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What is the general solution of the equation?

$$f(x) \cdot f(-x) = 1$$

I know that $f(x) = A^{k \cdot x}$ is a solution, and I am feeling this is the general solution, but I don't have any proof.

EDIT: After I read Didier's answer, I realised I might not have clarified the question. Didier's answer is correct, but I was mainly looking of other than the exponential functions. Vhailor's seem to have given the correct answer, because basically I can get any function, limit it to only x >= 0, then define g(-x) as 1/f(x). So I will mark this question as answered, but if you can think of some examples satisfying the conditions, please share them.

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Do you want your function $f$ to be continuous? – Zev Chonoles Dec 25 '11 at 14:58
You say it is a differential equation, so you assume that $f$ is differentiable. Is that right? – Paul Dec 25 '11 at 14:59
let $f(0)=1$, and let $f$ be an arbitrary positive continuous function on $\mathbb{R}^+$. then define $f(-x)$ by $1/f(x)$. (similarly for $f(0)=-1$). – yoyo Dec 25 '11 at 15:08
Hmm... good point about the continuity. Well, I was mainly looking for general solutions, but forgot that differential equations are concerned about continuous functions. – Rafid Dec 25 '11 at 16:07
Your edit is odd. In particular your description of my post is unfaithful (and I wonder where you saw that it was restricted to, or hinted at, the exponential function). – Did Dec 25 '11 at 17:11
up vote 5 down vote accepted

This equation has an awful lot of solutions. To wit, every function $f$ such that $f(x)=s(x)\mathrm e^{g(x)}$ for every $x$, where $s(x)=+1$ or $s(x)=-1$ and where $s$ is an even function and $g$ is an odd function, is a solution. And a moment of thought shows that every solution can be written like that. For some basics about (even functions and) odd functions, see there.

Edit Since it seems that the moment of thought alluded to above should have been described more explicitly, here it goes. Assume that $f$ is a solution. Then $f(x)\ne0$ for every $x$, hence $$ f(x)=s(x)\mathrm e^{g(x)}, $$ for some real numbers $g(x)$ and $s(x)$, where $s(x)=+1$ or $-1$ is the sign of $f(x)$, and where $g(x)=\log|f(x)|$. Thus, the condition on $f$ reads as $s(x)s(-x)=+1$, which means that $s(x)=s(-x)$, and as $\mathrm e^{g(x)+g(-x)}=1$, which means that $g(x)+g(-x)=0$, or equivalently, that $g$ is an odd function. One sees in particular that, as is the case for every odd function, $g(0)=0$, a fact which was obvious from the condition that $f(0)^2=1$.

A simple example outside of the scope of the formula given by the OP is $f(x)=\mathrm e^{\sin(x)}$.

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OK, your answer here is kind of similar to mine (except the sign, and the addition of the odd function g), but I was mainly interested in knowing whether there is another function (i.e. not the exponential function) which also satisfies the condition. – Rafid Dec 25 '11 at 16:08
The previous version answers exactly that question. You might not realize that EVERY function $f$ such that $f(x)\ne0$ for every $x$ can be written as $f=s\cdot\mathrm e^g$ for some real functions $s$ and $g$ with $s(x)$ in $\{+1,-1\}$ for every $x$. – Did Dec 25 '11 at 17:17
Ahhh... I see. Thanks for the clarification indeed. – Rafid Dec 25 '11 at 17:28

Well, if you impose no additional restrictions on the function (continuity, etc.) then you can extend any injection $f$ defined on the positive reals to $\mathbb{R}$ so that it satisfies this condition, just define

$g(x) = f(x)\quad$ if $x \geq 0$

$g(x) = 1/f(-x)\quad$ if $x<0$.

Edit : If you want your function to be continuous just add the requirements that $f$ be continuous, $f(0)=±1$ and $f(x)\neq 0$.

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The conditions that $f(0)=\pm1$ and that $f(x)\ne0$ for every $x$ are not related to $f$ being continuous or not. // Unrelated: I fail to see the relevance of an injectivity condition here. – Did Dec 25 '11 at 17:19
indeed I was wrong with the injectivity condition (I don't even remember why I put it there), but the conditions on $f(0)$ and non-negativity of $f$ are required for $g$ to be continuous. – Vhailor Dec 27 '11 at 22:43
No, these are both implied by the condition that $f(x)f(-x)=1$ for every $x$. – Did Dec 28 '11 at 0:23
Right, what i was trying to say was that the condition $g(x)g(-x)=1$ is equivalent to the condition that there exists a continuous $f$ defined on the positive reals satisfying $f(0)=±1$, $f(x)\neq 0$, and such that the relations between $g$ and $f$ given in my answer hold. (indeed, if $f$ doesn't satisfy $f(0)^2=1$ then $g$ defined as in my answer won't be continuous) – Vhailor Dec 28 '11 at 1:06

According to, the general solution of $f(x)f(-x)=1$ should be $f(x)=\pm e^{\Phi(x,-x)}$, where $\Phi(x,-x)$ is any antisymmetric function of $x$ and $-x$.

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Since the notion of antisymmetric function of two arguments $(x,z)$ restricted to the line $z=-x$ coincides with the notion of odd function of one argument, this seems to be an unnecessarily complicated rehash of (a partial version of) the result already on this page. – Did Jun 24 '12 at 8:52

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