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Math Help - taylor expansion of arctan(x) at infinity

  1. #1
    Super Member Random Variable's Avatar
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    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?
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  2. #2
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    <br />
\displaystyle  \arctan \Big(\frac{1}{x}\Big) = \frac{\pi}{2} -\arctan(x) <br />

    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

    <br />
\displaystyle \arctan(y) = \frac{\pi}{2} - \arctan \Big(\frac{1}{y}\Big)<br />

    this is what you get.
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  3. #3
    Super Member Random Variable's Avatar
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    But for what values is the series valid? It would appear that it's valid for x=1.
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  4. #4
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    It is valid for

    <br />
| \; 1/y \; | \leq \; 1<br />

    and it follows that

    <br />
|y| \; \geq \; 1.<br />
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Random Variable View Post
     \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
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