Taylor's theorem
Explicit formulas for the remainder

Under stronger regularity assumptions on f there are several precise formulas for the remainder term R of the Taylor polynomial, the most common ones being the following.

Mean-value forms of the remainder — Let f : R → R be k + 1 times differentiable on the open interval with f continuous on the closed interval between a {\textstyle a} and x {\textstyle x}. Then

R k (x) = f (k + 1) (ξ L) (k + 1) ! (x − a) k + 1 {\displaystyle R_{k}(x)={\frac {f^{(k+1)}(\xi _{L})}{(k+1)!}}(x-a)^{k+1}}

for some real number ξ L {\textstyle \xi _{L}} between a {\textstyle a} and x {\textstyle x}. This is the Lagrange form of the remainder.

Similarly,

R k (x) = f (k + 1) (ξ C) k ! (x − ξ C) k (x − a) {\displaystyle R_{k}(x)={\frac {f^{(k+1)}(\xi _{C})}{k!}}(x-\xi _{C})^{k}(x-a)}

for some real number ξ C {\textstyle \xi _{C}} between a {\textstyle a} and x {\textstyle x}. This is the Cauchy form of the remainder.

Both can be thought of as specific cases of the following result: Consider p > 0 {\displaystyle p>0}

R k (x) = f (k + 1) (ξ S) k ! (x − ξ S) k + 1 − p (x − a) p p {\displaystyle R_{k}(x)={\frac {f^{(k+1)}(\xi _{S})}{k!}}(x-\xi _{S})^{k+1-p}{\frac {(x-a)^{p}}{p}}} for some real number ξ S {\textstyle \xi _{S}} between a {\textstyle a} and x {\textstyle x}. This is the Schlömilch form of the remainder (sometimes called the Schlömilch- Roche). The choice p = k + 1 {\textstyle p=k+1} is the Lagrange form, whilst the choice p = 1 {\textstyle p=1} is the Cauchy form.

These refinements of Taylor's theorem are usually proved using the mean value theorem, whence the name. Additionally, notice that this is precisely the mean value theorem when k = 0 {\textstyle k=0}. Also other similar expressions can be found. For example, if G (t) is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between a {\textstyle a} and x {\textstyle x}, then

R k (x) = f (k + 1) (ξ) k ! (x − ξ) k G (x) − G (a) G ′ (ξ) {\displaystyle R_{k}(x)={\frac {f^{(k+1)}(\xi)}{k!}}(x-\xi)^{k}{\frac {G(x)-G(a)}{G'(\xi)}}}

for some number ξ {\textstyle \xi } between a {\textstyle a} and x {\textstyle x}. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem. The Lagrange form is obtained by taking G (t) = (x − t) k + 1 {\displaystyle G(t)=(x-t)^{k+1}} and the Cauchy form is obtained by taking G (t) = t − a {\displaystyle G(t)=t-a}.

The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. However, it holds also in the sense of Riemann integral provided the (k + 1)th derivative of f is continuous on the closed interval.

Integral form of the remainder — Let f (k) {\textstyle f^{(k)}} be absolutely continuous on the closed interval between a {\textstyle a} and x {\textstyle x}. Then

R k (x) = ∫ a x f (k + 1) (t) k ! (x − t) k d t. {\displaystyle R_{k}(x)=\int _{a}^{x}{\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}\,dt.}

Due to the absolute continuity of f on the closed interval between a {\textstyle a} and x {\textstyle x}, its derivative f exists as an L -function, and the result can be proven by a formal calculation using the fundamental theorem of calculus and integration by parts.
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