I'm talking about this: $$f(x) = f(a)*f^1(a)^{(x-a)}*f^2(a)^{\frac{(x-a)^2}{2!}}*f^3(a)^{\frac{(x-a)^3}{3!}}*................$$ By $f^n(a)$, I mean the nth geometric derivative of $f(x)$ at x=a. Look for multiplicative calculus on wikipedia: https://en.wikipedia.org/wiki/Multiplicative_calculus For those who don't know, the first geometric derivative of a function is $$f^1(x)=e^{\frac{f^{'}x}{f(x)}}$$, where $f^{'}x$ is the usual derivative of the function. Similarly, The second geometric derivative of a function is the geometric derivative of the first geometric derivative of the function, etc.
Have I got this series correct?
Is this series of any use like the Taylor series? I think maybe it converges more quickly. I know that it involves exponents, so it won't always be defined. But I think the farther terms of this series get close to 1 more quickly than the higher power terms in Taylor series get closer to 0.
Indeed it is! Many functions are better approximated by their multiplicative Taylor polynomials than by their linear ones. Take a look at pages 4 and 5 (aka 67 and 68) of this paper detailing possible applications to biomedical image analysis.
And yes, your formula looks to be correct according to that paper.