- #1
JG89
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Homework Statement
Find the Taylor series about the point x = 0 for the function [tex] \frac{1}{3-2x^3} [/tex]
Homework Equations
The Attempt at a Solution
[tex] \frac{1}{3 - 2x^3} = \frac{1}{3(1 - \frac{2x^3}{3})} [/tex]. Let [tex] u = \frac{2x^3}{3} [/tex]. Then [tex] \frac{1}{3(1 - \frac{2x^3}{3})} = \frac{1}{3} \frac{1}{1 - u} = \frac{1}{3} (1 + u + u^2 + u^3 + ... )= \frac{1}{3} (1 + \frac{2x^3}{3} + \frac{4x^6}{9} + \frac{8x^9}{27} + ...) [/tex].I learned that if a function can be expanded in a power series with the point x = 0 in its interval of convergence, then that power series expansion is also the expansion for the Taylor series, so this seems correct to me. Any mistakes?
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