# Thread: taylor expansion of arctan(x) at infinity

1. ## taylor expansion of arctan(x) at infinity

$\displaystyle \displaystyle \arctan(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ... \ , \ |x| \le 1$

You can't simply say that the expansion at infinity equals the expansion of $\displaystyle \arctan(1/x)$ at $\displaystyle 0$. That wouldn't make any sense.

But $\displaystyle \displaystyle \arctan(x) + \arctan \Big(\frac{1}{x}\Big) = \frac{\pi}{2}$

so $\displaystyle \displaystyle \arctan(x) = \frac{\pi}{2} - \arctan \Big(\frac{1}{x} \Big) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^{3}} - \frac{1}{5x^{5}} + ...$

Would that then be considered the expansion at infinity? And would it be valid for x >1?

2. $\displaystyle \displaystyle \arctan \Big(\frac{1}{x}\Big) = \frac{\pi}{2} -\arctan(x)$

RHS is true when |x|<1. Then LHS is true when 1/x goes to infinity.

Making substitution

y=1/x where y goes to infinity we get

$\displaystyle \displaystyle \arctan(y) = \frac{\pi}{2} - \arctan \Big(\frac{1}{y}\Big)$

this is what you get.

3. But for what values is the series valid? It would appear that it's valid for x=1.

4. It is valid for

$\displaystyle | \; 1/y \; | \leq \; 1$

and it follows that

$\displaystyle |y| \; \geq \; 1.$

5. Originally Posted by Random Variable
$\displaystyle \displaystyle \arctan(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ... \ , \ |x| \le 1$

You can't simply say that the expansion at infinity equals the expansion of $\displaystyle \arctan(1/x)$ at $\displaystyle 0$. That wouldn't make any sense.

But $\displaystyle \displaystyle \arctan(x) + \arctan \Big(\frac{1}{x}\Big) = \frac{\pi}{2}$

so $\displaystyle \displaystyle \arctan(x) = \frac{\pi}{2} - \arctan \Big(\frac{1}{x} \Big) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^{3}} - \frac{1}{5x^{5}} + ...$

Would that then be considered the expansion at infinity? And would it be valid for x >1?
The series...

$\displaystyle \displaystyle f(z)= \tan^{-1} (\frac{1}{z}) = \frac{1}{z} - \frac{1}{3 z^{3}} + \frac{1}{5 z^{5}} + ...$ (1)

... is a Laurent series and not a Taylor series!... it converges for $\displaystyle |z|>1$ and f(*) has a single essential singulatity in z=0...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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# taylor series of arctan

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