arithmetic sequence
number theory
relatively prime
number theory unsolved

I think it is not very hard problem ! . The first I will prove following result :
Lemma : Exist an arithmetic sequence of rational number $ \frac{a_i}{b_i},i\equal{}1,..,k$ such that $ \gcd(a_i,b_i)\equal{}1$ and $ b_j$ are pairwise distinct .
Let $ p$ is a prime greater than $ k$ . Consider following sequence :
$ \frac{1}{p},\frac{1}{p}\plus{}\frac{1}{k!},...,\frac{1}{p}\plus{}\frac{k\minus{}1}{k!}$
For all $ d\equal{}0,...,k\minus{}1$ we have : $ \frac{1}{p}\plus{}\frac{d}{k!}\equal{}\frac{\frac{k!}{d}\plus{}p}{p.\frac{k!}{d}}$
It is not hard to show that $ p\plus{}\frac{k!}{d}$ and $ p.\frac{k!}{d}$ is relative prime . Hence $ b_j\equal{}p.\frac{k!}{d}$ , and obvious they are different from each other .
Now let consider $ p_1,p_2$ are two different prime number , such that each of them is greater than $ \max\{a_1,b_1,..,a_k,b_k\}$ .Because the sequence $ \{\frac{a_i}{b_i}\}$ is an arithmetic sequence  so $ \{\frac{a_i}{b_i}\plus{}\frac{p_1}{p_2}\equal{}\frac{c_i}{d_i}\}$ is  also an arithmetic sequence . Because each of $ p_1,p_2$ greater than $ a_i,b_i$ so we have $ \gcd(a_i.p_2\plus{}b_i.p_1,b_i.p_2)\equal{}1$ . Therefore we must have $ c_i\equal{}a_i.p_2\plus{}b_i.p_1$ and $ d_i\equal{}b_ip_2$ . We will show that all $ c_i,d_j$ are different from each other .
i ) If $ c_i\equal{}c_j$ then we have $ a_ip_2\plus{}b_ip_1\equal{}a_jp_2\plus{}b_j p_1$ , it gives $ p_2|b_i\minus{}b_j$ , because $ p_2>max\{b_i,b_j\}$ and $ b_i,b_j$ are different so it can't hold .
ii) $ c_i\equal{}d_j$ then we have $ a_ip_2\plus{}b_ip_1\equal{}p_2b_j$ ,it gives $ p_2|b_1$ , it also can't hold .
iii ) $ d_i\equal{}d_j$ , this can't hold because $ b_i,b_j$ are different from each other.
Thus this sequence satisfies condition.

ya... i agree it's an easy problem in fact, many ways to construct them.
One is like how the official solution construct it, and another common one is $ \frac{a_i}{b_i} \equal{} \frac{k!\plus{}i}{k!}$

I believe that most candidates get this problem.

[quote="brianchung11"]ya... i agree it's an easy problem in fact, many ways to construct them.
One is like how the official solution construct it, and another common one is $ \frac {a_i}{b_i} \equal{} \frac {k! \plus{} i}{k!}$

I believe that most candidates get this problem.[/quote]

try $ k\equal{}3$ you will get

$ \frac{7}{6}$, $ \frac{4}{3}$, $ \frac{3}{2}$

I think better get this one:

$ \frac {a_i}{b_i} \equal{} \frac {(k!)^2 \plus{} i}{(k!)^2}$

Let $y_1,y_2, \ldots , y_{k-1}, x_1,x_2, \ldots , x_{k-1}$ be $2k-2$ distinct primes that are greater than $k$. By CRT, there exists $x$ so that $\gcd (x,k!)=1, x \equiv -i \pmod{x_iy_i}$ for all $1 \le i \le k-1$. Now, we pick $\frac{a_1}{b_1}= \frac{1}{x_1x_2 \cdots x_{k-1}}$ and $d= \frac{1}{xx_1x_2 \cdots x_{k-1}}$. Hence, it's not hard to show that $$\frac{a_{i+1}}{b_{i+1}}= \frac{(x+i)/x_i}{xx_1x_2 \cdots x_ix_{i+1} \cdots x_{k-1}} \; \quad \text{and} \quad \gcd \left( \frac{x+i}{x_i}, \frac{xx_1 \cdots x_{k-1}}{x_i} \right)=1.$$
Thus, $a_{i+1}= \frac{x+i}{x_i}, b_{i+1}= \frac{xx_1 \cdots x_{k-1}}{x_i}$ and $\gcd (a_{i+1},b_{i+1})=1$.

Note that $a_{i+1} \ne a_{j+1}$ for all $i \ne j$ because $y_i \mid a_{i+1}$ but $y_i \nmid a_{j+1}$ since $y_i \nmid x+j$. We also have $b_i \ne b_j$ and $a_i \ne b_j$ (since $y_{i-1} \mid a_{i}$ but $y_{i-1} \nmid b_j$). Thus, we've guaranteed that $a_1, \ldots , a_k,b_1, \ldots ,b_k$ are pairwise distinct.

We are done.

Nice  :)
[quote=brianchung11]Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i \equal{} 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ...,  a_k, b_k$ are all distinct.[/quote]

[size=150]Solution:[/size] Pick primes $P_1 , P{_2} , \dots , P_k$ such that $| P_i - P_j | >k$ and let $x$ be an integer such that :
$$\begin{array} {c}x  \equiv -1 \pmod {P_1} \\ x \equiv -2 \pmod {P_2} \\ \vdots \\ x \equiv -k \pmod {P_k} \end{array}$$
By [i]CRT[/i] the existence of such an $x$ is guaranteed. Now, let $M = \prod_{i = 1}^{k} $ and consider the rational numbers $$ \frac{x}{M} , \frac{x+1}{M} , \dots , \frac{x+k}{M}$$
Clearly, this sequence of rationals is an arithmetic progression. Moreover, observe that each of the numerator and denominators are distinct as $P_i$ divides no other number than $x+i$.

I hope it's correct
just put $a_1=k!.p$  and $b_1=q$ let $r$ the difference between the terms in the arithmetic sequence we will put $r=\frac{c}{d.n!}$ such that $p,q,c,d$ are different primes greater than $k!$
$\frac{a_2}{b_2}=\frac{(k!)^2pd+cq}{k!qd}$
$\frac{a_3}{b_3}=\frac{\frac{(k!)^2pd}{2}+cq}{\frac{k!qd}{2}}$
$\frac{a_4}{b_4}=\frac{\frac{(k!)^2pd}{3}+cq}{\frac{k!qd}{3}}$
in general
$\frac{a_i}{b_i}=\frac{\frac{(k!)^2pd}{i-1}+cq}{\frac{k!qd}{i-1}}$
it's clear that $(a_i,b_i)=1$
if we have $a_i=b_j$ then $\frac{(k!)^2pd}{i}+cq=\frac{k!qd}{j}$ but since $q $doesn't divide $\frac{(k!)^2pd}{i}$ that's impossible :D
and we're done :P

Consider $\frac{p(k! + 1)}{k!}, \frac{p(k!/2 + 1)}{k!/2}, \frac{p(k!/3 + 1)}{k!/3}, \dots, \frac{p(k!/k + 1)}{k!/k}$ for some prime $p > k$ so that all the $a$'s are greater than all the $b$'s and $\gcd(a_i, b_i) = 1$.

Solved with [b]nprime06, E_6[/b]

[i]Construction[/i]: Take sufficiently large $n$ and consider the sequence \[1+\frac{1}{n!}, 1+\frac{2}{n!}, \dots\] [i]Proof of Construction[/i]:
Note that for the denominators to be distinct, we need \[\gcd(a_1, b_1)\neq \gcd(a_1, b_1\cdot d+a_1) \neq \cdots \neq \gcd(a_1, d\cdot (k-1)b_k)\] which implies $b_i > b_i\cdot d+a_1$ by repeatedly applying the euclidean algorithm. For the numerators to be distinct, it is sufficient to let $n$ be sufficiently large.

If $k=1,2$ it is easy to verify an arithmetic sequence exists, so assume $k \geq 3$. Consider an increasing list of distinct primes $p_1, p_2, \dots p_k$, where all the $p_i>k$. Then consider the arithmetic sequence
\[\frac{x+1}{n}, \frac{x+2}{n}, \dots, \frac{x+k}{n}.\]
I will show that by simplifying the fractions, we produce the desired sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$.
Let $n=p_1\dots p_2, \dots, p_k$. Let $x$ be the minimal number that satisfies the following system of congruences: $x\equiv -1 \pmod {p_1},  x \equiv -2 \pmod {p_2}, \dots, x \equiv -k \pmod{p_k}$. A solution exists due to CRT.

Then, for each term $\frac{x+i}{n}$ of the arithmetic sequence, both the numerator and denominator are divisible by $p_i$. In fact, I claim that dividing the numerator and denominator of $\frac{x+i}{n}$ by $p_i$ results in the desired arithmetic sequence. I claim that the gcd of the numerator and denominator of $\frac{x+i}{n}$ is $p_i$. To see this, assume some other prime $p_j$ divides $x+i$, so $x+i \equiv 0 \pmod {p_j}$. But $x+j \equiv 0 \pmod  {p_j}$, so this means that
\[x+i \equiv x+j \pmod {p_j} \implies i \equiv j \pmod {p_j}, \]
which is a contradiction since $i$ and $j$ are both distinct and less than $p_j$. This establishes that $\gcd(a_i, b_i)=1$.

To see that no two $a_i$ are the same, we will proceed by contradiction. Assume $a_i=a_j$ and $i>j$. We also have
$$a_ip_i-a_jp_j=x+i-(x+j) \implies a_i(p_i-p_j)=i-j,$$ a contradiction since $p_i-p_j >i-j$.

To see that no $a_i=b_j$, we will again proceed by contradiction. Since $b_j=n/p_j$, we have that $a_i=p_1p_2\dots p_{j-1}p_{j+1}\dots p_k.$ But, $b_i=p_1p_2\dots p_{i-1}p_{i+1}\dots p_k$, so for $k\geq 3$, we have that $\gcd(a_i, b_i) \neq 1, $contradiction.

So, we've established that the arithmetic sequence we've constructed does indeed satisfy all the requirements.

Simplify the following fractions

$\dfrac{(n+1)!+2}{n+1}, \dfrac{(n+1)!+3}{n+1}, \dfrac{(n+1)!+4}{n+1}, \cdots, \dfrac{(n+1)!+n+1}{n+1}$

This clearly satisfies the conditions of the problem.

[quote=brianchung11]Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i \equal{} 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ...,  a_k, b_k$ are all distinct.[/quote]

Induct,Suppose that $$\frac{a_1}{b_i}=a \;\;\;\;\; \frac{a_{n-1}}{b_{n-1}}=a+kd$$ then at the $n$th step choose $$a_n=\left[\prod_{j=1}^{n-1} a_j \prod_{j=1}^{n-1} b_j \right] \cdot[a+(k+1)d]  \;\;\; b_j=\left[\prod_{j=1}^{n-1} a_j \prod_{j=1}^{n-1} b_j \right]$$ which clearly works.

Let
$$\frac{a_i}{b_i}=\frac{(k!)^{1000}+i}{(k!)^{1000}}=\frac{1+(k!)^{1000}/i}{(k!)^{1000}/i},$$
which is fully simplied. The absolute difference between $(k!)^{1000}/i$ and $(k!)^{1000}/j$ for $i \neq j$ is at least $\tfrac{(k!)^{1000}}{k(k-1)}\gg 1$, so the difference between $a_i-1$ and $b_j$ is massive and thus $a_i \neq b_j$ for $i \neq j$. Since $a_i \neq b_i$ clearly, we are done. $\blacksquare$

Choose primes $p_1>p_2>\ldots>p_k>k$. By the Chinese Remainder Theorem, consider a positive integer solution $x$ to the system of congruences:
\begin{align*}
x+1&\equiv 0\pmod{p_1}\\
x+2&\equiv 0\pmod{p_2}\\
&\vdots\\
x+k&\equiv 0\pmod{p_k}
\end{align*}
Since $p_i>k$ for all $i$, we have that $p_i\mid x+j$ if and only if $i=j$. Observe that $\gcd(a_i,b_i)=1$ because $p_j\neq a_i$ for $j\neq i$. Now, let $a_i=M\frac{x+i}{p_i}$ and $b_i=\frac{1}{p_i}\prod_{r=1}^k p_r$ for some integer $M>p_1p_2\ldots p_k$. Since $p_i$'s are distinct $b_i$'s are distinct. Since $p_1>p_2>\ldots>p_k$ we have $a_1<a_2<\ldots<a_k$ so the $a_i$'s are pairwise distinct. Since $M>p_1\ldots p_n$ we know $a_i\geq M>b_j$ for all $i,j$ so the $a_1,b_1,\ldots,a_k,b_k$ are all distinct. \\

It remains to check that the $\frac{a_i}{b_i}$'s form an arithmetic progression. But we have
$$\frac{a_i}{b_i}=\frac{M\frac{x+i}{p_i}}{\frac{1}{p_i}\prod_{r=1}^k p_r}=M\frac{x+i}{p_1p_2\ldots p_k}$$
so $\frac{a_i}{b_i}$'s indeed form an arithmetic progression with common difference $\frac{M}{p_1p_2\ldots p_k}$.

Consider the arithmetic sequence
\[\frac{1000!+1}{500!},\frac{1000!+2}{500!},\dots,\frac{1000!+100}{500!}.\]
I claim that when the fractions are simplified, we have a valid construction.

In fact,
\[\gcd(1000!+a,500!)=\gcd(a,500!)=a.\]

Note that no numerator can match a denominator as the numerators all have to be greater than $\dfrac{1000!}{500!}$ and the denominators all have to be less than $500!$. Obviously, no two numerators and no two denominators can match(we already know what the simplified fraction looks like), so we are done.

Suppose that $n=p_1p_2\cdots p_k$ for distinct primes such that $p_i>k$ for all $i$. Then, let $m$ be the smallest positive integer such that:\\

(1) $m\equiv -i\pmod {p_i}$ for all $1\leq i\leq k$,\\
(2) $m\geq G_{10^{1428k^{1434}}}$, where $G$ denotes the Graham sequence.\\

Then, let the $k$ fractions be $$\frac{m+1}{n},\frac{m+2}{n}\cdots +\frac{m+k}{n}.$$ Note that for any $1\leq i\leq k$, $m+i$ is divisible by $p_i$ but not by any of the other $k-1$ primes. Thus, after reducing this is $$\frac{(m+1)/p_1}{n/p_1},\frac{(m+2)/p_2}{n/p_2}\cdots +\frac{(m+k)/p_k}{n/p_k}.$$ The denominators are clearly all different, and the numerators are far larger than the denominators (e.g. by Bertrand's postulate), so we just have to show that the numerators are different from each other. However, this is clearly true as well, as $m$ is very large so the effect of the addition of a constant at most $k$ is dominated by the difference in the prime that it is being divided by. Thus, we are done.

Consider the sequence of numerators

$1 + d, 1 + 2d, ..., 1 + kd$ such that

$1 + d \equiv 0 \pmod p_1$
$1 + 2d \equiv 0 \pmod p_2$
\vdots

We chose the $p_i$ such that they are all larger than $k$ (since d is not 0 mod any of the primes, this makes it so that only 1 of the terms is divisible by each $p_i$). By CRT, exists a solution for $d$.  We then make the numerators $1 + d$, $1 + 2d$, etc. and the denominators for all of them $p_1 \cdot p_2 \dots p_k$ (then simplifying will produce k distinct denominators). Also, the numerators are distinct from each other because we can let $p_{a + 1} > 2 \cdot p_a$ so the numerators. always decrease as the primes increase.

We then multiply the denominator by the smallest prime greater than all of the $a$ to make sure the numerators are all distinct from the denomonators.

\begin{remark}
    Here was a fakesolve that i would've walked away with if not for $\textbf{incompleteusern/yaoaops}$; i decided to include it because its funny

    [hide]We establish a preliminary which is a problem ive been meaning to solve but putting it off: $\textit{Let a,b,c,d be integers, all not equal to 1, with }\gcd(a,b)=1=\gcd(c,d)=\gcd(b,d),\frac{a}{b}+\frac{c}{d}=\frac{e}{f}:\gcd(e,f)=1.\textit{ Is it always true that f=bd (and so e=ad+bc)?}$

    The answer is yes. $\textit{Proof.}$ Trivial. $\square$

    (Trivial by bottedimmaculator saying ``then it’s obviously true by vp".)

    Let the common difference be $\frac pq<1:q>n+2$, and establish a starting number $1<\frac{a_1}{b_1}<2$. We construct this inductively: Take $p_k$ to be pairwise relatively prime to the rest of the $a_i$s, and $q_k$ to be pairwise relatively prime to the rest of the $b_i$s. Then simply take the next term $\frac{a_{k+1}}{b_{k+1}}$ that follows this progression! By this construction each $a_i$ and $b_i$ are pairwise distinct from other $a_j$ and $b_j$ respectively (by the lemma since $b_{k+1}>b_k>...,a_{k+1}>a_k>...$, because the denominator and numerator never needs to be ``simplified"); all that remains is to remark that $a_{k+1}>b_{k+1}=qb_k>b_k>b_{k-1}>...$ and $b_{k+1}=qb_k>a_k>...$ ($\frac{a_k}{b_k}=k\frac{p}{q}+\frac{a_1}{b_1}<k+2<n+2<q$) so they're ALL pairwise distinct, as desired.

BUT. ``that only happens once
a/b + p/q = (aq+bp)/bq
then bq and q arent relatively prime"[/hide]
\end{remark}
    Ok now I got a hint, here's the correct attempt: Take the numbers $\frac{((10)^{10^{10^{10^{10}}}})!+i}{((1434)^{1434})!}\forall 1\le i\le n$, which works since numerators easily larger than denominators and obviously distinct since each numerator is divided by pairwise different numbers, and denominators are pairwise distinct since they're also divided by pairwise different numbers. (actually we'd need sufficiently larger numbers if n was even larger, but for the sake of generalizing i showed a funny case)

[i]Written by abeot with ccarolyn4[/i]

Choose primes $p_1, p_2, \dots, p_k$, such that $p_1 > p_2 > \dots > p_k$.
Let $N = p_1p_2\cdots p_k$. Then choose $x$ such that
$x \equiv -i \bmod p_i$ for $i = 1, 2, \dots, k$. By CRT this is always possible.

We show that the reduced sequence is valid. First, we examine the $b_i$. Note that
$b_i = \frac{N}{p_i}$, so all the $b_i$ are distinct.

Now, we just need to show that the $a_i$ are distinct. Suppose that $a_i = a_j$, where $i<j$.
Then since $a_i = \frac{x+i}{p_i}$, so then we must have that $p_j(x+i) = p_i(x+j)$. This is impossible.
Thus, the $a_i$ are all distinct. $\blacksquare$

[b]Remark:[/b] Our motivation for this solution was trying to get values in the form
$\frac{x+1}{N}, \frac{x+2}{N}, \dots$. However, there is one big problem;
this means that all $b_i$ are the same; we fix this by making
$N = p_1 p_2 \dots$ for some primes $p$ such that $x \equiv -i \pmod{p_i}$.
This means that the top and bottom will simplify, and the rest is written
in the proof above.

We claim that for "appropriate" positive integers $a,b$ and distinct primes $p_1,p_2,\ldots,p_k$, the following construction works:
\[\frac{a}{p_1p_2\cdots p_k},\frac{b}{p_2p_3\cdots p_k},\frac{\frac{2bp_1-a}{p_2}}{p_1p_3\cdots p_k},\cdots,\frac{\frac{kbp_1-(k-1)a}{p_k}}{p_1p_2\cdots p_{k-1}}\]
For this constuction to work, we need the primes and the numbers to follow:
[list]
[*] For each $1\leq i\leq k$, we have $p_i\nmid ab$, except $p_1\mid b$ is allowed.
[*] For each $1\leq i\leq k$, we have $p_i\mid ibp_1-(i-1)a$.
[*] For each $1\leq i\leq k$, we have $p_i\nmid bp_1-a$.
[*] For $1\leq i,j\leq k$ such that $i\ne j$, then $p_j\nmid ibp_1-(i-1)a$.
[/list]
This gives a sufficient condition to work with.

The first three properties can be satisfied by Chinese Remainder Theorem. For satisfying the fourth property, note that the common difference of the arithmetic progression is $\frac{bp_1-a}{p_1p_2\cdots p_k}$. Consider the numerators obtained $\pmod{p_i}$ when the denominators are set $p_1p_2\cdots p_k$. Two numerators (suppose numerator $i_1$ and $i_2$) can be congruent $\pmod{p_i}$ only if $p_i\mid (i_2-i_1)(bp_1-a)$. But $p_i\nmid bp_1-a$, so $p_i\mid i_2-i_1$. However, $-k<i_2-i_1<k$, so we can avoid this issue if we set $p_i>k$ for each $1\leq i\leq k$.

The construction is complete.

Consider the following $n$ fractions

\[\frac{n!+1}{n!} ,\frac{\left(\frac{n!}{2}\right)+1}{\left(\frac{n!}{2}\right)}, \frac{\left(\frac{n!}{3}\right)+1}{\left(\frac{n!}{3}\right)} ,\ldots ,\frac{\left(\frac{n!}{n}\right)+1}{\left(\frac{n!}{n}\right)}\]
They all satisfies $\gcd(a_i,b_i)=1$, all numerators are distinct as well as all denominators


limit
ratio
algebra
polynomial
real analysis
calculus
calculus computations

For what values of $x$ do each of the following power series converge?

1. $\sum_{n=0}^{\infty}\frac{(2+(-1)^n)^n}{5^n}x^n$

2. $\sum_{n=0}^{\infty}\frac{(-1)^n(3x+4)^n}{\sqrt{n^2+1}}$

3. $1-{x^2}+\frac{1\cdot3\,x^4}{2!}- \frac{1\cdot3\cdot5\,x^6}{3!}+ \cdots + (-1)^k\frac{1\cdot3\cdots(2k-1)\,x^{2k}}{k!}+\cdots$

On #3: that is the correct radius of convergence. Now, why do you think it diverges at the endpoints?

[quote="amirhtlusa"]hmm
this was my reasoning...
$\lim_{k\rightarrow \infty} |\frac{a_{k+1}}{a_k}|>1$[/quote]
But you have to let $x$ into this. If $|x|=\frac1{\sqrt{2}}$ then it is [i]not[/i] true that $\frac{|a_{n+1}x^{n+1}|}{|a_nx^n|}>1.$

Also: You already got all the information you're ever going to get out of the ratio test when you found the radius of convergence. Here's an oversimplified overview: both the ratio test and the root test are automated comparisons to geometric series. They do a good job of balancing off all of the factors of the terms of series that grow or shrink geometrically fast or faster. Items that grow or shrink at a slower than geometric rate (like polynomials) are invisible to the root or ratio test. A power series always has a geometric item, in the power of $x.$ You use the root test or ratio test to find out the point at which all the geometric or faster pieces, including the power of $x,$ lie in perfect balance. If your exactly at that balance point, the fate of the series will be determined by things that are invisible to the root or ratio test.

Before this slips too far down the topics list, let me point out that the issue of convergence at the endpoints of the interval of convergence in #3 hasn't been settled yet, and no one has yet answered #1. amirhtlusa has completely answered #2 and has the correct radius of convergence for #3.

so the end points in my answer for 2 are wrong? ;)

To completely settle the case in #3, it would help to know a tiny bit about infinite products. Let $|x|=\frac1{\sqrt{2}}.$ The series becomes

$1-\frac12+\frac{1\cdot3}{2\cdot4}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+ \cdots +(-1)^n\frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots(2n)}+\cdots$

Now what?

You put a question mark on that, dheardie.

Checking the premises for the alternating series test:

Do the terms alternate in sign? Check.

Do the terms decrease in size? Check.

Do the terms go to zero? Hmmm...

$\lim_{n\rightarrow \infty} a_n \not = 0$
nth term test$\Rightarrow$ diverges at both endpoints...

(1)
Apply the Cauchy-Hadamard formula.
We know \[{\overline{\lim_{n\rightarrow\infty}}\sqrt[n]{\frac{(2+(-1)^{n})^{n}}{5^{n}}}} = \frac{3}{5}\] so the radius of convergence is $\frac{5}{3}$. Thus the power series is convergent for all $x\in \left( -\frac{5}{3}, \,\frac{5}{3} \right).$ Not sure about the end points.

#12 is not correct: the terms [i]do[/i] go to zero. Can you see why?

The "careful", or limsup version of the root test (or as jgor2861 called it, the Cauchy-Hadamard formula) is a useful thing to know. But let's be a little more elementary here. If the terms oscillate between bigger and smaller, it's the bigger terms that determine the convergence. When they're bigger, the coefficients are $\left(\frac35\right)^n$. If $|x|=\frac53$ do the terms go to zero?

Because $\log(2n-1)-\log(2n)=-\frac{1}{\nu}$, by the mean value theorem, and comparison to the harmonic series shows that the terms of the original series approach zero, so it converges by the alternating series test.

Rookworm is correct. The interval of convergence for #3 is $\left[-\frac1{\sqrt{2}},\frac1{\sqrt{2}}\right].$

Only thing left: what is the full answer to #1?

[quote="Kent Merryfield"]When they're bigger, the coefficients are $\left(\frac35\right)^n$. If $|x|=\frac53$ do the terms go to zero?[/quote]

I thought that was a rhetorical question  :) . The answer is "no".

By the way, Kent Merryfield, someone suggested that you might have insight into [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42682#]this question.[/url]  :lol:

-Rookworm out

[quote]By the way, Kent Merryfield, someone suggested that you might have insight into this question.  :lol: [/quote]

Man, you guys ask a lot. If you want to know [i]all[/i] of my secrets, how about coming to Cal State Long Beach as students?  :)

I am working on a response to that other question, but it's not ready yet.


Let $ x,$ $ y$ and $ z$ are positive numbers. Prove that
\[ \frac{x^3\plus{}y^3}{x^2\plus{}y^2}\plus{}\frac{x^3\plus{}z^3}{x^2\plus{}z^2}\plus{}\frac{y^3\plus{}z^3}{y^2\plus{}z^2}\geq\frac{2(x^2\plus{}y^2\plus{}z^2)}{x\plus{}y\plus{}z}\]

Multiply both sides by $ (x^2 \plus{} y^2 \plus{} y^2 \plus{} z^2 \plus{} z^2 \plus{} x^2)$

Cauchy tells us that it suffices to show that

$ (\sqrt {x^3 \plus{} y^3} \plus{} \sqrt {x^3 \plus{} z^3} \plus{} \sqrt {y^3 \plus{} z^3})\ge\frac {2(x^2 \plus{} y^2 \plus{} z^2)}{\sqrt {x \plus{} y \plus{} z}}$

Note that $ \sqrt {x^3 \plus{} y^3}\sqrt {x \plus{} y}\ge x^2 \plus{} y^2$

$ (x^3 \plus{} y^3)(x \plus{} y) \ge x^4 \plus{} 2x^2y^2 \plus{} y^4$

$ y^3x \plus{} x^3y \ge 2x^2y^2$

Which is just muirhead.

Result follows immediately by multiplying both sides by $ \sqrt{x \plus{} y \plus{} z}$ and summing cyclically.

All sums below are cyclic.

We will show that
\[ \sum\frac {x^3}{x^2 \plus{} y^2}\ge \frac {x^2 \plus{} y^2 \plus{} z^2}{x \plus{} y \plus{} z}\qquad(1)
\]
We have
\[ \sum\frac {x^3}{x^2 \plus{} y^2}\ge\frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum x^3 \plus{} \sum xy^2}
\]
Hence it remains to show that
\[ \left(\sum x^2\right)\left(\sum x\right)\ge \sum x^3 \plus{} \sum xy^2
\]
which is equivalent to
\[ \sum x^2y\ge 0
\]
which is obviously true. Now replace $ (x,y,z)$ with $ (y,x,z)$ in $ (1)$ and add the two inequalities to get the desired one.

Nice solutions, rofler and nayel!
What about the following:
For positive $ x,$ $ y,$ $ z$ and $ n\in\mathbb R$ prove that:
\[ \frac {x^3 \plus{} y^3}{x^2 \plus{} y^2} \plus{} \frac {x^3 \plus{} z^3}{x^2 \plus{} z^2} \plus{} \frac {y^3 \plus{} z^3}{y^2 \plus{} z^2}\geq\frac {2\left(x^{n \plus{} 1} \plus{} y^{n \plus{} 1} \plus{} z^{n \plus{} 1}\right)}{x^n \plus{} y^n \plus{} z^n}
\]

[quote="arqady"]Let $ x,$ $ y$ and $ z$ are positive numbers. Prove that
\[ \frac {x^3 \plus{} y^3}{x^2 \plus{} y^2} \plus{} \frac {x^3 \plus{} z^3}{x^2 \plus{} z^2} \plus{} \frac {y^3 \plus{} z^3}{y^2 \plus{} z^2}\geq\frac {2(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z}
\]
[/quote]

Can we prove the following inequality:
\[ \frac {x^3 \plus{} y^3}{x^2 \plus{} y^2} \plus{} \frac {x^3 \plus{} z^3}{x^2 \plus{} z^2} \plus{} \frac {y^3 \plus{} z^3}{y^2 \plus{} z^2}\geq\frac {3(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z}
\]

[quote="NguyenDungTN"]

Can we prove the following inequality:
\[ \frac {x^3 \plus{} y^3}{x^2 \plus{} y^2} \plus{} \frac {x^3 \plus{} z^3}{x^2 \plus{} z^2} \plus{} \frac {y^3 \plus{} z^3}{y^2 \plus{} z^2}\geq\frac {3(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z}
\]
[/quote]
No! Try $ x\rightarrow\plus{}\infty.$ :wink:

[quote="arqady"][quote="NguyenDungTN"]

Can we prove the following inequality:
\[ \frac {x^3 \plus{} y^3}{x^2 \plus{} y^2} \plus{} \frac {x^3 \plus{} z^3}{x^2 \plus{} z^2} \plus{} \frac {y^3 \plus{} z^3}{y^2 \plus{} z^2}\geq\frac {3(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z}
\]
[/quote]
No! Try $ x\rightarrow \plus{} \infty.$ :wink:[/quote]

Oh, thank you, argady. I tried to use SOS method, but I couldn't solve it!


The difference between two prime numbers is 17. What is their sum?

First couple of prime numbers: 2, 3, 5, 7, 11, 13, 17, 19

Note that $ 19\minus{}2\equal{}17$. (It must be between an even and odd number to get an odd difference)

$ 19\plus{}2\equal{}\boxed{21}$, giving our final answer.

[quote="JSGandora"][quote="AwesomeToad"]You don't need to list the primes.

All you need is one must be a $ 2$ so the other prime $ p$ satisfies $ p\minus{}2\equal{}17$ then add $ 4$ to both sides.[/quote]
It should be "then add $ 2$ to both sides", not $ 4$.[/quote]

Actually, if you added $2$ to both sides you would get the sum is $19$ instead of $\boxed{21}$.  AwesomeToad was just doing the "shortcut way".


Find all 4-tuples $(a,b,c,n)$ of naturals such that

$n^a + n^b = n^c$

After dividing through $n^{\min(a,b,c)}$ we can assume that one of $a,b,c$ is $0$, and it will for sure not be $c$, so WLOG $a=0$.
Then $1+n^b = n^c$ is impossible when $b,c >0$ since it gives $1 \equiv 0 \mod n$ (and it's also wrong for $n=1$).
Thus $bc=0$, and $c=0$ is impossible as above, thus $b=0$ and this yields $2=n^c$, having the only solutions $n=2 , c=1$, and this one works, giving that all solutions are of type $2^k + 2^k = 2^{k+1}$.

It is obviosly, that n>1,c>a,c>b. Therefore $n^a+n^b=n^c\ge n*n^{max(a,b)} \Longrightarrow n=2,a=b=c-1$.


You must replace the 3(on the right side) by a 2.

[quote="manlio"]For $a,b,c,d$ positive reals such that $abcd=1$ prove that

$\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(d+1)}+\frac{1}{d(a+1)}\geq 2$[/quote]

This one is easy but instructive. Since abcd = 1, we can make Harazi's substitution $a=\frac{x}{w}$, $b=\frac{y}{x}$, $c=\frac{z}{y}$, $d=\frac{w}{z}$, and the inequality becomes

$\frac{1}{\frac{x}{w}\left( \frac{y}{x}+1\right) }+\frac{1}{\frac{y}{x}\left( \frac{z}{y}+1\right) }+\frac{1}{\frac{z}{y}\left( \frac{w}{z}+1\right) }+\frac{1}{\frac{w}{z}\left( \frac{x}{w}+1\right) }\geq 2$,

or, equivalently,

$\frac{w}{x+y}+\frac{x}{y+z}+\frac{y}{z+w}+\frac{z}{w+x}\geq 2$.

But this is simply the Shapiro inequality for four variables. $\blacksquare$

  Darij


57. Very careful measurement showed that a 0.1000 m solution of potassium sulfate has a freezing point of -0.4315 °C .What is the percent dissociation of potassium sulfate (K2SO4) in this solution?

    (A) 23.20%
    (B) 66.00%
    (C) 77.33%
    (D) 100% (effectively)

The key says the answer is B.

[hide="Attempt At a Solution "]
The molality of dissolved ions = $ \frac {0.4135 ^{\circ} C }{5.12 ^{\circ} C/ m} = 0.08427m$. This should be three times the amount of potassium sulfate that dissolves since potassium sulfate gives 2 K + 's and one sulfate ion. $ 0.08427m/3 = 0.028m$. Dividing this by 0.1000m, we get an answer that is not in the answer choices.
[/hide]

Wait, what is the potassium sulfate in?  You might have the wrong solvent.
Water Kf is 1.86C/m

It seems like the solven is water because of the way the original poster attempted to solve it using $ \Delta T_f \equal{} iK_fm$. In that case you are right $ K_f \equal{} 1.86^{\circ}Cm^{\minus{}1}$.

When I use the correct Kf, 1.86 I get this:

[quote="Me"]

The molality of dissolved ions = $ \frac {0.4135 ^{\circ} C }{1.86 ^{\circ} C/ m} = 0.222311828m$. This should be three times the amount of potassium sulfate that dissolves since potassium sulfate gives 2 K + 's and one sulfate ion.

[/quote]

Why is that second sentence incorrect? This is 1/3 of the correct answer instead of three times the correct answer, meaning the above is wrong.

I'm confused by this because I would think that we would get the molality of dissovled ions from $ \frac {0.4135 ^{\circ} C }{1.86 ^{\circ} C/ m} = 0.222311828m$ instead of the moles/ kilogram solvent of salt dissolved.

I thought the freezing point was $ \minus{}0.4315^{\circ} C$.

In that case you should get $ \frac{0.4315}{1.86} \equal{} 0.231989247$.

Oh right.

Though when we divide 0.23198924 / 3 = 0.0773297467, we don't get an answer in the answer choices and if we didn't divide by 3, then the key would be wrong... hm..

Actually, aren't we supposed to divide by 3? Since we get three ions per K2SO4?
Anyone have any thoughts on this regarding whether the key is right and whether we must divide by 3?

Aren't you trying to find percentage of dissociation? Ideally it should be $ 3 \times 0.1 m$ but we calculated the actual value to be $ 0.2319 .... m$ so a division seems to be required.

It seems the answer is $ C$ though.

Yeah, I did this and I got C too.

We probably won't need to correct for the ion pull of the dissociated ions because, well, the answer would be even farther off.

Yeah, dividing .232.....by 3 gets you choice C.

Unless something else comes up, that's the answer that I would probably choose for now.

maybe they should have mentioned the solvent, or maybe that was an assumption the test makers made at the beginning of the test.

The answer is B.

deltaT= kf x m x i
.4315 = 1.86 x .1 x i
i=2.32

This is less than the expected van hoff factor, 3, which would be the case if K2SO4 was completely dissociated.  This means some K2SO4 is dissociating into ions, and some of it is not.  A dissociation produces 3 particles (K+, K+, SO4 2-, whereas a nondissociation produces 1 particle (K2SO4).

If we let x=percentage dissocation, then 100-x=percentage undissocated
We then have
2.32x100%=3x + 100 - x
x=66

B.


Find all real pairs $ (x;y)$ in this system:

$ (x+y)^{2}+3(x+y)=4$

$ \frac{1}{x}+\frac{1}{y}=-\frac{1}{6}$

Solving the first equation with respect to $ x+y$ gives $ x+y=-4,\ 1$, then from the second equation we have $ xy=24,-6$ respectively. Thus $ x,\ y$ are the roots of the quadratic equation with respect to $ t,\ t^{2}+4t+24=0,\ t^{2}-t-6=0$, yielding $ (x,\ y)=(-2+2\sqrt{5}i,-2-2\sqrt{5}i),\ (-2-2\sqrt{5}i,-2+2\sqrt{5}i),\ (-2,\ 3),\ (3,-2)$.
Since $ x,\ y$ are real, the answer is $ (x,\ y)=(-2,\ 3),\ (3,-2)$.


linear algebra
matrix
geometry
geometric transformation
reflection
vector
linear algebra unsolved

I'm trying to find a matrix transform that reflects a point across the line y = mx + b.

Do I first start by grouping x and y of the line into a 2D matrix?  It probably isn't as simple as what people normally do, such as reflecting over the y-axis or x-axis right?  Because I essentially don't know what across means.  That leads me to believe that it isn't as simple as representing x and y in a matrix and applying a simple matrix multiplication that reflects over the line.  Not really sure how i'm supposed to go about figuring it out.  Any help would be greatly appreciated

line:
y = mx + b
x = (y-b)/m

[code]
Basic reflection over both x-axis and y-axis...given a point P(x,y) plug it into the 2nd matrix that represents a line?
( -1   0 )  (  (y-b)/m  )
(  0  -1 )  (  mx + b   )
[/code][/code]

Linear transformations are completely determined by where they send the basis elements, so we only need to figure out where reflection across y=mx+b sends (1,0) and (0,1). Then concatenating the two (column) vectors gives the matrix.

LkNsngth- that only applies if you know the origin is fixed. Three points are needed for a general isometry such as this.

The transformation will be of the form $ f(X) \equal{} AX \plus{} B$, for some $ 2\times 2$ matrix $ A$ and $ 2\times 1$ vector $ B$. Start with the matrix: you want $ A$ to fix a vector in the direction of the line, and negate a vector perpendicular to that. Those eigenvectors and eigenvalues determine the matrix.

Once you have $ A$, use the fact that $ f$ fixes a point on the line to determine $ B$.

oh oops good catch (lol for some reason I pictured the line in my mind as passing through the origin) :blush:

btw... did you um... call me t0rajir0u?


I am working on a number square that is attached.  I can't seem to solve it.  It appears to be some sort of a Fibonacci sequence.  If you have three squares, the difference of the first two plus the value of the second square equals the third square.  This seems to be true of all the squares.  However, there seems to be something else, but I can't seem to find it.  I was wondering if anyone could give me some pointers on how I could finish the square.

Interesting puzzle

[hide="Click for solution"]
The left hand column and uppermost row are both the counting numbers.  This should be clear

Take any three squares like this:
\begin{tabular}[t]{|c|c|} \multicolumn{2}{c}{} \\
\hline a & b \\
\hline c & d \\
\hline \end{tabular}
Then it turns out $ a + b = d$.  We can then fill in the missing squares as follows
[hide="click again"]
\begin{tabular}[t]{|c|c|c|c|c|c|} \multicolumn{6}{c}{} \\
\hline 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 2 & 3 & 5 & 7 & 9 & 11 \\
\hline 3 & 5 & 8 & 12 & \boldsymbol{16} & 20 \\
\hline 4 & 8 & 13 & \boldsymbol{20} & 28 & 36 \\
\hline 5 & 12 & 21 & \boldsymbol{33} & 48 & 64 \\
\hline 6 & \boldsymbol{17} & 33 & \boldsymbol{54} & \boldsymbol{81} & \boldsymbol{112} \\
\hline \end{tabular}
[/hide]
[/hide]


probability
expected value
combinatorics proposed
combinatorics

[u]Problem Statement:[/u]

Assume that there are n items (numbered from 1 to n) in an urn.
We select b items from the urn and record their numbers.
We return the selected b items into the urn and perform another selection.
We do in total m such selections.

At the end of the m selections we check the recorded numbers. For every unit number chosen in the first selection, if it is also chosen in any of the rest m-1 selections we have a loss of 1 unit (the loss is [b]1 unit [/b]irrespectively if the same item has been re-chosen [b]more than 1 times[/b]).

[u][b]Example:[/b][/u] n=10,b=4,m=3

[b]Selection 1 | Selection 2| Selection 3[/b]
[i]item1         | item1        | item3
item5         | item7        | item1
item8         | item2        | item9
item7         | item3        | item10[/i]

In this example [i]item1[/i] is re-chosen in [b]Selection 2[/b] and [b]Selection 3[/b] so we loose 1 unit from it. Also[i] item7 [/i]is re-chosen in [b]Selection 2[/b] so we loose another 1 unit from it. In total we have lost 2 units.

[u][b]
We want to compute:[/b][/u]
[b]a) the formula that gives the average loss that we will have.
b) the average number of times that an item is re-chosen.[/b]

[i]Actually I think the formula must be of the form:

Probability(1 unit re-chosen in >=2 selections)*(-1)+Probability(2 units re-chosen in >=2 selections)*(-2)+...+Probability(b units re-chosen in >=2 selections)*(-b)= ???

Also I think that the probability that no item is re-chosen is as given in the attached image.[/i]

Thank you,

Christopher

Yes, $ \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}}$ is the probability that you have 0 loss.  The formula you suggested is correct, but would be tough to actually work out.  Instead, consider the probability that a given marked item appears again.  Since it appears with probability $ \frac{b}{n}$ on each choice, the probability it doesn't appear is $ 1 \minus{} \frac{b}{n}$, and so the probability it doesn't appear in $ m \minus{} 1$ choices is $ \left(1 \minus{} \frac{b}{n}\right)^{m \minus{} 1}$.  Then your expected loss is just $ b\left(1 \minus{} \left(1 \minus{} \frac{b}{n}\right)^{m \minus{} 1}\right)$.

[quote="JBL"]Yes, $ \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}}$ is the probability that you have 0 loss.  The formula you suggested is correct, but would be tough to actually work out.  Instead, consider the probability that a given marked item appears again.  Since it appears with probability $ \frac {b}{n}$ on each choice, the probability it doesn't appear is $ 1 \minus{} \frac {b}{n}$, and so the probability it doesn't appear in $ m \minus{} 1$ choices is $ \left(1 \minus{} \frac {b}{n}\right)^{m \minus{} 1}$.  Then your expected loss is just $ b\left(1 \minus{} \left(1 \minus{} \frac {b}{n}\right)^{m \minus{} 1}\right)$.[/quote]

JBL you are correct, I couldn't find a closed form of the suggested formula.
[b]But[/b] if $ \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}}$ is the probability that you have 0 loss, then would it be correct to say that the average loss can be given also by $ b(1\minus{}\frac {\binom{n\minus{}b}{b}^{m\minus{}1}}{\binom{n}{b}^{m\minus{}1}})$      ?
If this is correct then there is a [b]problem[/b] because the formula you proposed and this one are not equal...
I must be missing something...maybe using combinations to derive the probability of loss is wrong. But why?
Also, have you got any idea for the second question, i.e. the average times an item may be re-selected in general?

Christopher

[quote="Chrisverv"]if $ \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}}$ is the probability that you have 0 loss, then would it be correct to say that the average loss can be given also by $ b(1 \minus{} \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}})$      ?[/quote]

No.  $ \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}}$ is the probability that you have 0 loss total (taking all $ b$ of the marked balls into account simultaneously).  In order to apply linearity of expectation (the reason we get to just multiply by $ b$), we need to calculate instead the probability that a given marked ball contributes (or equivalently that it does not contribute) to our loss, then sum up over all marked balls.  This is the calculation I made above.

The second question is much easier -- on each draw, the probability of getting a given marked ball is $ \frac {b}{n}$, so the expected number of times you draw a given marked ball is $ \frac {b(m \minus{} 1)}{n}$.

I am trying to understand this a bit better.
So can we say that $ \left(1 \minus{} \left(1 \minus{} \frac {b}{n}\right)^{m \minus{} 1}\right)$ covers all possible outcomes of loss:
[b]1 specific ball is reselected 1 up to m-1 times
+
2 specific balls (of different number) are reselected 1 up to m-1 times
+
1 specific ball is reselected 1 up to m-1 times and a second (different) ball is reselected 1 up to m-1 times
+
3 specific balls (of different number) are reselected 1 up to m-1 times
+
2 specific balls (of different number) are selected 1 up to m-1 times and a third (different) ball is reselected 1 up to m-1 times
+
1 specific ball is reselected 1 up to m-1 times and a second (different) ball is reselected 1 up to m-1 times and a third (different) ball is reselected 1 up to m-1 times
+
and so on...[/b]

Is this correct?

I just want to be sure that all possible outcomes are taken into account in the computation of the average loss...

Thanx JBL!
Christopher

I'm not entirely sure what that list really means, so rather than answering "yes" or "no," let me elaborate a little on my previous calculation.  $ \frac{b}{n}$ is the probability that [i]a particular marked ball[/i] is chosen in a given pick.  Thus, $ 1 \minus{} \frac{b}{n}$ the probability that a particular marked ball is [i]not[/i] chosen in a given pick.  Thus, $ \left(1 \minus{} \frac{b}{n}\right)^{m \minus{} 1}$ is the probability that a particular ball is not chosen in any of the $ m \minus{} 1$ additional picks.  Thus, $ 1 \minus{} \left(1 \minus{} \frac{b}{n}\right)^{m \minus{} 1}$ is the probability that a particular marked ball is chosen in at least one of the $ m \minus{} 1$ additional picks.  Thus, $ b\left(1 \minus{} \left(1 \minus{} \frac{b}{n}\right)^{m \minus{} 1}\right)$ is the [i]expected number[/i] of marked balls to be selected again in $ m \minus{} 1$ picks.

Ok I think I got it now! Thanx!
And a final question is what does the following probability mean?:
$ (1 \minus{} \frac {\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m \minus{} 1}})$

Isn't it the probability that [i]at least[/i] 1 ball get re-selected [i]at least [/i]1 time?

If it means something else than I guess this is the reason why we cannot compute the average loss by using it.

Christopher

$ \frac{\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m\minus{}1}}$ is the probability that no ball is reselected.  Thus, $ 1 \minus{} \frac{\binom{n \minus{} b}{b}^{m \minus{} 1}}{\binom{n}{b}^{m\minus{}1}}$ is the complement: the probability that at least one ball is reselected at least once.

That's what I meant. Sorry ;-)

So we cannot derive an average when we look at the probability of loss taking into account all balls simultaneously.
Instead we should find the probability of loss for 1 ball and multiply by the total number of balls to get the average loss...

This must drive us to what you called "linearty of expectation"...I guess :-)

Christopher

[quote="Chrisverv"][b]Modification of the problem: [/b]

If we assume that in case of reselection of an item the loss is not certain but it depends on the number of times that the item has been reselected in the m-1 consecutive selections (i.e. the probability that the profit loss will happen is 1-1/(x+1), where x is the times that the item has been selected). So if an item is reselected 1 time than the probability of loss for it would be 1-1/2=1/2, if it is reselected twice than the loss probability is 1-1/3=2/3 and so on...until m-1 reselections where the loss probability becomes 1-1/m.

[i]In this case how can we compute the average loss? [/i][/quote]

[quote="Chrisverv"]Ok...a mistake in my previous post...the formula should be:
$ E \equal{} \sum_{k \equal{} 1}^rk{r\choose k}x^{k}(1 \minus{} x)^{r \minus{} k}*(1 \minus{} \frac {1}{k \plus{} 1})$
Since the probability of loss when we have 1 reselection (k=1) is 1-1/2=1/2, when k=2 it is 1-1/3=2/3 and so on...

[b]Unfortunately the above expression counts 1 loss per re-selection, however 1 re-selection may involve more than 1 (up to b) items and hence the average loss cannot be computed this way....Any ideas welcome![/b][/quote]

[b]Maybe a solution would be to multiply the expected number of items that are reselected by the expected number of reselections, which can be given by:[/b]

$ b\Bigl(1 \minus{} \bigl(1 \minus{} \frac bn\bigr)^{m \minus{} 1}\Bigr)\Bigl((m \minus{} 1)\frac bn\Bigr)$

[color=red][i][b]Is this correct?[/b][/i][/color]


[i]Prove that in any triangle the follow inequality holds:[/i]
\[ \sqrt{\sqrt{a^3b}\plus{}\sqrt{ab^3}\minus{}ab}\plus{}\sqrt{\sqrt{b^3c}\plus{}\sqrt{bc^3}\minus{}bc}\plus{}\sqrt{\sqrt{c^3a}\plus{}\sqrt{ca^3}\minus{}ca} \leq a\plus{}b\plus{}c\]
[b]Claudiu Mîndrilă[/b]

$ LHS = \sum_{cyc}\sqrt {\sqrt {ab}(a + b - \sqrt {ab})}\leq\sum {\frac {\sqrt {ab} + a + b - \sqrt {ab}}{2} = \sum {a}}$

 PS. Inequality is true for any a,b,c $ \geq$0.

Marius Mainea wrote:
[quote]LHS = [b]\sum_{cyc}\sqrt {\sqrt {ab}(a + b - \sqrt {ab})}\leq\sum {\frac {\sqrt {ab} + a + b - \sqrt {ab}}{2} [/b]= \sum {a}[/quote]

Are you sure this highlighted step is true? If yes then pls. carify.


Given distinct positive integers $a_{1},a_{2},...,a_{1}1$ with sum less than 2007. Write $1,2,...,2007$ on a blackboard in ascending order. 22 consecutive [i]operations[/i] are defined as an [i]operation group[/i]. The $i$th [i]operation[/i] in a [i]operation group[/i] is defined as follow: Choose any number on the blackboard, if $1\leq i \leq11$ then add $a_{1}$ to it, otherwise subtract $a_{i-11}$ from it.If the final result of an [i]operation group[/i] is an even permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as even.If the final result of an [i]operation group[/i] is an odd permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as odd. What is the difference between the number of even [i]operation groups[/i] and the number of odd [i]operation groups[/i]?
  [b]Remark.[/b] A permutation $x_{1},x_{2},...,x_{n}$ of $1,2,...,n$ is called an even permutation if $\prod_{i>j}(x_{1}-x_{j})$ is a positive number, otherwise, it is called a odd permutation.[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole paper[/url]

Wlog, $a_{1}<a_{2}<\dots<a_{11}$ and we make operations in other order: 1)+$a_{1}$; 2)$-a_{1}$; 3)$+a_{2}$; 4)$+a_{3}$,\dots,12)$+a_{11}$; then $-a_{2},$ $-a_{3},$\dots, $-a_{1}1$.

Lemma. If we get a permutation of 1,2,\dots,2007 at the end, then either two first operations $+a_{1},$ $-a_{1}$ were made with the same number, or there was a situation, when two numbers were equal.

Proof of the lemma.  Assume the contrary. Then the first operation was $x_{1}\rightarrow x_{1}+a_{1}$ for some $x>2007-a_{1}$, the second was $y\rightarrow y-a_{1}$ for some $y\le a_{1}$. The third operation was $z\rightarrow z+a_{2}$ for $z>2007-a_{2}$ or $z=x_{1}-a_{2}$, etc. Since $\sum a_{i}$<2007, we see that number $y-a_{1}$ will not be touched on operations 3,4,\dots,12. So, it will remain nonpositive and we will not have a permutation at the end.

So, if we get a permutation., there was the first operation, after which we have two equal numbers, either the first two operations cancel. In the first case, we may construct a bijection between odd and even permutations with fixed starting operations: just replace our two equal numbers and make all permutations with first of them with another and viceversa.

In the second case, the difference is $f(a_{2},\dots,a_{11})$ ($f$ is the answer in the problem), and multiplied by $a_{1}$ ways to cancel two first operations. SO, the final answer is that the number of even permutations minus the number of odd permutations equals $a_{1}a_{2}\dots a_{11}$

Yeah... I think that problem still needs to be fixed, otherwise Fedor's solution is incomplete.

[quote="easternlatincup"][quote="Fedor Petrov"]
So, if we get a permutation., there was the first operation, after which we have two equal numbers, either the first two operations cancel.

[/quote]

Well, but what you proved, is that either the first two operations cancel, or [b]at some stage [/b]we have to equal numbers. So, we don't need to have equal numbers at the first stage, do we?  :)[/quote]

We make some operations, and at some time we get two equal numbers. Now stop. Fix the made operations, call them "the start". Each operation group has unique start (either two first operations are $ x\rightarrow x\plus{}a\rightarrow x$). I claim that for each start, the numbers of odd and even permutations with this start are equal. Then sum up by all possible starts.

If I am not mistaken, this solution resembles the official solution given. Is Russian team invited to this competition? if they are, I hope they could come out with a different solution. The idea of bijection is nice and hard to think of...only five people get partial marks for this question from the official result, but I am not sure about the Russian team's performance.

Nobody in the Russian tema solved this porblem.


Eh... I'm a beginner in LaTeX, and I was trying to compile my first LaTeX file.

I copied and pasted

\documentclass{article}
\begin{document}
Hello, world!
\end{document}

in the window, and i pressed ctrl + F7.

but it says 'There was an error opening this document. This file cannot be found.' 3 times!!!!!

and i guess the LaTeX result should be

0 error, 0 warning, 0 bad box, 1 page, but mine is 0 page.

Is it a problem dealing with 'Define Output profiles'?

I NEED A HELP!!! PLEASE!!!

What does CTRL+F7 do? :huh:

Maybe it's "Open Document"? :roll:

[quote="desperado"]Is it a problem dealing with 'Define Output profiles'?[/quote]Who knows? You're not even clear what editor you are using (TexnicCenter??) or what profile you are using or what you did to set it up.
Was the tex document saved? If not, then save it and try again.

ctrl+F7 is 'build current file', and i saved file.

You have to give us much more information - we can't see what's going on on your computer:
Are you using
1. Windows?
2. MiKTeX?
3. TexnicCenter?

If yes to all three, did you tell TexnicCenter where to find latex binaries when you first started it?

a) Now go to Build, Define Output Profiles and post a screenshot of that page.
b) Then tell us where you have installed MikTeX
c) Was a log file created for the tex file (it will be in exactly the same place as the tex file)? What does it say?
d) Copy here [i]exactly[/i] what TexnicCenter says as the error message.


function
inequalities
geometry
geometric transformation
reflection
Olimpiada de matematicas

[color=blue]Hola a todos,

Como sabran en las IMO's , siempre se tiene una sección para este tipo de Problemas: Las Inecuaciones!!!..  :idea:

Lo interesante de este  tipo de problemas es que en su [b]simplicidad[/b] se encierran grandes concepciones teoricas en algunos casos, y destrezas operativas en otras.   :wink:
De una u otra manera siempre encontraremos grandes satisfacciones el resolver este tipo de problema....

Esperemos fomentar en latinoamerica una cultura de resolver este tipo de problemas, pues como veran, no la tenemos muy bien desarrollada ... en comparacion a paises como China, USA, Japon, Korea, Rumania, etc.

Carlos Bravo  :wink:
Lima - PERU[/color]

[color=blue][b]PROB. 01[/b]

Dados $a,b,c>0$ numeros reales, tales que: $abc = 1$.

Mostrar que:

$\frac{a}{b}+\frac{b}{c}+\frac{1}{a}\geq a+b+1.$

Saludos,

PD: Numerar los problemas y colorear en azul los enunciados para tener un orden!!! Gracias..

Carlos Bravo
Lima - PERU[/color]

Bueno, creo que la intencion de Carlos es generalizar el problema que esta en el reto de la semana .
Si $a_{1},a_{2},..,a_{n}$ numeros reales postivos talque $a_{1}a_{2}...a_{n}=1$
$= > a_{1}/a_{2}+a_{2}/a_{3}+..+a_{n-1}/a_{n}+a_{n}/a_{1}\geq a_{1}+a_{2}+...+a_{n}$
---
Y en el problema seria $n=4$, donde $a_{1}=a, a_{2}=b, a_{3}=c, a_{4}=1$.

[color=blue][b]PROB. 02[/b]

Dados $a,b,c>0$ numeros reales.

Probar:

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 3/2.$

Saludos,

PD: Numerar los problemas y colorear en azul los enunciados para tener un orden!!! Gracias..

Carlos Bravo
Lima - PERU[/color]

[color=blue][b]PROB. 03[/b]

Dados $a,b,c>0$ numeros reales.

Mostrar que:
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{a+b+c}{2}.$

Saludos,

PD: Numerar los problemas y colorear en azul los enunciados para tener un orden!!! Gracias..

Carlos Bravo
Lima - PERU[/color]

Por la desigualdad de Cauchy-Schwartz:
\[(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{b+a})((b+c)+(c+a)+(a+b))\geq (\sqrt{\frac{a^{2}}{b+c}(b+c)}+\sqrt{\frac{b^{2}}{a+c}(a+c)}+\sqrt{\frac{c^{2}}{b+a}(b+a)})^{2}=(a+b+c)^{2}\]
de donde se obtiene lo que se pide.

$Tipe$

$\boxed{Sp_{2}}$

[color=blue]Por[/color] $MA \ge MH$ [color=blue]:[/color]

$\frac{\frac{a+b}{b+c}+\frac{c+a}{b+c}+\frac{b+c}{c+a}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{c+a}{a+b}}{6}\ge \frac{6}{\frac{b+c}{a+b}+\frac{b+c}{c+a}+\frac{c+a}{b+c}+\frac{c+a}{a+b}+\frac{a+b}{b+c}+\frac{a+b}{c+a}}$

$\left(\frac{a+b}{b+c}+\frac{c+a}{b+c}+\frac{b+c}{c+a}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{c+a}{a+b}\right)^{2}\ge 6^{2}$

$\frac{a+b}{b+c}+\frac{c+a}{b+c}+\frac{b+c}{c+a}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{c+a}{a+b}\ge 6$

$\frac{2a+b+c}{b+c}+\frac{2b+c+a}{c+a}+\frac{2c+a+b}{a+b}\ge 6$

$\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+3 \ge 6$

$\boxed{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}}$

[color=blue]Saludos[/color]

$\boxed{Sp_{3}}$ [color=blue]una solución alternativa con otra desigualdad mas potente (para los que suelen manejarse con desigualdad de las medias):[/color]

[color=blue]Demostraremos que:[/color] $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge a+b+c-\left( \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \ge \frac{a+b+c}{2}$

[color=blue][u]Parte 1:[/u][/color] $MA \ge MG$

$\frac{\frac{(a+c)(a+b)}{(b+c)}+\frac{(b+c)(a+b)}{(a+c)}}{2}\ge \sqrt{(a+b)^{2}}=a+b$

$\frac{\frac{(b+c)(a+b)}{(a+c)}+\frac{(b+c)(a+c)}{(a+b)}}{2}\ge \sqrt{(b+c)^{2}}=b+c$

$\frac{\frac{(b+c)(a+c)}{(a+b)}+\frac{(a+c)(a+b)}{(b+c)}}{2}\ge \sqrt{(a+c)^{2}}=a+c$

[color=blue]Sumando:[/color]

$\frac{a^{2}+bc+a(b+c)}{b+c}+\frac{b^{2}+ac+b(a+c)}{a+c}+\frac{c^{2}+ab+c(a+b)}{a+b}\ge 2a+2b+2c$

$\frac{a^{2}+bc}{b+c}+\frac{b^{2}+ac}{a+c}+\frac{c^{2}+ab}{a+b}\ge a+b+c$

$\boxed{\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge a+b+c-\left( \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) }$

[color=blue][u]Parte 2:[/u][/color] $MA \ge MG$

$\frac{a^{2}+b^{2}}{2}\ge ab$

$a^{2}+b^{2}\ge 2ab$

$(a+b)^{2}\ge 4ab$

$\frac{a+b}{4}\ge \frac{ab}{a+b}$

[color=blue]Analogamente:[/color]

$\frac{b+c}{4}\ge \frac{bc}{b+c}$

$\frac{c+a}{4}\ge \frac{ca}{c+a}$

[color=blue]Sumando:[/color]

$\frac{a+b+c}{2}\ge \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}$

$\boxed{a+b+c-\left( \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \ge \frac{a+b+c}{2}}$

[color=blue]Saludos[/color]

[color=blue]Y como ando jugoso:[/color]

$\boxed{Sp_{3}''}$

[color=blue]Con lo demostrado en el problema 2:[/color]

$(a+b+c)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \ge \frac{3}{2}(a+b+c)$

$\frac{a^{2}}{b+c}+a+\frac{b^{2}}{c+a}+b+\frac{c^{2}}{a+b}+c \ge \frac{3}{2}(a+b+c)$

$\boxed{\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\ge \frac{a+b+c}{2}}$

[color=blue]Saludos[/color]

Bueno hay mas generalisaciones de la desigualdad de nesbitt.
[color=blue]
1.Si $a,b,c \in \mathbb{R}^{+}$  y $n$ es un entero positivo entonces se cumple que:
$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{c+a}\ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$

2. Si $a_{1}, a_{2}, ... a_{k}\in \mathbb{R}^{+}$ cumplen que $\sum_{i=1}^{k}a_{i}= s$ y $n$ es un entero positivo entonces se cumple que:
$\sum_{i=1}^{k}\frac{a_{i}^{n}}{s-a_{i}}\ge \frac{\sum_{i=1}^{k}a_{i}^{n-1}}{k-1}$[/color]

[quote="carlosbr"][color=blue][b]PROB. 02[/b]

Dados $a,b,c>0$ numeros reales.

Probar:

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 3/2.$

[/color][/quote]
SOLUCION.-
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{ac+bc}$, por el Lema de Tittu.
$\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{ac+bc}\geq \frac{(a+b+c)^{2}}{2(ab+ac+bc)}\geq 3/2$.
                                                       $Q.E.D$.

[quote="carlosbr"][color=blue][b]PROB. 03[/b]

Dados $a,b,c>0$ numeros reales.

Mostrar que:
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{a+b+c}{2}.$

[/color][/quote]
SOLUCION:
POR EL LEMA DE TITTU(VARIACION DE CAUCHY).
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{a+b+c}{2}.\geq \frac{(a+b+c)^{2}}{2(a+b+c)}=\frac{(a+b+c)}{2}$
                                                                                $Q.E.D.$
Nota.- el Lema de titu es el siguiente.
sea $x_{1},x_{2},...,x_{n}$, $y_{1},y_{2},..,y_{n}$, donde son numeros reales positivos.
se cumple que:
$\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+...+\frac{x_{n}^{2}}{y_{n}}\geq \frac{(x_{1}+x_{2}+...+x_{n})^{2}}{y_{1}+y_{2}+...+y_{n}}$.
 LA DEMOSTRACION ES CON CAUCHY, Y ES COMO INDICA TIPE.

[quote="aev5peru"][quote="carlosbr"][color=blue][b]PROB. 05[/b]

Dados $a,b,c>0$, tal que: $a+b+c = 4.$

Probar que:
$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 1.$

[/color][/quote]
1ero.
        Cuando se da la igualdad.
$< = >\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}=\frac{3+2(a^{2}+b^{2}+c^{2})+(ab)^{2}+(bc)^{2}+(ca)^{2}}{1+a^{2}+b^{2}+c^{2}+(ab)^{2}+(bc)^{2}+(ca)^{2}+(abc)^{2}}\geq 1$
$< = >3+2(a^{2}+b^{2}+c^{2})+(ab)^{2}+(bc)^{2}+(ca)^{2}\geq 1+a^{2}+b^{2}+c^{2}+(ab)^{2}+(bc)^{2}+(ca)^{2}+(abc)^{2}$
$< = >2+a^{2}+b^{2}+c^{2}\geq (abc)^{2}$, pero no encuentro, valores para los cuales se dan la igualdad.[/quote]

Buenos dias,
gracias por enviar tu sol aev5peru...
pero no se ve  muy elegante... jeje

posteare la solucion "ofiicial" de fabycv ..
pero les dare chance a q posteen alguna sol adicional.

carlos bravo
Lima - PERU.

[quote="carlosbr"][color=blue][b]PROB. 05[/b]

Dados $a,b,c>0$, tal que: $a+b+c = 4.$

Probar que:
$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 1.$

Creado por [b]Fabycv[/b]

Saludos,

PD: Numerar los problemas y colorear en azul los enunciados para tener un orden!!! Gracias..

Carlos Bravo
Lima - PERU[/color][/quote]

Buenos dias,
este es uno de varios problemas que poden encontrar en este post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91541

aunq aun no tengo la sol de [b]Fabycv[/b] en las manos, aki posteo la mia...

Hint: Uso de Jensen...
$f(x) = \frac{1}{1+x^{2}}$, con $x>0.$
Osea: $f(a)+f(b)+f(c) \geq 3f(\frac{a+b+c}{3}).$

En la inecuacion tendríamos:

$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 3f(4/3) = 27/25 >1.$

Saludos,

[quote="carlosbr"][color=blue][b]PROB. 05[/b]

Dados $a,b,c>0$, tal que: $a+b+c = 4.$

Probar que:
$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 1.$

[/quote]

Ok aki la solucion de $Fabycv$...

Dado $x>0$, tenemos: $\frac{1}{1+x^{2}}\geq \frac{2-x}{2}.$

Para la ineq: $\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 3-\frac{a+b+c}{2}= 1.$
Lqqd.

Saludos,

[quote="carlosbr"][quote="carlosbr"]

Hint: Uso de Jensen...
$f(x) = \frac{1}{1+x^{2}}$, con $x>0.$

[/quote][/quote]  :maybe: Cómo garantizas que se pueda aplicar Jensen a esa función....derivando??  :huh:

Otra cosa: Qué es el método AC ?

$Tipe$

[quote="tipe"][quote="carlosbr"][quote="carlosbr"]

Hint: Uso de Jensen...
$f(x) = \frac{1}{1+x^{2}}$, con $x>0.$

[/quote]  :maybe: Cómo garantizas que se pueda aplicar Jensen a esa función....derivando??  :huh:

Otra cosa: Qué es el método AC ?

$Tipe$[/quote]

Haber, ...
[b]
Respecto a lo 1ero ...[/b]
una expresion como estas ${\frac{1}{a+bx^{n}}}, b<>0$, nos genera una rama de hiperbola, simetrica al eje Y, y asintoticas al Eje X...
 que sabemos q es convexa, (para $x>0$)

O en caso contrario, toma dos puntos distintos $a>b>0$, que seran moviles sobre la curva y toma un tercero (a<k<b) en el intervalo q determinan estos dos primeros, por variaciones veran que se cumple:
$f(k) \leq \frac{1}{2}f(\frac{a+b}{2})$.

[b]Respecto al 2do:[/b]

[quote="Vasc"]I called the AC-Theorem the following statement:
[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $s>0$, and let $F(x_{1},x_{2},...,x_{n})$ be
a symmetrical continuous function on the compact set in $R^{n}$,
$S=((x_{1},x_{2},...,x_{n}): x_{1}+x_{2}+...+x_{n}=s, x_{1}\geq 0,..., x_{n}\geq 0)$.
If $F(x_{1},x_{2},...,x_{n}) \leq \max{(F(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2}, x_{3},...,x_{n}),F(0,x_{1}+x_{2},x_{3},...,x_{n}))}$
for all $(x_{1},x_{2},...,x_{n})\in S$ with   $x_{1}>x_{2}>0$, then
$F(x_{1},x_{2},...,x_{n}) \leq \max_{1\leq k\leq n}{F(\frac{s}{k},..., \frac{s}{k},0,...,0)}$  for all $(x_{1},x_{2},...,x_{n})\in S$.[/quote][/quote]

Revisar el link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82172

Saludos,

[quote="carlosbr"][quote="tipe"][quote="carlosbr"]

Hint: Uso de Jensen...
$f(x) = \frac{1}{1+x^{2}}$, con $x>0.$

[b]
Respecto a lo 1ero ...[/b]
una expresion como estas ${\frac{1}{a+bx^{n}}}, b<>0$, nos genera una rama de hiperbola, simetrica al eje Y, y asintoticas al Eje X...
 que sabemos q es convexa, (para $x>0$)

O en caso contrario, toma dos puntos distintos $a>b>0$, que seran moviles sobre la curva y toma un tercero (a<k<b) en el intervalo q determinan estos dos primeros, por variaciones veran que se cumple:
$f(k) \leq \frac{1}{2}f(\frac{a+b}{2})$.
[/quote][/quote][/quote]

No me convence lo de "sabemos que es convexa", te refieres a ese tipo de "hipérbolas" ?, y que te refieres con lo de variaciones?

$Tipe$

[quote="conejita"]Disculpen, pero para el problema 6, si elijo $a=b=c=0$ se cumple la primera desigualdad y sin embargo la segunda desigualdad me keda $3\leq (1/2)$.
Kreo k ay algun error en el problema.[/quote]
Parece ser q la restriccion es erronea
Revisare el enunciado....

saludos,

[quote="tipe"][/quote][quote="carlosbr"][quote="tipe"][quote="carlosbr"]

Hint: Uso de Jensen...
$f(x) = \frac{1}{1+x^{2}}$, con $x>0.$

[b]
Respecto a lo 1ero ...[/b]
una expresion como estas ${\frac{1}{a+bx^{n}}}, b<>0$, nos genera una rama de hiperbola, simetrica al eje Y, y asintoticas al Eje X...
 que sabemos q es convexa, (para $x>0$)

O en caso contrario, toma dos puntos distintos $a>b>0$, que seran moviles sobre la curva y toma un tercero (a<k<b) en el intervalo q determinan estos dos primeros, por variaciones veran que se cumple:
$f(k) \leq \frac{1}{2}f(\frac{a+b}{2})$.
[/quote]

No me convence lo de "sabemos que es convexa", te refieres a ese tipo de "hipérbolas" ?, y que te refieres con lo de variaciones?

$Tipe$[/quote][/quote][quote="tipe"][/quote]

ok Jorge,

en la seccion Ineqs, vemos q la convexidad de las funciones q se toman se analizan derivando,
puesto q la gente tiene claro los conceptos de funciones, cosa q incluve analizar convexidades, puntos de inflexion, asintotas, etc.
(talvez lo q quieres es no usar estos conceptos aki....)
 o buscar alguna manera de mostrar justamente la desigualdad que se desea tener..
$f(k) \leq \frac{1}{2}f(\frac{a+b}{2})$.
para todo valor dentro del intervalo de analisis.

talvez no haya encontrado las palabras adecuadas para expresar mis ideas, y quisiera q nos ayudes aki...

Gracias de antemano..

carlos Bravo :)
Lima - PERU

El problema puede ser resuelto así como está, pienso yo :roll: Primero, observemos que la condición $\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{abc}$ puede ser escrita como
\[\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\leq 1 \]
Creo que es por esta razón que el caso $a=b=c=0$ falla; necesitamos que $a$, $b$ y $c$ sean estrictamente positivos para poder hacer esto. Por MA-MG tenemos que
\[\frac{1+x}{2}\geq \sqrt{x}\Rightarrow \frac{1}{1+x}\leq \frac{1}{2\sqrt{x}}\]
Usando esto obtenemos
\[\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\leq \frac{1}{2}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\leq \frac{1}{2}\]

jeje, esta tuanis este topic... jeje, hasta hoy lo encuentro... que tal si intentan la siguiente desig mia q aparecio en mathematical reflections issue 1/2007...?

sean $a,b,c$ reales positivos tales que $a+b+c+1=4abc$. probar que

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}$.

pd: jutaro (gabriel) no postee su solucion jeje.. :D

[quote="carlosbr"][color=blue][b]PROB. 06[/b]

Dados $a,b,c \geq 0$, tal que: $\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{abc}.$

Probar que:
${\frac{1}{1+ab}+\frac{1}{1+bc}}+\frac{1}{1+ca}\leq \frac{1}{2}.$

Creado por [b]Fabycv[/b]

Saludos,

PD: Numerar los problemas y colorear en azul los enunciados para tener un orden!!! Gracias..

Carlos Bravo
Lima - PERU[/color][/quote]

Hola despues de mucho tiempo...

Llamemos $x^{2}=bc$, $y^{2}=ca$, $z^{2}=ab$ de alli que, despejando $\frac{1}{\sqrt[]{bc}}+\frac{1}{\sqrt[]{ca}}+\frac{1}{\sqrt[]{ab}}\leq 1$ se convierte en $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 1$ y la desigualdad objetivo se transforma en $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}\leq \frac{1}{2}$

Sea $\alpha=arctg(\frac{1}{x})$, $\beta=arctg(\frac{1}{y})$, $\theta=arctg(\frac{1}{z})$.  De ahi que, transformando las ecuaciones se sigue que:

$tg\alpha+tg\beta+tg\theta\leq 1$ con lo que se tiene que probar $sen^{2}\alpha+sen^{2}\beta+sen^{2}\theta\leq \frac{1}{2}$

Lo cual es inmediato $\blacksquare$

[quote="M4RI0"]Aqui les va una fácil :D

Demostrar que: $(a+b)(a+c) \geq 2\sqrt{abc(a+b+c)}$,

para todos los números reales positivos $a$,$b$,$c$.   $(Brazil MO, 2001)$[/quote]

bueno creo q si estaba facil! por MA-MG \[2\sqrt{bc \cdot a(a+b+c)}\leq bc+a(a+b+c)=a^{2}+ab+ac+bc=(a+b)(a+c)\]

Este esta másomenos :P  fácil también:

[color=blue]
Sean $ A$, $ B$ y $ C$ tres puntos distintos; $ AB\equal{}c$, $ BC\equal{}a$ y $ CA\equal{}b$. Provar que:

$ \frac{|a\minus{}b|}{\sqrt{a^2\plus{}b^2}}\plus{}\frac{|b\minus{}c|}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{|c\minus{}a|}{\sqrt{c^2\plus{}a^2}}< 2\sqrt{2}$

[/color]

 :)


Heres a problem from 2001 UNC Contest (University of Northern Colorado)

An open rectangular box can be formed by cutting identical squares off the four corners of a rectangular piece of cardboard and folding up the four sections that stick out. For a paticular piece of cardboard, the same volume results whether squares of size 1 or squares of size 2 have been cut out. Comput the resulting volume if squares of size 3 are cut out.

i believe the answer is [hide] 24.[/hide]

[hide]Let the rectangular piece have width x and length y; then

(x - 2)(y - 2) = 2(x - 4)(y - 4)

xy - 6x - 6y + 28 = 0

(x - 6)(y - 6) = 8

3(x - 6)(y - 6) = 24.

Therefore the box with depth 3 has volume 24.[/hide]

Yea, here is solution: [hide]24; Set V(1) = (x-2)(y-2) and V(2) = 2(x-4)(y-4) equal to each other, giving xy - 2x - 2y + 4 = 2xy - 8y - 8x + 32 or 0 = xy - 6x - 6y + 28. Then V(3) = 2(x-6)(y-6) = 3(xy - 6x - 6y + 36) = 3(-28+36) = 24[/hide]

Yea it prolly shouldnt be on advanced, my bad.


Im confused abacadea said tghat 2=1..I dont get it.

To prove that, you'd have to make the mathematical mistake of dividing by 0. Basically, it's just a bogus proof.

[quote="abacadaea"]this guy is just posting spam dont reply to his topics[/quote]
You posted that question first hypocrite so stfu

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158099

see...so why dont YOU stop spamming

just stop posting on his threads...simple solution..if he's ignored, he'll eventually stop.

[quote="lokesh"]No offense, but when abacadaea started the other thread, it was implied that we were supposed to be invalid proofs. You, on the other hand seem to already KNOW that 1 doesn't equal 2, but seem so serious to know WHY 1=2.[/quote]
I dont get it..


Are there two 2003 digit numbers A and B with the following properties: the digits of B are a rearrangement of those of A, and the digits of A + B are all 9s? Is there such a pair of numbers with 2004 digits?

[hide]

Answer to part A: no.

If there are two numbers such that this property holds, then there are two cases:

a) There are carries. If this is true, then the maximum sum of any two units' places is 10 to 18. In any case, there is an non-9 digit left in the sum somewhere, so there are no such cases.

b) There are no carries. If this is true, each unit's digit n is added to 9 - n, implying that n=1, 2, 3, 4 and 9-n = 8, 7, 6, 5 without loss of generality. However, since A and B are rearrangements of each other, and either number has an odd number of digits, we conclude there is no such set of numbers since odd numbers do not "pair up".

Therefore, we conclude that there are no numbers A and B.

Answer to Part B:  yes.

Return to case (b). If there are an even number of digits, then there  are cases where it holds. 81818181... + 18181818... and 27272727... + 72727272... are two such cases. Therefore, there are numbers A and B in pairs of numbers with 2004 digits.

[/hide]


function
search
algebra unsolved
algebra

Find all functions$ f: Z \to Z$ satisfying :\[ %Error. "eqalign" is a bad command.
{
  \& f(x^3  + y^3  + z^3 ) = f^3 (x) + f^3 (y) + f^3 (z)  \cr\],\[ \forall x,y,z \in Z \cr}\]

This is Problem 3 in Vietnam TST 2005

You can see here
[url]http://www.mathlinks.ro/viewtopic.php?search_id=262129995&t=14908[/url]


A perfect semicircular shell in 2 dimensions.  The radius of the shell is 6 cm.  The shell is between two parallel lines, where the top one is 9 cm high. The shell rocks back and forth.  When the semi-circle hits the top line, it reverses direction and hit teh top line somewhere else.  What is the distance between these two points?k

I think this is what you mean... if so, it should help

but that link just links back to here....

I don't think it [b]just[/b] links back to here ;)

Read Filip J and Pontios' posts :)


geometry
parallelogram
cyclic quadrilateral
IMO Shortlist
IMO Longlist

Show that the four perpendiculars dropped from the midpoints of the sides of a cyclic quadrilateral to the respective opposite sides are concurrent.

[b]Note by Darij:[/b] A [i]cyclic quadrilateral [/i]is a quadrilateral inscribed in a circle.

Let $a,b,c,d$ be the affixes of the vertices of the quadrilateral. Now let $m_1=\frac{a+b}2,\ m_2=\frac{b+c}2,\ m_3=\frac{c+d}2,\ m_4=\frac{d+a}2$ be the affixes of the midpoints of the sides. Furthermore, assume $O(0)$, the center of $(ABCD)$, is the origin of the plane. Let $\ell_i$ be the line dropped from $M_i$ to the opposite side of the quadrilateral.

We can see that the quadrilateral formed by the perpendicular bisectors of $AB,CD$ and $\ell_1,\ell_3$ form a parallelogram, having $O$ as one of its vertices. $\ell_1\cap\ell_3$ then has affix $m_1+m_3$. In the same manner we show that $\ell_2\cap\ell_4$ has affix $m_2+m_4$. However, $m_1+m_3=m_2+m_4=\frac{a+b+c+d}2$, which means that $\ell_1,\ell_2,\ell_3,\ell_4$ are concurrent, Q.E.D.

This is just a technically brilliant proof by grobber.

Dear Mathlinkers,
the point of concurs is the retrocenter of ABC.
This result is also known as the Bretschneider's theorem but it is much older.
For a synthetic proof, consider the circles with diameter the side of ABCD....
Sincerely
Jean-Louis


superior algebra
superior algebra unsolved

A small thing I was wondering about

One constructs a ring by demanding that $R,+$ is a commutative group. Can't we drop this axiom because this follows from the distributivity axioms.

Then we have that

$(1+1)(a+b)=a+b+a+b$
$(1+1)(a+b)=(2)(a+b)=2a+2b=a+a+b+b$

hence

$a+b+a+b=a+a+b+b$

yields

$b+a=a+b$

as wanted. Or are we in trouble now? Because, if we assume a ring doesn't necessarily requires a unit, this would not work? If someone finds a counterexample, please let me know :)

You are right, yannick.
If we drop the requirement of commutative addition, we get it for free, as you have shown.

But your proof requires a unit.
If we drop the unit, we can equip $S_{3}$ with the multiplication $x y : = \mathrm{id}$, which fulfils associativity and distribution.

There are examples where the multiplication is not always zero.

LiMa


calculus
integration
function
real analysis
real analysis unsolved

Define increasing functions $ \alpha$, $ \beta$, and $ \gamma$ on $ [\minus{}1,1]$ by $ \alpha\equal{}\chi_{(0,1]}$, $ \beta\equal{}\chi_{[0,1]}$, and $ \gamma\equal{}\frac{1}{2}(\alpha\plus{}\beta)$.  Given $ f\in B[\minus{}1,1]$, show that:
a) $ f\in\mathcal{R}_{\alpha}[\minus{}1,1]$ iff $ f(0\plus{})\equal{}f(0)$.
b) $ f\in\mathcal{R}_{\beta}[\minus{}1,1]$ iff $ f(0\minus{})\equal{}f(0)$.
c) $ f\in\mathcal{R}_{\gamma}[\minus{}1,1]$ iff $ f$ is continuous at $ 0$.
d) If $ f\in\mathcal{R}_{\gamma}[\minus{}1,1]$, then $ \int_{\minus{}1}^{1}f\,d\alpha\equal{}\int_{\minus{}1}^{1}f\,d\beta\equal{}\int_{\minus{}1}^{1}f\,d\gamma\equal{}f(0)$

I believe, although not 100% sure, that this is the same as problem 3 on page 138 of Rudin.  [url=http://www-math.mit.edu/~katrin/teach/18.100/18.100pset10.pdf]This[/url] page is from the course that JBL TA'ed last semester, and that problem from Rudin is solved here.  Is the solution applicable to the problem I have posted?

Yes, it's the same problem with some minor aesthetic differences (most notably, with $ (\beta_1, \beta_2, \beta_3)$ instead of $ (\alpha, \beta, \gamma)$).  I remember writing that solution -- it was relatively unpleasant :)

Glad to know I'm not the only one who finds that problem annoying.  Could you elaborate on your solution for part (d) a little, please?

That $ \int_{\minus{}1}^1 f \, d\alpha \equal{} f(0)$ is actually implicit in the solution of part (a): we showed that $ M(y)$ and $ m(y)$ approach the same limit by showing that the limit they approach is $ f(0)$, and these since these are precisely the upper and lower sums, we're done.  Then the result where we replace $ \alpha$ by $ \beta$ is completely analogous (hold your proof for $ \alpha$ up to a mirror ;)), and the result for $ \gamma$ is either a slight modification or a consequence of the result for $ \alpha$ and $ \beta$.  (I think there's a theorem in Rudin that lets you add and multiply by a constant in the function you're integrating against assuming you don't ever break integrability, right?)


What about the solution don't you understand? If you tell us what gives you trouble, we can clarify it for you.  :)

in the solution it says that if $ p>50$ it appears twice in denominator but even if $ p<50$ it should appear twice right? And after that i don't really understand anything :(

[quote="Poincare"]What is the largest 2-digit prime factor of the integer {200 \choose 100}?
I got this from Aops Wiki but i don't really understand the solution given there :([/quote]

Is there any problem in your latex?

$ \binom{200}{100}\equal{} \frac {200\times 199\times ... \times 101}{100!}$

That 2-digit $ p$ is at least one time in the denominator so it must appear at least 2 times in the numerator. That means $ 3p \le 200$ so $ p<67$ The biggest $ p$ is $ 61$.

Am I correct?

i don't understand why it has to appear twice in the numerator  :(

Becaus if it appears only once in the numberator and once n the denominator, it will cancel out.

thanks everybody! especially FantasyLover and Mewtwo!!  :D

[quote="Poincare"]thanks everybody! especially FantasyLover and Mewt[b][size=200]w[/b][/size]o!!  :D[/quote]

Ahem?


Here is basically what Dr Sonnhard Graubner did (or something similar, I think).  Let's try to get all the numbers in terms of $ 2^4$.
$ 2^5$ is $ 2 \times 2^4$, $ 2^6$ is $ 2 \times 2 \times 2^4$, or $ 4 \times 2^4$, and $ 2^7$ is $ 2 \times 2 \times 2 \times 2^4$, or $ 8 \times 2^4$.  Because we now have like terms, we may combine them.  $ 2^4$ is basically $ 1 \times 2^4$.  So we can do $ (1\plus{}2\plus{}4\plus{}8) \times 2^4$, which is $ 15 \times 2^4$.  Now we have $ \frac{15 \times 2^4}{4}$
We make $ 2^4$ into $ 2^2 \times 2^2$, which is $ 4 \times 4$.  Now we have $ \frac{15 \times 4 \times 4}{4}$.  We see that a $ 4$ on the top cancels out with the $ 4$ on the bottom.  We have $ 15 \times 4$, which gives us our answer, $ \boxed {60}$


invariant
vector
linear algebra
matrix
trigonometry
linear algebra unsolved

Let $ V$ be a real vector space.
Suppose we have a nonzero $ v \in V$ such that $ (T^2 \plus{} b T \plus{} c)v \equal{} 0$ where $ b^2<4c$.
Find a $ 2$-dimensional $ T$-invariant subspace of $ V$ and describe all bases for that subspace with the property that the matrix of $ T$ on the subspace with respect to these bases has the form
   \[ r \begin{bmatrix}
   \cos(\theta) & \sin(\theta) \\
   \minus{}\sin(\theta) & \cos(\theta)
   \end{bmatrix}\]
for some $ \theta,r$ with $ r \ge 0$.

Assuming that $ T$ denotes an endomorphism of $ V$, show that $ W\equal{}span\{v,T(v)\}$ is a 2-dimensional $ T$-invariant subvector space of $ V$. Then diagonalize the induced endomorphism $ T'$ of $ W$ over $ \mathbb C$ and create an appropriate basis of $ W$ over $ \mathbb R$ to get your result.


First, I want to say, that I'm sooo happy I found this site.

My name is Angelo.  I live in Miami, Fl. I'm currently a student studying computer programming at MDC. I will be reading through the forums and trying to learn as much as I can. I have one question about a problem I have in class.

It's really difficult for me to post fractions in the forum, or maybe I just don't know how, so I decided to post my problem in the form of a picture.

Please see attached picture question, and any you can give me will be greatly appreciated.

Thank you !
 :D

After reviewing my post. I see that the picture is not very clear.

In this problem, I'm asked to solve for n.
First is the problem, and the second is the answer.
I just wanted some tips on how this answer might have been solved.

$I = \displaystyle\frac{ne}{nr+R}$

$Inr + Ir = ne$

$Inr - ne = -Ir$

$n(Ir - e) = -Ir$

$n = \displaystyle\frac{-Ir}{Ir-e}$

just try to collect like terms and then factor.

thank you very much.
things are much clearer now.

 :D


A 50 L closed flask at 500 K has the following reaction  -
$2A_{(g)}$ + $B_{(}g)$  --> $3C_{(g)}$
A and B are inert towards water and C forms an alkaline solution with water.Initially 3 moles of A and 2 moles of B are taken.Then after the reaction is complete 0.5 L of water is poured into the flask.The flask is then cooled to 200 K.Determine the final pressure in the flask (in atm).

A mixture of propane and acetylene and just enough amount of oxygen are taken in the flask.th initial pressure is found to be 80 atm.then after combustion the carbon dioxide evolved is found to have a pressure of 42.5 atm.The determine the amount fraction of acetylene in the original mixture.

Hey Ashwin! Here are the working hints.Probably they would help you out..i hope. And if you seek any further problem, you can ask me anytime

For the first problem: Use PV=nRT.

Calculate n from the given reaction and put its value.. Take volume as 50+.5 Litre. Hence you will finally get the value of P'A

For the second one:Take the 2 equations of C2H2 + 5/2O2---->2CO2+H2O

The second equation: C3H8+5O2------>3CO2 + 4H2O

Now take the partial pressures and solve on further..

Ok take care
Hope i could help.. :maybe:
Shreya :lol:


given 101 coins where 1 coin has 2 heads and the rest of the coins have a head and a tail.find the probability of getting a head on flipping a random coin

Multiply and add probabilities.
\[ \frac{100}{101}*\frac12\plus{}\frac1{101}\equal{}\frac{51}{101}\]


I couldn't make it this year, sigh, so can some of you let me know what happened at each site?

Thanks...

TJ was first place team beating us (SFBA) by 2 points (argh). Chicago was 3rd, Montgomery Blair 4th, NYC and Phillips Exeter 5th.

As for individual, Anders Kaseorg from North Carolina was 1st, since he was the only one to get 8. Jonathan Lowd from SFBA (and my high school actually) was 2nd. I don't know who was 3rd. I got a 7 but botched up the tiebreaks really badly, making stupid mistakes on all three... too much math for one 24-hour period given that I was out of mineral water. Oh well. I'm happy with my 7... very happy.

Winning team score was 155, which is about 35 points below last year's 1st place. The entire contest was far more difficult than last year.

[quote="ComplexZeta"]I don't know who was 3rd.[/quote]

Third place individual was Bob Hough from my team (Michigan). :)

Hi this is my first time here.

This was my third ARML and it was really good. We got there on Friday morning and my team practiced a little bit in the afterenoon. At nihgt I went to the big party Eve plannned and I finally got to meet Eve (such a sweety !) and I also got to see Ander and Steve Byrne and other people with names Ive heard many many times. This was definately the biggest/best ARML bash with people from different teams ever.

On Sat. we woke up really early for breakfast which was very good. Then my team tried to find where we were doing our team and power rounds but we got a little lost but we got there on time, thank god. During team I did number four (the one with the set of integers with no diffs of squares) and I got it right and then I spent the rest of the time on number ten (the ARMLovia one) but I kept getting different anwers every time so that was wrong. Power was not as hard this year as it was other years but I still had a lot of trouble with it especially some ones later on. Then was individuals. Those were really hard this year. Really really REALLy hard. After that we had lunch and I waited on a long long line for famous Pennn State ice cream. Then was relays that I messed up for my team but not really because I got the wrong answer from the number one guy anyway so it wasnt really my fault. Then super relays that we got wrong. And then the AAST girls won the song contest even though I think the MA team got more louder applause. I liked the girls the best though so it was not bad. Then they did awards and lots and lots of people I met on Friday night got things and Ander won all of ARML with the only perfect score in the whole country. TJ won too for team (not Anders team) and this was their second year winning. After that I tried to go back to the cafeteria for more ice cream but I couldnt go back in. I had time to kill before our bus left to go home but lots of people were gone who Id just met so I spent time with my team but they were kind of mad because of the relays. On the bus we watched Pi which was very different. The weekend was very much fun.

what was that line from the MA song?  Can't live a normal life cause i was raised by two asian parents? or something like that?

[quote="l156steven"]wat is TJ? and how on earth the SFBA lose???!!! NO!!!!!!! owel.[/quote]

TJ is Thomas Jefferson High School of Science and Technology in Virginia. They're pretty good. Sorry we lost. It's all because of the relay round. We were leading before the relay round, but then we got 16/40 and some other places got as much as 24/40. I won't blame anyone on our team, but we really didn't practice relays enough, although I warned my team that we were really weak on them. Next year (although I won't be on the team)...

surprise! you liked the girls! I'm practically drowining with shock! Not really...

BTW, did you notice half the screaming for the AAST team was the girls themselves?

where was this place that all you people met, BMJ? Can I go next year? I'm not a very important guy or anything, but I'd like to meet all these people too.

hahhahahahhhha. yea. I liked the girls.  They were cute though.  It wanst just AAST who liked them.  Other people did too. "When a guy walks in and you know hes doing math / With a calculator and a graph" to the tune of Baby Got Back? Come on that was greatness!!!!!

The party was in the commons on friday night.  SG basicly invited everyone she knows which is like, a **really** impressive bunch of ppl.  i wasn't there for this part but i heard tiankai lu even was there for a little while. good times.  Eve graduates soon tho so next year will be harder to organize anything as good.

It seems perhaps I missed out by being at Western (California) site. It appears that all the real ARML fun happens at Penn State. That's not to say that Western site isn't fun, but basically Western site is SFBA teams plus a bunch of random people who don't know anyone. I would have liked to have met these people.

Argh... I'll say.  Next year we definitely will practice relays (16 freakin points!)  Bah...  I won't blame anyone either, but I wish that we had lost by more...  It's one thing to say, "I wish everyone on the team had gotten two points more..." that it is to say, "Two points!"

Well Grace, it would have been nice if everyone on our team had scored two more points. For one thing, Jon and I would have got 9s, which would have been somewhat amusing. Also, beating TJ by 28 points would have been rather pleasant, don't you think?

*patpat* Grace, don't hyperventilate. You've got next year. And the years after that. I might tag along next time, depending on how things work out.

Sorry to say this, but I don't see SFBA doing well again for at least 3 more years. (Yes, please, prove me wrong!)

Ah, ok. Hmm...high expectations if I get on...

yeah good luck.

Did you contact the guy i told you?

I will contact him later on in the year, since not much is probably planned out yet.

Hmm, does anyone know how the tiebreak works if more than one person gets a perfect score?

in 2002 multiple people got perfect scores.  they go by who gets it first.

how does that work? the problems are given out 2 at a time.

hmm?  They're put up 1 at a time on the screen.

Someone else would know more about this, as i have never taken the tiebreaker...

Edit: ok i see what you're saying... the 8 individual questions are handed out 2 at a time, then the tiebreaker is something different.  They take the top scorers and they do 1 question at a time until the tie is broken.

Oh, I see. And it is done simultaneously across all three sites? And so the first person who answers the question wins? So do they raise their hands or something?

they pass in an answer once they're done, so the first to pass in a correct answer wins.  And yeah its done at the same time across all 3 sites.

How do they pass it in? By running down the aisle? This process seems sorta shaky...

all the people taking it sit in front row and they just hand it in.  It's not like the questions are trivial... they do take some time to solve.  So the time it takes to pass it in really wont affect the results.

Oh, i see. How many perfect scorers were there this year?

For the tiebreaks, they use entirely new questions (that is, the first eight don't count except to determine who takes the tiebreaks). The fastest person then gets the highest place (2nd in 2003), etc. If people get it wrong, they give more questions. It's really scary to be doing the tiebreaks. There are hundreds of people watching you and laughing at you when you factor 800 into primes incorrectly (don't ask how I managed to do that).


calculus
integration
function
algebra
partial fractions
rational function
calculus computations

Hi, I just have a quick question concerning partial fractions
The equation is :

dy/dx = (1+y)^2

How come partial fractions don't work here??  :(

Thanks.

This is a differential equation - a first order, separable differential equation. We'll assume that you are proceeding with solving it by separation of variables, and you have reached the integral

$ \int \frac {dy}{(1 \plus{} y)^2}.$

All right, so what about partial fractions?

Understand this: partial fractions is [i]not[/i] a technique of integration. It is a technique of algebra that just happens to have some uses applied to integration. When is it called for? It's called for when we have a rational function (a proper fraction) whose (already factored) denominator contains at least two [i]different[/i] factors. If that is the case, we use the partial fraction - but it won't and can't be the final step in the integration because it isn't itself a technique of integration. We use partial fraction to pull our function apart into a sum of simpler pieces, understanding that we do know methods for dealing with each of those pieces.

So what happens in this case, with only one kind of factor? It already is one of those simpler pieces, and we should know methods for dealing with it. So deal with it and be done.

partial fractions are amazing

$ \sum_{i\equal{}1}^{\infty}\frac{1}{n(n\plus{}1)(n\plus{}2)\ldots(n\plus{}p)}$

So we want to compute $ \int \frac{dy}{(y\plus{}1)^2}$. Substituting $ u\equal{}y\plus{}1$, we have

$ \int u^{\minus{}2} du\equal{}\minus{}\frac{1}{u}\plus{}C\equal{}C\minus{}\frac{1}{y\plus{}1}$...

Check: $ x\equal{}C\minus{}\frac{1}{y\plus{}1}$

$ dx\equal{}\frac{dy}{(y\plus{}1)^2}$

You don't need partial fractions with this problem(I don't think I did partial fractions... :maybe: ).


The number of revolutions of a wheel, with fixed center and with an outside diameter of $ 6$ feet, required to cause a point on the rim to go one mile is:
$ \textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi  \qquad\textbf{(E)}\ \text{none of these}$

The circumference is $ 6\pi$ feet and it needs to travel 5280 feet, so the number of revolutions is $ \frac{5280}{6\pi}\equal{}\frac{880}{\pi}$.


algebra
polynomial
factorial
algebra unsolved

Let $n$ and $d$ be positive integers with $0\leq d<n$ .Find all polynomials $P$ of degree $d$ such that \[\sum_{k=0}^{n}(-1)^{k}{n\choose k}P(k)=\sum_{k=0}^{\infty}(-1)^{k}\frac{P(k)}{k!}.\] [i]Edited by Myth[/i]

Are you sure about this one? It is quite simple to show that $LHS=0$, so the question is for whick $P(x)$ we have $RHS=0$. I don't see any closed form for it, except one with recurrence.


There is a midle circle with a diameter $ AB\equal{}6cm$ ,$ BX$ is a tangent with this circle in $ B$. In this tangent is tooken a point $ M$ such $ BM<AB$ . The midle circle with diameter $ BM$ cut the midle circle with diameter $ AB$ in a point $ C$. Prove that $ A,M$ and $ C$ are  extended in the same line.

I don't understand... why you call "midle" every circle?
Moreover, I think you don't need to suppose AB=6 cm nor BM<AB. But it's very probable I've misunderstood your problem.
Can you explain it more precisely?

Thanks
GioMott

I called midle circle because there are only half circles .

What part of the problem did you misunderstand?

[quote="Elio (n)"]There is a midle circle with a diameter $ AB \equal{} 6cm$ ,$ BX$ is a tangent with this circle in $ B$. In this tangent is tooken a point $ M$ such $ BM < AB$ . The midle circle with diameter $ BM$ cut the midle circle with diameter $ AB$ in a point $ C$. Prove that $ A,M$ and $ C$ are  extended in the same line.[/quote]
[hide="Solution"]
Since $ AB$ is the diameter of the first circle, $ \angle ACB\equal{}90$.  Since $ BM$ is the diamter of the second circle, $ \angle MCB\equal{}90$.  Thus, \[ \angle ACM\equal{}\angle ACB\plus{}\angle MCB\equal{}90\plus{}90\equal{}180\] and it follows that $ A$, $ M$, and $ C$ are collinear. [/hide]

OK very nice . I like that            :)


[quote]Cyclic inequalities of degree 3 theorem[/quote]
I think it is P.I.D. Just try with (1,1,1),(1,1,0),(2,1,1).

[quote="hungkhtn"]It can be solve easy enough by EMV and Cyclic inequalities of degree 3 theorem (if true). But I post this one just to ask nayel about the origin of this problem. Do you the author, Nayel? I would like to add this nice problem to my English book.[/quote]

well i created it just this morning. :lol:  and my solution is really elementary. so to those who want to solve it, i suggest you use just elementary methods. :P

@the soul of rock: what's a P.I.D?

[quote="Vasc"]We can solve it by Cauchy-Schwarz Inequality.
Nice inequality, Nayel![/quote]

It is unexpected. But EMV is also simple and elementary. I will think about your way, VASC.  :lol:

[quote="Vasc"]We can solve it by Cauchy-Schwarz Inequality.
Nice inequality, Nayel![/quote]

yes, i think your solution is the same as mine. :D

[quote="nayel"]Let $a,b,c$ be the side lengths of a triangle. Prove that
\[\frac{a}{3a+b-c}+\frac{b}{3b+c-a}+\frac{c}{3c+a-b}\ge 1 \]
[/quote]
By Cauchy Schwarz inequality, we have
\[\sum\frac{4a}{3a+b-c}=\sum\frac{a-b+c}{3a+b-c}+3 \ge \frac{(a+b+c)^{2}}{\sum (a-b+c)(3a+b-c)}+3=4 \]
Hence
\[\sum\frac{a}{3a+b-c}\ge 1 \]
We are done. :)

[quote=nayel]Let $a,b,c$ be the side lengths of a triangle. Prove that \[\frac{a}{3a+b-c}+\frac{b}{3b+c-a}+\frac{c}{3c+a-b}\ge 1\][/quote]

Assume $a\neq \text{mid}\{a,b,c\}$. We need to prove:\[\left( b-c \right) ^{2} \left( b-a+c \right) + \left( a-c \right)
 \left( a-b \right)  \left( -a+b+3\,c \right) \geqq 0\]
Done.

[quote=gghx]You can just use Ravi transformation :)[/quote]
Let $a=x+y,b=x+z,c=y+z$, the inequality is equivalent to $$\sum \frac{x+y}{4x+2y} \ge 1$$ Multiplying 4 on both sides and simplifying, the inequality becomes $$\sum \frac{y}{2x+y}\ge1$$ which is equivalent to $$\sum \frac{y^2}{2xy+y^2}\ge1$$ which is Titu's


Find the greatest $ k \in R$  such that for every positive $ a$ with $ a\minus{}\frac 1a \geq 1$ we have :$ a^3\minus{}\frac  1{a^3} \geq k(a\minus{}\frac 1a)$

[quote="amparvardi"]Find the greatest $ k \in R$  such that for every positive $ a$ with $ a \minus{} \frac 1a \geq 1$ we have :$ a^3 \minus{} \frac 1{a^3} \geq k(a \minus{} \frac 1a)$[/quote]

$ a>0$ and $ a \minus{} \frac 1a \geq 1$ $ \iff$ $ a\ge a_0\equal{}\frac{1\plus{}\sqrt 5}2$ and we are looking for $ \min_{a\ge a_0}\frac{a^3\minus{}\frac 1{a_3}}{a\minus{}\frac 1a}$ $ \equal{}\min_{a\ge a_0}(a^2\plus{}1\plus{}\frac 1{a^2})$

And since $ a^2\plus{}1\plus{}\frac 1{a^2}$ is decreasing over $ (0,1)$ and increasing over $ (1,\plus{}\infty)$ and $ a_0>1$, we get $ \min_{a\ge a_0}(a^2\plus{}1\plus{}\frac 1{a^2})$ $ \equal{}a_0^2\plus{}1\plus{}\frac 1{a_0^2}\equal{}4$

Hence the result : $ \boxed{k\equal{}4}$

[quote="broniran"]It's far too easy problem for an Iran National Olympiad, isn't it?[/quote]

you are right.
but it's first round of olympiad in olympiad and 14-years-old children solve these problems!!
better??


geometry
3D geometry
sphere
vector
Functional Analysis
combinatorial geometry
real analysis

Let $E$ be a infinitely dimensional Banach space and $K$ a compact in it. Prove that the set of points that are not in $K$ is connected.

Let $\{e_i\}_{i \in I}$ be a basis of this Banach space. One can suppose for example that $|e_i-e_j|\geq 1$ for $i \neq j$. Note $C_r$ the cube of side $r$ centered at the origin, that is the set of point of the form $\sum_{i \in F} a_i e_i$, $\max_{i \in F} |a_i| = r$. Then clearly $C_r$ is connected.

Now, since $K$ is compact, $K$ is included in $C_r$ for $r$ large enough. If one can prove that for any $x \not \in K$, there exists a line that joins $x$ to this cube, the proof will be complete. So let's prove it:

Take a point $x \not \in K$ and $\alpha > 0$ such that the ball $B(x, \alpha)$ does not intersect $K$. One will prove that there exists $j \in I$ such that the line $x+te_j$ does not intersect $K$. If it were not true, for any $i \in I$ there would exist $b_i \in \mathbb{R}$, $|b_i|> \alpha$, such that $x+b_ie_i \in K$. Hence the compact $K'=K-x$ would contain $b_ie_i$ for any $i \in I$. $K' \cup (-K')$ is also a compact, and so is $C$ the closure of the convex hull of $K' \cup (-K')$. But $C$ is a compact that contains $\alpha e_i$ for any $i \in I$, which is impossible because the sequence $u_i=a_i e_i \in C$ has no convergent subsequence since $|e_i-e_j| \geq 1$.

[quote="alekk"]Let $\{e_i\}_{i \in I}$ be a basis of this Banach space.[/quote]
Hmmm... There exist Banach spaces in which there are no bases  :P

what do you mean ? this is just an algebraic basis, every element in $E$ is a finite linear combination of elements of this basis  :?

In this case you can say neither that $|e_i-e_j|\ge 1$ for $i\ne j$, nor that $K\subset C_r$ for large $r$.  At least, I do not see how to justify any of these claims for algebraic bases. :?

I started to post my proof, but when I saw that alekk had gotten here first, I gave up :).

Let $\widetilde K$ be the closed convex hull of $K$. As is well-known, $\widetilde K$ is compact as well. First notice that $E\setminus\widetilde K$ is path connected: given $x_1,x_2,y\not\in\widetilde K$ we can take two linearly independent linear functionals $f_1,f_2$ such that $\sup_{\widetilde K}f_i<f_i(x_i),\ i=1,2$. The sets $\{x\in E\ |\ f_i(x)\ge f_i(x_i)\}$ are path connected, intersect (because $f_i$ are linearly independent), and contain $x_i$ respectively, which means that their union is a path connected set which contains $x_i,\ i=1,2$, and does not cut $\widetilde K$ (and hence $K$).

This means that it suffices to prove that a point in $\widetilde K\setminus K$ (which we assume WLOG is the origin $0$ of $E$) can be connected to a point outside $\widetilde K$ through a path which cuts $\widetilde K$ only in $0$. To this end, we show that there must be a ray $R_x=\{tx\ |\ t\ge 0\}$ which cuts $\widetilde K$ only in $0$.

Suppose the contrary. This just means that for every $x\in E$, there is a $t_x>0$ such that $tx\in\widetilde K,\ \forall\ 0\le t\le t_x$, i.e. $\widetilde K$ contains a non-degenerate segment $[0,t_xx]$ on each ray $R_x$. We can also assume that $t_x$ is maximal with this property.

Let $A_n$ be the union of $[0,x]$ for all $\|x\|=1$ such that $t_x\ge\frac 1n$. Each $A_n$ is easily seen to be closed, and the union of all the $A_n$ is the unit ball of $E$, by assumption. By the Baire Category Theorem, there will be an $A_n$ with non-empty interior, but then, since $\frac 1nA_n\subset\widetilde K,\ \widetilde K$ will have non-empty interior as well. Since a compact subset of an infinite-dimensional normed space cannot have non-empty interior, we have a contradiction.

Just to finish this, here is the simplest solution I could find. Let $x\notin K$. Consider the projection $K_x$ of $K$ to the sphere of radius $R>\|x\|$ by rays emanating from $x$. Since such projection is a continuous mapping (some work is needed to justify it, but it is fairly routine), $K_x$ is compact. Analogously, we can take a second point $y\notin K$ and construct $K_y$ (we will have to choose $R>\|y\|$ for that). Now, $K_x\cup K_y$ is compact but the sphere is not, so there exists a point $z$ with $\|z\|=R$ such that $z\notin K_x\cup K_y$. Then the union of the intervals $xz$ and $zy$ is a path connecting $x$ with $y$ in $E\setminus K$. This approach is pretty much the same as grobber's or alekk's, just the technical details are a bit simpler this way. :)

:blush:  yes, you're right Fedja. I will just try to repair it:

It suffices to find an infinite set of independant vectors $e_i$ such that $|e_i-e_j|>1$ for $i \neq j$, and $|e_i|>1$ and to replace $C_r$ by the sphere of radius $r$. With the same proof as above (more complicated thatn yours  :) ) on can show that a point inside the ball of radius $r$, not in the compact, is connected to the sphere of radius $r$.


Hei, fekk dokke til geometri oppgåvene, eg fekk til alle:)  Noke dokke vil diskutere?
Prøv her er ei oppgåve til også! A, B og C er vinklar i ein trekant. Finn maks til sinA * sinB * sinC

Ingenting spesielt å diskutere, nei...og den oppgaven har vi jo løst via mobil allerede ;)

men kanskje dei andre kan få prøve den då!


Prove that for all real numbers $ a,b$ with $ ab>0$ we have:

$ \sqrt[3]{\frac{a^2 b^2 (a\plus{}b)^2}{4}} \le \frac{a^2\plus{}10ab\plus{}b^2}{12}$

and find the cases of equality. Hence, or otherwise, prove that for all real numbers $ a,b$

$ \sqrt[3]{\frac{a^2 b^2 (a\plus{}b)^2}{4}} \le \frac{a^2\plus{}ab\plus{}b^2}{3}$

and find the cases of equality.

[quote="moldovan"]Prove that for all real numbers $ a,b$ with $ ab > 0$ we have:

$ \sqrt [3]{\frac {a^2 b^2 (a \plus{} b)^2}{4}} \le \frac {a^2 \plus{} 10ab \plus{} b^2}{12}$

and find the cases of equality. Hence, or otherwise, prove that for all real numbers $ a,b$

$ \sqrt [3]{\frac {a^2 b^2 (a \plus{} b)^2}{4}} \le \frac {a^2 \plus{} ab \plus{} b^2}{3}$

and find the cases of equality.[/quote]

By Am-Gm we have:
$ \sqrt [3]{\frac {a^2 b^2 (a \plus{} b)^2}{4}} \le \frac {ab \plus{} ab \plus{} \frac {(a \plus{} b)^2}{4}}{3} \equal{} \frac {a^2 \plus{} 10ab \plus{} b^2}{12}\le \frac {a^2 \plus{} ab \plus{} b^2}{3}$
When a=b