Why has arctan in Taylor series a range from $|x|\le1$
as you can see here:
Arctan x has a range from $|x|\le1$ ,but normally it is R.
Why is it different. Thank you :)
$\endgroup$ 53 Answers
$\begingroup$$\mid x \mid \le 1$ means the Taylor representation for $\arctan{x}$ about $0$ converges only for any $x$ has absolute value less than $1$. Even if you only consider real number, domain of $\arctan{x}$ is the whole real line, only the range is from $-\pi/2$ to $\pi/2$.
$\endgroup$ $\begingroup$You've got $x^{2n+1}$ in the series. Consider how $x^n$ grows, as $n$ approaches $\infty$. Do different things happen when $-1<x<1$ or when $x>1$ or $x<-1$?. Also, consider that for a series to converge, its terms need to get smaller. How does the rate of growth of $x^{2n+1}$ compare with that of $(2n+1)$?
$\endgroup$ 2 $\begingroup$Suppose a function has a power series representation. What this means is for some domain, we can express the function as this power series. The domain of this function can be bigger, but our representation of the function as a power series may not be valid there. As for why the representation for the taylor series of $tan(x)$ is only true for $|x| \leq 1$, consider how we derive it.
It's only for when $|x| \leq 1$
$$\frac{1}{1-x} = 1 + x + x^2 + ...$$ $$\frac{1}{1+x^2} = 1 - x +x^2 -x^3 +x^4 ...$$ Integrating both sides $$arctan(x) = x -\frac{x^3}{3} + \frac{x^5}{5} ...$$
But again, this was only true for when $|x| \leq 1$
Now once you have that this true for when $|x| \leq 1$, it's natural to ask if the domain can be extended, and it turns out that it cannot.
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