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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|) $

Martin Argerami
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Smit Patel
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    Usually, I find the answer at the back of the book not some options – VIVID Jul 17 '20 at 13:39
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    The question is badly phrased. None of the choices is correct. All solutions $f:\mathbb{R}{\neq 0}\to\mathbb{R}$ that satisfy the given functional equation are of the form $$f(x)=\left{ \begin{array}{ll} 1+\phi(x)&\text{if }0<|x|\le 1,,\ 1+\dfrac{1}{\phi\left(\frac1x\right)}&\text{if }|x|>1,, \end{array} \right.$$ where $\phi:[-1,0)\cup(0,+1]\to\mathbb{R}{\neq 0}$ is an arbitrary function. You might need a new book. – Batominovski Jul 17 '20 at 13:49
  • @Batominovski And $f(1),f(-1)\in{0,2}$ – Zerox Jul 17 '20 at 13:57
  • @Zerox, I forgot about that. But it should be $\phi(\pm 1)=\pm 1$. – Batominovski Jul 17 '20 at 13:58
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    I should remark that, while all given functions satisfy the functional equation, they are not the only solutions. The phrasing "if $f$ satisfies [...], then $f$ is one of the following functions" is logically false. What should be asked is probably: "what are examples of functions $f$ such that $f$ satisfies [...]?" Each choice give solutions, but none of them give all the solutions, and even if you combine all of them, you still don't get all solutions. Even if you demand that the solution be analytic (as all the choices are analytic functions), there are still other solutions. – Batominovski Jul 17 '20 at 14:03

3 Answers3

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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 $n\in\mathbb{R}.$

Bumblebee
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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 $f(x)$.

Sameer Baheti
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enter image description here

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

Hope this helps.

Anonymous
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  • The conclusion $g(x)\cdot g\left(\dfrac{1}{x}\right)=1$ implies $g(x)=x^n$ is false. Read the comments under the question. – Batominovski Jul 17 '20 at 16:49
  • can you please explain why is $g(x)\cdot g\left(\dfrac{1}{x}\right)=1$ implies $g(x)=x^n$ i false(if i am not wrong in asking this) @Batominovski – Anonymous Jul 17 '20 at 17:04
  • There are counterexamples. Try $g(x):=\dfrac{\text{arccot}(x)}{\text{arctan}(x)}$. – Batominovski Jul 17 '20 at 17:06
  • oh, ok I understand, but I had only one of the possibilities of $g(x)$, i.e $g(x)=x^n$ could.The could be much better if they had asked the possible functions satisfying $g(x)$ – Anonymous Jul 17 '20 at 17:42