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I was given the function: $f(x) = 1/(1+x)^2$ and its Taylor series: $1 - 2x + 3x^2 - 4x^3 + \cdots$

In order to get the Taylor series for the closely related function $1/((1/2)+x)^2$, I simply substituted $(x-(1/2))$ for $x$ in the Taylor series given to me:

$$1-2(x-(1/2)) + 3(x-(1/2))^2 - 4(x-(1/2))^3 +\cdots$$

This seems straight forward enough, but xMaxima gives me a different answer for the Taylor series of $f(x)$, it gives this:

$$4-16\cdot x+48\cdot x^2-128\cdot x^3+320\cdot x^4+\cdots$$

I also get this answer if I represent $f(x)$ as $1/(1/2)^2 \cdot 1/(1+2x)$ and then substitute the $x$'s for $2x$'s in the Taylor series before multiplying each term by $1/(1/2)^2$

What's going on here...Why is the first simpler approach giving me something different, and probably incorrect..

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    $\begingroup$ Yours is the Taylor series of $1/(1/2+x)^2$ with centre $1/2$, the other is the Taylor series with centre $0$. $\endgroup$ – Daniel Fischer Aug 19 '15 at 18:03
  • $\begingroup$ Is it the Maclaurin series that you seek? $\endgroup$ – user230734 Aug 19 '15 at 18:05
  • $\begingroup$ Ah, yes - You're both correct. I can see now that I didn't achieve the Maclaurin series because it's of the form (x-a) where a is (1/2). Thanks for that. a should be zero. $\endgroup$ – COOLBEANS Aug 19 '15 at 18:09
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To see this, do:

$$(\frac12 + x)^2 = \frac14(1 + 2x)^2$$

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