Taylor's Theorem/One Variable/Integral Version

Proof
This proof requires $f^{\paren n}$ to be absolutely continuous on $\closedint a x$, so that the Fundamental Theorem of Calculus holds.

Except at the end when the Mean Value Theorem is invoked, differentiability of $f^{\paren n}$ need not be assumed, since absolute continuity implies:


 * differentiability almost everywhere
 * the validity of the Fundamental Theorem of Calculus

provided the integrals involved are understood as Lebesgue integrals.

Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on $f$.

We first prove Taylor's Theorem with the integral remainder term.

The Fundamental Theorem of Calculus states that:


 * $\ds \int_a^x \map {f'} t \rd t = \map f x - \map f a$

which can be rearranged to:


 * $\ds \map f x = \map f a + \int_a^x \map {f'} t \rd t$

Now we can see that an application of Integration by Parts yields:

Another application yields:
 * $\ds \map f x = \map f a + \paren {x - a} \map {f'} a + \frac 1 2 \paren {x - a}^2 \map {f} a + \frac 1 2 \int_a^x \paren {x - t}^2 \map {f'} t \rd t$

By repeating this process, we may derive Taylor's theorem for higher values of $n$.

This can be formalized by applying the technique of Principle of Mathematical Induction.

So, suppose that Taylor's theorem holds for a $n$, that is, suppose that:

We can rewrite the integral using Integration by Parts.

A primitive of $\paren {x - t}^n$ as a function of $t$ is given by $\dfrac {-\paren {x - t}^{n + 1} } {n + 1}$.

So:

The last integral can be solved immediately, which leads to


 * $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$