Coefficient of degree $-1$ in $df/f$

Question

Let $f=\sum_{i\ge m} a^i t^t$ (for $m\in \mathbb Z$) a formal Laurent power series and consider its formal differential $df$. What is the coefficient of degree $-1$ of the differential form $\frac{df}{f}$? Do we have an explicit expression?

Answer 1

Since $\frac{\mathrm{d}f}{f} = \mathrm{d} \log(f)$, you can factor out the overall factor of $t$:

$$ \log \left( \sum_{i \geq m} a^i t^i \right) = m \log(t) + \log \left( \sum_{i \geq 0} a^{i+m} t^{i} \right) $$

The second term only has coefficients in nonnegative degree, and $\mathrm{d}\log(t) = \frac{\mathrm{d}t}{t}$.

If you're uncomfortable with taking logarithms of formal power series, you can view the above as inspirational; once you know what the result "should" be you can prove it without invoking logarithms.

For example, you can show by the product rule that

$$ \frac{\mathrm{d}(fg)}{fg} = \frac{\mathrm{d}f}{f} + \frac{\mathrm{d}g}{g} $$

which lets you mimic the proof sketch above.