# A function that increases faster than $\ln(x)$ when $x$ is small and then slower than $\ln(x)$ when $x$ is big.

As the title indicate:

I am looking for a function that increases faster than $\ln(x)$ when $x$ is small and then slower than $\ln(x)$ when $x$ is big.

Here is the fig:

The red curve is the $\log(x)$, and the green curve is the function I am looking for.

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Do you want the green curve to have a vertical and a horizontal asymptote like the figure suggests? – Hurkyl Jul 16 '14 at 17:02
@Hurkyl Ideally the green curve approaches 1 when x is infinity. – Changwang Zhang Jul 16 '14 at 17:07
@Leo The error function is one such function. See en.wikipedia.org/wiki/Error_function. Although it is not an elementary function. – Eff Jul 16 '14 at 17:08
I think you should give more specifications: desired range, values and slopes at the ends of the range. – Yves Daoust Jul 16 '14 at 17:21

What about $\sqrt{\log x\log{10}}$ ?

Actually, any function mapping $0$ to $0$ and $\log10$ to $\log10$ and such that $f(x)>x$ in between can be used to compose $f(\log x)$ like you want.

You can also use Hermite cubic interpolation, giving you all freedom to adjust the slopes.

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Your requirements "when x is small" and "when x is large" are a bit vague.

However, try the function:

$$f(x)=x^{\frac{a}{x}}$$

and adjust $a>1$ to fine-tune the function's descent. Try with $a\sim 5$.

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I add a figure for the function u mentioned. However, it decreases after some x value. I need a increase function f(x) (when x>0). – Changwang Zhang Jul 16 '14 at 17:17
@Leo: I thought you said (in above comment) that the green function "ideally" should approach 1, when $x\to\infty$. – Yiannis Galidakis Jul 16 '14 at 17:21
This bears little resemblance with what the OP asked for. – Yves Daoust Jul 17 '14 at 16:36
@Yves Daoust: The OP changed his mind, midway after some of the answers were given. His answer to Hurkyl: "Ideally the green curve approaches 1 when x is infinity.", etc. Then, below my answer: "However...I need a increase function $f(x)$ (when $x>0$). – Yiannis Galidakis Jul 17 '14 at 17:23
I downvoted him. – Yves Daoust Jul 17 '14 at 21:53