Continuous Approximation for The Kelly Criterion I am trying to follow the derivation of Kelly Criterion, the continuous case. Dr. Thorp shows the basics of the derivation here, pg. 22. With initial capital $V_0$, betting fraction $f$, and $X$ is a random variable representing returns where 
$$ P(X = m+s) = P(X = m-s) = 0.5$$
The final capital is,
$$ V(f) = V_0 (1 + (1-f)r + fX)  $$
$$ V(f) = V_0 (1 + r + f(X - r))  $$
His eventual goal is to find the $f$ for the maximum $E[\log(V_f)]$, and do this on a continuous scale. So he subdivides the time into $n$ pieces, $m$, $s^2$, and $r$ are replaced by $m/n$, $s^2/n$ and $r/n$ respectively,
$$ P(X_i = m/n + s/\sqrt{n}) = P(X_i = m/n - m/\sqrt{n}) = 0.5$$
$$ V_n(f)/V_0 = \prod _{i=1}^{n} (1 + r + f(X_i - r))  $$
then says take log of both sides, and apply the expectation operator. I did that,
$$ \log V_n(f)/V_0 = \sum _{i=1}^{n} \log (1 + r + f(X_i - r))  $$
$$ E[\log V_n(f)/V_0] = \sum _{i=1}^{n} E[\log (1 + r + fX_i - fr)]  $$
This is where I get stuck, Thorp mentions "we expand the result in a power series", and I've seen a similar trick in a different book, pg 137, where the author reaches a statement like $1/1+fg$ after the derivative on a log, and he turned that into $1 - fg + ..$. However I am not able to reach a similar statement.
$$  = n E[\log (1 + r + fX_n - fr)]  $$
Thorp eventually reaches a formula like 
$$ g(f) = r + f(m-r) - s^2f^2/2 + O(n^{-1/2})$$
Any ideas? 
Thanks,
 A: Per @Did's comment, the derivation goes somewhat like this:
$$ P(X = m+s) = P(X = m-s) = 0.5$$
which is compatible with the fact that returns are normal. 
With capital $V_0$, allocation $f$, risk-free return $r$
$$ V(f) = V_0 (1 + (1-f)r + fX)  $$
Rearrange
$$ V(f) = V_0 (1 + r + f(X - r))  $$
Divide time into $n$ pieces,
$$ P(X_i = m/n + s/\sqrt{n}) = P(X_i = m/n - m/\sqrt{n}) = 0.5$$
$$ V_n(f)/V_0 = \prod _{i=1}^{n} (1 + r/n + f(X_i - r/n))  $$
We are trying to maximize the log expetation,
$$ E[\log V_n(f)/V_0] = g(f) = n E[\log (1 + (r/n) + f(X_n - (r/n)))]  $$
In order to get rid of the log, we use power series
$$ \log(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} + .. $$
It's sufficient to use the first two terms. 
$$ \log \bigg( 
1 + \frac{r}{n} + f \big( \frac{m}{n} + \frac{Us}{\sqrt{n}} - \frac{r}{n} \big) 
\bigg)  $$
Let's look inside the log, the $u$ part
$$ u = \frac{r}{n} + f \big( \frac{m}{n} + \frac{Us}{\sqrt{n}} - \frac{r}{n} \big)  $$
.. and what happens to the square of $u$
$$ u^2 = \frac{r^2}{n^2} + f^2(..)^2 + 2 \frac{2}{n}f(...) $$
We are not interested in terms smaller than $O(1/n \sqrt{n})$. So from the statement below for example only $\frac{f^2U^2s^2}{n}$ will "make it". The rest is gone. 
$$ \big( \frac{m}{n} + \frac{Us}{\sqrt{n}} - \frac{r}{n} \big)^2 
= \frac{m^2}{n^2} + \frac{Usm}{\sqrt{n}n} - ...
$$
We end up with
$$ 
g(f)/n = 
E\big[
\frac{r}{n} + f \big( \frac{m}{n} + \frac{Us}{\sqrt{n}} - \frac{r}{n} \big) + 
\frac{f^2U^2s^2}{2n} + 
O(1/n \sqrt{n})
\big]
$$
$$ 
=
\frac{r}{n} + f \big( \frac{m}{n} + \frac{E[U]s}{\sqrt{n}} - \frac{r}{n} \big) + 
\frac{f^2E[U^2]s^2}{2n} + 
O(1/n \sqrt{n})
\big]
$$
Since $E[U] = 0, E[U^2] = 1$ 
$$ 
=
\frac{r}{n} + f \big( \frac{m}{n} - \frac{r}{n} \big) + 
\frac{f^2s^2}{2n} + 
O(1/n \sqrt{n})
$$
$$ g(f) = r + f(m-r) + f^2s^2/2 + O(1/n \sqrt{n}) $$
As  $n \to \infty$ 
$$ g(f) = r + f(m-r) + f^2s^2/2 $$
Sources
The Kelly Criteron
Blog
