= Solution
Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The <chain rule> gives
$$
f(0)=\ln s,\qquad f'(0)=\frac{t_1}{s}.
$$
The condition $t_3t_1<0$ ensures $t_1\ne0$. Since $t_0\ne0$, also $s\ne1$ and $\ln s\ne0$. Hence the first two nonzero terms are the constant and linear terms:
$$
\boxed{f(x)=\ln(1+t_0)+\frac{t_1}{1+t_0}x+O(x^2).}
$$
The inequality $s>0$, together with <continuity> of $t$, makes the <logarithm> defined on a neighbourhood of zero. The remainder is written in <Big O notation>.
Back to article page
