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I believe that there is no Maclaurin Series for $\frac{1}{|1+x|}$ as the latter is not differentiable at $x=-1$.

However, would it be appropriate for me to refer $\frac{1}{|1+x|}$ as 'not a smooth' curve (discontinuity at $x=1$) hence the non-existence of a Maclaurin or in any case a Taylor Series?

In addition, I am aware that we could use the definition of a modulus and have a Maclaurin Series for $x>0$ and $x\le 0$. Still, this does not constitute a Maclaurin Series for the entire function, no?

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  • $\begingroup$ if $x$ is close to $0$, one has $\frac{1}{|1+x|}=\frac{1}{1+x}=1-x+x^2-x^3+...$ and this is a Maclaurin series $\endgroup$ – user2097 Aug 30 '14 at 11:48
  • $\begingroup$ Why is this possible? When $x$ is close to $0$, the modulus sign would still have to be accounted for. Unless I am missing something here. $\endgroup$ – Joe Aug 30 '14 at 11:59
  • $\begingroup$ $x$ being close to $0$ means $1+x>0$ so $|1+x|=1+x$ $\endgroup$ – user2097 Aug 30 '14 at 12:01
  • $\begingroup$ Ok got you! Thanks! $\endgroup$ – Joe Aug 30 '14 at 12:03
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In order to write down a Taylor series for $f$ centered at $a$, we only need $f$ to have derivatives of all orders at $a$. (This, in turn, requires that for every $n$ the derivative $f^{(n)}$ be defined in some neighborhood of $a$, so that $f^{(n+1)}(a)$ makes sense.) The function $f(x)=1/|1+x|$ meets this requirement at every point $a\ne -1$.

For example, with $a=0$ the series is $$\sum_{n=0}^{\infty} (-1)^n x^n\tag1$$ which, not coincidentally, is the same series as for $1/(1+x)$. (As user2097 said, near $0$ we have $|1+x|=1+x$.)

I have not yet considered the question of whether the series actually converges to $f$. This can be addressed using the geometric series formula: when $|x|<1$, the series (1) has the sum $1/(1+x)$, which in this range agrees with $f(x)$.

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