
Let denote the restriction of to . Then , , and are representable by
presenting no further challenge to us. We proceed to a example which is more interesting. Let be the operator for differentiation. The following shift relations are known:
Also, the following contiguity relations are known:
Using these shift and contiguity relations, we can start from almost any representation to obtain any representation where , , . The denominators appearing in the shift relations and contiguity relations are troublesome since we can't let them become zero. We show how to compute by starting from the known formula for . Known:
Shift:
Contiguity:
Contiguity:
Shift:
Contiguity:
Shift:
Hence, we conclude:
This formula is not explicitly listed in [7]. Neither Mathematica 2.2 nor Maple 5.3 will compute it. Macsyma 419.0 does compute it, but returns a wrong answer. We will develop a strategy using shift relations and contiguity relations for where and are arbitrary (subject to ) to compute from . First, we will study shift relations and contiguity relations for general . 

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