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Posted 2018-12-06, 09:49 AM
in reply to Chruser's post starting "Well, to quote some..."
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For the example above,  seems to be the operation "differentiate, divide by x, integrate" (with some minor technical tweaks), but it's probably more complicated than that in general...
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Actually, this seems to be it (modulo some minor technical stuff). My previous post was getting a bit lengthy, so I'll post some calculations in here instead.
We have that
}(x) = -\sum_{n=1}^\infty\frac{x^{g(n)}}{g(n)},)
so
}(x)}{x} = -\sum_{n=1}^\infty x^{g(n)-2}.)
Consequently, by integrating both sides of the equation above and dropping the constant of integration, and by assuming that the order of integration and summation is interchangeable, we see that
}(x)}{x}\,\mathrm{d}x = -\sum_{n=1}^\infty \int x^{g(n)-2}\,\mathrm{d}x = -\sum_{n=1}^\infty\frac{x^{g(n)-1}}{g(n)-1} = f_{g(n)-1}(x).)
In other words, a differential operator (in the fractional calculus sense, where integration is considered differentiation of order -1) such that }(x)) = f_{g(n)-1}(x)) is
I can't say that I've seen it previously, but I like its symmetry...
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
Last edited by Chruser; 2018-12-10 at 11:32 AM.
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