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(1) I was trying to down an equation in $$S,D$$ (these are the shift and differential operators, respectively) satisfied by

$$\sum_{n=0}^{\infty}n!x^n$$

Mainly, I am trying to interpret these and how they are relevant in solving this problem. Thanks in advance

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up vote 1 down vote accepted

Denote by $P$ your power series.

I'm not sure exactly what the shift operator is. Let $L$ denote shift left and $R$ denote shift right, e.g. $$ LP = \sum_{n=0}^\infty (n+1)!x^n, \quad RP = \sum_{n=0}^\infty n!x^{n+1}. $$ We also have $$ DP = \sum_{n=0}^\infty ((n+1)!-n!)x^{n-1}. $$ So, for example, $RDP = (L-1)P$. If you want something involving only the left shift, you can apply the left shift on both sides to get $DP = (L^2-L)P$.

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