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 [quote="daniel801"]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: [b]f(x) = 6x^4[/b] f1(x) = 24x^3.........works f2(x) = 72x^2.........works f3(x) = 144x..........works f4(x) = 144...........[b]this one comes up undefined[/b] even... f-1(x)= (6x^5)/5......works[/quote]
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daniel801
Posted: Wed Mar 06, 2002 10:19 pm    Post subject: 1

great; it works then; thanks
Posted: Wed Mar 06, 2002 9:27 pm    Post subject: 0

*~agrees~*
tigg
Posted: Wed Mar 06, 2002 8:56 pm    Post subject: -1

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.
mikegoo
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.
daniel801
Posted: Wed Mar 06, 2002 7:16 pm    Post subject: -3

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