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By modifying the Maclaurin series of $\frac{1}{1-x}$, how can I obtain the Maclaurin series for $\frac{-2}{1+4x^3}$?

coolkid3127
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1 Answers1

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We know $$\frac{1}{1-t} = 1 + t + t^2 + \cdots$$

Just take $t = -4x^3.$ Then

$$\frac{1}{1+4x^3} = \frac{1}{1-(-4x^3)} = 1 + (-4x^3) + (-4x^3)^2 + \cdots = 1 - 4x^3 + 16x^6 -\cdots.$$

Multiplying by $-2$ gets your series.

  • Thanks! Would this mean that 1 is the first term, (-4x^3) is the second term and so on? – coolkid3127 May 27 '21 at 00:47
  • That's a little bit ambiguous because it's not clear whether the zero terms in the power series are included, so it's not clear whether they want the first four nonzero terms or if they want a Maclaurin polynomial of order $3.$ That said I would assume they mean the first, so yes that would be correct. (but remember to include the extra factor of $-2$) – Stephen Donovan May 27 '21 at 00:51
  • So -2, (-2)(-4x^3), (-2)(-4x^3)^2, (-2)(-4x^3)^3? – coolkid3127 May 27 '21 at 00:55