meemoe_uk
Jan16-04, 04:37 AM
Seem to remember reading about RF experimenting with non-integer differentiation. I found it quite interesting to play with.
I started with 'half' differentiation.
e.g. f(x) = x^2 , D[1]f = 2x , D[2]f = 2
but what about non-integer diff?
e.g. D[0.5]f = px^1.5
what is p?
Clearly D[0.5]D[0.5]f = D[1]f and D[1](x^n) = nx^n-1
So D[0.5](x^n) = p(n)x^(n-0.5) and
D[0.5]( p(n)x^(n-0.5) ) = p(n)p(n-0.5)x^(n-1)
which is equal to D[1](x^n) = nx^(n-1)
so p(n)p(n-0.5) = n
From this I figure the key to finding p is solving the formula...
p(n)*p(n-0.5) = n .
It looks abit like the gamma function
G(n+1)/G(n) = n
and considering the solution to the gamma function, I decide that solving my non-integer diff formula is beyond me.
I've got a pretty good approximation though.
p ~= ((x+0.5)^0.5 + x^0.5)/2 for x>1
hmm. Is this 'genaralized' diff alreadly established math? Since calculus is so central to math,it seems to me that any generalization has broad applications.
I started with 'half' differentiation.
e.g. f(x) = x^2 , D[1]f = 2x , D[2]f = 2
but what about non-integer diff?
e.g. D[0.5]f = px^1.5
what is p?
Clearly D[0.5]D[0.5]f = D[1]f and D[1](x^n) = nx^n-1
So D[0.5](x^n) = p(n)x^(n-0.5) and
D[0.5]( p(n)x^(n-0.5) ) = p(n)p(n-0.5)x^(n-1)
which is equal to D[1](x^n) = nx^(n-1)
so p(n)p(n-0.5) = n
From this I figure the key to finding p is solving the formula...
p(n)*p(n-0.5) = n .
It looks abit like the gamma function
G(n+1)/G(n) = n
and considering the solution to the gamma function, I decide that solving my non-integer diff formula is beyond me.
I've got a pretty good approximation though.
p ~= ((x+0.5)^0.5 + x^0.5)/2 for x>1
hmm. Is this 'genaralized' diff alreadly established math? Since calculus is so central to math,it seems to me that any generalization has broad applications.