Day 1 (September 21, 2024)

[mpham26uchicago]

Today, we look at Corollary 4.12. It states that for $t \geq 0$, we have

$$|p_1(t)-\mu + \nu| \leq \begin{cases} Ce^{-\nu t}, &\quad \nu <1, \ C\langle t\rangle ^N e^{-t}, &\quad \nu \geq 1\end{cases}.$$

It suffices to notice that $\mu - p_1(t)$ can be recovered from $v$ via

$$\mu - p_1(t) -\nu = \frac{v\prime(t)}{v(t)} + \log v(t)-\nu .$$

We are interested in the right hand side. We conjecture that $v(t) = e^\nu + A_1(t)e^{-t}+B_1(t)e^{-(\nu \wedge 2)t} +o(e^{-(\nu \wedge 2)t})$ and $v\prime(t) = A_2(t)e^{-t}+B_2(t)e^{-(\nu \wedge 2)t} +o(e^{-(\nu \wedge 2)t})$. Numerical experiments seem to confirm the above. Actually, numerical experiments shows that the bound in Proposition 4.11 is not optimal. We determine that $v\prime(t)/v(t)$ must decays to 0. A detailed calculations shows that

Numerical experiments seem to agree with the above. Next, we move our attention to $\log v(t) - \nu$. The general form of the function is as expected (the same as $\frac{v'}{v}$). However, the constants behind the $e^{-t}$ term seems to be half that of $v'/v$. I'm starting to doubt whether $\mu - p_1(t) - \nu$ actually grows $e^{-2t}$.

[mpham26uchicago]

So now what's happening is that the LHS decays $e^{-2t}$ but the RHS is only decaying $e^{-t}$.

[mpham26uchicago]

The reason for the above is that the constants of the $e^{-t}$ term in front of $\frac{v\prime}{v}$ and $\log v(t) - \nu$ don't match up. The constant of $\frac{v\prime}{v}$ is twice as large as $\log v(t) - \nu$. This is due to a mistake in computing the derivative where the denominator of the difference quotience was interval_length and not 2*interval_length.