How has this approximate relation been derived? I am currently looking at some lecture notes which state the following approximate relation:

Where $\mu$ is the expectation and $\sigma^2$ is the variance of a random variable.
Can anyone help me to see why these 2 expressions are approximately equal?
 A: The "quickest" way to see this is to replace every exponential with its second order Taylor approximation, $\mathrm{e}^x \approx 1+x+x^2/2$, then simplify, discarding every term with $\delta$ degree $\geq 1$.  The expansion gives
[Edit:  After correcting an error involving a missing "${}/2$" when asking a computer to generate the blob of TeX automatically...]
\begin{align*}
&\hspace{-0.5in}\text{[your photograph of an expression]} \\
\qquad\qquad{} &\approx \frac{(1+r \delta+(r\delta)^2/2) - (1+\mu \delta - \sigma \sqrt{\delta} + (\mu \delta - \sigma \sqrt{\delta})^2/2)}
    {(1 + \mu \delta + (\mu\delta)^2/2)((1+\sigma \sqrt{\delta}+(\sigma \sqrt{\delta})^2/2) - (1 - \sigma \sqrt{\delta})+(\sigma \sqrt{\delta}))^2/2)}  \\
{}&= 
\frac{1}{2}
+\sqrt{\delta } \left(-\frac{\mu }{2 \sigma }+\frac{r}{2 \sigma }-\frac{\sigma }{4}\right)
+\delta ^{3/2} \left(\frac{\mu ^2}{4 \sigma
   }+\frac{\mu  \sigma }{4}+\frac{r^2}{4 \sigma }-\frac{\mu  r}{2 \sigma
   }\right)
-\frac{\delta ^2 \mu ^2}{4}
+\delta ^{5/2} \left(-\frac{\mu ^2 \sigma }{8}-\frac{\mu  r^2}{4 \sigma }+\frac{\mu ^2 r}{4 \sigma }\right)
+\frac{\delta ^3 \mu ^3}{4}
+\delta ^{7/2} \left(\frac{\mu ^2 r^2}{8 \sigma }-\frac{\mu ^4}{8
   \sigma }\right)
\end{align*}
and it should be easy to see the result you want in the first two terms.
(If you only go to first order in $x$, you do not get the "$\frac{-\sigma}{4} = \frac{-\sigma^2/2}{2\sigma}$".)
