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Maclaurin series similar to those of e^x

I was playing around with functions of the form $\reverse f_{g(n)}(x) = \sum_{n=0}^{\infty} \frac{x^{g(n)}}{(g(n))!}$ (based on the trivial Maclaurin expansion of $\reverse e^x = f_n(x)$ ), and noticed, for example, that

$\reverse f_{2n}(x)=\cosh(x),$
$\reverse f_{2n+1}(x)=\sinh(x),$
$\reverse f_{2n+2}(x)=\cosh(x)-1,$
$\reverse f_{2n+3}(x)=\sinh(x)-x,$
$\reverse f_{4n}(x)=\frac{\cosh(x) + \cos(x)}{2},$
$\reverse f_{4n+1}(x)=\frac{\sinh(x) + \sin(x)}{2},$
$\reverse f_{4n+2}(x)=\frac{\cosh(x) - \cos(x)}{2},$
$\reverse f_{4n+3}(x)=\frac{\sinh(x) - \sin(x)}{2},$
$\reverse f_{5n}(x)=\frac{e^x}{5} + \frac{2}{5}\left(e^{-\varphi x/2}\cos\left(\frac{1}{2}\sqrt{\sqrt{5}\varphi^{-1}}x\right)+e^{\varphi^{-1}x/2} \cos\left(\frac{1}{2}\sqrt{\sqrt{5}\varphi}x\right )\right),$

where $\reverse \varphi$ is the golden ratio. If you're sufficiently bored, you should help me find other, interesting functions $\reverse g(n)$ , or similar series expansions, e.g. by using WolframAlpha. I suspect that the series above have been studied in some more general context since they're so "obvious", but I haven't seen anything along those lines previously.

(Edit: I should note that $\reverse f'_{g(n)}(x) = f_{g(n)-1}(x)$ if you throw away any terms such that $\reverse g(n) = 0$ .)

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)

So at first glance it just looks like a huge fucking decimal but if you count the numbers between the zeros, they are the same as the number that immediately proceeds the zero (including the first 0 before the decimal point)

0.845749350

I wish I could remember what I was doing, I have it in a notebook somewhere but since I've moved who knows where that's at

Quote:

Originally Posted by Chruser (Post 708745)

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.

I'm also not sure I understand what you're looking for. Sin, sinh, etc are all essentially special functions, and you could always define special functions based on whatever series you come up with by replacing n with 2n or what have you. So are you essentially looking for series where such replacements produce things in terms of known special functions? Or am I misunderstanding?

Quote:

Originally Posted by -Spector- (Post 708748)

What you're talking about is basically a repeating decimal. For instance $\reverse \frac{1}{81} = \overline{.0123456789} = .012345678901234567890123456789 .....$

Another interesting example of a repeated fraction is $\reverse \frac{1}{7}=\overline{.142857}$ . It has the additionaly property of it being cyclic. So 2/7 is the same as 1/7 except you move the 14 to the end and start with 2:

$\reverse \frac{2}{7} = \overline{.285714}$

$\reverse \frac{3}{7} = \overline{.428571}$

$\reverse \frac{4}{7} = \overline{.571428}$

$\reverse \frac{5}{7} = \overline{.714285}$

$\reverse \frac{6}{7} = \overline{.857142}$

As you can see, not only are they all repeating decimals, but they are all cyclic permutations of the original 1/7 string.

This kind of thing is related to cyclic numbers. These are numbers where successive multiples of that number are also cyclic permutations of that number. The smallest such number in decimal is the same string of numbers as 1/7: 142,857, since 142,857 * 2 = 285,714, etc (same as what is shown above).

Cyclic numbers are always of the form

$\reverse \frac{b^{p-1}-1}{p}$

where b is the base you're working in (10 in our case) and p is a prime that does not divide b. Not all primes satisfying this formula will produce a cyclic number, but all cyclic number will satisfy this formula. According to wikipedia, the following are the first to primes produce cyclic numbers:

7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983

142,857 have b = 10 and p = 7.

The next lowest cyclic number has b = 10 and p = 17: 0,588,235,294,117,647

You have to allow for leading zeros, otherwise 142857 is the only number that is cyclic.

I found those to be a cool math facts, so I thought I would share.

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

$\reverse f_n(x) = \ln \left(1 - x \right)$

$\reverse f_{2n}(x) = \frac{1}{2} \ln \left( 1-x^2 \right)$

$\reverse f_{2n+1}(x) = x- \arctan (x)$

For $\reverse f_{2n+k}(x)$ where k is odd an arctan term shows up, sometimes added to a polynomial and/or logarithmic terms. For even k I see only polynomial and logarithmic terms. k = 7 has a much neater form than the others that I've looked at

$\reverse f_{2n+7}(x) = \frac{x^7}{7} + \frac{x^5}{5} + \frac{x^3}{3} + x - \arctan (x)$

$\reverse f_{kn}(x) = \frac{1}{k} \ln \left( 1-x^k \right)$

There may be some other stuff that exists.

Curious that the "base" series is that for the natural log, but an arctan shows up when modifying that series. I remember seeing another relation between ln and arctan as well:

$\reverse \arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)\ + \frac{\pi}{2}$ (for x>0...switch the sign on the pi/2 for negative x.)

Quote:

Originally Posted by -Spector- (Post 708748)

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,

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

Quote:

Originally Posted by Demosthenes (Post 708755)

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.

Quote:

Originally Posted by Demosthenes (Post 708755)

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.

Quote:

Originally Posted by Demosthenes (Post 708757)

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

Quote:

Originally Posted by Chruser (Post 708764)

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...

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

$\reverse f_{g(n)}(x) = -\sum_{n=1}^\infty\frac{x^{g(n)}}{g(n)},$

so

$\reverse \frac{f'_{g(n)}(x)}{x} = -\sum_{n=1}^\infty x^{g(n)-2}.$

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

$\reverse \int\frac{f'_{g(n)}(x)}{x}\,\mathrm{d}x = -\sum_{n=1}^\infty \int x^{g(n)-2}\,\mathrm{d}x = -\sum_{n=1}^\infty\frac{x^{g(n)-1}}{g(n)-1} = f_{g(n)-1}(x).$

In other words, a differential o p e r a t o r $\reverse D$ (in the fractional calculus sense, where integration is considered differentiation of order -1) such that $\reverse D(f_{g(n)}(x)) = f_{g(n)-1}(x)$ is

$\reverse D = \int x^{-1}\frac{\mathrm{d}}{\mathrm{d}x}.$

I can't say that I've seen it previously, but I like its symmetry...

I still have PTSD from 4 years of doing proofs in undergrad. Interesting stuff though.