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 agree to our terms of service, privacy policy and cookie policy

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