My last question is related.
At the top of p. 529, it says,
"From the Taylor series expansion for $Z$ we find that the yield to maturity is given by
$$-\frac{log Z(r,t;T)}{(T-t)}\approx-a+\left(\frac{1}{2}a^{2}-b\right)(T-t)+\left(ab-c-\frac{1}{3}a^{3}\right)(T-t)^{2}+\dots$$
for short times to maturity."
We know that we derive yield to maturity from the inverse of the zero coupon bond equation
$$Z(r,t;T)=e^{-r(T-t)}$$
by taking logs and dividing by $(T-t)$ and multiplying both sides by $-1$.
If we plug the solutions for $a(r)$, $b(r)$, and $c(r)$ into the series expansion of $Z$ from p.528 we have that $Z$ is equal to
$$ Z\approx-r(T-t)+(\frac{1}{2}r^2-\frac{1}{2}(u-\lambda w))(T-t)^2+... $$
But the solution provided on p. 529 shows that no such substitutions for the values of a, b, and c have yet been made.
Taking logs, dividing by $(T-t)$, and multiplying both sides by $-1$ on our equation does not give us
$$-\frac{log Z(r,t;T)}{(T-t)}\approx-a+\left(\frac{1}{2}a^{2}-b\right)(T-t)+\left(ab-c-\frac{1}{3}a^{3}\right)(T-t)^{2}+\dots$$
and neither does starting with
$$Z\approx 1+a(r)(T-t)+b(r)(T-t)^2+c(r)(T-t)^3$$
and taking logs, etc..
So, obviously I am starting from the wrong place. Can you help me see which $Z$ to start with? I don't believe it is the computations that I am having trouble with, I am having trouble seeing the plan, as Polya would say.
Thanks in advance.