ok, so that's why it's weird: While $\epsilon_0=\omega^{\omega^{\omega^⋰}}=\omega [4] \omega =\omega [5] 2$, incrementing the subscript of the epsilon numbers one by one is NOT incrementing the pentation. To see this note the following:

$\epsilon_1=\sup(\epsilon_0,\epsilon_0^{\epsilon_0},\epsilon_0^{\epsilon_0^{\epsilon_0}},...)=\sup(\omega [4] \omega,(\omega [4] \omega)[3]2, (\omega [4] \omega)[3]3,...)=(\omega [4] \omega) [3] \omega \neq \omega [4] (\omega +1)$

Therefore what is happening when going from $\epsilon_0$ to $\epsilon_{\omega}$ is we are exponentiating the previous term $\om…

Guys, I'm a little bit confused by the way differential equations are solved. Say we have the following:
$$
\theta^2\,d\theta=\sin t\,dt.
$$
I'm guessing this actually meant:
$$
\frac{d\theta}{dt}=\frac{\sin t}{\theta^2}.
$$
Say we consider finite $\Delta t$ and $\Delta\theta$. So we can have the following approximation:
$$
\theta^2\Delta\theta\approx\sin t\Delta t,
$$
hence
$$
\sum_i\theta^2\Delta\theta_i\approx\sum_i\sin t\Delta t_i,
$$
where our $t$ starts at $t_0$ and ends at $t_1$. For each $i$, we can take a $t^*\in[t_1,t_{i+1}]$. However, how about $\theta$? We can't take $\theta(t^*…

Guys, say we have
$$
i\hbar\psi\frac{d\phi}{dt}=-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}\phi+V\psi\phi.
$$
Griffiths says that we can divide by $\psi\phi$. However, how do we know these functions are not zero at certain points? The functions were given as follows, btw:
$$
\Psi(x,t)=\psi(x)\phi(t).
$$
I understand that if $\psi$ or $\phi$ vanishes everywhere, then obviously $\Psi(x,t)=0$. But what happens is for instance $\psi$ vanishes at certain points only? Are we just saying then that our equation only holds for points where $\psi$ and $\phi$ don't vanish?