Taylor Series#

Definition: Taylor Polynomial

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is \(m\)-times differentiable at \(p \in \mathcal{D}\).

The \(m\)-th Taylor polynomial of \(f\) at \(p\) is the following function:

\[\sum_{k = 0}^m \frac{f^{(k)}(p)}{k!}(x-p)^k\]

Notation

\[T_m(p; x)\]

The Taylor polynomial \(T_m(p, x)\) has the following polynomial expansion:

\[T_m(p; x) = f(p) + f'(p)(x-p)+\frac{f''(p)}{2!}(x-p)^2 + \cdots + \frac{f^{(m)}(p)}{m!}(x-p)^m\]

Taylor's Theorem

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \((a,b) \subseteq \mathcal{D}\) and let \(p \in (a,b)\). Let \(R_{m+1}(p;x)\) denote the remainder of \(f\) and its \(m\)-th Taylor polynomial at \(p\):

\[R_{m+1}(p; x) = f(x) - T_m (p; x)\]

If \(f\) is \((m+1)\)-times continuously differentiable on \((a,b)\), then \(R_{m+1}(p;x)\) can be obtained by the following integral from \(p\) to \(x\) :

\[R_{m+1}(p; x) = \frac{1}{m!}\int_p^x (x-t)^m f^{(m+1)}(t) \,\mathrm{d}t\]

Moreover, for each \(x \in (a, b)\), there exists some \(\xi_x \in (p, x)\) or \(\xi_x \in (x,p)\) such that

\[R_{m+1}(p; x) = \frac{f^{(m+1)}(\xi_x)}{(m+1)!}(x-p)^{m+1}.\]

Notation

Sometimes, we substitute \(h = x - p\):

\[R_{m+1}(p; p + h) = f(p + h) - T_m (p; p + h)\]

\[R_{m+1}(p; p + h) = \frac{f^{(m+1)}(\xi_x)}{(m+1)!}h^{m+1}\]

Example: \(\mathrm{e}^x\)

For the exponential function \(f(x) = \mathrm{e}^x\) we have \(f^{(m)}(x) = \mathrm{e}^x\). For its \(m\)-th Taylor polynomial at \(0\):

\[\begin{aligned}f(x) & = T_m(0; x) + R_{m+1}(0; x) \\ \mathrm{e}^x & = \sum_{k=0}^m \frac{x^k}{k!} + R_{m+1}(0; x)\end{aligned}\]

The theorem tells us that there is some \(\xi_x \in (0, x)\) with

\[R_{m+1}(0; x) = \frac{\mathrm{e}^{\xi_x}}{(m+1)!}x^{m+1}.\]

Proof

TODO

Theorem: Remainder and Bachmann-Landau

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \((a,b) \subseteq \mathcal{D}\) and let \(p \in (a,b)\). Let \(R_{m+1}(p;p + h)\) denote the remainder of \(f\) and its \(m\)-th Taylor polynomial at \(p\):

\[R_{m+1}(p; p + h) = f(p + h) - T_m (p; p + h)\]

If \(f\) is \(m\)-times differentiable at \(p\), then \(R_{m+1}(p; p+h)\) is little O of \(h^m\):

\[R_{m+1}(p; p + h) = o(h^m) \qquad \text{for} \qquad h \to 0\]

If \(f\) is \((m+1)\)-times differentiable at \(p\), then \(R_{m+1}(p; p+h)\) is big O of \(h^{m+1}\):

\[R_{m+1}(p; p + h) = O(h^{m+1}) \qquad \text{for} \qquad h \to 0\]

Proof

TODO

Definition: Taylor Series

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is infinitely differentiable at \(p \in \mathcal{D}\).

The Taylor series of \(f\) at \(p\) is the following power series

\[\sum_{k = 0}^{\infty} \frac{f^{(k)}(p)}{k!}(x-p)^k\]

Notation

\[T(p; x)\]

The Taylor series of a function \(f\) is useful because it can sometimes allow us to express \(f\) as a power series.

Theorem: Taylor Series-Function Equality

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is infinitely differentiable at \(p \in \mathcal{D}\) and let \(x \in \mathcal{D}\).

The Taylor series \(T_{f}(p; x)\) converges to \(f(x)\) if and only if the sequence of the remainders of its Taylor polynomials at \(p\) converges to \(0\).

\[f(x) = T_{f}(p; x) \iff \lim_{n \to \infty} R_n(p; x) = 0\]

Proof

TODO

Theorem: Taylor Series-Function Equality via Derivatives

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function, let \(p, x \in \mathcal{D}\) and let \(I \subset \mathcal{D}\) be the compact interval whose endpoints are \(p\) and \(x\).

If \(f\) is infinitely differentiable on an open interval containing \(I\) and there exist \(A, B \in \mathbb{R}\) such that

\[|f^{(n)}(t)| \le A\cdot B^n\]

for all \(n \in \mathbb{N}\) and all \(t \in I\), then \(f(x)\) is equal to the Taylor series \(T_{f}(p; x)\):

\[f(x) = T_f (p; x)\]

Proof

From Taylor's theorem, we know that there exists some \(\xi_x\) in the open interval whose endpoints are \(x\) and \(p\) such that

\[R_n (p; x) = \frac{f^{(n)}(\xi_x)}{(n)!}(x-p)^n.\]

We have

\[|R_n (p; x)| = \left\vert \frac{f^{(n)}(\xi_x)}{(n)!}(x-p)^n \right\vert \le \frac{AB^n}{n!}|x-p|^n\]

for all \(n \in \mathbb{N}\). Since \(\frac{AB^n}{n!}|x-p|^n\) [converges](./Limits%20(Real%20Functions.md) to \(0\) for \(n \to \infty\) and since

\[0 \le |R_n (p; x)| \le \frac{AB^n}{n!}|x-p|^n,\]

the [sandwich theorem](./Limits%20(Real%20Functions.md) tells us that

\[\lim_{n \to \infty} |R_n (p; x)| = 0\]

and so

\[\lim_{n \to \infty} R_n (p; x) = 0.\]