Given that
$e^{r\Delta t}(u+d)-ud-e^{2r\Delta t} = \sigma^2\Delta t$
I would like to show that
$u=e^{\sigma\sqrt{\Delta t}}$
I know I must somehow use Taylor's approximation $e^x = 1 + x + \frac{x^2}{2}+...$ and ignore terms of $\Delta t$ higher than 1, but I can't seem to get to the value of $u$. Can someone show me this derivation?
If we assume that $ud=1$(Pretty standard approach) We find that $$ u^2 -Ru +1 =0 $$ Where $$ R = \mathrm{e}^{-r\Delta t}+\mathrm{e}^{r\Delta t}+\sigma^2\Delta t \mathrm{e}^{-r\Delta t} $$ We know we can't taylor expand now since the volatility will disappear straight away!
So solutions to the quadratic for $u$ is $$ u =\frac{R\pm\sqrt{R^2 -4}}{2} $$ Inserting $R$ back in we find $$ u =\frac{\mathrm{e}^{-r\Delta t}+\mathrm{e}^{r\Delta t}+\sigma^2\Delta t \mathrm{e}^{-r\Delta t}\pm \sqrt{ \mathrm{e}^{-2r\Delta t} + \mathrm{e}^{2r\Delta t} +2 + \sigma^4(\Delta t )^2 \mathrm{e}^{-2r\Delta t} +2\sigma^2\Delta t \left( \mathrm{e}^{-2r\Delta t} + 1\right) -4}}{2} $$ After that rather ugly expression we can taylor expand and under the root we ignore $(\Delta t)^2$ terms and outside $\Delta t$ terms.
So we end up having $$ u =\frac{2 + \sqrt{4+4\sigma^2\Delta t -4}}{2} = 1 + \pm\sigma \sqrt{\Delta t} $$ Since we are looking at the up and we require $\sigma \geq 0$ and $ud =1$ it is obvious we must take the positive sign. Thus $$ u = 1 + \sigma \sqrt{\Delta t}\approx \mathrm{e}^{\sigma \sqrt{\Delta t}} $$