# How does Mathematica take a Series expansion at Infinity?

I have a program that requires a series expansion at infinity and at a finite horizon, and I have posted two questions about simplifying these expressions or obtaining the series expansion recently. However, I now realize that I am a little confused about how this expansion even works in Mathematica, since I do not understand some of the test results I have, and I have gone through a few questions such as 1, 2 but I am still a little confused.

Consider the following examples I am obtaining (running Mathematica 11.3 on Windows 10)

In:  Series[Sin[x],{x,x0,3}]
Out: Sin[x0]+Cos[x0] (x-x0)-1/2 Sin[x0] (x-x0)^2-1/6 Cos[x0] (x-x0)^3+O[x-x0]^4

In:  Series[f[x],{x,x0,3}]
Out: f[x0]+(f^\[Prime])[x0] (x-x0)+1/2 (f^\[Prime]\[Prime])[x0] (x-x0)^2+1/6 (f^(3))[x0] (x-x0)^3+O[x-x0]^4


These examples I understand. However, at infinity:

In:  Series[Sin[x],{x,\[Infinity],3}]
Out: Sin[x+O[1/x]^5]


Since sin(x) is bounded, everywhere, I do not understand what this output means.

In:  Series[f[x],{x,\[Infinity],3}]
Out: f[x]


At infinity, I do not get symbolic series expansions as well. I understand that the regular Taylor series expansion definition around infinity is not meaningful, however, if you look at the second Mathematica notebook in this link (titled Amplification factors of the superradiant scattering of a neutral bosonic wave of generic spin off a Kerr BH, obtained by solving the Teukolsky equations), there is a section of the program that performs a symbolic series expansion at infinity, and I am able to run it and get results as well. So I'm confused about the conditions under which series expansions at infinity return meaningful results, since I think I do not know how to either give the right input, or understand the output from Mathematica. I'm quite new to this.

For reference, the actual code I need help with is here where the Series command just runs indefinitely with no results, or errors.

• The sine function has an essential singularity at infinity. Not much Series can do about that. Feb 10, 2022 at 20:16

Let us consider an example, where the power expansion at infinity exists, e.g.

Normal[Series[(x - 1)/(3 x + 1), {x, x0, 3}]]


$$\frac{36 (x-\text{x0})^3}{(3 \text{x0}+1)^4}-\frac{12 (x-\text{x0})^2}{(3 \text{x0}+1)^3}+\frac{4 (x-\text{x0})}{(3 \text{x0}+1)^2}+\frac{\text{x0}-1}{3 \text{x0}+1}$$

The above makes no sense at infinity because no arithmetical operation with Infinity is defined. The limit of the above at infinity, i.e. 1/3, is too poor. I think Mathematica follows usual way of calculus by the change of a variable

Normal[Series[(x - 1)/(3 x + 1) /. x -> 1/y, {y, 0, 3}]] /. y -> 1/x


$$-\frac{4}{81 x^3}+\frac{4}{27 x^2}-\frac{4}{9 x}+\frac{1}{3}$$

The above is identical with the result of

Normal[Series[(x - 1)/(3 x + 1), {x, Infinity, 3}]]

• What about symbolic expansions, as in the examples I have cited? Because in the code I have, the kernel just resets automatically after running for about an hour. Feb 10, 2022 at 15:18
• How long did you run it for, because for me it just runs indefinitely and I have tested it on another Windows machine as well, running Mathematica 12.3. If I set \[Epsilon] = \[delta] =0 then, I get sensible results (which is a physically well known base case). But for any other value of \[Epsilon],\[delta] or if I run it directly as in the code, I do not get the error message you seem to be getting. It runs for an hour, and the kernel resets. Feb 10, 2022 at 16:14
• @newtothis: I deleted my previous comment because I didn't execute ORDINF = 4; in your code. Your eqnaos is huge and serinf = Series[{eqni} //. ruleINF, {r, \[Infinity], ORDINF}] // Chop is running on my comp without any response for a long time. Try to evaluate some parameters in eqnaos. Feb 10, 2022 at 16:23
• For any non zero values of the parameters, it still takes too long. The physical ranges of the values are between 0 and 1. So I gave decimal values for \[Epsilon],\[delta],\[Gamma]1. That does not seem to help. Simplifying eqnaos also takes too long. Feb 10, 2022 at 16:36
• I should also point out that I do a similar series expansion at a symbolic (finite) value of r, and the whole series expansion runs smoothly and gives me an output in around 10-15 mins, without having to evaluate the various parameters in eqnaos Feb 10, 2022 at 16:40