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Demosthenes · 07:42 PM

-Spector- said: [Goto]

I fucking suck at math, but I love it. I genuinely wish I had the time to understand it as profoundly as you.

Edit: I'm very curious about math and fuck with numbers in my free time but its not nearly as complex as what you post. A year or so ago I found something cool, I can't remember exactly how/what I was doing but the result would look like something similar: 0.845749350254069765210543.... (infinitely)

Well, to quote some guy who broke the "known for" list in the sidebar of Wikipedia by making it too long, "young man, in mathematics you don't understand things. You just get used to them."

As for that number you referred to, MJ already mentioned that it's a repeating decimal, but I might as well write the (elementary) way to convert them into a fraction, since the educational system seems to miss stuff like that sometimes. Let's say that

$\reverse x = 2.00750750750750750\dots$

Here, the repeating part is obviously the string 075. We multiply both sides of the equation above by powers of 10 so we first get the repeating part directly after the ., and then so that we also get one string on the left side of the . as follows:

$\reverse 10x = 20.0750750750750750\dots$

$\reverse 10000x = 20075.0750750750750\dots$

Now note that $\reverse 10000x - 10x = 20055$ , or equivalently, $\reverse 9990x = 20055$ , and thus,

$\reverse x = \frac{20055}{9990} = \frac{1337}{666}.$

Demosthenes said: [Goto]

Minor quibble, but shouldn't $\reverse f'_{g(n)}(x)$ have a g(n) factor in each term? I haven't looked at this with pen and paper, so I might be missing some simplification, but that derivative formula looks wrong at first glance in the general case.

Well,

$\reverse f'_{g(n)}(x) = \sum_{n=0}^{\infty} \frac{g(n)\cdot x^{g(n)-1}}{(g(n))!} = \sum_{n=0}^{\infty} \frac{x^{g(n)-1}}{(g(n)-1)!} = f_{g(n)-1}(x),$

so I think it's fine, and I think it generalizes to replacement of the factorial with the gamma function as well, although I haven't considered the technical details of that carefully.

Demosthenes said: [Goto]

So are you essentially looking for series where such replacements produce things in terms of known special functions? Or am I misunderstanding?

Yes, the more elegant the better, in some sense, but you seem to be on the same track in your latest post, so this was probably a superfluous justification. I was inspired by $\reverse \varphi$ showing up in the closed form of $\reverse f_{5n}(x)$ (which isn't really a huge surprise since $\reverse \varphi = (\sqrt{5}+1)/2$ ), and by the curious average of the cosine and hyperbolic cosine functions in $\reverse f_{4n}(x).$ Also, the series converges in many cases (e.g. when $\reverse g(n)$ belongs to a large class of nonconstant and nonlinear polynomials), so I thought it might be interesting to try to find closed forms for some of those cases, since the series hasn't previously been studied in any generality, AFAIK.

Demosthenes said: [Goto]

$\reverse f_{g(n)}(x) = -\sum_{n=1}^{\infty} \frac{x^{g(n)}}{(g(n))}$ has some interesting ones.

I like that function! The $\reverse f_{2n+k}(x)$ series can be explained with the well-known closed forms of the series $\reverse f_{2n}(x) = \frac{1}{2}\cdot\log{(1-x^2)}$ and $\reverse f_{2n+1}(x) = x-\mathrm{arctanh}\,x$ that you already mentioned (it should be the hyperbolic arctangent, not the regular one), in addition to a shift of the index of summation to deal with the $\reverse k$ term.

Explicitly, for $\reverse k=2m,\,m\in\mathbb{N}$ (i.e. when $\reverse k$ is even),

$\reverse f_{2n+2m}(x) = -\sum_{n=1}^\infty \frac{x^{2(n+m)}}{2(n+m)} = -\sum_{n=m+1}^\infty \frac{x^{2n}}{2n} = -\sum_{n=1}^\infty \frac{x^{2n}}{2n} + \sum_{n=1}^m \frac{x^{2n}}{2n} = f_{2n}(x) + \sum_{n=1}^m \frac{x^{2n}}{2n}.$

Similarly, for $\reverse k=2m+1,\,m\in\mathbb{N}$ (i.e. when $\reverse k$ is odd),

$\reverse f_{2n+2m+1}(x) = f_{2n+1}(x) + \sum_{n=1}^m \frac{x^{2n+1}}{2n+1}.$

Note that the larger the value of $\reverse k$ (and thus, of $\reverse m$ ), the better the last sums approximate $\reverse -f_{2n}(x)$ and $\reverse -f_{2n+1}(x)$ , respectively. I haven't investigated it yet, but intuition tells me that $\reverse f_{2n}(x)$ and $\reverse f_{2n+1}(x)$ are thus "basis vectors" (in lack of a better term) for the "space" spanned by $\reverse f_{2n+k}(x),\,k=1,2,3,\dots$ . Perhaps you get some similar "space" with three basis vectors for $\reverse f_{3n+k}(x)$ (which hints at the basis vectors being particularly well-behaved), but again, I haven't explored the idea yet.

To go off on another tangent, if we look at the relation $\reverse f'_{g(n)}(x) = f_{g(n)-1}(x)$ from my original series and assume that it holds for your series, it's implied that the "derivative" of $\reverse f_{2n+1}(x) = x-\mathrm{arctanh}\,x$ is $\reverse f_{2n}(x) = \frac{1}{2}\cdot\log{(1-x^2)}$ . Obviously this isn't the standard derivative, but $\reverse f'_{2n+1}(x) = x^2/(x^2-1)$ and $\reverse f'_{2n}(x) = x/(x^2-1)$ are both pretty similar expressions, so I'm thinking there's some type of differential operator $\reverse D$ (perhaps it could conveniently be called an "exponential derivative") such that $\reverse D(f_{g(n)}(x)) = f_{g(n)-1}(x)$ for your series, which might have some very interesting properties.

For the example above, $\reverse D$ 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...