Applications of Differentiation

Abstract

The following theorem gives a useful method for determining critical limits.

Appendices

13.7 First Appendix: Proof of Theorem 13.20

Proof.

Let $f \in C[0,1]$ and $\varepsilon > 0$ be fixed. We begin with the equality

$$\displaystyle{ f(x) - B_{n}(x;f) =\sum _{ k=0}^{n}\left (f(x) - f\left (\frac{k} {n}\right )\right )\binom{n}{k}x^{k}(1 - x)^{n-k}, }$$

(13.43)

which is clear from the definition of B _n (x; f) and from

$$\displaystyle{\sum _{k=0}^{n}\binom{n}{k}x^{k}(1 - x)^{n-k} =\big (x + (1 - x)\big)^{n} = 1}$$

from an application of the binomial theorem. By Theorem 10.52, f is bounded, and so there exists an M such that $\vert f(x)\vert \leq M$ for all $x \in [0,1]$. On the other hand, by Heine’s theorem (Theorem 10.61), there exists a $\delta > 0$ such that if $x,y \in [0,1]$ and $\vert x - y\vert <\delta$, then $\vert f(x) - f(y)\vert <\varepsilon /2$.

Let $n \in \mathbb{N}^{+}$ and $x \in [0,1]$ be fixed. Let I and J denote the set of the indices $0 \leq k \leq n$ for which $\vert x - (k/n)\vert <\delta$ and $\vert x - (k/n)\vert \geq \delta$ respectively. The basic idea of the proof is that if k ∈ I, then by the definition of $\delta$ we have $\vert f(x) - f(k/n)\vert <\varepsilon /2$, and so on the right-hand side of (13.43), the sum of the terms with indices k ∈ I is small. We show that the sum of the terms corresponding to indices k ∈ J is also small for the function f ≡ 1, and that the given function f can only increase this sum to at most M times that. Let us see the details. If $\vert x - (k/n)\vert <\delta$, then $\vert f(x) - f(k/n)\vert <\varepsilon /2$, and so

$$\displaystyle\begin{array}{rcl} & & \sum _{k\in I}\left \vert f(x) - f\left (\frac{k} {n}\right )\right \vert \cdot \binom{n}{k}x^{k}(1 - x)^{n-k} < \\ & & \qquad <\sum _{k\in I} \frac{\varepsilon } {2} \cdot \binom{n}{k}x^{k}(1 - x)^{n-k} \leq \\ & &\qquad \leq \frac{\varepsilon } {2} \cdot \sum _{k=0}^{n}\binom{n}{k}x^{k}(1 - x)^{n-k} = \frac{\varepsilon } {2}.{}\end{array}$$

(13.44)

To approximate the sum $\sum _{k\in J}\binom{n}{k}x^{k}(1 - x)^{n-k}$ we will need the following identities:

$$\displaystyle{ \sum _{k=0}^{n}\frac{k} {n}\binom{n}{k}x^{k}(1 - x)^{n-k} = x }$$

(13.45)

and

$$\displaystyle{ \sum _{k=0}^{n}\frac{k^{2}} {n^{2}}\binom{n}{k}x^{k}(1 - x)^{n-k} = x^{2} + (x - x^{2})/n }$$

(13.46)

for all $x \in \mathbb{R}$ and n = 1, 2, …. Both identities follow easily from the binomial theorem (see Exercises 13.11 and 13.12). For the bound, we actually want the identity

$$\displaystyle{ \sum _{k=0}^{n}\left (\frac{k} {n} - x\right )^{2}\binom{n}{k}x^{k}(1 - x)^{n-k} = \frac{x - x^{2}} {n}, }$$

(13.47)

which we get by taking the difference of (13.46) and 2x times (13.45), and then adding it to the equality $\sum _{k=0}^{n}x^{2} \cdot \binom{n}{k}x^{k}(1 - x)^{n-k} = x^{2}$. If k ∈ J, then $\vert x - (k/n)\vert \geq \delta$, and so

$$\displaystyle\begin{array}{rcl} & & \sum _{k\in J}\binom{n}{k}x^{k}(1 - x)^{n-k} \leq \frac{1} {\delta ^{2}} \cdot \sum _{k\in J}\left (\frac{k} {n} - x\right )^{2}\binom{n}{k}x^{k}(1 - x)^{n-k} < \\ & & \qquad \qquad < \frac{1} {\delta ^{2}} \cdot \sum _{k=0}^{n}\left (\frac{k} {n} - x\right )^{2}\binom{n}{k}x^{k}(1 - x)^{n-k} = \\ & & \qquad \qquad = \frac{x - x^{2}} {n\delta ^{2}} \leq \frac{1} {4n\delta ^{2}}. {}\end{array}$$

(13.48)

Since $\vert f(x)\vert \leq M$ for all x,

$$\displaystyle{ \sum _{k\in J}\left \vert f(x) - f\left (\frac{k} {n}\right )\right \vert \cdot \binom{n}{k}x^{k}(1 - x)^{n-k} < \frac{M} {2n\delta ^{2}}. }$$

(13.49)

Thus by (13.43), using inequalities (13.44) and (13.49), we get that

$$\displaystyle{\left \vert f(x) - B_{n}(x;f)\right \vert < \frac{\varepsilon } {2} + \frac{M} {2n\delta ^{2}}.}$$

Since $x \in [0,1]$ was arbitrary, for $n > M/(\varepsilon \delta ^{2})$ we have $\left \vert f(x) - B_{n}(x;f)\right \vert < (\varepsilon /2) + (\varepsilon /2) =\varepsilon$ for all $x \in [0,1]$, which concludes the proof of the theorem. □ 

13.8 Second Appendix: On the Definition of Trigonometric Functions Again

The definitions of trigonometric functions (Definition 11.20) are mostly based on geometry. These definitions directly use the concept of an angle and indirectly of arc length, and to inspect their important properties, we needed to introduce rotations and its properties. At the same time, the trigonometric functions are among the most basic functions in analysis, so the need might arise to separate our definitions from the geometric ideas. Since we defined arc length precisely and proved its important properties, for our purpose only rotations and their role cause trouble. The addition formulas followed from properties of rotations, and the differentiability of the functions $\sin x$ and cosx used the addition formulas. Our theory of trigonometric functions is not yet complete without a previous background in geometry. We will outline a construction below that avoids this shortcoming.

We will use the notation of Remark 11.21. Let $c(u) = \sqrt{1 - u^{2}}$, and let $S(u) = s\big(c;[u,1]\big)$ for all u ∈ [−1, 1]. (Then S(u) is the length of the arc of the unit circle K that connects the points $\big(u,c(u)\big)$ and (1, 0).) We know that the function S is strictly monotone decreasing on [−1, 1], and by Remark 11.21, the function cosx on the interval [0, π] is none other than the inverse of S. We also saw that if we measure an arc of length x +π onto the unit circle, then we arrive at a point antipodal to $(\cos x,\sin x)$, so its coordinates are $(-\cos x,-\sin x)$. Thus it is clear that the following definition is equivalent to Definition 11.20.

Definition 13.48.

• (i) Let $c(u) = \sqrt{1 - u^{2}}$ and $S(u) = s\big(c;[u,1]\big)$ for all u ∈ [−1, 1]. The inverse of the function S, which is defined on the interval [0, π], is denoted by cosx. We extend the function cosx to the whole real line in such a way that $\cos (x+\pi ) = -\cos x$ holds for all real x.

• (ii) We define the function $\sin x$ on the interval [0, π] by the equation $\sin x = \sqrt{1 -\cos ^{2 } x}$. We extend the function $\sin x$ to the whole real line in such a way that $\sin (x+\pi ) = -\sin x$ holds for all real x.

The definition above uses the fact that if for a function $f: [0,a] \rightarrow \mathbb{R}$ we have $f(a) = -f(0)$, then f can be extended to $\mathbb{R}$ uniquely so that $f(x + a) = -f(x)$ holds for all x. It is easy to check that

$$\displaystyle{f(x + ka) = (-1)^{k} \cdot f(x)\ \ (x \in [0,a],\ k \in \mathbb{Z})}$$

gives such an extension, and that this is the only possible extension.

Now with the help of the above definition, we can again deduce identities (11.25)–(11.32), (11.38), and (11.39) as we did in Chapter 10. Although the argument we used there used the concept of reflection, it is easy to exclude them from the proofs.

The proof of inequality (11.42) is based on geometric facts in several points, including Lemma 10.82 (in which we used properties of convex n-gons) and the concept of similar triangles. We can prove Theorem 11.26 without these with the following replacements.

Theorem 13.49.

If x ≠ 0, then

$$\displaystyle{ 1 -\vert x\vert \leq \frac{\sin x} {x} \leq 1. }$$

(13.50)

Proof.

Since the function $(\sin x)/x$ is even, it suffices to consider the case x > 0. The inequality $(\sin x)/x \leq 1$ is clear by (11.38).

The inequality $1 - x \leq (\sin x)/x$ is evident for x ≥ π∕2, since if $(\pi /2) \leq x \leq \pi$, then $(\sin x)/x \geq 0 \geq 1 - x$, and if x ≥ π, then $(\sin x)/x \geq (-1)/\pi \geq -1 > 1 - x$.

Finally, suppose that 0 < x ≤ π∕2. Let cosx = u and $\sin x = v$. Then—again by the definition of cosx and $\sin x$—the arc length of the graph of the function $c(t) = \sqrt{1 - t^{2}}$ over [u, 1] is x. Since the function c is monotone decreasing on the interval [u, 1], by Theorem 10.79,

$$\displaystyle{x \leq (1 - u) +\big (c(u) - c(1)\big) = (1 - u) + (v - 0) = (1 -\cos x) +\sin x.}$$

Moreover, by (11.39), we have 1 − cosx ≤ x ², so $x \leq x^{2} +\sin x$, that is, the first inequality of (13.50) holds in this case too. □ 

The function c(x) is concave on [−1, 1] and differentiable on (−1, 1), where its derivative is $-x/\sqrt{1 - x^{2}}$. Since

$$\displaystyle{S(u) = s\big(c;[u,1]\big) =\pi -s\big(c;[-1,u]\big)}$$

for all u ∈ [−1, 1], by Theorem 13.41 it follows that S is differentiable on (−1, 1), and its derivative there is

$$\displaystyle{-\sqrt{1 + \left ( \frac{-x} {\sqrt{1 - x^{2}}}\right )^{2}} = - \frac{1} {\sqrt{1 - x^{2}}}.}$$

By the differentiation rule for inverse functions, it follows that the function cosx is differentiable on (0, π), and its derivative there is

$$\displaystyle{ \frac{1} {-1/\sqrt{1 -\cos ^{2 } x}} = -\sqrt{1 -\cos ^{2 } x} = -\sin x.}$$

By identities (11.25), this holds for all points $x\neq k\pi$. Since cosx and $\sin x$ are both continuous, the equality $(\cos x)' = -\sin x$ holds at these points as well (see Exercise 12.75). Using this, it follows from identities (11.32) that $(\sin x)' =\cos x$ for all x.

Finally, we prove the addition formulas. Let $a \in \mathbb{R}$ be arbitrary. The function

$$\displaystyle{A(x) = \left [\sin (a + x) -\sin a\cos x -\cos a\sin x\right ]^{2}+}$$

$$\displaystyle{+\left [\cos (a + x) -\cos a\cos x +\sin a\sin x\right ]^{2}}$$

is everywhere differentiable, and its derivative is

$$\displaystyle\begin{array}{rcl} & & \quad 2 \cdot \left [\sin (a + x) -\sin a\cos x -\cos a\sin x\right ] \cdot \left [\cos (a + x) +\sin a\sin x -\cos a\cos x\right ] + {}\\ & & 2 \cdot \left [\cos (a + x) -\cos a\cos x +\sin a\sin x\right ] \cdot \left [-\sin (a + x) +\cos a\sin x +\sin a\cos x\right ] = 0. {}\\ \end{array}$$

Thus the function A is constant. Since A(0) = 0, we have A(x) = 0 for all x. This is possible only if for all x and a,

$$\displaystyle{\sin (a + x) =\sin a\cos x +\cos a\sin x}$$

and

$$\displaystyle{\cos (a + x) =\cos a\cos x -\sin a\sin x,}$$

which is exactly what we wanted to show.