Talk:Formal power series: Difference between revisions
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(→Multiplication and division: Still no division) |
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expanding cos/sin as you know. Then you will see that having zero coeffs is not really a problem; doing it by hand is easier than thinking about a general algo that should work for every ''situation'' —I'm thinking about but not this wine-poisoned-evening (likely someone else more talented has already done it, but there's no fun without selfmade (re)discovering). --[[User:ShinTakezou|ShinTakezou]] 22:45, 10 March 2009 (UTC) |
expanding cos/sin as you know. Then you will see that having zero coeffs is not really a problem; doing it by hand is easier than thinking about a general algo that should work for every ''situation'' —I'm thinking about but not this wine-poisoned-evening (likely someone else more talented has already done it, but there's no fun without selfmade (re)discovering). --[[User:ShinTakezou|ShinTakezou]] 22:45, 10 March 2009 (UTC) |
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:No, it does not work. As I said above, divide 1 by ''x'': |
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<math>\frac{1}{x}</math> |
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</blockquote> |
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:It just does not have Taylor series in ''x''<sub>0</sub>=0. As for the method you refer to. It is based on solving a linear equations system. The system for 1/''x'' does not have a solution (''c''<sub>0</sub>=1, ''b''<sub>0</sub>=0). That's it, it could not be otherwise. In other cases, like also mentioned above |
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<blockquote> |
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<math>\frac{1}{1-x}=\sum_{n=0}^\infty x^n</math> (sum of a geometric progression |x|<1) |
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</blockquote> |
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:the solution is infinite. And we ignored the convergence issue, so far. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 09:36, 11 March 2009 (UTC) |
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==About solving equations== |
==About solving equations== |
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