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I attempt to do the series expansion for the following function

Simplify[Series[((-5)^(1/
        6) (2 + (-2 + ξ3) ξ3 + (-2 + ξ4) ξ4) a[
      1])/((-1 + ξ3)^2 ξ3^4 (-6 + 5 ξ3)^(4/
        5) (-1 + ξ4)^2 ξ4 (-6 + 5 ξ4)^(1/5)), {ξ3, 0, 
   1}], Im[ξ3] == 0]

Note that this function is expected to expand this function to order $\xi_3$. However, it fails to capture order $\frac{1}{\xi_3}$ term

enter image description here

If I slightly modify the code to expand the function to order $\xi^2$, then I could get the correct order $\frac{1}{\xi_3}$ term enter image description here

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  • $\begingroup$ I'm getting the same result for each. One is perhaps less simplified than the other, but they are equivalent. $\endgroup$ Commented May 31 at 18:32
  • $\begingroup$ @DanielLichtblau I attached my mma result. Maybe it's because my version is a bit old? It's 12 $\endgroup$
    – Vayne
    Commented Jun 3 at 6:55
  • $\begingroup$ It is possible this improved in the last few releases. I know some issues with getting correct order for Series were addressed in this time frame. $\endgroup$ Commented Jun 3 at 17:02
  • $\begingroup$ @DanielLichtblau Thanks for this info. Perhaps the best way is to update my Mathematica... $\endgroup$
    – Vayne
    Commented Jun 4 at 0:19

2 Answers 2

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I am not sure why this is happening, but here's a quick fix. Do a replacement $\xi_3 \rightarrow \tfrac{1}{x_3}$ and then expand around $\infty$.

expr = ((-5)^(1/
        6) (2 + (-2 + ξ3) ξ3 + (-2 + ξ4) ξ4) a[
      1])/((-1 + ξ3)^2 ξ3^4 (-6 + 5 ξ3)^(4/
        5) (-1 + ξ4)^2 ξ4 (-6 + 5 ξ4)^(1/5));
expr /. ξ3 :> 1/xx3;
Series[%, {xx3, Infinity, 1}]
Coefficient[%, xx3]

Blockquote

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Use Assuming.

Assuming[\[Xi]3 \[Element] Reals, 
 Series[((-5)^(1/
         6)  (2 + (-2 + \[Xi]3)  \[Xi]3 + (-2 + \[Xi]4)  \[Xi]4)  a[
       1])/((-1 + \[Xi]3)^2  \[Xi]3^4  (-6 + 5  \[Xi]3)^(4/
         5)  (-1 + \[Xi]4)^2  \[Xi]4  (-6 + 5  \[Xi]4)^(1/
         5)), {\[Xi]3, 0, 1}] // Simplify]

Blockquote

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  • $\begingroup$ Do you know why does Assuming help here? $\endgroup$
    – Vayne
    Commented May 30 at 14:27
  • $\begingroup$ @Vayne In your code, you told Simplify that Im[ξ3] == 0, but you didn't tell Series that. Assuming let both of them know that. $\endgroup$
    – Jie Zhu
    Commented May 30 at 15:09
  • $\begingroup$ Thanks! But why does the additional assumption could help series expansion capture 1/\xi term? $\endgroup$
    – Vayne
    Commented May 31 at 5:05

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