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It is obvious that the integration of *R*_{q}(t) gives *R*_{q+1}(t):
and in general:
Does it imply that: (where is the inverse
of )? If we apply the operator to the sum *R*_{p+q}(*t*)+*P*_{n}(*t*),
we get for :
But , so:
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(14) |

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Stanford Exploration Project

1/13/1998