Functional equation: $f(x) = x + 1 + \frac{f(x+1)}{f(x+2)}$ , $f(t) = 0$?

Question

Consider

$$f(x) = x+1+\dfrac{x+2+\dfrac{x+3+\dfrac{x+4+\cdots}{x+5+\cdots}}{x+4+\dfrac{x+5+\cdots}{x+6+\cdots}}} {x+3+\dfrac{x+4+\dfrac{x+5+\cdots}{x+6+\cdots}}{x+5+\dfrac{x+6+\cdots}{x+7+\cdots}}} $$

This implies

$$f(x) = x + 1 + \frac{f(x+1)}{f(x+2)}$$

And for large positive real $x$ , $f(x)$ is close to $x + 2$

We know the inequality for $f(0)$

$$\sqrt 3 > 1+\dfrac{2+\dfrac{3+\dfrac{4+\cdots}{5+\cdots}}{4+\dfrac{5+\cdots}{6+\cdots}}} {3+\dfrac{4+\dfrac{5+\cdots}{6+\cdots}}{5+\dfrac{6+\cdots}{7+\cdots}}} $$

See : Simplifying a complicated continued fraction expression.

What I wonder about is this

$f(t) = 0$

where $t$ is the largest solution.

** edit : with large(st) I mean the usual ordering of the reals ; so $-1$ is larger than $-2$ , not some kind of absolute value thing. Just to be clear. **

I estimate $t = - \frac{1 + \gamma }{2} $ or about $- 0,53860783245..$ but that is a very brute estimate.

Do we know a closed form for $t$ ?

edit

I found this :

Functional equation: $f(x)=x+\dfrac{f(2x)}{f(3x)}$

Maybe it relates ?

Answer 1

Let's define the approximation function:

$$\tilde{f}(x) = \begin{cases} x + 1 + \frac{f(x+1)}{f(x+2)} & x < 10^6 \\ x + 2 & x \ge 10^6 \end{cases}$$

Here's an iterative Python implementation:

def f(x):
    if x >= 1000000:
        return x + 2
    ipart, fpart = divmod(x, 1)
    n = 999999
    next_values = (n + fpart + 3, n + fpart + 4)
    while True:
        y = n + fpart + 1 + next_values[0] / next_values[1]
        next_values = (y, next_values[0])
        if n == ipart:
            return y
        else:
            n -= 1

Using this definition to estimate $f$ for small values of $x$ (multiples of $0.01$ with $|x| \le 10$), the function appears to have a vertical asymptote between $-4.52$ and $-4.51$, and another one between $-3.52$ and $-3.51$. There's also a sign change (but not at asymptote) between $-1.52$ and $-1.51$.

By gradually narrowing the search interval, I was able to find the exact (to IEEE double-precision) $x$ values:

• $x_1 \approx -4.518596026586683$
• $x_2 \approx -3.5185960265866827$
• $x_3 \approx -1.5185960265866827$

That all three “special” $x$ values have the same fractional part is surely no coincidence, but relates directly to the integer shifts of $x + 1$ and $x + 2$ in the definition of $f$.

So, the numerical answer to your question is $\boxed{t \approx -1.5185960265866827}$, as @Yimin previously noted in a comment.

---

I'm still trying to find a closed-form expression for $f$ or its root, but so far, no luck.