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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.

  1. Have I got this series correct?

  2. 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.

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    I think if you let $g(x) = \log f(x)$, then the standard Taylor series for $g$ will be the same as the "multiplicative" Taylor series of $f$. So in a sense they're the same, and I think anything you can do with the multiplicative series can also be done with the additive series using this transformation. – 2017-02-20
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    @TMM: I'm sorry, I don't see how that's true. I've used a different kind of derivative in my product expansion. It's the geometric derivative, not the usual derivative. Maybe you should write it as an answer about how both the series are the same. – 2017-02-20
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    @TMM Okay, now I get what you're saying. So, you're saying that the product series of $e^{f(x)}$ is same as the Taylor series of $f(x)$. But, still, for the same function, we have two different product series and Taylor series. – 2017-02-20
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    Sure, for a given $f$ this gives you two options - I'm just saying you can also get this other option by first transforming $f$ into $g$ and then applying the standard Taylor series to $g$. So on the one hand, the geometric Taylor series may be useful for cases where it's essentially more convenient to work with $g = \log f$, but on the other hand you can get the same expressions by a standard Taylor series applied to $g$. – 2017-02-20

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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.