# An interesting recurrent equality, possibly easier to solve in its differential form?

I encountered an interesting inequality that I'm not sure how to approach. Here $$c$$ is a positive constant. $$f(n+1) - f(n) = c f(n)\sum_{m=0}^n f(m)$$

I am not familiar with techniques to solve recursive equations, so I thought about a similar differential equation of it $$\frac{\partial f}{\partial x} = cf(x)\int_{z=0}^x f(z)dz.$$

I am quite new to the topic of differential equations though, so I might be missing some obvious techniques to solve this kind of equality. With my limited knowledge I tried applying Laplace transform or the Fourier transform to this equation to no avail, but the expression looks nice enough that I suspect there is an analytical solution to it. Any help to getting an analytical expression of this equation is much appreciated!

(This is coming from a real-world problem, so for now, any "nice" assumption on $$f$$ and the initial conditions and so on can be placed)

• An inequality does not define a sequence, so you can't "solve" it.
– dxiv
May 20 at 22:40
• @dxiv Do you mean I am missing initial conditions? Or assumptions on $f$? This is coming from an engineering problem so any nice assumption on $f$ to make the math go through is fine.
– Nen
May 20 at 22:43
• What I meant is that the inequality does not define the sequence univocally, so it is pointless to try to "solve" it. Take for example a trivial inequality $\,f_{n+1} - f_n \ge 1\,$, then either of $\,f_n = n\,$, $\,f_n = n^2\,$ or $\,f_n = 2^n\,$ satisfies it. There is no unique solution, so you can't "solve" it.
– dxiv
May 20 at 22:49
• @dxiv Ok, but I'm only interested in the expression of $f_n$ that grows the slowest, is that impossible to obtain in the current context?
– Nen
May 20 at 22:53
• Replace the $\,\ge\,$ with an $\,=\,$, if that's what you meant to ask. Then the question becomes a legitimate one, though it's unilkely that a closed form solution exists.
– dxiv
May 20 at 22:58

Your question comes in two variants. One is discrete. The other continuous.

The discrete variant asks about the equation $$f(n+1) - f(n) = c f(n)\sum_{m=0}^n f(m). \tag{1}$$ A nicer version of this is to define $$a_n := c f(n), \qquad b_n := 1 + \sum_{m=0}^n a_m. \tag{2}$$ Multiply equation $$(1)$$ by $$\,c\,$$ to get $$a_{n+1} = a_n + a_n\sum_{m=0}^n a_m = a_n b_n = a_0\prod_{m=0}^n b_m \tag{3}$$ where the initial value $$\,a_0\,$$ is arbitrary. A simple example is if $$\,a_0=1.\,$$ This is OEIS sequence A001697. The growth rate is given as $$\,a_n \sim c^{2^n}\,$$ where $$\,c \approx 1.335245.\,$$ This is the typical growth rate with a different constant for each choice of $$\,a_0.\,$$ I don't think there is a known closed form.

The continuous variant asks about the related differential equation $$\frac{\partial f}{\partial x} = cf(x)\int_{z=0}^x f(z)dz. \tag{4}$$ This is more interesting. The equation is equivalent to $$\frac{\partial g}{\partial x} = c\int_{z=0}^x e^{g(z)}dz \;\; \text{ where } \;\; g(x) := \log(f(x)).$$ Assuming $$\,g(x)\,$$ has a power series, let $$g(x) = a_0 + a_2\frac{x^2}{2!} + a_4\frac{x^4}{4!} + \cdots$$ where $$\,c = a_2/e^{a_0}.\,$$ For simplicity assume that $$\,a_0=0, a_2=1.\,$$ The solution of equation $$(4)$$ is $$f(x) = \sec\left(\frac{x}{\sqrt{2}}\right)^2 = 1\frac{x^2}{2!} +4\frac{x^4}{4!} +34\frac{x^6}{6!} + \cdots$$ where the coefficients are OEIS sequence A002105.

• very clever approach ! May 21 at 23:25
• +1 Shouldn't equation (2) be $a_n:=cf(n)$? May 22 at 18:38
• @EliBartlett Oops! Thanks for that correction. May 22 at 19:45

Being a combination of product and sum, it looks difficult that there might be a closed form.

The best I can suggest is to make the substitution

$$f(n) = 2^{g(n)}$$ I am using $$2$$ as a base because $$2^n$$ has a simple difference.

Then \eqalign{ & f(n + 1) - f(n) = 2^{g(n + 1)} - 2^{g(n)} = 2^{g(n)} \left( {2^{g(n + 1) - g(n)} - 1} \right) \cr & 2^{g(n)} \left( {2^{g(n + 1) - g(n)} - 1} \right) = c2^{g(n)} \sum\limits_{k = 0}^n {2^{g(k)} } \cr & 2^{g(n + 1) - g(n)} = 1 + c\sum\limits_{k = 0}^n {2^{g(k)} } \cr & 2^{g(n + 2) - g(n + 1)} - 2^{g(n + 1) - g(n)} = c2^{g(n + 1)} \cr & 2^{g(n + 2) - 2g(n + 1)} - 2^{ - g(n)} = c \cr & \left( {2^{g(n + 2) - 2g(n + 1) + g(n)} - 1} \right)2^{ - g(n)} = c \cr & 2^{\Delta ^2 g(n)} = 1 + c2^{g(n)} \cr & 2^{\Delta ^2 g(n)} = 1 + 2^{\ln c + g(n)} \cr & \approx \Delta ^2 g(n) + g(n) = \ln c\quad \left| {\;1 < < 2^{g(n)} } \right. \cr}

So the asymptotics is clear.