How does Mathematica find a series expansion of expressions containing logarithms when there is a singularity at the expansion point?

Question

I am looking for a good approximation to a function containing logarithms, especially at values close to zero. When I used Mathematica's Series command I found something unexpected.

Mathematica seems to expand expressions containing a logarithm differently when there is a singularity at the expansion point. For example, the function $(a+\log[x^3+x^7])/(b+x)$ has a singularity at $x=0$. If I use Series with an expansion point greater than zero, I get the expected Taylor expansion. However, at the expansion point zero I get

$$ \frac{a + 3 \log(x)}{b} + \frac{(-a - 3 \log(x)) x}{b^2} + O(x)^2. $$

It strikes me that the logarithm is retained as a function in $x$. And this result approximates the function much better than that of a simple Taylor series. Mathematica somehow separates the logarithm from the expansion, which I find quite intelligent. I would like to know what technique Mathematica uses and how it is called. I never came across such a thing and would be interested in learning more about it.

I understand that Series can use several expansions and choses an appropriate one depending on the problem, so maybe this is a special procedure to treat expressions with logarithms.

I also found (with some trial and error) that Mathematica's result equals the one I would get if I first did a Taylor expansion of the expression, but treating all logarithms as constants, and then reduced all logarithmic arguments to the lowest power of $x$. Why is this justified?

Answer 1

One good thing about Mathematica is that when doing Series[], Mathematica understands that the singularity of $log(x)$ is different from $x^{\alpha}$ (for any $\alpha$) (for $x \rightarrow 0$), thus treated separately.

Like this:

Series[Log[x^2 + x], {x, 0, 1}]

Output

Log[x]+x+O[x]^2

Because mathematically, $$ \log(x^2+x) = \log(x (x+1)) = \log(x) + \log(x+1) = \log(x) + x + O(x^2) $$

The same procedure applies to your slightly complex expression. i.e. take out the common factor in $\log(\cdot)$, (conceptually) expand everything except $\log(x)$, truncate the Taylor series based on the hierarchy (for $x \rightarrow 0$)

$$ \cdots,\quad x,\quad x\log(x),\quad 1,\quad\log(x),\quad x^{-1},\quad \cdots $$