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Math Help - Inequality from a Taylor series

  1. #1
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    Inequality from a Taylor series

    Hey everyone.

    I am working through a Taylor expansion of sin(x). Can anyone explain to me how we can draw this inequality from the Taylor series?

    The Taylor series of sin(x), around c=0, is sin(x) = x - x^3/3! + x^5/5! +...

    So, for x >= 0 : x - x^3/3! <= sin(x) <= x.

    Thanks!
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  2. #2
    ynj
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    Quote Originally Posted by cgiulz View Post
    Hey everyone.

    I am working through a Taylor expansion of sin(x). Can anyone explain to me how we can draw this inequality from the Taylor series?

    The Taylor series of sin(x), around c=0, is sin(x) = x - x^3/3! + x^5/5! +...

    So, for x >= 0 : x - x^3/3! <= sin(x) <= x.

    Thanks!
    Taylor series cannot prove this inequality.The right way to solve this problem is to contruct a function like g(x)=\sin x-x and use the derivation.
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