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Innequality prove question

  1. Feb 21, 2009 #1
    prove that..
    [tex]
    x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\
    [/tex]
    [tex]
    R_5=\frac{f^{5}(c)x^5}{5!}
    [/tex]
    i need to prove that the remainder is negative .
    [tex]
    \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5
    [/tex]
    [tex]
    R_5=\frac{cos(c)x^5}{5!}
    [/tex]
     
  2. jcsd
  3. Feb 21, 2009 #2
    I'm going to guess that you will use the Maclaurin series for sin(x) and cos(x).
     
  4. Feb 21, 2009 #3
    no?????/////

    i need to show that the remainder is negative??
     
  5. Feb 22, 2009 #4

    Mark44

    Staff: Mentor

    [tex]x-\frac{x^3}{6}+\frac{x^5}{120}[/tex]
    These are the first three terms of the Maclaurin series for sin(x). Do you know what the terms immediately following them look like? Do you know how successive terms in this Maclaurin series compare in absolute value?
     
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