Problem 2 - Taylor Expansion

For a function $f(x)$ on the closed interval $[a,b]$ which is differentiable $n+1$ times, there exists the Taylor expansion

$$f(b) = f(a) + (b-a)f'(a) + \frac{(b-a)^2}{2!}f''(a) + \cdots + \frac{(b-a)^n}{n!}f^{(n)}(a) + R_{n+1}.$$

1.

Compute $F'(x)$ for $F(x) = f(b) - \sum_{k=0}^n \frac{(b-x)^k}{k!}f^{(k)}(x).$

2.

Show $R_{n+1} = \frac{(b-a)^{n+1}}{(n+1)!}f^{(n+1)}(\xi_1)$, $(a < \xi_1 < b)$.

3.

Show $R_{n+1} = \frac{(b-a)(b-\xi_2)^n}{n!}f^{(n+1)}(\xi_2)$, $(a < \xi_2 < b)$.

4.

Prove, by induction concerning $n$, $R_{n+1} = \frac{1}{n!}\int_a^b (b-x)^n f^{(n+1)}(x)dx$.

1.

$$F(x) = f(b) - f(x) - \sum_{k=1}^n \frac{(b-x)^k}{k!}f^{(k)}(x)$$

$$F'(x) = - f'(x) + \sum_{k=1}^n \left( \frac{(b-x)^{k-1}}{(k-1)!}f^{(k)}(x) - \frac{(b-x)^k}{k!}f^{(k+1)}(x)\right)$$

$$F'(x) = - \frac{(b-x)^n}{n!}f^{(n+1)}(x)$$

2.

We first notice that $R_{n+1} = F(a)$. Let us assume that we are given the following function:

$$\phi(x) = F(x) - \frac{(b-x)^{n+1}}{(n+1)!}\frac{F(a)(n+1)!}{(b-a)^{n+1}}$$

We have both $\phi(a)=0$ and $\phi(b)=0$ (because $F(b)=0$). Thus, by the Rolle theorem, $\exists \xi_1$ such that $a<\xi_1<b$ and $\phi(\xi_1) = 0$.

$$\phi'(x) = F'(x) + \frac{(b-x)^n}{n!}\frac{F(a)(n+1)!}{(b-a)^{n+1}}$$

$$\phi'(\xi_1) = - \frac{(b-\xi_1)^n}{n!}f^{(n+1)}(\xi_1) + \frac{(b-\xi_1)^n}{n!}\frac{F(a)(n+1)!}{(b-a)^{n+1}} = 0$$

$$f^{(n+1)}(\xi_1) = R_{n+1}\frac{(n+1)!}{(b-a)^{n+1}}$$

$$R_{n+1} = \frac{(b-a)^{n+1}}{(n+1)!}f^{(n+1)}(\xi_1)$$

3.

$F$ is differentiable because $f$ is $n+1$ times differentiable, which means $\exists \xi_2 \in ]a,b[$ such that:

$$\frac{F(b) - F(a)}{b-a} = F'(\xi_2)$$

$$R_{n+1} = \frac{(b-a)(b-\xi_2)^n}{n!}f^{(n+1)}(\xi_2)$$

4.

For $n=0$, we have by definition $f(b) = f(a) + R_1$ and $\int_a^b f'(x)dx = f(b)-f(a)$, and the property is true for $n=0$. Let us assume that the property is true for $n=k-1$, i.e. we have:

$$R_{k} = \frac{1}{(k-1)!}\int_a^b (b-x)^{k-1}f^{(k)}(x)dx$$

By definition, we have:

$$R_{k+1} = R_k - \frac{(b-a)^k}{k!}f^{(k)}(a)$$

which is merely the decomposition of the following integral:

$$\frac{1}{k!}\int_a^b (b-x)^kf^{(k+1)}(x)dx = \frac{1}{k!}\left( \left[(b-x)^kf^{(k)}(x)\right]_a^b + \int_a^b n(b-x)^{k-1}f^{(k)}dx\right)$$

Therfore:

$$R_{k+1} = \frac{1}{k!}\int_a^b (b-x)^kf^{(k+1)}(x)dx$$

and the property has been proved by induction.