Preliminaries

Abstract

Sets and functions are key concepts that play an important role in understanding probability and random variables. In this chapter, we discuss those concepts that will be used in later chapters.

Appendices

Appendices

1.1.1 Appendix 1.1 Binomial Coefficients in the Complex Space

For the factorial $n! = n(n-1) \cdots 1$ defined in (1.4.57), the number n is a natural number. Based on the gamma function addressed in Sect. 1.4.3, we extend the factorial into the complex space, which will in turn be used in the discussion of the permutation and binomial coefficients (Riordan 1968; Tucker 2002; Vilenkin 1971) in the complex space.

1.1.1.1 (A) Factorials and Permutations in the Complex Space

Recollecting that

$$\begin{aligned} \alpha ! \ = \ \varGamma (\alpha +1) \end{aligned}$$

(1.A.1)

from $\varGamma (\alpha +1) = \alpha \varGamma (\alpha )$ shown in (1.4.75), the factorial p! can be expressed as

$$\begin{aligned} p! \ =\ \left\{ \begin{array}{ll} \pm \infty , &{} p \in \mathbb {J}_{-},\\ \varGamma (p+1), &{} p \notin \mathbb {J}_{-} \end{array} \right. \end{aligned}$$

(1.A.2)

for a complex number p, where $\mathbb {J}_{-} = \{-1, -2, \ldots \}$ denotes the set of negative integers. Therefore, $0! = \varGamma (1)= 1$ for $p=0$.

Example 1.A.1

From (1.A.2), it is easy to see that $(-2)! = \pm \infty$ and that $\left( - \frac{1}{2} \right) ! = \varGamma \left( \frac{1}{2} \right) = \sqrt{\pi }$ from (1.4.83). $\diamondsuit$

For the permutation ${}_n \text{P}_k= \frac{n!}{(n-k)!}$ defined in (1.4.59), it is assumed that n is a non-negative integer and $k=0, 1, \ldots , n$. Based on (1.4.92) and (1.A.2), the permutation ${}_p \text{P}_q = \frac{\varGamma (p+1)}{\varGamma (p-q+1)}$ can now be generalized as

$$\begin{aligned} {}_p \text{P}_q= & {} \left\{ \begin{array}{ll} \frac{\varGamma (p+1)}{\varGamma (p-q+1)}, &{} p \notin \mathbb {J}_{-} \text{ and } p-q \notin \mathbb {J}_{-}, \\ (-1)^{q}\frac{\varGamma (-p+q)}{\varGamma (-p)}, &{} p\in \mathbb {J}_{-} \text{ and } p-q\in \mathbb {J}_{-},\\ 0, &{} p\notin \mathbb {J}_{-} \text{ and } p-q\in \mathbb {J}_{-}, \\ \pm \infty , &{} p\in \mathbb {J}_{-} \text{ and } p-q\notin \mathbb {J}_{-} \end{array} \right. \end{aligned}$$

(1.A.3)

for complex numbers p and q, where the expression $(-1)^{q}\frac{\varGamma (-p+q)}{\varGamma (-p)}$ in the second line of the right-hand side can also be written as $(-1)^{q}{}_{-p+q-1} \text{ P}_q$.

Example 1.A.2

For any number z, ${}_{z} \text{P}_{0} = 1$ and ${}_{z} \text{P}_{1} = z$. $\diamondsuit$

Example 1.A.3

It follows that

$$\begin{aligned} {}_{0} \text{P}_{z} \ = \ \left\{ \begin{array}{ll} 0, &{} z~\text{ is } \text{ a } \text{ natural } \text{ number },\\ \frac{1}{\varGamma (1-z)}, &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(1.A.4)

$$\begin{aligned} {}_{1} \text{P}_{z} \ = \ \left\{ \begin{array}{ll} 0, &{} z=2, 3, \ldots ,\\ \frac{1}{\varGamma (2-z)}, &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(1.A.5)

$$\begin{aligned} {}_{z} \text{P}_{z} \ = \ \left\{ \begin{array}{ll} \pm \infty , &{} z \in \mathbb {J}_{-},\\ \varGamma (z+1), &{} z \notin \mathbb {J}_{-}, \end{array} \right. \end{aligned}$$

(1.A.6)

and

$$\begin{aligned} {}_{-1} \text{P}_{z}= & {} \left\{ \begin{array}{ll} (-1)^z {}_{z} \text{P}_{z}, &{} z=0, 1 \ldots ,\\ \pm \infty , &{} \text{ otherwise } \end{array} \right. \nonumber \\= & {} \left\{ \begin{array}{ll} (-1)^z \varGamma (z+1), &{} z=0, 1 \ldots ,\\ \pm \infty , &{} \text{ otherwise } \end{array} \right. \end{aligned}$$

(1.A.7)

from (1.A.3). $\diamondsuit$

Using (1.A.3), we can also get ${}_{-2} \text{P}_{-0.3} = \pm \infty$, ${}_{-0.1} \text{P}_{1.9} = 0$, ${}_{0} \text{P}_{3} = \frac{\varGamma (1)}{\varGamma (-2)}=0$, ${}_{ \frac{1}{2} } \text{P}_{3} = \frac{\varGamma \left( \frac{3}{2} \right) }{\varGamma \left( -\frac{3}{2} \right) }=\frac{3}{8}$, ${}_{ \frac{1}{2} } \text{P}_{0.8} = \frac{\varGamma \left( \frac{3}{2} \right) }{\varGamma (0.7)}= \frac{ \sqrt{\pi } }{ 2 \varGamma (0.7)}$, ${}_{3} \text{P}_{ \frac{1}{2} } = \frac{\varGamma (4)}{\varGamma \left( \frac{7}{2} \right) } = \frac{16}{5\sqrt{\pi }}$, ${}_{3} \text{P}_{- \frac{1}{2} } = \frac{\varGamma (4)}{\varGamma \left( \frac{9}{2} \right) } =\frac{32}{35\sqrt{\pi }}$, ${}_{3} \text{P}_{-2} = \frac{\varGamma (4)}{\varGamma (6)}=\frac{1}{20}$, ${}_{- \frac{1}{2} } \text{P}_{3} = \frac{\varGamma \left( \frac{1}{2} \right) }{\varGamma \left( -\frac{5}{2} \right) }= -\frac{8}{15}$, and ${}_{-2} \text{P}_{3} = \frac{\varGamma (-1)}{\varGamma (-4)}=(-1)\frac{\varGamma (5)}{\varGamma (2)}= -24$. Table 1.3 shows some values of the permutation ${}_{p} \text{P}_{q}$.

Table 1.3 Some values of the permutation ${}_{p} \text{P}_{q}$ (Here, $*$ denotes $\pm \infty$)

Table 1.4 The binomial coefficient $\, _{p}\text{C}_q = \frac{\varGamma (p+1)}{\varGamma (p-q+1) \varGamma (q+1)}$ in the complex space

1.1.1.2 (B) Binomial Coefficients in the Complex Space

For the binomial coefficient ${}_n \text{C}_k= \frac{n!}{(n-k)!k!}$ defined in (1.4.60), n and k are non-negative integers with $n \ge k$. Based on the gamma function described in Sect. 1.4.3, we can define the binomial coefficient in the complex space: specifically, employing (1.4.92), the binomial coefficient $\, _{p}\text{C}_q = \frac{\varGamma (p+1)}{\varGamma (p-q+1) \varGamma (q+1)}$ for p and q complex numbers can be defined as described in Table 1.4.

Example 1.A.4

When both p and $p-q$ are negative integers and q is a non-negative integer, the binomial coefficient $\, _{p}\text{C}_q = (-1)^{q} \, _{-p+q-1}\text{C}_{q}$ can be expressed also as

$$\begin{aligned} \, _{p}\text{C}_q= & {} (-1)^{q} \, _{-p+q-1}\text{C}_{-p-1} . \end{aligned}$$

(1.A.8)

Now, when p is a negative non-integer real number and q is a non-negative integer, the binomial coefficient can be written as $\, _{p}\text{C}_q = \frac{\varGamma (p+1)}{\varGamma (p-q+1)} \frac{1}{ \varGamma (q+1)} = (-1)^{p-p+q}\frac{\varGamma (-p+q)}{\varGamma (-p)}\frac{1}{ \varGamma (q+1)} = (-1)^{q}\frac{(-p+q-1)!}{(-p-1)! q!}$ or as

$$\begin{aligned} \, _{p}\text{C}_q= & {} (-1)^{q} \, _{-p+q-1}\text{C}_q \end{aligned}$$

(1.A.9)

by recollecting $\frac{\varGamma (\alpha +1)}{\varGamma (\beta +1)} = (-1)^{\alpha -\beta } \frac{ \varGamma (-\beta ) }{ \varGamma (-\alpha ) }$ shown in (1.4.91) for $\alpha -\beta$ an integer, $\alpha < 0$, and $\beta <0$. The two formulas (1.A.8) and (1.A.9) are the same as $\, _{-r}\text{C}_x = (-1)^x \, _{r+x-1}\text{C}_x$, which we will see in (2.5.15) for a negative real number $-r$ and a non-negative integer x. $\diamondsuit$

Example 1.A.5

From Table 1.4, we promptly get ${}_{z} \text{C}_{0} = {}_{z} \text{C}_{z} = 1$ and ${}_{z} \text{C}_{1} = {}_{z} \text{C}_{z-1} = z$. In addition, ${}_{0} \text{C}_{z} = {}_{0} \text{C}_{-z}$ and ${}_{1} \text{C}_{z} = {}_{1} \text{C}_{1-z} = \frac{1}{\varGamma (2-z)\varGamma (1+z)}$ can be expressed as

$$\begin{aligned} {}_{0} \text{C}_{z}= & {} \frac{1}{\varGamma (1-z)\varGamma (1+z)} \nonumber \\= & {} \left\{ \begin{array}{ll} 1, &{} z=0,\\ 0, &{} z= \pm 1, \pm 2, \ldots , \\ \frac{1}{\varGamma (1-z)\varGamma (1+z)}, &{} \text{ otherwise } \end{array} \right. \end{aligned}$$

(1.A.10)

and

$$\begin{aligned} {}_{1} \text{C}_{z}= & {} \left\{ \begin{array}{ll} 1, &{} z=0, 1,\\ 0, &{} z= -1, \pm 2,\pm 3, \ldots ,\\ \frac{1}{\varGamma (2-z)\varGamma (1+z)}, &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(1.A.11)

respectively. $\diamondsuit$

We can similarly obtain²⁵ ${}_{-3} \text{C}_{-2} = {}_{-3} \text{C}_{-1} = \frac{(-3)!}{(-2)!(-1)!} =0$, ${}_{-3} \text{C}_{2} = {}_{-3} \text{C}_{-5} = \frac{(-3)!}{(-5)!2!} = (-1)^2\frac{4!}{2!2!} = {}_{4} \text{C}_{2}$, ${}_{-7} \text{C}_{3} = {}_{-7} \text{C}_{-10} = \frac{(-7)!}{(-10)!3!} = (-1)\frac{9!}{6!3!} = -{}_{9} \text{C}_{3}$, and ${}_{\frac{5}{2}} \text{C}_{2} = {}_{\frac{5}{2}} \text{C}_{ \frac{1}{2} } = \frac{\varGamma \left( \frac{7}{2} \right) }{\varGamma (3)\varGamma \left( \frac{3}{2} \right) } = \frac{15}{8}$. Table 1.5 shows some values of the binomial coefficient.

Table 1.5 Values of binomial coefficient ${}_{p} \text{C}_{q}$ (Here, $*$ denotes $\pm \infty$)

Example 1.A.6

Obtain the series expansion of $h(z) = (1+z)^p$ for p a real number.

Solution

First, when $p \ge 0$ or when $p < 0$ and $|z| < 1$, we have $(1+z)^p = \sum \limits _{k=0}^{\infty } \frac{h^{(k)} (0) }{k!} z^k = \sum \limits _{k=0}^{\infty } \frac{ 1}{k!}p(p-1) \cdots (p-k+1) z^k$, i.e.,

$$\begin{aligned} (1+z)^p= & {} \sum _{k=0}^{\infty }{}_{p} \text{C}_{k} z^k . \end{aligned}$$

(1.A.12)

Note that (1.A.12) is the same as the binomial expansion

$$\begin{aligned} (1+z)^p \ = \ \sum _{k=0}^{p} {}_{p} \text{C}_{k} z^k \end{aligned}$$

(1.A.13)

of $(1+z)^p$ because ${}_{p} \text{C}_{k} =0$ for $k= p+1, p+2, \ldots$ when p is 0 or a natural number. Next, recollecting (1.A.12) and $(1+z)^p = z^p \left( 1+ \frac{1}{z} \right) ^p$, we get

$$\begin{aligned} (1+z)^p= & {} \sum _{k=0}^{\infty } {}_{p} \text{C}_{k} z^{p-k} \end{aligned}$$

(1.A.14)

for $p < 0$ and $|z| >1$. Combining (1.A.12) and (1.A.14), we eventually get

$$\begin{aligned} (1+z)^p \ = \ \left\{ \begin{array}{ll} \sum \limits _{k=0}^{\infty } {}_{p} \text{C}_{k} z^{k}, &{}\text{ for } p \ge 0 \text{ or } \text{ for } p<0 , |z|< 1,\\ \sum \limits _{k=0}^{\infty } {}_{p} \text{C}_{k} z^{p-k}, &{} \text{ for } p<0 , |z| > 1 , \end{array} \right. \end{aligned}$$

(1.A.15)

$(1+z)^p = 2^p$ for $p<0$ and $z=1$, and $(1+z)^p \rightarrow \infty$ for $p<0$ and $z=-1$. Note that the term ${}_{p} \text{C}_{k}$ in (1.A.15) is always finite because the case of only p being a negative integer among p, k, and $p-k$ is not possible when k is an integer. $\diamondsuit$

Example 1.A.7

Because ${}_{-1} \text{C}_{k}=(-1)^k$ for $k=0,1, \ldots$ as shown in (1.E.27), we get

$$\begin{aligned} \frac{1}{1+z}= & {} \left\{ \begin{array}{ll} \sum \limits _{k=0}^{\infty } (-1)^k z^k, &{} |z|< 1, \\ \sum \limits _{k=0}^{\infty } (-1)^k z^{-1-k}, &{} |z|> 1 \end{array} \right. \nonumber \\= & {} \left\{ \begin{array}{ll} 1 - z + z^2 - z^3 + \cdots , &{} |z| < 1, \\ \frac{1}{z} - \frac{1}{z^2} + \frac{1}{z^3} - \cdots , &{} |z| > 1 \end{array} \right. \end{aligned}$$

(1.A.16)

from (1.A.15) with $p=-1$. $\diamondsuit$

Example 1.A.8

Employing (1.A.15) and the result for ${}_{-2} \text{C}_{k}$ shown in (1.E.28), we get

$$\begin{aligned} \frac{1}{(1+z)^2}= & {} \left\{ \begin{array}{ll} \sum \limits _{k=0}^{\infty } (-1)^k (k+1) z^k, &{} |z|< 1, \\ \sum \limits _{k=0}^{\infty } (-1)^k (k+1) z^{-2-k}, &{} |z|> 1 \end{array} \right. \nonumber \\= & {} \left\{ \begin{array}{ll} 1 - 2z + 3z^2 - 4z^3 + \cdots , &{} |z| < 1, \\ \frac{1}{z^2} - \frac{2}{z^3} + \frac{3}{z^4} - \cdots , &{} |z| > 1 . \end{array} \right. \end{aligned}$$

(1.A.17)

Alternatively, from $\frac{1}{(1+z)^2} = \frac{1}{1- \left( -2z -z^2 \right) } = \sum \limits _{k=0}^{\infty } \left( -2z -z^2 \right) ^k$ for $\left| -2z -z^2 \right| < 1$, we have

$$\begin{aligned} \frac{1}{(1+z)^2}&= 1+ \left( -2z-z^2 \right) + \left( 4z^2 + 4z^3 + z^4 \right) \nonumber \\&\quad + \left( -8z^3 + \cdots \right) + \cdots , \end{aligned}$$

(1.A.18)

which can be rewritten as

$$\begin{aligned}& 1 -2z + \left( -z^2 + 4z^2\right) + \left( 4z^3 - 8z^3 \right) + \cdots \nonumber \\&= \ 1 - 2z + 3z^2 - 4z^3 + \cdots \end{aligned}$$

(1.A.19)

by changing the order in the addition. The result²⁶ (1.A.19) is the same as (1.A.17) for $|z| <1$. 

1.1.1.3 (C) Two Equalities for Binomial Coefficients

Theorem 1.A.1

For $\gamma \in \{0, 1, \ldots \}$ and any two numbers $\alpha$ and $\beta$, we have

$$\begin{aligned} \sum \limits _{m=0}^{\gamma } {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= & {} {{\alpha +\beta }\atopwithdelims (){\gamma }} , \end{aligned}$$

(1.A.20)

which is called Chu-Vandermonde convolution or Vandermonde convolution. 

Theorem 1.A.1 is proved in Exercise 1.35. The result (1.A.20) is the same as the Hagen-Rothe identity

$$\begin{aligned} \sum \limits \limits _{m=0}^{\gamma } \frac{\alpha -\gamma c}{\alpha -mc} {{\alpha - m c}\atopwithdelims (){\gamma -m}} \frac{\beta }{\beta +mc} {{\beta +mc}\atopwithdelims (){m}} \ = \ \frac{\alpha +\beta -\gamma c}{\alpha +\beta } {{\alpha +\beta }\atopwithdelims (){\gamma }} \ \ \ \ \ \end{aligned}$$

(1.A.21)

with $c=0$ and Gauss’ hypergeometric theorem

$$\begin{aligned} _2 F_1 (a, b; c; 1)= & {} \frac{\varGamma (c) \varGamma (c-a-b)}{\varGamma (c-a) \varGamma (c-b)}, \quad \text{Re}(c) > \text{Re}(a+b) \end{aligned}$$

(1.A.22)

with $a=\gamma$, $b=\alpha +\beta -\gamma$, and $c=\alpha +\beta +1$. In (1.A.22),

$$\begin{aligned} _2 F_1 (a, b; c; z) \ = \ \sum \limits _{n=0}^{\infty }\frac{(a)_n (b)_n}{(c)_n}\frac{z^n}{n!}, \quad \text{Re}(c)> \text{Re}(b) >0 \end{aligned}$$

(1.A.23)

is the hypergeometric function²⁷, and can be expressed also as

$$\begin{aligned} {}_2 F_1 (a, b; c; z) \ = \ \frac{1}{B (b, c-b)} \int _{0}^{1} x^{b-1} (1-x)^{c-b-1} (1-zx)^{-a} dx \ \ \ \ \ \end{aligned}$$

(1.A.24)

in terms of Euler’s integral formula.

Example 1.A.9

In (1.A.20), assume $\alpha =2$, $\beta = \frac{1}{2}$, and $\gamma =2$. Then, the left-hand side is ${{2}\atopwithdelims (){2}} {{ \frac{1}{2} }\atopwithdelims (){0}} + {{2}\atopwithdelims (){1}} {{ \frac{1}{2} }\atopwithdelims (){1}} + {{2}\atopwithdelims (){0}} {{ \frac{1}{2} }\atopwithdelims (){2}} = 1 + 2 \frac{ \left( \frac{1}{2} \right) ! }{ \left( - \frac{1}{2} \right) ! 1! } + \frac{ \left( \frac{1}{2} \right) ! }{\left( -\frac{3}{2} \right) ! 2! } = 1 + 1 + \frac{ \left( \frac{1}{2} \right) \left( - \frac{1}{2} \right) }{ 2! } = \frac{15}{8}$ and the right-hand side is ${{\frac{5}{2}}\atopwithdelims (){2}} = \frac{\left( \frac{5}{2} \right) ! }{ \left( \frac{1}{2} \right) ! 2!} = \frac{15}{8}$. $\diamondsuit$

Example 1.A.10

Consider the case $\beta =n$ for $\sum \limits _{m=0}^n {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}$, where $\alpha$ is not a negative integer and $n \in \{ 0, 1, \ldots \}$. When $\gamma =0, 1, \ldots , n$, we have $\sum \limits _{m=0}^n {{\alpha }\atopwithdelims (){\gamma -m}}{{\beta }\atopwithdelims (){m}} = \sum \limits _{m=0}^\gamma {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} + \sum \limits _{m=\gamma +1}^n {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} = {{\alpha +\beta }\atopwithdelims (){\gamma }}$ noting that ${{\alpha }\atopwithdelims (){\gamma -m}} = \frac{\alpha !}{(\alpha -\gamma +m)!} \frac{1}{(\gamma -m)!}=0$ for $m = \gamma +1, \gamma +2, \ldots$ due to $(\gamma -m)! = \pm \infty$. Similarly, when $\gamma =n+1, n+2, \ldots$, we have $\sum \limits _{m=0}^n {{\alpha }\atopwithdelims (){\gamma -m}}{{\beta }\atopwithdelims (){m}} = \sum \limits _{m=0}^\gamma {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} - \sum \limits _{m=n+1}^\gamma {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= {{\alpha +\beta }\atopwithdelims (){\gamma }}$ because ${{\beta }\atopwithdelims (){m}} = {{n}\atopwithdelims (){m}} = \frac{n!}{(n-m)!m!}=0$ from $(n-m)! = \pm \infty$ for $m = n+1, n+2, \ldots$. In short, we have

$$\begin{aligned} \sum \limits _{m=0}^n {{\alpha }\atopwithdelims (){\gamma -m}}{{n}\atopwithdelims (){m}} = {{\alpha +n}\atopwithdelims (){\gamma }} \end{aligned}$$

(1.A.25)

for $\gamma \in \{ 0, 1, \ldots \}$ when $n \in \{ 0, 1, \ldots \}$. $\diamondsuit$

Theorem 1.A.2

We have

$$\begin{aligned} \sum \limits _{m=0}^{\gamma } [m]_{\zeta } {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= & {} [\beta ]_{\zeta } {{\alpha +\beta -\zeta }\atopwithdelims (){\gamma -\zeta }} \end{aligned}$$

(1.A.26)

for $\zeta \in \{0, 1, \ldots , \gamma \}$ and $\gamma \in \{0, 1, \ldots \}$.

Proof

(Method 1) Let us employ the mathematical induction. First, when $\zeta =0$, (1.A.26) holds true for any values of $\gamma \in \{0, 1, \ldots \}$, $\alpha$, and $\beta$ from Theorem 1.A.1. Assume (1.A.26) holds true when $\zeta = \zeta _0$: in other words, for any value of $\gamma \in \{0, 1, \ldots \}$, $\alpha$, and $\beta$, assume

$$\begin{aligned} \sum \limits _{m=0}^{\gamma } [m]_{\zeta _0} {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= & {} \sum \limits _{m=\zeta _0}^{\gamma } [m]_{\zeta _0} {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} \nonumber \\= & {} [\beta ]_{\zeta _0} {{\alpha +\beta -\zeta _0}\atopwithdelims (){\gamma -\zeta _0}} \end{aligned}$$

(1.A.27)

holds true. Then, noting that $(m+1) {{\beta }\atopwithdelims (){m+1}} = \beta {{\beta -1}\atopwithdelims (){m}}$ and $[m+1]_{\zeta _0+1} = (m+1)[m]_{\zeta _0}$, we get $\sum \limits _{m=0}^{\gamma } [m]_{\zeta _0+1} {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} = \sum \limits _{m=\zeta _0+1}^{\gamma } [m]_{\zeta _0+1} {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} = \sum \limits _{m=\zeta _0}^{\gamma -1} [m+1]_{\zeta _0+1} {{\alpha }\atopwithdelims (){\gamma -m-1}} {{\beta }\atopwithdelims (){m+1}} = \sum \limits _{m=\zeta _0}^{\gamma -1} [m]_{\zeta _0} {{\alpha }\atopwithdelims (){\gamma -1-m}} (m+1) {{\beta }\atopwithdelims (){m+1}}$ from (1.A.27), i.e.,

$$\begin{aligned} \sum \limits _{m=0}^{\gamma } [m]_{\zeta _0+1} {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= & {} \beta \sum \limits _{m=\zeta _0}^{\gamma -1} [m]_{\zeta _0} {{\alpha }\atopwithdelims (){\gamma -1-m}} {{\beta -1}\atopwithdelims (){m}} \nonumber \\= & {} \beta [\beta -1]_{\zeta _0} {{\alpha +\beta -1-\zeta _0}\atopwithdelims (){\gamma -1-\zeta _0}} \nonumber \\= & {} [\beta ]_{\zeta _0+1} {{\alpha +\beta -(\zeta _0+1)}\atopwithdelims (){\gamma -(\zeta _0+1)}} . \end{aligned}$$

(1.A.28)

The result (1.A.28) implies that (1.A.26) holds true also when $\zeta =\zeta _0+1$ if (1.A.26) holds true when $\zeta =\zeta _0$. In short, (1.A.26) holds true for $\zeta \in \{0, 1, \ldots \}$.

(Method 2) Noting (Charalambides 2002; Gould 1972) that $[m]_{\zeta } {{\beta }\atopwithdelims (){m}} = [m]_{\zeta } \frac{ [\beta ]_{\zeta } [\beta -\zeta ]_{m-\zeta } }{[m]_{\zeta } (m-\zeta ) !} = [\beta ]_{\zeta } {{\beta -\zeta }\atopwithdelims (){m-\zeta }}$, we can rewrite (1.A.26) as $\sum \limits _{m=\zeta }^{\gamma } {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta -\zeta }\atopwithdelims (){m-\zeta }} = {{\alpha +\beta -\zeta }\atopwithdelims (){\gamma -\zeta }}$, which is the same as the Chu-Vandermonde convolution $\sum \limits _{k=0}^{\gamma -\zeta } {{\alpha }\atopwithdelims (){\gamma -\zeta -k}} {{\beta -\zeta }\atopwithdelims (){k}} = {{\alpha +\beta -\zeta }\atopwithdelims (){\gamma -\zeta }}$. $\spadesuit$

It is noteworthy that (1.A.26) holds true also when $\zeta \in \left\{ \gamma +1, \gamma +2, \ldots \right\}$, in which case the value of (1.A.26) is 0.

Example 1.A.11

Assume $\alpha =7$, $\beta =3$, $\gamma =6$, and $\zeta =2$ in (1.A.26). Then, the left-hand side is $2{{7}\atopwithdelims (){4}} {{3}\atopwithdelims (){2}} +6 {{7}\atopwithdelims (){3}} {{3}\atopwithdelims (){3}} = 420$ and the right-hand side is $6 {{8}\atopwithdelims (){4}}= 420$. $\diamondsuit$

Example 1.A.12

Assume $\alpha =-4$, $\beta =-1$, $\gamma =3$, and $\zeta =2$ in (1.A.26). then, the left-hand side is $2{{-4}\atopwithdelims (){1}} {{-1}\atopwithdelims (){2}} +6 {{-4}\atopwithdelims (){0}} {{-1}\atopwithdelims (){3}} = 2 \times \frac{ (-4)! }{ (-5)! 1! } \frac{ (-1)! }{ (-3)! 2! } + 6 \times \frac{ (-4)! }{ (-4)! 0! } \frac{ (-1)! }{ (-4)! 3! } = -14$ and the right-hand side is $(-1)(-2) {{-7}\atopwithdelims (){1}} = 2 \times \frac{ (-7)! }{ (-8)! 1! } = -14$. $\diamondsuit$

Example 1.A.13

The identity (1.A.26) holds true also for non-integer values of $\alpha$ or $\beta$. For example, when $\alpha = \frac{1}{2}$, $\beta =- \frac{1}{2}$, $\gamma =2$, and $\zeta =1$, the left-hand side is ${{ \frac{1}{2} }\atopwithdelims (){1}} {{- \frac{1}{2} }\atopwithdelims (){1}} +2 {{ \frac{1}{2} }\atopwithdelims (){0}} {{- \frac{1}{2} }\atopwithdelims (){2}} = \frac{ \left( \frac{1}{2} \right) ! }{\left( - \frac{1}{2} \right) ! 1!} \frac{ \left( - \frac{1}{2} \right) ! }{ \left( -\frac{3}{2} \right) !1! } + 2\frac{ \left( \frac{1}{2} \right) ! }{ \left( \frac{1}{2} \right) ! 0! } \frac{ \left( - \frac{1}{2} \right) ! }{ \left( -\frac{5}{2} \right) ! 2! } = \frac{1}{2}$ and the right-hand side is $- \frac{1}{2} \times {{-1}\atopwithdelims (){1}} = - \frac{1}{2} \times \frac{ \varGamma \left( 0 \right) }{ \varGamma \left( -1 \right) }= - \frac{1}{2} \times \frac{(-1)!}{(-2)! 1!} = \frac{1}{2}$. $\diamondsuit$

Example 1.A.14

Denoting the unit imaginary number by $j=\sqrt{-1}$, assume $\alpha =e-j$, $\beta =\pi +2j$, $\gamma =4$, and $\zeta =2$ in (1.A.26). Then, the left-hand side is $0 + 0 + 2 \times {{\alpha }\atopwithdelims (){2}} {{\beta }\atopwithdelims (){2}} + 6 \times {{\alpha }\atopwithdelims (){1}} {{\beta }\atopwithdelims (){3}} + 12 \times {{\alpha }\atopwithdelims (){0}} {{\beta }\atopwithdelims (){4}}= \frac{1}{2} \alpha (\alpha -1) \beta (\beta -1) + \alpha \beta (\beta -1)(\beta -2) + \frac{1}{2} \beta (\beta -1)(\beta -2)(\beta -3) = \frac{1}{2} \beta (\beta -1) \left\{ \alpha (\alpha -1)+ 2 \alpha (\beta -2) + (\beta -2)(\beta -3) \right\} = \frac{1}{2} \beta (\beta -1) \left\{ \alpha ^2+ \alpha (2\beta -5) + (\beta -2)(\beta -3) \right\} = \frac{1}{2} \beta (\beta -1) (\alpha +\beta -2) (\alpha +\beta -3)$ and the right-hand side is also $\beta (\beta -1) {{\alpha +\beta -2}\atopwithdelims (){2}} = \beta (\beta -1) \frac{ \left( \alpha +\beta -2 \right) ! }{ \left( \alpha +\beta -4 \right) ! 2! } = \frac{1}{2} \beta (\beta -1) (\alpha +\beta -2)(\alpha +\beta -3)$. $\diamondsuit$

Example 1.A.15

When $\zeta =1$ and $\zeta =2$, (1.A.26) can be specifically written as

$$\begin{aligned} \sum \limits _{m=0}^{\gamma } m {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}}= & {} \beta {{\alpha +\beta -1}\atopwithdelims (){\gamma -1}} \nonumber \\= & {} \frac{\beta \gamma }{\alpha +\beta } {{\alpha +\beta }\atopwithdelims (){\gamma }} \end{aligned}$$

(1.A.29)

and

$$\begin{aligned}&\sum \limits _{m=0}^{\gamma } m(m-1) {{\alpha }\atopwithdelims (){\gamma -m}} {{\beta }\atopwithdelims (){m}} \ = \ \beta (\beta -1) {{\alpha +\beta -2}\atopwithdelims (){\gamma -2}} \nonumber \\&\quad = \ \frac{\beta (\beta -1)\gamma (\gamma -1)}{(\alpha +\beta )(\alpha +\beta -1)} {{\alpha +\beta }\atopwithdelims (){\gamma }}, \end{aligned}$$

(1.A.30)

respectively. The two results (1.A.29) and (1.A.30) will later be useful for obtaining the mean and variance of the hypergeometric distribution in Exercise 3.68. $\diamondsuit$

1.1.1.4 (D) Euler Reflection Formula

 

We now prove the Euler reflection formula

$$\begin{aligned} \varGamma ( 1-x )\varGamma ( x ) \ = \ \frac{\pi }{\sin \pi x} \end{aligned}$$

(1.A.31)

for $0< x < 1$ mentioned in (1.4.79). First, if we let $x= \frac{s}{s+1}$ in the defining equation (1.4.95) of the beta function, we get

$$\begin{aligned} \tilde{B}(\alpha , \beta ) \ = \ \int _{0}^{\infty } \frac{s^{\alpha -1}}{(s+1)^{\alpha +\beta }} ds . \end{aligned}$$

(1.A.32)

Using (1.A.32) and $\varGamma ( \alpha ) \varGamma ( \beta )= \varGamma ( \alpha + \beta )\tilde{B}(\alpha , \beta )$ from (1.4.96) will lead us to

$$\begin{aligned} \varGamma ( 1-x )\varGamma ( x ) \ = \ \int _{0}^{\infty } \frac{s^{x-1}}{s+1} ds \end{aligned}$$

(1.A.33)

for $\alpha =x$ and $\beta =1-x$. To obtain the right-hand side of (1.A.33), we consider the contour integral

$$\begin{aligned} \int _{C} \frac{z^{x-1}}{z-1} dz \end{aligned}$$

(1.A.34)

in the complex space. The contour C of the integral in (1.A.34) is shown in Fig. 1.26, a counterclockwise path along the outer circle. As there exists only one pole $z=1$ inside the contour C, we get 

$$\begin{aligned} \int _{C} \frac{z^{x-1}}{z-1} dz= & {} 2 \pi j \end{aligned}$$

(1.A.35)

from the residue theorem $\int _{C} \frac{z^{x-1}}{z-1} dz = 2 \pi j \ {\underset{z=1}{\text{ Res }}} \frac{z^{x-1}}{z-1}$. Consider the integral along C in four segments. First, we have $z= R e^{j\theta }$ and $dz= j R e^{j\theta } d \theta$ over the segment from $z_1 = R e^{j(-\pi + \epsilon )}$ to $z_2 = R e^{j(\pi - \epsilon )}$ along the circle with radius R. Second, we have $z= r e^{j(\pi - \epsilon )}$ and $dz= e^{j(\pi - \epsilon )} dr$ over the segment from $z_2= R e^{j(\pi - \epsilon )}$ to $z_3= p e^{j(\pi - \epsilon )}$ along the straight line toward the origin. Third, we have $z= p e^{j\theta }$ and $dz= j p e^{j\theta } d \theta$ over the segment from $z_3 = p e^{j(\pi - \epsilon )}$ to $z_4 = p e^{j(-\pi + \epsilon )}$ clockwise along the circle with radius p. Fourth, we have $z= r e^{j(-\pi + \epsilon )}$ and $dz= e^{j(-\pi + \epsilon )} dr$ over the segment from $z_4= p e^{j(-\pi + \epsilon )}$ to $z_1= R e^{j(-\pi + \epsilon )}$ along the straight line out of the origin. Thus, we have

$$\begin{aligned}&\int _{C} \frac{z^{x-1}}{z-1} dz \ = \ \int _{-\pi + \epsilon }^{\pi - \epsilon } \frac{ \left( R e^{j \theta } \right) ^{x-1} j R e^{j \theta }}{R e^{j \theta }-1} d\theta + \int _{R}^{p} \frac{ \left\{ r e^{j(\pi - \epsilon )}\right\} ^{x-1} e^{j(\pi - \epsilon )}}{r e^{j(\pi - \epsilon )}-1} dr \nonumber \\&\quad + \int _{\pi - \epsilon }^{-\pi + \epsilon } \frac{ \left( p e^{j \theta } \right) ^{x-1} j p e^{j \theta } }{p e^{j \theta }-1} d\theta + \int _{p}^{R} \frac{ \left\{ r e^{j(-\pi + \epsilon )}\right\} ^{x-1} e^{j(-\pi + \epsilon )}}{r e^{j(-\pi + \epsilon )}-1} dr , \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

(1.A.36)

which can be written as

$$\begin{aligned} \int _{C} \frac{z^{x-1}}{z-1} dz&= \int _{-\pi + \epsilon }^{\pi - \epsilon } \frac{ jR^{x} e^{j \theta x } }{ R e^{j \theta }-1 } d\theta + \int _{R}^{p} \frac{ r^{x-1} e^{j(\pi - \epsilon ) x}}{ r e^{j(\pi - \epsilon )} -1 } dr \nonumber \\&\quad + \int _{\pi - \epsilon }^{-\pi + \epsilon } \frac{ jp^{x} e^{j \theta x } }{ p e^{j \theta }-1 } d\theta + \int _{p}^{R} \frac{ r^{x-1} e^{j(-\pi +\epsilon ) x}}{ r e^{j(-\pi + \epsilon )} -1 } dr \ \ \ \ \ \ \ \ \ \end{aligned}$$

(1.A.37)

after some steps. When $x>0$ and $p \rightarrow 0$, the third term in the right-hand side of (1.A.37) is $\lim \limits _{p \rightarrow 0} \int _{\pi - \epsilon }^{-\pi + \epsilon } \frac{ jp^{x} e^{j \theta x } }{ p e^{j \theta }-1 } d\theta = 0$. Similarly, the first term in the right-hand side of (1.A.37) is $\left| \frac{ jR^{x} e^{j \theta x } }{ R e^{j \theta }-1 } \right| = \frac{ R^{x} }{\sqrt{ R^2 -2R\cos \theta +1}} \le \frac{ R^{x} }{ R-1} \rightarrow 0$ when $x<1$ and $R \rightarrow \infty$. Therefore, (1.A.37) can be written as

$$\begin{aligned} \int _{C} \frac{z^{x-1}}{z-1} dz= & {} 0 \ + \ \int _{\infty }^{0} \frac{ r^{x-1} e^{j \pi x} }{ - r -1 } dr \ + \ 0 \ + \ \int _{0}^{\infty } \frac{ r^{x-1} e^{- j \pi x}}{ - r -1 } dr \nonumber \\= & {} \left( e^{j \pi x} - e^{- j \pi x} \right) \int _{0}^{\infty } \frac{ r^{x-1} }{ r+1 } dr \nonumber \\= & {} 2j \sin \pi x \int _{0}^{\infty } \frac{ r^{x-1} }{ r+1 } dr \end{aligned}$$

(1.A.38)

for $0< x < 1$ when $R \rightarrow \infty$, $p \rightarrow 0$, and $\epsilon \rightarrow 0$. In short, we have

$$\begin{aligned} \int _{0}^{\infty } \frac{r^{x-1}}{r+1} dr \ = \ \frac{\pi }{\sin \pi x} \end{aligned}$$

(1.A.39)

for $0< x < 1$ from (1.A.35) and (1.A.38) and, subsequently, we have (1.A.31) for $0< x < 1$ from (1.A.33) and (1.A.39).

Fig. 1.26

The contour C of integral $\int _{C} \frac{z^{x-1}}{z-1} dz$, where $0< p< 1 < R$

1.1.2 Appendix 1.2 Some Results

1.1.2.1 (A) Stepping Stone

Consider crossing a creek via $n-1$ stepping stones with steps 0 and n denoting the two banks of the creek. Assume we can move only in one direction and skip either 0 or 1 step at each move.

• (1) Obtain the number of ways we can complete the crossing in k moves.

• (2) Obtain the number of ways we can complete the crossing.

Table 1.6 Number $a(n,k) = \, _k\text{C}_{n-k}$ of ways we can cross a creek via $n-1$ stepping stones in k moves when we can move only in one direction and skip either 0 or 1 step at each move

Solution

• (1) Let a(n, k) be the number of ways we can complete the crossing in k moves and let

  $$\begin{aligned} n_2 \ = \ \left\lceil \frac{n}{2} \right\rceil , \end{aligned}$$

  (1.A.40)

  where $\left\lceil x \right\rceil$ denotes the smallest integer not smaller than x and is called the ceiling function. Denote the number of moves in which we skip 0 and 1 step by $k_1$ and $k_2$, respectively. Then, from $k_1 + k_2 = k$ and $k_1 + 2k_2=n$, we get $k_1 = 2k-n$ and $k_2 = n-k$. The number a(n, k) of ways we can complete the crossing in k moves is the same as the number $\frac{k!}{k_1! k_2!}$ of arranging $k_1$ of 1’s and $k_2$ of 2’s. In short, we have $a (n,k) = \frac{k!}{(2k-n)! (n-k)!}$, i.e., 

  $$\begin{aligned} a (n,k)= & {} _k\text{C}_{n-k} \end{aligned}$$

  (1.A.41)

  for $k=n_2, n_2+1, \ldots , n$: some values of a(n, k) are shown in Table 1.6.

• (2) Denoting the number of ways we can complete the crossing by a(n), which is the same as the number of arranging k of 1’s and $n-k$ of 2’s for $k=1, 2, \ldots , n$, we get $a(n) = \sum \limits _{k=1}^{n} a(n,k)$, i.e.,

  $$\begin{aligned} a(n)= & {} \sum _{k=n_2}^{n} \, _k\text{C}_{n-k} . \end{aligned}$$

  (1.A.42)

  We also have $a(1)=1$ and $a (2)=2$. Therefore, a(n) is the sum of the number of ways from step $n-1$ to n and that from $n-2$ directly to n. We thus have $a(n) = a(n-1) + a(n-2)$ for $n=3, 4, \ldots$ because the number of ways to step $n-1$ is $a(n-1)$ and that to step $n-2$ is $a(n-2)$. Solving the recursion, we get²⁸

  $$\begin{aligned} a(n) \ = \ \frac{1}{\sqrt{5}} \left\{ \left( \frac{1+\sqrt{5}}{2}\right) ^{n+1} - \left( \frac{1-\sqrt{5}}{2}\right) ^{n+1} \right\} . \end{aligned}$$

  (1.A.43)

  Here, as it is shown later in (1.A.58), the number a(n) is an integer when $n \in \{ 0, 1, \ldots \}$. Table 1.7 shows a(n) for $n=1, 2, \ldots , 20$. Recollecting that $_k\text{C}_{n-k} =0$ for $k=0, 1, \ldots , n_2 -1$, we have

  $$\begin{aligned} \sum _{k=0}^{n} \, _k\text{C}_{n-k} \ = \ \frac{1}{\sqrt{5}} \left\{ \left( \frac{1+\sqrt{5}}{2}\right) ^{n+1} - \left( \frac{1-\sqrt{5}}{2}\right) ^{n+1} \right\} \end{aligned}$$

  (1.A.44)

  from (1.A.42) and (1.A.43). The result (1.A.44) is the same as the well-known identity (Roberts and Tesman 2009) $F_{n+1} = \sum \limits _{k=0}^{\lfloor \frac{n}{2}\rfloor } {} _{n-k}\text{C}_{k}$, where $\left\{ F_n \right\} _{n=0}^{\infty }$ are Fibonacci numbers. Here, $\lfloor x \rfloor$, also expressed as [x], denotes the greatest integer not larger than x and is called the floor function or Gauss function. Clearly, $\lfloor x \rfloor = \lceil x \rceil = x$ when x is an integer while $\lfloor x \rfloor = \lceil x \rceil - 1$ when x is not an integer.

  

Table 1.7 Number $a(n)=\frac{1}{\sqrt{5}} \left\{ \left( \frac{1+\sqrt{5}}{2}\right) ^{n+1} - \left( \frac{1-\sqrt{5}}{2}\right) ^{n+1} \right\}$ of ways we can cross a creek via $n-1$ stepping stones when we can move only in one direction and skip either 0 or 1 step at each move

If the condition ‘skip either 0 or 1 step at each move’ is replaced with ‘skip either 0 step or 2 steps at each move’, then we have

$$\begin{aligned} \sum _{m=n_3, n_3+2, \ldots }^{n} {} _m\text{C}_{\frac{n-m}{2}} \ = \ a_{13} r_{13}^n + 2 \text{Re}\left\{ c_{13} z_{13}^n \right\} \end{aligned}$$

(1.A.45)

for $n=0, 1, \ldots$, where $n_3 = n- 2 \left\lfloor \frac{n}{3} \right\rfloor$. In addition, the three numbers $r_{13} = \frac{1}{3} \left( 1+ 2\epsilon _{13} \right)$, $z_{13} = \frac{1}{3} \left\{ 1 -\epsilon _{13} - j\sqrt{3 \left( \epsilon _{13}^2 -1 \right) } \right\}$, and $z_{13}^{*}$ with the unit imaginary number $j= \sqrt{-1}$ are the solutions to the difference equation $a(n) = a(n-1) + a (n-3)$ for $n \ge 4$ with initial conditions $a(1) = 1$, $a(2) = 1$, and $a(3) = 2$. The three numbers are also the solutions to $\theta ^{3} - \theta ^{2} -1 =0$, and $\epsilon _{13} = \frac{1}{2} \left\{ \left( \frac{ 29-3\sqrt{93} }{2} \right) ^{\frac{1}{3}} + \left( \frac{ 29+3\sqrt{93} }{2} \right) ^{\frac{1}{3}} \right\} \approx 1.6984$, $a_{13} = \frac{\left( z_{13}-1 \right) \left( z_{13}^{*}-1\right) +1}{r_{13} \left( r_{13} -z_{13} \right) \left( r_{13} -z_{13}^{*} \right) }= \frac{\left| z_{13}-1 \right| ^2 +1 }{r_{13} \left| r_{13} -z_{13} \right| ^2 }$, and $c_{13} = \frac{ \left( r_{13}-1\right) \left( z_{13}^{*}-1 \right) +1}{z_{13} \left( z_{13} -r_{13} \right) \left( z_{13} -z_{13}^{*} \right) }$.

In addition, if the condition ‘skip either 0 or 1 step at each move’ is replaced with ‘skip either 1 step or 2 steps at each move’, we can obtain

$$\begin{aligned} \sum _{m=0}^{n} {} _m\text{C}_{n-2m} \ = \ a_{23} r_{23}^n + 2 \text{Re} \left\{ c_{23} z_{23}^n \right\} \end{aligned}$$

(1.A.46)

for $n=2, 3, \ldots$ by noting that $_m\text{C}_{n-2m}=\frac{m!}{ \left( 3m -n \right) ! \left( n-2m \right) !} =0$ when $m=0, 1, \ldots , \left\lceil \frac{n}{3} \right\rceil -1$ or $m=\left\lfloor \frac{n}{2} \right\rfloor +1 , \left\lfloor \frac{n}{2} \right\rfloor +2, \ldots , n$. Here, the three numbers $r_{23} = 2 \epsilon _{23}$, $z_{23} = - \epsilon _{23} - j \sqrt{3 \epsilon _{23}^2 -1 }$, and $z_{23}^{*}$ are the solutions to the difference equation $a(n) = a(n-2) + a (n-3)$ for $n \ge 5$ with initial conditions $a(2) = 1$, $a(3) = 1$, and $a(4) = 1$. The three numbers are also the solutions to $\theta ^{3} - \theta -1 =0$, and $\epsilon _{23} = \left( \frac{ 9-\sqrt{69} }{144} \right) ^{\frac{1}{3}} + \left( \frac{ 9+\sqrt{69} }{144} \right) ^{\frac{1}{3}} \approx 0.6624$, $a_{23} = \frac{\left( z_{23}-1 \right) \left( z_{23}^{*}-1\right) }{r_{23}^2 \left( r_{23} -z_{23} \right) \left( r_{23} -z_{23}^{*} \right) } = \frac{\left| z_{23}-1 \right| ^2}{r_{23}^2 \left| r_{23} -z_{23} \right| ^2}$, and $c_{23} = \frac{ \left( r_{23}-1 \right) \left( z_{23}^{*}-1 \right) }{z_{23}^2 \left( z_{23} -r_{23} \right) \left( z_{23} -z_{23}^{*} \right) }$.

The sequences of numbers calculated from (1.A.43), (1.A.45), and (1.A.46) can be written as

$$\begin{aligned} \left\{ \sum _{m=0}^{n} {} _m\text{C}_{n-m} \right\} _{n=0}^{\infty } \ = \ \left\{ 1, 1, 2, 3, 5, 8, \ldots \right\} , \end{aligned}$$

(1.A.47)

$$\begin{aligned} \left\{ \sum _{m=n_3, n_3+2, \ldots }^{n} {} _m\text{C}_{\frac{n-m}{2}} \right\} _{n=0}^{\infty } \ = \ \left\{ 1, 1, 1, 2, 3, 4, 6, 9, \ldots \right\} , \end{aligned}$$

(1.A.48)

and

$$\begin{aligned} \left\{ \sum _{m=0}^{n} {} _m\text{C}_{n-2m} \right\} _{n=2}^{\infty } \ = \ \left\{ 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, \ldots \right\} , \end{aligned}$$

(1.A.49)

respectively.

1.1.2.2 (B) Order of Operations

When some operations such as limit, integration, and differentiation are evaluated, a change of order does not usually make a difference. However, the order is of importance in certain cases. Here, we present some examples in which a change of order yields different results.

Example 1.A.16

(Gelbaum and Olmsted 1964) For the function

$$\begin{aligned} f(x,y) \ = \ \left\{ \begin{array}{ll} y^{-2}, &{} 0< x< y< 1, \\ -x^{-2}, &{} 0< y< x < 1, \\ 0,&{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(1.A.50)

we have $\int _0^1 \int _0^1 f(x,y) dx dy = \int _0^1 dy =1$ because $\int _0^1 f(x,y) dx = \int _0^y \frac{dx}{y^2} - \int _y^1 \frac{dx}{x^2} = \frac{1}{y} + \left( 1- \frac{1}{y} \right) =1$. On the other hand, $\int _0^1 \int _0^1 f(x,y) dy dx = \int _0^1 (-1) dx = -1$ because $\int _0^1 f(x,y) dy = - \int _0^x \frac{dy}{x^2} + \int _x^1 \frac{dy}{y^2} = - \frac{1}{x} - \left( 1- \frac{1}{x} \right) = -1$. In other words, $\int _0^1 \int _0^1 f(x,y) dx dy \ne \int _0^1 \int _0^1 f(x,y) dy dx$. $\diamondsuit$

Example 1.A.17

(Gelbaum and Olmsted 1964) Consider a sequence $\left\{ f_n (x) \right\} _{n=1}^{\infty }$ with $f_n (x) = n^2x e^{-nx}$ on the support [0, 1]. Then, $\lim \limits _{n \rightarrow \infty } \int _{0}^{1} f_n (x) dx = \lim \limits _{n \rightarrow \infty } \left[ - ( nx + 1 ) e^{-nx} \right] _{0}^{1} = - \lim \limits _{n \rightarrow \infty } \left\{ (n+1)e^{-n} -1 \right\}$, i.e.,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \int _{0}^{1} f_n (x) dx= & {} 1 . \end{aligned}$$

(1.A.51)

On the other hand, $\lim \limits _{n \rightarrow \infty } f_n (x)$ is 0 for $0 \le x \le 1$ because $f_n (x) =0$ for $x= 0$ and $\lim \limits _{n \rightarrow \infty } f_n (x) = \frac{1}{x} \lim \limits _{n \rightarrow \infty } \frac{(nx)^2}{e^{nx}} =0$ for $0 < x \le 1$. Thus, $\int _{0}^{1} \lim \limits _{n \rightarrow \infty } f_n (x) dx = \int _{0}^{1} 0 dx = 0$ and, therefore, $\int _{0}^{1} \lim \limits _{n \rightarrow \infty } f_n (x) dx \ne \lim \limits _{n \rightarrow \infty } \int _{0}^{1} f_n (x) dx$. $\diamondsuit$

Example 1.A.18

(Gelbaum and Olmsted 1964) For the function

$$\begin{aligned} f(x,y) \ = \ \left\{ \begin{array}{ll} \frac{x^2 - y^2}{x^2 + y^2},&{} x^2 + y^2 \ne 0 ,\\ 0,&{} x=y=0 , \end{array} \right. \end{aligned}$$

(1.A.52)

we have $\lim \limits _{x \rightarrow 0} \lim \limits _{y \rightarrow 0} f(x,y) \ne \lim \limits _{y \rightarrow 0} \lim \limits _{x \rightarrow 0} f(x,y)$ because $\lim \limits _{x \rightarrow 0} \lim \limits _{y \rightarrow 0} f(x,y) = \lim \limits _{x \rightarrow 0} \frac{x^2}{x^2} = 1$ and $\lim \limits _{y \rightarrow 0} \lim \limits _{x \rightarrow 0} f(x,y) = \lim \limits _{y \rightarrow 0} \frac{-y^2}{y^2} = -1$. $\diamondsuit$

Example 1.A.19

(Gelbaum and Olmsted 1964) Consider a sequence $\left\{ f_n (x) \right\} _{n=1}^{\infty }$ of functions in which $f_n (x) = \frac{x}{1+n^2x^2}$ for $|x| \le 1$. Then, $\frac{d}{dx} \left\{ \lim \limits _{n \rightarrow \infty } f_n (x) \right\} = 0$, but

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \left\{ \frac{d}{dx} f_n (x) \right\}= & {} \lim \limits _{n \rightarrow \infty } \frac{1-n^2x^2}{\left( 1+n^2x^2 \right) ^2} \nonumber \\= & {} \left\{ \begin{array}{ll} 1,&{} x= 0 \\ 0,&{} 0 < |x| \le 1, \end{array} \right. \end{aligned}$$

(1.A.53)

implying that $\frac{d}{dx} \left\{ \lim \limits _{n \rightarrow \infty } f_n (x) \right\} \ne \lim \limits _{n \rightarrow \infty } \left\{ \frac{d}{dx} f_n (x) \right\}$. $\diamondsuit$

Example 1.A.20

(Gelbaum and Olmsted 1964) Consider the function

$$\begin{aligned} f(x,y) \ = \ \left\{ \begin{array}{ll} \frac{x^3}{y^2}\exp \left( -\frac{x^2}{y}\right) ,&{} y > 0 ,\\ 0,&{} y=0 . \end{array} \right. \end{aligned}$$

(1.A.54)

For any value of x, we have $\int _{0}^{1} f(x,y) dy = x \exp \left( -x^2 \right)$. If we let $g(x)= x \exp \left( -x^2 \right)$, then we have $\frac{d}{dx} \int _{0}^{1}f(x,y) dy = g^{\prime }(x) = \left( 1-2x^2\right) \exp \left( -x^2 \right)$ for any value of x. On the other hand, $\frac{\partial }{\partial x} f(x,y) =0$ for any value of y when $x=0$ because

(1.A.55)

In other words,

$$\begin{aligned} \int _{0}^{1}~\frac{\partial }{\partial x} f(x,y) ~dy= & {} \left\{ \begin{array}{ll} \int _{0}^{1} \left( \frac{3x^2}{y^2} - \frac{2x^4}{y^3} \right) \exp \left( - \frac{x^2}{y} \right) &{} x \ne 0 ,\\ \int _{0}^{1} 0 ~dy, &{} x=0 \end{array} \right. \nonumber \\= & {} \left\{ \begin{array}{ll} (1-2x^2) \exp \left( -x^2 \right) , &{} x \ne 0 ,\\ 0,&{} x=0 , \end{array}\right. \end{aligned}$$

(1.A.56)

and thus $\frac{d}{dx} \int _{0}^{1}f(x,y) dy \ne \int _{0}^{1}\frac{\partial }{\partial x} f(x,y) dy$. $\diamondsuit$

Example 1.A.21

(Gelbaum and Olmsted 1964) Consider the Cantor function $\phi _{C}(x)$ discussed in Example 1.3.11 and $f(x) = 1$ for $x \in [0, 1]$. Then, both the Riemann-Stieltjes and Lebesgue-Stieltjes integrals produce $\int _{0}^{1} f(x) d\phi _{C}(x) = [ f(x)\phi _{C}(x)]_{0}^{1} -\int _{0}^{1} \phi _{C}(x) df(x) = \phi _{C}(1) - \phi _{C}(0) - 0$, i.e.,

$$\begin{aligned} \int _{0}^{1} f(x) ~d\phi _{C}(x)= & {} 1 \end{aligned}$$

(1.A.57)

while the Lebesgue integral results in $\int _{0}^{1} f(x) \phi _{C}^{\prime }(x) dx = \int _{0}^{1} 0 dx = 0$. $\diamondsuit$

1.1.2.3 (C) Sum of Powers of Two Real Numbers

Theorem 1.A.3

If the sum $\alpha +\beta$ and product $\alpha \beta$ of two numbers $\alpha$ and $\beta$ are both integers, then

$$\begin{aligned} \alpha ^n+\beta ^n \ = \ \text{ integer } \end{aligned}$$

(1.A.58)

for $n \in \{0, 1, \ldots \}$.

Proof

Let us prove the theorem via mathematical induction. It is clear that (1.A.58) holds true when $n=0$ and 1. When $n=2$, $\alpha ^2+\beta ^2=(\alpha +\beta )^2-2\alpha \beta$ is an integer. Assume $\alpha ^n+\beta ^n$ are all integers for $n=1, 2, \ldots , k-1$. Then,

$$\begin{aligned} \alpha ^k +\beta ^k= & {} (\alpha +\beta )^k - \, _{k}\text{C}_{1} \alpha \beta \left( \alpha ^{k-2} + \beta ^{k-2}\right) - \, _{k}\text{C}_{2} \left( \alpha \beta \right) ^2\left( \alpha ^{k-4} +\beta ^{k-4}\right) \nonumber \\&\ - \cdots - \left\{ \begin{array}{ll} _{k}\text{C}_{\frac{k-1}{2}} \left( \alpha \beta \right) ^{\frac{k-1}{2}} \left( \alpha +\beta \right) , &{} k \text{ is } \text{ odd },\\ \, _{k}\text{C}_{\frac{k}{2}} \left( \alpha \beta \right) ^{\frac{k}{2}}, &{} k \text{ is } \text{ even } , \end{array} \right. \end{aligned}$$

(1.A.59)

which implies that $\alpha ^n+\beta ^n$ is an integer when $n=k$ because the binomial coefficient $\, _{k}\text{C}_{j}$ is always an integer for $j =0, 1, \ldots , k$ when k is a natural number. In other words, if $\alpha \beta$ and $\alpha + \beta$ are both integers, then $\alpha ^n+\beta ^n$ is also an integer when $n \in \{0, 1, \ldots \}$. $\spadesuit$

1.1.2.4 (D) Differences of Geometric Sequences

Theorem 1.A.4

Consider the difference

$$\begin{aligned} D_n \ = \ \alpha a^n - \beta b^n \end{aligned}$$

(1.A.60)

of two geometric sequences, where $\alpha > 0$, $a> b > 0$, $a \ne 1$, and $b \ne 1$. Let

$$\begin{aligned} r \ = \ \frac{\ln \frac{(1-b)\beta }{(1-a)\alpha } }{\ln \frac{a}{b} }. \end{aligned}$$

(1.A.61)

Then, the sequence $\left\{ D_n \right\} _{n=1}^{\infty }$ has the following properties:

• (1) For $0< a < 1$, $D_n$ is the largest at $n=r$ and $n=r+1$ if r is an integer and at $n= \lceil r \rceil$ if r is not an integer.

• (2) For $a > 1$, $D_n$ is the smallest at $n=r$ and $n=r+1$ if r is an integer and at $n= \lceil r \rceil$ if r is not an integer.

Proof

Consider the case $0< a < 1$. Then, from $D_{n+1} - D_n = b^n \left( 1-b \right) \beta - a^n \left( 1- a\right) \alpha$, we have $D_{n+1} > D_n$ for $n < r$, $D_{n+1} = D_n$ for $n = r$, and $D_{n+1} < D_n$ for $n > r$ and, subsequently, (1). We can similarly show (2). $\spadesuit$

Example 1.A.22

The sequence $\left\{ \alpha a^n - \beta b^n \right\} _{n=1}^{\infty }$

• (1) is increasing and decreasing if $a >1$ and $0< a < 1$, respectively, when $\frac{(1-b)\beta }{(1-a)\alpha } < \frac{a}{b}$,

• (2) is first decreasing and then increasing and first increasing and then decreasing if $a >1$ and $0< a < 1$, respectively, when $\frac{(1-b)\beta }{(1-a)\alpha } > \frac{a}{b}$, and

• (3) is increasing and decreasing if $a >1$ and $0< a < 1$, respectively, when $\frac{(1-b)\beta }{(1-a)\alpha } = \frac{a}{b}$ or, equivalently, when $\alpha a - \beta b = \alpha a^2 - \beta b^2$.

$\diamondsuit$

Example 1.A.23

Assume $\alpha =\beta$. Then,

• (1) $\left\{ a^n - b^n \right\} _{n=1}^{\infty }$ is an increasing and decreasing sequence when $a >1$ and $a+b <1$, respectively,

• (2) $\left\{ a^n - b^n \right\} _{n=1}^{\infty }$ is a decreasing sequence and $a - b = a^2 - b^2$ when $a+b =1$, and

• (3) $\left\{ a^n - b^n \right\} _{n=1}^{\infty }$ is a sequence that first increases and then decreases with the maximum at $n= \lceil r \rceil$ if r is not an integer and at $n=r$ and $n=r+1$ if r is an integer when $a+b >1$ and $0< a < 1$.

$\diamondsuit$

Example 1.A.24

Assume $\alpha =\beta =1$, $a=0.95$, and $b=0.4$. Then, we have $0.95 - 4< 0.95^2 - 4^2 < 0.95^3 - 4^3$ and $0.95^3 - 4^3> 0.95^4 - 4^4 > \cdots$ because $\left\lceil r \right\rceil = \left\lceil \frac{\ln \frac{0.6}{0.05} }{\ln \frac{0.95}{0.4} } \right\rceil \approx \lceil 2.87 \rceil = 3$. $\diamondsuit$

1.1.2.5 (E) Selections of Numbers with No Number Unchosen

The number of ways to select r different elements from a set of n distinct elements is $_n\text{C}_r$. Because every element will be selected as many times as any other element, each of the n elements will be selected $_n\text{C}_r \times \frac{r}{n} = {}_{n-1}\text{C}_{r-1}$ times over the $_n\text{C}_r$ selections. Each of the n elements will be included at least once if we choose appropriately

$$\begin{aligned} m_1 \ = \ \left\lceil \frac{n}{r} \right\rceil \end{aligned}$$

(1.A.62)

selections among the $_n\text{C}_r$ selections. For example, assume the set $\left\{ 1, 2, 3, 4, 5 \right\}$ and $r=2$. Then, we have $m_1 = \left\lceil \frac{5}{2} \right\rceil =3$, and thus, each of the five elements is included at least once in the three selections (1, 2), (3, 4), and (4, 5).

Next, it is possible that one or more elements will not be included if we consider $_{n-1}\text{C}_r$ selections or less among the total $_n\text{C}_r$ selections. For example, for the set $\left\{ 1, 2, 3, 4, 5 \right\}$ and $r=2$, in some choices of $_{4}\text{C}_2 = 6$ selections or less such as (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), and (3, 4) among the total of $_5\text{C}_2 = 10$ selections, the element 5 is not included.

On the other hand, each of the n elements will be included at least once in any

$$\begin{aligned} m_2 \ = \ 1 \, + \, _{n-1}\text{C}_{r} \end{aligned}$$

(1.A.63)

selections. Here, we have

$$\begin{aligned} _{n-1}\text{C}_{r}= & {} \left( 1- \frac{r}{n}\right) \, _{n}\text{C}_{r} \nonumber \\= & {} _{n}\text{C}_{r} - \, _{n-1}\text{C}_{r-1}. \end{aligned}$$

(1.A.64)

The identity (1.A.64) implies that the number of ways for a specific element not to be included when selecting r elements from a set of n distinct elements is the same as the following two numbers:

• (1) The number of ways to select r elements from a set of $n-1$ distinct elements.

• (2) The difference between the number of ways to select r elements from a set of n distinct elements and that for a specific element to be included when selecting r elements from a set of n distinct elements.

1.1.2.6 (F) Fubini’s theorem

Theorem 1.A.5

When the function f(x, y) is continuous on $A= \{(x,y): a \le x \le b , \ c \le y \le d \}$, we have

$$\begin{aligned} \underset{A}{\int \int } f(x,y) dx dy= & {} \int _{c}^{d} \int _{a}^{b} f(x,y) dx dy \nonumber \\= & {} \int _{a}^{b} \int _{c}^{d} f(x,y) dy dx . \end{aligned}$$

(1.A.65)

In addition, we have

$$\begin{aligned} \underset{A}{\int \int } f(x,y) dx dy \ = \ \int _{a}^{b} \int _{g_1 (x)}^{g_2 (x)} f(x,y) dy dx \end{aligned}$$

(1.A.66)

if f(x, y) is continuous on $A= \{(x,y): \, a \le x \le b , \ g_1 (x) \le y \le g_2 (x) \}$ and both $g_1$ and $g_2$ are continuous on [a, b]. 

1.1.2.7 (G) Partitions of Numbers

A representation of a natural number as the sum of natural numbers is also called a partition. Denote the number of partitions for a natural number n as the sum of k natural numbers by M(n, k). Then, the number N(n) of partitions for a natural number n can be expressed as

$$\begin{aligned} N(n) \ = \ \sum \limits _{k=1}^{n}M(n,k) . \end{aligned}$$

(1.A.67)

As we can see, for example, from

$$\begin{aligned} \begin{array}{cl} 1: &{} \{1\}, \\ 2: &{} \{2\}, \ \{1,1\}, \\ 3: &{} \{3\}, \ \{2,1\}, \ \{1,1,1\}, \\ 4: &{} \{4\}, \ \{3,1\}, \ \{2,2\}, \ \{2,1,1\}, \ \{1,1,1,1\}, \end{array} \end{aligned}$$

(1.A.68)

we have $N(1) = M(1,1)=1$, $N(2) = M(2,1) + M(2,2) =2$, and $N(3) = M(3,1) + M(3,2) + M(3,3) =3$. In addition, $M(4,1)=1$, $M(4,2)=2$, $M(4,3)=1$, and $M(4,4)=1$. In general, the number M(n, k) satisfies

$$\begin{aligned} M(n,k) \ = \ M(n-1,k-1)+M(n-k,k) . \end{aligned}$$

(1.A.69)

Example 1.A.25

We have $M(5,3)=M(5-1,3-1)+M(5-3,3)=M(4,2)+M(2,3)=2+0=2$ from (1.A.69). $\diamondsuit$

Theorem 1.A.6

Denote the least common multiplier of k consecutive natural numbers $1,2,\ldots , k$ by $\tilde{k}$. Let the quotient and remainder of n when divided by k be $Q_{k}$ and $R_{k}$, respectively. If we write

$$\begin{aligned} n \ = \ \tilde{k}Q_{\tilde{k}}+R_{\tilde{k}}, \end{aligned}$$

(1.A.70)

then the number M(n, k) can be expressed as

$$\begin{aligned} M(n,k) \ = \ \sum _{i=0}^{k-1}c_{i,k} \left( R_{\tilde{k}} \right) Q_{\tilde{k}}^{i}, \quad R_{\tilde{k}}=0,1, \ldots ,\tilde{k}-1 \end{aligned}$$

(1.A.71)

in terms of $\tilde{k}$ polynomials of order $k-1$ in $Q_{\tilde{k}}$, where $\left\{ c_{i,k} ( \cdot ) \right\} _{i=0}^{k-1}$ are the coefficients of the polynomial.

Based on Theorem 1.A.6, we can obtain $M(n,1) = 1$,

$$\begin{aligned} M(n,2)= & {} \left\{ \begin{array}{ll} \frac{n-1}{2}, &{}n \text{ is } \text{ odd },\\ \frac{n}{2}, &{}n \text{ is } \text{ even }, \end{array} \right. \end{aligned}$$

(1.A.72)

$$\begin{aligned} 12 \, M(n,3)= & {} \left\{ \begin{array}{llll} n^2, &{}R_{6}=0; &{} \quad n^2-1, &{}R_{6}=1,5; \\ n^2-4, &{}R_{6}=2,4; &{} \quad n^2+3, &{}R_{6}=3 , \end{array} \right. \end{aligned}$$

(1.A.73)

and

$$\begin{aligned} 144 \, M(n,4)= & {} \left\{ \begin{array}{llll} n^3+3n^2, &{}R_{12}=0, \\ n^3+3n^2-20, &{}R_{12}=2, \\ n^3+3n^2+32, &{}R_{12}=4, \\ n^3+3n^2-36, &{}R_{12}=6, \\ n^3+3n^2+16, &{}R_{12}=8, \\ n^3+3n^2-4, &{}R_{12}=10, \\ n^3+3n^2-9n+5, &{}R_{12}=1,7, \\ n^3+3n^2-9n-27, &{}R_{12}=3,9, \\ n^3+3n^2-9n-11, &{}R_{12}=5,11 , \end{array} \right. \end{aligned}$$

(1.A.74)

for example. Table 1.8 shows the 60 polynomials of order four in $Q_{60}$ for the representation of M(n, 5).

Table 1.8 Coefficients $\left\{ c_{j,5}(r) \right\} _{j=4}^{0}$ in $M(n,5)=c_{4,5}(R_{60})Q_{60}^{4}+c_{3,5}(R_{60})Q_{60}^3 + c_{2,5}(R_{60})Q_{60}^2+c_{1,5}(R_{60})Q_{60}+c_{0,5}(R_{60})$

Exercises

Exercise 1.1

Show that $B^c \subseteq A^c$ when $A\subseteq B$.

Exercise 1.2

Show that ${\overset{\infty }{\underset{i=1}{\cap }}} A_i = \left( {\overset{\infty }{\underset{i=1}{\cup }}} A_i^c \right) ^c$ for a sequence $\left\{ A_i \right\} _{i=1}^{\infty }$ of sets.

Exercise 1.3

Express the difference $A- B$ in terms only of intersection and symmetric difference, and the union $A\cup B$ in terms only of intersection and symmetric difference.

Exercise 1.4

Consider a sequence $\left\{ A_i \right\} _{i=1}^{n}$ of finite sets. Show

$$\begin{aligned} \left| A_1 \cup A_2 \cup \cdots \cup A_n \right|= & {} \sum _i \left| A_i \right| - \sum _{i<j} \left| A_i \cap A_j \right| + \sum _{i<j<k} \left| A_i \cap A_j \cap A_k \right| \nonumber \\&- \cdots + (-1)^{n-1} \left| A_1 \cap A_2 \cap \cdots \cap A_n \right| \end{aligned}$$

(1.E.1)

and

$$\begin{aligned} \left| A_1 \cap A_2 \cap \cdots \cap A_n \right|= & {} \sum _i \left| A_i \right| - \sum _{i<j} \left| A_i \cup A_j \right| + \sum _{i<j<k} \left| A_i \cup A_j \cup A_k \right| \nonumber \\&- \cdots + (-1)^{n-1} \left| A_1 \cup A_2 \cup \cdots \cup A_n \right| , \end{aligned}$$

(1.E.2)

where $\left| A_i \right|$ denotes the number of elements in $A_i$.

Exercise 1.5

For a sequence $\left\{ A_i \right\} _{i=1}^{n}$ of finite sets, show

$$\begin{aligned} \left| A_1 \triangle A_2 \triangle \cdots \triangle A_n \right|&= \sum _{i} \vert A_i \vert \ - \ 2 \sum _{i< j} \left| A_i \cap A_j \right| \ + \ 4 \sum _{i< j< k} \left| A_i \cap A_j \cap A_k \right| \nonumber \\&\quad - \ \cdots \ + (-2)^{n-1} \left| A_1 \cap A_2 \cap \cdots \cap A_n \right| . \end{aligned}$$

(1.E.3)

(Hint. As observed in Example 1.1.35, any element in the set $A_1 \varDelta A_2 \varDelta \cdots \varDelta A_n$ belongs to only odd number of sets among $\left\{ A_i \right\} _{i=1}^{n}$.)

Exercise 1.6

Is the set of polynomials with integer coefficients countable?

Exercise 1.7

Is the set of algebraic numbers²⁹ countable? 

Exercise 1.8

Show that the sets below are countable.

• (1) The set of functions that map a finite subset of a countable set A onto a countable set B.

• (2) The set of convergent sequences of natural numbers.

Exercise 1.9

Is the collection of all non-overlapping open intervals with real end points countable or uncountable?

Exercise 1.10

Find an injection from the set $A$ to the set $B$ in each of the pairs $A$ and $B$ below. Here, $\mathbb {J}_0 = \mathbb {J}_{+} \cup \{0\}$, i.e.,

$$\begin{aligned} \mathbb {J}_0= & {} \{0, 1, \ldots \} . \end{aligned}$$

(1.E.4)

• (1) $A= \mathbb {J}_0\times \mathbb {J}_0$. $B= \mathbb {J}_0$.

• (2) $A= (-\infty ,\infty )$. $B= (0,1)$.

• (3) $A= \mathbb {R}$. $B=$ the set of infinite sequences of 0 and 1.

• (4) $A=$ the set of infinite sequences of 0 and 1. $B= [0,1]$.

• (5) $A=$ the set of infinite sequences of natural numbers. $B=$ the set of infinite sequences of 0 and 1.

• (6) $A=$ the set of infinite sequences of real numbers. $B=$ the set of infinite sequences of 0 and 1.

Exercise 1.11

Find a function from the set $A$ to the set $B$ in each of the pairs $A$ and $B$ below.

• (1) $A= \mathbb {J}_0$. $B= \mathbb {J}_0\times \mathbb {J}_0$.

• (2) $A= \mathbb {J}_0$. $B= \mathbb {Q}$.

• (3) $A=$ the Cantor set. $B= [0,1]$.

• (4) $A=$ the set of infinite sequences of 0 and 1. $B= [0,1]$.

Exercise 1.12

Find a one-to-one correspondence between the set $A$ and the set $B$ in each of the pairs $A$ and $B$ below.

• (1) $A= (a,b)$. $B=(c,d)$. Here, $-\infty \le a<b \le \infty$ and $-\infty \le c<d \le \infty$.

• (2) $A=$ the set of infinite sequences of 0, 1, and 2. $B=$ the set of infinite sequences of 0 and 1.

• (3) $A= [0,1)$. $B=[0,1) \times [0,1)$.

Exercise 1.13

Assume $f_1: A \rightarrow B$ and $f_2: C \rightarrow D$ are both one-to-one correspondences, $A\cap C=\emptyset$, and $B\cap D=\emptyset$. Show that $f=f_1\cup f_2$ or, equivalently, $f: A\cup C \rightarrow B \cup D$ is a one-to-one correspondence.

Exercise 1.14

Find a one-to-one correspondence³⁰ between $A= \mathbb {J}_0$ and $B= \mathbb {J}_0\times \mathbb {J}_0$.

 

Exercise 1.15

Is the collection of intervals with rational end points in the space $\mathbb {R}$ of real numbers countable?

Exercise 1.16

Show that $U-C \sim U$ when U is an uncountable set and C is a countable set.

Exercise 1.17

Is the sum of two rational numbers a rational number? If we add one more rational number, is the result a rational number? If we add an infinite number of rational numbers, is the result a rational number? (Hint. The set of rational numbers is closed under a finite number of additions, but is not closed under an infinite number of additions.)

Exercise 1.18

Here, $\mathbb {Q}$ denotes the set of rational numbers defined in (1.1.29) and $0<a<b<1$.

• (1) Find a rational number between a and b when $a \in \mathbb {Q}$ and $b \in \mathbb {Q}$.

• (2) Find an irrational number between a and b when $a \in \mathbb {Q}$ and $b \in \mathbb {Q}^c$.

• (3) Find an irrational number between a and b when $a \in \mathbb {Q}$ and $b \in \mathbb {Q}$.

• (4) Find an irrational number between a and b when $a \in \mathbb {Q}^c$ and $b \in \mathbb {Q}^c$.

• (5) Find a rational number between a and b when $a \in \mathbb {Q}^c$ and $b \in \mathbb {Q}^c$.

• (6) Find a rational number between a and b when $a \in \mathbb {Q}$ and $b \in \mathbb {Q}^c$.

Exercise 1.19

Consider a game between two players. After a countable subset $A$ of the interval [0, 1] is determined, the two players alternately choose one number from $\{0, 1, \ldots , 9\}$. Let the numbers chosen by the first and second players be $x_0$, $x_1$, $\ldots$ and $y_0$, $y_1$, $\ldots$, respectively. When the number $0. x_1 y_1 x_2 y_2 \cdots$ belongs to $A$, the first player wins and otherwise the second player wins. Find a way for the second player to win.

Exercise 1.20

For a set function $f: \varOmega \rightarrow \mathbb {R}$, show

$$\begin{aligned} f^{-1} \left( A^c \right)= & {} \left( f^{-1} (A)\right) ^c ,\end{aligned}$$

(1.E.5)

$$\begin{aligned} f^{-1} (A \cup B)= & {} f^{-1} (A) \cup f^{-1} (B) , \end{aligned}$$

(1.E.6)

and

$$\begin{aligned} f^{-1} (A \cap B)= & {} f^{-1} (A) \cap f^{-1} (B), \end{aligned}$$

(1.E.7)

where $A, B \subseteq \mathbb {R}$.

Exercise 1.21

Show that the function $f(x) = x^2$ is uniformly continuous on $S= \{x: \, 0< x < 4 \}$.

Exercise 1.22

Show that the function $f(x) = \frac{1}{x}$ is continuous but not uniformly continuous on $S= (0, \infty )$.

Exercise 1.23

Show that the function $f(x) = \sqrt{x}$ is uniformly continuous on $S= (0, \infty )$.

Exercise 1.24

Confirm

$$\begin{aligned} \lim \limits _{m \rightarrow \infty } \int _{-\infty }^{\infty } f(x) \frac{\sin mx}{\pi x} dx \ = \ f(0) \end{aligned}$$

(1.E.8)

shown in (1.4.24). (Hint. Consider the inverse Fourier transform of product.) 

Exercise 1.25

Recollect the definition of the unit step function.

• (1) Express $u(ax+b)$, $u(\sin x)$, and $u\left( e^x - \pi \right)$ in other formulas.

• (2) Obtain $\int _{-\infty }^x u(t-y) dt$.

Exercise 1.26

Obtain the Fourier transform $\mathfrak {F} \left\{ u(x) \right\}$ of the unit step function u(x) by following the order shown below. 

• (1) Let the Fourier transform of the function

  $$\begin{aligned} s_{\alpha } (x) \ = \ \left\{ \begin{array}{ll} e^{-\alpha x}, &{} x> 0, \\ -e^{\alpha x}, &{} x < 0 \end{array} \right. \end{aligned}$$

  (1.E.9)

  with $\alpha > 0$ be $S_{\alpha } (\omega ) = \mathfrak {F} \left\{ s_{\alpha } (x) \right\}$. Show

  $$\begin{aligned} \lim _{\alpha \rightarrow 0} S_{\alpha } (\omega ) \ = \ \frac{2}{j \omega } . \end{aligned}$$

  (1.E.10)

• (2) Show that the Fourier transform of the impulse function $\delta (x)$ is $\mathfrak {F} \left\{ \delta (x) \right\} =1$. Then, show

  $$\begin{aligned} \mathfrak {F} \left\{ 1 \right\} \ =\ 2 \pi \, \delta ( \omega ) \end{aligned}$$

  (1.E.11)

  using a property of Fourier transform.

• (3) Noting that $\mathrm {sgn}(x) = 2u(x) - 1$ and therefore $u(x) = \frac{1}{2} \{1+ \mathrm {sgn}(x)\}$, we have $u(x) = \frac{1}{2} \left\{ 1+ \lim \limits _{\alpha \rightarrow 0} s_{\alpha } (x)\right\}$ because $\mathrm {sgn}(x) = \lim \limits _{\alpha \rightarrow 0} s_{\alpha } (x)$ from (1.E.9). Based on this result, and noting (1.E.10) and (1.E.11), obtain the Fourier transform

  $$\begin{aligned} \mathfrak {F} \left\{ u(x) \right\}= & {} \pi \delta (\omega ) + \frac{1}{j \omega } \end{aligned}$$

  (1.E.12)

  of the unit step function u(x).

Exercise 1.27

Express $\delta ^{\prime } (x) \cos x$ in another formula.

Exercise 1.28

Calculate $\int _{-2\pi }^{2\pi } e^{\pi x} \delta \left( x^2 - \pi ^2 \right) dx$. When $0 \le x < 2 \pi$, express $\delta (\sin x)$ in another formula.

Exercise 1.29

Calculate $\int _{-\infty }^{\infty } \delta ^{\prime } \left( x^2 -3x+2 \right) dx$ and $\int _{-\infty }^{\infty } (\cos x + \sin x )\delta ^{\prime } \left( x^3+ \right. \left. x^2+x \right) dx$.

Exercise 1.30

The multi-dimensional impulse function can be defined as

$$\begin{aligned} \delta \left( x_1, x_2 , \ldots , x_n \right) \ =\ \prod _{i=1}^{n} \delta \left( x_i \right) . \end{aligned}$$

(1.E.13)

Show

$$\begin{aligned} \delta \left( x , y \right) \ = \ \frac{ \delta \left( r \right) }{\pi r} \end{aligned}$$

(1.E.14)

for $r = \sqrt{x^2 + y^2}$ and

$$\begin{aligned} \delta \left( x , y, z \right) \ = \ \frac{ \delta \left( r \right) }{2 \pi r^2} \end{aligned}$$

(1.E.15)

for $r = \sqrt{x^2 + y^2 + z^2}$.

Exercise 1.31

Obtain the limits of the sequences below.

• (1) $\left\{ \left[ 1+\frac{1}{n}, 2 \right) \right\} _{n=1}^{\infty } \ \ \ \$

• (2) $\left\{ \left[ 1+\frac{1}{n}, 2 \right] \right\} _{n=1}^{\infty } \ \ \ \$

• (3) $\left\{ \left( 1, 1+\frac{1}{n} \right] \right\} _{n=1}^{\infty }$

• (4) $\left\{ \left[ 1, 1+\frac{1}{n} \right] \right\} _{n=1}^{\infty } \ \ \ \$

• (5) $\left\{ \left[ 1-\frac{1}{n}, 2 \right) \right\} _{n=1}^{\infty } \ \ \ \$

• (6) $\left\{ \left[ 1-\frac{1}{n}, 2 \right] \right\} _{n=1}^{\infty }$

• (7) $\left\{ \left( 1, 2-\frac{1}{n} \right) \right\} _{n=1}^{\infty } \ \ \ \$

• (8) $\left\{ \left[ 1, 2-\frac{1}{n} \right) \right\} _{n=1}^{\infty }$

Exercise 1.32

Consider the sequence $\left\{ f_n (x) \right\} _{n=1}^{\infty }$ of functions with

$$\begin{aligned} f_n (x) \ = \ \left\{ \begin{array}{ll} 2n^2 x,&{} 0 \le x \le \frac{1}{2n} , \\ 2n - 2n^2x, &{} \frac{1}{2n} \le x \le \frac{1}{n} , \\ 0,&{} \frac{1}{n} \le x \le 1. \end{array} \right. \end{aligned}$$

(1.E.16)

By obtaining $\int _{0}^{1} \lim \limits _{n \rightarrow \infty } f_n (x) dx$ and $\lim \limits _{n \rightarrow \infty } \int _{0}^{1} f_n (x) dx$, confirm that the order of integration and limit are not always interchangeable.

Exercise 1.33

For the function

$$\begin{aligned} f_n (x) \ = \ \left\{ \begin{array}{ll} 2n^3 x,&{} 0 \le x \le \frac{1}{2n} , \\ 2n^2 - 2n^3 x, &{} \frac{1}{2n} \le x \le \frac{1}{n} , \\ 0,&{} \frac{1}{n} \le x \le 1, \end{array} \right. \end{aligned}$$

(1.E.17)

and a number $b \in (0, 1]$, obtain $\int _{0}^{b} \lim \limits _{n \rightarrow \infty } f_n (x) dx$ and $\lim \limits _{n \rightarrow \infty } \int _{0}^{b} f_n (x) dx$, which shows that the order of integration and limit are not always interchangeable.

Exercise 1.34

Obtain the number of all possible arrangements with ten distinct red balls and ten distinct black balls.

Exercise 1.35

Show³¹ the identities 

$$\begin{aligned} \, _{a-1}\text{C}_{b-1} + \, _{a-1}\text{C}_{b}= & {} _{a}\text{C}_{b} , \end{aligned}$$

(1.E.18)

$$\begin{aligned} \sum _{k=0}^n \, _{p}\text{C}_{k} \, _{q}\text{C}_{n-k}= & {} _{p+q}\text{C}_{n}, \end{aligned}$$

(1.E.19)

and

$$\begin{aligned} \sum _{k=0}^n \, _{k}\text{C}_{j} \, _{n-k}\text{C}_{m-j}= & {} _{n+1}\text{C}_{m+1} , \end{aligned}$$

(1.E.20)

where a and b are complex numbers excluding $\{a=0, b$ $\ne$integer$\}$, p and q are complex numbers, $n \in \{ 0, 1, 2, \ldots \}$, and j and m are integers such that $0 \le j \le m \le n$. (Hint. For (1.E.19), recollect $(1+x)^{a+b} = (1+x)^{a}(1+x)^{b}$.)

Exercise 1.36

Show

$$\begin{aligned} \sum \limits \limits _{k=0}^{n} {{n}\atopwithdelims (){k}}{{n}\atopwithdelims (){k}}= & {} {{2n}\atopwithdelims (){n}} \end{aligned}$$

(1.E.21)

and

$$\begin{aligned} \sum \limits \limits _{k=0}^{n} {{k}\atopwithdelims (){m}} \ = \ \sum \limits \limits _{k=m}^{n} {{k}\atopwithdelims (){m}}= & {} {{n+1}\atopwithdelims (){m+1}}. \end{aligned}$$

(1.E.22)

For two integers n and q such that $n \ge q$, show

$$\begin{aligned} \sum \limits \limits _{k=q}^{n} {{n}\atopwithdelims (){k}} {{k}\atopwithdelims (){q}}= & {} 2^{n-q} {{n}\atopwithdelims (){q}} . \end{aligned}$$

(1.E.23)

Exercise 1.37

Show

$$\begin{aligned} \sum \limits \limits _{k_1=1}^{k_{0}} \sum \limits \limits _{k_2=1}^{k_1} \cdots \sum \limits \limits _{k_n=1}^{k_{n-1}} 1 \ = \ \sum \limits \limits _{k_1=1}^{k_{0}} \sum \limits \limits _{k_2=k_1}^{k_{0}} \cdots \sum \limits \limits _{k_n=k_{n-1}}^{k_{0}} 1 \ = \ {{k_{0}+n-1}\atopwithdelims (){n}}. \end{aligned}$$

(1.E.24)

(Hint. Consider the number of combinations of choosing n from $k_{0}$ with repetitions allowed.)

Exercise 1.38

The identity

$$\begin{aligned} \tilde{B}(\alpha , \beta ) \ = \ \frac{\varGamma (\beta ) \varGamma (\alpha ) }{\varGamma (\alpha +\beta ) } \end{aligned}$$

(1.E.25)

is shown in Appendix 1.1. Now, based on

$$\begin{aligned} \varGamma (\alpha )\varGamma (\beta )= & {} \int _{0}^{\infty } t^{\alpha -1} e^{-t} dt \int _{0}^{\infty } t^{\beta -1} e^{-t} dt \nonumber \\= & {} \int _{0}^{\infty } \int _{0}^{\infty } t^{\alpha -1} s^{\beta -1} e^{-(t+s)} ds dt , \end{aligned}$$

(1.E.26)

confirm (1.E.25).

Exercise 1.39

Referring to Table 1.4, show

$$\begin{aligned} {}_{-1} \text{C}_{z}= & {} \left\{ \begin{array}{ll} (-1)^{z}, &{} z= 0, 1, \ldots , \\ (-1)^{z+1}, &{} z= -1, -2, \ldots , \\ \pm \infty , &{} \text{ otherwise } \end{array} \right. \end{aligned}$$

(1.E.27)

and

$$\begin{aligned} {}_{-2} \text{C}_{z}= & {} \left\{ \begin{array}{ll} (-1)^{z}(z+1), &{} z= -1, 0, \ldots , \\ (-1)^{-z+1}(z+1), &{} z= -2, -3, \ldots , \\ \pm \infty , &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(1.E.28)

which imply ${}_{-1} \text{C}_{a} = {}_{-1} \text{C}_{b}$ for $a+b=-1$ and ${}_{-2} \text{C}_{a} = {}_{-2} \text{C}_{b}$ for $a+b=-2$.

Exercise 1.40

Show

• (1) ${}_{ \frac{1}{2} } \text{C}_{k} = \frac{ (-1)^{k-1} }{ 2k-1 } \frac{(2k)!}{2^{2k}(k!)^2}$ for $k=0, 1, 2, \ldots$.

• (2) ${}_{- \frac{1}{2} } \text{C}_{k} = -(2k-1)\ {}_{ \frac{1}{2} } \text{C}_{k} =(-1)^{k}\frac{(2k)!}{2^{2k}(k!)^2}$ for $k=0, 1, 2, \ldots$.

• (3) ${}_{-2k-1} \text{C}_{m} = (-1)^{m}\frac{(2k+m)!}{ (2k)!m!}$ for $k=0, 1, 2, \ldots$ and $m = 0, 1, 2, \ldots$.

Exercise 1.41

Based on (1.A.15), obtain the values of ${}_{p} \text{C}_{0}- {}_{p} \text{C}_{1} + {}_{p} \text{C}_{2} - {}_{p} \text{C}_{3} + \cdots$ and ${}_{p} \text{C}_{0} + {}_{p} \text{C}_{1} + {}_{p} \text{C}_{2} + {}_{p} \text{C}_{3} + \cdots$ when $p > 0$. Using the results, obtain $\sum \limits _{k=0}^{\infty } {}_{p} \text{C}_{2k+1}$ and $\sum \limits _{k=0}^{\infty } {}_{p} \text{C}_{2k}$ when $p > 0$.

Exercise 1.42

Obtain the series expansions of $g_1 (z) = (1+z)^{ \frac{1}{2} }$ and $g_2 (z) = (1+z)^{- \frac{1}{2} }$.

Exercise 1.43

For non-negative numbers $\alpha$ and $\beta$ such that $\alpha + \beta \ne 0$, show that

$$\begin{aligned} \frac{\alpha \beta }{\alpha + \beta } \ \le \ \min ( \alpha , \beta ) \ \le \ \frac{2 \alpha \beta }{\alpha + \beta } \ \le \ \sqrt{\alpha \beta } \ \le \ \frac{\alpha + \beta }{2} \ \le \ \max ( \alpha , \beta ). \ \ \ \ \ \end{aligned}$$

(1.E.29)