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

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  • $\begingroup$ Is i the imaginary unit in your expression? $\endgroup$ – march Oct 22 '15 at 22:59
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    $\begingroup$ You can try raising the result of expanding the square root to I... $\endgroup$ – J. M. will be back soon Oct 22 '15 at 23:00
  • $\begingroup$ @march yes. It's $\sqrt{-1}$ $\endgroup$ – Physics Moron Oct 22 '15 at 23:14
  • $\begingroup$ @J.M. Actually I tried that. But didn't get anything useful. $\endgroup$ – Physics Moron Oct 22 '15 at 23:24
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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@%

enter image description here

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@%

enter image description here

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  • $\begingroup$ Thanks! It looks good. But probably I should check it by hand once.. $\endgroup$ – Physics Moron Oct 22 '15 at 23:41
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    $\begingroup$ Actually, Series[(1 + 1/x)^(1/2), {x, 0, 3}]^I // Normal is what I'd have done. ;) $\endgroup$ – J. M. will be back soon Oct 23 '15 at 0:14
  • $\begingroup$ @J.M. Well of course you'd have done something smarter than me. :) $\endgroup$ – march Oct 23 '15 at 1:18
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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) *)
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  • $\begingroup$ Then I need to replace c by x? $\endgroup$ – Physics Moron Oct 22 '15 at 23:35
  • $\begingroup$ @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. $\endgroup$ – thils Oct 22 '15 at 23:38

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