I'm studying on Ahlfors, complex analysis.
Let's fix $f\colon D \subseteq \mathbb{C} \to \mathbb{C}$ a holomorphic function in the open disk $D$.
Then $\forall \,n \in \mathbb{N} \quad f^{(n)}(z)=\frac{n!}{2\pi i}\oint_C \frac{f(w)}{(w-z)^{n+1}}\,\text{d}w$ for all $z$ inside the positive-oriented circle $C$ contained in $D$.
In order to prove this statement, my textbook uses the following lemma:
$\forall \,n \in \mathbb{N} \quad \forall \,z \in D \setminus C \quad \exists \,F_n'(z)=nF_{n+1}(z)$, where $F_n(z)=\oint_C \frac{f(w)}{(w-z)^n}\,\text{d}w$.
Let's prove it by induction.
Suppose that $\exists \,F_{n-1}'(z)=(n-1)F_n(z)$ with $z \in D \setminus C$ fixed.
Now, by using the identity $\frac{1}{(u-w)^n}=\frac{1}{(u-w)^{n-1}(u-z)}+\frac{w-z}{(u-w)^n(u-z)}$, we have:
$\forall \,w \in D \setminus C \quad \frac{F_n(w)-F_n(z)}{w-z}=\frac{1}{w-z}\left(\oint_C \frac{f(u)}{(u-w)^{n-1}(u-z)}\,\text{d}u-\oint_C \frac{f(u)}{(u-z)^{n-1}(u-z)}\,\text{d}u\right)+$
$+\oint_C \frac{f(u)}{(u-w)^n(u-z)}\,\text{d}u$.
Here I get stucked. The textbook says that the first term is equal by inductive hypotesis to $(n-1)F_{n+1}(z)$ as $w \to z$, while the second term is equal to $F_{n+1}(z)$ as $w \to z$. I really can't understand what kind of trick the book does.
Best Answer
Copying from Ahlfors, we have
$\displaystyle F_n(z)-F_n(z_0)=\int_{\gamma}\frac{\varphi(\xi)d\xi}{(\xi-z)^{n-1}(\xi-z_0)}-\int_{\gamma}\frac{\varphi(\xi)d\xi}{(\xi-z_0)^n}+(z-z_0)\int_{\gamma}\frac{d\xi}{(\xi-z)^n((\xi-z_0)}.$
Write the first two terms as
$\displaystyle F_n(z)-F_n(z_0)=\int_{\gamma}\frac{\varphi(\xi)/(\xi-z_0)d\xi}{(\xi-z)^{n-1}}-\int_{\gamma}\frac{\varphi(\xi)/(\xi-z_0)d\xi}{(\xi-z_0)^{n-1}}:=G_{n-1}(z)-G_{n-1}(z_0)$
Now, divide this by $z-z_0$ and with $\varphi/(\xi-z_0)$ in place of $\varphi$, apply the inductive hypothesis to $G$ to see that $F_n'(z_0)=G_{n-1}'(z_0)=(n-1)G_n(z_0)=(n-1)F_{n+1}(z_0).$
Finally, divide the third term by $z-z_0$, and show directly that
$\displaystyle\int_{\gamma}\frac{d\xi}{(\xi-z)^n((\xi-z_0)}\to F_{n+1}(z_0)$ as $z\to z_0$.
It follows that $F_n'(z_0)=(n-1)F_{n+1}(z_0)+F_{n+1}(z_0)=nF_{n+1}(z_0).$