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Math Help - Convergence of the Taylor series for the sine function

  1. #1
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    Convergence of the Taylor series for the sine function

    I would like to know if the Taylor series for the sine function,

    sin(x) = x - x^3/3! + x^5/5! -...

    is convergent if the argument of the function, x , is expressed in degrees instead of radians.
    Last edited by jfpeji; November 20th 2011 at 09:26 AM. Reason: Fix formula
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Convergence of the Taylor series for the sine function

    Well wouldn't that imply that one degree equals one radian?!?
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    Re: Convergence of the Taylor series for the sine function

    Quote Originally Posted by jfpeji View Post
    I would like to know if the Taylor series for the sine function,

    sin(x) = x - x^3/3! + x^5/5! -...

    is convergent if the argument of the function, $x$, is expressed in degrees instead of radians.
    That series converges for arbitrary complex x, to \sin(x) where in the \sin the argument is to be regarded as in radians.

    A different series converges to the sin where the argument is in degrees

    CB
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    Re: Convergence of the Taylor series for the sine function

    Ok. You are right. Thank you for your answer.
    Last edited by jfpeji; November 20th 2011 at 12:37 PM.
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    Re: Convergence of the Taylor series for the sine function

    What is that different series which converges to the sin, where the argument is in degrees, that you mention?
    Last edited by jfpeji; November 20th 2011 at 10:43 AM.
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    Grand Panjandrum
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    Re: Convergence of the Taylor series for the sine function

    Quote Originally Posted by jfpeji View Post
    What is that different series which converges to the sin, where the argument is in degrees, that you mention?
    The rather obvious:

    \sin_d(x)=\left(\frac{\pi}{180}\right)x- \left(\frac{\pi}{180}\right)^3\frac{x^3}{3!}+\left  (\frac{\pi}{180}\right)^5\frac{x^5}{5!}- ...

    CB
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    Re: Convergence of the Taylor series for the sine function

    Ok. You are right. Thank you for your answer.
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