Prove Theorem 1.13 using induction. That is, assume that

\[E_{1}(x, \delta)=\int_{x}^{x+\delta}(x+\delta-t) f^{\prime \prime}(t) d t\]

which has been shown to be true, and that

\[E_{p}(x, \delta)=\frac{1}{p !} \int_{x}^{x+\delta}(x+\delta-t)^{p} f^{(p+1)}(t) d t\]

and show that these imply

\[E_{p+1}(x, \delta)=\frac{1}{(p+1) !} \int_{x}^{x+\delta}(x+\delta-t)^{p} f^{(p+2)}(t) d t .\]

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Theorem 1.13 (Taylor). Let f be a function that has p+ 1 bounded and continuous derivatives in the interval (x, x+8). Then P Sk f(k) (x) f(x+6)= + Ep(x, 5), k! k=0 where 1 Ep(x, 6) = (x+8-t) f(p+1)(t)dt: SP+1 f(p+1)() = p! (p+1)! for some = (x, x+6).