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I am wondering if there is a function $f(x)$ "similar" to the exponential function $\exp(x)$ such that:

$-f(x) \approx f(-x)$

I would also like $f(x)$ to have the following property:

$\frac{{f(a)}}{{f(b)}} = f(a - b)$

Or alternately,

$\frac{{f(a)}}{{f(b)}} \approx f(a - b)$

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$-\exp (x) $ is *not* approximately $\exp(-x).$ Perhaps you mean't something else? Anyway, there are many functions fitting your description. For example, every odd function. – Ragib Zaman Jul 28 '12 at 14:52
$\exp(z) >0$ always, what do you mean by $-\exp(x) \approx \exp(-x)$??? – copper.hat Jul 28 '12 at 15:01
Do you mean $\sinh(x)$ by chance? – Sasha Jul 28 '12 at 15:15
Yes, every odd function would indeed fit the description, and I have now updated my original question to remove the spurious $-\exp(x) \approx \exp(-x)$. Thanks for these insightful comments. – Nicholas Kinar Jul 28 '12 at 17:22
If $f(a)/f(b)=f(a-b)$ for all $a$ and $b$, then for any $x$ we can let $a=x$ and $b=-x$ to get $f(x)/f(-x)=f(x-(-x))=f(2x)$. But if $f$ is odd, $f(-x)=-f(x)$, so $f(x)/f(-x)=-1$. Putting the two together means that $f(2x)=-1$ for any $x$, but then $f$ can't be odd: a contradiction. So it is not possible for $f$ to be odd and satisfy $f(a)/f(b)=f(a-b)$ at the same time. – Rahul Jul 28 '12 at 18:39
up vote 9 down vote accepted

You might be interested in the hyperbolic sine "sinh". It is antisymmetric and its asymptotic behaviour for $x\to\infty$ is similar to the one of the exponential function.

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It is worth noting that $\sinh(x)$ is, in its heart, a exponential function. – Ian Mateus Jul 28 '12 at 18:15
If $\sinh(x) = (1/2)(\exp(x) - \exp(-x))$, note that when $x \rightarrow +\infty$ it approaches $(1/2)\exp(x)$ in behaviour. – Ian Mateus Jul 28 '12 at 18:19
At first sight, yes. I think one can put more rigour at this and get an idea of the power of these approximations. – Ian Mateus Jul 28 '12 at 18:40
@NicholasKinar Another interesting fact that may interest you is that $\sinh(x) + \cosh(x) = \exp(x).$ – Ian Mateus Jul 28 '12 at 18:50
@IanMateus: Thanks Ian; that is also insightful. To correct an error that I made in the comments above, I meant $\frac{{\sinh (a)}}{{\sinh (b)}} \approx \exp (a - b)$. I've found that the approximation is sufficiently good for $a,b$ greater than 5. – Nicholas Kinar Jul 28 '12 at 19:34

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