Given $m, n$ positive integers either both even or both odd, we consider a $m \times n$ board. In the beginning, when $t=0$, each square is either white or black. When $t=1$, for each square X we consider the union of the column and row X is in, which has $m+n-1$ squares. If the number of black squares among them is greater than the number of white squares among them, X becomes black, and vice-versa. This happens for all squares X when $t=1, t=2, ...$

Obviously we eventually obtain a period, as there are $2^{mn}$ possible configurations and one must eventually repeat, which implies the following stages must also repeat...etc. The question is: Is there a constant $C$ such that for any $m, n$ the length of the period is at most $C$, regardless of the initial configuration?

Is this known? Does it have a solution?

The Secret treasure ring of the Red Kahili.
 I have a solution.

[quote="Charlydif"]
 I have a solution.[/quote]

Could you please post it? :)


If $ \displaystyle\frac{x}{y} \equal{} 3$, $ \displaystyle\frac{y}{z} \equal{} 8$, and $ \displaystyle \frac{z}{w} \equal{} \frac{1}{2}$, then what is $ \displaystyle\frac{w}{x}$?

I did it a different way, although gantasy's solution is much easier.

Since $ \frac{x}{y} \equal{} 3$, $ x : y \equal{} 3 : 1$.

Since $ \frac{y}{z} \equal{} 8$, $ y : z \equal{} 8 : 1$

Multiply the top ratio by $ 8$, to get $ x : y : z \equal{} 24 : 8 : 1$

And since $ \frac{w}{z} \equal{} \frac{1}{2}$, we can immediately say that:

$ w : x : y : z \equal{} 2 : 24 : 8 : 1$.

Thus, the answer is $ \frac{2}{24}$ or $ \boxed{\frac{1}{12}}$.


group theory
abstract algebra
induction
superior algebra
superior algebra theorems

1. Prove that for n>= 3 , An = < 3-cycles > .
2. Prove that for n>= 2, Sn = < (12),(13),...,(1n) >.

It seems to me that they can be done by using induction on n but in a messy fashion. Is there any other way to prove these statements ? Like using the fact that disjoint cycles commute?

For your second question: visualize $n$ billiard balls placed next to each other. Try to see that it is always possible to permute those $n$ balls by interchanging adjacent balls suitably.

(for me this is a correct proof; induction is not required, I suppose)


Suppose $a>b>0$, ${f(x)=\frac{2(a+b)x+2ab}{4x+a+b}}$. Show that there exists unique $x>0$, such that $f(x)= \left( \frac{a^{\frac13}+b^{\frac13}}{2}\right)^{3}$.

I think after doing a few step, the question is to prove exist unique $x>0$ such that $[2(a+b)-\frac{(a^{\frac13}+b^{\frac13})^{3}}{2}]x=(a+b)(\frac{a^{\frac13}+b^{\frac13}}{2})^{3}-2ab$. Since showing $2(a+b)-\frac{(a^{\frac13}+b^{\frac13})^{3}}{2}$ and $(a+b)(\frac{a^{\frac13}+b^{\frac13}}{2})^{3}-2ab$ are positive is easy, so it is done without Calculus.


On the back of a cereal box, there's a really cheesy board game. The way you move is dependant on the flip of a coin. If you flip heads, you move 2 spaces. If you flip tails, you move one space. There are 10 spaces total on the board game. The board game has no "Lose a turn" or "Go back 2 spaces" on any of the spaces. If you are at space number 9, and you flip a heads, you still win. How many possibilities of flips will enable you to win the game?

For example, here are two different possibilities:

H, H, H, H, H
T, T, T, T, T, T, T, T, T, T

I tried to figure it out... but I just sorta stopped because I fell asleep. (I was doing this problem yesterday at 12:40 AM.

[hide]there must always be an even number of tails, so HHHHH, HHHHTT, HHHTTTT, HHTTTTTT, HTTTTTTTT, TTTTTTTTTT are our possible setups.  We have one solution to 5 heads and 10 tails, so thats 2.  We add 2 choose 6, 4 choose 7, 6 choose 8, and 8 choose 9, which equals 2+15+35+28+9=89.   :idea: [/hide]


[quote="pvthuan"]Oh, really. It is a my problem in my book. I chose $a+b+c=3$.[/quote]

Dear Thuan, it was just Japan 1997!

One solution based on:

        $\frac{a(b+c)}{a^2+(b+c)^2} \le \frac{28a+b+c}{25(a+b+c)}$

Try to guess how to come to this effect estimation.

Namdung

Let $a+b+c=3$ and it is equivalent to prove that

$\sum\frac{1}{2a^2-6a+9}\leq \frac{3}{5}$

But we also have $\frac{1}{2a^2-6a+9}\leq \frac{2a+3}{25}$ $\Longleftrightarrow (a-1)^2(a+1)\geq 0$ which is true .Hence sum up the and done .

[quote="pvthuan"]Oh, really. It is a my problem in my book. I chose $a+b+c=3$.[/quote]
Also have:
\[ \frac{a(3-a)}{a^2+(3-a)^2} \le \frac{9}{25}a+\frac{1}{25}  \]

[quote="tranthanhnam"]Let $ a,b,c$ be real numbers such that  $ a \plus{} b \plus{} c\ne 0,$ prove that

$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5},$

with equality if and only if $ a \equal{} b \equal{} c.$ [/quote]$ \frac {6}{5} \minus{} \sum{\frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2}} \equal{} \frac {2F (a,b,c)}{5\prod{[a^2 \plus{} (b \plus{} c)^2]}},$

where $ F(a,b.c)$

$ \equal{} 3\sum{a^6} \plus{} \sum{a^5(b \plus{} c)} \minus{} \sum{a^4(b^2 \plus{} c^2)} \plus{} 3abc\sum{a^3} \plus{} 2\sum{b^3c^3}$

$ \minus{} 6abc\sum{bc(b \plus{} c)} \plus{} 12a^2b^2c^2$

$ \equal{} \frac {1}{72}\sum{\left[\frac {774 \minus{} 7\sqrt {4602}}{113}a^2(b \plus{} c) \minus{} \frac {17 \plus{} 7\sqrt {4602}}{113}a(b^2 \plus{} c^2) \minus{} 10abc \minus{} 7 (b^3 \plus{} c^3) \plus{} \frac {599 \plus{} 14\sqrt {4602}}{113}bc(b \plus{} c)\right]^2}$

$ \plus{} \frac {(a \plus{} b \plus{} c)^2}{72}\sum{(b \minus{} c)^2\left[\sqrt {59}(b \plus{} c) \minus{} \frac {88\sqrt {59} \minus{} 49\sqrt {78}}{113}a\right]^2}\geq0.$[quote="Namdung"]One solution based on:

        $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \le \frac {28a \plus{} b \plus{} c}{25(a \plus{} b \plus{} c)}$[/quote]A counterexample $ a\equal{}\minus{}1,b\equal{}c\equal{}1:$

$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2}\equal{}\minus{}\frac{2}{5}>\minus{}\frac{26}{25}\equal{} \frac {28a \plus{} b \plus{} c}{25(a \plus{} b \plus{} c)}.$

See also: http://www.mathlinks.ro/viewtopic.php?t=227459

[quote="Ji Chen"][quote="tranthanhnam"]Let $ a,b,c$ be real numbers such that  $ a \plus{} b \plus{} c\ne 0,$ prove that
$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5},$
with equality if and only if $ a \equal{} b \equal{} c.$ [/quote]$ \frac {6}{5} \minus{} \sum{\frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2}} \equal{} \frac {2F (a,b,c)}{5\prod{[a^2 \plus{} (b \plus{} c)^2]}},$
where $ F(a,b.c)$
$ \equal{} 3\sum{a^6} \plus{} \sum{a^5(b \plus{} c)} \minus{} \sum{a^4(b^2 \plus{} c^2)} \plus{} 3abc\sum{a^3} \plus{} 2\sum{b^3c^3}$
$ \minus{} 6abc\sum{bc(b \plus{} c)} \plus{} 12a^2b^2c^2$[/quote]
$ F(a,b.c)\equal{} 3\sum{a^6} \plus{} \sum{a^5(b \plus{} c)} \minus{} \sum{a^4(b^2 \plus{} c^2)} \plus{} 3abc\sum{a^3} \plus{} 2\sum{b^3c^3}\minus{} 6abc\sum{bc(b \plus{} c)} \plus{} 12a^2b^2c^2$
$ \equal{}3abc(\sum{a^3}\minus{}\sum{bc(b\plus{}c)}\plus{}3abc)\plus{}(2\sum{b^3c^3}\minus{}2abc\sum{bc(b\plus{}c)}\plus{}6abc)\plus{}(3\sum{a^6}\plus{}\sum{a^5(b\plus{}c)\minus{}\sum{a^4(b^2\plus{}c^2)\minus{}3a^2b^2c^2\minus{}abc\sum{bc(b\plus{}c)}}})$
$ \equal{}3abc\sum{a(a\minus{}b)(a\minus{}c)}\plus{}2\sum{bc(bc\minus{}ca)(bc\minus{}ab)}\plus{}(3\sum{a^6}\plus{}\sum{a^5(b\plus{}c)\minus{}\sum{a^4(b^2\plus{}c^2)\minus{}3a^2b^2c^2\minus{}abc\sum{bc(b\plus{}c)}}}) \geq 0$


$(\sum_{i=1}^{n}{a_i}{b_i}{c_i})^2 \leq (\sum_{i=1}^{n}{a_i}^2)(\sum_{i=1}^{n}{b_i}^2)(\sum_{i=1}^{n}{c_i}^2)$
Is the above inequality correct?
In what other directions can we modify the Cauchy Schwarz Ineqlaity?

Actually, the generalization of Cauchy is different. You may consider it as Holder inequality.
Consider $p_{1}, p_{2}, \ldots , p_{n}$ be positive real numbers such that $\sum_{i=1}^{n}\frac{1}{p_{i}}=1$.Let $x_{i}, y_{i},z_{i},\ldots$ be $n$ sequences of $n$ positive real numbers.
Then the following inequality holds:
$(\sum x_{i}^{p_{1}})^{\frac{1}{p_{1}}}(\sum y_{i}^{p_{2}})^{\frac{1}{p_{2}}}(\sum z_{i}^{p_{3}})^{\frac{1}{p_{3}}}\ldots \geq \sum x_{i}y_{i}z_{i}\ldots$.
Cauchy is a particular case of Holder, for $n=2$, and $p_{1}=p_{2}=2$.

[quote="luimichael"]$(\sum_{i=1}^{n}{a_i}{b_i}{c_i})^2 \leq (\sum_{i=1}^{n}{a_i}^2)(\sum_{i=1}^{n}{b_i}^2)(\sum_{i=1}^{n}{c_i}^2)$
Is the above inequality correct?
[/quote]

Actually, it is correct, although there isn't a very useful equality condition.

Such a generalization would be

\[ \left( \sum_{x=1}^n \prod_{k=1}^m a_k(x) \right)^2\leq \prod_{k=1}^m \sum_{x=1}^n [a_k(x)]^2  \]

where the $a_k(x)$ are real numbers. ($m$ groups of $n$ real numbers each)

[hide="the inductive step"]
Let $b_k(x)=|a_k(x)|$. By Triangle Inequality, $\left| \sum_{x=1}^n \prod_{k=1}^m a_k(x) \right| \leq \sum_{x=1}^n \left| \prod_{k=1}^m a_k(x) \right|=\sum_{x=1}^n \prod_{k=1}^m b_k(x)$. So $\left( \sum_{x=1}^n \prod_{k=1}^m a_k(x) \right)^2\leq \left( \sum_{x=1}^n \prod_{k=1}^m b_k(x) \right)^2$.

By Cauchy, $\left( \sum_{x=1}^n \prod_{k=1}^m b_k(x) \right)^2\leq \left( \sum_{x=1}^n \left( \prod_{k=1}^{m-1} b_k(x) \right)^2 \right) \left( \sum_{x=1}^n [b_m(x)]^2 \right)$.

Let $P(x)=\prod_{k=1}^{m-1} b_k(x)$. Since the $b_k(x)$ are nonnegative, so is $P(x)$; then $\sum_{1\leq x<y\leq n} P(x)P(y)\geq 0\Rightarrow \sum_{x=1}^n [P(x)]^2\leq \left( \sum_{x=1}^n P(x) \right)^2$.

$\left( \sum_{x=1}^n P(x) \right)^2\leq \prod_{k=1}^{m-1} \sum_{x=1}^n [b_k(x)]^2$ (using the assumption that it holds for $m-1$ to prove that it holds for $m$). So $\left( \sum_{x=1}^n P(x) \right)^2 \left( \sum_{x=1}^n [b_m(x)]^2 \right) \leq \prod_{k=1}^m \sum_{x=1}^n [b_k(x)]^2$.

Since $[b_k(x)]^2=[a_k(x)]^2$, we finally arrive at $\left( \sum_{x=1}^n \prod_{k=1}^m a_k(x) \right)^2\leq \prod_{k=1}^m \sum_{x=1}^n [a_k(x)]^2$.
[/hide]

Um, isn't that overkill?

Let $C_i = c_i / \sqrt{c_1^2+\cdots + c_n^2}$ so $C_i \leq 1$ and

$\left(\sum_{i=1}^{n}{a_i}{b_i}{C_i}\right)^2 \leq \left(\sum_{i=1}^{n}{a_i}{b_i}\right)^2$
$\leq \sum_{i=1}^{n}{a_i}^2 \sum_{i=1}^{n}{b_i}^2$

And the generalization can be proven the same way... but yeah, Holder's is generally more useful


Two years ago there were 20 trailer homes on Elm Street with an average age of 18 years. At that time, a group of brand new trailer homes was then added to Elm Street. Today, the average age of all the trailer homes on Elm Street is 14 years. How many new trailer homes were added two years ago?

We know that two years ago, 20 trailers were an average of 18 years old.
This means that right now, 20 trailers are 20 years old.
We also know that a certain amount of trailers, $ x$, are 2 years old.

Since the average age is $ 14$, we know that

$ 20\cdot 20 \plus{} 2x \equal{} 14(x\plus{}20)$; solving for $ x$,
$ 400\plus{}2x \equal{} 14x \plus{} 280$;
$ x \equal{} \boxed{10}$


If any of you took Chinese school, which book were you currently in when you took the Chinese SAT 2? :)

Personally, I never took any Chinese courses in here, but I did take a lot in China. Please vist http://www.pep.com.cn/
if you can find any useful Chinese book, by clicking "Chinese Course" (Yu Wen in Chinese) in all elementary, middle, or high school regions.  :D
I don't know whether it is helpful, but those books are really what Chinese students are studying now.

p.s. if you cannot see Chinese due to the computer system, just please click the "view" on top of your IE, then select encoding function to see what happens.
  If you cannot read Chinese, then...., you can ask a Chinese friend or teacher to help.  :D


Define the operation $ \diamondsuit$ for any two real numbers $ a$ and $ b$ as $ a\diamondsuit b \equal{} 2a(a\plus{}b)$.  What is the value of $ p$ such that $ 3\diamondsuit (4\diamondsuit p) \equal{} 1122$?


[i][u]Evaluate :[/u]

 $ \lim_{n \  \to \  \plus{} \infty } \left[ \frac{1^2}{ \pi} \ \plus{} \ \frac{2^2}{\pi^{2}} \ \plus{} \ \frac{3^2}{ \pi^{3}} \ \plus{} \ .....\ \plus{} \ \frac{n^2 }{ \pi^{n}} \right]$

  This is my homework , but I  cannot solve on my own . Can anybody help me ?[/i] :lol:

Let $ f(x) \equal{} \sum_{i \equal{} 1}^{\infty}i^2x^i \equal{} \sum_{i \equal{} 1}^{\infty}ix^i \plus{} \sum_{i \equal{} 1}^{\infty}i(i \minus{} 1)x^i \equal{} f_1(x) \plus{} f_2(x)$.
$ f_1(x) \equal{} x(\frac {1}{1 \minus{} x})' \equal{} \frac {x}{(1 \minus{} x)^2},f_2(x) \equal{} x^2(\frac {1}{(1 \minus{} x)})'' \equal{} \frac {2x^2}{(1 \minus{} x)^3}.$
Therefore $ f(x) \equal{} \frac {x^2 \plus{} x}{(1 \minus{} x)^3}$.
Your mean is $ f(\frac {1}{\pi}) \equal{} \frac {\pi (\pi \plus{} 1)}{(\pi \minus{} 1)^3}$.


If $q_{1}(x)$ and $r_{1}$ are the quotient and remainder, respectively when the polynomial $x^{8}$ is divided by $x+\frac{1}{2}$, and if $q_{2}(x)$ and $r_{2}$ are the quotient and remainder, respectively, when $q_{1}(x)$ is divided by $x+\frac{1}{2}$, then find $r_{2}$.

I can find the remainder, my only problem is that I can't find the quotient for the first part.

let $y=x+\frac{1}{2}$

$p(y)=\left(y-\frac{1}{2}\right)^{8}$

Note our divisions are just removing the constant and dividing by $y$. We can easily find our solution using the binomial theorem. i.e. $r_{n}$ is the $x^{n-1}$ coefficient of the binomial expansion of $p(y)$.

In this case, we have that: $r_{2}=\binom{8}{1}*\left(-\frac{1}{2}\right)^{7}=-\frac{1}{16}$.

This method clearly generalizes to arbitrary polynomials.

[quote="danthemanvt"]x^8 is not a polynomial :wink:[/quote]

...?

I'm not sure what you think a polynomial is.

[u][b]Definition of a Polynomial[/b]  [/u]

A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The general form of a polynomial is:

$a_{n}x^{n}+a_{n-1}x^{n-1}+.......+a_{0}$.

@danthemanvt:

If we let $a_{n-1}= a_{n-2}=....=a_{0}= 0$ and $a_{n}= 1$ and $n=8$ we get $x^{8}$. So $x^{8}$ is a polynomial  :P .


The last day of the school year was on a Wednesday. Students then have a 77-day summer break. On what day of the week will the next school year start?

Since Thursday was the first day of break, and $ 77 \equiv 0 \pmod{7}$, we have the school starting $ 0$ days after Thursday, which is obviously $ \boxed{\text{Thursday}}$.


LaTeX
trigonometry
logarithms
limit
calculus
calculus computations

bgbgbgbg: can you please learn enough $ \text{\LaTeX}$ to post that "normally" rather than relying on uploading an image?

As $ u\to 0,$ $ \tan\left(\frac{\pi}2-u\right)=\frac{\sin\left(\frac{\pi}2-u\right)}{\cos\left(\frac{\pi}2-u\right)}=\frac{1+O(u^2)}{u+O(u^3)}=\frac1u+O(u).$

Since $ \frac{\pi x}{2x+1}=\frac{\pi}2-\frac{\pi}{2x+1}%=\frac{\pi}2-\frac{\pi}{2x}+O(x^{-2}),
$ that means that as $ x\to\infty,$

$ \tan\left(\frac{\pi x}{2x+1}\right)=\frac{2x+1}{\pi}+O(x^{-1}).$

We're taking that to the $ \frac1x$ power. That means that the logarithm of the expression we seek behaves like $ \frac1x\ln(Cx)$ (plus smaller terms) as $ x\to\infty.$ (In this case, $ C=\frac2{\pi},$ but that's not really important.) The logarithm grows slower than any power of $ x,$ so this tends to $ 0.$ Undo the logarithm, and we find that the original limit is $ \boxed{1}.$

Basically, I saw this immediately - a classic $ \infty^0$ case with the exponent outweighing the base - but it takes a little while to write it up.

Define $ t \equal{} \frac {\pi x}{2x \plus{} 1}$ $ \Rightarrow$ $ x \equal{} \frac {t}{\pi \minus{} 2t}$

$ \lim_{x\rightarrow 0} (\tan (\frac {\pi x}{2x \plus{} 1}))^{t} \equal{} \lim_{t\rightarrow \frac {\pi}{2}} (\tan t)^{\frac {t}{\pi \minus{} 2t}}$

$ \equal{} \lim_{t\rightarrow \frac {\pi}{2}} (1 \plus{} \tan t \minus{} 1)^{\frac {1}{\tan t \minus{} 1}\frac {t(\tan t \minus{} 1)}{\pi \minus{} 2t}}$

$ \equal{} e^{\lim_{t\rightarrow \frac {\pi}{2}} \frac {t(\tan t \minus{} 1)}{\pi \minus{} 2t}}$

=$ e^{\infty }$

That's incorrect. The limit is $ 1.$

Where did you go wrong?

While it is true that $ \lim_{u\to 0}(1\plus{}u)^{\frac bu}\equal{}e^b,$ we can't say the same for $ \lim_{u\to\infty}(1\plus{}u)^{\frac bu}.$ The latter tends to $ 1.$

You weren't paying attention to which way the limit was going. In particular, as $ t\to\frac{\pi}2,$ $ (\tan t\minus{}1)\to\infty,$ not zero.


If the pattern shown is continued, what is the sum of the terms in row 12?
\begin{tabular}{ccl}
Row 1 & \dots & 2\\
Row 2 & \dots & 2+4\\
Row 3 & \dots & 2+4+6\\
Row 4 & \dots & 2+4+6+8\\
Row 5 & \dots & 2+4+6+8+10
\end{tabular}

Lets fix the latex up:
\[ \begin{tabular}{ccl} Row 1 & \dots & 2 \\
Row 2 & \dots & 2 \plus{} 4 \\
Row 3 & \dots & 2 \plus{} 4 \plus{} 6 \\
Row 4 & \dots & 2 \plus{} 4 \plus{} 6 \plus{} 8 \\
Row 5 & \dots & 2 \plus{} 4 \plus{} 6 \plus{} 8 \plus{} 10 \end{tabular}
\]

i think i showed this formula b4

but i 4got the even #s formula AGAIN   :oops:  :lol:

so the formula is n (a + z) / 2
               n = number of terms in sequence
               a = first number in sequence
               z = last term in sequence

1st term = 2, last term = 24 (2*12), 12 terms in sequence

12 (2 + 24) / 2 or 26*12/2

ans = 156

The sum of $ n$ even numbers is $ (n)(n\plus{}1)$.


calculus
integration
function
irrational number
calculus computations

Let f: [2,5] -> R be the function

f(x) = 0     if x is an irrational number
f(x) = 1/q  if x is a rational number, expressed as p/q in lowest terms

Prove that integral (from 2 to 5) of f(x) dx exists.

How do you show this as a formal proof?

You posted this question less than a week ago, and I gave you a reference to as formal a proof as you're going to get on this site. Please do not re-post questions that have already been answered.


Let $ r$ inradius, $ m_a,m_b,m_c$ the medians, $ h_a,h_b,h_c$ the heights, $ l_a, l_b, l_c$ the bisectors of an triangle $ ABC$. Prove that:

$ b^2\equal{}\frac{l^4_a(m^2_a\minus{}h^2_a)}{a^2(l^2_a\minus{}h^2_a)}$

------

Ligouras  :|

[color=darkblue]Hi Ligouras. I think your post is not true.

Because if $ b^{2}\equal{}\frac{l^{4}_{a}(m^{2}_{a}\minus{}h^{2}_{a})}{a^{2}(l^{2}_{a}\minus{}h^{2}_{a})}$, so $ c^{2}\equal{}\frac{l^{4}_{a}(m^{2}_{a}\minus{}h^{2}_{a})}{a^{2}(l^{2}_{a}\minus{}h^{2}_{a})}$

Hence $ b^2\equal{}c^2 \Longrightarrow b\equal{}c \Longrightarrow ABC$ always is an isosceles triangle[size=200]?[/size][/color]

[quote="Virgil Nicula"]Is correctly so : $ b^2 \equal{} \frac {l_a^4\left(m^2_a \minus{} h^2_a\right)}{c^2\left(l^2_a \minus{} h^2_a\right)}$[/quote]

Thanks Virgin and sorry my friends. It is an transcript mistake...
The correct is
  $ b^2 \equal{} \frac {l_a^4\left(m^2_a \minus{} h^2_a\right)}{c^2\left(l^2_a \minus{} h^2_a\right)}$

Sorry

Panagiote LIGOURAS

Nice proof, [b]Yetti[/b] !

[b][u]Remark[/u].[/b] Using the [u][b]Yetti[/b]'s notations[/u], $ \boxed {\ \frac {M_aP_a}{M_aL_a}\equal{}\left(\frac {b\plus{}c}{a}\right)^2\ }\implies$

$ \frac {M_aP_a}{(b\plus{}c)^2}\equal{}$ $ \frac {M_aL_a}{a^2}\equal{}\frac {L_aP_a}{4p(p\minus{}a)}\implies \frac {M_aP_a}{L_aP_a}\equal{}\frac {(b\plus{}c)^2}{4p(p\minus{}a)}\equal{}\frac {bc}{l_a^2}$ .

[quote="yetti"]$ M_a, L_a, H_a \in BC$ are feet of $ m_a, l_a, h_a.$ $ AL_a$ cuts the citcumcircle $ (O)$ again at $ X_a.$
$ \triangle ABX_a \sim \triangle AL_aC$ have same angles $ \Longrightarrow$ $ \frac {AB}{AX_a} \equal{} \frac {AL_a}{AC}$  (1).
$ \triangle AL_aH_a \sim \triangle X_aL_aM_a$ have same angles $ \Longrightarrow$ $ \frac {L_aH_a}{M_aH_a} \equal{} \frac {AL_a}{AX_a}$ (2).
Plug $ AX_a$ from (1) to (2)  $ \Longrightarrow$ $ \frac {L_aH_a}{M_aH_a} \equal{} \frac {AL_a^2}{AB \cdot AC}$ and square it $ \Longrightarrow$ $ \frac {l_a^2 \minus{} h_a^2}{m_a^2 \minus{} h_a^2} \equal{} \frac {l_a^4}{b^2c^2}.$[/quote]

Very nice Proof Yetti  :)


Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC,$ respectively. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that $\frac{AE}{ED} = \frac{BF}{FC}$ if and only if $\frac{SA}{SB} = \frac{TD}{TC}$.

Let $M$ be the intersection of $AD$ and $BC$. By applying Menelaus to triangle $MAB$ with triple $(S, E, F)$ and triangle $MDC$ with triple $(T, D, C)$, we obtain the result.


He said
$ AB \equal{} C^{2}\minus{}CD\plus{}D^{2}$
where
$ A \equal{}\sum a^{4}\minus{}\sum a^{2}b^{2},\ \ \ \ B \equal{}\sum a^{2}b^{2}\minus{}abc\sum a,$
$ C \equal{}\sum a^{3}b\minus{}abc\sum a,\ \ \ \ D \equal{}\sum ab^{3}\minus{}abc\sum a.$

[quote="Abdek"]Let $ a,b,c$ be a real numbers prove that:

$ \left(\sum_{cyc}a^2b^2 \minus{} abc(a \plus{} b \plus{} c)\right)\left(\sum_{cyc}a^4 \minus{} \sum_{cyc}a^2b^2\right) \ge$$ \frac {3}{4}\left(\sum_{cyc}a^3b \minus{} abc(a \plus{} b \plus{} c)\right)^2$[/quote]
It is not difficult, in fact,

$ LHS\minus{}RHS\equal{}\frac{1}{4}\left(\sum{a^3b}\minus{}2\sum{a^3c}\plus{}\sum{a^2bc}\right)^2.$


How many divisors of $ 3\cdot2^6$ are there?

Using the divisor-counting formula this is $ 2 \times 7\equal{}\boxed{14}$.

please dont consider this as spam

but can someone please tell me what is the divisor-counting formula ?

For a number $p_1^{a_1} \cdot p_2^{a_2} \cdot \hdots$, where $p_k$ are primes (this is just the prime factorization), there are $(a_1+1)(a_2+1)\hdots$ divisors. This is because each factor is in the form $p_1^{a} \cdot p_2^{b} \hdots$, and each $a,b,c,\hdots$ can be anything from $0$ to $a_k$.

Aren't there 28 divisors because we can also count negative numbers?

Yea, I think 28 because the problem never specified positive or negative.


We wrote down all the integers of 1 to 7 digits we can, using only the two digits 0 and 1. How many 1's did we write?

Do you mean you wrote them out in base two?

He said digits, not the numbers $ \{1,2,3\cdots7\}$

[hide="Solution + Generalization"]Let's consider a small case, where we have $ d$ digits and $ k$ ones available.
The first digit is a one, so now we have $ \binom{d\minus{}1}{k\minus{}1}$ ways to place the rest of the ones. Multiplying this by $ k$ gives the number of ones in the digit.

Summing over all possible values of $ d$ and $ k$ gives the sum: $ \sum_{d\equal{}1}^{7}\sum_{k\equal{}1}^{d}\binom{d\minus{}1}{k\minus{}1}k$.

[hide="Lemma 1"][b]Lemma:[/b] $ \sum_{k\equal{}1}^{d}\binom{d\minus{}1}{k\minus{}1}k\equal{} 2^{d\minus{}2}(d\plus{}1)$.
[i]Proof:[/i] Let's consider $ \sum_{k\equal{}1}^{d}\binom{d\minus{}1}{k\minus{}1}k$.
$ \int\sum_{k\equal{}1}^{d}\binom{d\minus{}1}{k\minus{}1}kx^{k\minus{}1}\; dx\equal{}\sum_{k\equal{}0}^{d}\binom{d\minus{}1}{k\minus{}1}x^k\equal{}x(x\plus{}1)^{d\minus{}1}$.

$ \frac{d}{dx}x(x\plus{}1)^{d\minus{}1}\equal{}(x\plus{}1)^{d\minus{}1}\plus{}(d\minus{}1)(x\plus{}1)^{d\minus{}2}x$ evaluated at $ 1$ gives the original sum: $ 2^{d\minus{}1}\plus{}(d\minus{}1)2^{d\minus{}2}\equal{}2^{d\minus{}2}(d\plus{}1)$. $ \Box$[/hide]

Applying Lemma 1, our sum becomes $ \sum_{d\equal{}1}^{7}2^{d\minus{}2}(d\plus{}1)$. Let's generalize and consider $ \sum_{d\equal{}1}^{n}2^{d\minus{}2}(d\plus{}1)$.

This is equivalent to:
$ \frac{1}{4}\sum_{d\equal{}1}^{n}(d2^{d}\plus{}2^{d})\equal{}\boxed{n\cdot 2^{n\minus{}1}}$, using the closed forms for arithmetico-geometric and geometric series.

Grand finale: Let $ n\equal{}7$ to get $ 7\cdot 2^{6}\equal{}\boxed{448}$.[/hide]

Unfortunately this got a little long and used a little calculus (sorry).
It's probably highly overcomplicated...

I shall dedicate myself to finding a quick solution based on simpler methods.

[hide="Found it"]
With an $ n$ digit number, there are $ 2^n$ ways to place ones and zeroes to form a number.
The units digit is a 1 half the time and a 0 half the time, so out of $ 2^{n}$ numbers, $ 2^{n}/2\equal{}2^{n\minus{}1}$ numbers have a 1 in the units digit.
Similarly, $ 2^{n\minus{}1}$ numbers have a 1 in the tenths digit, etc, so our answer is simply $ n\cdot 2^{n\minus{}1}$, or $ \boxed{448}$ in this case.
[/hide]

Should have seen that earlier  :lol:


How do you make those headers in the top left and right corner? Does it use some special package or something?

Use the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]fancyhdr package[/url]

In the first point on that web site is:

[quote]Open the MiKTeX Package Manager. (You can do this by clicking Start, then All Programs, then MiKTeX, then MiKTeX Package Manager.) A window should open with a long list.
[/quote]

But if I do it, there is written: "In order to install packages, it is necessary to define a package repository."

What folder it has to be?

thanks for help

conceive

What that means is that MiKTeX wants to know where you downloaded its compressed files to. Then it can uncompress the correct file and install fancyhdr. Look for lots of CAB files in the folder you specified when installing Miktex (see item 7 at [url=http://www.ctan.org/tex-archive/systems/win32/miktex/setup/install.html#id421158]The MiKTeX Installation Procedure[/url])

If you're not sure where that is look in texmf\miktex\config\setup-the_date.log (the_date will be the date you installed MiKTeX) for the line Package repository: followed by the folder's name.

If you have mistakenly deleted that folder then re-download MiKTeX and re-install it. If you installed MiKTeX from a CD then it's just asking for the CD.

The AoPS page does tell you how to install fancyhdr manually, but it's best to get MiKTeX sorted out.

[quote="stevem"]What that means is that MiKTeX wants to know where you downloaded its compressed files to. Then it can uncompress the correct file and install fancyhdr. Look for lots of CAB files in the folder you specified when installing Miktex (see item 7 at [url=http://www.ctan.org/tex-archive/systems/win32/miktex/setup/install.html#id421158]The MiKTeX Installation Procedure[/url])

If you're not sure where that is look in texmf\miktex\config\setup-the_date.log (the_date will be the date you installed MiKTeX) for the line Package repository: followed by the folder's name.[/quote]

Well, I tryed this but it still wrie me that "This folder does not seems to be a local repository folder."

What to do?

Thanks for help.

[quote="stevem"]If you have mistakenly deleted that folder then re-download MiKTeX and re-install it. If you installed MiKTeX from a CD then it's just asking for the CD.

The AoPS page does tell you how to install fancyhdr manually, but it's best to get MiKTeX sorted out.

I think that the simplest thing is to try re-installing MiKTeX. In the folder with the CAB files run setup.exe and install MiKTeX into the same place as before which should sort out any errors (probably due to a missing entry in the Registry).


function
integration
real analysis
real analysis unsolved

Let $n \in \mathbb N_0 \setminus \left\{ 1 \right\}$, $n$ odd. Determine all continuous functions $f : [0,1] \to \mathbb R$ such that \[ \int_0^1 \left( f \left( \sqrt[k] x \right) \right)^{n-k} \, dx = \frac{k}{n} \, , \]
for all $k \in \left\{ 1,2,\ldots,n-1 \right\}$.

[i]Titu Andreescu[/i]

What about this problem?
[b]*Necromancy Skill [color=blue]+1[/color]*[/b]

This might be an idea. Claim that $f(x)=x$. To see this observe that your condition is equivalent to:

$\int_{0}^{1}(f(\sqrt[k]{x})^{n-k})-(\sqrt[k]{x})^{n-k})dx=0$.

Use the substitution $x=y^{k}$ and you get that

$\int_{0}^{1}((f(y))^{n-k}-y^{n-k})y^{k-1}dy=0$. for all $k=1,\cdots, n-1$.

Now, claim that using all these conditions we get after some rearrangements, involving binomial expansions, that

$\int_{0}^{1}(f(x)-x))^{n-1}dx=0$ from which the result follows. To understand the calculations let us play with the case when $n=3$.

We have that

$\int_{0}^{1}(f(y)^{2}-y^{2})dy=0$ and

$\int_{0}^{1}y(f(y)-y)dy=0$.

Substract these two equalities and you get that

$\int_{0}^{1}f(y)(f(y)-y))dy=0$, and using one more substraction we get that

$\int_{0}^{1}((f(y)-y))^{2}dy=0$ from which the result follows.

A similar technique can be applied to prove that the general case holds.
It can be understood why $n$ is an odd integer.

I think it can also be solved with Cauchy Inequality.


vector
linear algebra
linear algebra theorems

I was asked to give a rigorous proof for the following:

prove : (-a)b = -ab for all a,b.

Hint : show axioms or theorem taking place in the proof.

I have no idea what to do.

You should have written where $ a$ and $ b$ belong to but here is probably what you expect:
The thesis is equivalent to:
\[ (\minus{}a)b\plus{}ab\equal{}0\]
\[ ((\minus{}a)\plus{}a)b\equal{}0\]
\[ 0  b\equal{}0\]

And that assumes that you have already proven that $ 0b\equal{}0.$ That, in turn, can also be proven from the distributive law.

$ ab\equal{}(a\plus{}0)b\equal{}ab\plus{}0b.$ Add $ \minus{}(ab)$ to both sides to get that $ 0\equal{}0b.$

The result makes sense whenever the symbols make sense. Hence this applies in any ring, (including the ring of $ n\times n$ matrices), and it even works when $ a$ and $ b$ belong to different spaces - for instance, if $ b$ belongs to a vector space and $ a$ to the field of scalars. (Or vice versa.)

The key axioms needed for both of these are the appropriate distributive law and the existence of additive inverses.

Thank you both but I do not follow.

How do we start the proof ?

with (-a)b = (-a)b +0 and then take it from there ,or what

The proof starts with the definition $ \minus{}(ab).$ That's the additive inverse of $ ab,$ and it means that $ \minus{}(ab)\plus{}ab\equal{}0.$ Hence if we show that $ c\plus{}ab\equal{}0,$ that is evidence that $ c\equal{}\minus{}(ab).$

dondingo proceeded from there by using the distributive property.

You can prove 0b = 0, it's not usually an axiom. 0 is defined as an additive identity; so you have 0b = (0+0)b = 0b+0b, so 0=0b.

Yes ,0x=0 is a theorem, but is the rest of the proof correct??
can  we  consider it as rigorous??

[quote="Nukular"]Looks good, though the steps are in an order that makes it a little weird to read.

Also the cancellation law is a theorem.[/quote]

Yes the cancellation law is a theorem.

Why the steps are weird to read??

How would you write a rigorous proof for the proof ,0x=0 ,that you did in your post # 7

There's nothing wrong with what you have; just a personal style thing I guess.

For 0b = 0 I'd probably do something like

0 = 0b + -(0b) (additive inverses)
   = (0+0)b + -(0b) (additive ident)
   = (0b + 0b) + -(0b) (distributivity)
   = 0b + (0b + -(0b)) (associativity)
   = 0b + 0 (additive inverses)
   = 0b (additive identity)

I suppose I'm using implicit transitivity -- oh well, it's a tradeoff for readability.


[quote="Ezbakhe Yassin"]Let $a,b>0$, $k\geq 0$ and $n\geq 1$. Prove this inequality:

\[ (k+\frac{a}{b})^n+(k+\frac{b}{a})^n\geq 2(1+k)^n  \][/quote]

Hi Yassin!

Well, quite easy...

Power mean ineq yields $(n\geq 1)$
$(k+\frac{a}{b})^n+(k+\frac{b}{a})^n\geq \displaystyle {2(\frac{2k+\frac{b}{a}+\frac{a}{b}}{2})^n}\geq 2(\frac{2k+2}{2})^n= 2(1+k)^n$

since $\frac{b}{a}+\frac{a}{b}\geq 2$ when $a,b>0$

 :)

But you can't do that with n $\in R+$
Can anyone prove AM-GM with n $\in R+$?

[quote="silouan"]I think n is natural number
Solution to the problem
By AM-GM first and Cauchy secondly we have

\[ (k+\frac{a}{b})^n+(k+\frac{b}{a})^n\geq 2\sqrt{[(k+\frac{a}{b})(k+\frac{b}{a})]^n}\geq 2(1+k)^n  \][/quote]

you are faster than me.... :P

$n$ is not necessarily a natural one.
If i am not wrong power mean holds for positives (more generally...)

I proved it using Jensen's inequality for $f(x)=x^n$ with $n\geq 1$.

Could anyone improve the inequality ($0<n<1$ or $n<0$)?

silouan have already prooved the case $0<n<1$


Let $ABC$ be a triangle such that $|AC|>|AB|$. Let $X$ be on line $AB$ (closer to $A$) such that $|BX|=|AC|$ and let $Y$ be on the segment $AC$ such that $|CY|=|AB|$. Intersection of lines $XY$ and bisector of $BC$ is point $P$. Prove that $\angle BPC+\angle BAC = 180^\circ$.

UK 2006, II round
http://www.mathlinks.ro/viewtopic.php?search_id=1183486400&t=72727

Wow, thanks... Now I know for almost all the problems on our TST where they came from. I'm missing the problems 2, 5, 7... do you maybe know where those came from?

[quote]I'm missing the problems 2, 5, 7... do you maybe know where those came from?
[/quote]

Where we can see problems?

Oh, here you go.
http://www.mathlinks.ro/viewtopic.php?t=154720

Zabavi se.


A child has a set of 96 distinct blocks. Each block is one of 2 materials (plastic, wood), 3 sizes (small, medium, large), 4 colors (blue, green, red, yellow), and 4 shapes (circle, hexagon, square, triangle). How many blocks in the set are different from the “plastic medium red circle” in exactly two ways? (The “wood medium red square” is such a block.)

$ \textbf{(A) } 29 \qquad \textbf{(B) } 39 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 56 \qquad \textbf{(E) } 62$

[hide] We consider the ways in which we can change the properties. Suppose that the material changes. This can be done in $ 1$ way. Now, each of the other properties can be changed in $ 2$, $ 3$, and $ 3$ ways respectively, for a total of $ 1(2)\plus{}1(3)\plus{}1(3)\equal{}8$ ways. Next suppose that the size changes (we can ignore the material since we've already considered those case). The size can change in $ 2$ ways, and the color and shape can eachc hange in $ 3$ ways. Hence, $ 2(3)\plus{}2(3)\equal{}12$ ways. We last consider the case where the color and shape(but nothing else) changes. Thye can each change in $ 3(3)\equal{}9$ ways. Summing $ 8\plus{}12\plus{}9\equal{}\boxed{29}$ ways. Thus, the answer is A.[/hide]


If a and b are two acute angles such that sin a = sin b. prove that angle a = angle b.

In the range $ \left[0,\frac{\pi}{2}\right]$, $ \arcsin(\sin\theta)\equal{}\theta$, so the result follows.
I don't know if this is very rigorous though.

Or you can use congruence of the right triangles.

Observe two right triangles with the common hypotenuse $ c$, and let $ \sin\alpha_1 \equal{} \sin\alpha_2$. Then $ {a_1\over c} \equal{} {a_2\over c}\implies a_1 \equal{} a_2$, hence the triangles have two sides and the angle opposite the longer one equal, meaning they're congruent, hence $ \alpha_1 \equal{} \alpha_2$.


calculus
derivative
calculus computations

how can you use the power rule to solve the derivative of y=2x?
x^0 can be either positive or negtive :maybe:

Note that $ x^0 \equal{} 1$ for all real $ x$ except $ 0$.
Then: $ y' \equal{} (2x)' \equal{} 2(x)' \equal{} 2(x^1)' \equal{} 2*1*x^0 \equal{} 2$

but x^0 can be negative 1 if x is a negative number

No, $ x^0$ is still $ 1$ if $ x$ is a negative real number, because $ x^0\equal{}x^{1\minus{}1}\equal{}\frac{x^1}{x^1}\equal{}1$


[quote="Mashimaru"]Solve the following equation: $ (4^x \plus{} 2)(2 \minus{} x) \equal{} 6$[/quote]

Let $ h(x)\equal{}\frac{6}{4^x\plus{}2}\minus{}(2\minus{}x)$.

$ h"(x)\equal{}6\ln(4)\frac{4^x(4^x\minus{}2)}{(4^x\plus{}2)^3}$ and has one unique zero.

So $ h'(x)\equal{}0$ has at most 2 roots and then $ h(x)\equal{}0$ has at most 3 roots ( $ h(x)$ is obviously $ C_{\infty}$)

But we have $ h(0)\equal{}h(\frac{1}{2})\equal{}h(1)\equal{}0$

So the equation $ (4^x\plus{}2)(2\minus{}x)$ has exactly 3 roots : $ \{0,  \plus{}\frac{1}{2}, \plus{}1\}$


calculus
integration
function
trigonometry
floor function
derivative
algorithm

Evaluate $ \int_a^b \frac {sin(x)}{x}$, for $ |a|,|b| < 10^5$ and precision $ 10^{ \minus{} 10}$ ?
I know simpson method, but isn't fast enough.

Hi;

You can use gaussian integration or romberg integration, they will probably do better than simpsons rule.

For small $ a,$ by which I mean $ |a|$ less than about 10, I can do pretty well by integrating the Taylor series centered at zero. In fact, I've used that to plot graphs of the function $ \int_0^x\frac{\sin\pi t}{\pi t}\,dt$ for some notes I wrote.

Whereas, for larger $ b,$ I'd be tempted to use repeated integration by parts to find an asymptotic series for $ \int_b^{\infty}\frac{\sin x}{x}\,dx.$

Asymptotic series can be a little tricky to manage, since they don't converge - the precision increases up to some point, and then gets worse if you use more terms. But I think that for large enough $ b,$ we can probably make this approach work.

As for the limitation $ |a|,|b|<10^5,$ I don't really understand the restriction - the larger $ b$ is, the easier the asymptotic series gets. The toughest case would be for $ a$ and $ b$ somewhere in the middle - on the big side for the Taylor series, on the small side for the asymptotic series.

Of course, to use my approaches, you'll also need to know the exact value of $ \int_0^{\infty}\frac{\sin x}x\,dx\equal{}\frac{\pi}2.$

Elaborating on my answer:

$ \int_x^{\infty}\frac{\sin t}t\,dt\equal{}\frac{\cos x}x\plus{}\frac{\sin x}{x^2}\minus{}\frac{2\cos x}{x^3}\minus{}\frac{6\sin x}{x^4}\plus{}\frac{4!\cos x}{x^5}\plus{}\frac{5!\sin x}{x^6}\plus{}\cdots$

As I said, the series diverges for all $ x.$ What you have to do with it is to add up terms until they reach a minimum and then stop. We can state it this way:

$ \int_x^{\infty}\frac{\sin t}t\,dt\equal{}\sum_{k\equal{}0}^{\lfloor x\rfloor}\frac{k!\cos(x\minus{}k\pi/2)}{x^{k\plus{}1}}\plus{}R,$ where $ |R|\le \frac{(\lfloor x\rfloor\plus{}1)!}{x^{\lfloor x\rfloor\plus{}1}}.$

To get $ |R|<10^{\minus{}10}$ as requested, we need $ x>20.$

As a trial run for this idea,

$ \int_{25}^{45}\frac{\sin x}x\,dx\equal{}\int_{25}^{\infty}\frac{\sin x}x\,dx\minus{}\int_{45}^{\infty}\frac{\sin x}x\,dx$

$ \approxeq 0.0393137758\minus{}0.0120813259\approxeq0.0272324499.$

I tried doing this with Romberg integration. I got those same 10 digits when I used 129 function evaluations (which is a lot of steps for Romberg, 8 columns in the extrapolation table). With one fewer column (65 function evaluations), the accuracy was only about $ 10^{\minus{}7}.$ So in this particular case, I like the asymptotic series better than Romberg. If we tried the integral from $ 25$ to $ 2000,$ the asymptotic series would work just as well, while Romberg would be completely unworkable.

Smaller values of the limits would call for different methods.

First of all, one does not need two parameters. It is enough to know $ \int_0^a\frac {\sin x}{x}\,dx$ for each $ a > 0$ with twice smaller error. Now, Kent already did $ a > 20$ (or, to be on the safe side, $ a > 30$), so we should not be concerned with too large values of $ a$.

It is tempting to use the standard Taylor series for $ \sin x$ for small $ a$ but one has to be careful: for $ a\approx 20$ the numbers that ultimately cancel are too large to work in the usual machine precision. It is still a reasonably fast way but one has to implement high precision arithmetic if following this approach literally..

One way I can think of for small $ a$ is to do each period of $ \sin$ separately. We'll need to find integrals like $ \int_0^{u}\frac {x^m}{p \plus{} x}\,dx$ where $ p$ is an integer multiple of $ \pi$, $ |u| < \frac\pi 2$, and $ m$ is not too large (up to $ 20$ will be about enough). This can be easily achieved by decomposing $ \frac 1{p \plus{} x} \equal{} \frac 1p\sum_{\ell \ge 0}( \minus{} x/p)^\ell$ after which it becomes a question about integrating  power functions.

Still, this is ugly and error-prone. So let's try something "completely different". Go to the complex plane and replace the integral over the horizontal line by one over the vertical line (we'll integrate $ e^{iz}/z$ now). There will be a problem with very small $ a$ but there we can use the classical Taylor series. Ohterwise, the problem reduces to evaluating $ \int_0^\infty \frac {e^{ \minus{} y}}{y \plus{} ia}\,dy$. We'll need to integrate up to about $ 30$ regardless of $ a$. The fourth derivative is about $ 80e^{ \minus{} y}$ when $ a > 1$, so the error for the good old Simpson with step $ h$ is about $ \frac {80}{2880}h^4$, which is less than $ 10^{ \minus{} 10}$ for $ h \equal{} \frac 1{500}$. Thus, $ 500\cdot 30 \equal{} 1,500$ values is what you need for the main part of this algorithm, which makes it somewhat slower than the previous one but still fast enough for occasional computations. The advantage is that this algorithm is completely straightforward and you do not need to consider many cases like in the previous algorithm or write loops with several parameters where one stupid mistake may destroy everything (provided that $ \exp$, $ \sin$, and $ \cos$ are already implemented and work with sufficient precision in the range $ ( \minus{} 50,50)$).

Actually, you can, probably, handle the last integral in a much better way than Simpson: there exists vast literature about numerical methods for $ \int_0^\infty f(y)e^{ \minus{} y}\,dy$ and you can try to figure a few things out by yourself too here.


Let $x,y>0$  integers. Find the maximum value of $M<20$  such that

$x\frac{117}{100}+y\frac{160}{100}=M$

[quote="silouan"]Noone ?I think it is easy.[/quote]

Yes, if you follow the like at my signature below and go to Number Theory problems section of my page, you can find the general problem with hint!

I didn't find something .But if you have a solution post it please , because my own solution is very ungly. try to prove it with a nice way. My result is

[hide]19,94[/hide]

[quote="silouan"]price of $M<20$[/quote]

How many times have people told you that "price" is incorrect English (in this context), Silouan? :mad:

I am so sorry :blush: . It is a bad habbit

Please give me a nice solution for this. I gave the result so I think it is easier....


geometry Hi ratio function calculus quadratics USAMO

For what real values of $k>0$ is it possible to dissect a $1 \times k$ rectangle into two similar, but noncongruent, polygons?

I have a hunch that it can be shown that th 2 polygons must be rectangles, and we get an equation $\frac1x=k-x$, so the equaton $x^2-kx+1=0$ must have real solutions different from 1, so we get k>2. I don't know if these are all of them, but I think they are.

What do you think? Is it true that the 2 polygons must be rectangles?

Can somebody please clarify for me what dissection means?  During the contest, I interpreted it to mean cutting up the rectangle and rearranging it.  But beyond 1 cut, that starts to get very complicated.

It's very easy to prove that if you have a single straight line cut that it must be perpendicular to one of the rectangles edges.  As Grobber pointed out, this gives k>2 as a solution.  But there are more k possible.  But making the cut in the other direction (perpendicular to the edges of length 1), the same argument that gets k>2 gets 0<k<1/2 as well.

I believe these are the only values.

indeed Michael is right. we (dzeta, blondu and me) have managed to produce a reasoning, but it's not quite rigurous (yet).

the main idea floats around the fact that the figures must cut a certain side of the rectangle, and then, a polygonal line will start from that point, and move on to another side.

then we consider the minimal and maximal rectangle that are contained within each of the figures. it's easy to check that they each one is the other figures' other rectangle (I will later add a drawing).

computation now shows that the two rectangles must concide, therefore the only covering is given for $ k \notin \left( \displaystyle \frac 12, \ 2\right)$.

however, this is just a sketch. more "evidence" needs to be brougth :)

I'm getting a rumour saying that the answer to the problem is $k \ne 1$.

Here's a counter-example

Note that there could be more than one "stair" thingies. So if we increase the number of "stairs", we might be able to reach anything arbitrarily close to 1. I think that might be the idea.

I think your example can be extended to every k>1, the case k<1 being analogous.
You don't need to increase the number of stairs.
Note that the example doesn't match the case k=1.
The key idea is to consider the sum of angles at every vertex of polygons.
If the polygon has n vertices, the sum of angles at them is 360n-l*90( the number of corners among them)-x*180(number of vertices on sides). Since the sum of angles of each polygon is 180(n-2), we get an even number of corners in each polygon.
Since a polygon can't contain 0 or 4 corners, it must contain 2 corners. We easily show this corners are on a side. Now a polygon can't have more than 1 vertex on a side, also it can't have vertices on sides it contains. So we reach the only example in the figure( One polygon contains point A,B; the other C,D; and they contain one point on AB and BC)
. Now let L be the longest side of polygons. Since a similarity maps 1 to a segment, we get L>1, so L lies in the interior. But now then this side must be common to both polygons, so the second polygon contains also L. the similitude maps L to a longer segmenet, contradiction.
The proof is finished.

P.S. Bill, do you know the results of the contest?

[quote="iura"]I think your example can be extended to every k>1, the case k<1 being analogous.
You don't need to increase the number of stairs. [/quote]

Actually, I believe you *do* need to increase the number of stairs.
With no stairs (a single cut), the limit on the aspect ratio is >2:1.
With one stair (3 cuts), the limit on the aspect ration is >3:2.
With two stairs (5 cuts), the limit is >4:3.
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=84504&highlight=#84504]my posting[/url] on AoPS.
What do you think about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12306]dissecting a square?[/url]

Just solve these equations for f and c, where k>=1.5 is the aspect ratio:

You have a system of 3 high-degree equations, how do you show it has a solution?
I think they don't!
( This is the reply to your post on artofproblemsolving.com)

[quote="iura"]You have a system of 3 high-degree equations, how do you show it has a solution?
I think they don't!
( This is the reply to your post on artofproblemsolving.com)[/quote]

For the 3-line dissection...  I'll choose 2 of the 3 equations:

[tex]\displaystyle cf+f=k[/tex]

[tex]\displaystyle cf^2+cf^4=1[/tex]

Solve the 2nd for c, then substitute c into the first.

[tex]\displaystyle c=\frac{1}{f^2+f^4}[/tex]

[tex]{\frac{1}{f^2+f^4}} \times f + f = k[/tex]

[tex]{\frac{f}{f^2+f^4} } + f = k[/tex]

[tex]\displaystyle \frac{1}{f+f^3} + f = k[/tex]

f was defined to be the scale factor of one polygon versus the other.

Now, we let f be the independent variable, and vary it to find what values of k are possible.  By inspection, as long as f is finite and greater than zero, then k is finite and greater than zero.

As f approaches 0+, k asymptotically approaches 1/f (+infinity).
As f approaches +infinity, k asymptotically approaches f (+infinity).
At f=1, [tex]k=\frac{1}{1+1}+1=\frac{3}{2}[/tex]

I can show that k(f) = k(1/f) by substituting g=1/f:

[tex]\displaystyle k = \frac{1}{f+f^3} + f[/tex]

[tex]\displaystyle k = \frac{1}{f+f^3} + \frac{f(f+f^3)}{f+f^3}[/tex]

[tex]\displaystyle k = \frac{f^0}{f+f^3} + \frac{f^2+f^4}{f+f^3}[/tex]

[tex]\displaystyle k = \frac{f^4+f^2+f^0}{f^1+f^3}[/tex]

[tex]\displaystyle k = \frac{(1/g)^4+(1/g)^2+(1/g)^0}{(1/g)+(1/g)^3}[/tex]

[tex]\displaystyle k = \frac{g^4}{g^4}\times\frac{g^{-4}+g^{-2}+g^0}{g^{-1}+g^{-3}}[/tex]

[tex]\displaystyle k = \frac{g^0+g^2+g^4}{g^3+g^1}[/tex]

Since the function is logarithmically symmetric about f=g=1, this strongly suggests we will have a minimum at f=g=1.  [See attached graph of k vs f.]

To prove this, I resort to calculus.  I find the maxima/minima of k by taking the derivative of k with respect to f, then setting the derivative equal to 0:

[tex]\displaystyle k = \frac{1}{f+f^3} + f[/tex]

[tex]\displaystyle \frac{d}{df} ( \frac{1}{f+f^3} + f)=0[/tex]

[tex]\displaystyle -\frac{1+3f^2}{(f+f^3)^2}+1=0[/tex]

[tex]\displaystyle -(1+3f^2)+(f+f^3)^2=0[/tex]

[tex]\displaystyle (f+f^3)^2=1+3f^2[/tex]

[tex]\displaystyle f^2\times(1+f^2)^2=1+3f^2[/tex]
[tex]\displaystyle f^2\times(1+2f^2+f^4)=1+3f^2[/tex]
[tex]\displaystyle f^2+2f^4+f^6=1+3f^2[/tex]
[tex]\displaystyle f^6+2f^4-2f^2-1=0[/tex]

[tex]\displaystyle (f^2-1)(f^4+3f^2+1)=0[/tex]

From the first term, we get two real roots:

[tex]\displaystyle f^2-1=0[/tex]

[tex]\displaystyle (f-1)(f+1)=0[/tex]

[tex]\displaystyle f=+1[/tex]
[tex]\displaystyle f=-1[/tex]

From the second term and the quadratic formula, we get two pairs of complex conjugate roots:
[tex]\displaystyle f^4+3f^2+1=0[/tex]

[tex]f^2=\frac{-3\pm \sqrt{3^2-4\times 1\times 1}}{2}[/tex]

[tex]\displaystyle f^2=-\frac{3+\sqrt{5}}{2}[/tex]

[tex]\displaystyle f^2=-\frac{3-\sqrt{5}}{2}[/tex]

So we have proven that for positive real values of f, the one and only minimum value of k occurs at f=1.  k=3/2.  For f>0, the function is continuous and differentiable.  So all rectangles of the shape 1xk, where k>=3/2 are dissectible with three lines as shown in the attached illustration.

I leave it to you to prove or disprove my equivalent claims with respect to 5-line dissections and above.

Ok, I understood myself what's the point.
I just made such stupid mistake!

The answer is all $k \neq 1$. Here a "dissection" does not need to be a single cut.

The construction for $k > 1$ is not so hard to find (my friends and I got it over dinner at Legal Seafoods yesterday, and billzhao basically got it in #6).  Pick $2n$ and $r \in (1,\infty)$ arbitrary.  Consider $n$ rectangles of dimensions $1 \times r$, $r \times r^2$, $(1+r^2) \times r$, $(r+r^3) \times r^4$, and so on, gluing them together in this order in the obvious fashion.  For example, the case of $n=3$ is shown below.
[asy]
size(8cm);
defaultpen(fontsize(10pt));

real r = 1.6;

pair A = (0,r^3+r);
pair B = (r^4,r^3+r);
pair C = (r^4+r^2,r^3+r);
pair D = (r^4+r^2+1,r^3+r);
pair E = (r^4,r^3);
pair F = (r^4+r^2, r^3);
pair G = (r^4+r^2+1, r^3);
pair O = (0,0);
pair H = (r^4,0);
pair I = (r^4+r^2+1,0);
pair W = (-r^6, A.y);
pair X = (W.x,-r^5);
pair Y = (0,-r^5);
pair Z = (I.x,-r^5);

pen p = black+1.3;
pen q = red;
fill(O--A--B--H--cycle, palecyan);
fill(B--C--F--E--cycle, palecyan);
fill(W--A--Y--X--cycle, palecyan);
fill(C--D--G--F--cycle, palegreen);
fill(E--G--I--H--cycle, palegreen);
fill(O--Y--Z--I--cycle, palegreen);

draw(O--A, q);
draw(B--E, q);
draw(F--G, q);
draw(H--I, q);
draw(W--D--Z--X--cycle, p);
draw(C--F--E--H--O--Y, p);

label("$1$", (C+D)/2, dir(90));
label("$r^2$", (B+C)/2, dir(90));
label("$r^4$", (A+B)/2, dir(90));
label("$r^6$", (W+A)/2, dir(90));

label("$r^1$", (D+G)/2, dir(0));
label("$r^3$", (G+I)/2, dir(0));
label("$r^5$", (I+Z)/2, dir(0));
[/asy]
Now, the aspect ratio of the completed rectangle is \[ \frac{1+r^2+\dots+r^{2n}}{r+r^3+\dots+r^{2n-1}} \] which for a fixed $n$ can goes from $(1+\tfrac1n, \infty)$ as $r > 1$.  Thus by taking $n$ large, we can get any $k > 1$.  Of course by symmetry we can also get any $k < 1$.

Now to show that $k = 1$ is impossible (the tricky part, IMO), consider a dissection into two similar polygons $\mathcal P \sim \mathcal Q$, and let $\mathcal B$ be their common boundary.  By counting the number of sides of $\mathcal P$ and $\mathcal Q$ we see $\mathcal B$ must run from one side of the square to an opposite side.  From this we can show that the [i]longest[/i] side of $\mathcal P$ and $\mathcal Q$ are either the sides of the square (of length $1$), or a side shared in $\mathcal B$.  In particular, they are equal, which means in fact $\mathcal P \cong \mathcal Q$.

[hide="solution"]
Starting from the origin, draw a line $1$ right, then $k$ up, then $k^2$ right... then $k^{2n-1}$ up, then $k^{2n}$ right. Take the rectangle with two vertices at $(0,0)$ and $\left(\sum\limits_{i=0}^nk^{2i}, \sum\limits_{i=1}^nk^{2i-1}\right)$, and the path walked as the cut. It is clear to see that the resulting two polygons are similar with ratio $k$.
Scale the entire rectangle to $1\times c$ to get $c= \dfrac{\sum\limits_{i=0}^nk^{2i}}{ \sum\limits_{i=1}^nk^{2i-1}}$. Note that taking $k\to 0$ brings $c\to\infty$, while taking $k\to 1$ brings $c\to \dfrac{n+1}{n}$. Take $n$ arbitrarily large, and we can construct any $1\times c$ rectangle where $c\ne 1$.

To prove that $c=1$ does not work;
Note that the cut must be from one side of the square to the other side, else the number of sides of the two resulting polygon are not equal. Consider the maximal side length of each polygon. If this maximal side length occurs as a cut (i.e the maximal side length is $>1$), then it is the maximal side length for both polygons. Thus, if they are similar then they must be congruent, contradiction. Otherwise, the maximal side length occurs as one of the sides of the square (i.e the maximal side length is $1$). But again, this means that both polygons have the same maximal side length, implying their congruency, contradiction.

Thus, our final solution is all positive real $k$ except $k=1$. $\Box$[/hide]

Doesn't this follow immediately from the Wallace-Gerwin-Bolyai Theorem?


number theory
least common multiple
group theory
superior algebra
superior algebra solved

this is probably easy but i cant see it.exhibit an element of order 20 in $S_{9}$ . also show that there exists no element in $S_{9}$ of order 18. any hints appreciated. thanks in advance

For this task you should know (or learn) how to calculate the order of an element of $S_{n}$:
Express your permutation as a product of disjoint cycles. The order is then the least common multiple of the cycle lengths.

So an element of order 20 in $S_{9}$ is a 4-cycle times a disjoint 5-cycle. (And $S_{9}$ is the smallest symmetric group containing an element of order 20.)
To reach order 18 in $S_{9}$, you need to express 18 as the lcm of natural numbers s.th. the sum of these numbers is $\le 9$. That is not possible.

LiMa


write the product. the product has the formula. C16H13NO7 .

Hell_ever, please try to learn to compare first your's product molecular formula with the one given. Don't forget also that a pyridine ring is electron deficient.

I don't know how to latexify.  :blush: sorry for inconvinience.
@ hell_ever . isn't the compound having  have less number of hydrogens than mentioned?

@ carcul, what's the answer ?


geometry
3D geometry
percent
percent increase

If each edge of a cube is increased by $ 60\%$, what is the percentage increase in its surface area?

Let the original edge be $ x$. The old surface area is $ 6x^2$. The new edges are $ 1.6x$. The new surface area is $ 6 (1.6x)^2 \equal{} 6x^2 \cdot 2.56$. Thus, the percent increase is $ 2.56 \minus{} 1 \equal{} \boxed{1.56}$.

EDIT: Oops. It should be $ 256\minus{}100\equal{}156$, like AIME15USAMO said.

Well, PowerOfPi did everything correctly but then forgot to multiply by 100 to make it a percent, rather than a decimal. It should be $ \boxed {156}$


[quote="knexpert"]I sent my solutions in for the second round a couple of weeks ago. Are all of the solutions logged after the deadline or are they logged sooner? Thank you in advance.[/quote]

We won't get around to logging those for at least another couple weeks.  We're still wading through round 1 ;)


algebra
polynomial
calculus
calculus computations

need to find the approximate value for ∫from 0 to 1 e^x^2 by replacing the integrand with an appropriate polynomial of degree 6.  And i need to express the answer as a single fraction in lowest terms

Since $ e^x\equal{}\sum_{n\equal{}0}^{\infty}\frac{x^n}{n!}$, it follows that $ e^{x^2}\equal{}\sum_{n\equal{}0}^{\infty}\frac{x^{2n}}{n!}\equal{}
1\plus{}x^2\plus{}\frac{x^4}{2}\plus{}\frac{x^6}{6}\plus{}\dots$
This is the polynomial you need to integrate.


function quadratics Euler vector trigonometry logarithms calculus

$x^2y''+xy'+y=0$. We are supposed to get $y$.

This is solved easily by making the substitution: $y = x^\alpha$

where $\alpha$ is a function depending on $x$??
I believe that the answer $y$ should be a function of $x$.
Can you please post a solution. It doesn't seem that easy for me  :oops: .

That $\alpha$ should be a constant. We guess a solution of the form $y=x^\alpha$ because each term in $x^2y''+xy'+y$ will be a multiple of $x^\alpha$. This is analogous to guessing $y=e^{\alpha x}$ for a constant-coefficient equation.

Let $y=x^{\alpha}$ where $\alpha$ belongs to $R$

then : $y''=x^{\alpha-2}*\alpha*(\alpha-1)$
          $y'=x^{\alpha -1}*\alpha$

So our equation becomes:

$x^{\alpha}*\alpha*(\alpha-1)+x^{\alpha}*\alpha+x^{\alpha} = 0$

and so you get a quadratic equation. find the roots and get the two solutions
 ;)

If i am not mistaken this is called the Euler substitution (it works for any differential equation where you have all terms of the form $x^{i}y^{(i)}$. ).

And in the above case we would get that $y_1=x^i$, $y_2=x^{-i}$, $y_3=x^i,x>1;y_3=x^{-i},x\leq1$, $y_4=x^{-i},x>1;y_4=x^{i},x\leq1$ are the solutions ?? :?

Or the last two are not possible because $y'$ is not differentiable.. :?:

Thanks. I guess this should be it, although I must look over diferrentiability of functiuns like $x^i$ ;)

[quote="RaduSorici"]which are $cos(ln(x))$ and $sin(ln(x))$[/quote]
This looks much nicer  :) . And how precisely did u get to this...  :?

Also linear combinations of those two- the set of solutions on an interval away from the singularity is a 2-dimensional vector space.

Actually, $x^i=\cos\ln x+i\sin\ln x$ and $x^{-i}=\cos\ln x-i\sin\ln x$; the choice of $\cos \ln x$ and $\sin\ln x$ is made simply to get a real basis.

On any interval containing zero, the only solution is the zero function.

jmerry thanx for completing it all. I assumed what you said without writing it

Thank's guys. I guess that sets it  all :) .

Let $x=e^t,$ we have $\frac{dx}{dt}=e^t=x,\ \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}=y'e^t,$ and

$\frac{d^2y}{dt^2}=\frac{d}{dt}\frac{dy}{dt}=\frac{d}{dt}(xy)'=\frac{dx}{dt}y'+x\frac{dy'}{dt}=xy'+x\frac{d^2y}{dx^2}\frac{dx}{dt}=xy'+x^2y''.$

Thus we can rewrite the given differential equation as follows.

\[ \frac{d^2y}{dt^2}+\frac{dy}{dt}+y=0.  \]


[color=green]If O is an interior point of triangle ABC.  Prove that OA+OB+OC
(a) is greater tha 1/2(perimeter of triangle ABC)
(b) is less than the perimeter of triange ABC[/color]

thanks very much guys!!! :blush:

[hide](a)
$OA+OB>AB$
$OB+OC>BC$
$OC+OA>CA$

Add all these equations and you get $2(OA+OB+OC)>AB+BC+CA \Rightarrow  OA+OB+OC>\frac{AB+BC+CA}{2}$

(b)
The triangle $OAB$ is inside the triangle $ABC$, so $perimeter(OAB)<perimeter(ABC)\Rightarrow$
$AB+OA+OB < AB+BC+CA \Rightarrow$
$OA+OB < BC+CA$

Similarly, we get two more equations from the triangles $OBC,OCA:$
$OB+OC<AB+AC$
$OC+OA<AB+BC$

adding again the three equations and dividing by 2 we get $OA+OB+OC<AB+BC+CA$[/hide]


quadratics
algebra
polynomial
function
group theory
abstract algebra
combinatorics unsolved

Convert the following sum:
\[ \sum_{i\equal{}0}^{n}\binom{n}{i}^{\alpha}
\]
where $ \alpha\equal{}3,4$.
For $ \alpha\equal{}2$,this sum can be converted like:
$ \sum_{i\equal{}0}^{n}\binom{n}{i}^{\alpha}\equal{}\binom{2n}{n}$.
I doubt this problem can be solved for other values of $ \alpha$,i.e if $ \alpha\geq 5$.

http://www.research.att.com/~njas/sequences/A000172
http://www.research.att.com/~njas/sequences/A005260

The best you will find here are ugly recursion formulas. Do you have something better?

  darij

hello, Maple 12 says that
$ \sum_{i\equal{}0 }^n\binom{n}{i}^4\equal{}hypergeom([\minus{}n,\minus{}n,\minus{}n,\minus{}n],[1,1,1],1)$.
Sonnhard.

[quote="darij grinberg"]
The best you will find here are ugly recursion formulas. Do you have something better?
darij[/quote]
I didn't manage to solve it for $ \alpha \equal{} 4$,but for the value $ \alpha \equal{} 3$,I guess I have essentially the same formula,don't know whether my solution is the same...

may I know what Dr.Sonnhard Graubner meant by $ hypergeom$?

I wonder if there exist any simple representation of the formula for $ \alpha \equal{} 3$,not recursive relation,well...though we may use some properties of quadratic polynomial and its roots...But it seems to be too ugly...

P.S:Thanks to Dr.Sonnhard Graubner for a such good info.

hello, you can find the answer here
http://mathworld.wolfram.com/GeneralizedHypergeometricFunction.html
Sonnhard.

well...any thoughts about the case $ \alpha\equal{}3$?...the case $ \alpha\equal{}4$ seems to be too hard and in my opinion only very awkward formula can be obtained...
sorry for bad and shortly typed message,my computer works pretty bad and I have to use my laptop...

hello, also in the case $ \alpha\equal{}3$ i only know  this formula
$ \sum_{i\equal{}0}^n\binom{n}{i}^3\equal{}\text{hypergeom}\left([\minus{}n,\minus{}n\minus{}n],[1,1],\minus{}1\right)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, also in the case $ \alpha \equal{} 3$ i only know  this formula
$ \sum_{i \equal{} 0}^n\binom{n}{i}^3 \equal{} \text{hypergeom}\left([ \minus{} n, \minus{} n \minus{} n],[1,1], \minus{} 1\right)$.
Sonnhard.[/quote]
hmm...probably this formula can be somehow converted?
In any case,I think that this problem should have an easy general formula,maybe,not as in the case $ \alpha\equal{}2$,but not much harder...

[quote="JBL"]Why?  The equality Dr. Sonhard Graubner has posted is essentially a definition, and hypergeometric sums are not, in general, easy to simplify (else we wouldn't need the hypergeometric function to shorten their names).[/quote]
Well...probably I have some misconceptions in hypergeometric sums,I need to read it again...though,the problem is that,firstly I meant to not use any college knowledges,but hypergeom seems to belong there...please pay attention to the topic where the problem was initially posted...combinatorics,it presumes that I was wondering if there exist any totally elementary solution,in spirit of how we prove the case $ \alpha \equal{} 2$ - to divide the set of $ 2n$ elementes into two subgroups of $ n$ elements,etc...
Hmm...though now I doubt it is possible to find a such proof...


good luck to the AL nats team! WOO HOO! I wish I could have gone, (aka not failed state, though technically I'm holding Mr. Crawford responsible for changing the time to afternoon {in which my brain is dead} and not making CD official. :P sarcasm, if you couldn't tell) but hey, i obviously proved myself not worthy/capable/deserving. All of you had better have quit your life and studied math non stop for the last month or so. I can see AL being most improved this year. I can also see one of you in CD...  :ninja: . but overall good luck to everyone, i know you guys will do awesome! (after all, a team I didnt make must be an awesome team. sarcasm again.) dont stay up late at night, and dont study the day of. Tony had better bring his Wii again.  :P
one of you had better run into Darryl again this year, I bumped him last year at the museum and his team shot me death glares. Wait, none of you are tall enough.... :D

PS please dont suck at the cheer this year as bad as we did last year. Just sayin'.

darryl is like two feet tall.... anyone can even BO can... maybe not bo but, at least i know tony can XD

[quote="dynamo729"]one of you had better run into Darryl again this year, I bumped him last year at the museum and his team shot me death glares.[/quote]

wohoah

how tall is tony?

someone actually posted on alabama. watch me own you guys in high school

Oh dang our cheer 2 years ago was pretty bad also...mostly because every one of us was embarrassed when we did it...

anyways, good luck.

sorry i decided to not to bring my wii this year....

I'm 5'8-5'9ish

dang i haven't even thought about our cheer... how about "HERE!" or something....
bam now alabama now has a "cheer"...

I'll do it for you nick, only you  :D

well ur still taller than me... especially nick... *sigh*

shouldn't you be getting some sleep for nats?

@ tony thats a shame the hotel this year is beast and probably will have a good tv
(at grissom you get put in homeroom based on last name, and a girl with a last name close to mine in my typing class said our homeroom next year was going to be beast. i laughed.)
and
HOOOOORAY!!!!!! A CHEER FOR ONLY ME!  :D   :D   :D  i ttly dont mind phailing now, as long as u shout out to me in cd when u make it.
and
are u doing ARML?

@yaofan sorry... im taller than everyone i know, so you def. arent alone :D

OH CRAP ITS TOO LATE FOR ME TO FORCE TONY TO ACKNOWLEDGE ME IN CD

*sigh* at least nick and gene have the same feeling of failure as i do (but only worse)...

i guess i can look ahead to mu alpha theta and maybe go to hawaii

yaofan- hawaii's our junior year :D (or at least its supposed too)
lol our cheer was roll tide
yeah I'll do arml, ask your coach about it

When I asked about other ppl (other than me) from birmingham going to arml, the teacher at grissom said she didn't have more room. You should ask, but she might have already registered the number of ppl she had down as going.

Of course, if you can go, that'd be great, and you can ride with me.


trigonometry
calculus
derivative
calculus computations

$ y$ satisfies the differential equation

$ xy'' \minus{}y' \plus{}4x^{3}y \equal{} 0$.

By substituting $ y \equal{} \sum_{n\equal{}0}^{\infty} {a_{n}x^{n}}$ and similar expressions for $ y'$ and $ y''$, show that $ a_{1}\equal{}0$.

Show that, for $ n\ge 4$,

$ a_{n} \equal{} \frac{\minus{}4}{n(n\minus{}2)}a_{n\minus{}4}$,

and that if $ a_{0} \equal{} 1$ and $ a_{2}\equal{}0$ then $ y\equal{}\cos (x^{2})$.

Find the corresponding result when $ a_{0} \equal{}0$ and $ a_{2}\equal{}1$.

Taking the derivatives gives $ y''\equal{}\sum_{n\equal{}2}^{\infty}n(n\minus{}1)a_{n}x^{n\minus{}2}$, $ y'\equal{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$ so inserting these yields:
$ \sum_{n\equal{}2}^{\infty}n(n\minus{}1)a_{n}x^{n\minus{}1}\minus{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}\plus{}4\sum_{n\equal{}0}^{\infty}a_{n}x^{n\plus{}3}\equal{}0$.  Rewriting the indices and collecting like terms gives
$ \sum_{n\equal{}2}^{\infty}n(n\minus{}2)a_{n}x^{n\minus{}1}\plus{}4\sum_{n\equal{}0}^{\infty}a_{n}x^{n\plus{}3}\equal{}a_{1}$.  By assumption, the stuff on the left is equal to 0, so $ a_{1}\equal{}0$ (although you should probably check that I didn't botch any power rules or combining of coefficients).  For the second part, further collect terms so you have everything in one sum.

Thanks for your response JRav.

I'm not sure about "By assumption, the stuff on the left is equal to 0".

Substituting $ n\equal{}1$, I get $ 4a_{1}x^{4} \equal{} a_{1}$ (for all real $ x$), so $ a_{1}\equal{} 0$.

Then for $ n\ge 4$, I get $ \sum_{n\equal{}4}^{\infty} {n(n\minus{}2)a_{n}x^{n\minus{}1}} \plus{} 4\sum_{n\equal{}4}^{\infty} {a_{n\minus{}4}x^{n\minus{}1}}\equal{}0$.

This gives $ a_{n} \equal{} \frac{\minus{}4}{n(n\minus{}2)}a_{n\minus{}4}$.

Next $ a_{0} \equal{} 1, a_{2} \equal{} 0$ gives $ y \equal{} 1 \minus{} \frac{x^{4}}{2!} \plus{} \frac{x^{8}}{4!} \minus{} ... \equal{} \cos (x^{2})$

and finally $ a_{0} \equal{} 0, a_{2} \equal{} 1$ gives $ y\equal{}x^{2} \minus{} \frac{x^{6}}{3!} \plus{} ... \equal{}\sin(x^{2})$.

[quote="AndrewTom"]Substituting $ n \equal{} 1$[/quote]  This is not a meaningful operation when you have more than one sum indexed by $ n$.  The appropriate operation is "extracting the coefficient of $ x^n$ from both sides."

Thanks JBL: I'd substituted $ n\equal{}1$ because the first sum, $ \sum_{n\equal{}2}^{\infty}{n(n\minus{}2)a_{n}x^{n\minus{}1}}$, starts when $ n\equal{}2$ and so I only had one summation when $ n\equal{}1$.

hello, the solution is
$ y(x)\equal{}C_1\sin(x^2)\plus{}C_2\cos(x^2)$
Sonnhard.

Thanks Sonnhard.

(Have you looked at my "stationary points" post in this forum: I'm annoyed with myself because it should be fairly easy and I can't see where I'm going wrong.)

Thanks chuong01. I was wondering if Sonnhard had solved the differential equation without using series.


calculus
integration
number theory
least common multiple
greatest common divisor

1. Let T be a positive integer whose only digits are 0s and 1s. If $ x\equal{}\frac{t}{12}$, and X is an integer, what is the smallest possible value of X?

2. What is the least positive common fraction that is an integral multiple of $ \frac{9}{28},\frac{12}{35},\frac{15}{56}$

3. How many positive three digit integers have the property that the tens digit is greater than or equal to the hundreds digit and the ones digit is greater than or equal to the tens digit?

i think number 1 is 925...

you just choose the last 2 digits that have to be divisible by 4 (only option is 00) then the sum of its digits must be 0 mod 3.  00 is not positive so it has to be 11100. (ifbyou add anymore 0s or 1s inbetween the number will be bigger).  So that the rest follows..

i think number 3 is 165....
its basically reworded to non-decreasing 3 digit numbers-however it can include like 112, and like 122 and stuff-you just have to do some nCk's..-4 cases..

too lazy to do number 2 but it should be the easieset..just prime factor..

[hide="solution"][hide="click me again"][hide="almost there"]haha i got youT has to end in a 0 because 1 is odd. we want the first 2 digits of T to be divisible by 4. 10 doesnt work but 00 does. we want x to be as small as posible so we want to use the least amount of digits. since T is also divisible by 3, then we want the sum of the digits of T equalling 3.(we want X as small as posible). so just fill the rest in with 1s. 11100. divide by 12.

sorry i couldnt resist[/hide][/hide][/hide]

[hide=" # 2"]
the LCM of 9,12, and 15 is 3^2*2^2*5=180 by prime factorization.  However, since 9/28, 12/35 and 15/56 each have denominators that divide 7, you have to divide 180 by 7 to get the least multiple.

[b]180/7[/b]

[/hide]

I agree with #1 and #2.

For #1, you simply imply the divisbillity rule of 4 and 3.

For #2, the formula is $ LCM\equal{}\frac{LCM \ of \ (9,12,15)  }{GCF \ of \ (26,35,56)}$

#3: Look for the pattern.

[hide="I think"]When $ h\equal{}1$, there are (8+7+6+5+4+3+2+1) numbers.
When $ h\equal{}2$, there are (7+6+5+4+3+2+1) numbers.
When $ h\equal{}3$, there are (6+5+4+3+2+1) numbers.
and so on.

So it's $ 8\plus{}2(7)\plus{}3(6)\plus{}4(5)\plus{}5(4)\plus{}6(3)\plus{}7(2)\plus{}8(1)\equal{}\boxed{120}\text{numbers}$[/hide]

Im pretty sure number 3 is 165.

For h=1, there are more than 8+7+6+...+1 numbers since 122 can work (and all numbers like that)

So just bump your answer up by 1 each (so 2 8s instead of 1) making

9+2(8)+3(7)+4(6)...+9(1)=165

[quote="DonkeyKong"] For #2, the formula is $ LCM\equal{}\frac{LCM \ of \ (9,12,15) }{GCF \ of \ (26,35,56)}$ [/quote]

Then you would get 25.71428571 as the LCM, because the LCM of 9, 12, and 15 is 180, and the GCF is 7.

[quote="happyme"][quote="DonkeyKong"] For #2, the formula is $ LCM \equal{} \frac {LCM \ of \ (9,12,15) }{GCF \ of \ (26,35,56)}$ [/quote]

Then you would get 25.71428571 as the LCM, because the LCM of 9, 12, and 15 is 180, and the GCF is 7.[/quote]
So is that right?  Since thats what infiity got.

Yeah, I didn't object to it (its right).

#3 is 165 by simple counting...

It's just the sum of the first 9 triangular numbers.


show that for every positive integer $ n$ there are polynomials $ P(x)$ of degree $ n$ and $ Q(x)$ of degree $ n\minus{}1$ such that
$ (P(x))^2\minus{}1\equal{}(x^2\minus{}1)(Q(x))^2$

Write as $ P^2 \minus{} (x^2 \minus{} 1) Q^2 \equal{} 1$.  Our base case is $ P_1(x) \equal{} x, Q_1(x) \equal{} 1$.  Now take

$ P_n(x) \plus{} \sqrt{x^2 \minus{} 1} Q_n(x) \equal{} (P_1(x) \plus{} \sqrt{x^2 \minus{} 1} Q_1(x))^n$.

It's the same argument as that used to solve the Pell's equation.  Take the conjugate of both sides and multiply.


AMC USA(J)MO USAMO Euler Stanford college MIT

If you've qualified for the USAMO or made it to the MOSP that should put you in the top 300 math studnets in the nation, correct?  So when USAMO/MOSP participants apply to college shouldn't they be almost shoe-ins?  Assuming nothing abnormal, like failing out of their other classes.

the effect of USAMO on college app has been discussed before.
It will be fairly easy to find the thread (excuse my laziness)

MOSP?  I'm not too sure.  but the results of usamo qualifiers haven't been very optimistic (purely the results that has been posted.  again, not sure what adcoms think)
and no, they're definately shoe-in.

The number of math majors at Harvard + Princeton + MIT + Stanford + Caltech + Duke + (other perennial top 10 colleges on Putnam) is << 300 (as long as we're not including the perennial Canadian schools in that list; UBC, UofT, and UWaterloo have huge numbers of math majors.

[quote="nodrog2002"] but the results of usamo qualifiers haven't been very optimistic [/quote]

This bears some qualification - the people haven't been shoo-ins, but they get accepted at a much higher rate than non-USAMO qualifiers.

Each year, some of the very tip-top math schools take applicants who have never even qualified for USAMO--I know that for sure this year in the case of Princeton. And, more rarely, there are USAMO qualifiers who don't get into the most famous schools. So the set {USAMO qualifiers} has a large intersection with the set {admittees to top math schools} but it is NOT a proper subset of the set of admittees. You maximize your chances by being a strong student in the field you are most passionate about, but you also maximize your chances by having an engaging personality that shows up in your essays and your interviews, and in having generally good grades in any courses you take for grades.

Bottom line: don't worry about it. DEFINITELY apply to more than one school if you are applying to highly selective schools, and definitely build up your college list first from a reasonable "safety" school that is essentially certain to admit you but still has a strong academic program.

Good luck in the next round of competitions.

definitely.

I know someone who had like 80 average on his freshman year, but worked really hard and reached 95+ average on junior year, and he got into Cornell.
Of course thats not the only reason why he got into ivy league school, but what Im trying to say is that it is too early to give up. If you have a dream, study hard and I'm sure youre smart enough to handle all that.

hmm if your grade was worse than that and still hoping for better school than that.. I dont kno.

Frankly, it seems that Cornell is not as selective (applicant pool and admissions ofice) as the top 5 or so. I know plenty of people who have gotten into Cornell and were rejected by, say, Harvard or MIT.

I think bad grades is just not good in general, and that starts as soon as you get into high school. Of course, you may do yourself some justification by achieving high marks in the next two years, but they'll still wonder what happened that first year.

I agree. cornell might not be as good as top 5 math schools or schools youre talking about.

but as i said, just work hard. dont worry about the past. then you'll get into somewhere good, if not the best.

by the way. I got rejected from cornell, but got into MIT and caltech.

[quote]I made USAMO this year and there's chance I'll make it to MOP since I'm in 9th grade and they invite 9th graders this year to MOP, but I get pretty bad grades this year at school. I probably won't be able to go to one of the top colleges but it's not really my goal anyway. Oh also if I get good grades in my sophomore and junior year, will colleges care?[/quote]

As Tokeadult and many have pointed out, unless you are an admissions officer and have the list of admitted students in front of you, no one   is likely to give a guarantee... But to put the things in perspective, I would say there is a very good chance ("good" as is 95%+ if I have to give a figure) that if someone makes to MOP (or, even  qualify for USAMO) he would    get in the college of his choice.

I say this based on from my, (admittedly small - but not so small) sample.  No one would say, but for good math schools it is almost an unwritten understanding --  one must have a good reason to reject a USAMO qualifier.

Only exception,   (based on my sample) is  Ivy  colleges where I  have seen USAMO qualifiers, (even with top grades0 getting rejected  (But then , I also have seen some very mediocre (In my opinion )  people getting admitted there) .  But other schools (Like MIT, Caltech, Stanford, Duke etc)   there were few, if any, surprises; I have not seen too many USAMO qualifier's getting rejected.   ..(In one case, I know, some one just "missed" the fact that the person was an USAMO qualifier - when that was pointed out to the admissions office, the decision was VERY quickly reversed)..  I don't have actual statistics, but my guess for  say a school like MIT, duke, Caltech etc , where  general acceptance rate may be  around 10-20% ,  for someone making to MOP would be hight 90%.... I could easily be wrong as my sample-size  is rather small but  I would be surprised if I am wrong by a huge margin.

Just some thoughts .. and   as they say   some time school becomes top ranked if top-ranked students go there...:)

Good luck any way, you  still have some time to decide,..

One thing which may help  much more than some people think is good recommendation letters.

Added later: [b]May be some one can run a poll here to see how many  here get  admitted to the college of their choice.[/b]

[quote]by the way. I got rejected from cornell, but got into MIT and caltech.[/quote]
I don't  like to beat up on  Cornell, because its an excellent  school, but some how I am not surprised. A few years ago  in a magazine article there were  actual transcripts  of a few "typical" students who applied to Cornell .  SAT scores, essay, full resume etc (except that names  and  few other details were removed).

Okay, obviously I   am not an expert on admissons, but still I was surprised to see some with
1600 SAT and excellent  grades getting rejected in the [i]first [/i]screening (From what I can tell, they did not like the person's essay)  while some one with 900 SAT being  accepted. I am sure there was some reason, its just that  it didn't  seem    logical to me.

Cornell (presumably) knows that some good students treat it as a safety school; they might thus reject students who have excellent statistics but who they think will not go there if accepted.

[quote="beta"]I made USAMO this year and there's chance I'll make it to MOP since I'm in 9th grade and they invite 9th graders this year to MOP, but I get pretty bad grades this year at school. I probably won't be able to go to one of the top colleges but it's not really my goal anyway. Oh also if I get good grades in my sophomore and junior year, will colleges care?[/quote]

Most colleges consider 10th-12th grade grades more closely than 9th grade.

Colleges will definitely care.  Get the grades up - MOP won't be enough alone to get into the schools you're looking at!  And now that you've excelled at problem solving, it shouldn't be too hard to tackle the other classes.  Also, learning how to budget your time and be organized will help you with the careless error bug if you have it.

every college has good students. colleges want a balanced campus. and it also kinda depends on who reads your essays

The correct basic point is that MOSP qualification is not, by itself, a guarantee of admission to your top choice school. Therefore USAMO qualification is even less of a guarantee. Therefore no one should worry about finding guarantees (as JBL has correctly said on various threads on this forum). Do what you enjoy, do it well, apply to more than one college, and be prepared to enjoy and thrive in whatever college you get admitted to.

Simfish is correct. I was accepted to Stanford, MIT, Caltech, and Harvey Mudd, and rejected from Harvard. I must say I am flattered that people talk about me among themselves.

[quote="Phelpedo"]Is it just me or do you think that colleges prefer jocks to highly ranked mathematicians?

Also, do you think that being Mathcounts National Champion will have a big impact on your college resume? It's only a middle school thing, but the very top honor.[/quote]

What colleges are you talking about? Of course colleges want people who are good at sports, because that's part of being well-rounded. But extremey few peope get into premier colleges just because of sports.

Just my opinion, but Mathcounts doesn't measure mathematical skill very much. It's a fun thing for middle schoolers, but a tougher test would measure skill better. Anyways, speed isn't that important in higher maths, and that's what MC is about.

MC champ sure sounds good, but don't use it as a free admission ticket.  It's important to have multiple awards--one very big one is great, but you shouldn't rely on only that (and if you can get a big one, you should have others...).  Not that I'm in a position to give advice :P.

I think math types like us get a little too elitist and "us vs. them" about athletes. Athletes, especially those good enough to get recruited, work [i]extremely[/i] hard. Much harder, I would wager, than any highly ranked math students. I doubt many math kids, even IMO kids, practice 2-4 hours or more every day, like athletes have to. Also, athletes are not necessarily stupid... you have to be pretty smart to be any good at most sports. The best athlete I know also has a 4.0, 1600, real intelligence (i.e. not the kind on tests), the works. Besidse, in my experience, the great athletes are smarter than the great students- they may not score as well on tests, but they have better social skills, a more balanced outlook on life, and are certainly a lot less arrogant (ironic, considering the "jock" stereotype).

Yes, I think recruiting may go a little too far. But I do think great athletes deserve a leg up in admissions, because they have demonstrated dedication, work ethic, etc. whereas being well-ranked in math often simply means that you won the genetic lottery. Some get there by working hard, yes, but many work hard and don't get there, too. The point is that math contests are a fun diversion and a great thing for getting kids motivated, but ultimately are just for fun and bragging rights, and don't guarantee that the student would be a good addition to a student body. Athletic success is a more likely indicator of the ability to accomplish something real (not perfect of course, but I am only making a comparative statement here).

I guess my main point (which does have to do with this thread as a whole) is that too many kids get too much a sense of entitlement from their academic honors. Just because you've won some awards does not mean you deserve admission to colleges- colleges are private institutions that can choose students how they please, and to suggest they have an obligation to choose you, for [i]any[/i] reason- academics, athletics, whatever- is rather insolent.

By the way, if you are wondering, I am not a great athlete myself. But I do swim, and swim enough to know just what it takes to stand out as an athlete. I might go so far as to say getting 7th place at my district swim meet was a greater accomplishment than making MOSP, but I think that's a bit of an exaggeration. I certainly consider it greater than any of the times I've gotten 1st place at local or state math contests. I have an immense admiration for my team-mates who have done greater things, and many of them I would consider much more deserving of college admission than myself.

The important thing here is to define a concrete distinction between an athlete and a jock. The  school-team "athletes" many of us know are not the hard-working ones that strive for individual success in their sport or activity. What many high schoolers commonly refer to as jocks are the kids that play sports because they aren't willing to do homework, it impresses girls, and bragging rights are commonly attached to being able to do a "Jordan from the free throw line" as opposed to being able to use Cauchy-Schwartz correctly. To succeed - really succeed - on a competitive level, sports is just as difficult, and perhaps, as you said, even more difficult, as an academically-related topic. Those individuals deserve the highest respect. But the other wannabes?...nah...

I agree that many athletes do deserve much respect.  But at least in the pro- scene, I find that arrogance is very common.  I recall a certain basketball player rejecting a who-knows-how-many-million dollar contract saying "I've got a family to feed."  I definitely agree that many of the athletes work just as hard or even harder than the top math students, but I disagree with what Eugiver said that athletes were a lot less arrogant.

Eugevir,

Very well said!!! You truly got it!!!  I can easily see that you will get the most out of your career and your life.

[quote="MithsApprentice"]bragging rights are commonly attached to being able to do a "Jordan from the free throw line" as opposed to being able to use Cauchy-Schwartz correctly.[/quote]

Lol. Hey, guess what? I can use Cauchy-Schwartz correctly! Everybody bow down to me! :lol:

[quote="MithsApprentice"]bragging rights are commonly attached to being able to do a "Jordan from the free throw line" as opposed to being able to use Cauchy-Schwartz correctly.[/quote]

Really now, being able to use Cauchy-Schwarz correctly is much easier than the other move. Sometimes, tremendous athleticism is just so amazing, and Jordan is a good example of that.

[quote="tetrahedr0n"]Really now, being able to use Cauchy-Schwarz correctly is much easier than the other move.[/quote]

The parenthesis used in conjunction with the title denotes that it is merely a sorry imitation. :P To think of it, even I can do a Jordan on an 8' goal. Of course, I don't get anything close to the ridiculous 10 feet of air he gets...but still. ^_^

Jordan was one amazing player who deserves respect beyond his game. Completely off topic, but anyways...I was watching some old East-West games a few weeks ago. Anyways, there's this one shot that MJ does...he's like right outside of the paint at the edge of the court and he's being guarded by 3 people. So he takes a jump shot...but he jumps (and naturally the guards jump with him) and waits to the other people to land before he shoot...and all this while he's still in the air...then like when the defenders land he's still like 3-feet in the air (this was on slow-mo) and he launches the ball with a BACKWARDS spin. And it swishes. I'm like *drops lower jaw*.

[quote="WarpedKlown1335"]Yet he misses a dunk at his last all-star game.

Hahaha.[/quote]

Eh, he hit a huge shot at the end of the first overtime, an insane jumper from the corner with someone's hand in his face to take a lead. Unfortunately, Jermaine O'Neal made a very smart decision to fould Kobe Bryant on a desperate 3 pointer attempt to help the West tie and eventually win the game.


Hi all. Need help showing that

$ \sum_{i\equal{}1}^4 a_i^2 \plus{} 7 \sum_{i\equal{}1}^4 b_i^2 > 2\sqrt{7} \sum_{i\equal{}1}^4 a_i b_i$

for all $ a_i, b_i \in \mathbb{Z}\cup\{0\}$. The standard inequalities don't seem to be much help. It seems like

$ \sum_{i\equal{}1}^4 a_i b_i < \sum_{i\equal{}1}^4 max(a_i, b_i)^2 < \sum_{i\equal{}1}^4 a_i^2 \plus{} \sum_{i\equal{}1}^4 b_i^2$

should be useful.

This inequality stems from the fact

$ \sum_{i\equal{}1}^4 (a_i \plus{} \sqrt{7} b_i )^2 \equal{} ( \sum_{i\equal{}1}^4 a_i^2 \plus{} 7 \sum_{i\equal{}1}^4 b_i^2 ) \plus{} (2 \sum_{i\equal{}1}^4 a_i b_i) \sqrt{7}$

in $ \mathbb{Q}(\sqrt{7})$.

Nevermind, this was trivial.

$ (a \minus{} \sqrt{7} b)^2 > 0$

so

$ a^2 \plus{} 7 b^2 > 2  \sqrt{7} a b$

from whence the above inequality follows.


The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is 5. The solution set for $x$ is

$\text{(A)} \ \{5/2 \} \qquad \text{(B)} \ \{x \ | \ 2 \leq x \leq 3 \} \qquad \text{(C)} \ \{x \ | \ 2 \leq x < 3 \}$ $\text{(D)} \ \{x \ | \ 2 < x \leq 3 \} \qquad \text{(E)} \ \{x \ | \ 2 < x < 3 \}$

[hide="Answer"]Since both functions give equality as well, no exact integer value can give a sum of $5$. Therefore, our answer must be $\boxed{E}$.[/hide]

[hide]If x were an integer, we would have the expression be 2x, which is always even. THus we eliminate all instances where x is an integer (and the one that makes no sense, A), and the only one remaining is E. [/hide]


Let $ a,b,c\ge0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$. Find the maximum value of

$ P\equal{}\sum\frac {1 \minus{} a^2}{b^2 \plus{} bc \plus{} c^2}$

[quote="dduclam"]Let $ a,b,c\ge0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$. Find the maximum value of

$ P \equal{} \sum\frac {1 \minus{} a^2}{b^2 \plus{} bc \plus{} c^2}$[/quote]
We will show that
$ P \le 1/2$
$ \Leftrightarrow \sum \frac{b^2\plus{}c^2\minus{}a^2}{b^2\plus{}bc\plus{}c^2} \le 1$
$ \Leftrightarrow \sum \frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2} \ge 2$
$ \Leftrightarrow \sum \frac{(a\minus{}b)(a\minus{}c)}{b^2\plus{}bc\plus{}c^2} \plus{}\sum \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 2$

By Vornicu Schur, we have
$ \sum \frac{(a\minus{}b)(a\minus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 0$
It suffices to show that
$ \sum \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 2$
It is from Darij's Inequality. :)
Equality holds for $ a\equal{}b\equal{}c\equal{}\sqrt{\frac{2}{3}}$. :)


By lattice structure, I mean
http://wb.chem.lsu.edu/htdocs/people/sfwatkins/MERLOT/lattice/03bcc.html

Given a salt or something else, how do we predict the lattice structure?

If there's not a good way to predict them at this level, is there some good place to find a list? I don't see a chart or list in my chemistry book. Google searching has not gone well so far http://www.google.com/search?q=predict+lattice+structure&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

wow... sorry for the late response

In order to determine the lattice structure you must first determine what kind of unit cell that the structure has

POST 100!!!!! WOOT =)

ah Thanks.

So there's no way to predict the lattice structure without knowing the kind of unit cell? We can't just be like:
"oh hey CuO has this structure because [reason not requiring knowing what the unit cell looks like]?"

well usually the question will supply you with enough information to determine what the unit cell would be.


Can anyone help me on this question because I've tried many times and still don't know how to get the solution.

Let [b]F [/b]be the gravitational force field of a mass M, located at the origin in R^3, on a particle of mass m, located at (x, y, z):

[b]F[/b]=-((GMm)/(x^2+y^2+z^2)^(3/2)))(x[b]i[/b]+y[b]j[/b]+z[b]k[/b])

Given that G, M and m are all constants, show that the work done by [b]F [/b]as a particle of mass m moves from [b]X1[/b]= (x1,y1,z1) to [b]X2[/b]= (x2,y2,z2) depends only on l[b]X1 [/b]l and l[b]X2 [/b]l

Thank you in advance

you can move on an equipotential surface without any work. This should help.

I don't understand, can you solve this problem by posting the working out ? This is because I learn better this way.

Thank you.


trigonometry
geometry
circumcircle
geometric transformation

[hide="solution :| "]$ \frac{EF}{DE} \equal{} \frac{BD}{DC} \equal{} \frac{\tan \gamma}{\tan \beta} \ \Longleftrightarrow \ \frac{EF}{AE} \equal{} \cot \beta \Longleftrightarrow \cot \angle EFA \equal{} \cot \beta \ \Longleftrightarrow$
$ \Longleftrightarrow \ \angle EFA \equal{} \angle ABC \ \Longleftrightarrow \ AFDB \ \mathrm{is\  cyclic} \ \Longleftrightarrow \ AF \perp BF$[/hide]

[quote="Valentin Vornicu"]Let $ ABC$ be an acute angled triangle with $ \angle B > \angle C$. Let $ D$ be the foot of the altitude from $ A$ on $ BC$, and let $ E$ be the foot of the perpendicular from $ D$ on $ AC$. Let $ F$ be a point on the segment $ (DE)$. Show that the lines $ AF$ and $ BF$ are perpendicular if and only if $ EF\cdot DC \equal{} BD \cdot DE$.[/quote]
[hide="Solution"]
If $ AF\perp FB$, then $ \angle AFB \equal{} 90 \equal{} \angle ADB$, so $ AFDB$ is cyclic and it follows that $ \angle FBA \equal{} \angle ADE$, so $ \triangle AFB\sim \triangle AED$.  From this, a spiral similarity maps $ \triangle AEF\sim \triangle ADB$, which means that $ \frac {EF}{BD} \equal{} \frac {AE}{AD}$.  Since $ \triangle ADC$ is right with altitude $ DE$, we find that $ \triangle ADE\sim \triangle DCE\implies \frac {EF}{BD} \equal{} \frac {AE}{AD} \equal{} \frac {ED}{DC}$, so $ EF\cdot DC \equal{} BD\cdot ED$.  Now, if $ EF\cdot DC \equal{} BD\cdot DE\implies \frac {EF}{BD} \equal{} \frac {DE}{DC}$.  As before, we notice that $ \triangle AED\sim \triangle DEC\implies \frac {EF}{BD} \equal{} \frac {DE}{DC} \equal{} \frac {AE}{AD}$.  Since $ \angle AED \equal{} 90 \equal{} \angle ADB$, we find that $ \triangle AEF\sim \triangle ADB$, and so a spiral similarity maps $ \triangle ADE\sim \triangle ABF\implies \angle AFB \equal{} \angle ADB\equal{}90$ and we conclude that $ AF\perp FB$.  [/hide]

[quote="Valentin Vornicu"] [color=darkred]Let $ ABC$ be an acute triangle with $ B > C$ . Denote $ \|\begin{array}{c} D\in (BC)\ ,\ AD\perp BC \\
\ E\in (AC)\ ,\ DE\perp AC\end{array}$

and a point $ F\in (DE)$ . Show that $ FA\perp FB\ \Longleftrightarrow\ EF\cdot DC = BD \cdot DE$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ \triangle EAD\ \sim\ \triangle EDC$ $ \implies$ $ \frac {EA}{ED} = \frac {AD}{DC}$ $ \implies$ $ \frac {ED}{DC} = \frac {EA}{AD}\ \ (*)$ . Therefore,

$ FA\perp FB$ $ \Longleftrightarrow$ $ \widehat {AFE}\equiv\widehat {ABD}$ $ \Longleftrightarrow$ $ \triangle AFE\ \sim\ \triangle ABD$ $ \Longleftrightarrow$ $ \frac {AE}{AD} = \frac {EF}{BD}$ .

Using the relation $ (*)$ obtain $ \boxed {\ FA\perp FB\ }$ $ \Longleftrightarrow$ $ \frac {ED}{DC} = \frac {EF}{BD}$ , i.e. $ \boxed {\ EF\cdot DC = BD\cdot DE\ }$ .

[b][u]Remark.[/u][/b] Since $ \widehat {BAD}\equiv\widehat {FAC}$ obtain that the altitude $ AD$ and the line $ AF$ are isogonals w.r.t. $ \widehat {BAC}$ ,

i.e. [b]the circumcenter of $ ABC$ belongs to the line $ AF\ !$[/b] The particular case $ A = 90^{\circ}$ is also interestingly.[/color]


calculus
integration
linear algebra
matrix
algebra
polynomial
complex analysis

Use Cauchy's Integral Formula to prove the Cayley-Hamilton Theorem: If A is an n x n matrix over C and $f(z)=det(z-A)$ is the characteristic polynomial of A then $f(A)=0.$  Please help prove and define ambiguous notation.

If you're in college, then it's on JSTOR: http://links.jstor.org/sici?sici=0002-9890%28197504%2982%3A4%3C390%3ATCT%3E2.0.CO%3B2-5


We very definitely do [i]not[/i] mean integer here. This is a statement about positive [i]real[/i] numbers (although we do not use completeness, and the same thing would be true of rational numbers.)

The statements are perfectly clear and you shouldn't apologize for using words rather than symbols. But the exercise is very much about getting you to think about and understand the meaning of "for every" and "there exists".

One hint: there may be a use for $ \frac{x}2.$ Or maybe $ \frac{y}2.$

One paraphrase of what's going on: "in the real numbers, there's no such thing as an infinitesimal."

I believe that "positive number" means "positive real number", since he explicitly states "positive integer" when he means "positive integer" in other examples in that section (I looked up the book). Try to prove statement (b) wrong by contradiction. You can check your answer in the hidden content.

[hide]Assume there does in fact exist a positive number $ y$, such that for every positive number $ x$ we have $ y < x$. Consider $ y/2$. Since $ y$ is positive, $ y/2$ is as well, and by our assumption, $ y < y/2$, which is a contradiction. Hence statement (b) must be false.[/hide]

What problems are you having writing the statements in symbolic logic form? Just translate "for all" and "for every" to $ \forall$ and "there exists" to $ \exists$. For example, statement (b) would translate to $ \exists y > 0$ s.t. $ \forall x > 0, y < x$. Following that example, can you translate part (a) into symbolic logic?

Packing the words "there exist" down to the symbol $ \exists$ is just a minor sideshow here. It's just shorthand, just a notation for writing the same thing in less space - it doesn't really make it any easier (or harder, either) to think about this problem. I don't want MathDaoist to spend his time thinking about the notation; I want him to think about the meaning. As I said, his statement of the two problems was perfectly clear (as long as you understand "number" to be "real number.")

(Of course, Zhylliolom's hidden proof is correct - and the argument for (a) is a relative of that.)

Thank you Zhylliolom and Kent,
I see Kent's point, putting them in symbolic forms didn't help much, I thought I must have made mistakes and not seen the difference between the 2 statements.  (I finally realize that I was just not used to these wordings and not understanding the meaning of reversing the order of the quantifiers)

a) For every positive number x, there exists a positive number y, such that y < x

$ \forall x > 0, \exists y>0, s.t. y< x$

b) There exists a positive number y, such that for every positive number x we have y < x

$ \exists y > 0, s.t. \forall x > 0, y < x$

(a) in essence, is selecting any x first, then, all I have to do is to find existence of any y less than x (which I can always find such y given x).
While (b)'s meaning is totally different. It states that such a y can be found and the order matters given the fact that fixing y and iterating through all possible x, and testing for y<x, one will encounter contradiction upon x=y/2. Any contradiction will invalidate the initial assumption that such y exists. As such, (b) is a false statement.

[quote="MathDaoist"]This is supposedly simple and yet, my mind is blank and I couldn't tell the difference between these two. Please help or provide hints.

(Taken directly from Jonathan Lewin's Mathematical analysis book)
a) For every positive number x, there exists a positive number y, such that y < x

b) There exists a positive number y, such that for every positive number x we have y < x

I assume positive number = positive integer.
One of them is true and the other is false, but I fail to see the trick of solving similar problems.
I also fail to write those in symbolic logic forms.

Thanks[/quote]

If statement (b) is false then the the negation of that statement must be true and hence provable.

Let us see>

let y>0 ,then choose ,x such that   0<x<y ,hence there exists x ,x>0 and $ y\geq x$ .

Thus the negation of (b) is true ,hence (b) false


[quote="cezar lupu"]Let $(a_{n})_{n\geq 1}$ be a sequence of positive integers so that $a_{n-1}$ divides $a_{n}$ and $\frac{a_{n}}{a_{n-1}}\geq\frac{n}{n-1},$ for all $n\geq 1.$ Prove that the sequence \[ x_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots +\frac{1}{a_{n}}  \] has an irational limit. ;)[/quote]
Counterexample: $a_n = 2^n$

[quote="cezar lupu"]Let $(a_{n})_{n\geq 1}$ be a sequence of positive integers so that $a_{n-1}$ divides $a_{n}$ and $\frac{a_{n}}{a_{n-1}}\geq n-\frac{1}{n},$ for all $n\geq 1.$ Prove that the sequence \[ x_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots +\frac{1}{a_{n}}  \] has an irational limit. ;)[/quote]
Since $\frac{a_{n}}{a_{n-1}}$  is an integer for all $n$ by assumption, we may assume $\frac{a_{n}}{a_{n-1}}\geq n$. Now modify the well-known proof that $e = \sum_{k=0}^\infty \frac{1}{k!}$ is irrational.


inequalities
quadratics
inequalities proposed

Let $a,b,c>0$.Prove or disaprove the inequality:

            $3(a^2+b^2+c^2)\geq(ab+bc+ca)\cdot(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$ ;)

Set $a=b=x$ and $c=1$. The LHS is a quadratic while the RHS is cubic. So the inequality is untrue when $x$ is large.