function
LaTeX
floor function
limit
number theory unsolved
number theory

Let $\mu$ is Mobius function.Prove that

$\frac{\mu(1)}{1}+\frac{\mu(2)}{2}+...+\frac{\mu(n)}{n}\leq 1$ forall $n>0$.

@to all:I can't use LaTex,Why?

[mod.: For me it works  :? ]

For any positive integer $n > 0$, \begin{eqnarray*}\sum_{d\mid n}\mu ( d ) & = & \{\begin{array}{l}1, n = 1\\ 0, n > 1 \end{array} \end{eqnarray*} By the Moebius inversion formula, for $x \geq 1$

\[
\sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n} \rfloor = \sum_{n \leq x}( \sum_{d\mid n}\mu ( d ) ) = 1+0+0+\ldots+0 = 1
\]

[b]The upper bound[/b]

If $\mu ( n ) \geq 0$ we have $\mu ( n )  \lfloor \frac{x}{n}\rfloor \geq \mu ( n ) \cdot \frac{x}{n}-\mu ( n )$.
If $\mu ( n ) < 0$ we have $\mu ( n )  \lfloor \frac{x}{n} \rfloor \geq \mu ( n ) \frac{x}{n}$.
Summing these relations we obtain \begin{eqnarray*} \sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n} \rfloor & \geq & \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}( \mu ( n ) \cdot \frac{x}{n}-\mu ( n ) \cdot 1 )+\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) < 0 \end{array}}}\mu ( n ) \cdot \frac{x}{n}\\  \sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n}\rfloor & \geq & \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x \end{array}}}\mu ( n ) \cdot \frac{x}{n}-\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n ) \cdot 1 \end{eqnarray*} Hence \begin{eqnarray*} \frac{1}{x}\cdot 1 = \frac{1}{x}\sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n}\rfloor & \geq & \frac{1}{x}\cdot ( \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x \end{array}}}\mu ( n ) \cdot \frac{x}{n}-\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n ) \cdot 1 )\\ \frac{1}{x}& \geq & \frac{1}{x}\cdot x \sum_{n \leq x}\frac{\mu ( n )}{n}-\frac{1}{x}\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n )\\ \frac{1}{x}\cdot ( \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n )+1 ) & \geq & \sum_{n \leq x}\frac{\mu ( n )}{n}\end{eqnarray*} We trivially have \[x-1 \geq \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n )\] Therefore \[1 = \frac{1}{x}\cdot ( x-1+1 ) \geq \frac{1}{x}( \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) \geq 0 \end{array}}}\mu ( n )+1 ) \geq \sum_{n \leq x}\frac{\mu ( n )}{n}\] [b]The lower bound.[/b]

By equation $( 1 )$, using a similar reasoning as for the uppper bound, we obtain \begin{eqnarray*}\frac{1}{x}\cdot ( \sum_{n \leq x}\mu ( n ) \frac{x}{n}-\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) < 0 \end{array}}}\mu ( n ) ) & \geq &  \frac{1}{x}\sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n}\rfloor = \frac{1}{x}\cdot 1\\ \sum_{n \leq x}\frac{\mu ( n )}{n}& \geq & \frac{1}{x}+\frac{1}{x}\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) < 0 \end{array}}}\mu ( n ) \end{eqnarray*} Now clearly, \[\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) < 0 \end{array}}}\mu ( n ) \geq \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x \end{array}}}(-1 ) \geq-( x+1 )\] Therefore \[\sum_{n \leq x}\frac{\mu ( n )}{n}\geq \frac{1}{x}+\frac{1}{x}\sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c}n \leq x\\ \mu ( n ) < 0 \end{array}}}\mu ( n ) \geq \frac{1}{x}+\frac{-( x+1 )}{x}=-1\] Also note that [b]1.[/b] $\sum_{n \leq x}\mu ( n ) = o(x)$ is equivalent to the prime number theorem
[b]2.[/b] you can show that $\sum_{n \in \mathbb{N}}\frac{\mu ( n )}{n}= 0$ knowing that $1 / \zeta ( s ) = \sum_{n = 1}^{\infty}\frac{\mu ( n )}{n^{s}}$ and that $\lim_{s \rightarrow 1^{+}}\zeta ( s ) =+\infty$.

[quote="N.T.TUAN"][quote="{x}"] \begin{eqnarray*} \frac{1}{x}\cdot 1 = \frac{1}{x}\sum_{n \leq x}\mu ( n ) \cdot  \lfloor \frac{x}{n} \rfloor & \geq & \frac{1}{x}\cdot ( \sum_{n \leq x}\mu ( n ) \cdot \frac{x}{n}-\sum_{n \leq x}\mu ( n ) \cdot 1 )\\ \end{eqnarray*} [/quote]
Why?  :roll:  if $\mu (n)<0$

Sorry,because probem is -1<=f(1)/1+f(2)/2+...+f(n)/n<=1  :([/quote]
I also don't know why.
But I think we can modify the solution of {x}'s.
$\frac{1}{x}*1=\frac{1}{x}*\sum_{x\leq n}{\mu (n)*[\frac{x}{n}]}\geq \frac{1}{x}(\sum_{n\leq x}{\mu(n)\frac{x}{n}}-\sum_{n\leq x,\mu (n)>0}\mu(n)*1)$
Notice that$\sum_{n\leq x,\mu (n)>0}\mu(n)*1\leq x-1$
then$\sum_{n\leq x}{\mu(n)\frac{1}{n}}\leq 1$

Hawk Tiger: Thank you. I corrected using your idea :)
N.T.TUAN: I simply made a mistake. I'm sorry. Now it's corrected and I proved the lower bound too :) (which is actually more or less the same thing) Enjoy !


I'm having really difficult time with conditional probability. I know what they are and understand them from my stat book but I can't seem to be able to do it. I can't find like:

$ P(A|B) \equal{} \frac{P(\text{A and B})}{P(\text{B})}$

I can't find $ P(\text{A and B})$. Here, can anyone help me how to find that in this problem (I read the book couple times but the book doesn't seem to tell me)

[hide="Problem"]
Batteries for use in ordinary flashlights are packaged with six to a package. Two batteries the research department had developed having longer life than the regular batteries are accidentally mixed with 4 regular batteries in a package.  Suppose a battery is selected at random for use in a flashlight.  Also, suppose a second battery is selected after the first battery is identified as a long battery.  What is the probability that the second selection will turn up in the long-life battery? A regular battery on the second selection? [/hide]

I think that $ P(\text{B}) \equal{} \frac{2}{6} \equal{} \frac{1}{3}$.  But I don't know what $ P(\text{A and B})$ is.

Thanks for your help!

your A should be the probability that you get a long battery on the second time.

So, your probability of (A and B) should be the probability that you get a long battery the first time AND you get a long battery on the second time.

that's for the first part.

[quote="Silverfalcon"]
$ P(A|B) \equal{} \frac {P(\text{A and B})}{P(\text{B})}$
[/quote]

I don't understand where that comes from.

I thought $ P(A \text{ \& } B) \equal{} P(A) \times P(B)$ so isn't that saying $ P(A|B) \equal{} \frac {P(A \text{ \& } B)}{P(B)} \equal{} \frac {P(A) \times P(B)}{P(B)} \equal{} P(A)$

I thought $ P(A|B) \equal{} P(A) \plus{} P(B)$

Please explain where I've gone wrong

Thanks

P(A and B) = P(A) x P(B)  IF  A and B are independent.

So, yes, if A and B are independent, then the probability of A happening GIVEN that B happened is equal to the probability of A, because they are independent of each other.

For part 2,

A should be the probability that you get a regular battery on the second one, and probability of B is the probability that you get a long battery on the first one.

So, P(Second-> regular | First-> Long) = P(Second -> regular AND First -> Long)/ P(First->Long)

EDIT: [quote="SimonM"][quote="Silverfalcon"]
$ P(A|B) \equal{} \frac {P(\text{A and B})}{P(\text{B})}$
[/quote]

I thought $ P(A|B) \equal{} P(A) \plus{} P(B)$

[/quote]

Uh, I don't think that's true.

So, P(First -> Long) = 4/6, right?

P(Second -> Regular AND First -> Long) = 4/6*3/5?

My bad. I meant 2/6.

I don't think P(Second -> Regular AND First -> LONG) was right either. Or was it? Is it just product of P(Long) x P(Regular) [2/6*4/5]?

Okay, in your first attempt, the only thing wrong was the denominator.  It should've been 2/6.

In your second attempt, you had 4/6 for both of them.

The answer is (2/6)(4/5)/(2/6) = 4/5.

So....

P(B|A) = P(B)?

Isn't that how it comes out if I simplify 2/6 on numerator and denominator?

And if so, I thought that it was only for independent case. But this is dependent.

If you want practice, here's another problem that requires conditional probability.

I roll two fair six sided dice.  Given that one of them rolled a $ 3$, what's the probability that the sum is $ 6$

EDIT: Ah, I see it now, P(A = probability that second is long) =  2/6 x 1/5 + 4/6 x 2/5.

So $ P(A)\neq P(A|B)$

Your problem is really similar to one of the examples in my textbook.

[hide="Hopefully this is right"]

Let P(A) = probability of one of the rolled is 3 and P(B) = probability that the sum is 6.

P(A|B) = P(A and B)/P(B)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1,3),(2,3),(4,3),(5,3),(6,3)

(1,5),(2,4),(3,3),(4,2),(5,1)

So P(B) = 5/36 and P(A and B) = 1/36

P(A|B) = (1/36)/(5/36) = 1/5 [/hide]

If that's right (hopefully), then I do get this bit. It's just those problems like this where I can't have a casework (I have to multiply and stuff) and that's where I get confused.

woops, the other way around, given the fact that the sum is 6, what is the probability that one of them is 3?

[quote="Silverfalcon"][quote="Zellex"]EDIT: Ah, I see it now, P(A = probability that second is long) =  2/6 x 1/5 + 4/6 x 2/5.
[/quote]

Woah, where that came from?[/quote]

Well, the probability that the second is long has two cases.

Case 1: first one is long (2/6 chance of this happening)

Probability second is long = 1/5.

Case 2: first one is short (4/6 change of this happening)

probability second is long = 2/5.

So, probability second is long = 1/5 x 2/6 + 2/5 x 4/6.

If you're still confused, maybe we should look at another problem from your textbook (I don't have one in front of me :P).


Putnam
linear algebra
matrix
inequalities
induction
algebra
binomial theorem

1. Show that for every positive integer n,

((2n-1)/e)^((2n-1)/2) < 1*3*5*...*(2n-1) < ((2n+1)/e)^((2n+1)/2).

2. For any square matrix A, we can define sin A by the usual power series

sin A = :Sigma: n=0 to :inf: (-1)^n/(2n+1)! A^(2n+1).

Prove or disprove: there exists a 2*2 matrix A with real entries such that

sin A =  (1, 1996; 0, 1).

(That is, 1 and 1996 on the first row, 0 and 1 on the second row.)

[color=cyan]Well, I've done half of the first one and I suspect the second half follows with similar logic.

We wish to show that for every positive integer n,
((2[i]n[/i]-1)/[i]e[/i])^((2[i]n[/i]-1)/2) < 1*3*5*...*(2[i]n[/i]-1)

We will proceed by induction.  First, clearly (1/[i]e[/i])^(1/2)<1.

Assume ((2[i]k[/i]-1)/[i]e[/i])^((2[i]k[/i]-1)/2) < 1*3*5*...*(2[i]k[/i]-1).  We wish to show that
((2[i]k[/i]+1)/[i]e[/i])^((2[i]k[/i]+1)/2) < 1*3*5*...*(2[i]k[/i]+1).  It is satisfactory to show that
[((2[i]k[/i]+1)/[i]e[/i])^((2[i]k[/i]+1)/2)] / [((2[i]k[/i]-1)/[i]e[/i])^((2[i]k[/i]-1)/2)] < 2[i]k[/i]+1

We know that for all positive reals [i]j[/i], (1+1/[i]j[/i])^[i]j [/i]< [i]e[/i].  (I don't know how to justify this, but it's a standard theorem from analysis.  Very well known.)

I don't want to write the rest, because it's very messy.  However, if you substitute (2[i]k[/i]-1)/2 for [i]j[/i] and rearrange terms, and multiply both sides of the inequality by (2[i]k[/i]+1)/[i]e[/i], you get exactly what you wanted.

Thus, by induction, we have proven the left half of the inequality.[/color]

[quote="JBL"][color=pink]We know that for all positive reals [i]j[/i], (1+1/[i]j[/i])^[i]j [/i]< [i]e[/i].  (I don't know how to justify this, but it's a standard theorem from analysis.  Very well known.)[/color][/quote]
Can't you get that just by like expanding via binomial theorem and comparing with the power series for e?

I don't think so because you'll get a bunch of small terms, and you have to compare them somehow with the rest of it in the expansion. The small terms all go to zero in the limit, so you can forget about them, but for the finite representation, you can't do that, as far as I can see.

One can show that lim_{n \to \infty} sup (1 + 1/n)^n <= e. Proceed as follows: Let s_n = \sum_{k=0}^n 1/k! and t_n = (1 + 1/n)^n, by the binomial theorem we have t_n = 1 + 1 + 1/2(1 - 1/n) + 1/3!(1 - 1/n)(1 - 2/n) + ... + 1/n!(1 - 1/n)*...*(1 - (n - 1)/n) s that we get t_n <= s_n. And the rest follows, just like teratomato said.


[quote="Mango 070"]How many decimal digits are there in the number 2005^2005? (ans: 6621)

and

How many integers between 1000 and 9999 are perfect squares and have sum of digits equal to 14? (ans: 0)[/quote]
For the first one you take the log of that to get $ 2005\log2005$. $ \log2005$ is approximately $ 3.3$ so $ 2005*3.3 \equal{} 6620.73...$ Rounding up gives $ 6621$.

Any number and the sum of the digits of that number are congruent in $ \bmod{9}$, thus if the sum of the digits of a number were to be 14, then that number would have to be congruent to $ 5\bmod{9}$. Since this is a perfect square, we check the quadratic residues in $ \bmod{9}$, and they are $ 0,1,4,7$. Thus we can conclude that there are no perfect squares that have the sum of its digits equal to 14.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315897
Don't double post  :mad:

[quote="ckck"]For the first one you take the log of that to get $ 2005\log2005$. $ \log2005$ is approximately $ 3.3$ so $ 2005*3.3 \equal{} 6620.73...$[/quote]

Actually $ 2005\cdot 3.3\equal{}6616.5$, so I don't think this is possible without a calculator. There isn't really a good approximation for $ \log 2$ as far as I can tell. ($ \ln 2$ on the other hand happens to be $ 1\minus{}\frac 12\plus{}\frac 13\minus{}\frac 14\plus{}\cdots$)

[quote="grn_trtle"]

 There isn't really a good approximation for $ \log 2$ as far as I can tell.[/quote]

$ 2^{10} \sim  10^3$

$ log(2^{10}) \sim  log(10^3)$

$ 10log(2) \sim  3log(10)$

$ log(2) \sim  \frac {3}{10}$

$ log(2) \sim  0.3$

@Arrange your tan: That's exactly what I did in the other post I linked to, but it's unfortunately not close enough to get the right answer. You need at least $ 3.302$.


trigonometry
inequalities
function
logarithms
inequalities proposed

Colleagues:

I pretty soon less than 24 hours can type up my solution ( I am positive it is right )

for the trig inequality I asked you a while ago. I think it is best to lay out some ideas of

the proof.

1. I used the relation: sin(x + pi/4) <-> sinx + cosx.
2. Expanding the product sinx.siny.sinz as a sum and use 1.
3. Use Cauchy-Schwartz and then the well-known: ab <= 1/4.(a + b)^2.
 :lol:

can you give the problem ? i can't find it  :? (nevertheless can't you just type your solution in a topic with the problem you given ?)

Colleagues:

Below is the solution I just got. I haven't checked thoroughly yet, but hope it is true.

Q: For x, y, and z be acute angles, we have:
(sinx + cosx)(siny + cosy)(sinz + cosz)(sinx.siny.sinz + cosx.cosy.cosz) <= 2. (*)

A: Since 2^(1/2).sin(a + pi/4) = sina + cosa, we obtain:
(sinx + cosx)(siny + cosy)(sinz + cosz) = 2^(3/2).sin(x + pi/4).sin(y + pi/4).sin(z + pi/4).
The product-to-sum identities for sin and cos gives:
sinx.siny.sinz = 1/4.[sin(x - y + z) + sin(x + y - z) + sin(y + z - x) - sin(x + y + z)], and
cosx.cosy.cosz = 1/4.[cos(x + y - z) + cos(y + z - x) + cos(x + z - y) + cos(x + y + z)]. Thus:
sin(x + pi/4).sin(y + pi/4).sin(z + pi/4) = 1/4.[sin(x + y - z + pi/4) + sin(x + z - y + pi/4) + sin(y + z - x + pi/4) - sin(x + y + z + 3pi/4]. From this we get:
sinx.siny.sinz + cosx.cosy.cosz  = 2^(1/2)/4.[

-I will finish it up in about an hour.
 :lol:

Mhm...I think  I have received some other inequality but for arbitrary triangle:

1.) First remark that function $f(x)=\ln (\sin x + \cos x)$ is concave so by Jensen we have:

$\sum \ln (\sin x + \cos x) \leq 3 \ln (\sin 60 + \cos 60) = \ln ( (\frac{1+\sqrt{3}}{2})^3)= \ln (\frac{10+5 \sqrt{3}}{8})$
so:

(*)  $\prod (\sin x + \cos x) \leq \frac{10+5 \sqrt{3}}{8}$

2.) we know that: $\sin x \sin y \sin z \leq \frac{3 \sqrt{3}}{8}$ and $\cos x \cos y \cos z \leq \frac{1}{8}$ adding them together we obviously get (**)  $\sin x \sin y \sin z + \cos x \cos y \cos z \leq \frac{3 \sqrt{3} + 1}{8}$

By multiplying (*) and (**) we get:

$(\sin x + \cos x)(\sin y + \cos y)(\sin z + \cos z)(\sin x \sin y \sin z + \cos x \cos y \cos z) \leq (\frac{10+5 \sqrt{3}}{8})(\frac{3 \sqrt{3} + 1}{8})=1,806590285389229769897822218011$


The F2L is more intuitive than anything else...so first, just make the upper cross, and then pair up edges and corners as moogra said (blocks of 2, each containing an edge and a corner, so you have 4 blocks total). Alternatively, first get 3 corners, and then use the remaining corner to get all of the other edges in place (there is a small chance that amidst this, you'll accidentally get the 4th corner in).

Yeah it is very self-intuitive. You'll have to figure it out yourself (or you can memorize the 41 cases). Luckily most of them are very similar and I think I figured out 30 something cases in 1 day.

Once you've identified the four standard cases (which you should try to do without looking them up - it helps you remember them more easily) try to reduce everything else to those cases.  (You'll recognize them by how few moves you need to put them in and how easy they are to fingertrick.)

[quote="AIME15"]Or, you could go [url=http://www.cubefreak.net/F2L.html]here[/url].[/quote]

Thank you for the site~  :lol:


probability
combinatorics proposed
combinatorics

Suppose Bob can predict the outcome of a binary event with probability p.

In an attempt to institute performance pay, Bob’s employer pays him 1 dollar per correct prediction; with a catch:

If Bob gets 1,2,...i,...k correct predictions in a row; then each i’th correct prediction rewards Bob i*X

For example, if Bob has the sequence T,F,T,F,T,F,T ; then his payoff is 4 dollars, but if it is T,T,T,T,F,F,F; then it is 1+2+3+4 dollars.

So if there are N events in 1 year that Bob is to predict; what is his expected yearly salary?

$ S_n$: Expected salary for $ n$ events.

According to the probability tree, the following recursion can be established:
$ S_1 \equal{} (1 \minus{} p)\cdot 0 \plus{} p\cdot 1 \equal{} p$.
$ S_n \equal{} (1 \minus{} p)\cdot S_{n \minus{} 1} \plus{} p\cdot (S_{n \minus{} 1} \plus{} 1 \plus{} p \plus{} p^2 \plus{} \dots \plus{} p^{n \minus{} 1})$.

Some algebra is needed to solve the recursion:
$ S_n \equal{} S_{n \minus{} 1} \plus{} p \plus{} p^2 \plus{} p^3 \plus{} \dots \plus{} p^n$.
$ S_n \equal{} np \plus{} (n \minus{} 1)p^2 \plus{} (n \minus{} 2)p^3 \plus{} \dots \plus{} p^n$.
If $ p \equal{} 1$, $ S_n \equal{} \frac {n(n \plus{} 1)}{2}$.
If $ p < 1$, $ S_n \equal{} kn \plus{} k^2(p^n \minus{} 1)$, with $ k \equal{} \frac {p}{1 \minus{} p}$.

(I edited some steps to make it easier to understand).

That is great, and surprisingly simple. Any idea how we could calculate the variance?


prove that 3^2008 +4^2009can be written as a product of two numbers which are greater than 2009^182

its the  Q2 of RMO 2009


The ratio of the measures of the shortest side of two similar triangles is $ 2: 3$. The smaller triangle has an area of $ 100$ square feet. What is the number of square feet in the area of the larger triangle?

Okay, if the side ratio is $ 2$:$ 3$, that means that the area ratio is $ 4: 9$.  Thus, the answer is $ \boxed{225}$.


abstract algebra
geometry
advanced fields
advanced fields theorems

I've learned the definition of a flat module, and I've heard that flat modules intuitively relate to things being "flat" when you look at algebraic geometry, but I don't have any intuition for this. Could someone give me some examples of modules where flatness corresponds to some geometric object looking flat?

You don't have to be too specific - it's even nice to try to work out on my own why flatness corresponds to a geometric object being flat once I know what examples to look at. (For example, I learned localization when someone told me to look at the localization at the cusp of $ y^2\equal{}x^3$ or the double intersection of $ y\equal{}x^2$ and $ y\equal{}\minus{}x^2$ and explain how the localization tells us what kind of point it is.)

Flatness corresponds to a "continuous family". See Hartshorne Thm. III.9.9.

Sorry, but could you please give something more concrete? Like a specific example where something is flat, and that corresponds to a geometric object being flat.

e.g. if you have a fibration $ \pi: X \to S$ with all fibres elliptic curves, then this is a flat morphism according to the criterion in Hartshorne above under the assumptions on $ S$. Further examples and counterexamples can be found in the chapter on flatness of Eisenbud.


function
logarithms
real analysis
real analysis unsolved

Hi dears,

I do have some questions :
1. if f :[a,b] ------->R is measurable then there exists a sequence $ f_{n}$ of simple functions such that $ f_{n}\rightarrow f$ for all x in X.
Can you guys explain why people invent $ A_{n}^{i} \equal{} \{x: (i \minus{} 1)2^{ \minus{} n}\leq f(x)\le i2^{ \minus{} n}$ and $ f_{n} \equal{} \sum_{i \equal{} 1}^{n2^{n}}2^{ \minus{} n}(i \minus{} 1)\chi_{A_{n}^{i}}$????
what is the motivation to invent such things?

if f : [a, b]---->R is measurable functions, can we approximate f by [b]a simple functions[/b]? if possible, how can we approximate it?  (we have known that f can be approximated by[b] a sequence [/b]of simple functions )

That sort of function is a very natural construction; it's obvious from the right frame of mind.What does it do? Where the function values are in $ [a,b)$, we estimate those values with $ a$, and repeat for all such intervals in a mesh depending on $ n$. As $ n$ increases, the mesh both becomes finer and extends to cover more of the real line. (Note: you need some negative indices with that scheme to cover negative values of $ f$)

I am reading Real analysis of H. L. Royden,  and getting stuck with the following problem :

Let f be a measurable function defined on an interval $ [a,b]$ and assume that f takes the values $ infinity$ only on the set of measure zero. Then given $ \epsilon> 0$, we can find a [b]step function[/b] g and a [b]continous function [/b]h such that :

$ |f \minus{} g|\leq\epsilon$ and   $ |f \minus{} h|\leq\epsilon$

except for a set of measure less than $ \epsilon$, i.e, $ m\{ |f \minus{} g|\geq\epsilon\} < \epsilon$ and $ m\{ |f \minus{} h|\geq\epsilon\} < \epsilon$. In addition if $ m\leq f\leq M$ then we can choose g and h st :$ m\leq g,h \leq M$

could you please give me some hints to prove this proposition ?
thanks

okay well if you use the intuition above lets try to approximate by step functions and see how we can fix the proof for continuous functions

if we let $ A_n^i$ be the set of values of x that get mapped into the interval $ ((i\minus{}1)2^{\minus{}n}, i2^{\minus{}n}]$ as you did above

then since $ f: [a,b] \to R\equal{} \cup_{i\in Z} A_n^i$

we fix $ \epsilon> 0$ and know that $ n> \minus{}\frac{\ln \epsilon}{\ln 2}$

gives us a fine enough partition of $ A_n^i$'s to be within epsilon of

$ g_n(x) \equal{} \sum_{i \in Z} i2^{\minus{}n}1_{A_n^i} (x)$

then $ g_n(x) \equal{} i2^{\minus{}n} \to x \in A_n^i \to f(x) \in ((i\minus{}1)2^{\minus{}n}, i2^{\minus{}n}] \to |g_n(x)\minus{}f(x)| < 2^{\minus{}n}< \epsilon$

this is how we construct our simple function g.

for the continuous version, we start with a fixed $ \epsilon > 0$

and find the  $ |g_n(x)\minus{}f(x)|< \frac{\epsilon}{2}$ simple

then we notice it has countably many points of discontinuity (namely $ x\equal{}i2^{\minus{}n}, \forall i \in Z$)

enumerating these points $ x_1,x_2,...$ we take

$ h(x)\equal{}g_n(x), \forall x \notin (x_k\minus{}\epsilon 2^{\minus{}k\minus{}2},x_k\plus{}\epsilon 2^{\minus{}k\minus{}2})$

and define h to be linear on $ (x_k\minus{}\epsilon 2^{\minus{}k\minus{}2},x_k\plus{}\epsilon 2^{\minus{}k\minus{}2})$ such that it is continuous

then the $ |h(x) \minus{}f(x)| \le |h(x) \minus{}g(x)| \plus{} \sum_{k\equal{}1}^\infty \epsilon 2^{\minus{}k\minus{}1}<\epsilon$

Kenn,
thanks for replying my question. But I am confused to get your idea  :

1. You did not use the fact that $ f: [a,b]\rightarrow R$
2. $ (g_{n})_{n\in N}$ is a sequence of simple functions , but what we need is $ g$ is a step function. ( I think we cannot let $ g=lim_{n\rightarrow\infty}g_{n}$ since the limit of simple function may not be simple function.
3. why  $ g_{n}$ has countable countably many points of discontinuity (namely $ x = i2^{ - n},\forall i\in Z$)?

could you please explain more? thanks

If you want one function out of that sequence, just pick one. It has an associated tolerance $ \epsilon$, and we can make $ \epsilon$ as small as desired by going far enough.

An outline of kenn4000's procedure:
Step 1: Approximate $ f$ by simple functions $ g_n$ (already done in earlier posts)
Step 2: Approximate $ g_n$ by step functions $ h_\epsilon$. We do this by finding finite unions of intervals which are close in measure to the inverse images of the various points in the range of $ g_n$
Step 3: Approximate the $ h_\epsilon$ by continuous functions, by cutting off jumps.

1. $ R\equal{} \cup_{i\in Z} A_n^i$ was the thing i was pointing out here. I partitioned the range of the function until it was in intervals smaller than epsilon.

2. I didnt pick the limit of the $ g_n$'s i just picked one with an $ 2^{\minus{}n}<\epsilon$ (of which there are infinite choices)

3. actually what i said was wrong, the discontinuities arent at $ x\equal{}i2^{\minus{}n}$, so you would have to justify this claim that discontinuities are countable =(

[quote="kenn4000"]...you would have to justify this claim that discontinuities are countable[/quote]This is why we went to a step function. Step functions have finitely many jumps.


Derive this formula for the area of a convex quadrilateral:

$ A \equal{} \frac{(b^2 \plus{} d^2 \minus{} a^2 \minus{} c^2)tan\theta}{4}$

where a,b,c,d are sides and $ \theta$ the angle between diagonals

If you know any other area formulas for quads in terms of its sides or diagonals, please post them too

$ \mbox{Let quadrilateral } ABCD \mbox{ and } \{O\} = [AC]\cap [BD]; \\
|AB| = a, |BC| = b, |CD| = c, |AD| = d; \\
|OA| = x, |OB| = y, |OC| = z, |OT| = t\Rightarrow |AC| = x + z \mbox{ and } |BD| = y + z.$
$ \mbox{Then: } \\
 \begin{array}{c} {a^2 = x^2 + y^2 - 2xy.cos\theta} & {b^2 = y^2 + z^2 + 2yz.cos\theta} & {c^2 = z^2 + t^2 - 2zt.cos\theta} & {d^2 = x^2 + t^2 + 2xy.cos\theta}\end{array}\}\Rightarrow b^2 + d^2 - a^2 - c^2 = \\
= 2.(xy + yz + zt + xt).cos\theta = 2.(x + z)(y + z).cos\theta = 2.|AC|.|BD|.cos\theta \Rightarrow \\
\Rightarrow |AC|.|BD|=\frac{b^2 + d^2 - a^2 - c^2}{2.cos\theta}\Rightarrow \\
\Rightarrow S_{ABCD} = \frac {|AC|.|BD|.sin\theta}{2} = \frac {(b^2 + d^2 - a^2 - c^2).tg\theta}{4}.$


Two very simple inequalities from a Hong Kong contest

1)Let x,y,z be real numbers  such that

   x+y+z=5   and    xy+yz+zx=3

   Find the greatest value of z.

2)Let x,y,z,w be real numbers such that

    x+y+z+w=8   and   x 2 +y 2 +z 2 +w 2 =16

  Find the greatest value of w

Just hard enough for me  :)

1)

x'=x-5/3, ....

x+y+z=5 and xy+yz+zx=3 becomes
x'+y'+z'=0, z' = 16/3 + x'y'

wolog x'<=y'<=z', z'>= 0,x'<=0, since we want z maximum

. if y'<0, then 0 <= x'y' <= z'/4  and z = z'+5/3 <= 13/3
. if y'>0, z' <= 4/sqrt(3) < 8/3

Maximum value for z is 13/3.


How many distinct numbers can be written as the sum of two or more distinct members of the set $ \{0,1,2,3,4,5\}$?

The smallest number that can be made is 0+1=1, and the largest number that can be made is 0+1+2+3+4+5=15.
Every other number in between can be made using two or more members of this set, so $ \boxed{15}$ distinct numbers can be made.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ 25$. Prove that $ G$ has at least one subgroup of order $ 5$, and that if it contains only one subgroup of order $ 5$, then it is a cyclic group.

Consider $ A \equal{} \{ a_1,a_2,...,a_{25} \}$ the set attached to this group. The order of each element must be in the set $ \{ 1,5,25 \}$. If all the elements have the order $ \le 5$ then there are more than one subgroup of order $ 5$. If there is an element whose order is $ 25$ then the group is cyclic. Hope it is correct. I'm waiting for correction.

More generaly, for the first part, Cauchy's theorem states that for every prime $ p$ which divides $ ord (G)$, we can find an element $ a \in G$ s.t. $ ord(a) \equal{} p$. So $ \{ e, a, \dots, a^{p \minus{} 1}\}$ is a subgroup of $ G$.

In our case, if this subgroup of order $ 5$ is unique then there is at least one element of order $ 25$ in $ G$. So, $ G$ is cyclic.


Let $ a,b,c > 0$. Prove that $ a^3 \plus{} b^3 \plus{} c^3\ge a^2\sqrt {bc} \plus{} b^2\sqrt {ac} \plus{} c^2\sqrt{ab}$

$ a \equal{} x^{2},b \equal{} y^{2},c \equal{} z^{2}$

$ \implies$  $ x^{6} \plus{} y^{6} \plus{} z^{6}\geq x^{4}yz \plus{} y^{4}zx \plus{} z^{4}xy$  which is obviously true by Muirhead :wink:

It is AM-GM since it is equivalent to
\[ \sum_{cyc}\frac{4a^3\plus{}b^3\plus{}c^3}{6}\geqslant\sum_{cyc}\sqrt[6]{a^{12}b^3c^3}\mbox{.}\]
 :P

[quote="thanhnam2902"]Let $ a,b,c > 0$. Prove that $ a^3 \plus{} b^3 \plus{} c^3\ge a^2\sqrt {bc} \plus{} b^2\sqrt {ac} \plus{} c^2\sqrt {ab}$[/quote]

\[ LHS\minus{}RHS \equal{} \frac{1}{2}\sum_{cyc} {\left( {\left( {\sqrt a  \plus{} \sqrt b } \right)^2 \left( {a \plus{} b} \right) \plus{} c^2 } \right)\left( {\sqrt a  \minus{} \sqrt b } \right)^2 } \ge 0.\]

[quote="thanhnam2902"]Let $ a,b,c > 0$. Prove that $ a^3 \plus{} b^3 \plus{} c^3\ge a^2\sqrt {bc} \plus{} b^2\sqrt {ac} \plus{} c^2\sqrt {ab}$[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3\ge 4(\sqrt3\minus{}1)\left(a^2\sqrt {bc} \plus{} b^2\sqrt {ac} \plus{} c^2\sqrt {ab}\right)\minus{}3(4\sqrt3\minus{}5)abc\]

This is my solution:

$ \forall a,b > 0$ we get $ a^2 \minus{} ab \plus{} b^2\ge ab \Longleftrightarrow a^3 \plus{} b^3 \ge ab(a \plus{} b)$ and we get $ b^3 \plus{} c^3 \ge bc(b \plus{} c), c^3 \plus{} a^3 \ge ca(c \plus{} a)$

$ \Longrightarrow 2(a^3 \plus{} b^3 \plus{} c^3) \ge ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \equal{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)$

But $ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)  \ge 2a^{2}\sqrt {bc} \plus{} 2b^{2}\sqrt {ac} \plus{} 2c^{2}\sqrt {ab}$

Hence we get $ 2(a^3 \plus{} b^3 \plus{} c^3) \ge 2a^{2}\sqrt {bc} \plus{} 2b^{2}\sqrt {ac} \plus{} 2c^{2}\sqrt {ab}$  $ \Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \ge a^{2}\sqrt {bc} \plus{} b^{2}\sqrt {ac} \plus{} c^{2}\sqrt {ab}$

Since $ a^{3}\plus{}b^{3}\plus{}c^{3}\ge 3abc$ and

By AM-GM
$ a^{3}\plus{}b^{3}\plus{}c^{3}\ge\frac{a^{3}\plus{}abc}{2}\plus{}\frac{b^{3}\plus{}abc}{2}\plus{}\frac{c^{3}\plus{}abc}{2}\ge a^{2}\sqrt{bc}\plus{}b^{2}\sqrt{ac}\plus{}c^{2}\sqrt{ab}$


algebra
polynomial
superior algebra
superior algebra unsolved

Let $E/F/K$ be a tower of fields with $E/K$ algebraic, and $F/K$ finite Galois, as well as $u \in E$ with minimal polynomial $f(x)$ over $F$. Show that the the minimal polynomial of $u$ over $K$ is given by  \[ \prod_{g(x): \exists \theta \in \mathrm{Gal}(F/K), g(x)=\theta(f(x))}g(x) \] (Note that this product only allows one copy of each polynomial, so that if $\theta_1(f(x))=\theta_2(f(x))$ for $\theta_1 \neq \theta_2$, this term only appears once in the product).

Denote by $P(x)$ this polynomial. First it is trivial that $P(u)=0$, and it is rather clear that $P$ has coefficients in $K$, since by definition it has coefficients in $F$, and if $f$ is irreducible over $F$, then $\sigma(f)$ is irreducible over $F$ for every $\sigma \in Gal(F/K)$, and in particular over $K$. Whence $P$ is fixed by any $\sigma \in Gal(F/K)$. But also $P$ is minimal since it must be divisible by $\sigma(f)$, otherwise it would not be fixed by $\sigma$ (here we used the fact that the $\sigma(f)$ are irreducible) and its coefficients would not be in $K$, which would be absurd.


abstract algebra
group theory
number theory
least common multiple
superior algebra
superior algebra unsolved

Let $\left( G, \cdot \right)$ be a finite group with the following property: the set of the orders of the elements of $G$ is composed of $n$ consecutive numbers, where $n \geq 2$.

a) Prove that $G$ is commutative if and only if $n=2$.

b) Prove that if $n \geq 3$, then the only element $a \in G$ such that $ax = xa, \, \forall x \in G$ is $e$.

Hi.

Since 1 is always the order of the neutral element, the set of the orders of elements of G is exactly {1, ..., n}.

a.) If n=2, then G is of exponent 2 and therefore abelian.
If G were abelian and $n\geq 3$, then there would be an element of order n and one of order n-1. Since n and n-1 are relative prime and G is abelian, the product of those two elements would have the order n(n-1) which is greater than n. So we get a contradiction.

b) Let a with the desired property such thta ord(a)=k>1. Then k has a prime factor p. So, by Cauchy, <a> has a subgroup of oprder p generated by a power of a that comutes with all the elements from G. Now it is simple to observe that (p,n)=1 or (p,n-1)=1. As a result we have an element or order pn or p(n-1). But both pn and p(n-1) are grater than n for n>2 and p>1, contradiction. Therefore ord (a)=1, so a=e. For part a) I gave the same soluton in the contest :)

i have a nice solution for your problem using exponant of ${G}$.
let $r$ the exponant of $G$ ie:$r=max(ordr(x)),x\in\mathbb{G}$
then $r-1/r$ then $r =2$. :wink:

The exponent of G is [b]not[/b] $max_{r\in G}(ord(r))$ it is defined as $lcm_{r\in G}(ord(r))$. Only in abelian groups this is the same.


logarithms
inequalities
probability and stats

Suppose {$ X_n, n \ge1$} are iid with $ E(X_n)\equal{}0$, and $ Var(X_n)\equal{}1$.
Show $ \frac{\sum_{i\equal{}1}^{n}X_i}{n^{1/2}\log{n}}$ converges to zero almost surely.

it is a consequence of the celebrated law of [url=http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm]iterated logarithm[/url]. Nevertheless I do not know any simple proof of this result. Anyone ?

This can be proved for bounded r.v's using Borel-Cantelli and classic inequalities. I tried to prove the general case with some tricky truncation, but didn't succeed.

You can just use Kolmogorov's favorite inequality $ P\Bigl\{\max_{k\le n}\Bigl|\sum_{j\equal{}1}^k X_j\Bigr|>t\Bigr\}\le  t^{\minus{}2}\sum_{j\equal{}1}^n Var(X_j)$ and notice that your problem is equivalent to showing that for every $ \delta>0$ you almost surely have $ \max_{k\le 2^m}\Bigl|\sum_{j\equal{}1}^k X_j\Bigr|\le 2^{m/2}m\delta$ for all but finitely many $ m\equal{}1,2,3,\dots$, which is immediate now from Borel-Cantelli.

[quote="The Wicker Man"]Suppose {$ X_n, n \ge1$} are iid with $ E(X_n) \equal{} 0$, and $ Var(X_n) \equal{} 1$.
Show $ \frac {\sum_{i \equal{} 1}^{n}X_i}{n^{1/2}\log{n}}$ converges to zero almost surely.[/quote]
$ \sum_{n \equal{} 1}^{\infty}\frac {X_n}{n^{1/2}\log{n}} < \infty$ a.s. by Kolmogorov convergence Criterion $ \equal{} > \frac {\sum_{i \equal{} 1}^{n}X_i}{n^{1/2}\log{n}} \minus{} > 0$ a.s.  by Kronecker's Lemma.


[quote="Rep123max"]Hey, ive got this humoungoiousousousous crossword puzzle and i need just a little help.
first what is a word for kidnapping of american soldiers? it is supposed to fit _ M _ _ _ _ _ M _ N T.
Second, who put down the whiskey rebellion.
and finally, was it the treaty of greenville that gave america the northwest territory?

THANK S[/quote]

Yay crossword puzzles!  The first one is "impressment", and I think Washington put down the Whiskey rebellion.  I'm not sure tho.

It wasn't Treaty of Greenville, I think it was the Treaty of Paris.

maybe it was Greenville, that's the one that made the indians give up territory.

We just learned about the Whisky Rebellion! Let me go get my notes...*flip flip*...*flippity-flippity-flop :)*...grrr...*papers flying in air*...*ripping noise* oops...well anyway :) according to the text book it was Washington and 12,000 men. There is a guy named Jay though who was a delegate to Britain to make them stop attacking American ships and to give up forts in Northwestern U.S. though but that's about it.


[quote="Mr.Mathematics"]
\[x^{2}-49 = 3 y \]
find x,y
y is prime number
x,y integers[/quote]

(8,5) brute force.

[hide]$x^{2}-49=(x+7)(x-7)=3y$

Because $y$ is a prime number, either $x-7=3\Rightarrow x=10$, or $x-7=1\Rightarrow x=8$.

Therefore, the only possible solutions are $(8,5)$ and $(10,17)$.[/hide]


Can someone help we with these tough probability questions?

1. People in this room are asked their birthday month and day, not their year. What are the chances that two ppl in the room have the same birthday?

2. The Ace Bubble Gum Factory includes on picture in each of its packages of gum. THe company printed the following numbers of pictures and randomly put them packages of gum:

45,000 of Bill Cosby, 45,000 of Michael J Fox, 90,000 of Bart Simpson, 180,000 of Alf the ALien. How many packages would you probably have to buy in order to obtain a complete set of all 4 pictures?

3. Design 2 identical circular spinners. Each spinner is divided into 4 equal sections. THe sections are numbered 1, 2, 3 and 3. Each spinner will be spun and the 2 numbers added together. WHich sum will occur most often?

4. Roll a regular 6-sided die as often as you wish. Your score is your total of all numbers rolled, but if roll a 1, you lose all the points for that round. What strategy would maximize your total score?

These problems are very hard to me and I would greatly appreciate any help on them Thanks.

On number 1, how many people are in the room?

[hide="Problem 1"]$\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\ldots\frac{347}{365}\times\frac{346}{365}\times\frac{345}{365}$ is the probability that no one has the same birthday.  Subtract from 1 to get the answer.[/hide]

Problem 2 depends on how you define "probably".  50%?  90%?  If you buy 315,001, you will definitely get all four.

[hide="Problem 3"]Find the probability for each sum, then compare.[/hide]

[hide="Problem 4"]If you have $x$ dollars, the average gain/loss for each roll is $\frac{2+3+4+5+6-x}{6}=\frac{20-x}{6}$

If you have more then 20 points, your average is negative, so you shouldn't roll.[/hide]

[hide="Solution 1"]
Probability that none have birthday on same day=P= $\frac{365}{365} \cdot \frac{364}{365} \cdot\cdot\cdot \frac{345}{365}$
hence required probability = 1-P[/hide]


Hopefully the order of the problem would not be as screwed as last years...

LOL

I hope to do better than what I got last year...

[quote="zepiphanus"]what score is generally considered "good enough" (or, what do the top 10 people in Canada usually get)?[/quote]

A 0/35 is a pretty cool score.

LMAO :rotfl: its true, a single-digit score is already pretty good. If it is double digit, you will probably get top 10 in Canada, in the high teens could be top 3.
But it varies with year. APMO is pretty random.
0 is pretty good too

Good Luck tomorrow!

Last year even a single-digit score could give you a top 10 finish in Canada (or better still, an HM certificate)

By the way the winning scores for the last two years are both 19.

[quote="jix"]LOLZ at question 2.

I hope I can get a single digit, nonzero score on that exam.

When can discussions start?[/quote]

Please don't discuss the exam. Not until the Korean Mathematical Society posted their answers on the website. (And please modify your post).

2009 APMO problems and solutions are now available at http://www.kms.or.kr/competitions/apmo/

APMO seriously killed me... what do you think the cutoff for gold/silver/bronze/HM will be this year?

I still think it will be teh same as last year despite difficutly.

I could get 2 and maybe partial on 1 and 3, so hopefully a 7.

i found out that australians posted their unofficial scores: http://www.amt.canberra.edu.au/apmo2009.html

funny, no one got more than a mark for #2 :D

[quote="mathgeniuse^ln(x)"][quote="Nawahd Werdna"]When do Canadians receive their scores??? (usually in the past?)[/quote]

I don't think I got mine till like May.[/quote]

They send your APMO scores after they announce the IMO team:
"Congratulations, you got a 0/35 on APMO 2008!"

I have the email from last year. It was dated May 14.

I assume for Danny, it would be something like,

"This is to inform you that your score in the 2008 APMO was -0/35. It was an exceptionally difficult exam this year."

=P

That's the best email you could get from those people. Imagine showing that off. :)

Although I hope I don't get that.

i haven't received my result yet either... i've been waiting for the past one month

[quote="Mathgod"][quote="soledad42"]Just curious, did anyone receive his/her APMO results yet? The score release is uncharacteristically late this year for some reason, and I'm just too eager to see whether I made top 7 in Canada on this...[/quote]

Nah na na.[/quote]

@Danny: Nah na na :P

High Five Jon!

And Chen, was I 1 or 2 points behind you? (27/35 pour moi).

[quote="soledad42"][quote="zepiphanus"]that's exactly twice my score xD[/quote]

Congrats zepiphanus. This post also terminates my attempt to guess who got one place higher than I did.  :)[/quote]

Ha ha.

[quote="soledad42"][quote="Mathgod"][quote="rofler"]Hey Danny, I found a REALLY nice solution to the stupid grade 9 Russian Problem on the pre-camp problem set. Did you solve it yet?

(Altough after finding the solution, I can see why it is Russian!)

Edit: 500th post![/quote]

Grade 9 Russian? What's that? You work on grade 9 Russian problems?  :rotfl:[/quote]

To Danny: Please refer to warm-up problem section number 3.

To Hunter: I agree that the problem's pretty easy; however, my solution is pretty standard and not really Rrussian-like.

BTW, I found a "cheating" solution to warm-up number 5 where I proved that Jacob can win at least 1505 games without identifying the non-trivial part of his optimal play.[/quote]

What pre-camp problem set?

[quote="rofler"]Guess. Read my message, and then deduce based on the evidence given which one I could be possibly talking about... :P[/quote]

Noob.  :)

Danny, so you solved it? I'm proud of you :P.

@Danny: Nǐ yā zhēn hěn

@Whoever taught me that: I hope that actually means what you said it meant.

@Jarno: Does it mean what I think it means?

@Adrian: I need 2 sources of verification on this, you could tell me also.

Edit: Also, how I can type in pinyin is a closely guarded secret :ninja:

I thought I complimented you in Mandarin, and you hurt me like this.

I'm feeling you  are growing distant Danny, come back to me... :(

[quote="Mathgod"][quote="rofler"]Danny, so you solved it? I'm proud of you :P.

@Danny: Nǐ yā zhēn hěn

[/quote]

Double-plus noob.[/quote]

Ok.. deduction time.

Ni = you

Zhen = really

Hen= mean

1) What's "ya"?

2) Something's wrong here with my chinese.

It's a back-handed compliment because of the way it is pronounced.

Or so I've heard...:roll:

K I really can't think of a way to pronounce it and make it sound like a compliment...

@ Chen: yā is pretty hard to translate... All I can say is that this makes the sentence much less like a compliment as Hunter claimed.

[quote="sunchips"]haha nice job tho Hunter, still  :P

Is "ya" like "ai ya!"?[/quote]

Idiot. No.

On a different topic... I got my username capitalized! Yay!

Back to the same topic...

Chen it's pretty hard to explain over the internet. We'll try to explain the meaning to you about 4 days later unless Danny can come up with a nice and accurate explanation before that...

P.S. @Danny: You should be able to come up with a good translation because you use this phrase somewhat often in daily talk...


Let $ b,n>1$ be two integers. Suppose that for every integer $ k>1$, there exists an integer $ a_k$ such that $ k$ divides $ b\minus{}a_k^n$. Prove that $ b$ is an integer raised to the $ n$-th power (of the form $ x^n$).

See solution of this problem in IMO Shortlist 2007 .


inequalities
geometry
circumcircle
inequalities proposed

[color=darkred]Prove that in any triangle $ ABC$ there is the boxed inequality $ 6\le6\cdot\sqrt [3]{\frac{R}{2r}}\le\boxed{\ \sum\frac{bc}{a(p-a)}\le 4+\frac{R}{r}\ }\le\frac{3R}{r}$,

where $ 2p = a+b+c$, $ R$- the circumradius and $ r$- the inradius.

[b]Remark.[/b] $ \{\begin{array}{c}a: = y+z\\ \ b: = z+x\\ \ c: = x+y\end{array}$, where $ \{\begin{array}{c}p-a = x > 0\\ \ p-b = y > 0\\ \ p-c = z > 0\end{array}$ $ \implies$ $ \frac{R}{r}=\frac{(x+y)(y+z)(z+x)}{4xyz}$ and the upper inequality becomes :

$ \boxed{\ \{x,y,z\}\subset(0,\infty)\implies\sum\frac{x^{2}+yz}{x(y+z)}\le 1+\frac{(x+y)(y+z)(z+x)}{4xyz}\ }$.[/color]

\[ \sum\frac{x^{2}\plus{}yz}{x(y\plus{}z)}\le 1\plus{}\frac{(x\plus{}y)(y\plus{}z)(z\plus{}x)}{4xyz}\\ \iff\sum\frac{(x\plus{}y)(y\plus{}z)}{x(y\plus{}z)}\le 4\plus{}\frac{(x\plus{}y)(y\plus{}z)(z\plus{}x)}{4xyz}\\ \iff\sum\frac{1}{x(y\plus{}z)^{2}}\le\frac{4}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\plus{}\frac{1}{4xyz}\]
Is now ready for some isolated fudging


What do you think about these books?
I am reading one of them about set theory                                      .
The good thing about the book is that it is very instructive and you learn quick.
Anybody else who has read some of these books?

I read some, but I didn't learn from them, I used it to supplement what I already knew.

For instance, I studied geometry on my own, then read the schaums geometry to clarify.  It really does help.

Ok. Well I am reading the book about set-theory because my university doesnt offer much in that direction, even in more advanced courses you need knowledge of it.
(Both analysis and algebra. Many books about analysis / abstract algebra contain a first chapter about set-theory, so why not read a book on it, and skip all those boring first chapters :-)

I used Intermediate Algebra version (suppose to be for college use but I used it when I'm in eighth grade.. :P ) and despite the fact that it's pretty easy, it is very good on instructions and lectures. I also like the supplement problems because some of them are sometimes actually pretty good.

I don't recommend for deep knowledge from their books though. They're just for basics.

[quote="4everwise"]That's interesting. Do any of you know a book specifically about trigonometry that is good? I'd like to study one of those and become proficient at trig.[/quote]

Nah, I wish I knew a good book to learn trig.  I learned all of mine from high school precal, AoPS V2 and AoPS Intermediate Trig/complex numbers.  I've been thinking about getting that book by Andreescu, [i]103 Trigonometry Problems[/i]...anyone have any comments on it?

i heard that it was worth it.... (basically the best trig book out there, since there are rarely good books dedicated to specifically trig)


Let $ x \equal{} y$

Then,

$ x^2 \equal{} xy$
$ x^2  \minus{} y^2 \equal{} xy \minus{} y^2$
$ (x\minus{}y)(x\plus{}y) \equal{} y(x\minus{}y)$
$ (x\plus{}y) \equal{} y)$
$ 2x \equal{} y$

Where is the error in this proof?

Guys, guys, guys.
$ x\minus{}y\equal{}0$.

[quote="TZF"]You can't divide something out of an equation without either knowing that it isn't zero or assuming the case in which it isn't zero.

Consider the following ridiculous "proof" that has the same mistake, except it is more obvious:

$ x \equal{} 0$
Thus, $ x^2 \equal{} x$
Dividing by $ x$, we get $ x \equal{} 1$
Transitively, $ 0 \equal{} 1$[/quote]

But how do you get from $ x \equal{} 0$ to $ x^2 \equal{} x$?


limit
logarithms
calculus
calculus computations

$ \lim_{x \to \infty} (\frac {2x \plus{} 5}{2x})^{ \minus{} x/3}$

$ \lim_{x \to \infty} ( 1 \plus{} \frac {5}{2x})^{ \minus{} x/3}$

Unsure from there. How do you evaluate this without graphing?

Well, to start, you've divided wrong and then done a bizarrely wrong algebraic step.  So perhaps you should deal with those problems first.  After you've fixed that, you might think about similar-looking limits you've done before, and possible substitutions you can make to arrange it into even more familiar-looking form.

Whoops, I am so sorry. I was typing up a classmates steps, and I didn't bother to read what he had written. I corrected it.

I'm an ap calc student, and I have never seen a limit of this form before, so I don't know how to approach this. I considered L'Hopital's rule, but that didn't make it any simpler.

You've never seen a limit that looks like, say, $ \lim_{n \to \infty} \left(1 \plus{} \frac {1}{n}\right)^n$?

For what it's worth, l'Hopital's does work, but you need to do some finagling.  If the original expression is $ y$, write

$ \ln y \equal{} \minus{} \frac {x}{3} \left( \ln (2x \plus{} 5) \minus{} \ln (2x) \right) \equal{} \minus{} \frac {1}{3} \cdot \frac {\ln (2x \plus{} 5) \minus{} \ln (2x) }{\frac {1}{x} }$

and [i]then[/i] apply l'Hopital's.

Cool. I was on the right track with ln y, but I did not think of putting $ 1/x$ in the denominator and I figured an AP MC question wouldn't need more than one step. I got too complacent @_@.


Hello

In legatura cu problema a treia de la locala Bucuresti clasa a 8-a 2005  are cineva vre-o idee  cum s-ar demonstra faptul ca radacina patrata din produsul a doua numere naturale consecutive are totdeauna prima cifra de dupa virgula 4 ?

Regards

Să se demonstreze că:
$\forall k \in \mathbb{N},k \ge 1 \Rightarrow k + \frac{4}{{10}} < \sqrt {k\left( {k + 1} \right)}  < k + \frac{5}{{10}}$


algebra
polynomial
absolute value
algebra unsolved

Determine the lowest positive integer n such that following statement is true:
If polynomial with integer coefficients gets value 2 for n different integers,
then it can't take value 4 for any integer.

The polynomial $ P(x)\equal{}4\minus{}x(x^2\plus{}2x\minus{}1)$ has $ P(1)\equal{}P(\minus{}1)\equal{}P(\minus{}2)\equal{}2$ and $ P(0)\equal{}4$.
Hence $ n>3$.

Next consider a polynomial $ P(x)$ with integer coefficients such that $ P(a)\equal{}P(b)\equal{}P(c)\equal{}P(d)\equal{}2$ for four distinct integers $ a,b,c,d$, and $ P(e)\equal{}4$ for some integer $ e$.
Consider the polynomial $ Q(x)\equal{}4\minus{}P(x\plus{}e)$:  Then $ Q(0)\equal{}0$, and the four integers $ x_1\equal{}a\minus{}e$, $ x_2\equal{}b\minus{}e$, $ x_3\equal{}c\minus{}e$, and $ x_4\equal{}d\minus{}e$ satisfy $ Q(x_i)\equal{}2$ for $ i\equal{}1,2,3,4$.
Since $ Q(0)\equal{}0$, we have $ Q(x)\equal{}x*R(x)$ for some integer polynomial $ R(x)$.
Since $ x\minus{}y$ divides $ Q(x)\minus{}Q(y)$, we have $ x_i\minus{}0$ divides $ Q(x_i)\minus{}Q(0)\equal{}2$ for $ i\equal{}1,2,3,4$.
This implies that the $ x_i$ are a permutation of $ 2,1,\minus{}1,\minus{}2$.
Then $ R(2)\equal{}1$ and $ R(\minus{}2)\equal{}\minus{}1$, and $ 4\equal{}2\minus{}(\minus{}2)$ must divide $ R(2)\minus{}R(\minus{}2)\equal{}2$.
Contradiction.
Hence $ n\le4$.

Under the notations above, it follows $ P(x) \equal{} (x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d)T(x) \plus{} 2$. But then $ 4 \equal{} P(e) \equal{} (e\minus{}a)(e\minus{}b)(e\minus{}c)(e\minus{}d)T(e) \plus{} 2$, so $ |(e\minus{}a)(e\minus{}b)(e\minus{}c)(e\minus{}d)| \leq 2$. However, the absolute value of the product of four distinct not-null integers is at least $ 4$.

Suppose that, for a given n, there exists such a polynomial.

We define Q(x) as P(x)-2 for simplicity. We know Q(x) has n integer zeros, so we write:

Q(x)=(x-x1)(x-x2)....(x-xn)*R(x)

We want to find an integer x such that Q(x)=2, so since each of {x-x1, x-x2, ..., x-xn} is an integer, so is R(x).

We also know that each of {x-x1, x-x2, ..., x-xn, R(x)} has to divide 2, and that each of {x-x1, x-x2, ..., x-xn} is distinct, but 2 can be written as a product of at most 3 different integers (e.g. (-2)*1*(-1)) so n<=3.

Now we need a construction for every n<=3, and that's not hard.

For n=3: Q(x)=(x-1)(x+1)(x-2)

For n=2: Q(x)=-2*(x-1)*(x+1)

For n=1: Q(x)=x.

Therefore, n=4.

We claim the answer is $n=4$. First, consider the polynomial $(x-1)(x-3)(x-4)+2$. It equals $2$ for the $3$ integers $1,3,4$ and it takes on the value of $4$ at $x=2$. Thus, $n>3$. Now, suppose $n\ge4$ and the property is false. Let $P(x)$ be a polynomial for which the property fails. Then, $P(x)=\alpha(x-r_1)(x-r_2)\cdots(x-r_n)+2$ where $r_1,\ldots,r_n$ are distinct integers. Since the property fails, there is some integer $m$ for which $P(m)=4$. Then, $2=\alpha(m-r_1)(m-r_2)\cdots(m-r_n)$. It follows $|m-r_i|\in\{1,2\}$ for all $i$. Indeed, at most $1$ of them can be $2$, so at least three are $1$. Hence, there exist $i\ne j$ for which $m-r_i=m-r_j$, contradicting $r_i\ne r_j$. Therefore, the property never fails for $n=4$, as desired. $\blacksquare$


[hide="hint"]The trick to doing this is to realize that if you add the first two equations and subtract the third one, the x,y,w and z variables will kindly go away. Doing that, we get 0=4+a, and so a=-4.[/hide]

I think the a is the only number that i dont need....a can be any number...so  i need x,y,z.. Each of them will contain the a.

$x-y+z+w=1$
$3x-y-z+w=a$
$4x-2y+2w=3$

Adding the first two and subtracting the third gives us $a=2.$

$x-y+z+w=1$
$3x-y-z+w=2$
$4x-2y+2w=3$

The third equation is just the first two added, so it is of no use. You have 4 variables, but only 2 equations.


I second the motion.

To all people like me, let's get nonzero scores! To the smart people, good luck for USAMO!

Good luck to all of you. I'm always cheering for the AoPSers.

yea, good luck to everyone. i give everyone the gift of >7 on the AIME.  :D


integration
calculus
calculus computations

Is it possible to find a primitive to $ \int x^x dx$ ?

Thanks in advance

There are no "nice" solutions or elementary antiderivates for that, but you could express it as a series.


inequalities
geometry
area of a triangle
Heron\u0027s formula
inequalities proposed

[quote="socrates"]Prove that in any triangle the following inequality holds:
$4S\leq\sqrt{a^4+b^4+c^4}$
where S denotes the area of it![/quote]
Substituting heron's formula and use amgm. All you need is now power mean.

Use  BCS  and  Weitzenbock's  ineq : [tex] a^4+b^4+c^4\geq\frac{(a^2+b^2+c^2)^2}{3}\geq\frac{(4\sqrt3A)^2}{3}=16A^2 [/tex]

There is also  a  stronger  ineq : [tex] a^2b^2+b^2c^2+c^2a^2\geq16A^2 [/tex]  which is also  easy  to prove  :)

Yes,indeed.Thanks for correcting my huge mistake.  :)


inequalities
trigonometry
geometry
perimeter
incenter
circumcircle
geometric inequality

[quote="Marius Mainea"]Prove that in any triangle ABC is true inequality:

   $ a\sin \frac {A}{2} \plus{} b\sin \frac {B}{2} \plus{} c\sin \frac {C}{2}\geq\frac {a \plus{} b \plus{} c}{2}$[/quote]
I don't see an easy proof.
Let $ a \equal{} \frac {y \plus{} z}{2},$ $ b \equal{} \frac {x \plus{} z}{2}$ and $ c \equal{} \frac {x \plus{} y}{2}.$ Then
$ \sum_{cyc}a\sin \frac {A}{2}\geq\frac {a \plus{} b \plus{} c}{2}\Leftrightarrow\sum_{cyc}a\sqrt {\frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)}{bc}}\geq a \plus{} b \plus{} c\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(y \plus{} z)\sqrt {(y \plus{} z)yz}\geq(x \plus{} y \plus{} z)\sqrt {(x \plus{} y)(x \plus{} z)(y \plus{} z)}\Leftrightarrow$ ( after squaring )
$ \Leftrightarrow\sum_{cyc}(x \plus{} y)(x \plus{} z)\sqrt {x(x \plus{} y)(x \plus{} z)}\geq\sum_{cyc}(3x^2 \plus{} 5xy)\sqrt {xyz}.$
But $ \sqrt {(x \plus{} y)(x \plus{} z)}\geq x \plus{} \sqrt {yz}.$  Thus, it remains to prove that
 $ \sum_{cyc}(x \plus{} y)(x \plus{} z)\sqrt {x}(x \plus{} \sqrt {yz})\geq\sum_{cyc}(3x^2 \plus{} 5xy)\sqrt {xyz},$ which is equivalent to
$ \sum_{sym}\left(x^{3.5} \minus{} 2x^{2.5}y^{0.5}z^{0.5} \plus{} x^{1.5}yz \plus{} 2x^{2.5}y \minus{} 2x^{1.5}y^{1.5}z^{0.5}\right)\geq0,$ which is obvious.

See here: [url]http://www.mathlinks.ro/viewtopic.php?p=462855#p462855[/url] (Read especially Virgil Nicula's nice proof).

[quote="pohoatza"]See here: http://www.mathlinks.ro/viewtopic.php?p=462855#p462855 (Read especially Virgil Nicula's nice proof).[/quote]
Unfortunately this link is not opening from my computer. I thought in pure trigonometric approach:
$ a\sin \frac {A}{2} \plus{} b\sin \frac {B}{2} \plus{} c\sin \frac {C}{2}\geq\frac {a \plus{} b \plus{} c}{2}$
$ \Leftrightarrow\sum_{cyc}{a}\sqrt\frac{(S\minus{}b)(S\minus{}c)}{bc}\geq{S}$
$ \Leftrightarrow\sum_{cyc}{a}\sqrt{(\frac{1}{b}\minus{}\frac{1}{S})(\frac{1}{c}\minus{}\frac{1}{S})}\geq{1}$............ :maybe:

[quote="arqady"][quote="Marius Mainea"]Prove that in any $ \triangle ABC$ the inequality

$ a\sin \frac {A}{2} \plus{} b\sin \frac {B}{2} \plus{} c\sin \frac {C}{2}\geq\frac {a \plus{} b \plus{} c}{2}$

is true.[/quote]Let $ a \equal{} \frac {y \plus{} z}{2},$ $ b \equal{} \frac {x \plus{} z}{2}$ and $ c \equal{} \frac {x \plus{} y}{2}.$ Then[/quote]$ (y \plus{} z)^2(z \plus{} x)(x \plus{} y) \minus{} yz(2x \plus{} y \plus{} z)^2 \equal{} x(y \minus{} z)^2(x \plus{} y \plus{} z)\geq0\Longrightarrow$

$ \sum{(y \plus{} z)\sqrt {\frac {yz}{(z \plus{} x)(x \plus{} y)}}}\geq\sum{\frac {yz(2x \plus{} y \plus{} z)}{(z \plus{} x)(x \plus{} y)}} \equal{} \sum{x}.$

This proof is due to Cezar Lupu : http://www.mathlinks.ro/viewtopic.php?t=80834

Another proof see here : http://www.mathlinks.ro/viewtopic.php?t=216136

This inequality is equivalent to a nice geometric inequality

$ \frac {a}{IA} \plus{} \frac {b}{IB} \plus{} \frac {c}{IC}\geq\frac {s}{r},$

where $ a,b,c$ are the sides $ BC,CA,AB$, $ s$ is the semi-perimeter, $ r$ is the radius of incircle, $ I$ is the incentre and $ IA,IB,IC$ are the distances from $ I$ to the vertices of $ \triangle ABC$.

We can write the inequality in the form

$ \sum\frac{sin^2\frac{A}{2}}_{{cos\frac{B}{2}}.{cos\frac{C}{2}}}$ $ \geq$ 1 $ \leftrightarrow$
$ \sum\frac{{(4Rsin\frac{A}{2})}^2}_{{4Rcos\frac{B}{2}}.{4Rcos\frac{C}{2}}}$ $ \geq$ 1 $ \leftrightarrow$
$ \sum\frac{(II_a)^2}_{(I_aI_c).(I_aI_b)}$ $ \geq$ 1 , where I is the incentre and $ I_a,I_b,I_c$  are the ex-centres opposite the vertices A, B and C respectively.Call this form of the inequality (*)

For any point P in the plane of a triangle ABC and for arbitrary real numbers x ,y, and z there is the inequality:

$ (x+y+z)(x.PA^2 +y.PB^2 +z.PC^2)$ $ \geq$ $ (yza^2 +zxb^2 +xyc^2)$

Apply this inequality to the triangle $ I_aI_bI_c$: taking  P at the point I and letting x= $ \frac{1}{(I_aI_c))(I_aI_b)}$,  y = $ \frac{1}{(I_aI_b)(I_bI_c)}$ and z =$ \frac{1}{(I_aI_c).(I_bI_c)}$ ,the truth of the inequality (*) now follows.

from equality ∑xy(y-z)=(x-y)(y-z)(z-x), x,y,z in C,  obtain the inequality: ∑AB/MC≥(AB∙BC∙CA)/(AM∙BM∙MC), where M is a point in the plane of the triangle ABC.For M=I, where I is the center of the circle inscribed in the triangle ABC, inequality from the statement is obvious.

[quote="Marius Mainea"]Prove that in any triangle ABC is true inequality:

   $ a\sin \frac{A}{2}\plus{}b\sin \frac{B}{2}\plus{}c\sin \frac{C}{2}\geq\frac{a\plus{}b\plus{}c}{2}$[/quote]

It is just my ineq.....(In 1998).

But you do stronger:

In $\triangle ABC$,prove that:

\[\frac{a\sin{\frac{A}{2}}+b\sin{\frac{B}{2}}+c\sin{\frac{C}{2}}}{s}\geq Q \geq 1\]

$Q=\sum_{cyc}{\frac{2\sin^2{\frac{A}{2}}}{\cos^2{\frac{B}{2}}+\cos^2{\frac{C}{2}}}}$.

$s=1/2(a+b+c)$

BQ

$\cos B +\cos C$ =  $2\cos \frac{B+C}{2}\cos\frac{B-C}{2}$ $\leq$ $2\sin\frac{A}{2}$ , with equality when $B=C.$  Similar inequalities hold for $2\sin\frac{B}{2}$ and $2\sin\frac{C}{2}$.

Therefore $\sum{a2\sin\frac{A}{2}}$ $\geq$ $\sum{a(\cos B +\cos C)}$, and the right side of this inequality is

$(a\cos B + a\cos C)$ + $(b\cos C + b\cos A)$ + $(c\cos A + c\cos B)$,which is

$(b\cos C +c\cos B)$  + $(c\cos A +a\cos C)$ + $(a\cos B +b\cos A)$ and this is just $a + b + c .$ ! ! !

Equality occurs only when the angles are equal.

[quote="Marius Mainea"]Prove that in any triangle ABC is true inequality:

   $ a\sin \frac{A}{2}\plus{}b\sin \frac{B}{2}\plus{}c\sin \frac{C}{2}\geq\frac{a\plus{}b\plus{}c}{2}$[/quote]

$ (O,R),(I,r)$ is circumcircle and incircle respectively. $ AI,BI,CI$ cut $ (O)$ at $ A_1,B_1,C_1$ correspondence.

$LHS = \frac{1}{2R}(a \cdot IA_1+b \cdot IB_1+c \cdot IC_1) \ge \frac{2S_{AC_1BA_1CB_1}}{2R} = s $

Equality occurs when $(a=b=c)$ or $(\angle A = \pi ...)$.

Do
\[f(m,n)\geq(\le0)\]

$f \left( m,n \right) = \left( \sin \left( A \right)  \right) ^{m}
 \left( \cos \left( A \right)  \right) ^{n}+ \left( \sin \left( B
 \right)  \right) ^{m} \left( \cos \left( B \right)  \right) ^{n}+
 \left( \sin \left( C \right)  \right) ^{m} \left( \cos \left( C
 \right)  \right) ^{n}-8/3\, \left( 1/2\,\sqrt {3} \right) ^{m}
 \left( 1/2 \right) ^{n}\sqrt {3}\sin \left( A \right) \sin \left( B
 \right) \sin \left( C \right)
$

In acute triangle,prove that:

\[\sin^2{A}\cos{A}+\sin^2{B}\cos{B}+\sin^2{C}\cos{C}\geq \frac{8}{3}\sin^2{A}\sin^2{B}\sin^2{C}\]
BQ

[quote="Marius Mainea"]Prove that in any triangle ABC is true inequality:

   $ a\sin \frac{A}{2}\plus{}b\sin \frac{B}{2}\plus{}c\sin \frac{C}{2}\geq\frac{a\plus{}b\plus{}c}{2}$[/quote]
I can prove the case for acute $\triangle ABC$.

WLOG, let $a \leq b \leq c$. Note that $a \leq b \leq c \iff A \leq B \leq C \iff \sin \frac{A}{2} \leq \sin \frac{B}{2} \leq \sin \frac{C}{2}$ for acute $\triangle ABC$. \[\therefore \sum_{cyc} a \sin \frac{A}{2} \geq \frac{1}{3} (a+b+c) \left(\sum_{cyc} \sin \frac{A}{2} \right) \geq \frac{1}{3} (a+b+c) \left(3 \cdot \sin 30^{\circ} \right) = \frac{a+b+c}{2},\] the first inequality from Chebyshev, the second from Jensen.


[img]http://pic40.picturetrail.com/VOL298/8929090/16470363/260622079.jpg[/img]
show that
$\left( \frac{b}{c}-\frac{c}{b}\right) cos alfa-\left( \frac{c}{a}+\frac{a}{c}\right) cos beta+\left( \frac{a}{b}+\frac{b}{a}\right) cos gamma = 3$

[quote="yrhc@eht"]
show that
$\left( \frac{b}{c}-\frac{c}{b}\right) cos alfa-\left( \frac{c}{a}+\frac{a}{c}\right) cos beta+\left( \frac{a}{b}+\frac{b}{a}\right) cos gamma = 3$[/quote]

Problem is not correct:

Let be $\triangle ABC$ equilateral.$a=b=c$ and $cosA=cosB=cosC=\frac12$
$LHS=0$ but, $RHS=3$. :wink:

Also, given relation is not symmetric according to the $a,b,c$

[quote="yrhc@eht"][img]http://pic40.picturetrail.com/VOL298/8929090/16470363/260622079.jpg[/img]
show that
$\left( \frac{b}{c}+\frac{c}{b}\right) \cos \alpha+\left( \frac{c}{a}+\frac{a}{c}\right) \cos \beta+\left( \frac{a}{b}+\frac{b}{a}\right) \cos \gamma = 3$[/quote]

It is truly because $2bc\cos \alpha=b^{2}+c^{2}-a^{2}$ a.s.o. and $\sum x(y^{2}+z^{2})=\sum x^{2}(y+z)$.


[quote="songoku2006"]Prove that :
$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}=<\frac{a+b+c}{4}$[/quote]
I think this ineq is very easy:
We have:$\frac{bc}{b+c+2a}\le\frac{bc}{4(a+b)}+\frac{bc}{4(a+c)}\Rightarrow \sum \frac{bc}{b+c+2a}\le\sum \frac{a(b+c)}{4(b+c)}=\frac{\sum a}{4}$


The 2009 Putnam exam will be administered this Saturday, Dec. 5, but a few people may be taking it on Sunday. Let us have no discussion of it until Sunday evening, Dec. 6, even in general terms. Let's just keep this forum quiet until then.

At some time Sunday evening (on or before 8:00 p.m. EST) I will post 13 new topics - one for each of the 12 questions, and another for general discussion.

Regulars of this forum: I'll see you back here Sunday evening!

Just bumping the announcement for more visibility. Come here at about 8:00 p.m EST (5:00 p.m. PST) on Sunday.

What time is it by your watch, Professor Kent?

[color=green][3:47 p.m. PST. - K.M.][/color]

I'm curious, under what circumstances would anyone take the Putnam on Sunday?

The most likely reason would be a religious restriction against working or traveling on the Sabbath (Saturday). The rules and instructions go into some detail about how such a case should be handled.


group theory
abstract algebra
number theory
greatest common divisor
algebra
function
domain

Let $\displaystyle\ m$ and $n$ be integers. By $mZ$ + $nZ$ is meant the  set $\{$ $a+b \mid a\in mZ,b \in nZ$$\}$ of all sums of an element of $mZ$ and and element of $nZ$ .

how can i show that $mZ$ + $nZ$ is a subgroup of $Z$
also how can i find a generator for the subgroup $mZ$ + $nZ$ ?

First question: Show it directly: $(ma+nb)+(ma'+nb')=m(a+a')+n(b+b').$ Inverses are obvious.
Second question: Standard number theory. The gcd is the smallest positive linear combination.

Explained better:
Let $d=gcd(m,n)$.
If a number is a multiple of $m$ it is also a multiple of $d$. The same holds for $n$.
So $m\mathbb{Z}\subset d\mathbb{Z}$ and $n\mathbb{Z}\subset\ d\mathbb{Z}$.
We deduce $m\mathbb{Z}+n\mathbb{Z}\subseteq d\mathbb{Z}$.

We prove that $d\mathbb{Z}\subseteq m\mathbb{Z}+n\mathbb{Z}$.
For this we prove first $d\in m\mathbb{Z}+n\mathbb{Z}$. This is equivalent to the existance of $x,y\in \mathbb{Z}$ such as $d=mx+ny$. [b][1][/b]

Let $m=dm_1$ and $n=dn_1$. Then $gcd(m_1,n_1)=1$. So to solve (1) we must find $x,y\in\mathbb{Z}$ such as $m_1x+n_1y=1$.

Let $f:\{0,1,\ldots m_1-1\}\to\{0,1\ldots m_1-1\}$, $f(a)=n_1a (\mod m_1)$.
I prove that $f$ is injective (into).
Let $a,b\in\{0,1\ldots m_1-1\}$ such as
$f(a)=f(b)\Leftrightarrow n_1a\equiv n_1b(\mod m_1) \Leftrightarrow n_1(a-b)\equiv 0(\mod m_1) \Leftrightarrow m_1|n_1(a-b)$. Because $m_1$ and $n_1$ are coprime the later is equivalent to $m_1|a-b$. Because $0\leq a,b\leq m_1-1$ we get $a=b$. So $f$ is injective.
Because it's domain and codomain coincide we get that $f$ is bijective. So $f$ is surjective. Thus there exists ${y\in \{0,1\ldots m_1-1}$ such as $f(y)=1\Rightarrow n_1y\equiv 1 (\mod m_1)\Rightarrow m_1|n_1y-1$. Therefore there exists $-x\in \mathbb{Z}$ such as $n_1y-1=m_1(-x)\Rightarrow m_1x+n_1y=1$.

So (1) is proved and so $d\in m\mathbb{Z}+n\mathbb{Z}$. This implies $d\mathbb{Z}\subseteq m\mathbb{Z}+n\mathbb{Z}$ (because $d\in G$, where $G$ is a subgroup of $\mathbb{Z}$ implies $kd\in G$ for all $k\in\mathbb{Z}$).

So $d\mathbb{Z}=m\mathbb{Z}+n\mathbb{Z}$.


11. A 13-g bullet traveling 230 m/s penetrates a 2.0-kg block of wood and emerges going 170 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

This is what I did:

m1v1 + m2v2 = (m1 + m2)vi
(.013 kg)(230 m/s) + (2.0 kg)(0m/s) = .013 kg + 2.0 kg)vi
2.99 kg m/s + 0 m/s = 2.013 kg)vi
vi = 1.48 m/s

Thats as far as I got. I don't know what the vi is for, and I don't know where to go from there. Please help me.

I don't get how to get 0.39 m/s as an answer.

First of all, you should re-read the problem.  Momentum is conserved.

After the collision, the bullet is STILL moving.  That means it still has momentum.

Add that into your equation, and it should work out.

I re-read the problem 10 times. I finally understand what it was trying to say. Now I have the answer.

mbvb + mwvw = mbvb' + mwvw'
(.013kg)(230m/s) + (2.0kg)(0m/s) = (.013kg)(170m/s) +(2.0kg)vw'
(2.99kg m/s) + 0 m/s = (2.21 kg m/s) + 2.0kg vw'
2.99 kg m/s = (2.21kg m/s) + 2.0 kg vw'
.78kg m/s = 2.0 kg vw'
vw' = .39 m/s

Yay!!! I did it.


Support videos FTW search percent ratio inequalities

Since random, irrelevant comments about SSBB being awesome are getting all too common (actually, ever since March 9th), this is the thread to debate and talk about smash. And when saying "SSBB is awesome," try to support your reasoning.

I could start the debate topic, but I kinda have to leave soon.

PS. Don't say SSBB is better because it has Subspace Emissary, better graphics, more stages, and more character. It's ridiculously easy to counter each of those.

1. Subspace Emissary is no fun at all. It's basically a side-scrolling hack and slash. If you liked Subspace Emissary, let me tell you that there are much better games for you.

2. Remember the rule of gaming: gameplay over graphics. Also, if you play for graphics, why do you play at all? I mean, real life has the best graphics, if you get what I mean.

3&4. Quality over quantity. Enough said.

[quote="dragon96"][quote="Nerd_of_the_Ages"]...
2. Remember the rule of gaming: gameplay over graphics. Also, if you play for graphics, why do you play at all? I mean, real life has the best graphics, if you get what I mean.
...
[/quote]

Life may have the best graphics, but nothing really exciting in life seems to happen. The main reason people made video games other than having something fun to do is because you can't do things that you do in video games in life.[/quote]
Yes, but the point is, people shouldn't be playing games for their graphics. And nothing really exciting in Brawl seems to happen either. Camp, camp, camp, camp, camp, yay I outcamped you. Why not play melee instead?

Alright, first topic to explain the inferiority of Brawl over Melee.

1. Attitude of the Developers
Ok, so they know that melee is an egame (played competitively), and is very successful. It has a very deep metagame, and players benefit from quick, aggressive play. So, the developers know all this. And you know what? They decide to

1. Make it so it's not as good as an egame
2. Completely change the physics engine
3. Improve shield and air dodge
4. Get rid of hit stun (thereby making combos near impossible(

Ok, just ridiculous. Even worse, many people who don't play competitively will play with smash balls, just because it's a cool animation. This promotes camping even more, AND smash balls are random, AND there are pity balls. Ridiculous. Also, hits have lower knockback when you repeatedly use them (in melee, they also got weaker, but only the damage, not the knockback), which also discourages attacking and encourages camping.

*Tripping sir, tripping. Tripping is random (and there's no option to remove it), therefore it is bad. You might be confused by this. But think like this: in competitive play, when thousands of dollars are at stake, randomness it not good.

Things that made Melee Successful

Melee is quick paced (actually, this is true whether or not you camp; the entire physics engine is different), and players gain from attacking rather than camping. Elements such as wavedashing (removed) **, l-cancel (removed), JC grab (removed), etc. made for an extremely complex and developed metagame. The characters are better balanced (sure, a Fox can own a mew2, but not as badly as Snake vs Ganon in brawl). Also, it has no random elements (assuming you turn off items, of course). It actually takes skill to play well. And of course, hit stun makes it possible to do complex and [b]improvised[/b] combos. I might continue this in another post.

**Don't tell me that wavedashing is an unintended glitch. One, it is not a glitch. It is a combination of air dodge and momentum, both not glitches. Second, it was not unintended. The developers knew of it before it's release, but they left it in. Third, even if it was an unintended glitch, that doesn't mean it didn't benefit the metagame.

Btw, I like to call myself a "quasi-competitive player." I attempt to play at a competitive level (and fail, lol), but don't play competitively.

Wow, NoTA, I thought you loved Brawl.

I personally suck at Brawl, which is probably why I hate it and why I stick to first person shooters on the Xbox 360 instead. CoD4 and Halo 3 FTW  :P

Brawl is fun, just not as fun as melee.

Lol, first thing I thought when I saw you posted here was that you locked this topic.  :P

[quote="mewto55555"]I have found with brawl that the final smash is somewhat cheap, but I like the fact that is has more aerial combat. It allows for more technique, as opposed to the run in and hit them hard strategy that I used in SSBM[/quote]
More aerial combat? Is this a joke? I mean, please. In brawl, there's barely any hitstun, making aerial combat mostly pointless, whereas in melee, the game is based entirely on the aerial game. Also, your strategy for melee doesn't seem very efficient (face a patient guy, and you'll get shield grabbed the crap out of yourself).

"Technique" is practically nonexistent in brawl (still a bit, though). In melee, there's wavedashing, l-canceling, hyphen smash, jc shield smash, jc grab, chain grab (and no, I'm not referring to link and samus), and a plethora of other techs. In brawl, let's see what you have... Maybe ISRJ, DLX cancel, foxtrotting, and that's pretty much all there is. Additionally, brawl has so few combos (because of the physics engine, air dodge, hit stun, etc.) that they can't be improvised, which causes players to be much more predictable, the combos aren't really combos, and playing isn't as fun. So for the technique one, I think it should be pretty obvious. Also, grab is useless for nearly all characters now (with a select few exceptions, like Falco and Dedede).

In a 1-up interview of people like gimpyfish and hugs describing the competitive nature of brawl, here's my favorite quote (reworded, but relatively precise):

"Melee is like rock paper scissors; attack beats grab, shield beats attack, grab beats shield. Whereas in brawl, you just have rock and scissors."

[quote="mewto55555"]WHat do all of these terms mean?[/quote]
Ok, time to learn the way of the jedi (actually, lifeisacircle is my apprentince  :P . He knows all these things too).

Melee Exclusive Techs

Wavedashing - Basically air dodge diagonally in to the ground as soon as you jump up. Try it, and you'll see why it's called "wave"dash. It might take some practice for you to be able to get it down consistently, though. Btw, doing the same thing except right before you land is called wavelanding, but it's the same concept, just at the end of the jump, not the beginning.

l-canceling - After doing an aerial, if you time it perfectly and press l, r, or z (im not sure if z works, but I dont use it) right before you land, you will cancel all lag. It's awesome. Really. Again, it takes practice to nail this down (still working on this one).

hyphen smash* - Basically up smashing right out of a dash (this can be done in brawl, but you can do it by just using the c-stick while dashing, so it's not really a tech). In order to do this, you need to jump cancel it; basically you begin to jump so you cancel the dashing, but before you even get off the ground, you up smash. Again, you might need to practice.

jc (jump cancel) shield smash - same thing as hyphen smash, except just out of shield.

jc grab - If you look closely, the animations of dash grabs and standing grabs is different. For every character but yoshi, the animation for standing grabs is faster with less lag. In order to standing grab out of a dash, again jump but before you even lift off, grab.

chain grab - Basically grabbing over and over; not all characters have inescapable chain grabs, and some you have to predict the directional influence.

And there's way more, but that's all i listed

Brawl Exclusive

ISRJ/DSRJ - Basically on land you can jump again after an aerial (only some aerials can do this) right before you land. If the aerial can be ISRJed, that means you can repeat it indefinitely, but DSRJ means it can only be done twice.

DLX (Dante Link X; lol, some dude named it after himself) Cancel - Basically you can up smash right after a dash attack hits for Falco. So like for example, you dash attack and as soon as it touches them you can upsmash.

Foxtrotting - Practically the brawl equivalent of dash dancing, but worse. Basically you tap the control stick repeatedly in the direction you want to run. You can pivot out of it. For most characters, fox trotting is slower than dashing, and there are no characters where fox trotting is faster than dashing.

*Bonus passage: Origin of the name "hyphen smash." The most intuitive name for it would be "dash smash," but that just sounds plain stupid. But hey, "-" can be called a dash, but it's also called a hyphen! Hence the name hyphen smash.

[quote="crazychinesemaniac"]Brawl is much better because like you said NotA, Quality over Quantity. Brawl is a quality game AND it has quantity. Therefore it is greater! Easy math.[/quote]
Could you explain WHY it has more quality rather than just saying "brawl is better, therefore it is better?"

[quote="7h3.D3m0n.117"]I believe Melee is better than Brawl because it was an old-school fighting game. In Brawl - they introduced special attacks, combos, etc. that makes SSBB a textbook game. It wasn't nearly as fun as Melee because everyone just played by the book, instead of doing fun stuff like running around and doing simple melee attacks. If a person did a special attack, another person just plays by the book and pushes the button for the counter. And with online play - people will just resort to "noob" tactics - I imagine it can get very frustrating. (I don't know why MLG considers it a Pro-Circuit game).[/quote]
Actually, they didn't introduce combos, they removed them, which is why it's worse. Ok, I mean there are still combos, but you can't improvise, and THAT's what makes it like a textbook game, as you said. And special attacks were in melee, you know. What do you mean by "noob" tactics, btw? Also, people dont play online for serious matches, just casual. The lag makes it like... i dont know, driving down the highway blindfolded.

[quote="dragon96"]I haven't played Melee since Brawl came out, so I forgot what there was and what there wasn't. I dont think there were any special attacks, but wouldn't the same move get boring over and over? And plus, if you were Jigglypuff (which I don't know why you would be) and I was Kirby, we could just keep floating up and down, which makes the game rather boring. I guess the main reason why people like smash balls because of its animation, but it does help you catch up a lot if you are Marth, Ike, Link, etc.[/quote]
There ARE special attacks (I'm assuming you guys are referring to b-attacks, which are called special attacks; or maybe demon was referring to smash balls).

About not knowing why you would be jigglypuff, statistically, tierwise, entertainmentwise (my own opinion), etc, jigglypuff is far superior to kirby in melee. Higher tier, wall of pain, rest (BEST GIMP KILL MOVE EVER, ahead of even shine), etc. Also, what do you mean "just keep floating up and down?" They have to land eventually, so all you need to do is be patient and wait until an opportunity presents itself; play smart, in other words.  Also, in melee, you probably use more moves than you use in brawl. For example, dedede, you probably just spam your side-b then approach with your forward tilt (godly range) and if they get too close use you down tilt (godly knockback). And if the opportunity presents itself, chain grab a bit. In melee, there is a much wider range of strategies; you can't be overly aggressive (you'll get shield grabbed) or overly defensive (you can't wind by defense, and you'll make a mistake sooner or later). Also, you need to mix it up to be unpredictable (like a said earlier, it's like rock paper scissors; if you always play rock, they will pick up on it and play paper, whereas in brawl you just have rock and scissors, obviously you use rock).

Another reason why brawl less fun is because melee is faster paced. This is caused by a combination of many factors; different physics, wavedashing, l-canceling, etc.

[quote="theprodigy"]um...never going to happen
now stop dreaming
and in melee the top characters sometimes ko'ed at 70-80%
doesn't happen in brawl (well..except for snake :mad: )

BTW, why does every1 say meta knight's broken but no1 ever said any1 in melee was broken?[/quote]
Because no one was that good in melee :!: Not even fox or marth.

[quote="moogra"][quote="coolts"]I really think it's a bad idea to put more Sonic characters in Smash.[/quote]

Isn't that the same as putting starfox characters in game? They're all the same in most ways, but they are all different in others ways.[/quote]

Yeah but Sonic is from Sega.

Lucas and Ness make a good team. So do Meta Knight and Ike.

Uh.. Sonic... Fail. Sonic is like a total loser and a cheap player. He really can't do anything special.

[quote="starkiller2"]Super cheap lucas final destination strategy: Use PK thunder 2 towards the middle. Rinse. repeat.

Also, about sonic.
Good dair, great recovery, can pseudo-combo well, and his final smash is so good, it's why smash balls are banned from tourneys,
But otherwise he's a "total loser".[/quote]
The lucas strategy is not cheap. There is something called
1. Jumping
2. Shielding

To me, sonic's dair isn't that's great (although it is fun to use  :) ). I'll give you the recovery one. And for the comboing thing, not really.

And the final smash part was where I got annoyed enough to think that this merited a post.  :P
First of all, Sonic's final smash is NOT very good. Even with perfect control, one kill at most, and if you don't control it well, you'll find yourself off the edge. More than likely (especially if the enemy's percent is low), you won't get a kill at all. Space animals and Marth have far better final smashes.
Second of all, that is NOT why smash balls are banned. Smash balls are banned because they are RANDOM. That's like saying that food is banned because it is too powerful. No.

yeah final smashes are really cheap
but olimar's....is SO ANNOYING!!!!!!!!!!!

You know what's more annoying? TWEET TWEET! TWEET TWEET! TWEET TWEET! (That's the whistle, in case you're an idiot. Which I doubt.).

Meta knight is amazing, but his final smash sucks.  terrible knockback if the opponent has low damage.  not like other final smashes that are always a 1-hit ko.

Pitt talks too much. And consider how unpleasant his voice is, and how talkative he is.

Lol pit...spammy projectile, deflecting device, regular a-button spam, wings for a super-cheap way to always get back on stage...and so, people hate him because of how cheap he can be.

[quote="coolts"][b]Meta knight is amazing[/b], but his final smash sucks.  terrible knockback if the opponent has low damage.  not like other final smashes that are always a 1-hit ko.[/quote]

stop there :)
only noobs use final smashes

we play no smash ball, which makes the game alot more dun.

NOW it is a test of skill

and i like to play with

all the starfox caracters, kirby, metaknight, ike, pokemon trainer, pikachu, ness, and dk.

maybe a fofgot a few. o well......


Social Skills : Bob reasons that when he asks a woman for a date, she can accept or reject his request, so he assumes that he has a 0.5 probability of getting a date.  If his assumption is correct, what is the probability of getting five rejections when Bob asks five different women for dates? Is that result the correct probability that Bob will get five rejections when he asks five different women for dates?  Why or why not?

[hide="Probability"]$ \left(\frac{1}{2}\right)^{5}\equal{}\boxed{\frac{1}{32}}$[/hide]

[hide="Correct or Not"]No, because Bob's original assumption is incorrect. :P

[hide="For Real"]Probability tells you how likely something is, so I'm not sure what you mean by "correct probability".[/hide][/hide]


ARML
Princeton
college
geometry
AMC
AMC 12
AIME

I'm currently a High School Senior attempting to put together an ARML team for the Delaware Valley, NJ region.  However, we're unsure if we're part of the "Central Jersey" region.  Our team, as it stands, will be made up of students in the schools participating in the Delaware Valley Math League.  Since there exists a separate Central Jersey Math League (and googling CJML ARML comes up with a lot of hits), I would assume that the Central Jersey team is formed from members of the CJML, and thus we'd be fine making a Delaware Valley Math Team.  However, I'd like to check to make sure that our region isn't already included in Central Jersey, lest our organization efforts go to waste.  If we're included, no big deal, we can merge, but I'd like to avoid any last-minute surprises when we register our team.

I know there's a state line in between, but how far away is your group from the Lehigh Valley team? And do you go to the Lehigh University contest?

Google maps puts us about 65 miles (1:30) away from LeHigh (we're centered around Princeton, NJ).  While we haven't participated in the LeHigh contest in the past, we could definitely make it out there (the more contests the merrier...).  My only worry about joining them is that they seem to already have a lot of members (they fielded three teams last year, apparently), though, again, if the regions work out so that we should join them, it wouldn't be a problem.  Thanks for the quick response.

I would suggest contacting Steve Adrian (sladrian1@aol.com), who is the executive director of the contest, and describe to him your situation.  ARML does have teams in that area, but it is entirely possible that the area you're describing is distinct from all of the other teams in the area, in which case you should have no trouble.  Good luck!

I know that Jersey teams do Lehigh every year. I forget how many people from Jersey we have on the team, but people come from all places... I live farther than you from Lehigh, but Lehigh is basically the main PA team which happens to cover more than just PA. Doesn't Delaware County extend to... like parts near Philly? I don't know.
Anyway, good luck forming your team.

[quote="eyefragment"]Thanks a bunch for the suggestions.  I'll contact Steve and see what he thinks.[/quote]
It's probably a good idea to contact the head coach of the Central Jersey Team, Ruth Kaselis (732-356-2428). Other than that, you could start a team of mathletes who participate in your league.

[quote="neelnal"]Teams are only high school students rt? Or are middle school students allowed to participate?[/quote]
In 2000, Tiankai Liu defeated Gabriel Carroll and Sasha Schwartz in a tiebreaker to be declared the ARML individual champion, preventing Carroll from having a 4-year run as individual champion. Tiankai Liu was an 8th grader at the time, competing as a member of the San Francisco Bay A team.

The fact that Liu was placed on the A team of a multi-team organization suggests that the organizers of that team knew of his capabilities. In our multi-team organization, we have had a number of middle school students compete, mostly as members of our C team, and occasionally as members of our B team. I haven't yet placed a middle school student on the A team.

ARML does not make any explicit effort to accomodate middle schoolers. The material of ARML is high school level, in the same sense that the AMC12 and AIME are high school level. You're certainly expected to know about trigonometry, logarithms, and so forth. There are as many different selection procedures for qualifying to be a member of a team as there are teams. My guess is that few if any of those selection procedures go out of the way to accomodate middle schoolers. But no rule prevents the participation of such a student who is up to the challenge.

Some highschools have middleschool teams, others do what Kent said, it varies by highschool.

~lyra

did this team get off the ground?  If you guys need any help.. feel free to email me.

mail at jackpo dot org.


I am curious as to how to prove the 3 variable AM-GM inequality using the following identity.

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^3\minus{}ab\minus{}ac\minus{}cb)$

I can prove the general AM-GM inequality. I am just curious as to how this proof works. I looked it up, but all I found was a hint to prove

$ 2(x^3\plus{}y^3\plus{}z^3) \ge x^2y\plus{}x^2z\plus{}y^2x\plus{}y^2z\plus{}z^2x\plus{}z^2y \ge 6xyz$

I realize that the outer expressions above states the 3 variable case for the AM-GM. I am not good with inequalities.

I would appreciate any help.

The identity is already a proof. AM-GM says $ \frac {x \plus{} y \plus{} z}{3} \ge \sqrt [3] {xyz}$. Replace x with $ a^3$, y with $ b^3$, and z with $ c^3$ to get $ a^3\plus{}b^3\plus{}c^3 \ge 3abc$

I have not proved

$ 2(x^3\plus{}y^3\plus{}z^3) \ge x^2y\plus{}x^2z\plus{}y^2x\plus{}y^2z\plus{}z^2x\plus{}z^2y \ge 6xyz$

I am trying to figure out how to prove this.

I am merely suggesting that maybe

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^3\minus{}ab\minus{}ac\minus{}cb)$

could be used to prove the inequality.

Sorry if I was not clear.

Sure that approach works. You know $ a\plus{}b\plus{}c>0$, so it suffices to show that $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$

To do so, double it:

$ \iff 2a^2\plus{}2b^2\plus{}2c^2 \minus{} 2ab\minus{}2bc\minus{}2ca \geq 0$
$ \iff (a^2\minus{}2ab\plus{}b^2)\plus{}(b^2\minus{}2bc\plus{}c^2) \plus{} (c^2\minus{}2ca\plus{}a^2) \geq 0$
$ \iff (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \geq 0$

In other words:

$ a^3\plus{}b^3\plus{}c^3 \minus{} 3abc \equal{} \frac{1}{2} (a\plus{}b\plus{}c)\left( (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \right) \geq 0$

[quote="hurdler"]I have not proved

$ 2(x^3 \plus{} y^3 \plus{} z^3) \ge x^2y \plus{} x^2z \plus{} y^2x \plus{} y^2z \plus{} z^2x \plus{} z^2y \ge 6xyz$

I am trying to figure out how to prove this.

[/quote]

Solution to this inequality:

[hide="Left inequality"]
As $ (x\minus{}y)^2(x\plus{}y)\ge0$ it follows that $ x^3\plus{}y^3\ge x^2y\plus{}y^2x$
Now it's easy...[/hide]
[hide="Right inequality"]
Add $ 2xyz$ to both sides and by factoring the left side, you get:
$ (x\plus{}y)(x\plus{}z)(y\plus{}z)\ge8xyz$ which follows from AM-GM. [/hide]

Your left side proof is very nice, [b]Bugi[/b]! Can you do something similar to prove the right side?

[i](the objective is to prove AM-GM, so you cannot use AM-GM, and probably should not use something more advanced)[/i]

[quote="Bugi"]
Add $ 2xyz$ to both sides and by factoring the left side, you get:
$ (x \plus{} y)(x \plus{} z)(y \plus{} z)\ge8xyz$ which follows from AM-GM.[/quote]
[quote="TZF"]
(the objective is to prove AM-GM, so you cannot use AM-GM, and probably should not use something more advanced)[/quote]

Well, the objective was to prove AM-GM for 3 variables, but he only used AM-GM for 2 variables, like this:
$ x\plus{}y\geq 2\sqrt{xy}$
$ y\plus{}z\geq 2\sqrt{yz}$
$ z\plus{}x\geq 2\sqrt{zx}$
Multiply them all together and you get the right inequality.
For AM-GM with 2 variables, $ a^2\plus{}b^2\geq 2ab \Leftrightarrow (a\minus{}b)^2\geq 0$

Though the same approach he used for the left one works.

It's equivalent to
$ (x^2z \minus{} 2xyz \plus{} y^2z) \plus{} (x^2y \minus{} 2xzy \plus{} z^2y) \plus{} (y^2x \minus{} 2yzx \plus{} z^2x) \geq 0$,
$ z(x\minus{}y)^2 \plus{} y(x\minus{}z)^2 \plus{} x(y\minus{}z)^2 \geq 0$.


A,B,C,D are four points in a line in order, then prove that AB*CD+AD*BC=AC*BD.

[quote="shobber"]$A,B,C,D$ are four points in a line in order, then prove that $AB*CD+AD*BC=AC*BD$.[/quote]

Let $AB=x$
Let $BC=y$
Let $CD=z$

$AB*CD+AD*BC=x(z)+(x+y+z)(y)=xz+xy+y^{2}+yz=(x+y)(y+z)=AC*BD$

This can be called the Ptolemy Theorem on a straight line.


find $ x, y\in(0,1),$ if
$ log_{x} \frac{3xy}{2xy\plus{}x\plus{}y}\plus{}log_{y}\frac{3xy}{2xy\plus{}x\plus{}y}\plus{}log_{\frac{1}{2}}\frac{3xy}{2xy\plus{}x\plus{}y}\leq 3$


[quote="peregrinefalcon88"]i dont know why this picture wont work plz help me out and btw its saved in the same folder[/quote]What do you mean by "won't work"? It's fine although your picture is called Triangle_png.jpg and not Triangle.png. So either rename the image or change the \includegraphics image name.

Grr!! I repeat - what doesn't work? The code does work so you must tell me what is going wrong. Are there any errors or warnings? You must be much more specific as I can't see what is happening on your computer.

When you save your image, save it as a PNG. Else, it will name your file triangle.png.jpg.

Don't just rename your image, either, or it will not work... go into an image converter or something.

[quote="peregrinefalcon88"]sorry about that stevem  :oops: i have renamed my file Triangle.png and i have put that in the latex code and i hit the build and view button but then an error message comes up and says adobe reader cannot open the file because it has been damaged or is not supported.

and now as i just posted thoose 2 files i see that i did not change the name correctly plz tell me how to do so because when i do right click rename it does not seem to change the .jpg and if i do save as jpg is my only option[/quote]
I must point out that you haven't given us enough information. The first thing is, which operating system are you using, since this can affect how filenames are viewed? How are you saving the file? Right-click where?

Let's assume you are using Windows XP with TexnicCenter as the editor. If not, then you will have to tell us.
Firstly, you must have file extensions turned on, otherwise triangle.png.jpg appears as triangle.png which is most confusing. See [url=http://teamtutorials.com/windows-tutorials/show-file-extension-in-windows-xp]Show File Extension in Windows XP[/url] for how to do this. Then in Windows Explorer find the file, right-click on it and rename as triangle.jpg.

In Texnicenter make sure you have \includegraphics{triangle.jpg} then Build (NOT Build and View). Are there any errors in the bottom window of TexnicCenter? If not, then try View and see what happens. If there are errors report them here.


[quote="Jauchy"]Given a,b,c positive numbers .Prove that
 $\frac{a^{3}}{2a^{2}-ab+2b^{2}}+\frac{b^{3}}{2b^{2}-bc+2c^{2}}+\frac{c^{3}}{2c^{2}-ca+2a^{2}}\geq \frac{a+b+c}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271


first of all, please be nice and do not mark this as a spam
and second of all, watch this EDIT- please watch the third one first  :D
 [youtube]xEBz8hIgXhc[/youtube]

Haha! that was funny...
I didnt mark it as spam :D ... but someone will

theres a first one too.

[youtube]qo1d6ttbAq8[/youtube]

lol ! it won't let me submit it by just saying that because it is to short. blah blah blah. ok i think it is long enough.

:)

haha i actually like this one better...
i like how he shakes the other stick ppls hands  :rotfl:


limit
topology
function
real analysis
real analysis solved

Hi guys.Here is a nice problem(I think).
Let we have functional sequence ${f_{n}}(x):[a,b]{\longrightarrow}R(n\in{N})$,such that $\forall$ $n\in{N} f_{n}$ is monotonious.Let $\lim_{n\rightarrow{\infty}}{{f_{n}}(x)}=f(x),\forall x\in[a,b].$Prove that if $f$ is continious on $[a,b]$ then $f_{n}$ converges to $f(x)$ [b]uniformly[/b](sorry if it is the wrong word).

Ignore this post. It's a reference to a different problem than the one asked.

[i]The key consequence of the convergence being monotone: for any $x$, if $m>n$, then $|f_m(x)-f(x)|\le|f_n(x)-f(x)|.$[/i]

[quote="Myth"][quote="Kent Merryfield"]The key consequence of the convergence being monotone: for any $x$, if $m>n$, then $|f_m(x)-f(x)|\le|f_n(x)-f(x)|.$[/quote]
Are you sure? We only know that $f_n(x)$ is monotonous on $[a,b]$.[/quote]

Actually, what I understand is that the sequence $(f_n)_n$ is monotonic. Maybe it'a not like this, but I encountered this (in a slightly more general setting) as an exercise in a topology book.

I understood the problem the way grobber did - that for each $x, f_n(x)$ is a monotone sequence. If that's the statement this is a fairly standard and well known exercise.

But what if the problem really is the following, whether Levon meant it that way or not?

For each $n, f_n(x)$ is a monotone function on $[a,b].$
$\lim_{n\to\infty}f_n(x)=f(x)$, pointwise.
$f$ is continous on $[a,b].$

Does that force the convergence to be uniform?

I don't know whether that's true or not; all I can say so far is that I tried briefly to construct a counterexample and didn't find one.

It was clear that Levon didn't mean well known Dini theorem. I was wondered by your reply.

Yes,I mean that for each $n\in{N} f_{n}$ is monotonious function on $[a,b]$.

Take $\epsilon>0$. Since $f$ is continuous on $[a,b]$, it follows that $f$ is uniformly continuous on $[a,b]$. So we can find such $m\in\mathbb{N}$ that $|f(x)-f(y)|<\epsilon$ if $|x-y|\leq\frac{b-a}{m}$. Let $x_i=a+i\cdot\frac{b-a}{m}$, $i=0,...,m$. Since $f_n(x_i)\stackrel{n\to\infty}{\longrightarrow} f(x_i)$ for all $i$, we can find such $N\in\mathbb{N}$ that $\forall\,n>N$ we have $|f_n(x_i)-f(x_i)|<\epsilon$.
Suppose $n>N$. If $x\in[x_i,x_{i+1}]$ then
\[|f_n(x)-f(x)|\leq|f_n(x)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|<\]
\[<|f_n(x_{i+1})-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|<3\epsilon+\epsilon+\epsilon.\]

That's all.

Yes,[b]Myth[/b] that's all. :) .I have almost the same solution.


analytic geometry
complex numbers
real analysis
real analysis solved

Find $y:R\rightarrow C$, $y=y(x)$ derivable such that

$y'=(\frac{1}{2}+i)y+\frac{1}{2}|y|^2y$

where $|y|^2=y.y*$ here $y*$ is the complex conjugate of $y$

Hint
[hide]find a differential equation verify by $u=|y|^2$[/hide]

I did something related to, but not identical to, moubinool's hint. Express $y$ in polar coordinates as $y=re^{i\theta}$, where $r,\theta \in \mathbb{R}.$ Differentiate this, put it into the original differential equation, divide by $e^{i\theta}$, and match real and imaginary parts, and we wind up with:

$r'=\frac12r+\frac12r^3$

$\theta'=1.$

So I get the following solution:

$y(x)=\sqrt{\frac{Ce^x}{1-Ce^x}}\,e^{i(x-a)}$ for real constants $C, a.$

Here what I did
prove that $u'=u+u^2$ solve it, then plug $u=|y|^2$ in the initial differential
equation wich linear of order one, and solve it.


We make an equilateral triangle using the centers of the circle. This has side lenght 2 and an area of (3)^.5. The area of one of the sectors that make up the triangle is pi/6. Thus if we subtract three of these from the equilateral trangle, we will get the center area. Therefore,
(3)^.5-pi/2 (I hope that makes sense  :lol: )

[quote="surge"]Three circles are externally tangent to each other (so that each circle is externally tangent to the other circles). The radius of each circle is 1. What is the area of the section bounded by the tangent points (the center area, not in one of the circles)?[/quote]

Can anyone post a draw? That'd help....

visualization would help  :idea: . The only way that all the circles are tangent to each other is that if they are in a kind of trangle formation.

[quote="Inspired By Nature"]Take the example of three congruent coins. Put them all in a way so that they are tangent to each other. that should help understand the question. :)[/quote]

Yes, but a good picture wolud really help


What is Tetragonal Distortion in relation to coordination chemistry

tetragonal distortion is commonly called Jahn-Teller distortion.

In an octahedral complex what happens is the high energy eg orbitals are not symeetrically occupied in all of the complexes.

Jahn teller theorem gives an idea that any nonlinear molecule having degenracy will try to break its degenracy and attain lower energy.

In this way due to unsymmetrical occupation of the eg orbitals,ther is splitting among the eg orbitals themselves.

eg orbitals-d z2 and dx2-y2...

there are 2 possibilites--tertagonal elongation and tetragonal compression.

when a single electron occupies the dz2 orbital--the liggands coming along the +z and -z direction are repelled more then the corresponding liggands which approach along the +x,+y,-x and -y.
As a result of this,there are 2 long bonds along +z and -z and 4 short bonds causin the complex to be elongated along the z axis.or the molecule is tetragonally elongated along the z axis.

now when there are electrons in dx2-y2 ,it results in 4 long bonds and 2 short bonds or the molecule is tetragonally compressde along z axis.

now a case arises when there are single elctrons in botht the eg orbitals.in this casewhat happens the electrons in dx2-y2 repell the liggands more then dz2 causing the splitting of dz2 to lower energy and dx2-y2 to higher energy.

now the 1 electron in the dx2-y2 orbital is transferred to the dz2 orbital therby pairing the elctrons in the dz2 orbital.As a result of this liggands can no longer attach along the z axis there by causing the formation of a 4 coordinate square planar complex.