Asymptotic equivalence and kth derivative Let $f(x)=\log h(x)$ be a real analytic function on the open set $(0,\infty)$.

My question: If $f(x)\sim g(x)$ as $x\to 1$ where $g(x)$ is real analytic on $(0,\infty)$. Prove that there is a constant $c$ such that $f^{(k)}(x)\sim c g^{(k)}(x)$ as $x\to 1$. Next how do we find the value of this constant $c$ ?

My try: Since $f(x)\sim g(x)$ as $x\to 1$, so we have $$\lim_{x\to 1}\frac{f(x)}{g(x)}=1$$
Since $f$ is analytic at $x=1$ so it can be written as a power series $$f(x)=f_0+f_1(x-1)+f_2(x-1)^2+... $$ in an appropriate neighbourhood of $1$.
Also since $g$ is analytic at $x=1$ so it can be written as a power series $$g(x)=g_0+g_1(x-1)+g_2(x-1)^2+... $$ in an appropriate neighbourhood of $1$.
Since $f(x)\sim g(x)$ as $x\to 1$, so $g_0=f_0$. Since $f$ and $g$ are analytic, their derivatives are analytic as well and near $x=1$ they are given by the term-wise differentiated power series. Thus we have $$\lim_{x\to 1}\frac{f'(x)}{g'(x)}=\lim_{x\to 1} \frac{f_1+2f_2(x-1)+...}{g_1+2g_2(x-1)+...} $$
So we have $$\lim_{x\to 1}\frac{f'(x)}{g'(x)}=\frac{f_1}{g_1}$$
Similarly we get $$\lim_{x\to 1} \frac{(f(x))^{(k)}}{(g(x))^{(k)}}=\lim_{x\to 1}\frac{\sum_{n=k}^{\infty}\frac{f^{(n)}(1)}{(n-k)!}(x-1)^{n-k}}{ \sum_{n=k}^{\infty}\frac{g^{(n)}(1)}{(n-k)!}(x-1)^{n-k}   } $$
So we have $$\lim_{x\to 1} \frac{(f(x))^{(k)}}{(g(x))^{(k)}}=\frac{f^{(k)}(1)}{g^{(k)}(1)}$$
So taking $c=\frac{f^{(k)}(1)}{g^{(k)}(1)}$ we have $f^{(k)}(x)\sim c g^{(k)}(x)$ as $x\to 1$.
How do I find $f^{(k)}(1)$ given that $f(x)=\log h(x)$?
 A: Let $w = x - 1$. Then
$$
\log \left( {\sum\limits_{k = 0}^\infty  {\frac{{h^{(k)} (1)}}{{k!}}w^k } } \right) = \sum\limits_{k = 0}^\infty  {\frac{{f^{(k)} (1)}}{{k!}}w^k }. 
$$
Differentiating both sides with respect to $w$ gives
$$
 \cfrac{{\displaystyle\sum\limits_{k = 1}^\infty  {\cfrac{{h^{(k)} (1)}}{{(k - 1)!}}w^{k - 1} } }}{{\displaystyle\sum\limits_{k = 0}^\infty  {\cfrac{{h^{(k)} (1)}}{{k!}}w^k } }} = \sum\limits_{k = 1}^\infty  {\frac{{f^{(k)} (1)}}{{(k - 1)!}}w^{k - 1} } 
$$
or
$$
\sum\limits_{k = 0}^\infty  {\frac{{h^{(k + 1)} (1)}}{{k!}}w^k }  = \left( {\sum\limits_{k = 0}^\infty  {\frac{{f^{(k + 1)} (1)}}{{k!}}w^k } } \right)\left( {\sum\limits_{k = 0}^\infty  {\frac{{h^{(k)} (1)}}{{k!}}w^k } } \right).
$$
Performing the Cauchy product of the series on the right-hand side and equating like powers of $w$ yields
$$
\frac{{h^{(k + 1)} (1)}}{{k!}} = \sum\limits_{m = 0}^k {\frac{{f^{(m + 1)} (1)}}{{m!}}\frac{{h^{(m - k)} (1)}}{{(m - k)!}}} 
$$
or
$$
f^{(k + 1)} (1)h(1) = h^{(k + 1)} (1) - \sum\limits_{m = 0}^{k - 1} {\binom{k}{m}f^{(m + 1)} (1)h^{(m - k)} (1)} .
$$
This gives a recurrence relation for the $f^{(k)}(1)$'s. If $h(1)=0$ ($ \Leftrightarrow f(1) = 0$) we have to solve the above equation for $f^{(k)}(1)$.
