6)$ The region $\mathcal{S}$ in the complex plane contains all $z$ such that:
\[\frac{30}{ |z| }= \frac{ 50}{ |z+1| }\ \text{ and }\ |z| \leq 3 \]
Find the area of $\mathcal{S}$.
[/hide]

[quote="Altheman"]Yay you have over a thousand posts...and as for the arml problems:

[hide="Comments:"]

Overall, it is good

1. denesting is your favorite...this is fine...i am not sure who would put no reals solutions as 0 is obviously a solution...

2. is too computational...like i know you can just divide, but that takes a while with 12 degree...it is fine conceptually

3. eww for that polynomial...i guess it is okay...this is a bit contrived....

in mod 23...x^4-3x^3+3x^2-x=(x)(x-1)^3

4. not reasonable. basically you have seen the problem or you have not. it is a nice problem and all...but it shows up way to often on aops

5. ok...i am not sure how you do this quickly...p^2q^3 or pqr^2...i believe that 2^2*29*43=4988 works...but like who is going to see that...

6. good ...it gets kind of messy

2,6 seem familiar [from aime?]...which is fine for practice, but i doubt they would show up on an actual contest[/hide]

also this could get lost pretty easily...the topic title is not really eyecatching[/quote]

Lol, yay, over a thousand. :D

[hide="Stuffs (for Alex, mostly)"]
Actually, for #1 i had meant to do the $f(x)= x \implies f(f(x))=x$ thing... i think i might've copied down the problem wrong. i'll look... but w/e.

And merr, i really like the idea behind #4. But yea, you're probably right, not that many ppl who haven't seen this before would be like, BAM, roots of unity filter.

Yea, I think #6 is similar to that one where its like 40/|z| = |z|/50 or something, where like there are two circles... idk. It's pretty similar. And #2, i think, is also similar to one that has the same idea in mind.  :)[/hide]

[hide="I am actually hiding my posts for once"]1 is $|x-\sqrt{x^{2}+12}|=2x+\sqrt{x^{2}+12}$

so i don't know where that idea comes in...

also after solving it, the only solution is x=0...so you need a frustum...it is too guessable...ya kno...step your game up
sam, i am curious, what was your solution to 5...[/hide]

Is the answer to the last one
(225/256)pi?


Find all $a,b \in \mathbb{N}$ such that $a^3=1+2+3+...+b-1+b$.

It is equavalent to solution elliptic equation:
$2x^3=y(y+1).$ Solutions may be find elementary metods. Since $(y,y+1)=1, or \ y=2z^3,y+1=t^3, or \ y=z^3,y+1=2t^3,$ It is equavalent to equation Fermats vieve $t^3-2z^3=-1,$ with have only trivial solutions $z=1,t=1\Longrightarrow x=1,y=1.$


If to the numerator and denominator of the fraction $ \frac{1}{3}$ you add its denominator, 3, the fraction will double.
Find a fraction which will triple when its denominator is added to its numerator and to its denominator; find one that will quadruple.

[i]from book "The Moscow puzzles" by Boris A. Kordemsky[/i]

[u]Please hide Your solution . Write inside Hide ... /Hide[/u]

[hide="Solution"]

$ \frac {1}{5}$ and $ \frac {1}{7}$

[/hide]

[hide="Generic Solution"]
Solving it generically, we start with a simple equation for fraction p/q and a number n for which it should multiply when the denominator is added to both the numerator and denominator:

$ \frac{p\plus{}q}{2q} \equal{} \frac{np}{2}$

This immediately simplifies

$ \frac{p\plus{}q}{2} \equal{} np \implies p\plus{}q \equal{} 2np$

Now we have

$ q \equal{} 2np \minus{} p \equal{} p(n\minus{}1)$

This allows us to choose any $ p$ and $ n$ to find the necessary denominator.

Example for 3: $ q \equal{} p(2n \minus{} 1) \equal{} 1(2 \cdot 3 \minus{} 1) \equal{} 6 \minus{} 1 \equal{} 5$, so the fraction is $ \frac{1}{5}$

Example for 4: $ q \equal{} p(2n \minus{} 1) \equal{} 1(2 \cdot 4 \minus{} 1) \equal{} 8 \minus{} 1 \equal{} 7$, so the fraction is $ \frac{1}{7}$

Note that we can select any number we choose for $ p$, and using $ 1$ makes everything evaluate quickly. There are infinitely many examples for each $ n$, but we wish to use one that reduces the amount of calculations. If we assume $ p$ as 1, we get

$ q \equal{} 2n \minus{} 1$

Which is easy to use.
[/hide]

[quote="mathepic"]
Solving it generically, we start with a simple equation for fraction p/q and a number n for which it should multiply when the denominator is added to both the numerator and denominator:

$ \frac {p \plus{} q}{2q} \equal{} \frac {np}{2}$

This immediately simplifies

$ \frac {p \plus{} q}{2} \equal{} np \implies p \plus{} q \equal{} 2np$

Now we have

$ q \equal{} 2np \minus{} p \equal{} p(n \minus{} 1)$

[/quote]

what did you do here?

$ \frac {p \plus{} q}{2q} \equal{} \frac {np}{2} \implies \frac {p \plus{} q}{q} \equal{} np\implies p \plus{} q \equal{} npq$

do you mean $ \frac{np}{q}$ for your second fraction?

Yes, it should be $ \frac {np}{q}$.

Any more of these questions,small_unicorn?

[hide="Solution"]
OK. Let x = the numerator and y equal the denominator.

$ x/y$(3)=(x+y)/(2y)
$ y(x\plus{}y)$=2y(3x)
$ x\plus{}y\equal{}6x$
y=5x

So that would mean our answer is 1/5. [/hide]


A circle is divided in $k$ sectors and a pawn is put in each sectors.
With each move, two pawns are moved, in two different direction, to their neighbour sectors. Is there a sequence of moves where, in the end, all pawns are in the same sectors?

For odd k it's obvious that if u chose a target sector and in each move put two pawns that are in the same distance from it 1 sector closer to it, in the end all pawns will end up in the target sector.
For even k it is not possible.

[quote="Kondr"]
For even k it is not possible.[/quote]

To prove this, notice that both arcs which separate two pawns with two sectors between them are even, no matter what moves we perform, so the two pawns could never end up on the same sector.


geometry
circumcircle
power of a point
radical axis
geometry solved

Given a quadrangle $ABCD$ with $AB|| CD$, $F$ is a point on $AB$ satisfying $FC=FD$, $AC \cap BD = E$. Let $O_1,O_2$ are the circumcenters of $AFD$ and $BFC$ respectively.
Prove that $O_1O_2$ is perpendicular to $EF$.

A minute before I thought that I solved this problem but I has just found a stupid mistake :(

[i]Edited by Myth[/i]

Let circles $O_1$ and $O_2$ intersect AC and BD at P and Q respectively. We will show DCPQ is cyclic.
$\angle ABD+\angle BAC=\angle EDC+\angle FDP=\angle FDC+\angle QDP$. Similarly, $\angle ABD+\angle BAC=\angle FCD+\angle PCQ$. But since FD=FC, $\angle FDC=\angle FCD$. It follows that $\angle QDP=\angle PCQ$. Therefore $D,C,P,Q$ are concyclic. Note that this implies $A,B,P,Q$ are also concylic. Consider the radical axis formed by the circumcircle of ABPQ, $O_1,O_2$. Since F lies on both AP and BQ, it also lies on the radical axis between $O_1,O_2$. The result follows since the radical axis is perpendicular to the line joining the centers.

Nice! :)
I also tried to use radical axes idea, but I couldn't invent proper circles :(

My new solution is depends on yours, mecrazywong, but it’s simplier than yours  :D
First we have $\angle APD=\angle AFD$
	          $\angle BQC=\angle BFC$
But $\angle AFD=\angle BFC$ so $\angle APD=\angle BQC$ => $\angle DQC=\angle DPC$ => DCPQ is concyclic.
And it’s easy to get ABPQ is concyclic too.
Now we have $EP.EA=EQ.EB$ and we get the result
 :)


algebra
function
domain
calculus
calculus computations

Let $ A$ be a set and $ f$ a continuous function. Find if the following statement is true or false: If  $ f(A)$ is a closed interval and has more than one element, then $ A$ is a closed interval.

I think that the statement is false. For example $ f: (\minus{}4\pi,4\pi)\to{R},f(x)\equal{}sinx$ has $ f(A)\equal{}[\minus{}1;1]$ which is a closed interval and the domain isn't

Then if $ f(A)$ is simply an interval that has at least one element, means that $ A$ is an interval?

Heh no. For $ f$ above we can adjoin any set to the interval.


Kindly note these rules and regulations before posting in this forum. These rules will be strictly implemented.

1) The Advanced Physics forum is strictly for [color=red]High School - AP(Calculus based) , College, Olympiad or Research level [/color]Physics discussion. Esoteric , but popular topics like Relativity, String Theory can be discussed here.

2) The Problems to be posted here MUST be accompanied by what the poster has done about it, how much headway he/she has made and what specific help he/she expects. Problem posted should be such that they require thought and not just application of a formula. [color=violet]Calculus[/color] based problems should be posted here. Kindly note that you should not expect your problem to be fully solved for you. [color=orange]NO Homework problems [/color]will be solved here. Those problems requring simple application of formulae should be posted in the [url=http://www.mathlinks.ro/Forum/index.php?f=405]Introductory Physics Forum[/url].

3) Discussion on [color=blue]Past Olympiad [/color]problems, details of Olympiad procedures etc. are welcome. But again, no Olympiad problem will be fully solved on this forum for anyone.

4) If you need clarification of topics taught in higher College Physics classes or want to present your ideas on advanced topics, please do so here. [color=green]Theoretical discussions [/color]on advanced topics of research are welcome. But keep in mind that since few of the members on this forum are engaged in Physics research, you might not find your answers here. You are directed to [url=http://www.physicsforums.com]www.physicsforums.com[/url]for more advanced help.

For more information or clarification, kindly contact the Moderator. Further Suggestions are welcome.

Thank you.

IMPortant:
DO hide all ur answers,Hints and suggestions that u post to a problem so that evryone  gets equal chance :)

Can we post Quantum Physics problems here (e.x. Bohr Model)?

The reason I ask this is: in my textbook, this comes right after relativity so I think it should be OK for this forum but it is not mentioned here.

Thanks in advance!

[quote="Rushil"]
2) The Problems to be posted here MUST be accompanied by what the poster has done about it, how much headway he/she has made and what specific help he/she expects.[/quote]

What if you made up the problem for others to do for fun? Surely you shouldn't post your own solution?

Also, for fun/recreational problems, should you post full solutions, or still give hints, partial solutions, etc., or should you post full solutions? If the answer is the second option, then how can you tell whether a problem is a homework problem or a recreational one?

All my questions also apply for the Introductory Physics Forum.


algebra
polynomial
Putnam
algebra solved

Let P be a real polinomial with n distinct real roots that are all > 1.
We introduce Q like this:
Q(x)=(x 2 +1)P(x)P'(x) + x[P(x) 2 + P'(x)  2 ] Prove that Q has at least 2n-1 real roots.

Q(x)=(xP(x)+P'(x))(xP'(x)+P(x)).
T(x)=xP(x) you have :T(x) have n+1 roots
=> T'(x) have n roots ,
K(x)= P(X)e <sup>(-1/2)*x*x</sup>
=> K'(X) have n-1 roots
THEN YOU HAVE RESULTS THAT WE CONSOLE

This is problem A5 on the 1981 Putnam: [url]http://www.kalva.demon.co.uk/putnam/putn81.html[/url]


Βρήκα ότι η διακρίνουσα της τριτοβάθμιας εξίσωσης $ az^3\plus{}bz^2\plus{}cz\plus{}d\equal{}0$ είναι η

$ \Delta\equal{} 4b^3d\minus{}b^2c^2\plus{}4ac^3\minus{}18abcd\plus{}27a^2d^2$

Πώς όμως βγαίνει αυτός ο τύπος?

Ορίζουμε τη συνάρτηση $ f(x)\equal{}ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$ με $ a>0$ και $ g(x)\equal{}f^{'}(x)\equal{}3ax^{2}\plus{}2bx\plus{}c$. Η διακρίνουσα της $ g(x)$ είναι $ s_{1}\equal{}b^{2}\minus{}3ac$. Αν $ s_{1}>0$ τότε η $ f^{'}(x)$ έχει 2 άνισες πραγματικές ρίζες $ x_{1},x_{2}$ με $ x_{1}\equal{}\frac{\minus{}b\minus{}\sqrt{s_{1}}}{3a}$ και $ x_{2}\equal{}\frac{\minus{}b\plus{}\sqrt{s_{1}}}{3a}$.

Αν θεωρήσουμε ότι $ f(x_{1})\equal{}\frac{2b^{3}\minus{}9abc\plus{}(2b^{2}\minus{}6ac)\sqrt{s_{1}}\plus{}27a^{2}d}{27a^{2}}$ και $ f(x_{2})\equal{}\frac{2b^{3}\minus{}9abc\minus{}(2b^{2}\minus{}6ac)\sqrt{s_{1}}\plus{}27a^{2}d}{27a^{2}}$, τότε με πολλαπλασιασμό κατα μέλη έχουμε $ f(x_{1})f(x_{2})\equal{}\frac{(2b^{3}\minus{}9abc\plus{}27a^{2}d^{2})^{2}\minus{}(2b^{2}\minus{}6ac)^{2}(b^{2}\minus{}3ac)}{(27a^{2})^{2}}$ και μετά τις πράξεις προκύπτει ότι $ f(x_{1})f(x_{2})\equal{}\frac{s_{2}}{27a^{2}}$ όπου $ s_{2}\equal{}4ac^{3}\plus{}4db^{3}\plus{}27a^{2}d^{2}\minus{}b^{2}c^{2}\minus{}18abcd$.

Σύμφωνα με τα παραπάνω, αν η εξίσωση $ f(x)\equal{}ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d\Rightarrow f(x)\equal{}0$ έχει 3 άνισες πραγματικές ρίζες τότε $ s_{1}>0$ και $ s_{2}<0$. Ισχύει και το αντίστροφο, δηλαδή αν ισχύουν ότι $ s_{1}>0$ και $ s_{2}<0$ τότε η εξίσωση $ f(x)\equal{}0$ έχει 3 άνισες πραγματικές ρίζες. Εάν τώρα ισχύουν $ s_{1},s_{2}>0$ τότε η $ f(x)$ έχει μία απλή πραγματική ρίζα και 2 μιγαδικές συζύγείς ρίζες. Τέλος αν $ s_{2}\equal{}0$ τότε η $ f(x)$ έχει μία απλή πραγματική ρίζα και μία διπλή πραγματική ρίζα. :wink:

Εάν θέλεις απόδειξη για την ύπαρξη των ριζών να μου στείλεις pm :)

Έστω $ r_{1},r_{2},r_{3}$ με $ r_{1} < r_{2} < r_{3}$ οι 3 άνισες πραγματικές ρίζες της εξίσωσης $ f(x) \equal{} 0$. Τότε από το θεώρημα του Rolle έχουμε ότι
a)$ \exists x_{1}\in (r_{1},r_{2}): f^{'}(x_{1}) \equal{} 0$

b)$ \exists x_{2}\in (r_{2},r_{3}): f^{'}(x_{2}) \equal{} 0$

Άρα η $ f^{'}(x) \equal{} 3ax^{2} \plus{} 2bx \plus{} c$ έχει 2 άνισες πραγματικές ρίζες $ x_{1},x_{2}$.

Αυτή είναι η βασική σχέση που συνδέουμε τα $ x_{1},x_{2}$ με την $ f(x)$.

Τίποτα το σπουδαίο, όπως σου είπα :P

Παιδιά πολύ αναλυτικά τα λέει η [url=http://en.wikipedia.org/wiki/Discriminant]Wikipedia[/url] για διακρίνουσες πολυωνύμων οποιδήποτε βαθμού! :o


These are the rules in this game:

1. Someone posts their hobby
2.The next person then checks if his hobby is the same as one of the previous posts. If it is, then he posts the name of the username of the person who posted the hobby [b]first[/b] in spoiler tags.(if not, then do the same thing as in step 1)

eg: boy1: my hobby is video-gaming
      boy2: [hide=hobby]boy 1[hide]
At this point the hobby that boy1 posted gets 1 point.

3. At the end of 2 days, I will check which hobby has the most points.(Math is NOT a hobby :P )

My hobby: Programming

Swimming
And can you have more than one hobby? If so [hide="my other hobby is"]batteredbutnotdefeated[/hide]

My hobby is MUSIC.

Isn't this going to be a bit complicated after there are 3 or 4 pages of hobbies to check?

[hide="Scores so far"]
batteredbutnotdefeated
[/hide]
I think when you post your hobby you should also post your score
And isn't it possible for people to lie about there hobby just to make sure somebody else doesn't get a point?

I think this thread is going to die soon...

@supernerd: I'm depending on the good will of most AoPSers :rotfl:  :rotfl:  :rotfl:

[b]doing math[/b] (it isn't just math...)

[b]AoPS[/b]

If those don't count, nothing else.


Inequality
three variable inequality
Muirhead
IMO Shortlist
IMO Longlist
algebra
inequalities proposed

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="orl"]Prove that for arbitrary positive numbers the following inequality holds
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3}.\][/quote]

Multiplying this inequality with $a^3b^3c^3$, we get

$a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \leq a^8+b^8+c^8$.

This is trivial using the Muirhead inequality (in fact, $\left(8;\;0;\;0\right)\succ\left(3;\;3;\;2\right)$). Without the Muirhead inequality, one can prove this by noting that AM-GM yields

$a^2b^3c^3 = aabbbccc = \sqrt[8]{a^8a^8b^8b^8b^8c^8c^8c^8} \leq \frac{a^8+a^8+b^8+b^8+b^8+c^8+c^8+c^8}{8}$
$ = \frac28 a^8 + \frac38 b^8 + \frac38 c^8$

and similarly $b^2c^3a^3 \leq \frac28 b^8 + \frac38 c^8 + \frac38 a^8$ and $c^2a^3b^3 \leq \frac28 c^8 + \frac38 a^8 + \frac38 b^8$.

Summing up, we get

$a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3$
$\leq \left(\frac28 a^8 + \frac38 b^8 + \frac38 c^8\right) + \left(\frac28 b^8 + \frac38 c^8 + \frac38 a^8\right)$
$ + \left(\frac28 c^8 + \frac38 a^8 + \frac38 b^8\right)$
$=a^8+b^8+c^8$,

qed..

  Darij

I got the same solution as Darij. Most competitions don't allow 1 line proofs by Muirhead's inequality, but the fact that the inequality holds by Muirhead guarantees a solution by AM-GM or weighted AM-GM

and if we can use amgm,we can also use sos form

Actually, this is very easy problem: WLOG let $a \geq b \geq c$. Then by Chebyshev's inequality we have $3(a^8+b^8+c^8) \geq (a^2+b^2+c^2)(a^6+b^6+c^6) \geq (ab+bc+ca)(3a^2b^2c^2)=3(ab+bc+ca)(a^2b^2c^2)$ Now after dividing everything with $3a^3b^3c^3$ we have $\frac{(a^8+b^8+c^8)}{a^3b^3c^3} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, and we're done.

Why isn't there some really troll solution yet? Ok, may I do that part...

We know that $a^2+b^2+c^2 \ge ab+bc+ca$ for any reals.

Thus, $a^8+b^8+c^8 \ge a^2b^4+b^4c^4+c^4a^4 \ge a^2b^4c^2+b^2c^4a^2+c^2a^4b^2 \ge a^2b^2c^2(a^2+b^2+c^2) \ge (abc)^2(ab+bc+ca)$ which is the desired result.

Too easy...

Multiply both sides by $a^3b^3c^3$, yielding $a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \leq a^8+b^8+c^8$. Muirhead on $\left(8,0,0\right)\succ\left(3,3,2\right)$ and dividing by $2$ yields the desired result.


Sorry, but i want to know why cant i type latex? Where is [tex] tags? Please send me some help!

Hi, you can type LaTeX by inputing LaTeX code between [tex] and [/tex] tags, or between usual $ ... $ tags.
More info for general LaTeX is located here http://www.mathlinks.ro/LaTeX/AoPS_L_About.php or, if you want a quick guide, read this topic:
http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

You can test LaTeX in the Test forum: http://www.mathlinks.ro/Forum/index.php?f=224 and you can ask further questions about LaTeX typing in the LaTeX forum:  http://www.mathlinks.ro/Forum/index.php?f=123


Find all positive integers $(x,y,z)$ such that:

$xy+x-z=1$
$xy+y+z=2006$

It looks correct to me, but then again this is Pre-Olympiad and I'm way out of my league.

[hide]$2xy+x+y=2007$
$4xy+2x+2y=4014$
$4xy+2x+2y+1=4015$
$(2x+1)(2y+1)=4015$[/hide]

oh haha...i did not read that carefully...ok


How many functions $f: \{1,2,...,n\}\rightarrow \{1,2,3,4,5\}$ are there such that for any $k=1,2,...,n-1$ it holds that$|f(k+1)-f(k)|\geq 3$?

Let $U_n$ be the desired number.
Direct checking shows that $U_1=5$ and $U_2 = 6.$

Now, assume that $n >2:$
Clearly, none of the numbers $f(k)$ can be equal to $3$.
Let $a_n,b_n,c_n,d_n$ be the number of such functions such that $f(n)=1,2,4,5$ respectively. Thus $U_n = a_n+b_n+c_n+d_n.$

Any 'good'function for $n+1$ is a good function when restricted to $\{1, \cdots, n \}.$ Conversely, starting with a good function $f$ for $\{1, \cdots, n \}$ we may $f(n+1)=4,5$ if $f(n)=1$, we may choose $f(n+1) = 5$ if $f(n)=2$, we may choose $f(n+1)=1$ if $f(n)=4$ and we may choose $f(n+1)=1,2$ if $f(n)=5.$
It follows that $a_{n+1} = c_n+d_n,b_{n+1} = d_n, c_{n+1} =a_n$ and $d_{n+1} = a_n + b_n.$
Then $a_{n+1} + d_{n+1} = U_n$ so that $U_{n+1} = U_n + a_n + d_n = U_n + U_{n-1}.$

Solving this linear recurrence, it easily follows (it may be convenient to define $U_0 = 1$) that $U_n = (\frac 1 2 + \frac 9 {10} \sqrt 5)( \frac {1 + \sqrt 5} 2)^n +  (\frac 1 2 - \frac 9 {10} \sqrt 5)( \frac {1 - \sqrt 5} 2)^n .$

Pierre.


[quote="1234567890"]In general, how do you grade computer programs that yield a solution?

Do you want us to print out the entire program, or just explain in pseudocode?[/quote]
Don't give pseudocode.

We graders want a description in mathematical English why you expect the algorithm in your program to find the answer. We are interested in your mathematicial reasoning. If your program has some programming error in it that gave the wrong answer, the more fanatical graders might look through the code to point out the error (and maybe give you a little more partial credit if the error is minor), but that would be the only reason to give the actual code. Pseudocode would probably lack the bug, and thus would be useless.

Erin Schram
USAMTS grader


The four-digit numeral 3AA1 is divisible by 9. What digit does A represent?

[hide=" click to reveal hidden content"]

Since the sum of the digits must be divisible by $ 9$, $ A\plus{}A\equal{} 5$ or $ 14$; therefore $ A\plus{}A\equal{}14$ and so $ A\equal{}7$.

[$ \frac{3771}{9} \equal{}419$.]

[/hide]

So you know that 3AA1 has to be divisible by 9..
Anything divisible by 9= all the digits added up have to be divisible by 9,
so then you know that 3+1=4..then if u want a multiple of 9, they have to be 9,18...
But, 9 won't work..and 18 does. You subtract 4 from 18=14..Then you divide that by 2, and get 7 as your answer.

ok. So
[hide] 3AA1 has to be divisible by 9. So, it has to be divisible by 3....right? If you add up the digits for 3AA1, you have to get a multiple of nine. So, you already have a 4, when you add the 3 and the 1. So then A, A have to be the same. 9 would not work, because 9-4=5, and you can't split up nine into 4.5's. So our next choice, is eighteen. Eighteen works ,because 18-4=14, and 14 divided by 2=7 [/hide]


Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x\plus{}y)\plus{}f(y\plus{}z)\plus{}f(z\plus{}x)\ge 3f(x\plus{}2y\plus{}3z)\]
for all $x, y, z \in \mathbb R$.

[hide="Hint 1"]$ f$ is constant.[/hide]
[hide="Hint 2"]Prove that $ f(0)\le f(t)$ and $ f(t)\le f(0)$.[/hide]
[hide="Big hint (do not open before you give up)"]Plug in $ x \equal{} t,y \equal{} 0,z \equal{} 0$; plug in $ x \equal{} \frac {t}{2},y \equal{} \frac {t}{2},z \equal{} \minus{} \frac {t}{2}$.[/hide]

[hide="solution"]
Setting $ y = 0, z = 0 $ yields \[ 2f(x) + f(0) \ge 3 f(x) \] or \[ f(0) \ge f(x) \] for all $ x \in \mathbb{R} $.

Now we let $ x = n, y = n $ and $ z = -n $ to obtain \[ f(2n) + 2f(0) \ge 3f(0) \] or \[ f(2n) \ge f(0) \] which is equivalent to \[ f(x) \ge f(0) \] for all $ x \in \mathbb{R} $.

Thus, $ f(x) = f(0) = c $ for any real $ c $. Plugging into the equation gives $ 3c \ge 3c $, which clearly always hold so the answer is $ \boxed{f(x) \equiv c} $ $ \forall c \in \mathbb{R} $.

[/hide]

$1 + 2 = 3$ again?!

The answer is $f(x)=c$ for all reals $c$; it's easy to see that these solutions work. Let $P(x,y,z)$ denote the given assertion.

Setting $P(x,0,0)$ gives $3f(x) \leq 2f(x) + f(0)$, so $f(x) \leq f(0)$ for all reals $x$. Setting $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2})$ gives $3f(0) \leq 2f(0) + f(x)$, so $f(0) \leq f(x)$ for all reals $x$. Therefore $f(x) \leq f(0) \leq f(x)$, or $f$ is constant.

[b]Remark.[/b] I really like this FE because both steps are motivated: you get the first from just doing stuff and you wonder, ``huh, how do I use the inequality?'' Then the solution is natural.

Really cute :)
As usual let $P(x,y,z)$ denote the given assertion.
[list]
[*]$P(x,0,0) \implies f(x) \le f(0)$ for all $x$.

[*]$P\left( \dfrac{-3x}{2} , \dfrac{3x}{2} , \dfrac {-x}{2} \right) \implies 2f(0) \le f(-2x) + f(x)$ which is also $\le 2f(0) \implies f(-2x) + f(x) = 2f(0)$.

[*]$P(-x , 2x , -x) \implies f(x) \ge f(0)$.
[/list]
So $f(x) = c$ for constant $c$.

Edit: wait bruh how did I miss the more natural substitution shown above.

$P(x,0,0)\Rightarrow f(x)\le f(0)$
$P(x,x,-x)\Rightarrow f(2x)\ge f(0)\Rightarrow f(x)\ge f(0)\Rightarrow\boxed{f(x)=c}$ for some $c\in\mathbb R$, which works.

$P(y,y,-y) \implies f(2y)+f(0)+f(0) \geq 3f(0) \implies f(x) \geq f(0) \forall x$
$P(-y,y,-y) \implies f(0)+f(0)+f(-2y) \geq 3f(-2y) \implies f(0) \geq f(-2y) \implies f(0) \geq f(x) \forall x$
Therefore, $f(x)=c$ where $c$ is a constant for all $x$

Only constant functions work.

Summing this inequality cyclically, $$f(x+2y+3z) + f(y+2z+3x)+f(z+2x+3y) \leq f(x+y) + f(y+z) + f(z+x).$$ On the other hand, substituting $P(2x+y, 2y+z, 2z+x)$, we obtain exactly the reverse inequality. As a result, equality must hold everywhere. Now, setting $y=z=0$, $$3f(x) = f(x) + 2f(0) \implies f(x) = f(0)$$ is constant, as needed.

We claim the only solution is $f(x)=c$, where $c$ is a real constant. Note that this satisfies the given conditions. Let $P(x,y,z)$ denote the given assertion and $P(x,0,0)$ yields $f(x)\le f(0)$. Then, $P(-z,-z,z)$ yields $f(0)\le f(-2z)$ so $f(x)\ge f(0)$. Hence, $f(x)=f(0)$ so we are done. $\square$

Plugging in $(x,y,z)=(x,0,0)$ to get

\[3f(x) \le 2f(x)+f(0)\]
\[\implies f(x) \le f(0)\]

Plugging in $(x,y,z)=(x/2,x/2,-x/2)$ gives

\[3f(0) \le f(x)+2f(0)\]
\[\implies f(x) \ge f(0)\]

Thus, $f(x)=f(0)$, so our answer is $\boxed{f(x)=c, \ c \in \mathbb{R}}$.

Substitute $(2k,0,0)$ and $(k,k-k)$ to find
\[f(2k) \leq f(0), \quad f(2k) \ge f(0).\]

Hence the constant $f(0)$ is the only value $f$ can take, so $\boxed{f(x) = c, \quad c \in \mathbb{R}}$, which evidently works. $\blacksquare$

$f\equiv c$ for some constant $c$ is the solution.

$P(x,0,0) \implies 3f(x) \le f(x) + f(0) + f(x) \implies f(x) \le f(0)$.

$P(-x,-x,x) \implies 3f(-x + 2(-x) + 3x) \le f(-2x) + f(0) + f(0) \implies 3f(0) \le f(-2x) + 2f(0) \implies f(-2x) \ge f(0)$.
Now changing $-2x \rightarrow x$, we get that $f(x) \ge f(0)$.

Thus $f(x) = f(0)$. :yoda:

Letting $y = z = 0$ gives $f(x) \leq f(0)$.Then $z = -x = -y$ gives $3f(0) \leq f(2x) + 2f(0)$ so $f(0) \leq f(2x)$, which is only possible if $f(x) = f(0) = c$.

The answer is $f \equiv c$ for any real constant $c$, which works. Let $P(x, y, z)$ denote the given assertion. From $P(x, 0, 0)$ we have $3f(x) \le 2f(x) + f(0)$, so $f(x) \le f(0)$. From $P(\tfrac{x}{2}, \tfrac{x}{2}, -\tfrac{x}{2})$ we have $3f(0) \le f(x) + 2f(0)$, so $f(x) \ge f(0)$. Hence $f(x) = f(0)$ for all $x$, which implies the solution set.

The answer is $f(x) = c$ only - clearly this works. Now
\[ P(x, 0, 0) \implies 3f(x) \le 2f(x) + f(0) \implies f(x) \le f(0) \]
so
\[ 3f(x+2y+3z) \le f(x+y) + f(y+z) + f(z+x) \le 3f(0). \]
Now $P(a, a, -a)$ implies that
\[ 3f(0) \le f(2a) + 2f(0) \le 3f(0) \]
so $f(0) \le f(a)$ for all $a$, and we're done.


1. In a game in which there are 52 cards, and I draw 5, what are the chances (fraction) that I can make a straight flush, if As can be the low one or the high one?

2. I roll 10 dices, in how many different ways can I row atleast one 1?

3. I have fourteen blocks labled 1,2,3,4...14.  I select one, then another, and then finally a third one, without replacement.  What are the chances that the blocks I select are in increasing order in the way I chose them (so if I select 4,7,2 I have selected that, I cannot rearrange them).

What are the chnaces that the blocks are in strictly increasing order ascending by 1 (so 1,2,3 or 2,3,4..)

Define flush :oops:
#2[hide]Use complimentry counting. There are a total of $6^{10}$ possibilities. There are $5^{10}$ ways to not roll a one. Just subtract.[/hide]

Hint on #3: [hide] There are a total of $\binom{14}{3}$ ways. Someone else can figure out the number of ways to have an increasing order.[/hide]

[quote="Art of Owna"][quote="xpmath"]Define flush :oops:
#2[hide]Use complimentry counting. There are a total of $6^{10}$ possibilities. There are $5^{10}$ ways to not roll a one. Just subtract.[/hide]

Hint on #3:  There are a total of $\binom{14}{3}$ ways. Someone else can figure out the number of ways to have an increasing order.[/quote]

actually there are a total of $\binom{14}{3}$ ways for the numerator, as there are a total of $\binom{14}{3}$ the numbers can be increasing :wink:[/quote]
Oh I see now. Every 3 numbers you choose have exactly 1 way to be increasing.


what is the radius of the sphere whose volume is 330 cubic meters?

[hide]
The volume of a sphere is given by $V = \frac{4 \pi r^{3}}{3}$.
Let $V=330$ and solve for r.  :maybe:
[/hide]

Do I use 3.14 for pi when solving for r?

Depends on how precise you want the answer.

Usually you can just leave the answer in terms of pi.
However, the directions on your homework or schoolwork might tell you to use 3.14 .

Why do you keep posting your homework problems anyway? AoPS isnt a place for people to do your work for you...

Yes that is right.
However, solving for $r$ yields

$r=\sqrt[3]{\frac{3V}{4\pi }}$

Since no one has the answer yet...

$r=\sqrt[3]{\frac{3V}{4\pi }}$

$r=\sqrt[3]{\frac{3(330)}{4\pi}}$

$r=\sqrt[3]{\frac{990}{12.56}}$

$r \approx \sqrt[3]{78.82}$

$r \approx 8.878$

$\approx 8.878 \, \text{meters}$

Lol not only is posting homework on AoPS wrong, but chances are we won't see the problem until it is too late. Try and to it yourself, won't you?  :wink:

Can 8.878 meters be rounded off to the nearest tenths?

[quote="color"]Depends on the requirements, but a good rule is to equal the significant numbers of the input
which are 3. Thus an appropriate round off would be $8.88$[/quote]

He meant nearest tenths. Therefore the answer would be $8.9$.

[quote="trivial34"]uh, the answer is not 8.878, its 4.288. You accidently took the square root of 78.82 rather than the cube root. I realize this is an old thread, but just in case anyone starts looking  :lol:[/quote]

Huh. So it is. My mistake. I must have messed up on my calculator  :blush: .


[quote="bye"]gcd(a,b)=1,a,b>0and there exists N∈Zsuch:aN+b is prime.
prove that:there exists infinite n s.t:an+b is prime.
I wanna a conformation.so please don't tell me  that using Dirichlet Theorem,it's obvious...
thx[/quote]

I'm sorry, but I dont quite well understand your exercise. Are you just asking for a Dirichlet theorem proof ?

If so, Google is your friend : Just look for "Dirichlet theorem proof" and the first result seems ok (and too long to be copied here).
If not, maybe you could clarify your question.

Sorry if my request bother you.

i see a proof in a book.
it says that if there exists n making the faction a prime.then dirichlet theorem is proved.i don't quite understand that.
e.g if there exists a prime with the form 4n +1 .like 5,then there exists infinite primes with the form 4n+1.
i think that there must be a conformation.you get a prime in the form,then you can get another prime in that form using the prime.i just don't know what the conformation is.

to make it more clear
suppose:an+b=p p is a prime
then there exists s satisfying:asp+b=q q is a prime
what's s?

[quote="bye"]e.g if there exists a prime with the form 4n +1 .like 5,then there exists infinite primes with the form 4n+1.[/quote]
No, that's not what the question is saying; $ a, b$ aren't fixed.  You need the full strength of the given result.  In this case, it would go as follows:

There exists a prime of the form $ 8n \plus{} 1$, which must be greater than $ 5$.  $ 17$ is such a prime.

There exists a prime of the form $ 20n \plus{} 1$, which must be greater than $ 17$.  $ 41$ is such a prime.

There exists a prime of the form $ 44n \plus{} 1$, which must be greater than $ 41$...

And so forth.  Hence there exist infinitely many primes of the form $ 4n \plus{} 1$.

well, i think i understand it...
yeah...
if an+b=p is a prime
then there exists l that:alp+b=q is a prime
and so on...
so there exists infinitive k s.t:
ak+b is a prime
it seems easy now
thanks t0rajir0u and pco...
 :lol:


Given a number whose division by 3 gives remainder 1,division by 7 leaves remainder 3.What is the remainder if it is divided by 21?

I don't know if there's any way to find this.
But this is my way  :blush:
If $n$ is divided by 7 gives 3 it could be $21k+3; 21k+10; 21k+17$
Moreover, $n$ is divided by 3 gived 1 so $n$ must be $21k+10$


During the electrolysis of $ Zn(NO_{3})_{2}$ was released 3.36 liter of a gass. Find the mass of the metal  emplaced in the     cathode  .

You could probably begin by writing the half reactions and determining what gas is evolving. Then you could use gas laws to go from volume to mass.

I can write the reacion :

$ 2Zn(NO_{3})_{2}\equal{}2Zn^{2\plus{}}\plus{}4(NO_{3})^{\minus{}}$
$ 8H_{2}O\equal{}4H_{3}O^{\plus{}}\plus{}4OH^{\minus{}}$

Cathode (-) | $ 2Zn^{\plus{}2}\plus{}4e^{\minus{}}\equal{}2Zn$
_________________________________

Anode (+)   | $ 4OH^{\minus{}}\equal{}2H_{2}O\plus{}O_{2}\plus{}4e^{\minus{}}$

So the reaction evolved is : $ 2Zn(NO_{3})_{2}\plus{}2H_{2}O\equal{}2Zn\plus{}O_{2}\plus{}4HNO_{3}$

Than what shuld I do ?

 [quote="feynman_wannabe"]Then you could use gas laws to go from volume to mass.
[/quote]

Can you explain it please

Since reaction conditions have not been mentioned, lets assume we carry out the process at STP.  :P

Now, as you yourself mentioned, Zn is deposited at the cathode. So we have to find its weight.

3.36 litres of oxygen is equivalent to $ n\equal{}\frac {3.36}{22.4}$ moles of oxygen and hence $ 32n g$ of oxygen.

According to Faraday's law, ratio of substances deposited/liberated at anode/cathode is in the ratio of their chemical equivalents. Note that one mole of $ Zn^{2\plus{}}$ ions gains 2 moles of electrons while out of four moles of $ OH^\minus{}$, 2 moles undergo change in oxidation but the change is equal to 2 (to form one mole of oxygen). Hence the equivalent weight of Zn is equal to $ \frac {63.5}{2}$ and that of oxygen is equal to $ \frac {32}{4}$.

So, $ \frac {32n}{32/4}\equal{}\frac {x}{63.5/2}$, x=weight of zinc

 :)


algebra
polynomial
superior algebra
superior algebra solved

Consider the polynomials :
$F_{0}=1,F_{1}=X,...,F_{p}=X\left( X-1\right) \left( X-2\right) ...(X-p+1)$
one can easily show that $\left( F_{0},F_{1},...,F_{n}\right)$ is a basis of $\mathbb R_{n}\left[ X\right]$
But how can we decompose an element $P$ of $\mathbb R_{n}\left[ X\right]$ in that basis ? anyone can help ??

The nicest way to do this is with differences. Let $\Delta f(n)=f(n+1)-f(n)$.

Then $P(n)=P(0)F_0+\Delta P(0)F_1+\frac12\Delta^2P(0)F_2+\cdots+\frac1{k!}\Delta^k P(0)F_k+\cdots$
This works because $\Delta F_k=kF_{k-1}$

If we used $\frac1{k!}F_k$ as the basis elements instead, this would be a very nice basis for the integer-valued (on the integers) polynomials.


calculus
integration
probability and stats

suppose that $ v$ is a signed measure,$ f$ is $ L$-measurable.If the intergral $ f$ with respect to $ v$ exists,$ f.v$$ <<v$ holds

Indeed,

Simply write $ \nu.f(A)$ for a $ \nu$-neglgeable set to see that.


calculus
integration
trigonometry
algorithm
function
real analysis
real analysis unsolved

Compute:
$ \int_a^{\plus{}\infty} \left| \frac{\sin x}{x^a} \right| \, \mbox{d}x$

to show a method how to integrate this and to write the final result.

If you are aiming at a nice formula, there exists none. If you are aiming at an efficient computation that allows you to get the answer with precision better than an inverse power of the number of steps in the algorithm, I don't know of any (though essentially it boils down to computing the values of the $ \zeta$-function with high precision fast, so I cannot guarantee that such an algorithm is impossible or unknown). If you just want to compute it for given $ a$ with not too high precision, split into intervals between $ \pi k$ and $ \pi{k\plus{}1}$ and use the Gaussian quadrature formula on each of them except the ones containing $ 0$ where the Taylor decomposition of $ \sin$ is more efficient. Needless to say, the fact that the lower limit is also $ a$ rather than a fixed number doesn't make anything easier. So, alas, what you are asking for (especially the "write the final result" part of it) is impossible.


If $A = (a_{1}, . . . a_{2006})$ is a sequence of positive integers, then let $m_{A}$ be the number of three element
subsequences $a_{i}, a_{j}, a_{k}$ of $A$ with $1 \ge i < j < k \ge 2006$ and $a_{j}= a_{i}+1$ and $a_{k}= a_{j}+1$. Among all such
sequences $A$, find the maximum possible value of $m_{A}$. :ninja:

$\text{Latex}$

I think you mean $1 \leq i < j < k \leq 2006$

If the sequence is not increasing, we rearrange it so that it becomes an increasing sequence. $m_{A}$ can change, but can't decrease.

Let $A = (x_{1}, \ldots, x_{1}, x_{2}, \ldots, x_{n})$. Because we want to find the maximum of $m_{A}$, we assume $x_{k+1}-x_{k}=1$ wlog $x_{k}= k$.

If $n \leq 2$, $m_{A}= 0$. Assume $n > 3$. Let $B = (b_{1}, \ldots , b_{2006}), a_{k}\leq n-1 \Rightarrow b_{k}= a_{k}, a_{k}= n \Rightarrow b_{k}= n-1$. We get $m_{B}> m_{A}$, hence $m_{A}$ can achieve the maximum value only if $n=3$.

Let $f(i) = \#\{ a_{k}| a_{k}= i\}$. Then $m_{A}= f(1) \cdot f(2) \cdot f(3) \leq 669 \cdot 669 \cdot 668=298970748$

$\text{\LaTeX}$  :wink:


Suppose you add a polynomial of degree m  to a polynomial of degree n, where m>n

Then the result is a polynomial of degree_____

Suppose you subtract a polynomial of degree m from a polynomial of degree n, where m>n

The result is a polynomial of degree______

thanks for the help  :)

Both the answers are m, subtracting or adding n from m just changes all of the terms from x^n down, but still keeps x^m


calculus
integration
limit
inequalities
real analysis
real analysis unsolved

Do you mean: $ \lim_{h\to 0^ \plus{} }h\sum_{n \equal{} 1}^{\infty}\frac {1}{1 \plus{} e^{nh}} \equal{} \int\limits_{0}^{\infty}\frac {dx}{1 \plus{} e^{x}} \; ?$

Let $ h>0$ and consider the integral $ I\equal{}\int_{1}^{\infty}\frac{dx}{1\plus{}e^{xh}}$.

We have that

$ I\equal{}\sum_{k\equal{}1}^{\infty}\int_{k}^{k\plus{}1}\frac{dx}{1\plus{}e^{xh}}$.

and note that

$ \frac{1}{1\plus{}e^{(k\plus{}1)h}}<\int_{k}^{k\plus{}1}\frac{dx}{1\plus{}e^{xh}}<\frac{1}{1\plus{}e^{kh}}$.

It follows, by summation, that

$ \int_{1}^{\infty}\frac{dx}{1\plus{}e^{hx}}<\sum\limits_{k\equal{}1}^{\infty}\frac{1}{1\plus{}e^{kh}}<\frac{1}{1\plus{}e^{h}}\plus{}\int_{1}^{\infty}\frac{dx}{1\plus{}e^{hx}}$

Multiply by h both sides of the preceding inequality and the problem reduces to calculating the limit

$ \lim\limits_{h\rightarrow 0}\int_{1}^{\infty}\frac{hdx}{1\plus{}e^{xh}}$.

However, the substitution $ xh\equal{}y$, solves the problem.


limit
topology
calculus
calculus computations

Can someone explain why the set of limit points $L\left[\mathbb{R}\right]=\varnothing$?

what is a limit point ?
after reading : http://mathworld.wolfram.com/LimitPoint.html

Every interval (a,b) from the reals (with a not equal to b) has an infinite number of real numbers in between them (1). Now suppose that L(R) is not the null set than tere exist such a point p that has the property that there exists an open set from R which contains the point p and only p but that contradicts (1).
I hope i am right!

I understand a limit point to be a point $p$ in a set $X$ if there exists a sequence of points $p_{n}$ with $p\neq p_{n}$ such that $\lim_{n\rightarrow\infty}p_{n}=p$.  I mean, if I had the interval $\left(0,1\right)$, then the set of limit points would be $L\left[\left(0,1\right)\right]=\left[0,1\right]$.  I don't see why this doesn't extend to the real numbers and becomes $L\left[\mathbb{R}\right]=\left[-\infty,\infty\right]$.

I think you understood the exact opposite... although.... are you sure about your  first post?

[quote="jmerry"]The set of limit points of $\mathbb{R}$ is $\mathbb{R}$. Maybe you've got this confused with "boundary"?[/quote]

What do you mean!

I'll have to double-check with my instructor, but I clearly remembering him saying that the set of limit points of the real numbers is the null set.


Let $ a$ and $ b$ be positive integers with $ (a,b) \equal{} 1$. Show that there exist integers $ m$ and $ n$ for which $ a^m \plus{} b^n\equiv1\mod{ab}$.

Since (a, b) =1 we need to find m and n such that $ a^{m}\equal{}1(mod b)$ and $ b^{n}\equal{}1(mod a)$. But clearly we can just take m = $ \phi(b)$ and $ n\equal{}\phi(a)$.

oh god, I was about to answer the question as same way you have answered


linear algebra
matrix
calculus
derivative
linear algebra unsolved

Let $A=\left(\begin{array}{ccc}3 & 1 & 1 \\ 6 & 2 & 2 \\ 3a & b & b \end{array}\right)$ a matrix and $B$ the set of pairs $(a,b)\in\mathbb{R}^{2}$ such that $\det(xA+yI_{3})$ is independent of $x$. Then $B$ is?

Let A be a (n,n) matrix and f:(x,y)->det(xA+yI); f is independent of x iff for all x,y  D1f(x,y)=0 or trace(A.adj(xA+yI))=0 (where adj(u).u=det(u).I); y=0 implies det(A)=0. y<>0 implies trace(A.(I-x.A)^-1)=0 for x rather small or                 0=trace(A.(I+xA+x^2.A^2+......)=trace(A)+xtrace(A^2)+.....; by differentiations in 0: 0=trace(A)=trace(A^2)=....=trace(A^n). One concludes from it that A is nilpotent.
       Conversely if A is nilpotent, we may suppose A upper triangular with diagonal 0; det(xA+yI)=y^n.
Conclusion: f is independent of x iff A is nilpotent.
   Here trace(A)=b=-5 and trace(A^2)=30+6a and a=-5.