2008-01-11, 06:31 | #1 |
Nov 2005
B6_{16} Posts |
Fractional Calculus
f(x,0)=x^a
f(x,1)=ax^(a-1) f(x,2)=a(a-1)x^(a-2) f(x,n)=a(a-1)(a-2)...(a-n)x^(a-n) This is of course the n'th derivative function for x^a. Now do you see why I wanted the previous two problems solved? ;) It should be very clear if you read the previous two topics that I want not only solutions for non-negative integer values of 'n' but also for non-negative real values. The idea is to extend my calculus tools' abilities. Note 1: Not all f(x) with f'(x) has more than the whole number solutions? I'll have to do more research before I can say that definitively! Note 2: I originally called this 'recursive derivatives' because of the frist paragraph but found references to 'fractional calculus' at Wikipedia. This matches the overall post. |
2008-01-11, 08:53 | #2 |
Cranksta Rap Ayatollah
Jul 2003
1201_{8} Posts |
http://en.wikipedia.org/wiki/Differintegral
specifically, I think this is what you want (where is your n) Last fiddled with by Orgasmic Troll on 2008-01-11 at 08:55 |
2008-01-11, 21:46 | #3 |
Nov 2005
2·7·13 Posts |
Holy calculators Batman - someone actually replied! ;)
Yeah, that's another good article on Wikipedia for this subject. 'Fractional calculus' is kind of a misnomer since real values are allowed as well. I was working from the derivative point of view, but according to those articles the derivatives are sometimes actually worse than integrals. |
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