= Solution
From $f''=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f'$ whenever those <derivatives> have been constructed. Starting with twice differentiability, this relation supplies the next two <derivatives> at each stage. Thus \b[$f$ is infinitely differentiable], and its origin series is
$$
\boxed{A\sum_{r=0}^\infty\frac{x^{2r}}{(2r)!}
+B\sum_{r=0}^\infty\frac{x^{2r+1}}{(2r+1)!}.}
$$
For a fixed $x$, both $f$ and $f'$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher <derivative> is one of these two functions. <Taylor's theorem>, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x|^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence \b[the series converges to $f(x)$ for every real $x$]. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.
At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g''=g$. Its even <derivatives> at zero are $f(a)$, and its odd ones are $f'(a)$. Consequently
$$
\boxed{f(a+h)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}h^k
=f(a)\sum_{r=0}^\infty\frac{h^{2r}}{(2r)!}
+f'(a)\sum_{r=0}^\infty\frac{h^{2r+1}}{(2r+1)!}.}
$$
This proves convergence at every centre without assuming that smoothness alone implies equality to a <Taylor series>.
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