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The following example is from Calculus, 7e by James Stewart:

Example 2, Chapter 11.9 (Representations of functions as Power Series)

Find a power series representation for $\frac{1}{x+2}$

My solution is:

$$\frac{1}{x+2}=\frac{1}{1-\left(-x-1\right)}=\sum_{n=0}^{\infty}\left(-x-1\right)^{n}=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(x+1\right)^{n}$$

Then to find the interval of convergence

$$\left|x+1\right|<1 \Rightarrow x\in\left(-2,0\right)$$

But the given solution is different:

“In order to put this function in the form of the left side of Equation 1, $\frac{1}{1-x}$ , we first factor a 2 from the denominator:

$$\frac{1}{2+x}=\frac{1}{2\left(1+\frac{x}{2}\right)}=\frac{1}{2\left[1-\left(-\frac{x}{2}\right)\right]}=\frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^{n}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{2^{n+1}}x^{n}$$

This series converges when $\left|-\frac{x}{2}\right|<1$, that is, $\left|x\right|<2$. So the interval of convergence is $\left(-2,2\right)$."

So my questions are:

  1. Did I make an error somewhere?
  2. If not, are the two representations equivalent? Can there be more than one representation of a function as a power series?
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    $\begingroup$ The representation as a power series at a given point is unique. Here, yours is at the point $-1$, while the solution gives the one at the point $0$. (By default, power series are usually taken at 0 when nothing is specified or obvious from context) $\endgroup$ – Clement C. Dec 29 '18 at 21:59
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I think your method is right but you made a power series around $$x_0=-1$$

while they did it around $$x_0=0$$ which is what they asked for probably

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Both solutions are correct: your series is a power series centered at $-1$, whereas the power series from the given solution is centered at $0$.

As a result, the question is ambiguous. Are you sure that the question contains no reference to the center of the series?

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  • $\begingroup$ The question doesn't contain any reference to the center of the series - they do discuss it a bit later in the chapter. I guess since it was an example, they figured the given information was adequate, but I try to do all the examples myself before seeing their solution $\endgroup$ – ghufran syed Dec 29 '18 at 22:27

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