Prove that the \(p^{\text {th }}\)-order Taylor expansion of a function \(f(x)\) has the same derivatives of order \(1, \ldots, p\) as \(f(x)\). That is, show that

\[\left.\frac{d^{j}}{d \delta^{j}} \sum_{k=0}^{p} \frac{\delta^{k} f^{(k)}(x)}{k !}ight|_{\delta=0}=f^{(j)}(x)\]

for \(j=1, \ldots, p\). What assumptions are required for this result to be true?