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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!

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  • $\begingroup$ The multiplicative identity is the power series $1$. $\endgroup$ Jun 4 '11 at 15:55
  • $\begingroup$ i) seems to have been stated in a somewhat confused way. $\endgroup$ Jun 4 '11 at 16:13
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HINT $\:$ for ii) $\:$ Using your previous exercise on composition of formal power series, you can reduce computation of general inverses to that of $\rm\displaystyle \frac{1}{1+x}\ $ by using $\rm\displaystyle\ \frac{1}{c+x\:f}\ =\ \frac{1}c\ \frac{1}{1+x\:f/c}\ $ i.e. substitute $\rm\ xf/c\ $ for $\rm\:x\:$ in the power series for $\rm\displaystyle\ \frac{1}{1+x}$

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  • $\begingroup$ That's quite nice. Maybe even a little too slick for the first time you do the problem, but quite a wonderful perspective to have. $\endgroup$
    – Aaron
    Jun 4 '11 at 16:38
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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.

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For (1), given a power series, what is its derivative? If two power series are equal, what can you say about the coefficients of $x^n$?

For (2), show that if $a_0\neq 0$, you can find a polynomial $P_n$ such that $F\cdot P_n = 1+0t + 0t^2+\ldots +0t^n + b_{n+1}t^{n+1}+ b_{n+2}t^{n+2}+\ldots$, and that you can choose $P_{n+1}=P_n + c_{n+1}x^{n+1}$. Then the "limit" of the $P_n$ will be a power series which is the multiplicative inverse of $F$.

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