REMARKS ON A FACTORIZATION OF X^{n} − Y^{n}

Richard E. Bayne, James E. Joseph, Myung H. Kwack, and Thomas H. Lawson

Introduction. Early on, elementary algebra students learn to factor expressions x^{2} − y^{2}, x^{3} − y^{3}, …, where x and y represent real numbers. Later, they learn the generalization

x^{n} − y^{n} = (x − y) ∑_{m=0}^{n−1} x^{m}y^{n−1−m}  (∗)

for each positive integer n and all real x and y. This identity has been shown to have various applications throughout the undergraduate mathematics curriculum and beyond. For example, Johnsonbaugh, using the inequality

\frac{b^{n+1} − a^{n+1}}{b − a} < (n + 1)b^{n}

for all positive integers n and real a and b with 0 ≤ a < b, an easy consequence of (∗), published an old and relatively simple proof of the monotonicity and boundedness of the sequence {(1 + 1/n)^{n}} in [2]. Evidently, the proof was discovered by Fort in 1864 (see [3]). In [1], there is a nice proof of the existence of nth roots, which is a good deal simpler than other proofs using the Fundamental Axiom of the Reals (see [3]) and proofs using the Intermediate Value Theorem (see [5]). The following characterization of C_{n,k} is an interesting by–product obtained by relating (∗) to the Binomial Theorem: If n and k are positive integers with k ≤ n, then

C_{n,k} = ∑_{m=k−1}^{n−1} C_{m,k−1}.

We obtain this relationship by combining (∗) and the Binomial Theorem to get

(x + 1)^{n} = 1 + x ∑_{m=0}^{n−1}(x + 1)^{m} = 1 + ∑_{m=0}^{n−1} ∑_{k=0}^{m} C_{m,k} x^{k+1}

for each real x and then equating coefficients of x^{m}.

The purpose of this note is to present other interesting applications of (∗) and to begin an investigation of a functional inequality, which we discovered while studying (∗). In Section 1, we show how inequalities, which are typically found in elementary analysis courses, as well as some which seem to be absent from the literature, flow easily from (∗) and how those in the second category may be used to advantage in elementary courses. In [1], identity (∗) is used to show that for each positive integer n and all reals x and y,

|x^{n} − y^{n}| ≤ (|x − y| + |y|)^{n} − |y|^{n}.  (1)

Another known inequality, which is easily deduced from (∗), is

(n^{1/n} − 1)^{2} ≤ 2/n

for each positive integer n (see [4]). We show in this article that it is possible to prove a stronger inequality, although we are unable to see how (∗) can be used to verify this inequality. We prove that

(n^{1/n} − 1)^{2} ≤ 1/n  (2)

for each positive integer n.

Following a suggestion of W. Rudin (private communication), we give a shorter proof than that given for (2) that

(x^{1/x} − 1)^{2} ≤ 1/x  (3)

for each positive real x.

In Section 2, we study real–valued functions f satisfying the functional inequality

|f(x) − f(y)| ≤ f(|x − y| + |y|) − f(|y|)  (4)

for all x and y in D(f), where D(f) is the domain of f. This study is motivated by the observation, from (1), that the function f, defined by f(x) = x^{n}, satisfies (4). We see also that the exponential function satisfies (4).

| exp(x) − exp(y)|

= \{ \begin{matrix}exp(y)(exp(x − y) − 1) ≤ exp(|x − y| + |y|) − exp(|y|), & if x ≥ y \\ exp(x)(exp(y − x) − 1) ≤ exp(|x − y| + |y|) − exp(|y|), & if x < y. \end{matrix}

We denote the class of functions satisfying (4) by Ω. It is obvious that constant functions are elements of Ω and fairly obvious that any f ∈ Ω is nondecreasing on the set of nonnegative elements of D(f). We show that any f ∈ Ω is continuous and is convex on [c, ∞) if c ≥ 0 and [c, ∞) ⊂ D(f). That is,

f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y)

for all x, y ∈ [c, ∞) and 0 ≤ λ ≤ 1. In addition, we show that if f, g ∈ Ω and λ ≥ 0, then f + g ∈ Ω and λf ∈ Ω. Moreover, if f, g ∈ Ω, f(0) ≥ 0, and g(0) ≥ 0, then fg ∈ Ω.

1. Some Applications. As an application of (1), for any fixed y and ϵ > 0, each solution to (|x − y| + |y|)^{n} − |y|^{n} < ϵ is a solution to |x^{n} − y^{n}| < ϵ. Since (|x − y| + |y|)^{n} − |y|^{n} < ϵ is equivalent to |x − y| < (ϵ + |y|^{n})^{1/n} − |y|,

if ϵ > 0 and 0 < δ ≤ (ϵ + |y|^{n})^{1/n} − |y|, then |x^{n} − y^{n}| < ϵ, when |x − y| < δ. (A_{1})

It should be readily obtainable for the reader that if P (x) = ∑_{m=0}^{n} a_{n−m}x^{n−m} is a polynomial function of degree n in x, then

for any ϵ > 0 and any fixed y, |P (x) − P (y)| < ϵ, if |x − y| < δ,  (A_{2})

where

0 < δ ≤ min \{(\frac{ϵ}{nK} + |y|^{m})^{1/m} − |y| : m = 1, …, n\}

and

K = max{|a_{n−m}| : m = 0, …, n − 1}.

The identity (∗), along with the statement (∗∗), offered below without proof, leads to another useful inequality (5).

Let x and y be real and let n be a positive integer such that xy > 0 (∗∗)
and x^{1/n} is real. Then x^{k/n}y^{(n−1−k)/n} > 0 for each integer k.

Let x and y be real and let n be a positive integer such that xy > 0 and x^{1/n} is real. Then the following inequality holds.

|x^{1/n} − y^{1/n}| ≤ \frac{|x − y|}{y^{(n−1)/n}} .  (5)

The verification of (5) comes from (∗) and (∗∗) as follows.

\begin{aligned}|x − y| & = |x^{1/n} − y^{1/n}| ∑_{m=0}^{n−1} x^{m/n}y^{(n−1−m)/n} \\ & ≥ |x^{1/n} − y^{1/n}|y^{(n−1)/n}. \end{aligned}

Utilizing (5) and (∗), we may establish that for any real x and y and any integer n for which y^{1/n} is real and any ϵ > 0,

any solution to |x−y| < min{ϵy^{(n−1)/n}, |y|} is a solution to |x^{1/n}−y^{1/n}| < ϵ. (A_{3})

The exercises below may be used to obtain more experience with applying (∗) to arrive at other elementary inequalities.

\underline{Exercise 1}. If p ≥ 1, show that p^{n} ≥ 1 + n(p − 1) for each positive integer n. \underline{Exercise 2}. If p ≥ 1, show that p^{n} − 1 ≥ (p − 1)^{2}n(n − 1)/2 for each positive

integer n.

\underline{Solution}. From (∗) and the result of Exercise 1, we obtain

p^{n} − 1 ≥ (p − 1) ∑_{m=0}^{n−1} p^{m} ≥ (p − 1)^{2} ∑_{m=1}^{n−1} m = (p − 1)^{2}n(n − 1)/2.

\underline{Exercise 3}. If p > 1, show that n/p^{n} < 2p/n(p − 1)^{2} for each positive integer n.

\underline{Solution}. From the result of Exercise 2, for each positive integer n,

p^{n+1} ≥ (p − 1)^{2}n(n + 1)/2 > (p − 1)^{2}n^{2}/2.

\underline{Exercise 4}. Show that (n^{1/n} − 1)^{2} ≤ 2/n for each positive integer n. \underline{Solution}. From the result of Exercise 2, for n > 1,

n − 1 = (n^{1/n})^{n} − 1 ≥ (n^{1/n} − 1)^{2}n(n − 1)/2.

\underline{Exercise 5}. If x > 0, show that |x^{1/n} − 1| ≤ max\{1, 1/x\}|x − 1|/n for each positive integer n.

\underline{Solution}. For x ≥ 1, we arrive at |x^{1/n} − 1| ≤ |x − 1|/n by applying the result of Exercise 1 with p = x^{1/n}. For x < 1, we apply the result of Exercise 1 with p = (1/x)^{n} to conclude that |x^{1/n} − 1| ≤ |x − 1|/(nx). Hence, for all x > 0,

|x^{1/n} − 1| ≤ max{|x − 1|/(nx), |x − 1|/n} = max{1, 1/x}|x − 1|/n.

Although we are unable to see how (∗) could be used to verify (2), we establish (2) below using the equivalent form (6) for (2):

n ≤ (1 + \frac{1}{\sqrt{n}})^{n}  (6)

for each positive integer n. If x ≥ 1, there is a positive integer m such that m ≤ x < m + 1. For such an m we have

(3/2)^{m} ≤ (1 + \frac{1}{m + 1})^{m^{2}} < (1 + \frac{1}{x})^{x^{2}},

since

(1 + \frac{1}{m + 1})^{m}

is strictly increasing, as can be seen easily by analyzing the expression obtained by multiplying and dividing

(1 + \frac{1}{m + 1})^{m} by (1 + \frac{1}{m + 1}).

We have by induction that (3/2)^{m} ≥ (m + 1)^{2} for all m ≥ 14. We note that 14 is the smallest positive integer satisfying this property. We conclude that

x^{2} < (1 + \frac{1}{x})^{x^{2}}

for all x ≥ 14. We find that (6) holds for every positive integer m ≥ 196. We have, by direct computation, that each of the integers 1, 2, …, 195 satisfies (6). (The difference (1 + 1/\sqrt{n})^{n} − n strictly increases from 1 at n = 1 to approximately 719727 at n = 195.)

We should point out that we have exhibited

x < (1 + \frac{1}{\sqrt{x}})^{x}

for all real x ≥ 196.

Following a suggestion of Rudin, we note that if f is the function defined by f(x) = ln(1 + x) − x + x^{2}/2, then

(a) f is increasing on [0, ∞) and

(b) f(0) = 0 (observe that x − x^{2}/2 is the sum of the second and third terms of the Taylor expansion of ln(1 + x) about 0). It follows that

x ln(1 + \frac{1}{\sqrt{x}}) − ln x > \sqrt{x} − \frac{1}{2} − ln x.

Since the minimum value of the expression on the right hand side of the last inequality is 3/2 − ln 4 which is positive, (3) holds.

2. Some Properties of Ω. The first item in this section is a verification that functions in Ω are continuous. This will be accomplished by establishing (1^{∘})

– (3^{∘}) in succession: (1^{∘}) If f ∈ Ω, then |y| + n|x − y| ∈ D(f) for all x, y ∈ D(f) and n ≥ 0. (2^{∘}) If f ∈ Ω and x_{0}, x_{1}, … is an increasing sequence of equally spaced nonnegative reals in D(f), i.e., {x_{n} − x_{n−1}} is a constant sequence, then

(a) f(x_{n}) − f(x_{n−1}) ≥ f(x_{1}) − f(x_{0}) for all n.

(b) f(x_{n}) − f(x_{0}) ≥ n(f(x_{1}) − f(x_{0})) for all n.

(3^{∘}) If f ∈ Ω, then f is continuous.

To show that (1^{∘}) and (2^{∘}) are valid, we use induction on n. As for (1^{∘}), let x, y ∈ D(f). Since f ∈ Ω, |y|, |y| + |x − y| ∈ D(f). If |y| + m|x − y| ∈ D(f) for each nonnegative m ≤ n, then |y| + (n + 1)|x − y| ∈ D(f), since

|y| + (n + 1)|x − y| = ||y| + n|x − y|| + ||y| + n|x − y| − (|y| + (n − 1)|x − y|)|.

Parts (a) and (b) of (2^{∘}) clearly hold for n = 1. Suppose n is an integer for which (a) is true, i.e., for which f(x_{n}) − f(x_{n−1}) ≥ f(x_{1}) − f(x_{0}). Then

f(x_{n+1})−f(x_{n}) = f(x_{n} +|x_{n} −x_{n−1}|)−f(x_{n}) ≥ f(x_{n})−f(x_{n−1}) ≥ f(x_{1})−f(x_{0}).

The proof of (a) is complete. If (b) holds for n, then

f(x_{n+1}) − f(x_{0}) = (f(x_{n+1}) − f(x_{n})) + (f(x_{n}) − f(x_{0})) ≥ (n + 1)(f(x_{1}) − f(x_{0}))

from part (a). This completes the demonstration that (1^{∘}) and (2^{∘}) are valid. Now for (3^{∘}), suppose that f is not constant and that y is a non–isolated point of D(f). We will show that f is continuous at y. We show first that for each ϵ > 0, there is a z ∈ D(f) satisfying z > |y| and f(z) < f(|y|) + ϵ. Assume ϵ > 0 and there is no such z. Since f is not constant, there is a v > |y| such that f(v) > f(|y|). Choose a positive integer m such that mϵ > f(v) − f(|y|) and an x ∈ D(f) such that x ≠ y and m|x − y| < v − |y|. Then |y| + m|x − y| ∈ D(f) from (1^{∘}) and (2^{∘}) yields

f(v) − f(|y|) ≥ f(|y| + m|x − y|) − f(|y|) ≥ m(f(|y| + |x − y|) − f(|y|)),

so mϵ > f(v) − f(|y|) ≥ mϵ, a contradiction. Now, if ϵ > 0, choose z ∈ D(f) such that z > |y|, f(z) < f(|y|)+ϵ and let δ = z −|y|. Then, if x ∈ D(f) and |x−y| < δ, it follows that

|f(x) − f(y)| ≤ f(|x − y| + |y|) − f(|y|) ≤ f(δ + |y|) − f(|y|) = f(z) − f(|y|) < ϵ.

Hence, f is continuous at y.

It is fairly obvious, from the arguments made to establish (3^{∘}) that, if f is strictly increasing and [f(0), ∞) is a subset of the range of f, for ϵ > 0, we may take 0 < δ ≤ f^{−1}[f(|y|) + ϵ] − |y|; in particular, for any ϵ > 0 and any fixed y, we have

| exp(x) − exp(y)| < ϵ, if |x − y| < ln[exp(|y|) + ϵ] − |y|.

Interestingly, any continuous function f satisfying the condition in (a) under 2^{∘} for any increasing sequence x_{0}, x_{1}, … of equally spaced nonnegative reals in D(f) also satisfies (4^{∘}) If w, x, y, and z are nonnegative reals satisfying w < x < z, x − w = z − y, and [w, x] ⊂ D(f). Then f(z) − f(y) ≥ f(x) − f(w).

We show first that this assertion is true when w, x, y, and z are rationals. To this end, let the function f and rationals w, x, y, and z satisfy the hypothesis of (4^{∘}) and let m be a positive integer such that mx, my, and mz are integers; let P be a partition of [w, z] into intervals of length 1/m. If P = {w + (j − 1)/m : j = 1, 2, …, m(z − w) + 1}, then for each integer 1 ≤ k ≤ m(x − w), the sequence in P (in order of magnitude) beginning at w + (k − 1)/m and terminating at y + k/m is an increasing sequence of equally spaced terms in D(f);

f(y + k/m) − f(y + (k − 1)/m) ≥ f(w + k/m) − f(w + (k − 1)/m),

since f satisfies part (a) of (2^{∘}). Thus,

∑_{k=1}^{m(x−w)}(f(y + k/m) − f(y + (k − 1)m)) ≥ ∑_{k=1}^{m(x−w)}(f(w + k/m) − f(w + (k − 1)/m)).

Since x−w = z−y, we get f(z)−f(y) ≥ f(x)−f(w). Now, let the real numbers w, x, y, z and the function f satisfy the hypothesis of (4^{∘}) and let {a_{n}}, {b_{n}}, and {c_{n}} be sequences of nonnegative rationals with w ≤ a_{n} < b_{n} ≤ x, y ≤ c_{n} ≤ y + (a_{n} − w), a_{n} → w, b_{n} → x. Let d_{n} = c_{n} + b_{n} − a_{n}. Then a_{n} < b_{n} < d_{n}, b_{n} − a_{n} = d_{n} − c_{n}, [a_{n}, d_{n}] ⊂ D(f), so a_{n}, b_{n}, c_{n}, d_{n} and f satisfy the hypothesis of (4^{∘}) and f(d_{n}) − f(c_{n}) ≥ f(b_{n}) − f(a_{n}). By continuity, we have f(z) − f(y) ≥ f(x) − f(w).

From part (b) of (2^{∘}), we observe that if f ∈ Ω, a < b and a, (a+b)/2, b ∈ D(f), then f((a + b)/2) ≤ (f(a) + f(b))/2. This is equivalent to f being convex on [c, ∞), if c ≥ 0 and [c, ∞) ⊂ D(f) (see [5]).

It is not difficult to show that f + g, λf ∈ Ω, whenever f, g ∈ Ω and λ ≥ 0. We will now establish that if f, g ∈ Ω, f(0) ≥ 0, and g(0) ≥ 0, then the product fg ∈ Ω. We observe first that f(0) ≥ 0 if and only if |f(x)| ≤ f(|x|) for each x ∈ D(f). Now if x, y ∈ D(fg), then

|fg(x) − fg(y)| ≤ |f(x)||g(x) − g(y)| + |g(y)||f(x) − f(y)| ≤ f(|x|)(g(|x − y| + |y|)) − g(|y|)) + g(|y|)(f(|x − y| + |y|) − f(|y|)) ≤ (f(|x − y| + |y|)(g(|x − y| + |y|) − g(|y|)) + g(|y|)(f(|x − y| + |y|) − f(|y|)) ≤ fg(|x − y| + |y|) − fg(|y|).

We close with two examples and an observation for complex–valued functions of a complex variable.

\underline{Example 1}. The function f defined by

f(x) = \{ \begin{matrix}x^{2}, & if x ≤ 0 \\ x^{3}, & if x ≥ 0 \end{matrix}

is differentiable, increasing, and convex on [0, ∞); however, f ∉ Ω, since if x = −1/2 and y = −1/4, we have |f(x) − f(y)| = 12/64 and f(|x − y| + |y|) − f(|y|) = 7/64.

\underline{Example 2}. If f and g are defined by f(x) = x^{3} − 1 and g(x) = x^{2}, then f, g ∈ Ω, while fg ∉ Ω.

We point out that (∗) and (1) are also valid if x and y are complex numbers and that any complex–valued function of a complex variable satisfying (4) is continuous.

We would like to thank Sterling Berberian for reading an earlier version of this paper and making several useful comments and suggestions.

\underline{References}

1. R. E. Bayne, J. E. Joseph, M. H. Kwack, and T. H. Lawson, “Exploiting a Factorization of X^{n} − Y^{n},” The College Mathematics Journal, 28 (1997) 206– 209.

2. R. F. Johnsonbaugh, “Another Proof of an Estimate for e,” American Mathematical Monthly, 81 (1974), 1011–1012.

3. R. F. Johnsonbaugh, Foundations of Mathematical Analysis, Marcel Dekker, Inc., New York, 1981.

4. P. P. Korovkin, Inequalities, Pergamon Press, London, 1961.

5. M. Spivak, Calculus, 2nd ed., Publish or Perish, Inc., Houston, Texas, 1980.

Richard E. Bayne James E. Joseph Myung H. Kwack Department of Mathematics Howard University Washington, D.C. 20059 email: bayne@scs.howard.edu

Thomas H. Lawson Center for Academic Reinforcement Howard University Washington, D.C. 20059