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Many real functions support a $\log$-series representation along with their usual power series representation. An example is the canonical

$$ x = 1 + \frac{1}{1!}\ln(x) + \frac{1}{2!}\ln(x)^2 + \frac{1}{3!}\ln(x)^3 + ... $$

To find the $\log$-series one has two strategies effectively but (the question being asked here) sometimes they conflict and it's not clear why.

The substitution and then taylor method...

You start with some $f(x)$ that you want a log expansion for. You then consider the function $f(e^u)$ and compute its taylor series centered at 0 the normal way in terms of powers of $u$, you then make the substitution $u = \ln(x)$ to get back where you started. Now sometimes you might encounter divergent sums as a result, if that happens, you can do some totally non rigorous ramanujan summation / other renormalization and still get away with it.

Example: to find the log series $x$ we consider $e^u$ and then look at it's taylor series

If $f(x) = x$ $$ e^u = 1 + u + \frac{1}{2!}u^2 + \frac{1}{3!}u^3 + ... $$ $$ x = 1 + \log(x) + \frac{1}{2!}\log(x)^2 + ... $$

If $f(x) = \frac{1}{1-x}$ $$ \frac{1}{1-e^u} = -\frac{1}{u} + \frac{1}{2} - \frac{1}{12}u + \frac{1}{720}u^3 ... $$

$$ \frac{1}{1-x} = -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 ... $$

The taylor then substitution method...

You can also go the other way. You start with a power series, replace each instance of $x^n$ with $e^{n \ln(x)}$ and now you expand these things into a massive power series and collect terms with common powers of $\ln(x)$. As usual if you encounter divergent series you ramanujan sum them (as long as they respect indices which is a handwavy idea i vaguely understand that probably has a rigorous interpretation)

if $ f(x) = x $

$$ f(x) = e^{\ln(x)} = 1 + \frac{1}{1!} \ln(x) + \frac{1}{2!}\ln(x)^2 + ... $$

if $ f(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 ... $

$$ f(x) = 1 + e^{\ln(x)} + e^{2 \ln(x)} + e^{3 \ln(x)} + ... $$

$$ f(x) = 1 + \frac{1}{0!} \left(1 + 1 + 1 + ... \right) + \frac{1}{1!} \left( 1 + 2 + 3 + .... \right) \ln(x) + \frac{1}{2!} \left(1^2 + 2^2 + 3^2 + ... \right)\ln(x)^2 + ... $$

Now if you know anything about divergent sums then you realize this is saying

$$ f(x) = 1 - \frac{1}{2} - \frac{1}{12} \ln(x) ... $$

But obviously

$$ \frac{1}{2} - \frac{1}{12} \ln(x) + \frac{1}{720}\ln(x)^3 ... \ne -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 .. $$

Theres a missing laurent term of $\frac{1}{\ln(x)}$. Why did that happen?

Inspired by some weirdness found in my answer here: Why does $1+2+3+\cdots = -\frac{1}{12}$?

  • You are rearranging series which may be conditionally convergent; by the Riemann series theorem, that could leave you with anything whatsoever. You are also assuming uniqueness of these “log-Laurent” series, which requires justification. Divergent summation techniques are also arbitrary and cannot be expected to give well-behaved results without further justification either! – FShrike Aug 22 '22 at 22:44
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    I realize how stupid this sounds but i'm gonna say it anyways: theres a way to make this all make sense, i don't think the results i'm arriving at are arbitrary but i definitely don't understand all the rules to the game i'm playing either. And perhaps by asking this question i'm indirectly asking "what are the most fundamental rules here?" – Sidharth Ghoshal Aug 22 '22 at 22:45
  • Hmm, I'm doing similar Sisyphus-work for a longer time already and do as well have not yet a firm answer. What I'm trying is to understand that the use of the exponential series might require to consider a leading (infinitesimal) term $\exp(x)=x^-1/(-1)! + 1 +x + x^2/2! + ...$ which does usually not interfere with common formulas including $\exp(x)$. But for formula where $1-exp(x)$ is in the denominator, or in context with summing Bernoulli-numbers/zetas at negative argument, this infinitesimals become significant when formally multiplied by $\zeta(1)$. Well that is one (... ) – Gottfried Helms Sep 19 '22 at 13:11
  • (...) thing I'm trying to find its appropriateness, its weak aspects and so on. Don't remember whether I have linked to my initial fiddling with this at one of your questions already, but if not here is the link: https://go.helms-net.de/math/divers/ProblemWithBellmatrix.pdf This only throws in the problem as it occured in some analysis involving series of $ 1/(\exp(-x)-1)$ terms... – Gottfried Helms Sep 19 '22 at 13:19

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