# $f(x)f(1/x)=f(x)+f(1/x)$

Find a function $f(x)$ such that:

$$f(x)f(1/x)=f(x)+f(1/x)$$

with $f(4)=65$.

I have tried to let $f(x)$ be a general polynomial: $$a_0+a_1x+a_2x^2+\ldots a_nx^n$$

which leaves $f(1/x)$ as:

$$a_0+a_1{1\over x}+a_2{1\over x^2}+\ldots + a_n{1\over x^n}$$

On comparing the coefficients of both sides, we see that:

$$2a_0=(a_0)^2+(a_1)^2+(a_2)^2+\ldots+(a_n)^2$$

And

$$a_1=(a_0a_1)+(a_1a_2)+ \ldots +(a_{n-1}a_n)$$

I don't know how to proceed further. I know I need to compare coefficients and come to a conclusion based on their values, but I don't see what to do next.

• See this page on how to write math formulas on this site. Apr 27, 2015 at 17:09
• Is $f$ defined at $0$?
– gary
Apr 27, 2015 at 17:15
• why do you let be $f(x)$ a polynomial? Apr 27, 2015 at 17:21
• The answer is actually $f(x) = x^(3) +1$ Apr 27, 2015 at 17:29
• (My friend used hit and trial to find it, but we require a solid proof) Apr 27, 2015 at 17:31

Given: $$\displaystyle f(x)+f\left(\frac{1}{x}\right) = f(x)\cdot f\left(\frac{1}{x}\right)\tag 1$$

Now we can write $$(1)$$ as $$\displaystyle f(x) = \frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right)-1}\tag 2$$

Now again we can write $$(1)$$ as $$\displaystyle f\left(\frac{1}{x}\right) = \frac{f(x)}{f(x)-1}\tag 3$$

Now multiplying these two equations, we get $$\displaystyle \left[f(x)-1\right]\cdot \left[f\left(\frac{1}{x}\right)-1\right]=1$$

Let $$f(x)-1 = g(x)$$. Then, $$f\left(\frac{1}{x}\right)-1 = g\left(\frac{1}{x}\right)$$

So the equation gets converted into: $$\displaystyle g(x)\cdot g\left(\frac{1}{x}\right) =1$$

Now if $$g(x)$$ is a polynomial, then $$g(x) = \pm x^n$$. So $$f(x) = 1\pm x^n$$.

We are given that $$f(4) = 65$$. So, $$f(x)=1+x^n$$ On putting $$x=4$$, we get $$n=3$$.

Thus, we get, $$f(x)=1+x^3$$

• Suppose $f(x)$ is not a polynomial, then can more functions be concluded from $g(x).g(1/x)=1$? Nov 19, 2017 at 16:59

As other answers show, there is a large class of solutions to your functional equation if one does not restrict $$f$$ more. As per your request, we shall try the restriction that $$f$$ should be a polynomal.

Two polynomial solutions jump into our eyes, namely the zero polynomial $$f(x)=0$$ and $$f(x)=2$$. Neither of these has $$f(4)=65$$, though.

If $$f$$ is a polynomial of degree $$n>0$$, say $$f(x)=a_0+a_1x+\ldots +a_nx^n$$ with $$a_n\ne 0$$, then $$\hat f(x) = x^nf(1/x)$$ is also a polynomial, namely $$\hat f(x)=x^nf(1/x)=a_n+a_{n-1}x+\ldots+a_0x^n.$$ Observe that $$a_0$$ could be $$0$$, so possibly $$\deg\hat f. The functional equation becomes after multiplication with $$x^n$$ $$f(x)\hat f(x)=x^nf(x)+\hat f(x)$$ or $$\tag1 f(x)\cdot(\hat f(x)-x^n)=\hat f(x).$$ But if we multiply the degree $$n$$ polynomial $$f(x)$$ with the polynomial $$\hat f(x)-x^n$$ and obtain the polynomial $$\hat f(x)$$ that has degree $$\le n$$, we conclude that $$\hat f(x)-x^n$$ must be a constant. This implies that $$a_0=1$$ and $$a_1=\ldots=a_{n-1}=0$$, so $$f(x)=a_nx^n+1$$ and $$(1)$$ simplifies to $$(a_nx^n+1)\cdot a_n = x^n+a_n$$ This allows only $$a_n=\pm1$$. To meet the constraint $$f(4)=65$$, we need $$\pm 4^n=65-1$$, so the positive sign and exponent $$3$$. We finally conclude that $${f(x)=x^3+1}.$$

• I fail to understand why $a_n$ could be equal to -1. Could you explain? I feel it can only be +1. Jun 23, 2020 at 12:25
• look at the second last equation. open up the brackets to see why May 16, 2021 at 14:33

I just observed that a non polynomial solution $$\frac{\pi}{2\tan^{-1}(x)}$$ also exists . How can we find other non polynomial solutions ?

• Does $f(4)=65$?
– robjohn
Jun 30, 2019 at 21:56
• @Samyak But i think it fails for $x<0$. Nov 24, 2021 at 8:53

The equation is equivalent to $$\frac1{f(x)}+\frac1{f(1/x)}=1$$ so we can set $$f(x)=\frac1{\frac12-\frac{63\,g(\log(x))}{130\,g(\log(4))}}$$ where $$g(x)$$ is any odd function.

$$g(x)=\tanh(2x)$$ gives $$f(x)=\frac{11050\left(1+x^4\right)}{10922+128x^4}$$ Not a polynomial, but at least a rational function.

$$g(x)=\tanh\left(\frac32x\right)$$ gives $$f(x)=1+x^3$$ which is the answer given by juantheron.

Note that plugging in $x=1$, we obtain $$f(1)^2 = 2f(1) \implies f(1) = 0 \text{ or }2$$ Similarly, plugging in $x=-1$, we obtain $$f(-1) = 0 \text{ or }2$$ We have $$f(1/x)(f(x) - 1) = f(x) \implies f(1/x) = \dfrac{f(x)}{f(x)-1}$$ Hence, for $\vert x \vert > 1$ define $f(x) = g(x)$ such that $g(x) \neq 1$ for all $\vert x \vert >1$. Then define $$f(1/x) = \dfrac{g(x)}{g(x)-1}$$ Hence, there exists a wide class of possible functions given by $$f(x) = \begin{cases} 0 \text{ or }2& x = \pm1\\ g(x) & \vert x \vert > 1 \text{ such that }g(x) \neq 1\\ \dfrac{g(1/x)}{g(1/x)-1} & \vert x \vert < 1 \text{ and }x \neq 0 \end{cases}$$

• I'm not sure I understand the last part of your answer. Exactly how are you defining $g(x)$ Apr 27, 2015 at 17:49
• @tennispro1213 $g$ can be any function. Apr 27, 2015 at 17:52
• Okay, so basically $g$ is the same thing as $f$ and it just makes $f(x)$ easier to read. Apr 27, 2015 at 17:58

Assuming that $f$ is nonzero, we can rewrite this equation as: $$\frac{1}{f(x)} + \frac{1}{f(1/x)} = 1$$

Making the substitution $g(y) = -\frac{1}{2} + \frac{1}{f(e^y)}$, we get the functional equation $g(y) + g(-y) = 0$, which simply says that $g$ is odd.

This equation can be solved, with the initial condition, by setting $g(y) = cy$ for a suitable constant $c$.

Following up on @user17762's answer, suppose $f(x)=\sum_{i=0}^n a_i x^i$ where $a_n\ne 0$ and $n\ge 1$, then $f=g$ for $|x|>1$, we have for $|x|<1$,

$$\sum_{i=0}^n a_i x^i=\frac{\sum_{i=0}^n a_i x^{-i}}{\sum_{i=0}^n a_i x^{-i}-1}=\frac{\sum_{i=0}^n a_i x^{n-i}}{\sum_{i=0}^n a_i x^{n-i}-x^n}$$

For RHS to be a polynomial you have $a_0=1$ or $a_0\ne 1$ and $a_i=0$ for $i>0$ (which contradicts $a_n\ne 0$). Now the numerator has highest power $n$ and so does LHS, so the denominator has to be constant, i.e. $a_i=0$ for $0<i<n$, now $f(x)=c x^n+1$ for some $c\ne 0$. Putting this back in you have $c=\pm 1$. Solving this for $x=4$ gives your result.