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
\[\vert \phi _ i(x) \vert \le K\left( \frac{R _ 1}{R _ 0} \right)^i\]and likewise
\[\vert \phi _ i'(x) \vert \le \frac{K}{R _ 0} i \frac{R _ 1}{R _ 0}^{i-1}\]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.