Let $$\mathbb{C}[[x]] := \left\{ \sum_{n\geq 0} a_n x^n \;\colon\; a_n \in \mathbb{C} \right\}$$ be the set of formal power series of $x$ and $$F(x) = \sum_{n\geq 0} a_n x^n \in \mathbb{C}[[x]], \; F(0) = a_0 = 1.$$
Exercise
i) Reason that the ring of formal power series of $F'(x) = F(x)$ leads to $a_n = \frac{1}{n!}$, $n\geq 0$ and $F(x) = \exp(x)$.
ii) Prove that $F(x)$ has a multiplicative inverse in $\mathbb{C}[[x]] \Leftrightarrow a_0 \neq 0$.
i) Do you have any hint for me how to approach this one best?
ii) I searched our lecture but didn't find any further information about the ring of formal power series. What is it's multiplicative identity?
As I understand the definition of rings there is a multiplicative identity $1$ with $x \cdot 1 = 1 \cdot x = x$ and the multiplicative inverse $x'$ of $x$ is defined by $x \cdot x' = x' \cdot x = 1$.
Where is my error in reasoning or what is that multiplicative inverse?
Thank you in advance!
Best Answer
To prove (i), consider a power series $F=\sum_{n\geq0}a_nx^n$ and suppose it is such that $F'=F$. Since $F'$ is $\sum_{n\geq1}na_nx^{n-1}=\sum_{n\geq0}(n+1)a_{n+1}x^{n}$, that $F$ and $F'$ are equal means exactly that for all $n\geq0$ we have $a_n=(n+1)a_{n+1}$ or, equivalently, $$a_{n+1}=\frac1{n+1}a_n.$$ If we fix $a_0$, there is exactly one sequence $(a_n)$ that verifies this recurrence relation, $$a_n=\frac{1}{n!}a_0, \quad\forall n\geq0.$$
If we set $a_0=1$, we get the solution you want.