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I must admit that my understanding of Taylor and power series is flimsy at best. I am supposed to find the Taylor series for the following function centred at $x=0$:

$$\frac{1}{(1-5x)^2}$$

I honestly have no idea as to the approach to take from here. Should I look to relate this to the function $\frac{1}{1-x}$, attempt to determine the first few terms of the series and notice a pattern, or something else entirely? I'm not necessarily looking for an answer but rather an explanation as to how to best tackle problems of this nature.

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  • $\begingroup$ Hint : Differentiate $\frac{1}{1-5x}$ to get the desired relation $\endgroup$ – Peter Dec 11 '16 at 22:50
  • $\begingroup$ If one applies generalized binomial expansion on $f(x)=(1-5x)^{-2}$... $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 22:53
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HINT:

Let $f(x)=\frac{1/5}{1-5x}$ so that $f'(x)=\frac{1}{(1-5x)^2}$. Now, differentiate the geometric series for $f(x)$ term by term.

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  • $\begingroup$ So my first step is to integrate the given function? $\endgroup$ – Lanier Freeman Dec 11 '16 at 22:57
  • $\begingroup$ The first step is to find the geometric series representation for $f(x)=\frac{1/5}{1-5x}$. The second step is to differentiate that series term by term. $\endgroup$ – Mark Viola Dec 11 '16 at 22:59
  • $\begingroup$ And how might I do that? $\endgroup$ – Lanier Freeman Dec 11 '16 at 23:00
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    $\begingroup$ Lanier, the index $n$ is a discrete dummy index. After you sum, do you see its appearance? If not, then does that help answer the question? $\endgroup$ – Mark Viola Dec 11 '16 at 23:24
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    $\begingroup$ Not quite. So, $$\frac{1}{(1-5x)^2}=1+10x+75x^2+500x^3 +\cdots =\sum_{n=1}^\infty (5x)^{n-1}$$ $\endgroup$ – Mark Viola Dec 12 '16 at 2:58

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