The Exponential Anti-hero

Any Taylor-based calculus joke links to the Taylor series. This is an incredibly powerful technique to approximate non-polynomial functions like $\exp(x) \equiv e^x$ using polynomials.

For this post, we will establish the existence of a Taylor series centred at $0$ , known as a Maclaurin series.

Definition. Define the $n$ -th derivative of $f$ by

$\displaystyle f^{(n)} := \begin{cases} f & \text{if}\, n = 0, \\ (f^{(n-1)})' & \text{if}\, n \geq 1. \end{cases}$

Theorem 1 (Taylor’s Theorem). Suppose $f : (-r, r) \to \mathbb R$ is a real-valued function that is $n$ -times differentiable (that is, $f^{(k)}$ exists for $k=0,1,\dots,n$ on $(-r, r)$ ). Then for any $x \in (0, r)$ , there exists $c \in (0, x)$ such that

$\displaystyle f(x) = f(0) + f'(0)x + \cdots + \frac{f^{(n-1)}(0)}{(n-1)!} x^{n-1} + \frac{f^{(n)}(c)}{n!}x^{n}$ .

More concisely by using summation notation,

$\displaystyle f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} x^k + \frac{f^{(n)}(c)}{n!} x^{n}.$

Before we prove the result, we observe that the case $n=1$ is the vanilla mean value theorem. Furthermore, we actually need to use this special case to prove the general case.

Proof. Fix $x \in (0, r)$ . Define $M$ by

$\displaystyle M := \frac 1{x^n} \left(f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} x^k\right).$

This ensures that

$\displaystyle f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} x^k + M x^{n}.$

Define the $n$ -differentiable function $g : [0, x] \to \mathbb R$ by

$\displaystyle g(t) := f(t) - \left(\sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} t^k + Mt^n\right)$

Direct substitutions yield $g(0) = g'(0) = \cdots = g^{(n-1)}(0)= g(x) = 0$ . Apply the mean value theorem to $g(0) = g(x) = 0$ to find $c_1 \in (0, x)$ such that

$g'(0) = g'(c_1) = 0.$

Apply the mean value theorem $(n-2)$ more times to find $c_{n-1} \in (0, x)$ such that

$g^{(n-1)}(0) = g^{(n-1)}(c_{n-1}) = 0.$

Apply the mean value theorem one last time to find $c_n \in (0, c_{n-1}) \subseteq (0, c_n) \subseteq (0, x)$ such that

$\displaystyle f^{(n)}(c_n) - n! M = g^{(n)}(c_n) = 0 \quad \iff \quad M = \frac{f^{(n)}(c_n)}{n!}.$

Back-substituting yields

$\displaystyle f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} x^k + \frac{f^{(n)}(c)}{n!} x^{n}.$

To use this theorem we need to ensure that $(f^{(n)}(c) / n!) x^{n} \to 0$ as $n \to \infty$ . One way that works is if there exists $M > 0$ such that for any $n$ and for any $t \in (-r, r)$ , $|f^{(n)}(t)| \leq M$ , since

$\displaystyle \left|\frac{f^{(n)}(c)}{n!} x^{n}\right| \leq M \cdot \frac {r^n}{n!} \to 0,$

where $r^n / n! \to 0$ can be established using real-analytic techniques. This works well in the case of the exponential function

Theorem 2. For any $x \in \mathbb{R}$ ,

$\displaystyle \exp(x) \equiv e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}.$

This is the Maclaurin series of $\exp$ .

Proof. We will suppose $x > 0$ for simplicity. Using $\exp' = \exp$ , for any integer $n \geq 0$ , $f^{(n)}(0) = 1$ and for any $t \in (0, 2x)$ , $|f^{(n)}(t)| \leq e^{2x}$ .

Similarly, we can establish the Maclaurin series for $\sin$ and $\cos$ .

Theorem 3. For any $x \in \mathbb{R}$ ,

$\displaystyle \sin(x) = \sum_{k=0}^{\infty} \frac{{(-1)}^k x^{2k+1}}{(2k+1)!},\quad \cos(x) = \sum_{k=0}^{\infty} \frac{{(-1)}^k x^{k}}{(2k)!}.$

Finally, we can formulate the general Taylor series based on the Maclaurin series.

Theorem 4 (Taylor’s Theorem). Suppose $f : (a-r, a+r) \to \mathbb R$ is a real-valued function that is $n$ -times differentiable. Then for any $x \in (a, a+r)$ , there exists $c \in (a, x)$ such that

$\displaystyle f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!} (x-a)^k + \frac{f^{(n)}(c)}{n!} (x-a)^{n}.$

Proof. Apply the vanilla Taylor’s theorem to the transformed function $g:= f(\cdot + a) : (-r, r) \to \mathbb{R}$ .

—Joel Kindiak, 22 Oct 24, 1517H

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