Analysis II HT23, Taylor’s theorem

Created: March 22, 2023 | Read markdown | About these notes | View in context | Study these flashcards |

Flashcards

@State Taylor’s theorem in full.

taylors-theorem

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 $f(x _ 0 + h)$?

taylor-polynomial-form

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

\[\begin{aligned} f(x _ 0 + h) &= f(x _ 0) \\ &+ f'(x _ 0)h \\ &+ \frac{f''(x _ 0)}{2!} h^2 \\ &+ \cdots \\ &+ \frac {f^{(n)}\space(x _ 0)} {n!}h^n \\ &+ \frac {f^{(n+1)}\space(x _ 0 + \theta h)}{(n+1)!}h^{n+1} \end{aligned}\]

@Define what it means for a function $f$ to be analytic at a point $x _ 0$.

analytic-function-definition-a2
\[\exists \delta > 0 \text{ s.t. } \forall \vert h \vert < \delta, f(x _ 0 + h) = \sum^\infty _ {k=0} \frac{f^{(k)}(x _ 0)}{k!}h^k\]

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?

taylor-proof-remainder-function

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?

taylor-proof-remainder-justification
\[\begin{aligned} \frac{r _ n(b)}{(b-a)^{n+1}\,} &= \frac{r _ n(b) - r _ n(a)}{(b-a)^{n+1} - (a - a)^{n+1}\,} \\ &= \frac{r _ n'(\xi _ 1)}{(n+1)(\xi _ 1 -a)^n} \\ &= \cdots \\ &= \frac{r^{(n)} _ n(\xi _ n)}{(n+1)!(\xi _ n - a)} \\ &= \frac{r^{(n+1)} _ n(\xi _ {n+1})}{(n+1)!} \end{aligned}\]

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

Todo.

@todo~



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