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Homework Help: Identifing an interval

  1. Dec 6, 2008 #1
    Identify an interval [-a,a] so that
    |arctan(x)-x+x^3/3| < or = 10^-15
    for all x E [-a,a]. Your result needs to be supported by a convincing proof

    I'm not sure how to go about this... can someone help me out?
    all i could probably do is... since.. 10^-15 is close to zero, i can find out what the x's are equal to... but, i don't know what else i can do.

    How would i be able to find the x's if there's an arctan? Is there any trig identities that i use?

  2. jcsd
  3. Dec 6, 2008 #2


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    Hint: as you say that 10^{-15} is so close to zero, so probably will the boundary x's have to be. So try Taylor-expanding the arctan around x = 0. You might want to go up to 5th order :wink:
  4. Dec 6, 2008 #3
    I'm not sure, how i'm suppose to find the boundaries of x... i got down to
    |x^5/5| [tex]\leq[/tex] 10^-15

    Is -a and a equal to.. -10^-15 and 10^-15 respectively?
  5. Dec 6, 2008 #4


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    If x^5/5 has to be smaller than 10^(-15), i.e. [tex]\frac{x^5}{5} \in [-10^{-15}, 10^{-15}][/tex], then which interval can x itself lie in?
  6. Dec 6, 2008 #5


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    So you want to solve the inequality -10^{-15}< x^5/5< 10^{-15}.
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