Define 0 < x! < 1 to be 1, and define all fractional x!s above 1 recursively. For example, (5/2)! becomes (5/2)(3/2)(1) = 15/4. This method won't result in any number greater than another having a lesser factorial - the proof is left as an exercise to the reader.

Of course, now that factorials are defined continuously, the question arises... what's the derivative of x‽

## A definition for fractional factorials

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### Re: A definition for fractional factorials

LSK wrote:Define 0 < x! < 1 to be 1, and define all fractional x!s above 1 recursively. For example, (5/2)! becomes (5/2)(3/2)(1) = 15/4. This method won't result in any number greater than another having a lesser factorial - the proof is left as an exercise to the reader.

Of course, now that factorials are defined continuously, the question arises... what's the derivative of x‽

The justification that you ask the reader to supply would show that your function is increasing, so the question of continuity is unresolved. Also Continuity does not imply differentiability so keep that in mind.

and as a final note...

Just to make sure you're aware of it the convensional way to "extend" ! is by the gamma function. http://en.wikipedia.org/wiki/Gamma_function

- NathanielJ
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### Re: A definition for fractional factorials

The function you gave isn't the usual extension of the factorial to the real line, as Yesila pointed out, but it's not hard to see that it is indeed continuous, and differentiable except at x = 1, 2, 3, ...

This is because the function breaks down in the following way on the following intervals:

x \in [0, 1): 1

x \in [1, 2): x

x \in [2, 3): x(x-1)

x \in [3, 4): x(x-1)(x-2)

...and so on. The derivatives are easy to calculate on each of these intervals, but the derivative does not exist at any natural number. The gamma function fixes this problem, as it is differentiable for all x > 0.

This is because the function breaks down in the following way on the following intervals:

x \in [0, 1): 1

x \in [1, 2): x

x \in [2, 3): x(x-1)

x \in [3, 4): x(x-1)(x-2)

...and so on. The derivatives are easy to calculate on each of these intervals, but the derivative does not exist at any natural number. The gamma function fixes this problem, as it is differentiable for all x > 0.

### Re: A definition for fractional factorials

fix'd... for some users. We have tags for rendering LaTeX now... for some users. Also, QFT. I apologise in advance for the next line.NathanielJ wrote:The function you gave isn't the usual extension of the factorial to the real line, as Yesila pointed out, but it's not hard to see that it is indeed continuous, and differentiable except at x = 1, 2, 3, ...

This is because the function breaks down in the following way on the following intervals:

[math]x \in [0, 1): 1[/math][math]x \in [1, 2): x[/math][math]x \in [2, 3): x(x-1)[/math][math]x \in [3, 4): x(x-1)(x-2)[/math]

...and so on. The derivatives are easy to calculate on each of these intervals, but the derivative does not exist at any natural number. The gamma function fixes this problem, as it is differentiable for all x > 0.

So the factorial has been extended this way, but what is the interrobang (‽) supposed to do?

LOWA

### Re: A definition for fractional factorials

Methinks you want this.

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### Re: A definition for fractional factorials

The OP might be interested in the Bohr-Mollerup theorem as well, since it justifies the Gamma function as the extension of factorials

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### Re: A definition for fractional factorials

Okay, so I tried taking the derivative of x! using Mathematica, and it gave me the following as the answer:

[math]\Gamma (x+1) \psi ^{(0)}(x+1)[/math]

After looking this up on Wikipedia:

[math]\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt[/math]

and

[math]\psi^{(m)}(z)= (-1)^{(m+1)}\int_0^\infty \frac{t^{m}e^{-zt}}{1-e^{-t}}\,dt[/math]

After solving for these at z = 1, I get the negative of the Euler constant as the answer.

So naturally, I look up this constant, and it's defined as:

What I'm wondering is how come this integral (or alternatively, this derivative) at x=1 evaluates to this number? It doesn't really say anything about it on the Wiki page.

[math]\Gamma (x+1) \psi ^{(0)}(x+1)[/math]

After looking this up on Wikipedia:

[math]\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt[/math]

and

[math]\psi^{(m)}(z)= (-1)^{(m+1)}\int_0^\infty \frac{t^{m}e^{-zt}}{1-e^{-t}}\,dt[/math]

After solving for these at z = 1, I get the negative of the Euler constant as the answer.

So naturally, I look up this constant, and it's defined as:

Its numerical value to 50 decimal places is 0.57721 56649 01532 86060 65120 90082 40243 10421 59335 93992 …

What I'm wondering is how come this integral (or alternatively, this derivative) at x=1 evaluates to this number? It doesn't really say anything about it on the Wiki page.

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### Re: A definition for fractional factorials

Equivalent formulas for the gamma function:

[math]\eqalign{ \Gamma(s) &= \int_0^\infty x^{s-1}e^{-x}\,dx & \quad \text{for } \Re (s) > 0 \\ &= \lim_{k \to \infty}{k!k^s \over s (s + 1) \cdots (s + k)} &\quad \text{for } s \ne 0, -1, -2, \ldots \\ &= e^{-\gamma s} s^{-1} \prod_{n = 1}^\infty \left( 1 + {s \over n} \right)^{-1} e^{s \over n} &\quad \text{for } s \ne 0, -1, -2, \ldots }[/math]

Additionally it has the properties:

[math]\Gamma (s + 1) = s \Gamma (s)[/math]

[math]\Gamma (n + 1) = n![/math]

[math]\Gamma (1) = 0! = 1[/math]

In the integral form [imath]\Gamma ' (1)[/imath] is difficult to calculate. But in product form:

[math]\log \Gamma (s) = \lim_{k \to \infty} \sum_{i = 1}^k (\log i - \log (s + i)) + s \log k - \log s[/math]

[math]{\Gamma ' (s) \over \Gamma (s)} = (\log \Gamma(s))' = \lim_{k \to \infty} \log k - \sum_{i = 0}^k {1 \over s + i}[/math]

[math]\Gamma ' (1) = {\Gamma ' (1) \over \Gamma (1)} = - \lim_{k \to \infty} \sum_{i = 1}^k {1 \over i} - \log k = -\gamma[/math]

[math]\eqalign{ \Gamma(s) &= \int_0^\infty x^{s-1}e^{-x}\,dx & \quad \text{for } \Re (s) > 0 \\ &= \lim_{k \to \infty}{k!k^s \over s (s + 1) \cdots (s + k)} &\quad \text{for } s \ne 0, -1, -2, \ldots \\ &= e^{-\gamma s} s^{-1} \prod_{n = 1}^\infty \left( 1 + {s \over n} \right)^{-1} e^{s \over n} &\quad \text{for } s \ne 0, -1, -2, \ldots }[/math]

Additionally it has the properties:

[math]\Gamma (s + 1) = s \Gamma (s)[/math]

[math]\Gamma (n + 1) = n![/math]

[math]\Gamma (1) = 0! = 1[/math]

In the integral form [imath]\Gamma ' (1)[/imath] is difficult to calculate. But in product form:

[math]\log \Gamma (s) = \lim_{k \to \infty} \sum_{i = 1}^k (\log i - \log (s + i)) + s \log k - \log s[/math]

[math]{\Gamma ' (s) \over \Gamma (s)} = (\log \Gamma(s))' = \lim_{k \to \infty} \log k - \sum_{i = 0}^k {1 \over s + i}[/math]

[math]\Gamma ' (1) = {\Gamma ' (1) \over \Gamma (1)} = - \lim_{k \to \infty} \sum_{i = 1}^k {1 \over i} - \log k = -\gamma[/math]

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