I was recently given an example using this asymptotic expansion of the digamma function where:
$$\frac{d}{dx}(\ln\Gamma(x)) = \psi(x) \sim \ln(x) - \frac{1}{2x} - \frac{1}{12x^2}$$
Here's the example:
$$\frac {\psi\left(\frac x4\right)}4 - \frac {\psi\left(\frac x5 + \frac 12\right)}5 - \frac {\psi\left(\frac x{20} + \frac 12\right)}{20}\sim -\frac {\ln(4)}4 + \frac{\ln(5)}5 +\frac {\ln(20)}{20}-\frac 1{2\,x}-\frac {11}{8\,x^2}$$
I'm unclear on the following points:
What happened to each $x$ term?
Why is the first term negative and the rest of the terms positive? Why isn't the signs of the original terms would be preserved?
I would have expected something like this:
$$\frac{\ln(\frac{x}{4})}{4} - \frac{\ln(\frac{x}{4})}{4} - \frac{\ln(\frac{x}{4})}{4} - \ldots $$
- How is $\frac{11}{8x^2}$ being determined? Why does $-\frac{1}{12x^2}$ change but $-\frac{1}{2x}$ stays the same?
Sorry for the elementary questions. The explanation will really help! :-)
Thanks,
-Larry
The $\log(x)$ terms cancel because $\dfrac{1}{4} - \dfrac{1}{5} - \dfrac{1}{20} = 0$. In somewhat more detail, $$ \eqalign{\frac{\psi(x/4)}{4} &= -\frac{1}{2}\,\ln \left( 2 \right) +\frac{1}{4}\,\ln \left( x \right) -\frac{1}{2\,x} -\frac{1}{3\, x^2}+O \left( {x}^{-4} \right)\cr \frac{\psi(x/5+1/2)}{5} &= -\frac{1}{5}\,\ln \left( 5 \right) +\frac{1}{5}\,\ln \left( x \right) +{\frac {5}{24 \,x^2}}+O \left( {x}^{-4} \right) \cr \frac{\psi(x/20+1/2)}{20} &= -\frac{1}{20}\,\ln \left( 20 \right) +\frac{1}{20}\,\ln \left( x \right) +\frac{5}{6\,x^2}+O \left( {x}^{-4} \right) }$$