2
$\begingroup$

I'm reading a paper where they use a discrete approximation of a logarithmic mass growth rate as follows:

$$ \frac{d \log M}{d \log t} \approx \frac{(t_B + t_A)(M_B - M_A)}{(t_B - t_A)(M_B + M_A)}$$

I'm trying to derive this approximation, but wasn't successful thus far. I tried forward, centered and backward differencing, using Taylor expansions and $\log(x) \approx (x-1) - \frac{(x-1)^2}{2} + ...$ for $x \rightarrow 1$, but haven't gotten that result. In particular, I don't see how the factors $(M_A + M_B)$ and $(t_A + t_B)$ come into play.

Can anyone help me out?

$\endgroup$
3
$\begingroup$

This should be coming from a chain rule (assuming $M$ is a function of $t$, $t_A<t_B$):

$$ f(t)=\frac{d \log\big(M(t)\big)}{d \log t} = \frac{d\log\big(M(t)\big)}{dt}t=\frac{dM(t)}{dt}\frac{t}{M(t)} $$

First, consider the backward difference, $$ f(t_B) \approx \frac{M_B-M_A}{t_B-t_A}\frac{t_B}{M_B} $$ Second, consider the forward difference, $$ f(t_A) \approx \frac{M_B-M_A}{t_B-t_A}\frac{t_A}{M_A} $$ Clearly, we are getting expressions similar to the given in the paper, but something is a little bit off.

Now, if we are looking for the central difference, $$ f\left(\frac{t_A+t_B}{2}\right) \approx \frac{M_B-M_A}{t_B-t_A}\frac{\frac{t_A+t_B}{2}}{M_*} $$ Here, $M_*$ is the value of $M$ at $t=\frac{t_A+t_B}{2}$. Let $M_*=\frac{M_A+M_B}{2}$ - an average, which is exact if $M$ is linear between $t_A$ and $t_B$ or a very close approximation for a fine enough grid step. Then, for central differencing $$ f\left(\frac{t_A+t_B}{2}\right) \approx \frac{M_B-M_A}{t_B-t_A}\frac{\frac{t_A+t_B}{2}}{\frac{M_A+M_B}{2}}=\frac{(M_B-M_A)}{(t_B-t_A)}\frac{(t_A+t_B)}{(M_A+M_B)} $$ that is exactly the original expression from the paper.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.