When you explain something deep using terms somewhat crude, that's recursion.


When you fly through the sky, the simple terms, they won't lie, that's recursion.

Will just go ahead and ask.. So, I have these:
f:R->R, where f(x)=-(3^-x), A={n+1-1/(k+1) : n,k natural numbers}, and B=f[A]. I have to prove that f is the isomorphic function from A to B, and that it is the only one. The first part is easy, I have no clue, however, how to go for the second part. I tried by stating that it is true and get to a contradiction, however I can't figure out how to get to one...

v(n) denotes the index of n relative to a fixed primitive root g, that is the exponent v for which g^v \equiv n.

I can't understand this, can some body explain me this definition. I mean g^v \equiv n wrt to what

http://math.stackexchange.com/questions/663101/for-n-geq-2-prove-that-1-frac141-frac191-frac116
so confused towards the end :/!

@robjohn, I have this theorem, "Let $X$ and $Y$ have a continuous joint distribution. Suppose that $\{(x, y) :f(x, y) > 0\}$
is a rectangular region $\mathbb R$ (possibly unbounded) with sides (if any) parallel to the
coordinate axes. Then $X$ and $Y$ are independent if and only if $f(x,y)=h_1(x)\cdot h_2(y)$ holds for
all $(x, y) ∈ \mathbb R$, where $h_1$ and $h_2$ are nonnegative functions." which I can't prove. Can I ask it on main without showing any efforts? Because I can't prove it?

ok for $2n$
$P(0) = 0$
$P(1) = 2$
$P(2) = 4$
$P(3) =6$

$2n^2 -2n$
$2(n+2)^2-2(n+2)$ is even more people giving handshakes

$2n^2-2n$
$2(n-1)^2-2(n-1)$

so what happens if there is 6n....
$P(6n) = 2(6n)^2 -2(6n)$ tons of handshakes..........

a) $u$ is the real part of $ze^z$
b) $v=xe^x\sin(y)+ye^x\cos(y)$
c) $\int_0^{2\pi}h(\theta)\,\mathrm{d}\theta=\mathrm{Re}\left(\int_0^{2\pi}e^{i\theta}e^{e^{i\theta}}\frac{\mathrm{d}e^{i\theta}}{ie^{i\theta}}\right)$