Notes - Analysis II HT23, Taylor’s theorem

Created: March 22, 2023

Flashcards

Can you state Taylor’s theorem in full?

Suppose:

  • $f : [a, b] \to \mathbb R$
  • $f, f’, \ldots, f^{(n)}$ exist and are continuous on $[a, b]$.
  • $f^{(n+1)}$ exists on $(a, b)$.
  • $n \ge 0$

Then $\exists \xi \in (a, b)$ such that

\[\begin{aligned} f(b) &= f(a) \\\\ &+ f'(a)(b - a) \\\\ &+ \frac{f''(a)}{2!} (b-a)^2 \\\\ &+ \cdots \\\\ &+ \frac {f^{(n)}\space(a)} {n!}(b-a)^n \\\\ &+ \frac {f^{(n+1)}\space(\xi)}{(n+1)!}(b-a)^{n+1} \end{aligned}\]

Assuming all the conditions are met, what does the Taylor polynomial look like for $f(x _ 0 + h)$?

For some $\theta \in (0, 1)$

\[\begin{aligned} f(x_0 + h) &= f(x_0) \\\\ &+ f'(x_0)h \\\\ &+ \frac{f''(x_0)}{2!} h^2 \\\\ &+ \cdots \\\\ &+ \frac {f^{(n)}\space(x_0)} {n!}h^n \\\\ &+ \frac {f^{(n+1)}\space(x_0 + \theta h)}{(n+1)!}h^{n+1} \end{aligned}\]

What does it mean for a function $f$ to be analytic at a point $x _ 0$?
\[\exists \delta > 0 \text{ s.t. } \forall |h| < \delta, f(x_0 + h) = \sum^\infty_{k=0} \frac{f^{(k)}(x_0)}{k!}h^k\]

Taylor’s theorem states that:

Let $f : [a, b] \to \mathbb R$. Let $n \ge 0$ be such that

  • $f, f’, \ldots, f^{(n)}$ exist and are continuous on $[a, b]$.
  • $f^{(n+1)}$ exists on $(a, b)$. Then there exists $\xi \in (a, b)$ such that
\[\begin{aligned} f(b) &= f(a) \\\\ &+ f'(a)(b - a) \\\\ &+ \frac{f''(a)}{2!} (b-a)^2 \\\\ &+ \cdots \\\\ &+ \frac {f^{(n)}\space(a)} {n!}(b-a)^n \\\\ &+ \frac {f^{(n+1)}\space(\xi)}{(n+1)!}(b-a)^{n+1} \end{aligned}\]

What function do you consider for the proof, and what do you want to show?

Consider

\[r_n(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\]

(i.e. the remainder), and you want to show

\[\frac{r_n(b)}{(b-a)^{n+1}\\,} = \frac{r_n^{(n+1)}(\xi)}{(n+1)!}\]

Taylor’s theorem states that:

Let $f : [a, b] \to \mathbb R$. Let $n \ge 0$ be such that

  • $f, f’, \ldots, f^{(n)}$ exist and are continuous on $[a, b]$.
  • $f^{(n+1)}$ exists on $(a, b)$. Then there exists $\xi \in (a, b)$ such that
\[\begin{aligned} f(b) &= f(a) \\\\ &+ f'(a)(b - a) \\\\ &+ \frac{f''(a)}{2!} (b-a)^2 \\\\ &+ \cdots \\\\ &+ \frac {f^{(n)}\space(a)} {n!}(b-a)^n \\\\ &+ \frac {f^{(n+1)}\space(\xi)}{(n+1)!}(b-a)^{n+1} \end{aligned}\]

For the proof, you consider

\[r_n(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\]

(i.e. the remainder), and you want to show

\[\frac{r_n(b)}{(b-a)^{n+1}\\,} = \frac{r_n^{(n+1)}(\xi)}{(n+1)!}\]

Can you justify this, given that $r _ n$ has a certain property about its $i$-th derivative?
\[\begin{aligned} \frac{r_n(b)}{(b-a)^{n+1}\\,} &= \frac{r_n(b) - r_n(a)}{(b-a)^{n+1} - (a - a)^{n+1}\\,} \\\\ &= \frac{r_n'(\xi)}{(n+1)(\xi_1 -a)^n} \\\\ &= \cdots \\\\ &= \frac{r^{(n)}_n(\xi_n)}{(n+1)!(\xi_n - a)} \\\\ &= \frac{r^{(n+1)}_n(\xi_{n+1})}{(n+1)!} \end{aligned}\]

Proofs

Prove Taylor’s theorem:

Let $f : [a, b] \to \mathbb R$. Let $n \ge 0$ be such that

  • $f, f’, \ldots, f^{(n)}$ exist and are continuous on $[a, b]$.
  • $f^{(n+1)}$ exists on $(a, b)$. Then there exists $\xi \in (a, b)$ such that
\[\begin{aligned} f(b) &= f(a) \\\\ &+ f'(a)(b - a) \\\\ &+ \frac{f''(a)}{2!} (b-a)^2 \\\\ &+ \cdots \\\\ &+ \frac {f^{(n)}\space(a)} {n!}(b-a)^n \\\\ &+ \frac {f^{(n+1)}\space(\xi)}{(n+1)!}(b-a)^{n+1} \end{aligned}\]

Todo.



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