Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 81

$p(x+1)=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$

$p(x+1)=\Sigma_{i=0}^{\infty}\frac{p^{i}({a})}{i!}(x+1-a)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(x+1-x)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(1)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$ Hence, $p(x+1)=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$