function
limit
logarithms
ratio
inequalities
calculus
real analysis

\[ a_1\geq a_2\geq\cdot\cdot\cdot\geq a_n >0,p_1>p_2 >\cdot\cdot\cdot >p_n,p_1\plus{}p_2\plus{}\cdot\cdot\cdot \plus{}p_n\equal{}1\] \[ F(x)\equal{}(p_1a_1^x\plus{}p_2a_2^x\plus{}\cdot\cdot\cdot\plus{}p_na_n^x)^\frac{1}{x}\]
\[ (1)a_n\leq F(x)\leq a_1(2)\lim_{x\rightarrow0\plus{}}F(x)(3)\lim_{x\rightarrow\plus{}\infty}F(x)(4)\lim_{x\rightarrow\minus{}\infty}F(x)\]

I can do part two for you, I think.

[hide]
Let

$ L = \lim_{x\to{0^ + }}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdot + p{}_na{}_n^x)^{\frac {1}{x}}$

We can see that assuming that $ L$ exists that

$ \ln(L) = \lim_{x\to{0^ + }}\frac {\ln(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x\right_}{x}$

Now seeing that direct substitution yields $ \frac {0}{0}$ since $ \ln(p{}_1a{}_1^0 + p{}_2a{}_2^0 + p{}_3a{}_3^0 + \cdots + p{}_na{}_n^0) = \ln(p_1 + p_2 + p_3 + \cdots + p_n) = \ln(1) = 0$

I used the condition on the last part.

So L'hopital's may be applied, so

$ L = \lim_{x\to{0^ + }}\frac {\ln(a_1)p{}_1a{}_1^x + \ln(a_2)p{}_2a{}_2^x + \ln(a_3)p{}_3a{}_3^x + \cdots + \ln{}_a{}_n{}_p{}_na{}_n^x}{p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x}$

Now direct substitution on the denominator yields one for the same reason as it caused an indeterminacy in the previous step.

So I could be really wrong but

$ \ln(L) = \sum_{k = 1}^{n}\ln(a_k)p_k\Rightarrow{L=\prod_{k=1}^{n}a_k^{p_k}}$
-------------------------------------------------------------------------------------------------------------------------------------------------

Now for the second part

$ L_1\lim_{x\to\infty}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^ + \cdots + p{}_na{}_n^x)^{\frac {1}{x}}$

We know by the relationship between the Ratio and Root tests that this particular limit is equivalent to

$ L_1 = \lim_{x\to\infty}\frac {a{}_1p{}_1a{}_1^x + a{}_2p{}_2a{}_2^x + a{}_3p{}_3a{}_3^x + \cdots + a{}_np{}_na{}_n^x}{p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x}$

Now dividing everything by $ a_1^x$ we get

$ L_1 = \lim_{x\to\infty}\frac {a{}_1p{}_1 + a{}_2p{}_2(\frac {a_2}{a_1})^x + a{}_3p{}_3(\frac {a_3}{a_1})^x + \cdots + a{}_np{}_n(\frac {a_n}{a_1})^x}{p_1 + p_2(\frac {a_2}{a_1})^x + p_3(\frac {a_3}{a_1})^x + \cdots + p_n(\frac {a_n}{a_1})^x}$

Now seeing that $ a_1 > a_2 > a_3 > \cdots > a_n$ we see that all but the first terms of the numerator and denominator go to zero giving us

$ L_1 = \frac {a{}_1p{}_1}{p_1} = a_1$
------------------------------------------------------------------------------------------------------------------------------------------------

Now for the last one we have that

$ L_2 = \lim_{x\to - \infty}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x)^{\frac {1}{x}}$

Now firstly letting $ \varphi = - x$ we see that

$ \text{As }x\to - \infty\Rightarrow\varphi\to\infty$

So we have that

$ L_2 = \lim_{\varphi\to\infty}(p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi})^{\frac { - 1}{\varphi}}\Rightarrow\frac {1}{L_2} = \lim_{\varphi\to\infty}(p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi})^{\frac {1}{\varphi}}$

So once again by the relation of the Ratio and Root test we have that

$ \frac {1}{L_2} = \lim_{\varphi\to\infty}\frac {\frac {p{}_1a{}_1^{ - \varphi}}{a_1} + \frac {p{}_2a{}_2^{ - \varphi}}{a_2} + \frac {p{}_3a{}_3^{ - \varphi}}{a_3} + \cdots + \frac {p{}_na{}_n^{ - \varphi}}{a_n}}{p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi}}$

Now multiplying top and bottom by $ \frac {a_n^{\varphi}}{a_n{\varphi}}$ and once again noticing the bounding inequality you should get

$ \frac{1}{L_2} = \frac {1}{a_n}\Rightarrow{L_2=a_n}\quad\blacksquare$

That is my best attempt

Any criticism welcome.

[/hide]

For the limit at zero, your work is generally correct, but you forgot that you were computing the logarithm of the limit, not the limit itself. In fact,

$ \lim_{x\to0}F(x)\equal{}a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.$ In other words, the (weighted) geometric mean.

And $ \lim_{x\to\minus{}\infty}F(x)$ is simply $ a_n,$ the minimum value. I think you have a similar issue with that one.

Note that these are all means. $ F(1)$ is the weighted arithmetic mean, $ F(\minus{}1)$ is the weighted harmonic mean. Item (1) is saying that the behave as means should behave, being somewhere between the minimum and the maximum of the $ a$'s.

For more, the classic work is Hardy, Littlewood, and Polya's [i]Inequalities[/i].

[quote="Kent Merryfield"]For the limit at zero, your work is generally correct, but you forgot that you were computing the logarithm of the limit, not the limit itself. In fact,

$ \lim_{x\to0}F(x) \equal{} a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.$ In other words, the (weighted) geometric mean.

And $ \lim_{x\to \minus{} \infty}F(x)$ is simply $ a_n,$ the minimum value. I think you have a similar issue with that one.

Note that these are all means. $ F(1)$ is the weighted arithmetic mean, $ F( \minus{} 1)$ is the weighted harmonic mean. Item (1) is saying that the behave as means should behave, being somewhere between the minimum and the maximum of the $ a$'s.

For more, the classic work is Hardy, Littlewood, and Polya's [i]Inequalities[/i].[/quote]

Wow, thank you. I did that too quick and made stupid errors.

For the first one $ \ln(L) \equal{} \sum_{k \equal{} 1}^{n}\ln\left(a_k^{p_k}\right)\Rightarrow{L \equal{} e^{\ln(a_1^{p_1}) \plus{} \cdots} \equal{} \prod_{k \equal{} 1}^{n}a_k^{p_k}}$

And even more stupidly

I forgot that I had

$ \frac {1}{L_2} \equal{} \frac {1}{a_n}\Rightarrow{L_2 \equal{} a_n}$

I didn't make any careless mistakes on the limit to positive infinity did I?

As for the inequality stuff, sorry I am ignorant to such things, I always, how do you say on this site? "Calculus Bash" everything.

Thanks Mr. Merryfield

Yes, $ F( \plus{} \infty) \equal{} a_1$ - in other words, the maximum of the $ a$'s. You did that correctly.

The statement $ F(0)\le F(1)$ is (a weighted version of) the well-known arithmetic mean - geometric mean inequality, always abbreviated here as AM-GM. In this setting, it's a special case of the statement that $ F$ is a nondecreasing function on $ [ \minus{} \infty,\infty].$

[quote="mathwizarddude"][quote="Mathstud28"]I can do part two for you, I think.

[hide]
Let

$ L = \lim_{x\to{0^ + }}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdot + p{}_na{}_n^x)^{\frac {1}{x}}$

We can see that assuming that $ L$ exists that

$ \ln(L) = \lim_{x\to{0^ + }}\frac {\ln(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x\right_}{x}$

Now seeing that direct substitution yields $ \frac {0}{0}$ since $ \ln(p{}_1a{}_1^0 + p{}_2a{}_2^0 + p{}_3a{}_3^0 + \cdots + p{}_na{}_n^0) = \ln(p_1 + p_2 + p_3 + \cdots + p_n) = \ln(1) = 0$

I used the condition on the last part.

So L'hopital's may be applied, so

$ L = \lim_{x\to{0^ + }}\frac {\ln(a_1)p{}_1a{}_1^x + \ln(a_2)p{}_2a{}_2^x + \ln(a_3)p{}_3a{}_3^x + \cdots + \ln{}_a{}_n{}_p{}_na{}_n^x}{p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x}$

Now direct substitution on the denominator yields one for the same reason as it caused an indeterminacy in the previous step.

So I could be really wrong but

$ \ln(L) = \sum_{k = 1}^{n}\ln(a_k)p_k\Rightarrow{L = \prod_{k = 1}^{n}a_k^{p_k}}$
-------------------------------------------------------------------------------------------------------------------------------------------------

Now for the second part

$ L_1\lim_{x\to\infty}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^ + \cdots + p{}_na{}_n^x)^{\frac {1}{x}}$

We know by the relationship between the Ratio and Root tests that this particular limit is equivalent to

$ L_1 = \lim_{x\to\infty}\frac {a{}_1p{}_1a{}_1^x + a{}_2p{}_2a{}_2^x + a{}_3p{}_3a{}_3^x + \cdots + a{}_np{}_na{}_n^x}{p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x}$

Now dividing everything by $ a_1^x$ we get

$ L_1 = \lim_{x\to\infty}\frac {a{}_1p{}_1 + a{}_2p{}_2(\frac {a_2}{a_1})^x + a{}_3p{}_3(\frac {a_3}{a_1})^x + \cdots + a{}_np{}_n(\frac {a_n}{a_1})^x}{p_1 + p_2(\frac {a_2}{a_1})^x + p_3(\frac {a_3}{a_1})^x + \cdots + p_n(\frac {a_n}{a_1})^x}$

Now seeing that $ a_1 > a_2 > a_3 > \cdots > a_n$ we see that all but the first terms of the numerator and denominator go to zero giving us

$ L_1 = \frac {a{}_1p{}_1}{p_1} = a_1$
------------------------------------------------------------------------------------------------------------------------------------------------

Now for the last one we have that

$ L_2 = \lim_{x\to - \infty}(p{}_1a{}_1^x + p{}_2a{}_2^x + p{}_3a{}_3^x + \cdots + p{}_na{}_n^x)^{\frac {1}{x}}$

Now firstly letting $ \varphi = - x$ we see that

$ \text{As }x\to - \infty\Rightarrow\varphi\to\infty$

So we have that

$ L_2 = \lim_{\varphi\to\infty}(p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi})^{\frac { - 1}{\varphi}}\Rightarrow\frac {1}{L_2} = \lim_{\varphi\to\infty}(p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi})^{\frac {1}{\varphi}}$

So once again by the relation of the Ratio and Root test we have that

$ \frac {1}{L_2} = \lim_{\varphi\to\infty}\frac {\frac {p{}_1a{}_1^{ - \varphi}}{a_1} + \frac {p{}_2a{}_2^{ - \varphi}}{a_2} + \frac {p{}_3a{}_3^{ - \varphi}}{a_3} + \cdots + \frac {p{}_na{}_n^{ - \varphi}}{a_n}}{p{}_1a{}_1^{ - \varphi} + p{}_2a{}_2^{ - \varphi} + p{}_3a{}_3^{ - \varphi} + \cdots + p{}_na{}_n^{ - \varphi}}$

Now multiplying top and bottom by $ \frac {a_n^{\varphi}}{a_n{\varphi}}$ and once again noticing the bounding inequality you should get

$ \frac {1}{L_2} = \frac {1}{a_n}\Rightarrow{L_2 = a_n}\quad\blacksquare$

That is my best attempt

Any criticism welcome.

[/hide][/quote]

I didn't know that in applying both the ratio and root test to a series, they yield the same limit.  However, here's a counterexample that I come up with:

Consider the divergent harmonic series $ \sum_x 1/x.$  Applying the ratio test yields the limit as $ 1$, but applying the root test yields the limit as $ \infty.$[/quote]

You misinterpreted

$ \lim_{n\to\infty}f(n)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{f(n+1)}{f(n)}$

So $ \lim_{n\to\infty}(\frac{1}{n})^{\frac{1}{n}}=1=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}$

[quote="Mathstud28"]

You misinterpreted

$ \lim_{n\to\infty}f(n)^{\frac {1}{n}} \equal{} \lim_{n\to\infty}\frac {f(n \plus{} 1)}{f(n)}$

So $ \lim_{n\to\infty}\left(\frac {1}{n}\right)^{\frac {1}{n}} \equal{} 1 \equal{} \lim_{n\to\infty}\frac {\frac {1}{n \plus{} 1}}{\frac {1}{n}}$[/quote]

How would you prove that

$ \lim_{n\to\infty}f(n)^{\frac {1}{n}} \equal{} \lim_{n\to\infty}\frac {f(n \plus{} 1)}{f(n)}$?

EDIT:
By the way, sorry; I boobed on the limit in the root test; the limit is definitely NOT $ \infty$.

[quote="mathwizarddude"]I didn't know that in applying both the ratio and root test to a series, they yield the same limit.  However, here's a counterexample that I come up with:

Consider the divergent harmonic series $ \sum_x 1/x.$  Applying the ratio test yields the limit as $ 1$, but applying the root test yields the limit as $ \infty.$[/quote]
That's simply not correct. $ \lim_{x\to\infty}x^{1/x}\equal{}1$ and there is no counterexample here.

[quote="Kent Merryfield"][quote="mathwizarddude"]I didn't know that in applying both the ratio and root test to a series, they yield the same limit.  However, here's a counterexample that I come up with:

Consider the divergent harmonic series $ \sum_x 1/x.$  Applying the ratio test yields the limit as $ 1$, but applying the root test yields the limit as $ \infty.$[/quote]
That's simply not correct. $ \lim_{x\to\infty}x^{1/x} \equal{} 1$ and there is no counterexample here.[/quote]

Yes, I realized my stupid error and have already acknowledged it in the post before yours...

(I used the sub u = 1/x and then don't know why was I so stupid as to write the new limit as $ \lim_{x \rightarrow 0} \exp(\ln|u|/u)$ which should be infinity; the correct denominator should be 1/u, of course!  Perhaps I always make these kind of silly mistakes whenever right after my dinner :blush:  :D  )

EDIT:
Now the main issue is, how are we gonna to prove the limits in using the ratio and root test are equal, if it is true indeed?

[quote="Kent Merryfield"]For the limit at zero, your work is generally correct, but you forgot that you were computing the logarithm of the limit, not the limit itself. In fact,

$ \lim_{x\to0}F(x) \equal{} a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.$ In other words, the (weighted) geometric mean.

And $ \lim_{x\to \minus{} \infty}F(x)$ is simply $ a_n,$ the minimum value. I think you have a similar issue with that one.

Note that these are all means. $ F(1)$ is the weighted arithmetic mean, $ F( \minus{} 1)$ is the weighted harmonic mean. Item (1) is saying that the behave as means should behave, being somewhere between the minimum and the maximum of the $ a$'s.

For more, the classic work is Hardy, Littlewood, and Polya's [i]Inequalities[/i].[/quote]

The main difficulty I have here is when $ p_i<0$ but in Hardy, Littlewood, and Polya's [i]Inequalities[/i] it is supposed that $ p_i>0$

If any $ p_i<0,$ then none of it works. In particular, question (1) - the one that says this acts like a mean because it's between the max and the min - that one could be false.

[quote="Kent Merryfield"]If any $ p_i < 0,$ then none of it works. In particular, question (1) - the one that says this acts like a mean because it's between the max and the min - that one could be false.[/quote]
but I don't think to question (2)(3)(4),the proofs don't work if $ p_i < 0$
to question(1),can you give an counterexample?

[quote="GaloisEva"]to question(1),can you give an counterexample?[/quote]  Basically any chose of values with $ n \equal{} 2, x \equal{} 1$ works: $ a_2 \equal{} 1, a_1 \equal{} 2, p_2 \equal{} \minus{} 1, p_1 \equal{} 2$ gives $ F(1) \equal{} 3 > 2$.


Show that if $E$ is a linear operator on the vector space V such that $E^{2}=E$ then
\[V=V_{1}\oplus V_{2}\]
where $V_{1}: =\{v\in V |Ev=0\}$ and $V_{2}: =\{v\in V |Ev=v\}$.

Both $V_{1}$ and $V_{2}$ are clearly subspaces, and their intersection is clearly trivial since the only way we could have both $Ev=v$ and $Ev=0$ is for $v=0.$ What remains to be shown is that the sum of these two subspaces is all of $V.$

Let $x\in V.$ Define $y=Ex$ and $z=x-y.$

Then $Ey=E(Ex)=E^{2}x=Ex=y$ so $y\in V_{2}.$

$Ez=E(x-Ex)=Ex-E^{2}x=Ex-Ex=0$ so $z\in V_{1}.$

There's nothing clever about this proof - just grind away at what needs to be done.


Find the condition for $u_0>0$ such that $\exists limu_n$:
  i)  $u_{n+1}=\frac{u_n}{u_{n+1}} +1$
 ii)  $u_{n+1}=\frac{u_{n+1}}{u_n} +1$

Hello Pigfly !
Do you assume the first sequence is defined so that it is positive ? I will assume it.
Let n be a natural.
1 if U(n) >= 2, then U(n+1) >= 2 and U(n+1) <= U(n)
2 if U(n) < 2, then U(n+1) < 2 and U(n + 1) > U(n).

If U takes a value >= 2, relation 1 shows that the sequence becomes decreasing and has thus a limit.
If all values of the series are <  2, relation 2 shows that the sequence is increasing; since we assume the sequence is bounded from above (by 2), it has a limit.
So the first sequence has always a limit.

I leave the second sequence to other amateurs.

Panurge.

Hello.

For the second sequence, we have always U(n + 2) = U(n).
Thus, the second sequence is U(0), U(1), U(O), U(1) ...
Thus it has a limit if and only if U(1) = U(0), i. e. U(0) = 2.

Panurge.

Hi panurge,
 Why $u_n\geq 2 \Rightarrow u_{n+1} \geq 2$ ? Can you prove it for me?
Thanks, ;)

Hello Pigfly.

So I said, I have assumed that the values of the first sequence are defined as positive.
Now, the definition implies U(n + 1) = (1 + sqroot(1 + 4 U(n)))/2.
If U(n) > 2, then the square root is > 3 and so U(n + 1) > 2.

Am I wrong ?

Panurge.

Oh, you assumed that the values of the first sequence are defined as positive. You aren't wrong . I know your ideas,$a_i >0$ it's easy.Sorry for my stupid question.Can you solve it if$ a_i<0$?


[quote="clinkmath"]Prove that if a,b,c are positive and distinct then (a+b+c)/3>(a+b+c)^(1/3)[/quote]

Let $x=a+b+c$, then we have to show that $\frac{x}{3}>\sqrt[3]{x}\leftrightarrow x^3>27x\leftrightarrow x(x^2-27)>0\leftrightarrow x>3\sqrt{3}$ which obviously is not always true. Take for example $a=\frac12.a=\frac13, c=\frac14$. But if $a,b,c$ are integers, then it is true, since $x=a+b+c\geq 1+2+3=6$ ($a,b,c$ are distinct). cheers! :D

He made a typo... He originally wrote $\frac{a+b+c}{3}>\sqrt[3]{a+b+c}$, but then he changed it to $\frac{a+b+c}{3}>\sqrt[3]{abc}$, which is AM-GM for 3 variables.

[quote="clinkmath"]Prove that if a,b,c are positive and distinct then (a+b+c)/3>(abc)^(1/3)[/quote]

Oops! Now this you get simply from AM-GM: $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$, where equality holds iff $a=b=c$. Since $a,b,c$ are distinct, strict inequality holds only. Cheers again! :D

[quote="clinkmath"]How do you prove that in general that the arithmetic mean of a set of positive numbers is greater than or equal to their geometric mean[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=69204

There's a discussion going on in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90723]this thread[/url] about various proofs for AM-GM.

Just take the log of each and use jensen's...


function
induction
algebra proposed
algebra

find all monotonic $f:R\longrightarrow R $ such that $ f(x)+f^{-1}(x)=2x$

You may find a solution, in french, here :

http://www.animath.fr/tutorat.html#Les%20cours%20de%20l'Olympiade%20fran%E7aise%20de%20math%E9matiques

See 'Les quations fonctionnelles'. This is ex. n 38.

Pierre.

Hey Sam-n, where did you find the reference 'France 1979'?

Pierre.

Dear pbornsztein I don't know French if the   solution is nice plz post it
(about refrence there is old material in iran which is collection of problem from around the world :maybe: )

First it easy to see that $f$ is a solution then $f$ is increasing and is a bijective function from $R$ onto $R$(since $f^{-1}$ is defined on $R$). It follows that $f$ is continuous on $R$.
It is easy too to verify that each function of the form $f(x) = x+d$ where $d$ is a real constant, is a solution of the problem. We will prove that there is no other.

Let $f$ be a solution. For each real number $d$ define $S_d =\{x \in R / f(x) = x+d\}$. Our goal is to prove that there is only one of the sets $S_d$ which is non-empty.
Note that it is clear that at least one of these sets is non-empty (choose $d=f(0)$).

a) First we prove that if for some $d$, we have $x \in S_d$ then $x+d \in S_d$ :
This comes from the fact that, if $x \in S_d$ then $f(x) = x+d$ and $f^{-1}(x+d) = x$. Since $f(x+d) + f^{-1}(x+d) = 2(x+d)$, it follows that $f(x+d) = x+2d$, which proves the result.

b) Now, I claim that if $d,x,y$ are real numbers, with $x \in S_d$ and $x \leq y$ then for each real number $d'<d$, we have $y \notin S_d'$ :
- If $x \leqq < x + (d-d')$, since $f$ is increasing, we deduce that $f(y) \ f(x) = x+d > y + d'$ so that $y \notin S_{d'}$.
- Then by induction, on $n$, we deduce that for each integer $n>0$, if $x \in S_d$, if $x+(n-1)(d-d') \leq y < x + n(d-d')$ then $y \notin S_{d'}$. This is true for $n=1$. Suppose it holds for $n-1$, thus since $x+d+(n-1)(d-d') \leq y+d' < x+d+n(d-d')$, and since from a) we know that $x+d \in S_d$, the induction hypothesis gives that $y+d' \notin S_{d'}$, thus from a) again $y \notin S_{d'}$, which ends the induction steps.
Since $d-d'>0$, for each each $y \geqq$ there is an integer $n>0$ such that $x+(n-1)(d-d') \leq y < x + n(d-d')$, which proves the claim.

c) Now, we prove that if $d'<d''<d$ are real numbers such that $S_d$ and $S_{d'}$ are non empty then $S_{d''}$ is no-empty too :
Let $x \in S_d$ and $x' \in S_{d'}$. Then $f(x=-)-x = d$ and $f(x') - x' = d'$. From the intermediate values theorem, we deduce that there is some $y$ such that $f(y)-y =d''$, which proves the claim.

d) We now prove our initial claim :
Suppose that there are $d'<d$ such that $S_d$ and $S_{d'}$ both are non-empty. From c), we may assume that $d'<d<0$ o $0<d'<d$. But, in the first case and from a), the set $S_{d'}$ contains elements as big as we want which is impossible from b). And in the second case, the set $S_d$ contains elements small as we want, which contradicts b) again, and we are done.

Pierre.

Oh boy, oh boy, what a terrible solution. I have a shorter one: clearly f is strictly increasing and so is its inverse. By a trivial induction we have $ f^n_{(x)}=nf(x)-(n-1)x$ for any natural n and also $ f^{(-n)}_{(x)}=(n+1)x-nf(x)$ . Take any x<y. Then from the monotony we must have $ nf(x)-(n-1)x<nf(y)-(n-1)y, (n+1)x-nf(x)<(n+1)y-nf(y) $ for any natural n. From here it follows that f(x)-x=f(y)-y and we are done.


trigonometry
algebra
polynomial
function
modular arithmetic
induction
binomial theorem

[quote="n0vakovic"]Let $\cos (\alpha)=\frac 35$. Proove that $\alpha$ is not $r\cdot \pi$ with $r$ rational.

Note that the statement is true when $\frac 35$ is replaced with any rational number different than $0, \pm \frac 12, \pm 1$.[/quote]

looks like a nice problem, but i have no idea how to tackle it :(

[quote="Rafal"]Maybe using de'Moivre's formula ?[/quote] seems like nice a idea, but i didn't succeded in proving; i tried something with the binomial theorem but came up empty

[quote="tetrahedr0n"]Here's a start:

$\cos \alpha = 3/5$ so $\sin \alpha = 4/5$.

Then, $e^{i\alpha}= 3/5 + 4/5i$. Then if $\alpha = r(\pi)$ where $r=a/b$ where $a,b$ are integers, then $e^{i\alpha}^b = (-1)^a$. So we get one of

$(3/5+4/5i)^b = (-1)^a$

and hopefully, we can show that the left hand side is always real.[/quote]

that is where i got, too

btw, you have to show that lhs is always not real

How about the following idea?  For every positive integer $n$, we know that $2 \cos(n \theta)$ is a polynomial function of $2 \cos \theta$; furthermore that polynomial is a monic polynomial with integer coefficients.  Thus, by the Rational Root Theorem, we have $2\cos(n \alpha) \neq 2$.  That means $n \alpha$ is not a multiple of $2 \pi$.  That implies the conclusion.

Why does the rational root theorem imply that $\displaystyle 2\cos(n \alpha) \neq 2$?

The function $2 \cos(n\theta) - 2$ is a monic polynomial of $2 \cos(\theta)$ with integer coefficients.  Thus by the Rational Root Theorem, if $\theta$ is a root of $2 \cos(n \theta) - 2 = 0$ with $\cos(\theta)$ rational, then $2 \cos(\theta)$ must be an integer.  But $2 \cos(\alpha)$, which is 6/5, is not an integer.  Thus $2 \cos(n \alpha) - 2$ can't be zero.

Very nice, but I don't think you can assume that the participants in this test know about Chebyshef polynomials - there must be a more elementary solution.

How about the following idea?  Show that
\[
  {(3 + 4i)}^n \equiv 3 + 4i \pmod 5
\]
for every positive integer $n$.  (Induction on $n$ should work fine.)  Thus the left-hand side can't be real.


1. Find all x,y when y^2 = x^3 +16

2. Find all x,y when 7^x - 3^y =4

Help me~~

The first one is from Italian (96' ?) and the second is from India I believe , ok solution to 2. :
$ 7^x \minus{} 3^y \equal{} 4$ ,note that $ 7^k\equiv ( \minus{} 1)^k (mod4)$ and $ 3^l\equiv ( \minus{} 1)^l ( mod 4)$
1) $ k,l$ are even
$ 7^{2k} \minus{} 3^{2l} \equal{} 4$ , $ (7^k \minus{} 3^l)(7^k \plus{} 3^l) \equal{} 4$ , it has no solution ,
2) $ k,l$ are odd
$ 7^{2k \plus{} 1} \minus{} 3^{2l \plus{} 1} \equal{} 4$< -> $ (7 \minus{} 3)(7^{2k} \plus{} 7^{2k \minus{} 1}*3 \plus{} ... \plus{} 7*3^{2l \minus{} 1} \plus{} 3^{2l}) \equal{} 4$
it's easy to notice that the eqation has only solution when $ k \equal{} l \equal{} 0$ <=> $ x \equal{} y \equal{} 1$ because for $ k,l > \equal{} 1$ term $ (7^{2k} \plus{} 7^{2k \minus{} 1}*3 \plus{} \plus{} 7*3^{2l \minus{} 1} \plus{} 3^{2l}) > 1$

The solution of the first one you can find in my first post.

edit: ah , what a stupid mistake ..  I don't know what i was thinking., so sorry  . :(

[quote="Pablo09"]
$ 7^{2k \plus{} 1} \minus{} 3^{2l \plus{} 1} \equal{} 4$< -> $ (7 \minus{} 3)(7^{2k} \plus{} 7^{2k \minus{} 1}*3 \plus{} ... \plus{} 7*3^{2l \minus{} 1} \plus{} 3^{2l})$
[/quote]this is wrong...
for solution,see: [url]http://www.mathlinks.ro/viewtopic.php?t=150672[/url]


[quote="worthawholebean"]The sum of $ 18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$ \textbf{(A)}\ 169 \qquad \textbf{(B)}\ 225 \qquad \textbf{(C)}\ 289 \qquad \textbf{(D)}\ 361 \qquad \textbf{(E)}\ 441$[/quote]

This might not be the fastest way, but this is how the equation would look like: $ 18x \plus{} 153 \equal{} y^2$.  You can guess and check for $ y$, so lets start with the first one to get the least one.

(A) $ 18x \plus{} 153 \equal{} 169 \implies 18x \equal{} 16$ which makes $ x$ a decimal, so it is not this.

(B) $ 18x \plus{} 153 \equal{} 225 \implies 18x \equal{} 72 \implies x \equal{} 4$.  So the answer is $ \boxed{B}$

[hide] We let the first number be x so the last number is x+17. Summing the arithmetic series gives the sum as $ \frac{(2x\plus{}17)(18)}{2}\equal{}9(2x\plus{}17)$. For this to be a perfect square, 2x+17 must be equivalent to $ y^2$ for some value of y. Clearly $ 2x\plus{}17>17$ so the smallest value that works is 25, for y=5 and x=4. Our answer is $ 25\cdot9\equal{}225$, so $ \text{(B) 225}$.[/hide]

Let first number be n-9, and last number n+8. The sum will be 18n-9=p. n has to be an integer. Check the answers. :lol:


algebra
polynomial
superior algebra
superior algebra unsolved

find the degree of extension of Q($ \sqrt{3\plus{}\sqrt(8)}$

 for simplicity let the above be denoted as $ Q(\alpha)$. it is easy to see that $ \alpha$ is algebraic over Q. infact a polynomial which has the same as a root is $ p(x)\equal{}x^4\minus{}6.x^2\plus{}1$. since $ \alpha$ is algebraic , i concluded that the degree of the extension is equal to the degree of the minimal polynomial of $ \alpha$. but over Q the polynomial factors as $ (x^2\plus{}2.x\minus{}1).(x^2\minus{}2.x\minus{}1)$. so minimal polynomial has degree less than 4. it cant be 1 and it cant be 2 also, and i think it cant be 3 since p doesnt have roots in Q. i find this very strange..may be i made a stupid mistake but can someone tell me where i went wrong or what is the answer? thanks

$ \sqrt{3 \plus{} \sqrt{8}} \equal{} \sqrt{3 \plus{} 2 \sqrt{2}} \equal{} 1 \plus{} \sqrt{2}$.  The degree of the extension is $ 2$ and the correct minimal polynomial is $ x^2 \minus{} 2x \minus{} 1$.


inequalities geometry rectangle perimeter

Can someone please explain in what sense the AM-GM inequality, which asserts $ \frac {x \plus{} y}{2} \geq \sqrt {xy}$, is equivalent to the statement, "Of all rectangles with a given perimeter, the square has maximal area?"  Thanks.

I'm sorry, I'm not following your argument.  When I see a claim that two statements, say $ p$ and $ q$, are equivalent, I expect the equivalence proof to show first that $ p \implies q$ and then that $ q \implies p$.

If you could structure your explanation that way, or even provide the underlying intuition for why each implies the other, that would be great.  Thanks.

[quote="t0rajir0u"]Suppose you are given a rectangle with side lengths $ x, y$; thus its perimeter is $ 2x \plus{} 2y$.  Among all rectangles with this perimeter, the claim that the one with side lengths $ \frac {x \plus{} y}{2}, \frac {x \plus{} y}{2}$ maximizes the area is equivalent to the inequality $ \left( \frac {x \plus{} y}{2} \right)^2 \ge xy$ since the LHS is the area of the maximal rectangle and the RHS is the area of some other rectangle.[/quote]

That works, thank you.


[quote="SorrowMan"][quote="mailo"][b][u]Ασκηση 1[/u][/b]

Να βρεθει το οριο της ακολουθιας $ a_n \equal{} \frac {n}{3} \minus{} \left[\frac {n}{3}\right]$

[b][u]Ασκηση 2[/u][/b]

Να βρεθει το οριο της ακολουθιας $ a_n \equal{} \left(1 \plus{} \frac {3}{n}\right)^n$

[b][u]Ασκηση 3[/u][/b]

Να βρεθει το οριο της ακολουθιας $ a_n \equal{} \left(4 \plus{} \frac {1}{n}\right)^n \left(\frac {1}{4} \plus{} \frac {1}{n}\right)^{2n}$[/quote]

Xoris molivi kai xarti apla kanontas kati prakseis sto mialo mou , to prwto den prepei na iparxei...Diorthose me an kanw lathos...Pira 2 ipakolouthies  n = 3k , n = 3k+1  . Gia n=3k vgenei oti tinei sto 0 enw gia n=3k+1 vgenei oti pigenei sto 1/3 ..Ta alla den ta kitaksa akoma![/quote]
Στηρίξου στο γεγονός ότι
$ x_n \equal{} \frac {\left(1 \plus{} \frac {x}{n}\right)^n}{e^x}$καθώς και κάθε υπακολουθία της και θα βρεις για όριο της [2] το $ e^3$ και για την [3] αφου βγάλεις κοινό παράγοντα το $ 4^n$ και το $ \frac {1}{4^{2n}}$ όριο το $ 0.e^{\frac {1}{4}}.e^8 \equal{} 0$


Let $\large a;b;c;d > 0$.
$\large \sum \frac{a}{a^2+ab} \leq\ \sum \frac{1}{a+b}$

yes it's true...it's very true.
because $\frac{a}{a^2+ab}=\frac{1}{a+b}$
so it's very very true :blush: .


inequalities
LaTeX
algebra unsolved
algebra

[quote="Rijul saini"]Prove that there exist two infinite sequences(An)n>=1 and (Bn)n>=1 of posive integers such that the following conditions hold simutaneously :

(i)1<a1<a2<a3<.......
(ii)an<bn<(an)^2, for all n>=1
(iii)an-1 divides bn -1, for all n>=1
(iv)(an)^2-1 divides (bn)-1 for all n>=1[/quote]

how can (ii) and (iv) hold together? an<bn<(an)^2 implies an-1<bn-1<(an)^2-1 and (an)^2-1 is a factor of bn-1 despite being greater than bn-1 and both being +ve integers?

pardon me if i misread or misinterpreted your question.

the sum is given wrong.
the last condition is that (an)^2-1divides(bn)^2-1.

Think of an example in which the conditions r applicable.......
But simplify the datas first :)
I ve completed it u try and post the solution otherwise i will post it :D

$ \text{\LaTeX}$-ed

Prove that there exist two infinite sequences$ (a_{n})_{n\ge1}$ and $ (b_{n})_{n\ge1}$ of positve integers such that the following conditions hold simutaneously :

$ (i)1<a_{1}<a_{2}<a_{3}<.......$
$ (ii)a_{n}<b_{n}<a_{n}^{2}$, for all $ n\ge1$
$ (iii)a_{n}-1$ divides $ b_{n} -1$, for all $ n\ge1$
$ (iv)a_{n}^{2}-1$ divides $ b_{n}^{2}-1$ for all $ n\ge1$

[quote="rt_08"]hey could this solved by weirstrass inequality[/quote]

It will be good if u elaborate


Notice how you can pair up the terms like this:
\[ (1 \minus{} 2) \plus{} (3 \minus{} 4) \plus{} (5 \minus{} 6) \plus{} \dots
\]
Each term in the parentheses is just $ \minus{} 1$, and if $ n$ is even, then the sum is simply $ \minus{} \frac n2$.

If $ n$ is odd, then the sum is the sum of the previous $ n \minus{} 1$ terms plus $ n$, or $ \minus{} \frac {n \minus{} 1}2 \plus{} n \equal{} \frac {n \plus{} 1}2$.

$ 1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \minus{} 6 \plus{} 7 \minus{} 8 \plus{} 9\cdots\pm{n} \equal{} (1 \minus{} 2) \plus{} (3 \minus{} 4) \plus{} \cdots \equal{} 1 \plus{} 1 \plus{} \cdots$.

So we have two cases:
Case 1: n is even
In this case we can see that there are just $ \frac {n}{2}$ pairs of positive and negative integers (1 and -2, 3 and -4, etc.). Thus we have $ \boxed{ \minus{} \frac {n}{2}}$.
Case 2: n is odd
In this case, we can see that there are $ \frac {n \minus{} 1}{2}$ pairs, and then an $ n$ following. Thus we have $ n \minus{} \frac {n \minus{} 1}{2} \equal{} \boxed{\frac {3n \minus{} 1}{2}}$.
So to generalize this, we have $ \boxed{ \minus{} \left\lfloor{\frac {n}{2}}\right\rfloor \plus{} 2n\left\{\frac {n}{2}\right\}}$ where $ \{\}$ denotes the fractional part of a number and $ \lfloor\rfloor$ denotes the greatest integer less than the number.

edit: beaten...

[quote="math154"]$ 1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \minus{} 6 \plus{} 7 \minus{} 8 \plus{} 9\cdots\pm{n} \equal{} (1 \minus{} 2) \plus{} (3 \minus{} 4) \plus{} \cdots \equal{} \color{red} \minus{} 1 \minus{} 1 \minus{} \cdots$.

So we have two cases:
Case 1: n is even
In this case we can see that there are just $ \frac {n}{2}$ pairs of positive and negative integers (1 and -2, 3 and -4, etc.). Thus we have $ \boxed{ \minus{} \frac {n}{2}}$.
Case 2: n is odd
In this case, we can see that there are $ \frac {n \minus{} 1}{2}$ pairs, and then an $ n$ following. Thus we have $ n \minus{} \frac {n \minus{} 1}{2} \equal{} \boxed{\frac {\color{red}{n \plus{} 1}}{2}}$.
So to generalize this, we have $ \boxed{\color{red}( \minus{} 1)^{n \plus{} 1}\left \lceil \frac n2 \right \rceil}$ where $ \lceil x \rceil$ denotes the least integer greater than or equal to $ x$.
edit: beaten...[/quote]

Not only beaten, but also incorrect. The fixes are shown in red above.

Meh, sometimes it helps to be sane.

Those mistakes were very very sad... :P

:furious: You are implying that I'm insane? :furious:

OK, maybe I am! :D

so maybe you are.   Could somebody help me out what is the name of that formula
(n*n+1)/2


Did you [b]NOT[/b] vote for Bush?  Check out the gallery in
http://www.sorryeverybody.com/

 :D  :D  :D

They're all sorry that they didn't vote for Bush even though Bush is better than Kerry.  ;)

I didnt start this topic to debate whether or not bush is better than kerry.

while i dont necessarily agree, this is a quite entertaining site. i encourage all kerry-ites to see and add to it, and i encourage all bush-fans to view it and poke fun as what they might interpret as over-zealousness.

Hehe, that's actually pretty funny, regardless of one's political opinion.

meh i did a search on this board before i posted, but your post didnt show up

I looked at it breifly but didn't understand what it was.

[quote="ComplexZeta"]Kerry supporters are apologizing for the collective stupidity of Bush supporters who happen to inhabit the same 9,629,091 square kilometers.[/quote]And they do a really funny job :) Elections are soon cooming up in Romania, and people have started already preparing a site like this in case the same governing party is preserved :) (although I doubt anyone in the world would actually care that much :D:D).

[quote="ComplexZeta"]Kerry supporters are apologizing for the collective stupidity of Bush supporters who happen to inhabit the same 9,629,091 square kilometers.[/quote]

Come now, this has been a fun-loving thread so far.  We don't need another debate thread; we have plenty already, not counting those that started as one topic or another and degenerated into a "my candidate was better than yours, despite/because of the election's outcome" shout-fest.

Notice that I didn't actually say anything that wasn't absolutely true, at least if read as I had intended it.


A set $M$ of points in the plane will be called obtuse, if any 3 points from $M$ are the vertices of an obtuse triangle.

a.) Prove: For each finite obtuse set $M$ there is a point in the plane with the following property: $P$ is no element from $M$ and $M \cup \{P\}$ is also obtuse.

b.) Determine whether the statement from a.) will remain valid, if it is replaced by infinite.

a) Note that if $A,B$ are two distinct points of the plane, and $P$ is any point which does not belong to the (closed) band $D(A,B)$ delimited by the lines through $A$ and $B$ respectively, and both perpendicular to the line $(AB)$ then the triangle $ABP$ is obtuse (in $A$ or in $B$ according to the greatest distance between $AP$ and $BP$).

Let $M$ be an obtuse finite set. Wlog, we may assume that $|M| \geq 2$.
Let $A_1,...,A_n$ be the elements of $M$.
Since there are a finite number of points in $M$, they determine by pair a finite number of directions. Thus, there are a finite number of directions each of which is orthogonal to the direction of some line $(A_iA_j)$ where $i<j$.
Let $ \Delta$ be a line which is orthogonal to none of the lines $(A_iA_j)$ for all $i<j$
Then, each of the bands $D(A_i,A_j)$ defined as above can cover at most a bounded segment of $\Delta$. Thus, since there are a finite number of bands, we may find a point $P \in \Delta$ which belongs to none of these bands. It follws that $P$ does not belong to $M$ and, from the initial remark, that each of the triangles $PA_iA_j$ is obtuse in $A_i$ or in $A_j$.
Thus $M \cup \{P \}$ is an obtuse set.

b) No. Choose $M$ to be all the points of the line $d$.
Thus, any three elements of $M$ form a degenerated triangle, which is obviously obtuse....and if $P$ is any point of the plane which does not belong to $d$, then we may find $A,B \in d$ such that $ABP$ is equilateral, from which we deduce that $M \cup \{ P \}$ is not an obtuse set.

Pierre.


real analysis
calculus
integration
function
topology
complex analysis
rudin

There's a substantial difference in level between the two books. My understanding of the place of "green Rudin" is that it is intended for a course that presumes that a student has already seen an undergraduate level one-semester introduction to the Lebesgue integral. Bartle might be a book appropriate for that earlier course.

As for the structure of Rudin: the first chapter of that book is a fairly slick construction of the integral, and the usual convergence theorems (MCT, Fatou's Lemma, DCT) - on the assumption that you have a measure space. What's logically missing from that first chapter is the existence of any nontrivial examples of measure spaces, or any connection between measure spaces and topological spaces.

That's left to the second chapter. The heart of that chapter is a many-page proof of a version of the Riesz Representation theorem: given a positive linear functional on the continuous compactly supported functions on a locally compact topological space, there exists a Borel regular measure on that topological space such that the linear functional can be realized by integration against that measure. As a corollary, the Riemann integral provices such a positive linear functional on $C_{c}(\mathbb{R}),$ and hence there exists Lebesgue measure.

The one thing I'm sure of about that presentation - that shouldn't be your first introduction to Lebesgue measure on the line. Far better that you have some familiarity with a more concrete construction before you tackle that.

For my own sake, when I teach such a course, I choose a different book, one truer to my own roots: Wheeden and Zygmund. The unique spin of Wheeden and Zygmund is that the theory is developed specifically in $\mathbb{R}^{n}$ to begin with, and the integral of a positive function is defined to be the measure in $\mathbb{R}^{n+1}$ of the set beneath the graph.

Another commonly used book that's been around for a long time is Royden. You might want to look at that, as well.

"Real Analysis" by Folland is a good, although terse, textbook.

halmos -- measure theory is also very good

although it's a bit old, so it is possible you see very different things than given in class :)

If you want a new book, you could try Stein's "Measure Theory, Integration, and Hilbert Spaces." The presentation in that book seems to be more like what Prof Merryfield would suggest as an intro to Lebesgue integration theory: it begins with a construction of the Lebesgue measure in $\mathbb{R}^{n}$, followed by construction of the Lebesgue integral and some results about differentiation. Then he gives an intro to Hilbert spaces by talking about $L^{2}(\mathbb{R}^{n})$, and the book concludes with a couple of chapters on general measure theory and Hausdorff measure & dimension.

The book has very good problems too. I recommend it. (I have to admit that I'm a bit biased, though.)

[quote="abdul"]I'm unexposed to this subject. I am thinking of working with these:
Real & Complex Analysis - Rudin
Theory of Measure and integration - Bartle.
What book do you recommend?[/quote]
Here is 4 books i recommend you:
http://ndmanh.wordpress.com/mathematics/


Given $2n$ cards labeled from $1,2,\cdots,2n$. We divide it half $(1,2,\cdots,n+1)$ and $(n+2,n+3,\cdots 2n)$ and combine these cards as follows $(1,n+2,2,n+3,3,n+4,\cdots,n+1,2n)$ the same suffling method has been repeated until the who deck is again in order $(1,2,3,\cdots,2n)$. Determine how many ways we have to split the deck and collect it.

Original problem was to determine the cases $2n=52$ and $2n=234$.

[quote="MeKnowsNothing"]Given $2n$ cards labeled from $1,2,\cdots,2n$. We divide it half $(1,2,\cdots,n+1)$ and $(n+2,n+3,\cdots 2n)$ and combine these cards as follows $(1,n+2,2,n+3,3,n+4,\cdots,n+1,2n)$ the same shuffling method has been repeated until the whole deck is again in order $(1,2,3,\cdots,2n)$. Determine how many ways we have to split the deck and collect it.

Original problem was to determine the cases $2n=52$ and $2n=234$.[/quote]

The first and last cards in the deck are stationary, so we may ignore them. Let us label the positions of the other cards from 1 to 2n-2. After one shuffle, the card in position p moves to position 2p (mod 2n-1). So, after s shuffles, the card in position p will move to position (2^s)*p (mod 2n-1). Thus, the number of shuffles required to return the deck to its original order is equal to the least positive integer s such that 2^s = 1 (mod 2n-1). If the deck has 52 cards then 8 shuffles are required, and if it has 234 cards then 29 shuffles are required.


Prove that for any natural numbers m,n the following quantity is an integer: gcd(m,n).C(n,m)/n


The first term of an arithmetic series of consecutive integers is $ k^2 \plus{} 1$. The sum of $ 2k \plus{} 1$ terms of this series may be expressed as:

$ \textbf{(A)}\ k^3 \plus{} (k \plus{} 1)^3\qquad
\textbf{(B)}\ (k \minus{} 1)^3 \plus{} k^3\qquad
\textbf{(C)}\ (k \plus{} 1)^3\qquad \\
\textbf{(D)}\ (k \plus{} 1)^2\qquad
\textbf{(E)}\ (2k \plus{} 1)(k \plus{} 1)^2$

[hide]

$ (k^2\plus{}1)\plus{}(k^2\plus{}2)\plus{}\cdots \plus{} (k^2\plus{}2k\plus{}1) \equal{} (2k\plus{}1)(k^2)\plus{}(2k\plus{}1)(k\plus{}1)\equal{}(2k\plus{}1)(k^2\plus{}k\plus{}1)$, which is the factorization of $ k^3\plus{}(k\plus{}1)^3$. $ \boxed{\textbf{A}}$
[/hide]

Kinda similar
[hide]The $ 2k\plus{}1$th term is $ k^2\plus{}2k\plus{}1$
So the sum is $ \frac {[(k^2\plus{}1)\plus{}(k^2\plus{}2k\plus{}1)](2k\plus{}1)}{2}$
$ \equal{}\frac {(2k^2\plus{}2k\plus{}2)(2k\plus{}1)}{2}$
$ \equal{}(k^2\plus{}k\plus{}1)(2k\plus{}1)$
$ \equal{}2k^3\plus{}3k^2\plus{}3k\plus{}1$
$ \equal{}k^3\plus{}(k\plus{}1)^3$

or $ \boxed A$[/hide]


Please post A proof for Muirhead inequality .
I by myself tried to prove it By using Karamata inequality but i couldnt reach a satisfactory proof.

I think we can use the idea of compression.
http://www.mathlinks.ro/Forum/viewtopic.php?p=140365#140365


How do I prove that some powers of 2 exist such that the first term is 2002 or 7, or 77777, or 8?

Can someone help?

[hide="Hint"]Consider what it means when you take the logarithm.[/hide]


function
inequalities proposed
inequalities

Find the maximum of $ x_{1}^{3} \plus{} \ldots \plus{} x_{10}^{3}$ for $ x_{1},\ldots,x_{10}\in[ \minus{} 1,2]$ such that $ x_{1} \plus{} \ldots \plus{} x_{10} \equal{} 10.$

D. Mitin

Maximum will be $ 47,5$. When $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}\minus{}0.5$ and $ x_5\equal{}x_6\equal{}x_7\equal{}x_8\equal{}x_9\equal{}x_{10}\equal{}2$. Because function $ f(x_1,\ldots,x_{10})\equal{}x_1^3\plus{}\ldots\plus{}x_{10}^3$ has minimum when $ |x_1|\equal{}\ldots\equal{}|x_{10}|$. This means that maximum will be when some of the $ x_{i}$ takes it's critical value. In our case this is $ \minus{}1$ or $ 2$. Obviously We should maximize the number of $ x_{i}\equal{}2$.


Yo I got an SAT score of 2120. Can I get into MIT? I'm an eighth grader looking into skipping college. I got a 3 this year on the AIME.

I made AIME.

I got 5th place at mathcounts states.

Will I get in? Can I send in a late application?

if you mean skipping high school, i hope ur joking

and, no offense, a 3 on the AIME in 8th grade is nothing spectacular

I can't tell to what extent this post is a joke, but I'm going to take it seriously in case anyone out there seriously thinks this is feasible.

First, selective schools are [b]very[/b] reluctant to accept younger applicants.  Even juniors looking to skip senior year must work harder to demonstrate that they have the mental and emotional maturity necessary for college.  Even extremely exceptional students are rejected on maturity grounds.

Second, a 2120 and a 3 are (sorry to say) not that impressive in the grand scheme of things.  A local 7th grader around here recently got a 10, and there are 8th graders who get 2400s.

Finally, in any case, late applications aren't accepted under any circumstances, as far as I know.

If you really are looking to skip high school (and I would not recommend it), contact nearby state schools or community colleges; some of them may have early-entrance programs targeted directly at students your age (although the application deadlines have probably passed).

I'm currently finishing Precalc. That has to be worth something.

What about my fifth place finish in Mathcounts?

I got a 24 on the AMC 8.

I got a 30 on CML and a perfect on MOEMS.

I thought I actually could skip high school. :o

I'll probably go to high school next year and apply for MIT next fall

Didn't you read what DPatrick said? Sarcasm just doesn't work on the Internet...

ok, i got 1st place written and cd this yr at mc states, this will be my third yr going to mc nats, 2100 SAT, 6 AIME.

I'm not even thinking about skipping a grade, let along high school altogether.  I doubt you're academically ready for college, you must think about other subjects too.  and also, are you ready socially?

I still really hope this a joke...

[quote="DonkeyKingKong"]I'm currently finishing Precalc. That has to be worth something.[/quote]

My 8th grade precalc class had 11 kids in it, one of whom attended college after his junior year and one of whom entered an early-entrance program at the University of Washington.  The other 9 (including myself) are graduating on time.  The 7th grader I told you about is currently studying AP Calculus.

You're well on your way to being a fine candidate [i]in four years.[/i]  Slow down a bit.

I hope this isn't a serious question. No, there are a lot of other AoPSers with similar or even better scores at the same age, and many of them wouldn't dream of skipping high school. Instead, they plan carefully to make high school challenging and worthwhile.

See

[url=http://www.mitadmissions.org/topics/before/recommended_high_school_preparation/many_ways_to_define_the_best.shtml]The Many Ways to Define the Best[/url]

for a helpful statement about MIT's admission policies, including the important concept of the "next logical step."

No one can tell if you're serious, but in case you are, the fundamental question is:

Why on earth would you want to skip high school?  Whatever reasons you might have, I'm sure there are less drastic ways to address them.

Imho, the only good reason to skip high school is that there is absolutely nothing you can learn from high school, and even that might be a necessary but not sufficient reason.

Am I the only one amused that one 8th grader is asking about being admitted to MIT *right now*, while other 8th graders with similar credentials are asking about being admitted 4 years from now?  (Or maybe that's DonkeyKingKong's joke?  I can't tell.)

[quote="Quickster94"]this was a joke right?[/quote]I'm not sure if it was or not, but it had died out, at least until you revived it for no reason.  So now it's locked to make sure that it doesn't happen again.


Bob can type at a constant rate of 50 words per minute and Rob can type at 100 words per minute. If they work together to efficiently type a document with 1500 words, how long will it take them (in minutes) to type the document?

[hide]$10$ minutes:
Bob does $500$ and Rob does $1000$. [/hide]

[hide]Add how much they can do in a minute. Thats 150. 1500/150=10 so it would be 10 minutes.[/hide]

[quote="ckck"][hide]Add how much they can do in a minute. Thats 500. 1500/500=10 so it would be 10 minutes.[/hide][/quote] Actually, added up, their work is 150 words per minute but your answer is still right.

[hide="you know you want to click here, you know you want to..."]10mins[/hide]lol

[hide="answer"]10 min. because you just add there word typing speeds together, which is 150. Then you take 1500 and divide 150. Answer is 10[/hide]


Somebody recently claimed that the majority believed that "crap" is a swear word.  However, she provided no factual evidence, so I have taken the responsibility of running an actual poll.

Keep in mind that while 'crap' may not be the best thing to say to your first-grade english teacher, the question is whether or not the word falls into the same category as certain unspeakable words that get changed into - s when you type them here.

glad this is at the round table. I posted it in fun and games and tare instantly deleted it.  Either hes really offended by it or was afraid to see the results come in.

Oh by the way, i believe that certain person is a he.

I think he's a she, and no, crap is not a swear word. It's not even vulgar.

Whos the person that your talking about?

hint: it is a moderator.  I remember somewhere he said he was a he.

[quote="Tare"]haha very funny BHorseMath...

I [b]was [/b]offended by it along with mathfanatic's persistant editing of his posts so it would say c***, and I'm getting really irritated, and you guys probably never seen me when I get [b]really [/b]irritated...

Hmph I see that you're now using your signature space to insult other people, go ahead, you're just hurting your own integrity.

Might I mention no matter what the outcome of any poll c*** will still be considered a swear unless you get the concent of [b]Fun and Games[/b] mods or the admins? Might I also mention you have every right to say any swear in other boards? (well that may not be the case but it;s not up to me to decide what is a swear and what isn't outside of F&G and MATHCOUNTS)[/quote]

None of them said they supported you tare.  O well.  Maybe they havent seen that thread yet.  By the way, you're a boy right?

What did I do Tare?  I was completely serious.

DOESN'T THIS THREAD HAVE A TOPIC?  STICK TO IT!

while i don't think that the word in question is so bad, it is not a word that anyone would use in the classroom, and I think it should be considered a swear word.

[quote="BHorseMath"]What did I do Tare?  I was completely serious.[/quote]
Oh, sorry, they were just talking about me.

This topic is not worthy of being stuck to, after all this poll means nothing, absolutely [b]nothing[/b].

If we taunt her enough, perhaps she'll get so cross she'll make a mistake.
-Monty Python

Why do you need to say "crap" anyways? It won't kill you (I hope) to avoid it. Not that I think that its a swear.

I usually say another word, but that would get censored.

I see many more interesting ideas to discuss in this world that what the grand list of swear words includes.

I'll name a few:
Is 2 an optimal number of eyes?
Why have no new body types evolved on Earth for hundreds of millions of years?
If life developed on another planet, would it have dna?
Is it worth the cost of sending a human to Mars anytime soon?
How long will it be before computers have the power of a human brain?
How do I learn how to solve every IMO problem in an hour and a half (er, I still can't)?

crap is listed as sometimes vulgar for the definition of stuff and vulgar for the defition of turd.

In any event,

[quote]a.) The majority of us thought that the "c-word" is a swear[/quote]

is deemed a falsehood.

Fine, it is incorrect, I admit I was wrong. Does this change anything? No. but if it gives you self-satisfaction then I guess our time and effort have gone into a good cause...I guess :)

[quote="Tare"]Does this change anything?[/quote]

Yeah, it undermines your argument that the word 'crap' should be banned because most AoPSers believe it's a swear word.  But I guess undermined arguments aren't that important, eh?

Back to work.

Fun list, MCrawford. I especially found the DNA one interesting.

What I want to ask all of you guys is, what's so bad about using "swear words"? I know that people can lose respect for you due to the mere vulgarness and meanness that comes with using them. However, that can happen with any sequence of words, so why define a specific set that may not be used?

The only reason I can think of is that some people's parents do not like to see Bad Words online, and therefore will not let their kids use the site. That's a perfectly good reason to ban them.

But is that the reason?

Mike Nolan, Missouri.

i was doing this report on idioms and i found a ton with s*** and some of them are considered vulgar and some of them aren't so does that mean if i say them to someone i'm not swearing.

[quote="mpn"]But is that the reason?[/quote]
I believe so, yes, and this should be especially enforced in boards where younger members might go to like Getting Started or MATHCOUNTS. Besides swearing is unnecessary and there's lot's of acceptable alternatives for each one like "poo" or "dumb" :lol:


I checked a few values, and it's not true.  Maybe there are more conditions.

[quote="borislav_mirchev"]I would like to ask you how to prove the following conditional identity:

If A,B,C,D,E >0 and A+B+C+D+E=180
Then:
sinB.sinC.[sin(B+C).sin(E+C) - sin(A+D).sinC] =
sinA.sinE.[sin(A+E).sin(E+C)-sin(B+D).sinE]

All the angles are in degrees.[/quote]

I'm sorry A,B,C,D,E are angles of an inscribed pentagram in this order.
I used trigonometry form of Cheva to prove a problem posted here related to an pentagram "Pentagram open problem 1" and leads me to such equality.


(Before you read this, you better read the former thread, in this round table forum, under the same tittle)
Continuing the first thread, now I'll give you more tables that are filled with numbers only 1 to 40, so it won't much hurt your eyes. Thus many rows are blank. I attach the file here in .xls format.
Tables 1-20 in sheet #1 & #2, are same as before and have been corrected some (of mistyping), Tables 21-32 in sheet #3, Tables 33-44 in sheet #4,...,Tables 273-284 in sheet #24, Table Total Amount in sheet #25.
I name and arrange the tables just like that, though you can do else.
[b]HAPPY PUZZLING..[/b]

The clues of this puzzle had been written at:
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1192863624

I need to write again here what I’ve been doing to this puzzle, though this had been written partly in other forum. I assume you guys have seen the file sept09A.xls.
I’m not a puzzle mania, this puzzle just disturbs my mind. So to those whoever interested in this puzzle, I [b]THINK [/b]the tables in my explanation below are the “keys” of this puzzle because:
1).Their shapes are quite different from other tables, where numbers 1 to 40 are only in certain rows. Especially in Table 45, numbers 1 to 40 are only in rows 1 & 10. (This can also approve that this is not a random case. Do you guys agree with that?). Meanwhile for other tables, numbers 1 to 40 spread almost in rows 1 to 10.
2).These tables have similar shapes each other.
E.g:
Table 45 has similar shape with Table 49 & Table 53.
Table 57 has similar shape with Table 61 & Table 65.
Here below I think tables that have similar shapes:
45,57,69,81,93,105,117,129,141,153,165,177,189,201,213,225,237,249,261,2 73.
with
49,61,73,85,97,109,121,133,145,157,169,181,193,205,217,229,241,253,265,2 77.
and
53,65,77,89,101,113,125,137,149,161,173,185,197,209,221,233,245,257,269, 281.

These tables “themselves” look similar.
E.g:
Table 49 looks similar with Table 169.
Table 73 looks similar with Table 193.
Here below I think tables look similar:
49,73,97,121,145,53,77,101,125,149,57,81,105,129,153,61,85,109,133,157,6 5,89,113,137,161,69,93,117,141,45
with
169,193,217,241,265,173,197,221,245,269,177,201,225,249,273,181,205,229, 253,277,185,209,233,257,281,189,213,237,261,165

Next time I’ll try to explain more details about the gray areas (my approaching method) as shown in Table 1, sheet #1, file sept09A.xls.
Yeah, for its big sizes, this puzzle is really tough. Needs geniuses to crack it, at least above averages…

Here I attach the file name #26&#27.xls where in sheet #26 & #27 I rearrange the tables that I [b]think[/b] could be the "keys" of this puzzle.

Now I'd like to post this problem in a different format. I re-numbered and re-arranged the tables like the file that I attach below, name enigma.xls.
So there are now 284 tables with ten rows (0,1,2,..9) each, which every tables contain of numbers 1 to 40. The problem is how to put the next larger numbers 41,42,43 and so on, in the precise rows for every tables.
Thx.


[color=green][b]  Suppose  $- \infty <a \le b \le c \le d \le e <\infty$ with  $a+b+c+d+e =0$ and $a^2+b^2+c^2+d^2+e^2=1 .$
Show that   $| b+d| < 0.77133 \; .$
[/b]  [/color]


Is it possible to place $1995$ different natural numbers along a circle so that for any two of these numbers, the ratio of the greatest to the least is a prime?

I feel that my solution's wording and notation is awkward (and perhaps unnecessarily complicated), so please feel free to critique it:
[hide]
Suppose that we do have such a configuration $a_{1},a_{2},...a_{1995}$. WLOG, $a_{2}=p_{1}a_{1}$. Then
\[\frac{a_{2}}{a_{3}}= p_{2}, \frac{1}{p_{2}}\]
\[\frac{a_{3}}{a_{4}}= p_{3}, \frac{1}{p_{3}}\]
\[... \]
\[\frac{a_{1995}}{a_{1}}= p_{1995}, \frac{1}{p_{1995}}\]
Multiplying these all together,
\[\frac{a_{2}}{a_{1}}= \frac{\prod p_{k}}{\prod p_{j}}= p_{1}\]
Where $\prod p_{k}$ is some product of the elements in a subset of $\{ p_{2},p_{3}, ...p_{1995}\}$. We clear denominators to get
\[p_{1}\prod p_{j}= \prod p_{k}\]
Now, by unique prime factorization, the set $\{ p_{j}\}\cup \{ p_{1}\}$ is equal to the set $\{ p_{k}\}$. However, since there are a total of $1995$ primes, this is impossible. We conclude that no such configuration exists.
[/hide]

is your problem same as select any 1995 natural no.s in such a way so that if we take any 2 no.s then the ratio of the greatest to the least is a prime?Are you from russia ? Can you suggest me some good physics problems(by russian authors) books written / translated in english?Iam from india and cuurently using books by s.s.krotov and i.e. irodov

It's not quite the same: we can't select any two, we can only select adjacent ones.

can't you just say that consider three numbers, $a_1$, $a_2$, and $a_3$.

Then $\frac{a_2}{a_1} =\, prime\, p$, so $a_2\,=\,a_1*p$
$\frac{a_3}{a_2} = \,prime\,q$, so $a_3\,=\,a_2*q$
$\Longrightarrow\,a_3\,=a_1*p*q$. Since $p$ and $q$ are primes, their product cannot be, so as long as you have n different natural numbers along a circle, where n >= 3, this problem is impossible.

But I read somewhere that this problem is possible for all even n, though...

A more formal way is to take $\dfrac {a_{k+1}} {a_k} = p_k^{\varepsilon_k}$, where $a_k$ are $n$ distinct natural numbers around the circle, indexed by $0,1,\ldots,n-1$ in $\mathbb{Z}_n$ (so $a_n = a_0$), $p_k$ are primes, and $\varepsilon_k \in \{-1,+1\}$. Then $1 = \prod_{k=0}^{n-1} \dfrac {a_{k+1}} {a_k} =  \prod_{k=0}^{n-1} p_k^{\varepsilon_k}$, therefore $\prod_{\varepsilon_k = -1} p_k = \prod_{\varepsilon_k = +1} p_k$. When $n$ is odd, this is impossible, since if we define by $\Omega(m)$ the number of the prime factors of $m$ (including multiplicities), we have $\Omega \left (\prod_{\varepsilon_k = -1} p_k\right ) = \left | \{k \mid \varepsilon_k = -1\} \right |$ and $\Omega \left (\prod_{\varepsilon_k = +1} p_k\right ) = \left | \{k \mid \varepsilon_k = +1\} \right |$, so $\left | \{k \mid \varepsilon_k = -1\} \right | = \left | \{k \mid \varepsilon_k = +1\} \right |$, but $\left | \{k \mid \varepsilon_k = -1\} \right | + \left | \{k \mid \varepsilon_k = +1\} \right | = n$.

When $n$ is even, this is possible, with the easy example $a_{2k} = 1$, $a_{2k+1} = p$ prime.


I can't make that box in the login that says "Log me in automatically" work.  Please help.


These numbers are known as the highly composite numbers.

http://mathworld.wolfram.com/HighlyCompositeNumber.html

Yes, there is a logical way to do it; consider the prime factorization.

I look like an idiot once again... :maybe:

[quote="anirudh"][quote="i_like_pie"]What number, less than 100, has more factors than any other? Also solve for 1000, 10000, 100000, and 1000000.

Is there a logical way to do this?[/quote]

[hide]
I think:

for 100: 60
for 1000: 840
[/hide][/quote]
Correct.


Find all 2 digit positive integers $N$ for which the sum of the digits of the $10^{N}-N$ is divisible by $170$.

Circle $S$ lies inside circle $T$ and touches it at $A$. From a point $P$ (distinct from $A$) on $T$, chords $PQ$ and $PR$ of $T$ are drawn toucing $S$ at $X$ and $Y$ respectively. Show that $\angle{QAR}=2\angle{XAY}$.

[hide]Let $N=10a+b$, for digits a and b.  We have two cases:

b=0:
$10^{N}-N=99999....9999(10-a)0$, with the first N-2 digits being 9's, the second to last digit equal to 10-a, and the last digit 0.  Thus the digit sum is $9(10a-2)+10-a=89a-8$.  It is easy to observe that, when a=2, this sum is equivalent to 170.  Then, since 89 and 170 share no common factors, there are clearly no other digit solutions for a.

$b\neq 0$: $10^{N}-N=9999.....9999(9-a)(10-b)$, with the first N-2 digits being 9's, the second to last being 9-a, and the last being 10-b.  Then the digit sum is $9(10a+b-2)+9-a+10-b=89a+8b+1$.  Through experimentation,we find that a=3, b=9 gives a digit sum of 340, which is divisible by 170.  Then, remembering that $89\times 2-8=170$ from our last case, we see that if we increase a by 2 and decrease b by 1, the digit sum will still be divisible by 170.  Therefore the pairs a=5, b=8; a=7, b=7; and a=9, b=6 also give digit sums divisible by 170.

Therefore N=20, 39, 58, 77, and 96.[/hide]

Yeah nice solution.. quite similar to mine :lol:

[hide]$\angle PXY=\angle PYX=\delta$ and $\angle PXY=\angle XAY$ (periferial and inscribed angles). $\angle XPY=180^{o}-2\delta$. As $PRAQ$ is inscribed $\angle QAR=180^{o}-\angle XPY =2\delta =2\angle XAY$[/hide]


[quote="tinadoty"]A bookstore buys a book for 10$. They add x% to make the cost 12.50. They decide not to sell the book, and apply a discount of y% to make the cost 10
$ again. What is x-y.[/quote]

You forgot to put the money signs around the ten. There has to be one before the ten and after the ten.

[quote="tinadoty"]A bookstore buys a book for 10 dollars.. They add x% to make the cost 12.50. They decide not to sell the book, and apply a discount of y% to make the cost 10 again. What is x-y.[/quote]

[hide]Well, in increasing the book's cost to 12.50, they added 25 percent of the book's orginal value. That is x. Then, to make it ten again, the decreased the value by 20 percent. That is y. Thus, $x-y = 5$[/hide]

[quote="snipefreak"][quote="tinadoty"]A bookstore buys a book for 10$. They add x% to make the cost 12.50. They decide not to sell the book, and apply a discount of y% to make the cost 10
$ again. What is x-y.[/quote]

"Unparseable or potentially dangerous latex formula. Error 6"  What is this about?[/quote]

to write in mathy font, or latex, you put the math stuff in dollar signs, resulting in $10$ instead of 10. only putting one dollar sign is assumed as an error in latex, almost like a person forgetting to type in closing quotes, or something like that


Hoi iedereen,
Ik ga mezelf hier eventjes voorstellen, zou leuk zijn moesten jullie dat ook es doen. (geen verplichting aan uiteraard, als je liever anoniem blijft ofzo doe maar :))

Ik ben Peter, 18 jaar, ik woon in Torhout (zo'n 20 km onder brugge), en studeer momenteel wiskunde in Gent. Mijn hobbies (naast wiskunde :roll: ) zijn vooral muziek beluisteren, websites maken, programmeren, sport (vooral hockey), en uitgaan. Binnen wiskunde interesseer ik mij vooral aan combinatoriek, eindige meetkunde, groepentheorie, getaltheorie, speltheorie, ongelijkheden, algemene problem-solving, grafentheorie en andere dingen op de grens wiskunde-informatica.

Ik ben door omstandigheden niet naar IMO geweest, wel 2 jaar naar de Vlaamse trainingsstages voor IMO uitgenodigd.

Ik ben Stijn Vermeeren, 18 jaar, wonend in Aarschot (Vlaams-Brabant). Ik begin in september Wiskunde te studeren aan de KU Leuven.

Ik behaalde een aanmoedigingsprijs op de JWO 2003, maar bleef verder een min of meer onontgonnen talent tot Arne Smeets daar begin 2005 verandering in bracht. (Nuja, een 150 op de 1e ronde en de selectie voor het stageweekend haalde ik ook al zonder zijn training, maar toch.) Dankzij zijn training kwam ik als een duiveltje uit een doosje de VWO-wereld verbazen met een eerste prijs en IMO-selectie voor Mexico. Daar behaalde ik zoals bekend een honorable mention, nota bene op het zesde probleem.

Hobby's: muziek (luisteren: alt/indie in een hele brede zin; spelen: bugel), internet/webmastering ([url=http://stijnshome.be]StijnsHome.be[/url], [url=http://admiral.stijnshome.be]Admiral Freebee Info Site[/url], [url=http://tijdloze.stijnshome.be]Tijdloze Website[/url]), lezen...

Hallo iedereen,

Ik heet Johannes Steenstra, ben 18 jaar oud en woon in Gouda (Zuid-Holland). In september hoop ik te beginnen aan de twee studies Wiskunde en Natuur- en Sterrekunde aan de Universiteit Utrecht.

Mijn hobby's zijn vooral programmeren, piano spelen en wiskunde. Ik heb aan verschillende wiskundewedstrijden meegedaan. Mijn mooiste resultaten zijn een vijfde plaats bij de Nederlandse Wiskunde Olympiade 2004 en een tweede plaats bij de Wiskunde Kangoeroewedstrijd Nederland 2004. In 2005 heb ik deelgenomen aan de 46e Internationale Wiskunde Olympiade in Mexico.

Ik ben Thomas Decru en woon in Wielsbeke (relatief dicht bij kortrijk). In oktober volg ik Peter op en zal burgerlijk ingenieur en wiskunde studeren in Gent.  :P

Mijn hobby's zijn volleybal, rubik-kubussen verzamelen/oplossen, manga tekenen,... en natuurlijk wiskunde. Ik ben voornamelijk geïnteresseerd in getaltheorie, en meetkunde is mijn minst favoriete onderwerp (ik ben niet de enige ;) ).
Twee jaar geleden een derde prijs behaald in de JWO. Vorig jaar enkel een finaleplaats VWO en dit jaar opnieuw een derde prijs, dit keer in de VWO...
Hier op mathlinks ken ik Stijn en Arne van zien (toen je naar voor werd geroepen tijdens de proclamaties), en Peter heb ik al eens ontmoet... Morgen opnieuw, don't forget it Peter  :D

Ik ben Fokko van de Bult, de deputy van Nederland in Athene en Merida en daarvoor deelnemer aan de IMO in 1997 en 1998 (wat ben ik dus al oud). Nu help ik met de training van het Nederlandse team en in het dagelijks leven ben ik promovendus wiskunde.

Hey alle,

mijn naam is Jinbi Jin, 16 jaar en woon in Lichtenvoorde (8 km onder Groenlo, 40 km onder Enschede). Ik ga nu naar de zesde klas VWO op het Marianum in Groenlo.

Mijn hobby's zijn tafeltennis, programmeren, en wiskunde. Binnen de wiskunde vind ik vooral ongelijkheden en getaltheorie leuk. Wat mijn programmeren betreft: Mijn enige creatie tot nu toe: Defender (als bijlage toegevoegd) :P (2/3 jaar geleden).
Volledig Nederlandstalig spel ^^

Ah, bij ons op de unief circuleert er ook zo'n spelleke... van Evil Freddy, een prof die probeert decaan/rector te worden en daarom de plaatsen waar zijn tegenstanders zitten probeert te bombarderen. :lol:

maar eum... is dat mijn gedacht, of is uw background ongeveer 85% van de totale grootte van uw spel? ;)

Ja dat klopt ja. Had de achtergrond niet in bmp moeten opslaan ^^;

Ik kom me ook eventjes voorstellen...
Ik ben Jan Vonk, 16 jaar en ik woon in het onwaarschijnlijk leuke en interessante Lede (ergens in de buurt van Aalst)
Ik interesseer me sterk voor Wiskunde (jawel) , maar ook voor Astronomie (is zelfs een grote hobby) en Fysica, en dat laatste ben ik ook van plan te gaan studeren over 2 jaar. Alhoewel Wiskunde ook een kanshebber is.
Mijn grote hobby's zijn lopen, scouts, Sirius (Een jongerenvereniging voor sterrenkunde. Jazeker, dat bestaat ook al), lezen, heel erg veel en intensief naar goeie muziek luisteren, gitaar spelen, gewoon met vrienden weggaan, blokfluit spelen met m'n neus (begin er goed in te worden) en nog veel meer.

In de wiskunde hou ik van alles, geen voorkeuren.

Ja, ik ben er dus ook nog - Arne, 18 jaar oud, en al heeeeeeeeel lang lid van dit forum (juli 2003 denk ik). Ik ben naar twee IMO's geweest (2003 en 2004) en ik haalde zilver in 2004. Vorig jaar heb ik doorgebracht in Zuid-Afrika als uitwisselingsstudent, een fantastische ervaring was dat. (Door omstandigheden kon ik niet aan IMO 2005 deelnemen, wat theoretisch gezien wel mogelijk was, maar ja.) Ik ga nu wiskunde studeren aan de KU Leuven, en ik hou me ook bezig met het trainen van jong Vlaams talent, met het modereren van deze site en het creëren en oplossen van problemen :) (Ik was trouwens ook trainer van het Zuid-Afrikaanse team dit jaar.) Euhm, wat nog - verdere hobbies: tja, sport (atletiek, lange afstand) (en ook een beetje zwemmen), en muziek - ik ben pianist :)

Haj iedereen,
Ik ben dus Xander, 17 jaar jong. Ik woon in Moen, een verwaarloosbaar dorpje onder Kortrijk. Ik ga nu naar mn 2e jaar wiskunde. Ik ben naar de IMO in 2004 geweest samen met oa Arne. Algebra blijft wel mn favoriete tak in de wiskunde. In mijn vrije tijd ben ik vooral bezig met chiro, crefi, eigenlijk jeugdwerking in het algemeen. De betere film kan er ook altijd bij, of een avondje met vrienden. Ik ben ook iedere dag een paar uur met muziek bezig. Zelf speel ik niets (buiten djembé) maar alles in het genre alternatief, indie, etc. staat me meestal wel aan... Euhm, verder ben ik nog behoorlijk sportief. Lopen, zwemmen en pingpong zijn mijn favoriete sporten. Voor vragen, 1 adres :-) xanderverbeke@gmail.com

Ja, iedereen heeft hier haast zijn verhaal gedaan zeker :) dit zijn waarschijnlijk alle actieve nederlandssprekende leden...

Tja, als het hier echt zo stil is , zit er niks anders op dat dat ik hier ook even me laat horen

Ik ben Frédéric Vanhove, ik ben geboren in 1984 (eigenlijk ben ik dus ook oud) en ik woon in Wetteren (een gemeente ten oosten van Gent).

Ik deed in 2000-2001 mee aan de VWO voor de eerste keer en behaalde een aanmoedigingprijs, in 2001-2002 heb ik een derde prijs behaald en had ik het fantastische genoegen om naar de IMO in Glasgow te mogen, waar ik drie puntjes heb behaald (tja het is EEN score  :oops: ).

Ik heb in dat schooljaar een finaleplaats in de VFO (fysica behaald) bovendien.

Ik studeer in Gent maar heb eerst een jaar ingenieur gedaan, nadien mijn eerste en tweede jaar wiskunde (volgend jaar ga ik dus naar mijn eerste licentie zuivere wiskunde)

Ik ben waanzinnig zuiver ingesteld, mijn favoriete interesse is getaltheorie, met daarnaast groepentheorie, algebra (velden en zo ) , en ook meetkunde (geen grafische!)
Ik heb ook een sterke interesse in de geschiedenis van de ruimtevaart, en hoewel ik niet zo een fysicus ben , ben ik heel af en toe geinteresseerd in de wiskunde / fysica daarvan.

Ik hou niet echt van computertoestanden (wat verklaart waarom Valentin me al es op de vingers heeft getikt omdat ik nooit Latex gebruik, wat ik probeer te veranderen) en optica, maar ben volledig los daarvan ook geinteresseerd in geschiedenis (algemene).

Ik ben ook (zoals sommigen misschien zijn aan mijn nick) geïnteresseerd in manga/anime (respectievelijk Japanse strips/tekenfilms  :ninja: ) zoals Rurouni Kenshin, Akira, NGE, en een flinke discussie (wat bij dat laatste zeker kan voorkomen) zie ik meestal wel zitten.

Voila ik hoop dat ik nu een beetje een bijdrage heb kunnen leveren

heeee nerds!

ik ben youssef 20 jr en woon in TIlburg.
ik doe maar 1e hbo bedrijfswiskunde en wil als ik me propedeuse haal wisk gaan studeren..
ik twijfel tussen Leiden en  Utrecht!

Kan iemand mij een goede raad geven welke van de twee het best is ..?
vooral als je later je bezig wilt houden met wiskunde...!!!!
succes en dank je!
ik wacht p reactie!

propedeuse? :oops:  de taalgrens uit zich weer pijnlijk ;)  ?

neen serieus, tja ik weet er niets van hoor, kan enkel welkom zeggen, ik heb me laten wijsmaken dat leiden een goeie naam heeft voor getaltheorie?

Absoluut, Leiden is ongelofelijk sterk op het vlak van getaltheorie, met oa professores Lenstra, Stevenhagen, Tijdeman, ... :)

Maar ik wil ook wel weten wat een "propedeuse" is ;)

Ff Googlen geeft dit [url]http://wiki.student.utwente.nl/wiki/Propedeuse[/url]

propedeuse = (w-vl.) proppendoos = doos waarin men proppen gooit. :) alle, bij ons toch.

[quote="Peter VDD"](w-vl.)[/quote][quote="fredbel6"]westvlaams? :?[/quote] :arrow: ja dus. :)

:D Ik ben er ook

:bye: Dag Arne, dag Peter, dag al de rest :bye:

Naam: Koen Van den Eeckhout
Woonplaats: Overmere, bij Lokeren, tussen Gent en Antwerpen, 2km van het Donkmeer ;)
Interesses: natuurkunde, wiskunde, muziek luisteren, af en toe eens een fietstochtje, :icecream: ijsjes eten
Studies: burgerlijk ingenieur, 2e jaar, toegepaste natuurkunde, Universiteit Gent
IMO: nooit :( , wel 1 keer voorbereidende stage

Dag.

ben Leslie, 18 jaar, woon in Zoelen (in de buurt van Tiel, Gelderland). Ga nu nog naar school op het RSG Lingecollege maar binnenkort studeer ik wis- en natuur- en sterrenkunde aan de uni van Utrecht. Wil uiteindelijk Master theoretische natuurkunde doen (dat is eik meer wiskunde dan natuurkunde, maar goed).

Hobby's zijn wiskunde, programmeren, muziek luisteren (vanalles, vooral hardrock), sport (vooral basket bal).

Favoriete gebied binnen de wiskunde is analyse, favoriete gebied binnen de analyse is complexe functietheorie. Vind complexe functietheorie vooral zo leuk omdat je met 4 dimensies werkt, dat maakt het allemaal een stuk abstracter en dus uitdagender. Onderwerpen als de Gauchy-Riemann vergelijkingen en de residuenstelling vind ik fascinerend. Momenteel bestudeer ik vooral de Riemann Zeta-Functie.

Hallo, ik ben dus Gert-Jan Dugardein, bijna 16 jaar oud wonende in Bikschote, tussen Ieper en Diksmuide in West-Vlaanderen. Ik studeer Wetenschappen-wiskunde(8 uur) in het Lyceum van Ieper.


There are $ n$ points on the circumference of a circle. The $ \frac{n(n\minus{}1)}{2}$ chords that connect any two points are drawn, and no three chords are concurrent. Into how many parts is the circle divided?

[hide]
There are $ n$ regions between the circle and the $ n$-gon. To find the number of regions formed by the diagonals in the $ n$-gon, we do so by induction; with each new vertex we add, we add a number of regions equal to the sum of the number of new diagonals and the number of new intersection points of diagonals. For a triangle, we only have one region. It follows that the $ n$-gon is partitioned into a number of regions equal to one more than the sum of the number of diagonals and number of intersections of diagonals.

The number of diagonals is $ {n \choose 2} \minus{} n$, and the number of intersections of diagonals is determined by the number of quadruples of vertices, or $ {n \choose 4}$.

So, the answer is $ n \plus{} {n \choose 2} \minus{} n \plus{} {n \choose 4} \plus{} 1 \equal{} {n \choose 2} \plus{} {n \choose 4} \plus{} 1$.
[/hide]


How did u guys do in state last year?
Me: 28 :(

[quote="happyme"]I had an epic phail at states but what comforts me is that...

THE BRAINS WERE SO AWESOME.[/quote]
LOL those brains eh? Some people I knew kneaded them so much that they were all messed up. I couldn't bear messing it up, so I kept it perfectly preserved. It's still sitting in a drawer, and it actually looks like a brain still :D I totally failed though, got a 32 :( I made a lot of dumb mistakes on target which is what killed me.

ehehe lol. This is sad... My sprint score at state is less than mine at nationals!... Ok ready... seriously ready? this is really sucky...
[hide="Score"][hide][hide="err... sorry"]
21+16=37[/hide][/hide][/hide]

23+14=37=4th place in MA= :D =/=  :( or  :mad:
Stupid trapezoid problem...

Hmmm I believe I got 42, which I did not get on the USAMO.

Hmm.... I think I got 25 on sprint and 12 on the Target, for a total of 37.

Second place, Michigan. (First after countdown though!  :D)

No matter how badly you scored, you couldn't have scored worse than me....why, you ask? I wasn't there.  :wink:

Never took it plus I never got my scores from chapter.  Maybe they will send them this year...
 :|

[hide]24/14 = 38 = 3rd[/hide]

Kudos to mathcrazed for being the only one in the state to get perfect on target!

yeah i got a 29/14 on state last year and omg what a coincidence that is also what i got on USAMO.  :rotfl: , jk i got much worse and i'd rather not a talk about it. i got:

[hide][hide][hide]less than 46[/hide][/hide][/hide]

I was clicking for soooo long cognos, and u tricked me  :)

anyways, I didn't do MC last year.

[quote="cognos599"]r u in 6th, 7th, or 8th grade? or higher.[/quote]

8th, I just learned about MC in August, barely made the deadline.

[quote="cognos599"]nice, second to the AMC's isn't mathcounts one of the most prominent and prestigious math contests?[/quote]

Gap is pretty large though, yeah its one of the more famous math competitions.

Maybe Putnam is pretty prominent for undergrads.

Gah... 21+6= 27, and I wasn't ranked in the top 25%....

Because I sucked. And I still suck. I misread BOTH of the first two target problems.

WHO CARES ABOUT LAST YEAR?

who's in for it "this" year?  :D

i have an odd history of doing better in competitions where i dont try as hard  :|

I sucked at sprint last year, although target was surprisingly easy (only realized how to do the trapezoid problem after it was over)

17+14=31=16th place in PA

Although I'm going again this year and I hope to do better.


So how did you guys do on the B? Walton's best was a 118.5 (Bill Qiao). I think we have almost 20 that are qualified for the AIME now. I thought that the test was a little easier to get 100, but maybe a little harder to get a high score.

Fulton

Our school didn't take the AMC B, except for me, where I took it at Lassiter. I'm sure I qualified with ~121 or something like that.

I think it was harder than the A...so was the 10

The tests were SO MUCH FREAKING HARDER THAN LAST YEAR!!!
I got (7th grade)
10B: 136.5
12B (I totally got screwed over): 118.5?!? (*cough* *hack*...)
Was anyone of the same opinion?

I'm such a slow worker.  I don't even think I qualified.   :(
(Which means USAMTS next year; I can handle a full month of working time!)

I was disappointed because I guessed so horribly. On the AMC 10A I guessed 2 out of 4 correct, but on the 12B, I guessed 0 for 6 right. I guess they kind of evened each other out (2 for 10 is the expected value). I ended up with 126 on the 10 and 114 on the 12. If I hadn't guessed on the 12, I would have had a 123.

126.  (Who'd have thought number 3 was so hard?)

This might or might not be Lassiter's highest score.

And yes, guessing is bad.

And I think this was harder.

I got too lucky. I qualified for AIME but only by percentage, not by score. I missed the qualification score by 1 point.

I had TERRIBLE luck...I took the AMC 12A and then, since I thought I would have a better chance of making AIME on the 10, I took the AMC 10B...turns out I should have taken the A instead...i guess i'll just have to try again next year...


function
integration
calculus
calculus computations

Let $ f: [a,b]\to\mathbb R$ be an integrable function on $ [a,b].$ Let $ \epsilon>0$ be a given value. Prove that does exist a continuous function $ \varphi: [a,b]\to\mathbb R$ such that $ \left|\int_a^bf\minus{}\int_a^b\varphi\right|<\epsilon.$

Too easy? Boring? Hard?

I want to learn how to do it, can you guys provide a solution?  :)

Utterly trivial, to the point that I don't think it's the question you wanted to ask. With the absolute values outside, we can just choose the right constant function.

we know that every non-negative measurable function $ f$ can be approximated by an increasing sequence $ < \phi_{n} >$of nonnegative simple functions. Moreover, since then we have $ \int f \equal{} sup\int \phi$ over all simple functions $ \phi\leq f$.  For f is integrable then $ f \equal{} f^{ \plus{} } \minus{} f^{ \minus{} }$, by applying the above argument for $ f^{ \plus{} }, f^{ \minus{} }$ given $ \epsilon > 0$ we can find a simple function $ \phi$ st : $ |\int f \minus{} \int\phi| < \epsilon$.

Similarly we can also approximate $ f$ by a continuous functions $ \psi$ by modifying the step function $ \phi$ at points it is not continous, hence given $ \epsilon >0$ : $ |\int f \minus{} \int\psi| < \epsilon$


[b]Tom takes a PVC pipe and rubs it vigorously with a piece of silk cloth.  He then places his finger 1 cm from the pipe and a crackling noise is heard.  Explain.[/b]

I know there is a net negative charge on the PVC, but what is the cause for the noise? Thanks in advance.

What I think is due to the breakdown of the atmosphere.

I mean that the electric field reaches its limiting value for the air(I think it is called breakdown electri field, I'm not sure about the term).

The actual term is Dielectric Breakdown!


Αν θεωρήσουμε $ \sigma{(n)} \equal{}$ το άθροισμα των διαιρετών του $ n$, να δείξετε ότι για κάθε ακέραιο $ n$:
.
$ \displaystyle \frac {\sigma{(1)}}{1} \plus{} \frac {\sigma{(2)}}{2} \plus{} ... \plus{} \frac {\sigma{(n)}}{n}\leq 2n$

Πολυ ωραια ασκηση Δημητρη (δικια σου ειναι?)

Λοιπον, αρχικα χρησημοποιουμε το παρακατω λημμα:

Λημμα: Ισχυει $ \frac {\sigma(m)}{m} \equal{} \sum_{i \equal{} 1}^{k}{\frac {1}{p_i}}$, οπου $ p_i, i \equal{} 1,2,...,k$ ειναι οι διαιρετες του $ m$

Αποδειξη: Προφανος ισχυει $ \sum_{i \equal{} 1}^{k}{\frac {1}{p_i}} \equal{} \sum_{i \equal{} 1}^{k}{\frac {1}{\frac {m}{p_i}}} \equal{} \frac {\sum_{i \equal{} 1}^{k}{p_i}}{m} \equal{} \frac {\sigma(m)}{m}$

Πισω στην ασκηση τωρα
Επηδη απο τους αριθμους $ 1,2,3,...,n$ ακριβως $ \left[\frac {n}{p_i}\right]$ διαιρουνται με $ p_i$, με βαση το παραπανω λημμα, το αθροισμα μας ειναι ισο με:
$ \sum_{i \equal{} 1}^{n}{\left[\frac {n}{i}\right]\frac {1}{i}} \leq \sum_{i \equal{} 1}^{n}{\frac {n}{i}\frac {1}{i}} \equal{} n\sum_{i \equal{} 1}^{n}{\frac {1}{i^2}} \leq 2n$

Αφου $ \sum_{i \equal{} 1}^{n}{\frac {1}{i^2}} \leq 2$
Το τελευταιο ισχυει διοτι προκειτε για σειρα θετικων αριθμων που συγκλινει στο $ \frac {{\pi}^2}{6} < 2$.

Μπορει ομως να δειχθει και ετσι:
$ \sum_{i \equal{} 1}^{n}{\frac {1}{i^2}} \leq 1 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{\frac {1}{i(i \plus{} 1)}} \equal{} 1 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{\frac {1}{i} \minus{} \frac {1}{i \plus{} 1}} \equal{} 1 \plus{} 1 \minus{} \frac {1}{n} \equal{} 2 \minus{} \frac {1}{n} < 2$

Υπαρχει αλλη μια αποδειξη (κατα Bernoulli) οπου φρασουμε τη συγκεκριμενη παρασταση απο το αθροισμα των ορων μιας φθινουσας Γεωμετρικης Προοδου

[quote="bilot4"]Πολύ σωστός Νίκο!! Την άσκηση την είδα στο number theory book του Αndreescu![/quote]

Thanx. Παντος οπως προκειπτει απο την αποδειξη που δινω στο $ \sum_{i\equal{}1}^{n}{\frac{1}{i^2}} \leq 2$, ισχυει $ \sum_{i\equal{}1}^{n}{\frac{1}{i^2}} \leq 2 \minus{} \frac{1}{n}$, αρα η αρχικη παρασταση φρασεται και απο το $ 2n \minus{} 1$.


$ \frac{s^2}{2r_a}\equal{}p\minus{}a$ and $ \frac{s^2}{2r} \equal{} p$ so you can cosctruct the incircle and the A-excircle. The side BC one of the 2 internal tangents of these circle.


conics
hyperbola
rotation
analytic geometry
trigonometry
Asymptote

I don't really know where to put this, but here goes:

Give the equation of the rectangular hyperbola with A(0,3) and B(2,-1) as tops.

I always make some kind of mistake and can't get a right solution :blush: , can somebody help me?

No need for help anymore, I found a solution.

The solution I found is wrong too, I think, so help is welcome again...

before i start, just wanna say i've always hated conic sections, so.... (by the way BC, people should have answered this by now.... ;) )
i belive it is obvious that you are dealing with a rotated axis here... so you have to define a new axis $x'y'$
you know that the axis $x'y'$ is made by rotating the $xy$ axis for an angle of $\theta$, based on the coordinates of the $A$ and $B$ you can find the $\theta$
you also know (hopefully... :P ) that:
$x=x'\cos\theta+y'\sin\theta$
$y=x'\sin\theta+y'\cos\theta$
based on the coordinates of the $A$ and $B$, by creating a right triangle you can find the value of $\theta$ $(\theta=26.56 degrees)$ but i'm not gonna use this number, because you can use the triangle you made to find the values of $\sin\theta, \cos\theta$
plugging in values for $x'$, and $y'$ you get:
$x=\frac{2x'-y'}{\sqrt{5}}$
$y=\frac{x'-2y'}{\sqrt{5}}$
now find the points $A$ and $B$ in your new axis of $x'y'$
you get $A(\sqrt{5},2\sqrt{5})$, and $B(\sqrt{5},0)$ now the problem is very easy....

now you can easily find the center of hyperbola, it would be $C(\sqrt{5},\sqrt{5})$,
you can find the equation of the hyperbola, in $x'y'$ then you can plug in valus for $x'y'$ to get the equation in $xy$,
you do the rest,
.... :)

The solution I found yesterday was right after all, I just made a mistake in a calculation.

I used the tangents in the top together with the line connecting the top to form a bundle of conic sections.
Then I used the directions of the asymptotes to get the equation of the hyperbola out of the equation of the bundle of conic sections.


probability
function
combinatorics unsolved
combinatorics

Write down the probability generating function for the score on a standard, fair six-faced die whose faces are labelled 1, 2, 3, 4, 5, 6. Hence show that the probability generating function for the sum of the scores on two standard, fair six-faced dice, rolled independently, can be written as \[ \frac {1}{36} t^2(1 \plus{} t)^2 \left(1 \minus{} t \plus{} t^2\right)^2\left(1 \plus{} t \plus{} t^2\right)^2\text{  }.\]

Write down, in factorised form, the probability generating functions for the scores on two fair six-faced dice whose faces are labelled with the numbers $ 1, 2, 2, 3, 3, 4$ and $ 1, 3, 4, 5, 6, 8$, and hence show that when these dice are rolled independently, the probability of any given sum of the scores is the same as for the two standard fair six-faced dice.

Standard, fair four-faced dice are tetrahedra whose faces are labelled $ 1, 2, 3, 4$, the score being taken from the face which is not visible after throwing, and each score being equally likely. Find all the ways in which two fair four-faced dice can have their faces labelled with positive integers if the probability of any given sum of the scores is to be the same as for the two standard fair four-faced dice.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191595&search_id=412920648]this thread[/url].

As for the second question:  the relevant generating function is $ (t \plus{} t^2 \plus{} t^3 \plus{} t^4)^2 \equal{} t^2(1 \plus{} t^2)^2 (1 \plus{} t)^2$.  We want to split this up into two functions whose value at $ t \equal{} 1$ is $ 4$, both of whom have a factor of $ t$.  This can be done in the following ways:

$ t(1 \plus{} t^2)(1 \plus{} t) \cdot t(1 \plus{} t^2)(1 \plus{} t)$; this is two tetrahedral die.
$ t(1 \plus{} t^2)(1 \plus{} t^2) \cdot t(1 \plus{} t)(1 \plus{} t)$; these are the die $ 1, 3, 3, 5$ and $ 1, 2, 2, 3$.


induction
combinatorics unsolved
combinatorics

There are eight different symbols designed on $n\geq 2$ different T-shirts. Each shirt contains at least one symbol, and no two shirts contain all the same symbols. Suppose that for any $k$ symbols $(1\leq k\leq 7)$ the number of shirts containing at least one of the $k$ symbols is even. Determine the value of $n$.

Let us consider a set $S$ of symbols with $|S| = n> 1$, and a set $\mathcal{T}$ of T-shirts with  $|\mathcal{T}|=t>0$ having the stated property, which clearly establishes an injection $i\colon \mathcal{T} \to \mathcal{P}(S)\setminus \{\emptyset\}$ given by $i(T) = \{s\in S \mid s\in T\}$. We will also define $j\colon S \to \mathcal{P}(\mathcal{T})$ by $j(s) = \{T \in \mathcal{T} \mid s\in T\}$ (as a byproduct, we will prove that $j(s) \neq \emptyset$ for any $s\in S$, and that $j$ is an injection also, but for the moment we don't need all that).

We are given that for any set $\emptyset \neq K\subsetneq S$ we have $\displaystyle \left | \bigcup_{s\in K} j(s) \right |$ even. In particular $|j(s)|$ is even for any $s\in S$, and so, by induction on $|K|$, $\displaystyle \left | \bigcap_{s\in K}j(s) \right |$ is shown to be even for all $1\leq |K| \leq n-1$, since by PIE
\[\left | \bigcup_{s\in K} j(s) \right | = \sum_{\ell=1}^{|K|-1} \left ( (-1)^{\ell-1} \sum_{\stackrel {|L| = \ell} {\stackrel {L\subseteq K} {}}} \left | \bigcap_{s\in L} j(s) \right |\right ) - (-1)^{|K|} \left | \bigcap_{s\in K} j(s) \right |.\]
Let us now consider a $M\in \mathcal{T}$ of maximal $|i(M)|>0$. If it were $|i(M)|< n$, then since $\displaystyle M \in \bigcap_{s\in i(M)}j(s)$, with $\displaystyle \left |\bigcap_{s\in i(M)}j(s)\right |$ even, it follows $\displaystyle \bigcap_{s\in i(M)}j(s)$ contains another $T\neq M$, with $i(M) \subsetneq i(T)$, thus $|i(T)| > |i(M)|$, absurd, therefore $|i(M)| = n$, i.e. $i(M) = S$, or equivalently $i^{-1}(S) \in \mathcal{T}$.

Assume now that for some $1\leq k < n$ we have $i^{-1}(S') \in \mathcal{T}$ for all $S'\subseteq S$ with $|S'|>k$; we just proved $k=n-1$ is such. Take any set $K\subset S$ with $|K|=k$, and notice that $\displaystyle \bigcap_{s\in K}j(s)$ contains the $2^{n-k} - 1$ T-shirts $i^{-1}(S')$ for all $K\subsetneq S'\subseteq S$. For parity reasons, it must contain yet at least another element, which can only be $i^{-1}(K)$. Therefore $i^{-1}(K) \in \mathcal{T}$ for all such $K\subset S$ with $|K|=k$, and by induction we proved our assumption for all $1\leq k < n$. The end result of all this is that $i$ is also surjective, thus $\boxed{t = |\mathcal{T}| = |\mathcal{P}(S)\setminus \{\emptyset\}| = 2^n - 1}$. Conversely, this unique $\mathcal{T} = \{i^{-1}(K) \mid \emptyset\neq K \subseteq S\}$, the set of all possible (non-empty) T-shirts, having $2^n-1$ elements, has the stated property, since for any set $\emptyset \neq K\subsetneq S$ of symbols with $1\leq |K| = k \leq n-1$, the number of T-shirts containing at least one symbol from $K$ is $(2^n-1) - (2^{n-k} - 1) = 2^{n-k}(2^k-1)$, an even number.


This year the Perseid meteor shower is suppose to be spectacular with the a new moon.

The peak of the shower is during the very early morning of August 13th, approximated at 4 am, with optimal nights being the 12th and 13th.  The best time to look the latest possible time to look, around midnight.  That is when the "radiant", or center it seems the meteors come from in the sky, is overhead so the horizon does not interfer.  The expected rate of meteors is around 60 meteors per hour in the darkest sky conditions.   For more information on "dark skies" check [url=http://www.mathlinks.ro/viewtopic.php?t=157030]here[/url].

How meteor showers are named is by the constellation where the radiant appears, in this case in Perseus, "The King".
Another "trait" each meteor shower has is a comet, whose trail and dust cause meteors.  The comet associated with the Perseids is Comet Swift-Tuttle.

A Sky Map from New Scientist
[img]http://space.newscientist.com/data/images/ns/cms/dn12396/dn12396-1_500.jpg[/img]

[url=http://upload.wikimedia.org/wikipedia/commons/0/09/Perseus_constellation_map.png]Wikipedia Picture of the Perseus Constellation[/url]

The most I have seen in one night from the Perseids is $ 15$ meteors.

[b]Links[/b] - As always to further explain
[url=http://en.wikipedia.org/wiki/Perseids]Wikipedia Article About Perseids[/url]
[url=http://en.wikipedia.org/wiki/Category:Perseus_constellation]Wikipedia Article About Perseus[/url]
[url=http://stardate.org/nightsky/meteors/]Star Date - Information and Tips for Viewing[/url]
[url=http://www.theskyscrapers.org/meteors/index.php/year/2007]Skyscrapers - Information About This Year's Meteor Showers[/url]
[url=http://www.imo.net/calendar/2007]International Meteor Organization - In Depth Imformation About This Year's Meteor Showers[/url]
[url=http://space.newscientist.com/article.ns?id=dn12396&feedId=online-news_rss20]New Scientist - Article About the Perseids[/url].

The most I have seen is 33. But I had to stay up all night.

Some more information from [url=http://SpaceWeather.com]SpaceWeather.com[/url]

[quote="SpaceWeather.com"]
Space Weather News for August 9, 2007
http://spaceweather.com

The Perseid meteor shower has begun.  Although the display will not peak until August 12th and 13th, sky watchers are already counting as many as a dozen meteors per hour (including some nice fireballs) during the darkest hours before dawn.  Rates could increase 5- to 10-fold when the shower reaches maximum on Sunday night and Monday morning. Advice: Get away from city lights. While many Perseids will be bright enough to see from light-polluted urban areas, the shower's full grandeur is reserved for places with dark and starry skies. This is a good weekend to go camping.

Visit http://Spaceweather.com for photos, observing tips and a sky map.  Full coverage of the shower begins today and will continue until Perseid activity subsides.
[/quote]

That is good maybe I can stay out and look.

I started watching at around 10:15 central time Sunday night and saw around a dozen in an hour, though I think I only caught some of the brightest ones. There were some really nice ones, though. A couple had trails that glowed even after the  meteor burned up. Things like that keep me excited about astronomy. :)

Well I went Saturday night and only saw 17 meteors after close to two hours.  :)
Sunday and Monday were kind of cloudy do I did not look.

Lately the neighborhood I am in has gangster problems so my parents won't let me go outside.


Hey there everybody. I'm new here and I've been looking for a message board like this for a long time so I'm excited. Anyway, I've got a question. I am trying to teach my self basic Calculus, nothing too serious, Intergals, etc. Anyway, Every resource that I find online to learn from all of a sudden jumps into a bunch of crap that I don't know what they're talking about. Does anybody know of any sites that are, more or less, easy going when providing information about Calc. Thanks in advanced!

I"m assuming he doesn't have access to a textbook?

I'm sort of skipping to calculus (I know, I shouldn't be doing that...), and I really don't know any good calculus (because like you said, they jump around). I do have a book that teaches basic concepts like deriatives and integrals, but it's an outline. It's not too bad, I can understand it well.

I can't say that I have really looked online for a self-contained place to learn calculus.  But, perhaps try [url=http://www.karlscalculus.org/#enter]this site[/url] or [url=http://archives.math.utk.edu/visual.calculus/]this[/url] one.

Also keep in mind that you are trying to [i]learn[/i] calculus...thus you probably won't be familiar with all the terminology and symbols.

i really do recomend you getting a calculus textbook.

mine is by giovanne or something

it has a skier on the front cover

good luck

oh yeah...and you should also try posting in the calculus forum

I would agree.  A textbook would be preferable if possible.  If you want to get a textbook I recommend asking one of your teachers if they have one they can loan to you.  Calculus textbooks usually aren't cheap but teachers usually have some old ones they can lend you.  If you're lucky, they might even be able to give you some that they would otherwise throw out ;)

Your local library might have some calculus text books too.