Question about the Taylor formula and its applications

Question

My author isn't too specific about a certain variable in the expression of the Taylor formula, $u$.

$$ f(x) = T_{n}(x) + (x-a)^{n+1} \int^{1}_{0} \frac{(1-u)^{n}}{n!} f^{(n+1)}[a+u(x-a)] \, du $$

My guess is that the integral is related to the radius of convergence, such that it computes the additional area/volume required to make the approximation complete.

I'm struggling to determine where this integral comes from, whether the limits he has written are always between $0$ and $1$, for all $n$, and where and how "$u$" is introduced in Taylor approximation.

Also, how is this applied to obtain the power series for transcendent functions such as:

$$ \sin(x) := \sum_{k=1}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!} $$

Answer 1

It simply comes from integration by parts:

\begin{align*} \int_{a}^{x} \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dt &= \left[ -\frac{(x-t)^{n}}{n!} f^{(n)}(t) \right]_{a}^{x}+ \int_{a}^{x} \frac{(x-t)^{n}}{n!} f^{(n+1)}(t) \, dt \\ &= \frac{(x-a)^{n}}{n!} f^{(n)}(a)+ \int_{a}^{x} \frac{(x-t)^{n}}{n!} f^{(n+1)}(t) \, dt \\ \end{align*}

Since $\displaystyle f(x) = f(a)+\int_{a}^{x} f'(t) \, dt \, ,\,$ by induction on $n$ and letting $\displaystyle u=\frac{t-a}{x-a}$ it follows that

\begin{align*} f(x) &= \sum_{k=0}^{n} \frac{(x-a)^{k}}{k!} f^{(k)}(a)+ \int_{a}^{x} \frac{(x-t)^{n}}{n!} f^{(n+1)}(t) \, dt \\ &= \sum_{k=0}^{n} \frac{(x-a)^{k}}{k!} f^{(k)}(a)+ (x-a)^{n+1} \int_{0}^{1} \frac{(1-u)^{n}}{n!} f^{(n+1)}[a+u(x-a)] \, du \end{align*}

Applications:

We can estimate the error bound of the Taylor series using mean value theorem:

$$R_{n} = \frac{(x-a)^{n+1}}{(n+1)!} f^{(n+1)} (\xi)$$ for some $a<\xi <b$.