Taylor Series Expansion of $\frac{|\sin(x)|}{x}$

Question

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.

Answer 1

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.

Answer 2

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 $(-\pi, \pi)$.

Answer 3

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.

Answer 4

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 $a=0$.