3
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ $f(x) = x^2$ grows quadratically with $x$. $f(x) = e^x$ grows exponentially. $\endgroup$
    – Aryabhata
    Jan 23, 2011 at 22:38
  • $\begingroup$ @Moron: Thanks, you pointed that out just as I started to realize my mistake (see my edits—I've now corrected that part). $\endgroup$
    – Dan Tao
    Jan 23, 2011 at 22:40
  • $\begingroup$ I think you mean $e^x$ where you wrote $e^2$. $\endgroup$
    – user856
    Jan 23, 2011 at 23:03
  • $\begingroup$ @Rahul: Yes I did, thanks for pointing that out! (Man, clearly it pays to proofread your math.) $\endgroup$
    – Dan Tao
    Jan 23, 2011 at 23:04

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ @Moron: You forgot about the $e^{-n}$-term in Stirling's formula. $\endgroup$
    – Rasmus
    Jan 23, 2011 at 22:45
  • $\begingroup$ @Rasmus: Yes, was editing :-)Thanks for pointing that out. $\endgroup$
    – Aryabhata
    Jan 23, 2011 at 22:46
  • 1
    $\begingroup$ 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 :) $\endgroup$ Jan 23, 2011 at 23:13
  • $\begingroup$ @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. $\endgroup$
    – Aryabhata
    Jan 24, 2011 at 0:38
  • $\begingroup$ I'd say this, combined with Will's comment, answers my question. Thanks! $\endgroup$
    – Dan Tao
    Jan 24, 2011 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.