Existence of a differentiable function $f:\mathbb R\to\mathbb R$ satisfying $f(0)=1$ and $f(x+1)=\exp\left(3x^2+1\right)f(x)$

Question

Let $f\colon\mathbb{R}\to\mathbb{R}$ be differentiable function that satisfies $$ f(0)=1 \\ \forall x \in \mathbb{R}, \quad f(x+1)=\exp\left(3x^2+1\right)f(x)$$

I think a function $f$ exist which satisfies these conditions, but I don't have any evidence justify my opinion.

Can anyone prove or disprove this?

Answer 1

As pointed out by Kavi Rama Murthy and other contributors, I have rewritten my answer.

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Note that $f(0)=1$ implies $f(x)>0$ for any $x\in\mathbb{Z}$.

Taking logarithm gives $\ln\frac{f(x+1)}{f(x)}=3x^2+1$.

For any $x\in\mathbb{Z}^+$,

$$ \begin{align*} \sum_{k=0}^{x-1}\ln\frac{f(k+1)}{f(k)} & =\sum_{k=0}^{x-1}(3k^2+1) \\ \ln\frac{f(x)}{f(0)} & =3\cdot\frac{x(x-1)(2x-1)}{6}+x \\ \ln f(x) & =\frac{2x^3-3x^2+3x}{2} \\ \end{align*} $$

For any $x\in\mathbb{Z}^-$,

$$ \begin{align*} \sum_{k=x}^{-1}\ln\frac{f(k+1)}{f(k)} & =\sum_{k=x}^{-1}(3k^2+1) \\ \ln\frac{f(0)}{f(x)} & =3\cdot\frac{(-x)(-x+1)(-2x+1)}{6}+(-x) \\ \ln f(x) & =\frac{2x^3-3x^2+3x}{2} \\ \end{align*} $$

Therefore, $\ln f(x)=\frac{2x^3-3x^2+3x}{2}$ for any $x\in\mathbb{Z}$.

To fill up the gap between consecutive integral values of $x$ with the differentiability requirement fulfilled, it can be assumed that $\ln f(x)=\frac{2x^3-3x^2+3x}{2}+\pi(x)$ for $x\in\mathbb{R}$ where $\pi:\mathbb{R}\to\mathbb{R}$ is a differentiable function with the following conditions.

1. $\pi(x)=0$ for any $x\in\mathbb{Z}$.
2. $\pi(x+1)=\pi(x)$ for any $x\in\mathbb{R}$, i.e. $\pi$ is periodic.

Therefore, $f(x)=\exp{\left(\frac{2x^3-3x^2+3x}{2}+\pi(x)\right)}$ for $x\in\mathbb{R}$.

Answer 2

Let $g$ be any differentiable function on $[0,1]$ with $g(0)=1, g(1)=e$ and $eg'(0)=g'(1)$. Extend $g$ to the whole line using the given equation. For example, $f=g$ on $[0,1]$, $f(x)=g(x-1)e^{3(x-1)^{2}+1}$ for $x \in [1,2]$ etc.

Answer 3

Taking the logarithm, $f$ satisfies the linear recurrence

$$\log f(x+1)=\log f(x)+3x^2+1.$$

The solution of the homogeneous equation is just

$$\log f(x)=c.$$

A particular solution can be obtained by indeterminate coefficients, noting that the leading term must be $x^3$, and the constant term doesn't matter (it is absorbed in the homogeneous solution).

Hence,

$$(x+1)^3+a(x+1)^2+b(x+1)-x^3-ax^2-bx=3x^2+(2a+3)x+a+b+1=3x^2+1$$

gives

$$-a=b=\frac32.$$

Finally, using $\log f(0)=0$, $$f(x)=e^{x^3-3x^2/2+3x/2}.$$

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But the story doesn't end here. As the recurrence only relates values that are one unit apart, the constant $c$ can vary as a function of the fractional part of $x$, and the general solution is of the form

$$f(x)=e^{x^3-3x^2/2+3x/2}e^{c(\{x\})}.$$

For differentiability of $f$, $c$ must be differentiable in $[0,1)$, $c(0^+)=c(1^-)=0$ must hold, as well as $c'(0^+)=c'(1^-)$.

An example solution is

$$e^{x^3-3x^2/2+3x/2+\sin(2\pi x)}.$$

Answer 4

This is a functional difference equation with solution

$$ f(x) = \Phi(x) e^{x^3-\frac 32 x^2+\frac 32 x-1} $$

Here $\Phi(x)$ is any periodic function with period $1$