Talk:Formal power series: Difference between revisions

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 <math>\frac{1}{x}</math>
 </blockquote>
-: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
+: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 (''a''<sub>0</sub>=1, ''b''<sub>0</sub>=0). That's it, it could not be otherwise. In other cases, like also mentioned above
 <blockquote>
 <math>\frac{1}{1-x}=\sum_{n=0}^\infty x^n</math> (sum of a geometric progression |x|<1)