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Talk:Formal power series: Difference between revisions
→Multiplication and division
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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)
:: There are several real problems. One is that we can't expand in 0 if the function (or any of the derivative) diverges in 0. Instead of sin, use the 1/x function itself to find its coefficients "directly" the same way you use for sin... Another problem is that we expand in powers of x where the exponent is positive, so it exists no expansion for x<sup>-1</sup>. We should drop Taylor series and use Laurent series (always with care I suppose). If we can "generate" the Taylor series in 0 for r(x), being r(x) = f(x)/g(x), then we can divide the expansion of f(x) (its formal power series) by the one for g(x). There are still problems anyway; e.g. if it exists N so that f<sub>i</sub> = 0 for i>N, and similar for g<sub>i</sub> but with M, and N<M, what happens? (Similar to 1/x case I suppose), at a glance this situation gives problem too. But this still does not say that division is '''always''' impossible. It is "sometimes"! --[[User:ShinTakezou|ShinTakezou]] 19:15, 11 March 2009 (UTC)
==About solving equations==
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