fascinating average

[tommy1729]

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.

again with respect , but that is trivial and already known to me and i believe all regular posters.

then again it might not have been explictly posted and some newbies might learn from it.

( i dont wanna sound unthankfull for a well-intented and true fact posted )

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

if i got the notation right , i think i know this as well , but im not sure because of the notation.

as for the generalization of the taylor series i had some ideas , possibly similar possibly not , of generalizing taylor series too in the hyperoperator context.

so im all ear for your taylor series - i intended to post about it but canceled it for some reasons ( e.g. doubt of use ) - but i cannot promise i will like it or not.

its easier for me if you work with more commen notation.

regards

tommy1729