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<DIV>What you need, in general, is to compare as many members as the sum of the degree of the numerator and the denominator, plus 1. The numerator provides an initial condition, and the denominator provides a recurrence.</DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: Mitch Harris <Harris.Mitchell@mgh.harvard.edu><BR>
<DIV class=AOLPlainTextBody id=AOLMsgPart_0_7e2c5fd2-4305-4b16-bf6c-3cedc2a29052>>- if we can be sure both sequences have a rational g.f., it suffices <BR>> to compare about twice as much members as the degree of the g.f. <BR>denominator. <BR>> I think one can prove that. (It also applies to your example) <BR> <BR>how about exactly as many members as the degree? the degree of the <BR>denominator is the degree of the corresponding recurrence, which only <BR>requires degree # of base cases. <BR> <BR></DIV><!-- end of AOLMsgPart_0_7e2c5fd2-4305-4b16-bf6c-3cedc2a29052 --></DIV></DIV>
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