January 18

• ${\displaystyle f(x)=x!\!}$
• ${\displaystyle f'(x)=?\!}$
• ${\displaystyle \int _{\,}^{\,}f(x)dx=?\!}$

Thanks. --Ķĩřβȳ♥ŤįɱéØ 07:15, 18 January 2007 (UTC)[reply]

The factorial in its strictest form can't be "calculused", because it only applies to integers. What you need is the gamma function. yandman 07:38, 18 January 2007 (UTC)[reply]

Also, the factorial can be approximated with Stirling's approximation, which can be easily differentiated. Dunno about an integral though. --Spoon! 07:40, 18 January 2007 (UTC)[reply]

Mathematica gives me an error when I try to put in Stirling's approximation. =( --Ķĩřβȳ♥ŤįɱéØ 07:45, 18 January 2007 (UTC)[reply]

What did you input? When I input Sqrt[2*Pi*x]*x^x*Exp[-x] I get basically the same indefinite integral back, not an error message. For Gamma[x+1] the system reports: Mathematica could not find a formula for your integral. Most likely this means that no formula exists. That is what I expected, and I don't think the Stirling formula has an analytically expressible primitive either.  --Lambiam^Talk 10:00, 18 January 2007 (UTC)[reply]

Just to clarify something, if f(x) is the Gamma function, which equals (x-1)! for positive integers, then

• ${\displaystyle f(x)=\Gamma (x)=\int _{0}^{\infty }t^{z-1}e^{-t}\,\mathrm {d} t}$
• ${\displaystyle f'(x)=x^{z-1}e^{-x}\!}$

(from the Fundamental theorem of calculus) Dugwiki 21:50, 18 January 2007 (UTC)[reply]

No. ${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt}$ But notice how the integral is in ${\displaystyle t}$ and not in ${\displaystyle z}$. —The preceding unsigned comment was added by Spoon! (talk • contribs) 21:57, 18 January 2007 (UTC).[reply]

Yeah, be careful about the way you've written the Gamma function. There is no x in the expression; are you saying the function is constant?

Lastly, the derivative of the Gamma function is complicated and involves something called the polygamma function. –King Bee ^{(T • C)} 22:01, 18 January 2007 (UTC)[reply]

But that's only because the polygamma functions are defined in terms of the derivatives of the gamma function. So saying the derivative involves the polygamma functions is not useful. --Spoon! 00:03, 19 January 2007 (UTC)[reply]

It is useful in pointing out that the derivative of the above is incorrect. In any event, the derivative of the gamma function is not a simple application of the fundamental theorem of calculus. –King Bee ^{(T • C)} 00:17, 19 January 2007 (UTC)[reply]

D'oh! - "I am smart! S-M-R-T!" My bad, guys, sorry. :/ Dugwiki 18:40, 19 January 2007 (UTC)[reply]

i am just a beginner how do you do fraction notation i am really stuck i am on online school so please just tell me the basics. —The preceding unsigned comment was added by 12.210.136.29 (talk) 19:08, 18 January 2007 (UTC).[reply]

Fraction (mathematics)? x42bn6 Talk 21:26, 18 January 2007 (UTC)[reply]

A fraction is one number divided by another. The numerator and denominator may be separated by a slanting line called a solidus or slash, for example 3⁄4, or may be written above and below a horizontal line called a vinculum, like so: ${\displaystyle \textstyle {\frac {3}{4))}$. X [Mac Davis] (DESK|How's my driving?) 23:24, 18 January 2007 (UTC)[reply]

I have a feeling the question is on the mixed fraction notation, such as 5 ¹/₄ for 5.25. Then it's very easy, really. When you encounter a number in mixed notation, just add a + before the fraction, so a ^b/_c becomes a + ^b/_c. Then just use the arithmetic rules to get the actual number. — Kieff 03:56, 19 January 2007 (UTC)[reply]