# Find a non constant function that is a quotient of two polynomial, for which: $f\left(x+\frac{1}{f(x)}\right)=f(x)f(-x)$

Before I post the problem, I want to ask what is wrong with the exactly same problem I posted three days ago, 'cause nobody seemed willing to answer it.

The non constant function must satisfy the equation: $$f\left(x+\frac{1}{f(x)}\right)=f(x)f(-x).$$

The function could also be written as: $f[x(x+\frac{x}{xf(x)})]$

• How about $f(x)=1$? – barak manos Jun 1 '14 at 17:34
• a) there's no evident answer. b) you provided no indications that you tried to solve the problem yourself. c) there's no motivation to solve it (no sources, no application of the result, no evident pedagogical benefit, etc). In my books this question falls in the "close as missing context and details" category. – TZakrevskiy Jun 1 '14 at 17:38
• After some playing around, I discovered that $f(x)=\frac{x}{1-x^2}$ satisfies $f\left(x+\frac{1}{f(x)}\right)=f(-x)$. Close but no cigar. – David H Jun 1 '14 at 17:54
• I don't it is worth closing this question. a) The user is new here and probably just need to learns whats expected when asking questions, b) I find this problem interesting, c) this does not seem to me to be a homework question. – abnry Jun 1 '14 at 18:25
• In other words, if you can show that $f(x)$ is invertible, then a condition necessary for solving the above is that $$\frac{1}{f(x)}-\frac{1}{f(-x)} = -2x$$ – abnry Jun 1 '14 at 18:31

Edited. My conjecture is as follows:

Conjecture. The only non-constant solutions are of the form $f(x) = 1/(x^{2} - x + c)$ for some $c \in \Bbb{C}$.

By plugging $g(x) = 1/f(x)$ instead, the functional equation reduces to

$$g(x+g(x)) = g(x)g(-x). \tag{1}$$

Step 1. Since $\text{(1)}$ is true for any real $x$, by the Identity Theorem this relation continues to hold on the Riemann sphere $\hat{\Bbb{C}} = \Bbb{C} \cup \{\infty\}$, where $g$ is holomorphic as mapping $\hat{\Bbb{C}} \to \hat{\Bbb{C}}$. What we want to prove first is as follows:

Claim. $g(\infty) = \infty$.

Assume that $g(\infty) \neq \infty$. Since $g$ is non-constant, the Liouville's Theorem tells us that $g$ must have a pole at some $x_{0} \in \Bbb{C}$. Then $g(x_{0}) = \infty$ shows that

$$g(-x_{0})= \frac{g(x_{0}+g(x_{0}))}{g(x_{0})} = \frac{g(\infty)}{\infty} = 0$$

and hence $g(x) = (x+x_{0})^{m}h(x)$ for some $m \geq 1$ and rational $h$ with $h(-x_{0}) \neq 0$. Thus in view of the relation $\text{(1)}$, we have

\begin{align*} g(-x) = \frac{g(x+g(x))}{g(x)} &= \frac{(x+x_{0}+(x+x_{0})^{m}h(x))^{m}h(x+g(x))}{(x+x_{0})^{m}h(x)} \\ &= \frac{(1+(x+x_{0})^{m-1}h(x))^{m}h(x+g(x))}{h(x)}. \end{align*}

Taking $x \to -x_{0}$ we obtain $g(x_{0}) \neq \infty$, which contradicts our assumption. Therefore the claim follows.

Step 2. Now we have two cases:

1. $x + g(x) \to \infty$ as $x \to \infty$.
2. $x + g(x) \to \alpha$ for some $\alpha \in \Bbb{C}$ as $x \to \infty$. In other words, $g(x) = -x + \alpha + h(x)$ for some $h(x)$ vanishing at $x = \infty$.

We consider the case 1. Our claim is the the conjecture is true in this case.

Claim. $g(x)$ is of the form $x^{2} - x + c$ in the case 1.

By the Step 1, $g(x) \sim A x^{d}$ as $x \to \infty$ for some $d \geq 1$ and $A \neq 0$. Then the assertion of the case 1 says that $x + g(x) \sim A'x^{d}$ (where $A' = A$ if $d \geq 2$ and $A' = A+1 \neq 0$ if $d=1$). So we have

$$AA'^{d}x^{d^{2}} \sim g(x+g(x)) = g(x)g(-x) \sim A^{2}(-1)^{d} x^{2d}.$$

Comparing both the exponent and the coefficient, we have $d = 2$ and $A = 1$. This allows us to expand $g$ as the Laurent series at $x = \infty$ by

$$g(x) = x^{2} + bx + c + \sum_{n=1}^{\infty} \frac{a_{n}}{x^{n}} \quad \text{as } x \to \infty.$$ Plugging this to $\text{(1)}$ and comparing both sides up to degree 3, we have

$$0 = g(x+g(x)) - g(x)g(-x) = 2(b+1)x^{3} + \mathcal{O}(x^{2}) \quad \text{as } x \to \infty$$

and hence $b = -1$. So it remains to show that $a_{n} = 0$ for all $n \geq 1$. To this end, assume otherwise and let $m$ be the smallest positive integer for which $a_{m} \neq 0$. Then we can write

$$g(x) = x^{2} - x + c + \frac{h(x)}{x^{m}}, \quad \text{where} \quad h(x) = \sum_{n=0}^{\infty} \frac{a_{n+m}}{x^{n}}.$$

Then $h(x)$ is a rational function with $h(\infty) = a_{m}$ and plugging this to $\text{(1)}$,

\begin{align*} 0 &= x^{2n}\{ g(x+g(x)) - g(x)g(-x) \} \\ &= (x^{2} - x + c)\{ h(x) - (-1)^{m} h(-x) \} - h(x) + \mathcal{O}(x^{-1}). \end{align*}

Expanding the last expression and comparing the coefficients shows that in any cases we must have $a_{m} = 0$, a contradiction! Therefore $a_{n} = 0$ for any $n$ and the conjecture is true in the case 1. ////

Unfortunately, I have no idea how to deal with the exceptional case 2.

• How is that obvious to you? – pppqqq Jun 1 '14 at 19:26
• If you've seen sos440's other posts, you'd realize how. – heropup Jun 1 '14 at 19:30
• Looking at (*) one suspects that twice the degree equals the square of the degree, so quadratic $g$. And as the RHS is even, one is tempted to make $x+g(x)$ even, so $g(x)=x^2-x+c$ – Hagen von Eitzen Jun 1 '14 at 19:30
• Can you make it any easier and eliminate terms like holomorphic? – most venerable sir Jun 4 '14 at 2:50