Exercise 6.5.8

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that power series representations are unique. If we have
$$
\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n}
$$
for all $x$ in an interval $(-R, R)$, prove that $a_{n}=b_{n}$ for all $n=0,1,2, \ldots$
\item Let $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$ converge on $(-R, R)$, and assume $f^{\prime}(x)=f(x)$ for all $x \in(-R, R)$ and $f(0)=1$. Deduce the values of $a_{n}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item If we substitute $x = 0$ we get that $a_0 = b_0$. If we take the termwise derivative and then substitute $x = 0$, we get that $a_1 = b_1$. We can proceed inductively by taking the termwise derivative to show that $a_n = b_n$ for all $n$.
\item $f(0) = 1$ implies $a_0 = 1$. $f'(0) = f(0) = 1$ implies $n a_n = 1$ for $n = 1$, or $a_1 = 1$. $f^{\prime \prime}(0) = f'(0) = 1$ implies $(2) (2-1) a_2 = (2!) a_2 = 1$. We can use induction to show in general that $a_n = 1 / n!$.
 \end{enumerate}

Exercise 6.5.8

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Exercise 6.5.8

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