yes, actually I have.
I've proved, using logarithmic semi operator notation (0<= q < 1):
if {-q} is "lowered addition", {1-q} is "lowered multiplication", {2-q} is lowered "exponentiation"; All follow the law of recursion. Also }-q{ is "lowered subtraction", }1-q{ is "lowered division", }2-q{ is "lowered roots".
and if S(p) is the identity function, such that, p E R
x {p} S(p) = x
we find, just as 0 (the additive identity) multiplied by x is 0, S(-q) (the "lowered" additive identity) {1-q} multiplied by x is S(-q), so it behaves just like multiplication by zero. This means, the inverse function }1-q{ division has a pole at S(-q), just like division by zero.
Therefore we can create a new calculus as such
q:d/dx f(x) = lim h -> S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h
which is basically just the normal difference quotient with lowered operators and h approaching a lowered additive identity.
using this and a lot of further digging, I can prove that
q:d/dx e^x = e^x
If you care I can explain how (it involves a generalization of taylor series). I didn't mean to tell you to use my notation or anything, I just saw what you posted and that's what popped into my head. some of your ideas seem to gravitate around logarithmic semi operators
I've proved, using logarithmic semi operator notation (0<= q < 1):
if {-q} is "lowered addition", {1-q} is "lowered multiplication", {2-q} is lowered "exponentiation"; All follow the law of recursion. Also }-q{ is "lowered subtraction", }1-q{ is "lowered division", }2-q{ is "lowered roots".
and if S(p) is the identity function, such that, p E R
x {p} S(p) = x
we find, just as 0 (the additive identity) multiplied by x is 0, S(-q) (the "lowered" additive identity) {1-q} multiplied by x is S(-q), so it behaves just like multiplication by zero. This means, the inverse function }1-q{ division has a pole at S(-q), just like division by zero.
Therefore we can create a new calculus as such
q:d/dx f(x) = lim h -> S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h
which is basically just the normal difference quotient with lowered operators and h approaching a lowered additive identity.
using this and a lot of further digging, I can prove that
q:d/dx e^x = e^x
If you care I can explain how (it involves a generalization of taylor series). I didn't mean to tell you to use my notation or anything, I just saw what you posted and that's what popped into my head. some of your ideas seem to gravitate around logarithmic semi operators