Exercises from Section 1.2.5

2. [10] In the notation of Eq. (2), show that

p_{n (n - 1)} = p_{n n}

, and explain why this happens.

In the notation of Eq. (2), since

p_{n k} = \prod_{n - k + 1 \leq j \leq n} j

we have that

p_{n n} = \prod_{1 \leq j \leq n} j = \prod_{2 \leq j \leq n} j = \prod_{n - (n - 1) + 1 \leq j \leq n} j = p_{n (n - 1)} .

That is, after choosing the $(n - 1)$ th element, we have no choice left for the last element.

3. [10] What permutations of

{1, 2, 3, 4, 5}

would be constructed from the permutation

3

1

2

4

using Methods 1 and 2, respectively?

We can construct permutations of the set ${1, 2, 3, 4, 5}$ from the permutation $3$ $1$ $2$ $4$ using either method.

In Method 1, we insert $5$ in all possible positions to obtain

$5$ $3$ $1$ $2$ $4$ , $3$ $5$ $1$ $2$ $4$ , $3$ $1$ $5$ $2$ $4$ , $3$ $1$ $2$ $5$ $4$ , and $3$ $1$ $2$ $4$ $5$ .

In Method 2, we start with an intermediary set of permutations

$3$ $1$ $2$ $4$ $\frac{1}{2}$ , $3$ $1$ $2$ $4$ $\frac{3}{2}$ , $3$ $1$ $2$ $4$ $\frac{5}{2}$ , $3$ $1$ $2$ $4$ $\frac{7}{2}$ , and $3$ $1$ $2$ $4$ $\frac{9}{2}$ ,

which are ﬁnally renamed as

$4$ $2$ $3$ $5$ $1$ , $4$ $1$ $3$ $5$ $2$ , $4$ $1$ $2$ $5$ $3$ , $3$ $1$ $2$ $5$ $4$ , and $3$ $1$ $2$ $4$ $5$ .

▸

4. [13] Given the fact that

\log_{10} 1000! = 2567.60464 \dots

, determine exactly how many decimal digits are present in the number

1000!

. What is the most signiﬁcant digit? What is the least signiﬁcant digit?

Since the number of decimal digits in a number $n$ is given by $⌊ \log_{10} n ⌋ + 1$ , $\log_{10} 1000! = 2567.60464 \dots$ implies that $1000!$ has $2568$ decimal digits.

The most signiﬁcant digit is given by $⌊\frac{1000!}{1 0^{2568 - 1}}⌋$ , but since $\log_{10} 4 = 0.60206 \dots$ and $\log_{10} 5 = 0.69897 \dots$ ,

\begin{array}{l} \log_{10} 1000! = 2567.60464 \dots & \Rightarrow 2567 + \log_{10} 4 \leq \log_{10} 1000! < 2567 + \log_{10} 5 \\ \Rightarrow \log_{10} 1 0^{2567} + \log_{10} 4 \leq \log_{10} 1000! < \log_{10} 1 0^{2567} + \log_{10} 5 \\ \Rightarrow \log_{10} (4 \cdot 1 0^{2567}) \leq \log_{10} 1000! < \log_{10} (5 \cdot 1 0^{2567}) \\ \Rightarrow \log_{10} 4 \leq \log_{10} \frac{1000!}{1 0^{2567}} < \log_{10} 5 \\ \Rightarrow 4 \leq \frac{1000!}{1 0^{2568 - 1}} < 5, \end{array}

and so, the most signiﬁcant digit is $⌊\frac{1000!}{1 0^{2568 - 1}}⌋ = 4$ .

The least signiﬁcant digit is intuitively $0$ , since the factorial is some number multiplied by a factor of $1000$ . We can determine precisely how many zeros using Eq. (8), since

\begin{array}{l} μ_{2} & = \sum_{k > 0} ⌊\frac{1000}{2^{k}}⌋ \\ = \sum_{1 \leq k \leq 9} ⌊\frac{1000}{2^{k}}⌋ \\ = 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 \\ = 994 \end{array}

and

\begin{array}{l} μ_{5} & = \sum_{k > 0} ⌊\frac{1000}{5^{k}}⌋ \\ = \sum_{1 \leq k \leq 4} ⌊\frac{1000}{5^{k}}⌋ \\ = 200 + 40 + 8 + 1 \\ = 249 \end{array}

giving us that for some arbitrary integer $z$ not divisible by $10$ , $1000! = 2^{994} \cdot 5^{249} z = 2^{745} z \cdot 1 0^{249}$ ; that is, $1000!$ is a number that ends with 249 zeros.

________________________________________________________________________________

[Scripta Mathematica 21 (1955), 266–267]

5. [15] Estimate

8!

using the more exact version of Stirling’s approximation:

We may estimate $8!$ as

\begin{array}{l} 8! & \approx \sqrt{2 π 8} {(\frac{8}{e})}^{8} (1 + \frac{1}{12 (8)}) \\ = 4 \sqrt{π} \frac{16777216}{e^{8}} \frac{97}{96} \\ = \frac{203423744 \sqrt{π}}{3 e^{8}} \\ \approx 40318 . \end{array}

Since the primes that divide $20$ are $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , and $19$ , we may determine the multiplicity of the prime factors of $20!$ as

\begin{array}{l} μ_{2} & = \sum_{k > 0} ⌊\frac{20}{2^{k}}⌋ \\ = \sum_{1 \leq k \leq 4} ⌊\frac{20}{2^{k}}⌋ \\ = 10 + 5 + 2 + 1 \\ = 18, \end{array}

\begin{array}{l} μ_{3} & = \sum_{k > 0} ⌊\frac{20}{2^{k}}⌋ \\ = \sum_{1 \leq k \leq 2} ⌊\frac{20}{2^{k}}⌋ \\ = 6 + 2 \\ = 8, \end{array}

\begin{array}{l} μ_{5} & = \sum_{k > 0} ⌊\frac{20}{5^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{5^{k}}⌋ \\ = 4, \end{array}

\begin{array}{l} μ_{7} & = \sum_{k > 0} ⌊\frac{20}{7^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{7^{k}}⌋ \\ = 2, \end{array}

\begin{array}{l} μ_{11} & = \sum_{k > 0} ⌊\frac{20}{1 1^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{1 1^{k}}⌋ \\ = 1, \end{array}

\begin{array}{l} μ_{13} & = \sum_{k > 0} ⌊\frac{20}{1 3^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{1 3^{k}}⌋ \\ = 1, \end{array}

\begin{array}{l} μ_{17} & = \sum_{k > 0} ⌊\frac{20}{1 7^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{1 7^{k}}⌋ \\ = 1, \end{array}

and

\begin{array}{l} μ_{19} & = \sum_{k > 0} ⌊\frac{20}{1 9^{k}}⌋ \\ = \sum_{1 \leq k \leq 1} ⌊\frac{20}{1 9^{k}}⌋ \\ = 1, \end{array}

so that $20! = 2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$ .

7. [M10] Show that the “generalized terminal” function in Eq. (10) satisﬁes the identity

x ? = x + (x - 1) ?

for all real numbers

x

Proposition. $x ? = x + (x - 1) ?$ for all real numbers $x$ .

Proof. Let $x$ be an arbitrary real number. We must show that

x ? = x + (x - 1) ? .

But by Eq. (10)

\begin{array}{l} x ? & = \frac{1}{2} x (x + 1) \\ = x - x + \frac{1}{2} x (x + 1) \\ = x + \frac{- 2 x}{2} + \frac{1}{2} x (x + 1) \\ = x + \frac{1}{2} (x^{2} + x - 2 x) \\ = x + \frac{1}{2} (x^{2} - x) \\ = x + \frac{1}{2} (x - 1) x \\ = x + \frac{1}{2} (x - 1) ((x - 1) + 1) \\ = x + (x - 1) ? \end{array}

as we needed to show. □

8. [HM15] Show that the limit in Eq. (13) does equal

n!

when

n

is a nonnegative integer.

When $n$ is a nonnegative integer, we have

\begin{array}{l} \lim_{m \to \infty} \frac{m^{n} m!}{\prod_{1 \leq k \leq m} (n + k)} & = \lim_{m \to \infty} \frac{m^{n} m!}{(n + m)! ∕ n!} \\ = \lim_{m \to \infty} \frac{n! m^{n} m!}{(m + n)!} \\ = \lim_{m \to \infty} \frac{n! m^{n}}{(m + n)! ∕ m!} \\ = \lim_{m \to \infty} \frac{n! m^{n}}{\prod_{1 \leq k \leq n} (m + k)} \\ = \lim_{m \to \infty} n! \prod_{1 \leq k \leq n} \frac{m}{m + k} \\ = n! \lim_{m \to \infty} \prod_{1 \leq k \leq n} \frac{m}{m + k} \\ = n! . \end{array}

9. [M10] Determine the values of

Γ (\frac{1}{2})

and

Γ (- \frac{1}{2})

, given that

(\frac{1}{2})! = \sqrt{π} ∕ 2

Given that $(\frac{1}{2})! = \sqrt{π} ∕ 2$ , we have that

\begin{array}{l} Γ (\frac{1}{2}) & = (\frac{1}{2})! ∕ (\frac{1}{2}) \\ = \frac{\sqrt{π}}{2} \frac{2}{1} \\ = \sqrt{π} \end{array}

and that

\begin{array}{l} Γ (\frac{- 1}{2}) & = \frac{2}{- 1} Γ (\frac{1}{2}) \\ = - 2 \sqrt{π} \end{array}

since $Γ (n + 1) = n Γ (n)$ .

▸

10. [HM20] Does the identity

Γ (x + 1) = x Γ (x)

hold for all real numbers

x

? (See exercise 7.)

The identity $Γ (x + 1) = x Γ (x)$ holds for all real numbers $x$ , except when $x$ is zero or a negative integer, since

\begin{array}{l} Γ (x + 1) & = \lim_{m \to \infty} \frac{m^{x} m!}{\prod_{1 \leq k \leq m} (x + k)} \\ = \lim_{m \to \infty} \frac{m x (x + m)}{x (x + m)} \frac{m^{x - 1} m!}{\prod_{2 \leq k \leq m + 1} ((x - 1) + k)} \\ = \lim_{m \to \infty} \frac{m x}{x + m} \frac{m^{x - 1} m!}{\prod_{1 \leq k \leq m} ((x - 1) + k)} \\ = x \lim_{m \to \infty} \frac{m}{m + x} \frac{m^{x - 1} m!}{\prod_{1 \leq k \leq m} ((x - 1) + k)} \\ = x \lim_{m \to \infty} \frac{m^{x - 1} m!}{\prod_{1 \leq k \leq m} ((x - 1) + k)} \\ = x Γ (x) . \end{array}

11. [M15] Let the representation of

n

in the binary system be

n = 2^{e_{1}} + 2^{e_{2}} + \dots + 2^{e_{r}}

, where

e_{1} > e_{2} > \dots > e_{r} \geq 0

. Show that

n!

is divisible by

2^{n - r}

but not by

2^{n - r + 1}

Given $n = \sum_{1 \leq j \leq r} 2^{e_{j}}$ , we may ﬁnd the exact multiplicity of the prime factor $2$ from Eq. (8) as:

\begin{array}{l} μ & = \sum_{i > 0} ⌊\frac{n}{2^{i}}⌋ \\ = \sum_{1 \leq i \leq r} ⌊\frac{\sum_{1 \leq j \leq r} 2^{e_{j}}}{2^{e_{i}}}⌋ \\ = \sum_{1 \leq i \leq r} ⌊\sum_{0 \leq j \leq r} \frac{2^{e_{j}}}{2^{e_{i}}}⌋ \\ = \sum_{1 \leq j \leq r} \sum_{1 \leq i \leq j} \frac{2^{e_{j}}}{2^{e_{i}}} \\ = \sum_{1 \leq j \leq r} 2^{e_{j}} \sum_{1 \leq i \leq j} \frac{1}{2^{e_{i}}} \\ = \sum_{1 \leq j \leq r} 2^{e_{j}} \frac{2^{- e_{j}} (2^{e_{j}} - 1)}{2 - 1} \\ = \sum_{1 \leq j \leq r} \frac{2^{e_{j}}}{2^{e_{j}}} (2^{e_{j}} - 1) \\ = \sum_{1 \leq j \leq r} (2^{e_{j}} - 1) \\ = \sum_{1 \leq j \leq r} 2^{e_{j}} - \sum_{0 \leq j \leq r} 1 \\ = n - \sum_{1 \leq j \leq r} 1 \\ = n - r . \end{array}

That is, $n!$ is divisible by $2^{n - 1}$ , but not by $2^{n - r + 1}$ .

▸

12. [M22] (A. Legendre, 1808.) Generalizing the result of the previous exercise, let

p

be a prime number, and let the representation of

n

in the

p

-ary number system be

n = a_{k} p^{k} + a_{k - 1} p^{k - 1} + \dots + a_{1} p + a_{0}

. Express the number

μ

of Eq. (8) in a simple formula involving

n

p

, and

a

’s.

Given $n = \sum_{0 \leq j \leq k} a_{j} p^{j}$ , we may express the number $μ$ from Eq. (8) as:

\begin{array}{l} μ & = \sum_{i > 0} ⌊\frac{n}{p^{i}}⌋ \\ = \sum_{1 \leq i \leq k} ⌊\frac{\sum_{0 \leq j \leq k} a_{j} p^{j}}{p^{i}}⌋ \\ = \sum_{1 \leq i \leq k} ⌊\sum_{0 \leq j \leq k} \frac{a_{j} p^{j}}{p^{i}}⌋ \\ = \sum_{0 \leq j \leq k} \sum_{1 \leq i \leq j} \frac{a_{j} p^{j}}{p^{i}} \\ = \sum_{0 \leq j \leq k} a_{j} p^{j} \sum_{1 \leq i \leq j} \frac{1}{p^{i}} \\ = \sum_{0 \leq j \leq k} a_{j} p^{j} \frac{p^{- j} (p^{j} - 1)}{p - 1} \\ = \sum_{0 \leq j \leq k} a_{j} \frac{p^{j}}{p^{j}} \frac{p^{j} - 1}{p - 1} \\ = \sum_{0 \leq j \leq k} a_{j} \frac{p^{j} - 1}{p - 1} \\ = \frac{\sum_{0 \leq j \leq k} a_{j} (p^{j} - 1)}{p - 1} \\ = \frac{\sum_{0 \leq j \leq k} a_{j} p^{j} - \sum_{0 \leq j \leq k} a_{j}}{p - 1} \\ = \frac{n - \sum_{0 \leq j \leq k} a_{j}}{p - 1} . \end{array}

13. [M23] (Wilson’s theorem, actually due to Leibniz, 1682.) If

p

is prime, then

(p - 1)! mod p = p - 1

. Prove this, by pairing oﬀ numbers among

{1, 2, \dots, p - 1}

whose product modulo

p

is 1.

Proposition. If $p$ is prime, $(p - 1)! mod p = p - 1$ .

Proof. Let $p$ be an arbitrary prime. We must show that

(p - 1)! mod p = p - 1 .

By exercise 1.2.4-19, for each integer $k$ , $1 < k < p - 1$ , there is another integer $k^{'}$ such that

k k^{'} \equiv 1 (\mod p)

allowing us to ’pair oﬀ numbers’ as

(p - 2)! ∕ 1! \equiv 1 (\mod p) .

Also, clearly,

1 \equiv 1 (\mod p)

and

(p - 1) \equiv p - 1 (\mod p) .

And so,

(p - 1)! \equiv p - 1 (\mod p)

or equivalently

(p - 1)! mod p = p - 1

as we needed to show. □

▸

14. [M28] (L. Stickelberger, 1890.) In the notation of exercise 12, we can determine

n! mod p

in terms of the

p

-ary representation, for any positive integer

n

, thus generalizing Wilson’s theorem. In fact, prove that

n! ∕ p^{μ} \equiv {(- 1)}^{μ} a_{0}! a_{1}! \dots a_{k}! (modulo p)

Proposition. $n! ∕ p^{μ} \equiv {(- 1)}^{μ} \prod_{0 \leq i \leq k} a_{i}! (\mod p)$ .

Proof. Let $n$ and $p$ be arbitrary positive integers such that $p$ is prime, $n = \sum_{0 \leq i \leq k} a_{i} p^{i}$ , and $μ = \sum_{1 \leq j \leq k} ⌊\frac{n}{p^{j}}⌋$ . We must show that

n! ∕ p^{μ} \equiv {(- 1)}^{μ} \prod_{0 \leq i \leq k} a_{i}! (mod p) .

First consider the trivial case $k = 0$ , so that $n = a_{0}$ and $μ = 0$ . Then clearly

\begin{array}{l} a_{0}! \equiv a_{0}! (\mod p) & \Leftrightarrow a_{0}! ∕ p^{0} \equiv {(- 1)}^{0} \prod_{0 \leq i \leq 0} a_{i}! (mod p) \\ \Leftrightarrow n! ∕ p^{μ} \equiv {(- 1)}^{μ} \prod_{0 \leq i \leq k} a_{i}! (mod p) . \end{array}

Then, assuming as an inductive hypothesis for $n_{k} = \sum_{0 \leq i \leq k} a_{i} p^{i}$ and $μ_{k} = \sum_{1 \leq i \leq k} ⌊\frac{n_{k}}{p^{i}}⌋$ that

n_{k}! ∕ p^{μ_{k}} \equiv {(- 1)}^{μ_{k}} \prod_{0 \leq i \leq k} a_{i}! (mod p),

we must show for $n_{k + 1} = a_{k + 1} p^{k + 1} + n_{k}$ and $μ_{k + 1} = ⌊\frac{n_{k + 1}}{p^{k + 1}}⌋ + μ_{k}$ that

n_{k + 1}! ∕ p^{μ_{k + 1}} \equiv {(- 1)}^{μ_{k + 1}} \prod_{0 \leq i \leq k + 1} a_{i}! (mod p) .

(Note that the equality for $μ_{k + 1}$ holds since $n_{k + 1}$ has grown by a multiple of $p$ ; namely, $a_{k + 1} p^{k}$ .)

But by Wilson’s theorem, all the terms between $n_{k} + 1$ and $n_{k + 1}$ that are not multiples of $p$ may be collected into sets of size $p - 1$ whose product is congruent to $- 1$ modulo $p$ , and there are precisely $⌊\frac{n_{k + 1}}{p^{k + 1}}⌋$ of these products, with $a_{k + 1}$ left over, congruent to $a_{k + 1}!$ modulo $p$ . That is

\prod_{n_{k} + 1 \leq i \leq n_{k + 1}} i / p^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋} \equiv {(- 1)}^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋} a_{k + 1}! (mod p) .

Multiplying this by the inductive hypothesis yields

\begin{array}{l} (\prod_{n_{k} + 1 \leq i \leq n_{k + 1}} i / p^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋}) n_{k}! ∕ p^{μ_{k}} \equiv {(- 1)}^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋} a_{k + 1}! {(- 1)}^{μ_{k}} \prod_{0 \leq i \leq k} a_{i}! (mod p) \\ \Leftrightarrow n_{k}! \prod_{n_{k} + 1 \leq i \leq n_{k + 1}} i / p^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋ + μ_{k}} \equiv {(- 1)}^{⌊\frac{n_{k + 1}}{p^{k + 1}}⌋ + μ_{k}} a_{k + 1}! \prod_{0 \leq i \leq k} a_{i}! (mod p) \\ \Leftrightarrow n_{k + 1}! ∕ p^{μ_{k + 1}} \equiv {(- 1)}^{μ_{k + 1}} \prod_{0 \leq i \leq k + 1} a_{i}! (mod p) . \end{array}

Therefore

n! ∕ p^{μ} \equiv {(- 1)}^{μ} \prod_{0 \leq i \leq k} a_{i}! (mod p)

as we needed to show. □

15. [HM15] The permanent of a square matrix is deﬁned by the same expansion as the determinant except that each term of the permanent is given a plus sign while the determinant alternates between plus and minus. Thus the permanent of

The permanent of a square matrix may be deﬁned recursively as

perm ({[a_{i j}]}_{n}) = \{\begin{matrix} a_{11} & if n = 1 \\ \sum_{1 \leq j \leq n} a_{i j} + perm (submatrix (a_{i} j)) & otherwise. \end{matrix}

In the case that $a_{i j} = i \times j$ , we may simply add

(1 \times 1) + (2 \times 2) + \dots + (n \times n) = {(n!)}^{2},

and we do this for $n!$ terms, yielding a total sum of

n! {(n!)}^{2} = {(n!)}^{3} .

16. [HM15] Show that the inﬁnite sum in Eq. (11) does not converge unless

n

is a nonnegative integer.

The inﬁnite sum in Eq. (11),

\sum_{k \geq 0} ((\sum_{0 \leq j \leq k} \frac{{(- 1)}^{j}}{j!}) (\prod_{0 \leq j < k} (n - j))),

does not converge unless $n$ is a nonnegative integer, since if $n < 0$ , the product $\prod_{0 \leq j < k} (n - j)$ never vanishes as the coeﬃcients

\lim_{k \to \infty} \sum_{0 \leq j \leq k} \frac{{(- 1)}^{j}}{j!} = \frac{1}{e} .

(In the case that $n \geq 0$ , the product eventually vanishes with a factor of zero.)

equals

Γ (1 + β_{1}) \dots Γ (1 + β_{k}) ∕ Γ (1 + α_{1}) \dots Γ (1 + α_{k})

, if

α_{1} + \dots + α_{k} = β_{1} + \dots + β_{k}

and if none of the

β

’s is a negative integer.

Proposition. $\prod_{n \geq 1} \prod_{1 \leq i \leq k} \frac{n + α_{i}}{n + β_{i}} = \prod_{1 \leq i \leq k} \frac{Γ (1 + β_{i})}{Γ (1 + α_{i})}$ if $\sum_{1 \leq i \leq k} α_{i} = \sum_{1 \leq i \leq k} β_{i}$ and $β_{i} \geq 0$ for $1 \leq i \leq k$ .

Proof. Let $α_{i}$ and $β_{i}$ be arbitrary integer sequences such that $\sum_{1 \leq i \leq k} α_{i} = \sum_{1 \leq i \leq k} β_{i}$ and $β_{i} \geq 0$ for $1 \leq i \leq k$ . We must show that

\prod_{n \geq 1} \prod_{1 \leq i \leq k} \frac{n + α_{i}}{n + β_{i}} = \prod_{1 \leq i \leq k} \frac{Γ (1 + β_{i})}{Γ (1 + α_{i})} .

But

\begin{array}{l} \prod_{1 \leq i \leq k} \frac{Γ (1 + β_{i})}{Γ (1 + α_{i})} & = \lim_{m \to \infty} \prod_{1 \leq i \leq k} \frac{m^{1 + β_{i}} m! \prod_{0 \leq j \leq m} (1 + α_{i} + j)}{m^{1 + α_{i}} m! \prod_{0 \leq j \leq m} (1 + β_{i} + j)} \\ = \lim_{m \to \infty} \prod_{1 \leq i \leq k} \frac{m^{β_{i}} \prod_{0 \leq j \leq m} (1 + α_{i} + j)}{m^{α_{i}} \prod_{0 \leq j \leq m} (1 + β_{i} + j)} \\ = \lim_{m \to \infty} \frac{m^{\sum_{1 \leq i \leq k} β_{i}}}{m^{\sum_{1 \leq i \leq k} α_{i}}} \prod_{1 \leq i \leq k} \frac{\prod_{0 \leq j \leq m} (1 + α_{i} + j)}{\prod_{0 \leq j \leq m} (1 + β_{i} + j)} \\ = \lim_{m \to \infty} \prod_{1 \leq i \leq k} \prod_{0 \leq j \leq m} \frac{1 + α_{i} + j}{1 + β_{i} + j} \\ = \lim_{m \to \infty} \prod_{0 \leq j \leq m} \prod_{1 \leq i \leq k} \frac{1 + α_{i} + j}{1 + β_{i} + j} \\ = \lim_{m \to \infty} \prod_{1 \leq n \leq m} \prod_{1 \leq i \leq k} \frac{n + α_{i}}{n + β_{i}} \\ = \prod_{n \geq 1} \prod_{1 \leq i \leq k} \frac{n + α_{i}}{n + β_{i}} \end{array}

as we needed to show. □

18. [M20] Assume that

π ∕ 2 = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \dots

. (This is “Wallis’s product,” obtained by J. Wallis in 1655, and we will prove it in exercise 1.2.6-43.) Using the previous exercise, prove that

(\frac{1}{2})! = \sqrt{π} ∕ 2

Proposition. $(\frac{1}{2})! = \sqrt{π} ∕ 2$ .

Proof. We must show that

(\frac{1}{2})! = \sqrt{π} ∕ 2 .

But according to exercise 17 with $α_{1} = α_{2} = 0$ , $β_{1} = - 1 ∕ 2$ , and $β_{2} = 1 ∕ 2$ so that $\sum_{1 \leq i \leq 2} α_{i} = \sum_{1 \leq i \leq 2} β_{i}$ and $β_{i} \geq 0$ for $1 \leq i \leq 2$ ,

\begin{array}{l} \frac{\sqrt{π}}{2} & = \sqrt{\frac{1}{2} \frac{π}{2}} \\ = \sqrt{\frac{1}{2} \prod_{n \geq 1} \frac{2 n}{2 n - 1} \frac{2 n}{2 n + 1}} \\ = \sqrt{\frac{1}{2} \prod_{n \geq 1} \frac{n}{n - \frac{1}{2}} \frac{n}{n + \frac{1}{2}}} \\ = \sqrt{\frac{1}{2} \prod_{n \geq 1} \frac{n + α_{1}}{n + β_{1}} \frac{n + α_{2}}{n + β_{2}}} \\ = \sqrt{\frac{1}{2} \prod_{n \geq 1} \prod_{1 \leq k \leq 2} \frac{n + α_{k}}{n + β_{k}}} \\ = \sqrt{\frac{1}{2} \prod_{1 \leq k \leq 2} \frac{Γ (1 + β_{k})}{Γ (1 + α_{k}}} \\ = \sqrt{\frac{1}{2} \prod_{1 \leq k \leq 2} \frac{Γ (1 + β_{k})}{Γ (1 + α_{k})}} \\ = \sqrt{\frac{1}{2} \frac{Γ (1 + β_{1})}{Γ (1 + α_{1})} \frac{Γ (1 + β_{2})}{Γ (1 + α_{2})}} \\ = \sqrt{\frac{1}{2} \frac{Γ (1 - \frac{1}{2})}{Γ (1)} \frac{Γ (1 + \frac{1}{2})}{Γ (1)}} \\ = \sqrt{\frac{1}{2} \frac{Γ (\frac{1}{2})}{Γ (1)} \frac{Γ (\frac{3}{2})}{Γ (1)}} \\ = \sqrt{\frac{1}{2} Γ (\frac{1}{2}) Γ (\frac{3}{2})} \\ = \sqrt{\frac{2}{2} Γ (\frac{3}{2}) Γ (\frac{3}{2})} \\ = \sqrt{Γ {(\frac{3}{2})}^{2}} \\ = Γ (\frac{3}{2}) \\ = (\frac{1}{2})! \end{array}

as we needed to show. □

______________________________________________________________________________________________________________________________

[Wallis’s own heuristic “proof” can be found in D. J. Struik’s Source Book in Mathematics (Harvard University Press, 1969), 244–253.]

19. [HM22] Denote the quantity appearing after “

\lim_{m \to \infty}

” in Eq. (15) by

Γ_{m} (x)

. Show that

Proposition. $Γ_{m} (x) = \int_{0}^{m} {(1 - \frac{t}{m})}^{m} t^{x - 1} d t = m^{x} \int_{0}^{1} {(1 - t)}^{m} t^{x - 1} d t$ if $x > 0$ .

Proof. Let $x$ be an arbitrary positive real number so that $x > 0$ . We must show that

Γ_{m} (x) = \int_{0}^{m} {(1 - \frac{t}{m})}^{m} t^{x - 1} d t = m^{x} \int_{0}^{1} {(1 - t)}^{m} t^{x - 1} d t .

But by substituting $m t$ for $t$

\begin{array}{l} \int_{0}^{m} {(1 - \frac{t}{m})}^{m} t^{x - 1} d t & = \int_{0}^{t ∕ t} {(1 - \frac{m t}{m})}^{m} {(m t)}^{x - 1} d m t \\ = \int_{0}^{1} {(1 - t)}^{m} m^{x} t^{x - 1} d t \\ = m^{x} \int_{0}^{1} {(1 - t)}^{m} t^{x - 1} d t . \end{array}

Then, let

f_{m} (x) = \int_{0}^{1} {(1 - t)}^{m} t^{x - 1} d t .

We may prove by induction on $m$ that

f_{m} (x) = \frac{m!}{\prod_{0 \leq k \leq m} (x + k)} .

If $m = 0$ , clearly

\begin{array}{l} f_{0} (x) & = \int_{0}^{1} {(1 - t)}^{0} t^{x - 1} d t \\ = \int_{0}^{1} t^{x - 1} d t \\ = {\frac{t^{x}}{x}|}_{0}^{1} \\ = \frac{1^{x}}{x} - \frac{0^{x}}{x} \\ = \frac{1 - 0}{x} \\ = \frac{1}{x} \\ = \frac{0!}{\prod_{0 \leq k \leq 0} (x + k)} . \end{array}

Then, assuming

f_{m} (x) = \frac{m!}{\prod_{0 \leq k \leq m} (x + k)},

we must show that

f_{m + 1} (x) = \frac{(m + 1)!}{\prod_{0 \leq k \leq m + 1} (x + k)} .

But by integration by parts, since $\frac{d}{d x} \frac{t^{x}}{x} = t^{x - 1}$ and $\frac{d}{d t} {(1 - t)}^{m + 1} = - (m + 1) {(1 - t)}^{m}$ ,

\begin{array}{l} f_{m + 1} (x) & = \int_{0}^{1} {(1 - t)}^{m + 1} t^{x - 1} d t \\ = {{(1 - t)}^{m + 1} \frac{t^{x}}{x}|}_{0}^{1} - \int_{0}^{1} - (m + 1) {(1 - t)}^{m} \frac{t^{x}}{x} d t \\ = {(1 - 1)}^{m + 1} \frac{1^{x}}{x} - {(1 - 0)}^{m + 1} \frac{0^{x}}{x} + \frac{m + 1}{x} \int_{0}^{1} {(1 - t)}^{m} t^{x} d t \\ = 0 - 0 + \frac{m + 1}{x} \int_{0}^{1} {(1 - t)}^{m} t^{x} d t \\ = \frac{m + 1}{x} f_{m} (x + 1) . \end{array}

And so, by the inductive hypothesis,

\begin{array}{l} f_{m + 1} (x) & = \frac{m + 1}{x} f_{m} (x + 1) \\ = \frac{m + 1}{x} \frac{m!}{\prod_{0 \leq k \leq m} (x + 1 + k)} \\ = \frac{(m + 1) m!}{x \prod_{1 \leq k \leq m + 1} (x + k)} \\ = \frac{(m + 1)!}{\prod_{0 \leq k \leq m + 1} (x + k)}, \end{array}

so that

f_{m} (x) = \frac{m!}{\prod_{0 \leq k \leq m} (x + k)} .

Finally

\begin{array}{l} \int_{0}^{m} {(1 - \frac{t}{m})}^{m} t^{x - 1} d t & = m^{x} \int_{0}^{1} {(1 - t)}^{m} t^{x - 1} d t \\ = m^{x} f_{m} (x) \\ = m^{x} \frac{m!}{\prod_{0 \leq k \leq m} (x + k)} \\ = \frac{m^{x} m!}{\prod_{0 \leq k \leq m} (x + k)} \\ = Γ_{m} (x) \end{array}

as we needed to show. □

20. [HM21] Using the fact that

0 \leq e^{- t} - {(1 - t ∕ m)}^{m} \leq t^{2} e^{- t} ∕ m

, if

0 \leq t \leq m

, and the previous exercise, show that

Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t

, if

x > 0

Proposition. $Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t$ , if $x > 0$ .

Proof. Let $x$ be an arbitrary positive real number such that $x > 0$ . We must show that

Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t .

Let

g (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t .

It is suﬃcient to show that $g (x) - Γ (x) = 0$ .

Note that if $0 \leq t \leq m$ ,

\begin{array}{l} 1 + x \leq e^{x} & \Leftrightarrow 1 \pm t ∕ m \leq e^{\pm t ∕ m} \\ \Leftrightarrow {(1 \pm t ∕ m)}^{m} \leq e^{\pm t} \end{array}

and from exercise 1.2.1-9,

\begin{array}{l} e^{- t} & \geq {(1 - t ∕ m)}^{m} \\ = e^{- t} {(1 - t ∕ m)}^{m} e^{t} \\ \geq e^{- t} {(1 - t ∕ m)}^{m} {(1 + t ∕ m)}^{m} \\ = e^{- t} {(1 - t^{2} ∕ m^{2})}^{m} \\ \geq e^{- t} (1 - t^{2} ∕ m) \end{array}

so that

0 \leq e^{- t} - {(1 - t ∕ m)}^{m} \leq t^{2} e^{- t} ∕ m .

Then since $x + 1 \geq 2$ ,

\begin{array}{l} 0 \leq e^{- t} - {(1 - t ∕ m)}^{m} \leq t^{x + 1} e^{- t} ∕ m \\ \Leftrightarrow 0 \leq \int_{0}^{m} e^{- t} - {(1 - t ∕ m)}^{m} d t \leq \frac{1}{m} \int_{0}^{m} t^{x + 1} e^{- t} d t < \frac{1}{m} \int_{0}^{\infty} t^{x + 1} e^{- t} d t \\ \Leftrightarrow 0 \leq \int_{0}^{\infty} e^{- t} - {(1 - t ∕ m)}^{m} d t \leq \frac{1}{m} \int_{0}^{\infty} t^{x + 1} e^{- t} d t \\ \Leftrightarrow 0 \leq \int_{0}^{\infty} e^{- t} - {(1 - t ∕ m)}^{m} d t \leq 0 \\ \Leftrightarrow \int_{0}^{\infty} e^{- t} - {(1 - t ∕ m)}^{m} d t = 0 \\ \Leftrightarrow \int_{0}^{\infty} e^{- t} t^{x - 1} d t - \int_{0}^{\infty} {(1 - t ∕ m)}^{m} t^{x - 1} d t = 0 \\ \Leftrightarrow g (x) - Γ (x) = 0 \end{array}

as we needed to show. □

21. [HM25] (L. F. A. Arbogast, 1800.) Let

D_{x}^{k} u

represent the

k

th derivative of a function

u

with respect to

x

. The chain rule states that

D_{x}^{1} w = D_{u}^{1} w D_{x}^{1} u

. If we apply this to second derivatives, we ﬁnd

D_{x}^{2} w = D_{u}^{2} w {(D_{x}^{1} u)}^{2} + D_{u}^{1} w D_{x}^{2} u

. Show that the general formula is

Proposition. $D_{x}^{n} f = \sum_{0 \leq j \leq n} \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} n! D_{g}^{j} f \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}}$ .

Proof. Let $f$ be an arbitrary function of $g$ , $g$ an arbitarary function of $x$ , and $n$ a positive integer. We must show that that the $n$ th derivative of $f$ with respect to $x$ is

D_{x}^{n} f = \sum_{0 \leq j \leq n} \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} n! D_{g}^{j} f \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} .

As done by T. A.¹ , it suﬃces to analyze the coeﬃcients for any function $f$ . Let $f = e^{p g (x)}$ . By Taylor’s theorem for an arbitrary $h$ ,

e^{p g (x + h)} = \sum_{n \geq 0} \frac{h^{n}}{n!} D_{x}^{n} f .

But also, by expanding $g (x + h)$ and developing the product,

\begin{array}{l} e^{p g (x + h)} & = e^{p g (x)} \prod_{n \geq 1} e^{p D_{x}^{n} g \frac{h^{n}}{n!}} \\ = \sum_{n \geq 0} h^{n} \sum_{0 \leq j \leq n} p^{j} e^{p g (x)} \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} \\ = \sum_{n \geq 0} h^{n} \sum_{0 \leq j \leq n} D_{g}^{j} f \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} . \end{array}

Equating these two yields

\begin{array}{l} \sum_{n \geq 0} \frac{h^{n}}{n!} D_{x}^{n} f = \sum_{n \geq 0} h^{n} \sum_{0 \leq j \leq n} D_{g}^{j} f \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} \\ \Leftrightarrow \frac{1}{n!} D_{x}^{n} f = \sum_{0 \leq j \leq n} D_{g}^{j} f \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} \\ \Leftrightarrow D_{x}^{n} f = \sum_{0 \leq j \leq n} \sum_{\begin{array}{c} k_{1} + k_{2} + \dots + k_{n} = j \\ k_{1} + 2 k_{2} + \dots + n k_{n} = n \\ k_{1}, k_{2}, \dots, k_{n} \geq 0 \end{array}} n! D_{g}^{j} f \prod_{1 \leq i \leq n} \frac{{(D_{x}^{i} g)}^{k_{i}}}{k_{i}! {(i!)}^{k_{i}}} \end{array}

and hence the result. □

______________________________________________________________________________________________________________________________

[Bull. Amer. Math. Soc. 44 (1938), 395–398]

[Du Calcul des Dérivations (Strasbourg: 1800), §52]

[Quarterly J. Math. 1 (1857), 359-360]

$\dots$ see the paper by I. J. Good, Annals of Mathematical Statistics 32 (1961), 540–541.

▸

22. [HM20] Try to put yourself in Euler’s place, looking for a way to generalize

n!

to noninteger values of

n

. Since

(n + \frac{1}{2})! ∕ n!

times

((n + \frac{1}{2}) + \frac{1}{2})! ∕ (n + \frac{1}{2})!

equals

(n + 1)! ∕ n! = n + 1

, it seems natural that

(n + \frac{1}{2})! ∕ n!

should be approximately

\sqrt{(} n)

. Similarly,

(n + \frac{1}{3})! ∕ n!

should be

\approx \sqrt[3]{n}

. Invent a hypothesis about the ratio

(n + x)! ∕ n!

n

approaches inﬁnity. Is your hypothesis correct when

x

is an integer? Does it tell anything about the appropriate value of

x!

when

x

is not an integer?

Observing that

\begin{matrix} \frac{(n + \frac{1}{2})!}{n!} \approx \sqrt{n} and \\ \frac{(n + \frac{1}{3})!}{n!} \approx \sqrt[3]{n}, \end{matrix}

we might hypothesize that

\lim_{n \to \infty} \frac{(n + x)!}{n! n^{x}} = 1 .

When $x$ is an integer, the equality holds, as

\begin{array}{l} \lim_{n \to \infty} \frac{(n + x)!}{n! n^{x}} & = \lim_{n \to \infty} \frac{\prod_{1 \leq k \leq x} (n + k)}{n^{x}} \\ = \lim_{n \to \infty} \frac{n^{x} \prod_{1 \leq k \leq x} (1 + \frac{k}{n})}{n^{x}} \\ = \lim_{n \to \infty} \prod_{1 \leq k \leq x} (1 + \frac{k}{n}) \\ = 1 . \end{array}

It tells us something about the appropriate value of $x!$ when $x$ is not an integer as well, since

\begin{array}{l} 1 = \lim_{n \to \infty} \frac{(n + x)!}{n! n^{x}} = x! \lim_{n \to \infty} \frac{\prod_{1 \leq k \leq n} (x + k)}{n! n^{x}} \\ \Leftrightarrow x! = \lim_{n \to \infty} \frac{n^{x} n!}{\prod_{1 \leq k \leq n} (x + k)} = Γ (x + 1) . \end{array}

23. [HM20] Prove (16), given that

π z \prod_{n = 1}^{\infty} (1 - z^{2} ∕ n^{2}) = \sin π z

Proposition. $(- z)! Γ (z) = \frac{π}{\sin π z}$ for $z$ not an integer.

Proof. Let $z$ be an arbitrary real number, not an integer. We must show that

(- z)! Γ (z) = \frac{π}{\sin π z} .

But given

\begin{array}{l} π z \prod_{m \geq 1} (1 - \frac{z^{2}}{m^{2}}) = \sin π z \\ \Leftrightarrow \frac{π z}{\sin π z} = \frac{1}{\prod_{m \geq 1} (1 - \frac{z^{2}}{m^{2}})} \end{array}

we have

\begin{array}{l} z (- z)! Γ (z) & = z \lim_{m \to \infty} \frac{m^{- z} m!}{\prod_{1 \leq k \leq m} (- z + k)} \lim_{m \to \infty} \frac{m^{z} m!}{z \prod_{1 \leq k \leq m} (z + k)} \\ = \lim_{m \to \infty} \frac{m^{- z} m! m^{z} m!}{\prod_{1 \leq k \leq m} (- z + k) (z + k)} \\ = \lim_{m \to \infty} \frac{{(m!)}^{2}}{\prod_{1 \leq k \leq m} k^{2} (1 - \frac{z}{k}) (1 + \frac{z}{k})} \\ = \lim_{m \to \infty} \frac{{(m!)}^{2}}{{(m!)}^{2} \prod_{1 \leq k \leq m} (1 - \frac{z}{k}) (1 + \frac{z}{k})} \\ = \lim_{m \to \infty} \frac{1}{\prod_{1 \leq k \leq m} (1 - \frac{z^{2}}{k^{2}})} \\ = \frac{1}{\prod_{m \geq 1} (1 - \frac{z^{2}}{k^{2}})} \\ = \frac{π z}{\sin π z} . \end{array}

Finally, dividing both sides by $z$ yields

(- z)! Γ (z) = \frac{π}{\sin π z}

as we needed to show. □

[Hint:

1 + x \leq e^{x}

for all real

x

; hence

(k + 1) ∕ k \leq e^{1 ∕ k} \leq k ∕ (k - 1)

Proposition. $\frac{n^{n}}{e^{n - 1}} \leq n! \leq \frac{n^{n + 1}}{e^{n - 1}}$ for integer $n \geq 1$ .

Proof. Let $n$ be an arbitrary integer such that $n \geq 1$ . We must show that

\frac{n^{n}}{e^{n - 1}} \leq n! \leq \frac{n^{n + 1}}{e^{n - 1}} .

Note that since $1 + x \leq e^{x}$ for all real $x$ ,

\begin{array}{l} 1 + \frac{1}{k} \leq e^{\frac{1}{k}} & \Leftrightarrow \frac{k + 1}{k} \leq e^{\frac{1}{k}} \\ \Leftrightarrow \frac{{(k + 1)}^{k}}{k^{k}} \leq e, \end{array}

and

\begin{array}{l} 1 - \frac{1}{k + 1} \leq e^{- \frac{1}{k + 1}} & \Leftrightarrow \frac{k}{k + 1} \leq e^{- \frac{1}{k + 1}} \\ \Leftrightarrow \frac{k^{k + 1}}{{(k + 1)}^{k + 1}} \leq e^{- 1} \\ \Leftrightarrow \frac{{(k + 1)}^{k + 1}}{k^{k + 1}} \geq e . \end{array}

Then

\begin{array}{l} \frac{n^{n}}{n!} & = \frac{\prod_{1 \leq k \leq n} n}{\prod_{1 \leq k \leq n} k} \\ = \frac{\prod_{1 \leq k \leq n} n k^{k - 1}}{\prod_{1 \leq k \leq n} k k^{k - 1}} \\ = \frac{\prod_{1 \leq k \leq n} n k^{k - 1}}{\prod_{1 \leq k \leq n} k^{k}} \\ = \frac{n^{n}}{n^{n}} \frac{\prod_{1 \leq k \leq n} k^{k - 1}}{\prod_{1 \leq k \leq n - 1} k^{k}} \\ = \frac{\prod_{2 \leq k \leq n} k^{k - 1}}{\prod_{1 \leq k \leq n - 1} k^{k}} \\ = \frac{\prod_{1 \leq k \leq n - 1} {(k + 1)}^{k}}{\prod_{1 \leq k \leq n - 1} k^{k}} \\ = \prod_{1 \leq k \leq n - 1} \frac{{(k + 1)}^{k}}{k^{k}} \\ \leq \prod_{1 \leq k \leq n - 1} e \\ = e^{n - 1}, \end{array}

and so

\frac{n^{n}}{e^{n - 1}} \leq n! .

Then also

\begin{array}{l} \frac{n^{n + 1}}{n!} & = \frac{\prod_{1 \leq k \leq n + 1} n}{\prod_{1 \leq k \leq n} k} \\ = \frac{\prod_{1 \leq k \leq n + 1} n k^{k}}{\prod_{1 \leq k \leq n} k k^{k}} \\ = \frac{\prod_{1 \leq k \leq n + 1} n k^{k}}{\prod_{1 \leq k \leq n} k^{k + 1}} \\ = \frac{n^{n + 1}}{n^{n + 1}} \frac{\prod_{1 \leq k \leq n} k^{k}}{\prod_{1 \leq k \leq n - 1} k^{k + 1}} \\ = \frac{\prod_{2 \leq k \leq n} k^{k}}{\prod_{1 \leq k \leq n - 1} k^{k + 1}} \\ = \frac{\prod_{1 \leq k \leq n - 1} {(k + 1)}^{k + 1}}{\prod_{1 \leq k \leq n - 1} k^{k + 1}} \\ = \prod_{1 \leq k \leq n - 1} \frac{{(k + 1)}^{k + 1}}{k^{k + 1}} \\ \geq \prod_{1 \leq k \leq n - 1} e \\ = e^{n - 1}, \end{array}

and so

n! \leq \frac{n^{n + 1}}{e^{n - 1}} .

Therefore,

\frac{n^{n}}{e^{n - 1}} \leq n! \leq \frac{n^{n + 1}}{e^{n - 1}}

as we needed to show. □

25. [M20] Do factorial powers satisfy a law analogous to the ordinary law of exponents,

x^{m + n} = x^{m} x^{n}

Factorial powers satisfy laws analogous to the ordinary law of exponents. In particular,

\begin{matrix} x^{\underset{̲}{m + n}} = x^{\underset{̲}{m}} {(x - m)}^{\underset{̲}{n}} \\ x^{\bar{m + n}} = x^{\bar{m}} {(x + m)}^{\bar{n}}, \end{matrix}

since

\begin{array}{l} x^{\underset{̲}{m + n}} & = \frac{x!}{(x - (m + n))!} \\ = \frac{x!}{(x - m - n)!} \\ = \frac{x!}{(x - m - n)!} \frac{(x - m)!}{(x - m)!} \\ = \frac{x!}{(x - m)!} \frac{(x - m)!}{(x - m - n)!} \\ = x^{\underset{̲}{m}} {(x - m)}^{\underset{̲}{n}} \end{array}

and

\begin{array}{l} x^{\bar{m + n}} = & = \frac{Γ (x + m + n)}{Γ (x)} \\ = \frac{Γ (x + m + n)}{Γ (x)} \frac{Γ (x + m)}{Γ (x + m)} \\ = \frac{Γ (x + m)}{Γ (x)} \frac{Γ (x + m + n)}{Γ (x + m)} \\ = x^{\bar{m}} {(x + m)}^{\bar{n}} . \end{array}