Talk:Formal power series: Difference between revisions
→Multiplication and division
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: I hope multiplication of the Ada solution is correct. As for division, it is impossible to implement because the result can infinite (example: 1-''x'') or non-existent (example: ''x''). --[[User:Dmitry-kazakov|Dmitry-kazakov]] 15:14, 10 March 2009 (UTC)
:: Division introduces two problems. One problem has to do with remainders (as in the 1/(1-x) or 1/x cases suggested by [[User:Dmitry-kazakov|Dmitry-kazakov]]), and the task specification did not specify how they should be treated. A related but deeper problem is that the quotient is not knowable unless every element can be inspected (or treated in some symbolic fashion). In other words, the Kth element of the result depends on the Jth element of the divisor and the (J+K)th element of the numerator for arbitrary K and J. In the general case, significant examples of both J and J+K can be infinite for any finite K. This problem becomes tractable when we deal with finite sequences (which, in essence, is what lazy evaluation gives us, though without the tractability). --[[User:Rdm|Rdm]] 19:32, 22 January 2010 (UTC)
It is like normal multiplication of polynomials, but done on (potentially) infinite elements (we shall truncate the series anyway at some point...) It is enough to do the right grouping according to the ''power'', so say you have (a<sub>0</sub> + a<sub>1</sub>x + ...) and (b<sub>0</sub> + b<sub>1</sub>x + ...) then you do what you would normally do: a<sub>0</sub> with all the b<sub>i</sub>x<sup>i</sup> and so on; but of course you must group equal powers. You could write it shortly as
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