I am curious about incorporating absolute values into Taylor Series expansions and the function I want to do this for is $$f(x) = \frac{|\sin(x)|}{x}~.$$Does anyone know how to do this? Thanks.
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2$\begingroup$ I think there's no series expansions for functions which contains absolute value. $\endgroup$– FFjetCommented Sep 9, 2019 at 1:30
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4$\begingroup$ This isn't even continuously definable at $x=0$... $\endgroup$– Simply Beautiful ArtCommented Sep 9, 2019 at 1:31
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$\begingroup$ Did you even plot this function ? $\endgroup$– user65203Commented Sep 9, 2019 at 6:40
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4 Answers
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If you assume by contradiction that there exists a Taylor series $\sum_{n=0}^\infty a_n x^n$ such that $f(x)=\sum_{n=0}^\infty a_n x^n$ then $f$ is differentiable at $x=0$ and
$$a_1=f'(0)$$
But in your case $$\lim_{x \to 0} \frac{ \frac{|\sin(x)|}{x}-1}{x-0}$$ does not exist.
On another hand $\left| \frac{\sin(x)}{x} \right|$ has a Taylor series on $(-\pi, \pi)$.
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Hint: You may also write $|x|$ as $\sqrt{x^2},$ but then you need to be very careful not to forget to be tempted to write $\sqrt{x^2}=x,$ except you're sure $x$ is not negative.
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Your function has a jump discontinuity at $x=0$ so whatever you define $f(0)$ it can't be continuous. Therefore it does not have any Taylor series.
answered Sep 9, 2019 at 6:29
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The expansion around $x=a$ is
$$\text{sgn}(a)\sum_{k=1}^\infty\Re(i^ke^{ia})\frac{(x-a)^{k-1}}{k!}.$$
(The $\Re$ coefficients are one of $\pm\cos a,\pm\sin a$.) It converges to $f(x)$ for all $x$ in $(-\infty,0)$ or $(0,\infty)$, depending on the sign of $a$.
It doesn't exist for $a=0$.