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The MacLaurin series expansion of the function

$$f(x) = \frac{1}{1 - x}$$

is

$$f(x) = 1 + x^2 + x^3 + x^4 + \ldots = \sum_{i = 0}^{+\infty} x^i, \ (x \neq 0)$$

so all the powers of $x$ taken with a positive sign.

We have

$$\lim_{x \to 1^-} \frac{1}{1 - x} = + \infty$$

while

$$\lim_{x \to 1^+} \frac{1}{1 - x} = - \infty$$

The first limit can be easily predicted from the series expansion: when $x$ is slightly greater than $1$, all the terms sum: the final result of their sum should be $+\infty$. The series expansion becomes in fact

$$f(x)_{x = 1} = \left( \sum_{i = 0}^{+\infty} x^i \right)_{x = 1} = \lim_{i \to +\infty} i \cdot 1 = +\infty$$

But this is true for both $x \to 1^+$ and $x \to 1^-$ and it shouldn't!

1) How can the limit for $x \to 1^+$ be correctly predicted from the series expansion?

The above series is centered in $x = 0$; $x = 1$ is quite far away from that point; but with more terms included, the series should provide the exact function values over a wider range, so even $x = 1$. But the result, even with more terms, is the same: the series limit is always $+\infty$.

2) Is this a correct perspective, or the problem should be dealt with in another way?

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    $\begingroup$ geometric series works when $|x|<1$ $\endgroup$
    – E.H.E
    Oct 17 '15 at 9:48
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To get a series for values with $|x|>1$ you use a Laurent series $$ \frac1{1-x}=\frac1x·\frac1{\frac1x-1}=-\frac1x·\sum_{k=0}^\infty \frac1{x^k} $$

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    $\begingroup$ One should note that this is not a Laurent Series; Laurent series have only finitely many terms with negative exponents. In a sense, what you are doing here is expanding about $\infty$, when considering the function $f(z) = 1/(1 - z)$ as a function on $\mathbb{CP}^1$. $\endgroup$
    – Simon Rose
    Oct 17 '15 at 15:17
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The issue here is that the series expansion isn't valid for $x \geq 1$. Remember that any infinite series has a radius of convergence $R$; that is, $R$ is the value such that $$ f(x) = \sum_{n=0}^\infty a_nx^n $$ converges for $|x| < R$. In the case of the geometric series that you state, we have that $R = 1$. So attempting to evaluate the series for $x > 1$ doesn't make sense.

The point is that the equality $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$ isn't always true. It only works as long as $|x| < 1$, and so for any other values of $x$, the given statement is false.

One should note that while you get the "right" answer for $x = 1$, you should still be careful about that. In this case they agree, but it isn't always true that they will (e.g. evaluating both sides at $x = -1$ yields something false).

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  • $\begingroup$ Ok, thank you! But so, how to represent the function $1/(1 - x)$ with MacLaurin series for $x \geq 1$? If this could be too long to explain in a comment, don't worry, I will make a separate question. $\endgroup$
    – BowPark
    Oct 17 '15 at 9:56
  • $\begingroup$ As a MacLaurin series, you can't. But as Taylor series, you can. You would have to expand it around a different point, say 2 for example, and you would write $$\frac{1}{1-x} = \sum_{n=0}^\infty b_n (x - 2)^n$$ for some coefficients $b_n$. $\endgroup$
    – Simon Rose
    Oct 17 '15 at 15:15

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