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Math Help - Simple issue using Taylor's Thm

  1. #1
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    Simple issue using Taylor's Thm

    I'm trying to figure out how to show
    1-x^2\leq e^{-x^2}

    I am using f(x) = e^{-x^2} and x_0 = 0.

    I'm getting the 1 for the first term but then I keep getting zeroes, am I approaching this wrong?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    I'm trying to figure out how to show
    1-x^2\leq e^{-x^2}

    I am using f(x) = e^{-x^2} and x_0 = 0.

    I'm getting the 1 for the first term but then I keep getting zeroes, am I approaching this wrong?
    e^{-x^2}=1-x^2+R_2. What's the remainder term and why is it greater than zero?
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  3. #3
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    I don't know why I'm not getting that -x^2 term, but then R(x) should then be \frac{2c^2 e^{-c^2}}{2!} (x^2), which is obviously a positive term
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    I don't know why I'm not getting that -x^2 term, but then R(x) should then be \frac{2c^2 e^{-c^2}}{2!} (x^2), which is obviously a positive term
    Taylor series for e^x is 1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \dots.

    Just sub -x^2 into that. Second term is therefore \frac{-x^2}{1!} = -x^2.
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  5. #5
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    I was given a hint to split this into 2 cases, |x| \leq 1 and |x|>1 , but I can't figure out why to do this?
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