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Given that the Maclaurin series for $g(x) = \frac{1}{1+x}$ is $1 - x + x^2 - x^3 + x^4 ... $, I'm told that the Maclaurin series for $\frac{1}{1+x^2}$ is $1 - x^2 + x^4 - x^6 ... $, by substituting in $x^2$.

What I'm confused about is why this is valid for this case specifically and what rules govern when you can and can't do something like this -- for example, you couldn't do it with the series for $x$ itself and substitute in, say, $cos(x) = x$, right?

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  • $\begingroup$ Technically, $\cos(x)$ is a "series" in the variable $\cos(x)$ which sums up to $\cos(x)$. But it isn't a power series. $\endgroup$
    – Ian
    Commented May 20, 2015 at 23:08
  • $\begingroup$ In general I think you are asking when you can assume a power series form of solution. The mathematicians' answer is that you have to know that the result is analytic (which in practice is a very strong assumption). The physicists' answer is more complicated and usually more ad hoc. (Either they just assume it blindly, they check it rigorously, or actually it doesn't work but there is some regularization technique that makes it "morally" work.) $\endgroup$
    – Ian
    Commented May 20, 2015 at 23:09
  • $\begingroup$ So that wouldn't even be a Maclaurin series, right? Which means the substitution doesn't work there -- which ties back to my question of when is and isn't a substitution like that valid? $\endgroup$ Commented May 20, 2015 at 23:09
  • $\begingroup$ It's just as valid as ever...but it doesn't get you a power series. Are you asking about when the result will get you a power series? $\endgroup$
    – Ian
    Commented May 20, 2015 at 23:10
  • $\begingroup$ Yeah, I suppose -- more generally, when will performing such a substitution get a result that is a correct power series, and when will such a substitution be useful in solving problems? $\endgroup$ Commented May 20, 2015 at 23:13

1 Answer 1

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The point is that $$ \sum_{n=0}^{\infty} x^n= \frac{1}{1-x} $$ for all $\lvert x\rvert < 1$. So, whenever $x$ is something that has an absolute value less than $1$, then you have this equality. If you replace $x$ by $x^2$, then $$ \sum_{n=0}^{\infty} (x^2)^n = \frac{1}{1-x^2} $$ and this is true when $\lvert x^2\rvert < 1$. You also have $$ \sum_{n=0}^{\infty} (\cos(x))^n = \frac{1}{1 - \cos(x)} $$ for all $x$ such that $\lvert \cos(x)\rvert < 1$.

And you have $$ \sum_{n=0}^{\infty} (-x)^n = \frac{1}{1 + x} $$ for all $x$ where $\lvert -x\rvert = \lvert x \rvert < 1$ (This is the series that you are considering. These all "come from" the first series.

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