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When I ask SeriesCoefficient[] to give me a general term, it outputs a response with head "Piecewise" that contains the general term in the first line and (every time I tried it) "0 True" in the second line. SeriesCoefficient[][[2]] brings back the '0' but not 'True'. There is no [[3]].

My hunch is that this device is used for many functions (answering, say, "Is there valid output at input '0'?) but tbh I have no idea. Thanks, TB

Example (from documentation):

SeriesCoefficient[Exp[-x], {x, 0, n}]

Output: enter image description here

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    $\begingroup$ Are you asking what the second entry in piecewise means? It means for any other n, the result is zero. i.e. for n<0 the result is zero. True means for everything else. So in this case, everything else is n<0. To find the internal structure and levels of piecewise result, you can look at its FullForm. $\endgroup$
    – Nasser
    Commented 15 hours ago
  • $\begingroup$ In general, a series can start or end with any term. The Piecewise covers all possibilities. Look at f[x_] = Exp[-x]; Sum[SeriesCoefficient[f[x], {x, 0, n}] x^n, {n, -Infinity, Infinity}] == f[x] $\endgroup$
    – Bob Hanlon
    Commented 14 hours ago

1 Answer 1

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Second argument stands for default value of the Piecewise function when none of the conditions holds. Such a behavior of SeriesCoefficient[...] might be helpful, e.g., for the successive summation

Sum[SeriesCoefficient[Exp[-x], {x, 0, n}], {n, -\[Infinity], \[Infinity]}] (*=>1/E*)

Still, this output of SeriesCoefficient[...] is a bit silly as the Piecewise evaluates to non-zero value even when n is non-integer positive. E.g. try

Sum[SeriesCoefficient[Exp[-x], {x, 0, n}] /. n -> k + 1/2, {k, -\[Infinity], \[Infinity]}]

If you want to avoid Piecewise, you can use

SeriesCoefficient[Exp[-x], {x, 0, n}, Assumptions->n>0] (*=> (-1)^n/n!*)
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  • $\begingroup$ Thank you! I will explore these. TB $\endgroup$
    – Tom Barson
    Commented 14 hours ago

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