10
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I know that it is possible to expand StirlingS2[n, 10] in terms of elementary functions of n. I tried FunctionExpand but it returned unevaluated:

In[1]:= FunctionExpand[StirlingS2[n, 10]]
Out[1]= StirlingS2[n, 10]

Is there a way to do this in Mathematica?

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  • 8
    $\begingroup$ Try FunctionExpand[StirlingS2[n, 10], Element[n, Integers] && n > 0]. $\endgroup$ – chyanog May 9 '13 at 0:48
10
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You need to specify assumptions:

In[1]:= FunctionExpand[StirlingS2[n, 10], n > 0 && Mod[n, 1] == 0]
Out[1]= -(1/362880) + 2^(-8 + n)/315 + 1/135 2^(-7 + 2 n) + 
 1/315 2^(-8 + 3 n) - 3^(-3 + n)/1120 + 
 1/5 2^(-7 + n) 3^(-3 + n) - 5^(-2 + n)/576 + 
 1/567 2^(-8 + n) 5^(-2 + n) - 7^(-1 + n)/4320 - 9^(-2 + n)/4480
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