# Notes - Analysis II HT23, Differentiation theorem

### Flashcards

Can you state the differentiation theorem in full?

Let the real of complex power series $f(x) = \sum^\infty _ {k=0} c _ k x^k$ have radius of convergence $R \in (0, \infty]$. Then $f$ is differentiatble in $\{x : \vert x \vert < R\}$ and $f’$ is given term-by-term differentiation:

\[f'(x) = \sum^\infty_{k=1} kc_k x^{k-1}\]Suppose the power series $\sum _ {k=0}^\infty c _ k x^k$ has radius of convergence $R = [0, \infty]$. Then what is true about the radius of convergence of $\sum _ {k = 1}^\infty kc _ k x^{k-1}$?

It also has radius of convergence $R$.

Can you quickly prove that the series $\sum _ {i=1}^\infty i \lambda^{i-1}$ converges for $ \vert \lambda \vert < 1$?

Consider

\[\sum^n_{i=0} \lambda^i = \frac{1-\lambda^n}{1-\lambda}\]Differentiating both sides (the sum is finite)

\[\begin{aligned} \sum^{n}_{i=1} i\lambda^{i-1} &= \frac{(1-\lambda)(-n\lambda^{n-1})-(-1)(1-\lambda^n)}{(1-\lambda)^2} \\\\ &\to \frac{1}{(1-\lambda)^2} \end{aligned}\]for $ \vert \lambda \vert < 1$.

Suppose you’re proving the differentiation theorem for power series and you have $0 < R _ 1 < R _ 0 \le R$ and are considering $\phi _ i(x) = a _ ix^i$ on $[-R _ 1, R _ 1]$. Given that you know the series converges for $x = R _ 0$, how do you bound both $\phi _ i(x)$ and $\phi’ _ i(x)$?

and so

\[|\phi_i(x)| \le K\left( \frac{R_1}{R_0} \right)^i\]and likewise

\[|\phi_i'(x)| \le \frac{K}{R_0} i \frac{R_1}{R_0}^{i-1}\]### Proofs

Prove the differentiation theorem for power series:

Let the real of complex power series $f(x) = \sum^\infty _ {k=0} c _ k x^k$ have radius of convergence $R \in (0, \infty]$. Then $f$ is differentiatble in $\{x : \vert x \vert < R\}$ and $f’$ is given term-by-term differentiation:

\[f'(x) = \sum^\infty _ {k=1} kc _ k x^{k-1}\]

Let the real of complex power series $f(x) = \sum^\infty _ {k=0} c _ k x^k$ have radius of convergence $R \in (0, \infty]$. Then $f$ is differentiatble in $\{x : \vert x \vert < R\}$ and $f’$ is given term-by-term differentiation:

Todo.