A series expansion

It might be a silly question. Actually I’m facing a problem expanding (1+2x)i2(1+\frac{2}{x})^{\frac{i}{2}} for small xx. Mathematica can not expand it. But it can of course expand (1+2x)12(1+\frac{2}{x})^{\frac{1}{2}} for small xx. Is it related to properties of the expression ( like convergence etc) or I need to use mathematica a bit carefully?

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Is i the imaginary unit in your expression?
– march
Oct 22 ’15 at 22:59

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You can try raising the result of expanding the square root to I…
– J. M.♦
Oct 22 ’15 at 23:00

  

 

@march yes. It’s √−1\sqrt{-1}
– Physics Moron
Oct 22 ’15 at 23:14

  

 

@J.M. Actually I tried that. But didn’t get anything useful.
– Physics Moron
Oct 22 ’15 at 23:24

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2 Answers
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Perhaps this, from J.M.’s hint:

Series[(1 + 1/x)^(1/2), {x, 0, 3}]
Series[(Normal@%)^I // Expand, {x, 0, 3}]
Expand@Normal@%

Or, perhaps this: (per another of J.M.’s hints):

Series[(1 + 1/x)^(1/2), {x, 0, 3}]
Series[%, {x, 0, 3}]^I
Normal@Expand@%

  

 

Thanks! It looks good. But probably I should check it by hand once..
– Physics Moron
Oct 22 ’15 at 23:41

3

 

Actually, Series[(1 + 1/x)^(1/2), {x, 0, 3}]^I // Normal is what I’d have done. 😉
– J. M.♦
Oct 23 ’15 at 0:14

  

 

@J.M. Well of course you’d have done something smarter than me. 🙂
– march
Oct 23 ’15 at 1:18

Is this acceptable where you introduce the factor “c”?

Series[(1 + 2/x)^(I/3), {x, c, 4}]

(* (1 + 2/c)^(I/3) – (2 I (1 + 2/c)^(-1 + I/3) (x – c))/(3 c^2) + (
2 I (1 + 2/c)^(I/3) ((3 + I) + 3 c) (x – c)^2)/(9 c^2 (2 + c)^2) – (
2 I (1 + 2/c)^(I/
3) ((34 + 18 I) + (54 + 18 I) c + 27 c^2) (x – c)^3)/(
81 c^3 (2 + c)^3) + ….*)

Limit[%, x -> 0]

(* (((2 + c)/
c)^(I/3) ((3260 +
2640 I) + (6912 + 4704 I) c + (5508 + 2916 I) c^2 + (1944 +
648 I) c^3 + 243 c^4))/(243 (2 + c)^4) *)

  

 

Then I need to replace c by x?
– Physics Moron
Oct 22 ’15 at 23:35

  

 

@pinu The meaning of “c” here is the point at which you expand the series on “x”. So you reach singularity if you put “c=0”, maybe try a small number for c.
– thils
Oct 22 ’15 at 23:38