Hjort and Pollard write in their article Asymptotics for minimisers of convex processes that the following expansion holds for all $u$ and $u+h$, in terms of $\pi(u) = \exp(u)/\{1+\exp(u)\}$:
$$\log \frac{1+\exp(u+h)}{1+ \exp(u)} = \pi(u)h + \frac{1}{2}\pi(u)\{1-\pi(u)\}h^{2} + \frac{1}{6}\pi(u)\{1-\pi(u)\}\gamma(u,h)h^{3}$$
where $|\gamma(u,h)| \le \exp(|h|)$.
I try to show this with Taylor expansion. I interpret $h$ to be a constant. However, I then get that the derivative of the expression at $u$ is $\pi(u+h) -\pi(u)$. This is wrong, as it should be $\pi(u)$ according to the first term on the right hand side of the equation above. I'm not able to detect the error. Could someone help me?
Let $f (u)=\log (1+\exp (u)) $. Then $f '(u)=\pi (u) $, $f''(u)=\pi (u)(1-\pi (u)) $ and $f'''(u)=\pi(u)(1-\pi (u))(1-2\pi (u))$ hence, by Lagrange theorem, we have $f (u+h)=f (u)+h\pi (u)+\frac 12h^2\pi (u)(1-\pi (u))+\frac 16f'''(v)$ for some $u <v <u+h$. If $\gamma(u,h)=\frac{f'''(v)}{f''(u)}$, then \begin{align} \log\frac{1+\exp(u+h)}{1+\exp(u)} &=f(u+h)-f(u)\\ &=h\pi (u)+\frac 12h^2\pi (u)(1-\pi (u))+\frac 16\pi (u)(1-\pi (u))\gamma(u,h) \end{align}
Since $0<\pi(u)<1$ we have $\pi'(u)=f''(u)>0$ and $0<\pi(u)\leq\pi(v)\leq\pi(u+h)<1$ and \begin{align} |\gamma(u,h)| &=\left|\frac{f'''(v)}{f''(u)}\right|\\ &=\frac{\pi(v)(1-\pi (v))\left|1-2\pi (v)\right|}{\pi(u)(1-\pi (u))}\\ &\leq\frac{\pi(u+h)\left|1-2\pi (v)\right|}{\pi(u)}\\ &=\frac{\exp(h)(1+\exp(u))\left|1-2\pi (v)\right|}{(1+\exp(v))}\\ &\leq\exp(h)\left|1-2\pi (v)\right|\\ &\leq\exp(h) \end{align}