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I did not know that you could find an answer for that. However, I can only use Excel so far to do it. How to calculate 3.6! by hand?

Loan Vu
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    @LoanVu Factorial is only defined for integers! Unless you want to extend the function in some way like the $\Gamma$ function. – azarel Oct 13 '14 at 00:17
  • Although the Gamma function is the standard extension of the factorial, you could define whatever function you want. Say, n! for non-integers equals 0, so $3.6!=0$ is quite easy to calculate by hand. – jinawee Oct 13 '14 at 09:25

3 Answers3

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One way to do this is to notice that: $$(n+t)!\approx n!n^t$$ when $n$ is large. For example, we have $(n+3)!=n!(n+1)(n+2)(n+3)\approx n!n^3$.

Also notice that: $$t!=\frac{(t+n)!}{(t+1)(t+2)\dotsb(t+n)}$$

Putting these together, we have: $$t!=\lim_{n\to\infty}\frac{n!n^t}{(t+1)(t+2)\dotsb(t+n)}$$ which is valid even when $t$ is not an integer!

In case you haven't seen it before: $\lim_{n\to\infty}$ means, roughly, that we let $n$ become larger and larger, and we see what value the expression gets closer and closer to. For example, $\lim_{n\to\infty}\frac1n=0$, because as $n$ gets larger and larger (i.e. as $n$ "goes to infinity"), $\frac1n$ becomes closer and closer to $0$.

EDIT: What Excel does, however, is to simply truncate it. So, instead of being bothered to find out what 3.6! is (roughly 13.38129), it just gives you 3! instead (which is 6).

EDIT EDIT: I just realized that using this, we can prove the reflection formula $(-z)!z!=\pi z\csc(\pi z)$. (We need Euler's infinite product for the sine, though.)

Akiva Weinberger
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    That first expression can be written more precisely as:$$\lim_{n\to\infty}\frac{(n+t)!}{n!n^t}=1,\quad\textrm{for all }t$$Also note that $(-1)!$ is undefined (there is a division by zero), as well as $(-2)!$, $(-3)!$, etc. – Akiva Weinberger Oct 13 '14 at 03:58
  • It's not obvious, but that definition gives $(1/2)!=\frac{\sqrt{\pi}}2$. I think you can show that using the Wallis Product or something. ... It's also not obvious, but this is equivalent to the integral definition given by Zack. I don't know how to prove this. – Akiva Weinberger Oct 13 '14 at 04:02
  • By the way, a very similar method gives us $\sin(i)$ and $\cos(i)$. In my answer, I had (A) an approximation and (B) a recurrence relation. So, first note that (A) $\sin(x)\approx x$ and $\cos(x)\approx1$ when $x$ is small. For (B), use De Moive (or just the trig sum rules). So, to find $\sin(i)$, we have $\cos(i)+i\sin(i)=\lim_{n\to\infty}(1+i(\frac in))^n$ and $\cos(i)-i\sin(i)=\lim_{n\to\infty}(1-i(\frac in))^n$. Simplifying, we end with $\sin(i)=i\sinh(1),\cos(i)=\cosh(1)$. – Akiva Weinberger Oct 13 '14 at 21:15
  • ...Same for finding $e^i$. For (A), use $e^x\approx1+x$ when $x$ is small (look at a graph). (B) is fairly obvious. You end up with $e^i=\lim_{n\to\infty}(1+\frac in)^n$, which, coincidentally, looks similar to what we got in my last comment. Using that, we can simplify this to $e^i=\cos(1)+i\sin(1)$. – Akiva Weinberger Oct 13 '14 at 23:32
  • Recommend you check out Guass' Duplication formula as well as this link. – Simply Beautiful Art Nov 29 '16 at 00:18
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The factorial function $n! = \prod\limits_{i=1}^n i$ is defined only for positive integers $n$. But a standard move in function theory is to extend such functions to the reals, and then the complex plane. To do this, you have to decide what the logical "generalization" of the factorial function would be. An obvious place to start is the observation that the factorial obeys the recurrence relation $n! = n(n-1)!$, which has an obvious generalization in the reals, $f(x) = x\,f(x-1)$. (I'm calling this function $f$ now because it's no longer the factorial function.) We also need a fixed point; $1! = 1$, so set $f(1) = 1$. And it's convenient to shift it left by one: $f(x+1) = x\,f(x)$ (so now $f(x) = (x-1)!$ when $x$ is a positive integer.

There are still a great many functions that fit these constraints. But if you add just one more, for the function $f$ to be logarithmically convex over the positive reals, it's a theorem of Bohr and Mollerup (not that Bohr) that this uniquely defines the gamma function,

$$ \Gamma(t) = \int\limits_0^\infty x^{t-1} e^{-x} \;\text{d}x $$

So $\Gamma(t+1)$ is what mathematicians generally think of when you ask "what is the factorial of a real number that isn't an integer". However, what Excel does if asked to compute the factorial of a non-integer real is truncate it.

zwol
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  • "And it's convenient to shift it left by one." I've never understood why that is. Can you shed some light on it for me? – Akiva Weinberger Oct 13 '14 at 04:04
  • As for what logarithmic convexity means: it just means that its logarithm is convex. – Akiva Weinberger Oct 13 '14 at 04:18
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    Not that Bohr, but his brother! – Micah Oct 13 '14 at 05:22
  • An additional constraint is needed to rule out trivial generalisations such as what Excel does. For 2D shapes, "concave" is intuitively definable as "if you stretch a rubber band and put it around the shape, it takes the form of that shape", so a circle, square, trapezoid are all concave, but a crescent isn't (you might get a half disk, for example). For functions, an intuitive definition is similar to that, but harder to find accurate wording for. Hopefully, the meaning for shapes is sufficient to help you get a picture of what it means for functions. – hvd Oct 13 '14 at 09:46
  • @columbus8myhw I think the offset by 1 is just to make the first pole of the function appear at 0 instead of −1. I never formally studied this stuff, though, I could be wrong. (You can see from the formula that the definition would actually be simpler without that shift!) – zwol Oct 13 '14 at 13:02
  • @columbus8myhw It's not true that the logarithm is convex! See http://www.wolframalpha.com/input/?i=plot+log%28Gamma%28x%2B1%29%29 for graphical demonstration. Each interval between two poles is convex, but the function as a whole isn't. – zwol Oct 13 '14 at 13:08
  • @hvd I know what the calculus definition of convexity is, but I was too tired last night to understand what http://en.wikipedia.org/wiki/Logarithmically_convex_function was saying (it appeared to be something much more complicated than what it actually is) and even now, per what I said to columbus8myhw, it appears to me that $\log \Gamma(x)$ isn't convex. – zwol Oct 13 '14 at 13:11
  • Well, as a whole, I suppose $\ln;\Gamma$ technically isn't convex, but each connected part is. (Furthermore, the entire positive part is.) I think that this is what they meant. – Akiva Weinberger Oct 13 '14 at 13:18
  • Yes, like @columbus8myhw suggests, the Wikipedia article you link to explicitly states "the only function f on the interval x > 0 that simultaneously has the three properties [...] f is logarithmically convex". – hvd Oct 13 '14 at 13:50
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You can use gamma function with the property that $\Gamma(n+1)=n!$ in particular look here http://en.wikipedia.org/wiki/Gamma_function. $\Gamma(4.6)=3.6!=13.383....$

mcmat23
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    Look up the gamma function on wikipedia it looks cool in the complex plane. – Matthew Levy Oct 13 '14 at 00:47
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    This answer incorrectly states that the gamma function is defined by the equation $\Gamma(n+1)=n!$, which would be problematic since we don't know in the first place what $n!$ means. Perhaps something else is intended? – Jonas Meyer Oct 13 '14 at 02:48
  • That is not how $\Gamma$ is defined. This s a bad answer, I can't believe it got five upvotes! – 5xum Oct 13 '14 at 13:52
  • @5xum: 8 upvotes. The Hot Questions list is part of the reason. – Jonas Meyer Oct 13 '14 at 17:07