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.

Blue = g_2, red = g_3, yellow = g_5

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.

Oscillations are more rapid when lambda is smaller

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.

The waves get smaller but then start growing again, moving to the left

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.