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I just wrote a blog post and wasn't sure how to word a particular sentence. Say I have the following function:

\begin{equation} f(x) = x^2 \end{equation}

Then I can say that the value of f(x) grows quadratically with x*. Similarly with this function:

\begin{equation} f(x) = e^x \end{equation}

...I could say that f(x) grows exponentially with x. But what about this?

\begin{equation} f(x) = x! \end{equation}

Do I say that f(x) grows "factorially"? What's the proper term?


*I worded this wrong at first. It should be right now. Wait, is that even right? Or would "exponentially" imply f(x) = kx? Should the first term be "quadratically" instead?

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$f(x) = x^2$ grows quadratically with $x$. $f(x) = e^x$ grows exponentially. –  Aryabhata Jan 23 '11 at 22:38
    
@Moron: Thanks, you pointed that out just as I started to realize my mistake (see my edits—I've now corrected that part). –  Dan Tao Jan 23 '11 at 22:40
    
I think you mean $e^x$ where you wrote $e^2$. –  Rahul Jan 23 '11 at 23:03
    
@Rahul: Yes I did, thanks for pointing that out! (Man, clearly it pays to proofread your math.) –  Dan Tao Jan 23 '11 at 23:04

1 Answer 1

up vote 3 down vote accepted

Using Stirling's formula,

$$n! \times e^n \approx Cn^{n + 1/2}$$

I am not sure if there is a name for that kind of growth. It is super-exponential and might be enough to get the point across, I suppose.

btw, $f(x) = x^2$ is said to grow quadratically, not exponentially.

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@Moron: You forgot about the $e^{-n}$-term in Stirling's formula. –  Rasmus Jan 23 '11 at 22:45
    
@Rasmus: Yes, was editing :-)Thanks for pointing that out. –  Aryabhata Jan 23 '11 at 22:46
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Well, $n^n$ can be described as tetration, in which case it could be denoted ${}^2n$. Not sure how that word could be turned into an adjective though, and it's not exactly widely used I guess :) –  Will Vousden Jan 23 '11 at 23:13
    
@Will: Yes, but $n!$ is $n^{n+1/2}$ divided by $e^n$, so it is not exactly tetration... There is also that $\sqrt{n}$ factor. –  Aryabhata Jan 24 '11 at 0:38
    
I'd say this, combined with Will's comment, answers my question. Thanks! –  Dan Tao Jan 24 '11 at 2:56

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