Analysis I MT22, Power series
Flashcards
Let $\sum c _ k z^k$ be a power series. What’s the full definition of its radius of convergence?
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
\[R := \begin{cases}
\sup \\{ |z| : \sum |c_k z^k| \text{ converges} \\} &\text{if the supremum exists} \\\\
\infty &\text{otherwise}
\end{cases}\]
if $R > 0$, then what two things follow from this about the convergence of $\sum c _ k z^k$?
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
When proving part (1) of the following
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
\[R := \begin{cases}\sup \\{ \vert z \vert : \sum \vert c _ k z^k \vert \text{ converges} \\} &\text{if the supremum exists} \\\\\infty &\text{otherwise}\end{cases}\]
then
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
Why can’t you use the comparison test with $\sum \vert c _ k R^k \vert $?
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
then
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
Because it’s the supremum, it might actually converge when $z = R$.
When proving part (2) of the following
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
\[R := \begin{cases}\sup \\{ \vert z \vert : \sum \vert c _ k z^k \vert \text{ converges} \\} &\text{if the supremum exists} \\\\\infty &\text{otherwise}\end{cases}\]
then
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
you use contradiction and assume there exists some $z$ where $ \vert z \vert > R$ and $\sum c _ k z^k$ converges. What do you know about $c _ k z^k$ that you exploit later?
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
then
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
$(c _ k z^k)$ is bounded and hence $\exists M \ge 0$ such that $ \vert c _ k z^k \vert \le M$.
Proofs
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$, defined by
\[R := \begin{cases}
\sup \\{ |z| : \sum |c_k z^k| \text{ converges} \\} &\text{if the supremum exists} \\\\
\infty &\text{otherwise}
\end{cases}\]
prove that
- $ \vert z \vert < R$ implies $\sum c _ k z^k$ converges absolutely (and hence normally).
- $ \vert z \vert > R$ implies $\sum c _ k z^k$ diverges.
Todo.