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I've come up with the following variation of a logarithm, intended to work on values that can be 0, or can grow exponentially from zero in either positive or negative direction.

$$myLog(x) = \begin{cases} \log(x+1), & \mbox{if }x \geq 0 \\ -\log(-x+1), & \mbox{if }x \lt 0 \end{cases} $$

Is there a name for this formula or something similar to it? Has it been studied before? I'd like to read up on it if there is anything out there, but without knowing a name for it I haven't had much success with Google.

Here's a graph of the function. It resembles arctangent, but it doesn't have asymptotes.

alt text

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I don't have anything useful to say, other than Hi! How's life been? –  Jason DeVito Nov 16 '10 at 15:54
    
The log function has been extended to complex numbers, perhaps you can consider studying that. en.wikipedia.org/wiki/Complex_logarithm –  Aryabhata Nov 16 '10 at 17:25
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I doubt this particular formula would have a name; it's simply two logarithmic functions glued together. However, it's quite similar to the inverse hyperbolic sine; in particular, it has the same logarithmic behaviour at infinity. –  Rahul Nov 16 '10 at 17:49
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Rahul has a good point. The inverse hyperbolic sine has the advantage of being analytic, whereas your function isn't twice differentiable. –  Jonas Meyer Nov 16 '10 at 18:35

1 Answer 1

up vote 9 down vote accepted

Given your requirements, here's how you could have invented the inverse hyperbolic sine, even if you had never heard of it.

You want a monotonic function that grows logarithmically at both infinities. You try adding, subtracting, and multiplying logarithmic functions, but that doesn't work. However, since the function you want is monotonic, you can think about the inverse of your function, which must grow exponentially at both infinities. And adding together exponentials does do the job for that.

Consider the inverse of the function you invented, which I'll call $f$. Its inverse is $$f^{-1}(y) = \begin{cases}e^y - 1 & \text{if } y \ge 0, \\ 1 - e^{-y} & \text{if } y \lt 0.\end{cases}$$ Since $e^{-y} \le 1$ when $e \ge 0$, and $e^y < 1$ when $y < 0$, the above inverse function is quite close to the much more nicely-behaved $$\begin{align}g(y) &= e^y - e^{-y} \\ &\equiv 2 \sinh y.\end{align}$$

So in some sense, $f(x)$ behaves like $\sinh^{-1} (x/2)$. As it turns out, they agree at $x = 0$ (although their derivatives differ by a factor of two), and as $x \to \infty$, they both approach $\ln x$ and the difference between them decreases as $1/x$.

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thanks, i think asinh may work for my purposes –  Kip Nov 16 '10 at 19:53

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