1
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For example,

SeriesCoefficient[ArcCos[d], {d, 0, n}] returns:

enter image description here

,while

SeriesCoefficient[ArcCos[d]^2, {d, 0, n}]

returns:

enter image description here

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1
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Workaround:

func = InverseZTransform[D[ArcCos[a x]^2, a] /. x -> 1/x, x, n]
func2 = FullSimplify[Limit[Integrate[func // FullSimplify, a], a -> 1], 
Assumptions -> {n >= 1, n \[Element] Integers}]

(*((1 + (-1)^n) Sqrt[\[Pi]] Gamma[1 + n/2])/(n^2 Gamma[(1 + n)/2]) + (
I^(1 + n) \[Pi] Binomial[-(1/2), 
1/2 (-1 + n)] UnitStep[-Mod[1 + n, 2]])/n*)

UnitStep[-Mod[1 + n, 2]] is equal to: 1/2 (1 + (-1)^(1 + n)) then:

func3 = ((1 + (-1)^n) Sqrt[\[Pi]] Gamma[1 + n/2])/(n^2 Gamma[(1 + n)/2]) + (
I^(1 + n) \[Pi] Binomial[-(1/2), 1/2 (-1 + n)] 1/2 (1 + (-1)^(1 + n)))/n;

func4 = FullSimplify[func3, Assumptions -> {n >= 1, n \[Element] Integers}]

(*((-1)^n Sqrt[\[Pi]] Gamma[n/2])/(n Gamma[(1 + n)/2])*)(*Solution*)

Check:

Pi^2/4 + Sum[func4*x^n, {n, 1, Infinity}] /. x -> 1/2 // N
(*1.09662*)
ArcCos[x]^2 /. x -> 1/2 // N
(*1.09662*)
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  • $\begingroup$ Thank you for your answer! Is there a more general way that one can use for more complicated expressions ? $\endgroup$ – Pao Sep 27 '19 at 9:01
  • $\begingroup$ For more complicated expressions, probably, no way. $\endgroup$ – Mariusz Iwaniuk Sep 27 '19 at 15:28
  • 1
    $\begingroup$ @Pao Do you have examples of the more complicated expressions? $\endgroup$ – nilo de roock Nov 19 '19 at 8:36

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