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Question: asymptomatic series expansion

Question:asymptomatic series expansion

Adam Ledger 360 Maple
series riemann-zeta
November 26 2016
1

so yep pretty self explainatory, i was just wondering why it works when i use the asymptomatic expansion but not the iterated reciprocal substitution code shown in the that help interface for asympt.

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Zeta(1/2+I*y)

Zeta(1/2+I*y)

 (1)
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subs(y = 1/y, series(subs(y = 1/y, Zeta(1/2+I*y)), y = 0, 6))

Error, (in series/Zeta) unable to compute series

  
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asympt(Zeta(1/2+I*y), y, 6)

1+exp(-I*ln(2)*y)+exp(-I*ln(3)*y)+exp(-(2*I)*ln(2)*y)

 (2)
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