Question 7.T.18: Let the function f and all its derivatives up to order n be ...
Let the function f and all its derivatives up to order n be continuous on [a, b], and suppose f^{(n)} is differentiable at the point x_{0} ∈ [a, b]. If x ∈ [a, b] then
f(x) = f(x_{0}) + \frac{f^{′}(x_{0})}{1!}(x − x_{0}) + \frac{f^{′′} (x_{0})}{2!}(x − x_{0})^{2} + · · ·
+\frac{f^{(n)}(x_{0})}{n!}(x − x_{0})^{n} + \frac{f^{(n+1)}(x_{0})}{(n + 1)!}(x − x_{0})^{n+1} + E_{n}(x), (7.22)
where E_{n}(x)/(x − x_{0})^{n+1} → 0 as x → x_{0}.
Let p be the polynomial
p(x) = f(x_{0})+ \frac{f^{′}(x_{0})}{1!}(x−x_{0})+ \frac{f^{′′}(x_{0})}{2!}(x−x_{0})^2 +· · ·+ \frac{f^{(n)}(x_{0})}{n!}(x−x_{0})^{n},
and define
P (x) = f(x) − p(x), Q(x) = (x − x_{0})^{n+1}.
To prove the theorem, we have to show that
\underset{x→x_{0}}{\lim }\frac{P (x)}{Q(x)}=\frac{f ^{(n+1)}(x_{0})}{(n + 1)!}.
Observe that
p(x_{0}) = f(x{_{0}}),
p^{(k)}(x_{0}) = f^{(k)}(x_{0}), k = 1, 2, · · · , n
which implies
P(x_{0}) = P^{′}(x_{0}) = · · · = P ^{(n)}(x_{0}) = 0.
Also
Q^{(k)}(x) =\frac{(n + 1)!}{(n + 1)!(n + 1 − k)!}(x − x_{0})^{n+1−k}
⇒ Q(x_{0}) = Q^{′}(x_{0}) = · · · = Q^{(n)}(x_{0}) = 0.
To evaluate the limit of P/Q at x_{0}, we apply L’Hôpital’s rule n times:
\underset{x→x_{0}}{\lim }\frac{P(x)}{Q(x)}=\underset{x→x_{0}}{\lim }\frac{P^{′}(x)}{Q^{′}(x)}
=\underset{x→x_{0}}{\lim }\frac{P^{′′}(x)}{Q^{′′}(x)}
= · · ·
=\underset{x→x_{0}}{\lim }\frac{P^{(n)}(x)}{Q^{(n)}(x)}
=\underset{x→x_{0}}{\lim }\frac{f^{(n)}(x) − p^{(n)}(x)}{(n + 1)!(x − x_{0})}
=\underset{x→x_{0}}{\lim }\frac{f^{(n)}(x) − f^{(n)}(x_{0})}{(n + 1)!(x − x_{0})}
=\frac{f ^{(n+1)}(x_{0})}{(n + 1)!}.