Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all non-constant function $g$ such that $$\big(g(x)-1\big)\big(g(-x)-1\big)=1.$$

I started with some special functions like $g(x)=1+e^x$. Then later: $g(x)=1+a^{bx}$ or $g(x)=1-a^{bx}$. I wonder there are more, or how to find all of them.

Edit:

At first, I didn't mentioned that I am interested only in those continuous function.

Follows from Marvis's post, I think I got the answer.

Let $h(x)=|g(x)−1|$. Then $h(x)h(−x)=1$ and hence $\ln(h(x))+\ln(h(−x))=0$. This implies that $\ln(h(x))=−\ln(h(−x))$. Now, $\ln(h(x))=k(x)$ where $k(x)$ is an odd function. So $h(x)=e^{k(x)}$ and hence $g(x)=1±e^{k(x)}$, where $k(x)$ is an odd function.

Note that the base e can be changed to any base a with positive $a≠1$.

Do I miss out anything?

By the way, this questions was due to the following post which I wonder why the function $1+e^x$ is so special. And now, we can replace the $1+e^x$ with $1±a^{k(x)}$, where $k(x)$ is an odd function.

share|improve this question
    
@Hurkyl thanks. I edited. –  pipi Mar 3 '13 at 14:09
add comment

1 Answer

up vote 3 down vote accepted

Let $f(x) = g(x) -1$. Then we have $$f(x) f(-x) = 1$$ From this we get $f(0) = \pm 1$. Now consider a function $h(x): (0,\infty) \to \mathbb{R} \backslash \{0\}$. Then $$f(x) = \begin{cases}h(x) & x > 0\\ \dfrac1{h(-x)} & x < 0\\ \pm1 & x=0\end{cases}$$ Hence, $$g(x) = \begin{cases}1+h(x) & x > 0\\ 1+\dfrac1{h(-x)} & x < 0\\ 0 \text{ or } 2 & x=0\end{cases}$$ where $h(x)$ is any function with $h(x): (0,\infty) \to \mathbb{R}$.

share|improve this answer
    
@copper.hat As Ross has also pointed out, it should read $f(0) = \pm 1$. –  user17762 Oct 17 '12 at 4:23
    
$h(x)$ above cannot be $0$. –  Stefan Smith Oct 19 '12 at 0:26
    
@bogus Thanks. Updated accordingly. –  user17762 Oct 19 '12 at 0:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.