Solving functional equation $f(x)+f(1/x)=f(x)\cdot f(1/x)$

Question

Let a function $f(x)$, and $x$ not equal to zero be such that: $f(x)+f(1/x)=f(x)\cdot f(1/x)$ then $f(x)$ is ?

I tried differentiating it but did not find any useful outcome. answer given at back of book is

A)$1+x^n$, $n$ belongs to $\mathbb R$

B)$1-x^n$, $n$ belongs to $\mathbb R$

C)$\pi/(2\arctan|x|)$

D)$2/(1+k\ln|x|) $

Answer 1

Solve the functional equation $f(x)+f(1/x)=f(x)\cdot f(1/x), x\in\mathbb R-\{0\}$.

Assuming $f(x)$ to be a polynomial, \begin{align*} (f(x)-1)\left(f\left(\frac1x\right)-1\right)&=1\\ (a_0^\prime+a_1x+\cdots a_nx^n)\left(a_0^\prime+\frac{a_1}x+\cdots \frac{a_n}{x^n}\right)&=1\\ \end{align*} is only possible when there is only one term in the brackets, i.e. $f(x)-1=\pm x^n$.

PS: I think this might help to generalize $f(x)$.

Answer 2

Therefore,$$f(x)=1+x^n$$.

Hope this helps.

Answer 3

The given functional equation factored as $$(f(x)-1)(f(1/x)-1)=1.$$ Now if we let $f(x)=g(x)+1$ for some other function $g,$ then the only constrain on $g$ is $$g(x)g(1/x)=1.$$ Note that $g$ can never be zero and $(g(\pm1))^2=1.$ Also, if we know $g$ on $[-1,0)\cup(0,1],$ then this last condition determine it everywhere on on $\mathbb{R}_{\neq0}.$ Hence the general solution to this functional equation is $$f(x)=\left\{ \begin{array}{ll} 1+g(x), &\text{if }0<|x|\lt 1\,,\\ 0\text{ or }2, &\text{if }|x|=1\,,\\ 1+\dfrac{1}{g\left(1/x\right)},&\text{if }|x|>1\,. \end{array} \right.$$ as mentioned in a comment. However this doesn't mean that $f$ is always piecewise or non-smooth. The condition $g(x)g(1/x)=1$ on $g$ is equivalent to say that the composition $$h(x)=\ln |g(\exp(x))|$$ is an odd-function. So, always we can take any odd function $h$ and create a nice solution to this functional equation via $f(x)=\pm\exp( h(\ln |x|))+1$ on $\mathbb{R}_{\neq0}.$ For some examples:

\begin{array}{|c|c|} h(x) & f(x) \\ \hline nx & \pm x^n+1 \\ n\ln\left(\dfrac{\cot^{-1}(e^x)}{\tan^{-1}(e^x)}\right) & \left(\dfrac{\pi}{2\tan^{-1}(x)}-1\right)^n+1\\ -2\tanh^{-1}(nx) & \dfrac{2}{1+n\ln |x|}, \dfrac{2n\ln |x|}{1+n\ln |x|}\\ n\sinh^{-1}(x) & \pm(\ln|x|+\sqrt{(\ln |x|)^2+1})^n+1\\ \tanh (nx) & \exp\left(\dfrac{x^{2n}-1}{x^{2n}+1}\right)+1\\ \sinh (nx) & \exp\left(\dfrac{x^{2n}-1}{2x^n}\right)+1\\ \end{array}

for any $n\in\mathbb{R}.$