taylor expansion of arctan(x) at infinity

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• Jan 12th 2011, 02:22 PM
Random Variable
taylor expansion of arctan(x) at infinity
$\displaystyle \arctan(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ... \ , \ |x| \le 1$

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

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

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

Would that then be considered the expansion at infinity? And would it be valid for x >1?
• Jan 12th 2011, 02:55 PM
zzzoak
$
\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 \arctan(y) = \frac{\pi}{2} - \arctan \Big(\frac{1}{y}\Big)
$

this is what you get.
• Jan 12th 2011, 03:22 PM
Random Variable
But for what values is the series valid? It would appear that it's valid for x=1.
• Jan 12th 2011, 03:59 PM
zzzoak
It is valid for

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

and it follows that

$
|y| \; \geq \; 1.
$
• Jan 12th 2011, 07:58 PM
chisigma
Quote:

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

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

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

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

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

The series...

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

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

Kind regards

$\chi$ $\sigma$