# Completely monotone imitation of 1/x

I wanted an example of a function ${f}$ that behaves mostly like ${1/x}$ (the product ${xf(x)}$ is bounded between two positive constants), but such that ${xf(x)}$ does not have a limit as ${x\rightarrow 0}$.

The first thing that comes to mind is ${(2+\sin(1/x))/x}$, but this function does not look very much like ${1/x}$.

Then I tried ${f(x)=(2+\sin\log x)/x}$, recalling an example from Linear Approximation and Differentiability. It worked well:

In fact, it worked much better than I expected. Not only if ${f'}$ of constant sign, but so are ${f''}$ and ${f'''}$. Indeed, $\displaystyle f'(x) = \frac{\cos \ln x - \sin \log x - 2}{x^2}$

is always negative, $\displaystyle f''(x) = \frac{4 -3\cos \log x + \sin \log x}{x^3}$

is always positive, $\displaystyle f'''(x) = \frac{10\cos \log x -12}{x^4}$

is always negative. The sign becomes less obvious with the fourth derivative, $\displaystyle f^{(4)}(x) = \frac{48-40\cos\log x - 10\sin \cos \ln x}{x^5}$

because the triangle inequality isn’t conclusive now. But the amplitude of ${A\cos t+B\sin t}$ is ${\sqrt{A^2+B^2}}$, and ${\sqrt{40^2+10^2}<48}$.

So, it seems that ${f}$ is completely monotone, meaning that ${(-1)^n f^{(n)}(x)\ge 0}$ for all ${x>0}$ and for all ${n=0,1,2,\dots}$. But we already saw that this sign pattern can break after many steps. So let’s check carefully.

Direct calculation yields the neat identity $\displaystyle \left(\frac{1+a\cos \log x+b\sin\log x}{x^n}\right)' = -n\,\frac{1+(a-b/n)\cos\log x+(b+a/n) \sin\log x}{x^{n+1}}$

With its help, the process of differentiating the function ${f(x) = (1+a\cos \log x+b\sin\log x)/x}$ can be encoded as follows: ${a_1=a}$, ${b_1=b}$, then ${a_{n+1}=a_n-b_n/n}$ and ${b_{n+1} = b_n+a_n/n}$. The presence of ${1/n}$ is disconcerting because the harmonic series diverges. But orthogonality helps: the added vector ${(-b_n/n, a_n/n)}$ is orthogonal to ${(a_n,b_n)}$.

The above example, rewritten as ${f(x)=(1+\frac12\sin\log x)/x}$, corresponds to starting with ${(a,b) = (0,1/2)}$. I calculated and plotted ${10000}$ iterations: the points ${(a_n,b_n)}$ are joined by piecewise linear curve.

The total length of this curve is infinite, since the harmonic series diverges. The question is, does it stay within the unit disk? Let’s find out. By the above recursion, $\displaystyle a_{n+1}^2 + b_{n+1}^2 = \left(1+\frac{1}{n^2}\right) (a_n^2+b_n^2)$

Hence, the squared magnitude of ${(a_n,b_n)}$ will always be less than $\displaystyle \frac14 \prod_{n=1}^\infty \left(1+\frac{1}{n^2}\right)$

with ${1/4}$ being ${a^2+b^2}$. The infinite product evaluates to ${\frac{\sinh \pi}{\pi}a\approx 3.7}$ (explained here), and thus the polygonal spiral stays within the disk of radius ${\frac12 \sqrt{\frac{\sinh \pi}{\pi}}\approx 0.96}$. In conclusion, $\displaystyle (-1)^{n} \left(\frac{1+(1/2)\sin\log x}{x}\right)^{(n)} = n!\,\frac{1+a_{n+1}\cos\log x+b_{n+1} \sin\log x}{x^{n+1}}$

where the trigonometric function ${a_{n+1}\cos\log x+b_{n+1} \sin\log x}$ has amplitude strictly less than ${1}$. Since the expression on the right is positive, ${f}$ is completely monotone.

The plot was generated in Sage using the code below.

a,b,c,d = var('a b c d')
a = 0
b = 1/2
l = [(a,b)]
for k in range(1,10000):
c = a-b/k
d = b+a/k
l.append((c,d))
a = c
b = d
show(line(l),aspect_ratio=1)

## 2 thoughts on “Completely monotone imitation of 1/x”

1. What was the motivation for finding such a function? Interesting stuff!

1. L says:

A Math.SE question

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