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

Question

Consider the functional equation:

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

where $x>0$, together with the boundary condition $f(x)=x+O(1)$ as $x\to\infty$ (that is, $f(x)-x$ is bounded). Furthermore suppose that $f(x)$ is continuous.

What is the value of $f(1)$?

One approach is to expand $f(x)$ as a Laurent series of the form $$ f(x)=x+a_0 + \frac{a_1}{x}+\frac{a_2}{x^2}+\cdots $$ Set $f_n(x)=x+a_0+\cdots+a_n x^{-n}$, and observe that $f_{n+1}(x)=x+\frac{f_n(2x)}{f_n(3x)}+o(x^{-n-1})$. This allows one to recursively compute the coefficients of the Laurent series very easily; using this technique I obtained that $$ f(x)=x+\frac{2}{3}+\frac{2}{27x}-\frac{7}{729x^2}+\frac{227}{236196 x^3}-\frac{34925}{459165024 x^4}+o(x^{-4}). $$ While I was unable to discern a pattern for these coefficients, it does appear that the $3$-adic valuation of $a_n$ is $n(n+1)/2$.

Source: This question is a variant of a question posed by Jacob Lance on the mathriddles subreddit involving a continued fraction. He asked whether $f(1)<\sqrt{3}$. (One may show that in fact $f(1)\approx \sqrt{3}-2\cdot 10^{-5}$.)

Unfortunately I couldn't find a link to the original question... — pre-kidney, Jul 31 '16 at 08:13
Yes, although it does not contain the original question and neither does the link therein. — pre-kidney, Jul 31 '16 at 08:15
Some mention of function, but the core is the same problem as linked. — Kenny Lau, Jul 31 '16 at 08:33
$\displaystyle \lim_{x\to\infty} f(x) = \lim_{x\to\infty} x+\frac{f(2x)}{f(3x)} = \lim_{x\to\infty} x+\frac{2x+O(1)}{3x+O(1)} = \lim_{x\to\infty} x+\frac23$, therefore the constant is actually $\dfrac23$. — Kenny Lau, Jul 31 '16 at 08:39
Of course, but this is already posted above! (Take another look at the Laurent series.) — pre-kidney, Jul 31 '16 at 08:41
I don't understand how the coefficients are computed. If $f_{n + 1}(x) = x + f_n(2x) / f_n(3x)$, then $a_{n + 1} = x^{n + 1} \left(x + \frac{f_n(2x)}{f_n(3x)} - f_n(x)\right)$, since $f_{n + 1}(x) = f_n(x) + a_{n + 1}x^{-n-1}$. But $a_{n + 1}$ depends on $x$! For example, I get: $$a_2 = \frac{2x^2}{3(9x + 2)}$$ — rubik, Jul 31 '16 at 10:00

score 3 · Accepted Answer · answered Aug 01 '16 at 13:14

Write $f(x):=x+g(1/x)$, and let $y:={1\over x}$. The given functional equation then transforms into $$g(y)={2+y\>g(y/2)\over3+y\> g(y/3)}\ ,$$ or $$g(y)\bigl(3+y\>g(y/3)\bigr)-\bigl(2+y\>g(y/2)\bigr)=0\ .\tag{1}$$ Plugging the "Ansatz" $g(y):=\sum_{k\geq0} a_ky^k$ into $(1)$ leads to the recursion $$a_0={2\over3},\qquad a_r={1\over3}\left({a_{r-1}\over 2^{r-1}}-\sum_{k=0}^{r-1}{a_k\>a_{r-1-k}\over 3^k}\right)\quad(r\geq1)\ ,\tag{2}$$ which produces exactly the numbers you obtained. In order to check that this is not only formal we have to provide an estimate for the $a_k$. I claim that $$|a_r|\leq{4\over5}\cdot2^{-r}\qquad(r\geq0)\ .\tag{3}$$ This can be verified numerically for $r\leq3$. When $r\geq4$ we use $(2)$ to obtain $$|a_r|\leq{1\over3}\left({1\over 2^{r-1}}\cdot{4\over5}\>2^{-(r-1)}\>+{16\over25}\>2^{-(r-1)}\sum_{k=0}^\infty{1\over 3^k}\right)\leq{1\over3}\left({1\over2^{r-2}}+{8\over5}\cdot{3\over2}\right){4\over5}\>2^{-r}\ .$$ If $r\geq4$ then $${1\over3}\left({1\over2^{r-2}}+{8\over5}\cdot{3\over2}\right)\leq{1\over3}\left({1\over4}+{8\over5}\cdot{3\over2}\right)={53\over60}<1\ ,$$ so that $(3)$ follows by induction. The estimate $(3)$ shows that $g$ is analytic at least in the disc $|y|<2$, so that we may use the obtained expansion to compute $f(1)=1+g(1)$ approximatively. One obtains $f(1)-\sqrt{3}\approx-0.000023664$, as indicated by the OP.

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

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