combinatorics – KindiakMath

Problem 1. Let $f$ be differentiable on $\mathbb R$ and

$\displaystyle \lim_{x \to -\infty} f(x) = \lim_{x \to \infty} f(x)= 0.$

Prove that there exists $c \in \mathbb R$ such that $f'(c) = 0$ .

(Click for Solution)

Solution. The main idea is that $f(x) \approx 0$ for sufficiently large $|x|$ .

If $f = 0$ identically, then $f' = 0$ , and in particular, $f'(0) = 0$ . Now suppose $f \neq 0$ . Then there exists $x_0 \in \mathbb R$ such that $f(x_0) \neq 0$ . By applying subsequent arguments on $\pm f$ , we may assume without loss of generality that $f(x_0) > 0$ .

Define $\epsilon_0 := f(x_0)/2$ . Use the limits to find $N > |x_0| \geq 0$ such that

$|x| \geq N \quad \Rightarrow \quad f(x) \leq |f(x)| < \epsilon_0.$

In particular, $\max\{ f(-N), f(N)\} < \epsilon_0 < f(x_0)$ . Apply the intermediate value theorem on the intervals $[-N, x_0]$ and $[x_0, N]$ respectively to find $c_1 \in (-N, x_0)$ and $c_2 \in (x_0, N)$ such that

$f(c_1) = f(c_2) = \epsilon_0.$

Now apply Rolle’s theorem to obtain $c \in (c_1,c_2) \subseteq \mathbb R$ such that $f'(c) = 0$ , as required.

—Joel Kindiak, 25 Oct 24, 2153H

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 criterion 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