Editing Taylor's theorem (section)

=== Derivation for the mean value forms of the remainder ===

Let ''G'' be any real-valued function, continuous on the closed interval between <math display=inline>a</math> and <math display=inline>x</math> and differentiable with a non-vanishing derivative on the open interval between <math display=inline>a</math> and <math display=inline>x</math>, and define

<math display="block"> F(t) = f(t) + f'(t)(x-t) + \frac{f''(t)}{2!}(x-t)^2 + \cdots + \frac{f^{(k)}(t)}{k!}(x-t)^k.
</math>

For <math> t \in [a,x] </math>. Then, by [[mean value theorem#Cauchy's mean value theorem|Cauchy's mean value theorem]],

{{NumBlk|:|<math> \frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)}</math>|{{EquationRef|★★★}}}}

for some <math display="inline">\xi</math> on the open interval between <math display=inline>a</math> and <math display=inline>x</math>. Note that here the numerator <math display="inline">F(x)-F(a)=R_k(x)</math> is exactly the remainder of the Taylor polynomial for <math display="inline">y=f(x)</math>. Compute

<math display="block">\begin{align}
F'(t) = {} & f'(t) + \big(f''(t)(x-t) - f'(t)\big) + \left(\frac{f^{(3)}(t)}{2!}(x-t)^2 - \frac{f^{(2)}(t)}{1!}(x-t)\right)  +  \cdots \\
& \cdots + \left( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\right)
= \frac{f^{(k+1)}(t)}{k!}(x-t)^k,
\end{align}</math>

plug it into ({{EquationNote|★★★}}) and rearrange terms to find that

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)}.</math>

This is the form of the remainder term mentioned after the actual statement of Taylor's theorem with remainder in the mean value form.
The Lagrange form of the remainder is found by choosing <math> G(t) = (x-t)^{k+1} </math> and the Cauchy form by choosing <math> G(t) = t-a</math>.

'''Remark.''' Using this method one can also recover the integral form of the remainder by choosing

<math display="block"> G(t) = \int_a^t \frac{f^{(k+1)}(s)}{k!} (x-s)^k \, ds,</math>

but the requirements for ''f'' needed for the use of mean value theorem are too strong, if one aims to prove the claim in the case that ''f''{{i sup|(''k'')}} is only [[absolutely continuous]]. However, if one uses [[Riemann integral]] instead of [[Lebesgue integral]], the assumptions cannot be weakened.