Analyses with Powers and Transcendentals

Problem Introduction

Which is greater, 888988^{89} or 898889^{88}? Computing with straightfoward arithmetic, we have that 8889>898888^{89} > 89^{88}. But is there another way to look at the problem?

Notice that 89=88+189 = 88 + 1. Substituting 88+188 + 1 for 8989, we get 8888+1>(88+1)8888^{88 + 1} > (88 + 1)^{88}. To generalize the question further, we replace 8888 with a natural number xx, asking ourselves “under what condition does xx+1>(x+1)xx^{x+1} > (x+1)^x hold?”

Quick Observation

Consider the set A={0,1,,9}A = \{0, 1, \dots, 9\}. Computing the left and right expressions of the inequality for each xAx \in A, we have

x\mathbf{x}xx+1\mathbf{x^{x+1}}(x+1)x\mathbf{(x+1)^x}
001
112
289
38164
41,024625
515,6257,776
6279,936117,649
75,764,8012,097,152
8134,217,72843,046,721
93,486,784,4011,000,000,000

Observe that xA\forall x \in A such that x2x \leq 2, xx+1<(x+1)xx^{x+1} < (x+1)^x. Now observe that xA\forall x \in A such that x3x \geq 3, xx+1>(x+1)xx^{x+1} > (x+1)^x. Next, we perform similar observations for the set of all nonnegative real numbers.

Analysis with e

We prove that xx+1>(x+1)xx^{x+1} > (x+1)^x for any real number x3x \geq 3 by finding the stricter condition that xx+1>(x+1)xx^{x+1} > (x+1)^x for any real number x>ex > e.

Suppose xR>0x \in \mathbb{R_{> 0}}. Dividing xxx^x on both sides of the inequality, we get

xx+1xx>(x+1)xxxx>(x+1x)xx>(1+1x)x\begin{align*} \frac{x^{x+1}}{x^x} &> \frac{(x+1)^x}{x^x} \\ x &> \left(\frac{x+1}{x}\right)^x \\ x &> \left(1 + \frac{1}{x}\right)^x \end{align*}

Let h:R>0R>0h : \mathbb{R_{> 0}} \rightarrow \mathbb{R_{> 0}} be given by h(x)=(1+1x)xh(x) = (1 + \frac{1}{x})^x. Note that limxh(x)=e\lim_{x\to\infty} h(x) = e. To prove that ee is the least upper bound of hh, we show that hh always increases when x>0x > 0. We use the following method borrowed from an online thread:1

It is sufficient to show that ln(h(x))=ln(1+1x)xln(h(x)) = ln(1 + \frac{1}{x})^x is increasing. Let y=1xy = \frac{1}{x}. Substituting in yy, the expression can be rewritten as

ln(1+y)ln1y\begin{align} \frac{ln(1+y) - ln1}{y} \end{align}

Now we show that as yy decreases, (1) increases. Notice that hh is strictly positive when x>0x > 0, so ln(h(x))ln(h(x)) is concave for x>0x > 0. Also notice that (1) is the slope of the line through (1,ln1)(1, ln1) and (1+y,ln(1+y))(1+y, ln(1+y)). Since any concave function has the property that the slope of the line (1) increases as yy decreases, hh always increases for x>0x > 0.

Since hh converges to ee and always increases for x>0x > 0, we have that ee is the least upper bound of hh. Since ee is the least upper bound of hh, then for any xRx \in \mathbb{R} such that xex \geq e, we have that x(1+1x)xx \geq (1 + \frac{1}{x})^x. Since this inequality is equivalent to xx+1>(x+1)xx^{x+1} > (x+1)^x, we get

xe,xx+1>(x+1)x\forall x \geq e, x^{x+1} > (x+1)^x

\square

Analysis with Newton’s Method

Let f(x)=xx+1f(x) = x^{x+1} and g(x)=(x+1)xg(x) = (x+1)^x. Note that xx+1x^{x+1} and (x+1)x(x+1)^x are polynomials, so ff and gg are continuous. We prove a stronger result by finding the only point of intersection of ff and gg in the nonnegative reals.

Consider the interval (0, e). From the sections above, we know that the inequality between xx+1x^{x+1} and (x+1)x(x+1)^x flips when going from x=0x = 0 to x=ex = e. Since ff and gg are continuous, then with the Intermediate Value Theorem (IVT), we have that for some c(0,e)c \in (0, e), f(c)=g(c)f(c) = g(c).

To find cc, we evaluate the expression f(x)=g(x)f(x) = g(x), which is given by xx+1=(x+1)xx^{x+1} = (x+1)^x. Unfortunately, there seems to be no straightforward way to apply the Lambert W function.2 Instead, we consider the interval Newton’s method, a modification of Newton’s method to find all intersections in a given interval.3 Applying this method on (0, e), we approximate a single intersection c=2.293166c = 2.293166\dots, which (as an aside) is curiously a transcendental number.4

Now we show that x>c,xx+1>(x+1)x\forall x > c, x^{x+1} > (x+1)^x:

Suppose x>c,xx+1<=(x+1)x\exists x > c, x^{x+1} <= (x+1)^x to reach a contradiction.

Consider the case xx+1=(x+1)xx^{x+1} = (x+1)^x. Since xx+1>(x+1)xx^{x+1} > (x+1)^x for x>ex > e, then xx must be an intersection in the interval (c,e)(c, e). But from interval Newton’s method, we know there cannot be another intersection in (c,e)(c, e). \Rightarrow\Leftarrow

Consider the case xx+1<(x+1)xx^{x+1} < (x+1)^x. Since xx+1>(x+1)xx^{x+1} > (x+1)^x for x>ex > e, then xx must be an intersection in the interval (c,e)(c, e). From interval Newton’s method, we know there cannot be another intersection in (c,e)(c, e). But by IVT, there must be another intersection in (c,e)(c, e) since f(x)<g(x)f(x) < g(x) and f(x)>g(x)f(x') > g(x') for some x>ex' > e. \Rightarrow\Leftarrow

So by contradiction,

x>c=2.2931663,xx+1>(x+1)x\forall x > c=2.2931663\dots, x^{x+1} > (x+1)^x

Moreover, xx+1<(x+1)xx^{x+1} < (x+1)^x for 0x<c0 \leq x < c can be proved via similar “proof by cases” as above, which will be left as an exercise.

\square

Footnotes

  1. zhw. Answer to “How to prove (1 + 1/x)^x is increasing when x > 0?” https://math.stackexchange.com/questions/83035/how-to-prove-11-xx-is-increasing-when-x0

  2. Miquilino, Alexandre Ribeiro. Answer to “How do you solve x^(x+1) = (x+1)^x?” https://www.quora.com/How-do-you-solve-x-x-1-x-1-x

  3. Seidl, Daniel. “Formalisation of Interval Methods for Nonlinear Root-Finding” https://www21.in.tum.de/students/past/interval_newton/assets/newton.pdf

  4. Lord, Nick. “86.16 Two Other Transcendental Numbers Obtained by (Mis)Calculating e.” The Mathematical Gazette, vol. 86, no. 505, 2002, pp. 103–05. https://doi.org/10.2307/3621590