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daniel801
Daedalian Member

 Posted: Wed Mar 06, 2002 7:16 pm    Post subject: 1 For any function, f(x) = ax^y, with a being the coefficient, x being the variable, and y being the exponent, the certain derivative or integrative equation exluding constants and their integral progressions, and where b is the certain derivative or integral equation (integral represented as a negative number), then: fb(x)=[(y!ax^(y-b)]/(y-b)! I derived that two nights ago when I was bored, but the interesting thing is that it only works when b < y. Thus, it won't find the "last" derivative. I wonder how that can be; it just yields undefined. Any ideas how to perfect this formula? example: f(x) = 6x^4 f1(x) = 24x^3.........works f2(x) = 72x^2.........works f3(x) = 144x..........works f4(x) = 144...........this one comes up undefined even... f-1(x)= (6x^5)/5......works
mikegoo
Daedalian Member

 Posted: Wed Mar 06, 2002 8:03 pm    Post subject: 2 The only thing i can think of is that the division by (y-b)! which in the last derivative would be 0! is casuing problems. 0! is defined as equaling 1 (in my world at least) then your formula would work, but if what ever application you are using doesn't know 0!=1 and instead thinks 0!=0 then undefined is what would result (if it were using a loop to calculate factorials for example). Of course I'm just making this up as I go, but I think it is accurate.
tigg
Daedalian Member

 Posted: Wed Mar 06, 2002 8:56 pm    Post subject: 3 I'm with mikegoo. (n-1)! = n!/n. Let n=1 and you get 0! = 1. That is pretty standard. We didn't make that up.
Daedalian Member

 Posted: Wed Mar 06, 2002 9:27 pm    Post subject: 4 *~agrees~*
daniel801
Daedalian Member

 Posted: Wed Mar 06, 2002 10:19 pm    Post subject: 5 great; it works then; thanks
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