I'm trying to verify the following algebraic step from an old textbook, but I'm having trouble getting the same result.
The starting point is here:
\begin{equation} 1 = \frac{\rho_{si}a}{2\epsilon_0} \ln \frac{\frac{\Delta l_i}{2} + \sqrt{a^2 + \left( \frac{\Delta l_i}{2} \right )^2}}{-\frac{\Delta l_i}{2} + \sqrt{a^2 + \left( \frac{\Delta l_i}{2} \right )^2}} \end{equation}
Here I'm supposed to assume $a << \Delta l_i$. I see two ways to do this immediately:
(1) in the numerator, assume that $\sqrt{a^2 + \left( \frac{\Delta l_i}{2} \right )^2} \approx \sqrt{\left(\frac{\Delta l_i}{2} \right )^2}$
(2) in the denominator, perform a Taylor series expansion of $\sqrt{a^2 + \left( \frac{\Delta l_i}{2} \right )^2}$ and throw out the higher order terms to taste.
Doing this, I get
\begin{equation} 1 = \frac{\rho_{si}a}{2\epsilon_0} \ln \frac{\Delta l_i}{-\frac{\Delta l_i}{2} + \frac{\Delta l_i}{2} + \left(\frac{1}{2}\right) \left(\frac{a^2}{\frac{\Delta l_i}{2}}\right)} \end{equation}
which simplifies to
\begin{equation} 1 = \frac{\rho_{si}}{2\epsilon_0} \ln \frac{\Delta l_i^2}{a} \end{equation}
However, the book has
\begin{equation} 1 = \frac{\rho_{si}a}{\epsilon_0} \ln \frac{\Delta l_i}{a} \end{equation}
in other words, I've come up with one too many $\Delta l_i$ in the numerator. I've checked the Taylor expansion twice now, and also verified that the rest of the chapter uses $\Delta l_i$, not $\Delta l_i^2$ in subsequent steps.
What am I missing? (The book isn't clear on how the simplification was carried out, only that $a$ should be neglected versus $\Delta l_i$.)
@mathguy refuses to take credit, but they're the one who spotted it.
The simplification I performed erroneously tries to factor an $a^2$ term from outside the log operator. The expression I should have gotten was
\begin{equation} 1 = \frac{\rho_{si}a}{2\epsilon_0} \ln \frac{\Delta l_i^2}{a^2} \end{equation}
which you can extract the power from and move outside the log:
\begin{equation} 1 = \frac{\rho_{si}a}{2\epsilon_0} \ln \left(\frac{\Delta l_i}{a}\right)^2 = \frac{\rho_{si}a}{\epsilon_0} \ln \frac{\Delta l_i}{a} \end{equation}
This lines up with the book's derivation.