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I am trying to get my head around the definition of this function (that I concocted as an exercise in defining a function). Let $f$ denote the function satisfying:

$f(0) = +\infty$, and $f(+\infty) = 0$ and is a logarithmic curve.

I know this should be elementary to define but I am stumped.

jsj
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    What exactly do you mean by "has a logarithmic curve"? –  Jun 21 '14 at 03:15
  • Okay maybe the restrictions I have put on it disqualify it as logarithmic.. but I mean that I am looking for $-log(x)$, but with the function approaching $0$ instead of $-\infty$ – jsj Jun 21 '14 at 03:21
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    Does $\dfrac{1}{x}$ where $x>0$ help you? – Jika Jun 21 '14 at 03:27
  • Somewhat relevant to the definition of logarithm: http://math.stackexchange.com/a/109491/118109 – Vibhav Jun 21 '14 at 04:06

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It's not clear what you mean by "logarithmic curve," but one interesting question is to look for functions of the form $$f(x) = -\log(g(x))$$ that satisfy your properties. Do there exist such $g(x)$?

You want $$+\infty = -\log(g(0))$$ so $g(0) = e^{-\infty} = 0$. You also want $0 = -\log(g(\infty))$ so $g(\infty) = 1$. (And of course, the above equations involving $\infty$ are really just a shorthand for taking limits.)

Finally, while you didn't say so in your question, presumably you want $f$ to decrease monotonically and so $g$ must increase monotonically.

The simplest $g$ would be a polynomial, but you can check that no polynomial can possibly exist with $g(0) = 0, g(\infty) = 1.$ What about rational functions? Here there are many possible solutions, for instance $g(x) = \frac{x}{x+1}$. What about exponentials? $g(x) = a^{-1/x}$ will do the trick for any $a>1$... but here we're cheating, because composing this exponential with $\log$ gives you: $$f(x) = -\log g(x) = -\log a^{-1/x} = -\frac{\log_a a^{-1/x}}{\log_a e} = -\frac{\frac{-1}{x}}{\frac{\log e}{\log a}} = \frac{\log a}{x}$$ which I wouldn't really call "logarithmic curve."

user7530
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