Exercise 6.5.4

Question

[Term-by-term Antidifferentiation]
Assume $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$ converges on $(-R, R)$.
\begin{enumerate}[label=(\alph*)] 
\item Show
$$
F(x)=\sum_{n=0}^{\infty} \frac{a_{n}}{n+1} x^{n+1}
$$
is defined on $(-R, R)$ and satisfies $F^{\prime}(x)=f(x)$.
\item Antiderivatives are not unique. If $g$ is an arbitrary function satisfying $g^{\prime}(x)=f(x)$ on $(-R, R)$, find a power series representation for $g$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Let $N \in \mathbf{N} > R$ and split the function into
    $$
        \begin{aligned}
    F(x) &= \sum^{N-1}_{n=0} \frac{a_n}{n+1}x^{n+1} + \sum^N_{n=N} a_nx^n \left(\frac{x}{n+1}\right)\
    &\leq \sum^{N-1}_{n=0} \frac{a_n}{n+1}x^{n+1} + \sum^N_{n=N} a_nx^n \left(\frac{x}{R}\right)\
    &= \sum^{N-1}_{n=0} \frac{a_n}{n+1}x^{n+1} + \left(\frac{x}{R}\right)\sum^N_{n=N} a_nx^n \
        \end{aligned}
$$
The first term is finite, while the second term converges by the original assumption. This shows that $F(x)$ is defined on $-R,R$, at which point we can use Theorem 6.5.7 to conclude $F'(x) = f(x)$.
\item From Corollary 5.3.4, $g(x) = F(x) + k$ for some constant $k$; $k$ gets folded into the constant term of the power series.
 \end{enumerate}

Exercise 6.5.4

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Exercise 6.5.4

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