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Math Help - Proving the Taylor Series of Sin(x)

  1. #1
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    Proving the Taylor Series of Sin(x)

    Could someone help me with this problem? Thanks.

    "Prove that the Taylor series of sin(x) at a=pi/2 represents sin(x) for all x."

    I already know how to find the Taylor series, but I have no idea how to prove anything.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Leaf View Post
    Could someone help me with this problem? Thanks.

    "Prove that the Taylor series of sin(x) at a=pi/2 represents sin(x) for all x."

    I already know how to find the Taylor series, but I have no idea how to prove anything.
    Use Taylors' remainder theorem

    Taylor's theorem - Wikipedia, the free encyclopedia

    Note that for all x \in \mathbb{R}

    |\sin(x)|\le 1 and |\cos(x)|\le 1. Also that for each x that the remainder (the error) goes to 0 \text{ as } n \to \infty
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    I don't think it works. The Taylor Series only represents sin(x) when x is between (pi/2,pi/2] when the problem asks for all x. I can't prove something that's not true.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Leaf View Post
    I don't think it works. The Taylor Series only represents sin(x) when x is between (pi/2,pi/2] when the problem asks for all x. I can't prove something that's not true.
    I think you are mistaken. Both \sin(x) and \cos(x) are real analytic entire functions.

    Consider the remainder term from the link above it gives

    \displaystyle R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}\left(x-\frac{\pi}{2} \right)^{n+1}

    Since all derivatives of \sin(x) are less than 1

    \displaystyle |R_n(x)| \le \frac{1}{(n+1)!}\left|x-\frac{\pi}{2} \right|^{n+1}

    now for any y \in \mathbb{R} we can choose an n so large that the remainder is as small as we wish

    \displaystyle |R_n(y)| \le \frac{1}{(n+1)!}\left|y-\frac{\pi}{2} \right|^{n+1} < \epsilon
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  5. #5
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    Alright, thanks for the help. I think I got it now.
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