# Functions with horizontally scaled derivative

Exponential functions ${f_\lambda(x)=e^{\lambda x}}$ are characterized by the property that the graph of the derivative ${f_\lambda'}$ is the graph of ${f_\lambda}$ scaled by the factor of ${\lambda}$ in the vertical direction: ${f_\lambda'(x) = \lambda f_\lambda(x)}$. We also have ${f_\lambda(0)=1}$ for the sake of normalization.

What happens if we replace “vertical” by “horizontal”? That is, let ${g_\lambda}$ be the function such that ${g_\lambda'(x) = g_\lambda(x/\lambda)}$, still with ${g_\lambda(0)=1}$ for normalization. Clearly, ${g_1(x)=e^x=f_1(x)}$. Let’s consider ${\lambda > 1}$ from now on. Assuming we can express ${g_\lambda}$ as a power series ${g_\lambda(x) = \sum\limits_{n=0}^\infty c_n x^n}$, the comparison of ${g'(x) =\sum\limits_{n=0}^\infty c_{n+1} (n+1) x^n }$ and ${g(x/\lambda) =\sum\limits_{n=0}^\infty c_{n} \lambda^{-n}x^n }$ yields the recurrence relation ${c_{n+1} = \lambda^{-n} c_n/(n+1)}$. This relation leads to an explicit formula: ${c_n = \lambda^{-\binom{n}{2}}/n!}$ where ${\binom{n}{2}=n(n-1)/2}$. That is,

${\displaystyle g_\lambda(x) = \sum_{n=0}^\infty \lambda^{-\binom{n}{2}} \frac{x^n}{n!}}$

The behavior of ${g_2, g_3}$, and ${g_5}$ in the vicinity of ${0}$ is shown below.

A few questions come up. Is ${g_\lambda}$ strictly increasing? Does it have a finite limit at ${-\infty}$? If so, is this limit negative? If so, where does ${g_\lambda}$ cross the ${x}$-axis? What is its rate of growth as ${x\to \infty}$? Does the number ${g_\lambda(1)}$ have any significance, considering that ${g_1(1) = e}$?

First of all: if ${g_\lambda}$ does become negative at some point then its derivative also becomes negative further to the left, which can make ${g_\lambda}$ rise to positive values again, and then the process will probably repeat as shown below. Also, ${g_\lambda}$ is negative at each point of local minimum, and positive at each point of local maximum. This is because both the function and its derivative are positive at ${0}$, and moving to the left, the derivative cannot change the sign until the function itself changes the sign.

When ${\lambda}$ is close to ${1}$, the oscillating pattern takes longer to develop: here it is with ${\lambda=1.1}$. Note the vertical scale: these are very small oscillations, which is why this plot does not extend to the zero mark.

For any ${\lambda > 1}$ the alternating series estimate gives ${g_\lambda (x)>1+x}$ when ${x\in [-1, 0]}$, hence ${g_\lambda(x)>0}$ for ${x\ge -1}$. It follows that ${g_\lambda}$ is strictly increasing when ${x\ge -\lambda}$. We have

${\displaystyle g_\lambda(-\lambda) = \sum_{n=0}^\infty (-1)^n \lambda^{n-\binom{n}{2}}/n! = \frac{5}{6} - \frac{\lambda}{2} + \sum_{n=4}^\infty (-1)^n \lambda^{n-\binom{n}{2}}/n! }$

Since ${n-\binom{n}{2}}$ is decreasing for ${n\ge 4}$, the alternating series estimate applies and shows that ${\displaystyle g_\lambda(-\lambda) < \frac{5}{6} - \frac{\lambda}{2} + \frac{\lambda^{-2}}{24}}$

So, when ${\lambda \ge 2}$ we have ${g_{\lambda}(-\lambda) < 0}$ and therefore there is a unique point ${r\in (-\lambda, -1)}$ where ${g_\lambda(r)=0}$. Specifically for ${g_2}$, this root is ${r \approx -1.488}$. Its significance is that ${\lambda r}$ is the critical point closest to ${0}$, meaning that ${[\lambda r, \infty)}$ is the largest interval of monotonicity of ${g_\lambda}$.

In general it is not true that ${g_\lambda(-\lambda) < 0}$. Indeed, ${g_\lambda(x)\to e^x}$ uniformly on bounded sets as ${\lambda \to 1}$. I do not have a proof that ${g_\lambda}$ has a real root for every ${\lambda > 1}$.

When ${\lambda=2}$, the sequence of denominators ${2^{\binom{n}{2}}n!}$ is A011266 in OEIS (it is related to counting the evil and odious numbers). But the sum of its reciprocals, ${g_2(1) \approx 2.2714925555}$, did not show up anywhere.

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