03/02/2019, 02:50 AM
Hi Micah,
As promised, I took a look at the product-based derivative in your equation (3), and it's very interesting. As I was working out the derivation, I found that, to a first order approximation, it appears to be resemble the logarithmic derivative. Of course, those higher order differences really add it. Interestingly enough, I think this ends up being the exponentiation of the logarithmic integral. At first, that seems like the exponentiation and the logarithm would cancel, being functional inverses of each other. But alas, the order of operations prevents them from directly cancelling.
Forgive my lack of TeX expertise here. I couldn't remember how to do some of the formatting, so it's a bit rough.
Repeating your equation (3):
\( \partial f^{(\partial x)^{-1}}=\lim\limits_{\Delta x \to 0}\left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}\qquad\qquad(3) \)
Let's compare the logarithmic derivative:
\( \frac{\partial \ln(f)}{\partial x} = \frac{1}{f}\frac{\partial f}{\partial x} \)
\( {}={\frac{1}{f(x)}}\lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \)
\( {}=\lim\limits_{\Delta x \to 0} \frac{1}{\Delta x}\frac{f(x + \Delta x) - f(x)}{f(x)} \)
\( {}=\lim\limits_{\Delta x \to 0}{\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)} \)
Already, there should be a sense of similarity, as both expressions have the term \( \frac{f(x+\Delta x)}{f(x)} \).
We can take a binomial expansion of (3):
\( \lim\limits_{\Delta x \to 0}\left(1+\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right)^{\frac{1}{\Delta x} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left({{\Delta x}^{-1} \choose {k}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)} \)
The "choose" notation here is shorthand for a product:
\( {{\Delta x}^{-1} \choose {k}}=\prod_{n=1}^{k}\left(\frac{{\Delta x}^{-1}-(n-1)}{n}\right) \)
Only the highest power of (1/dx)^k matters, alternatively the lowest power of dx^{-k}, so eliminate the neglible terms:
\( {{\Delta x}^{-1} \choose {k}} \approx \frac{{\Delta x}^{-k}}{k!} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{{\Delta x}^{-k}}{k!}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right]^{k}\right)} \)
And now we see that we have the power series of the exponential function, taking the logarithmic derivative as its argument:
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{\partial \ln(f)}{\partial x}\right]^{k}\right)} \)
Therefore:
\( \partial f^{(\partial x)^{-1}}=\exp\left(\frac{\partial \ln(f)}{\partial x}\right) \)
I hope my derivation was correct. I get so caught up in editing the TeX code, that I sometimes miss mistake in my maths. However, I did some crude testing in Excel, and my results do seem to bear out this result.
If this is true, could we conjecture that the derivative based on the next-higher hyperoperation would be sexp(derivative(slog(f(x))))?
As promised, I took a look at the product-based derivative in your equation (3), and it's very interesting. As I was working out the derivation, I found that, to a first order approximation, it appears to be resemble the logarithmic derivative. Of course, those higher order differences really add it. Interestingly enough, I think this ends up being the exponentiation of the logarithmic integral. At first, that seems like the exponentiation and the logarithm would cancel, being functional inverses of each other. But alas, the order of operations prevents them from directly cancelling.
Forgive my lack of TeX expertise here. I couldn't remember how to do some of the formatting, so it's a bit rough.
Repeating your equation (3):
\( \partial f^{(\partial x)^{-1}}=\lim\limits_{\Delta x \to 0}\left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}\qquad\qquad(3) \)
Let's compare the logarithmic derivative:
\( \frac{\partial \ln(f)}{\partial x} = \frac{1}{f}\frac{\partial f}{\partial x} \)
\( {}={\frac{1}{f(x)}}\lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \)
\( {}=\lim\limits_{\Delta x \to 0} \frac{1}{\Delta x}\frac{f(x + \Delta x) - f(x)}{f(x)} \)
\( {}=\lim\limits_{\Delta x \to 0}{\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)} \)
Already, there should be a sense of similarity, as both expressions have the term \( \frac{f(x+\Delta x)}{f(x)} \).
We can take a binomial expansion of (3):
\( \lim\limits_{\Delta x \to 0}\left(1+\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right)^{\frac{1}{\Delta x} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left({{\Delta x}^{-1} \choose {k}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)} \)
The "choose" notation here is shorthand for a product:
\( {{\Delta x}^{-1} \choose {k}}=\prod_{n=1}^{k}\left(\frac{{\Delta x}^{-1}-(n-1)}{n}\right) \)
Only the highest power of (1/dx)^k matters, alternatively the lowest power of dx^{-k}, so eliminate the neglible terms:
\( {{\Delta x}^{-1} \choose {k}} \approx \frac{{\Delta x}^{-k}}{k!} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{{\Delta x}^{-k}}{k!}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)} \)
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right]^{k}\right)} \)
And now we see that we have the power series of the exponential function, taking the logarithmic derivative as its argument:
\( {}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{\partial \ln(f)}{\partial x}\right]^{k}\right)} \)
Therefore:
\( \partial f^{(\partial x)^{-1}}=\exp\left(\frac{\partial \ln(f)}{\partial x}\right) \)
I hope my derivation was correct. I get so caught up in editing the TeX code, that I sometimes miss mistake in my maths. However, I did some crude testing in Excel, and my results do seem to bear out this result.
If this is true, could we conjecture that the derivative based on the next-higher hyperoperation would be sexp(derivative(slog(f(x))))?
~ Jay Daniel Fox