Discretisation of logarithmic derivative: Deriving the formula

Question

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?

Answer 1

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.