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Taylor inequality

  1. Dec 5, 2007 #1
    We are supposed to use taylor's inequality to estimate the accuracy of the approximation of the taylor polynomial within the interval given.

    so, f(x) = cos x , a = pi/3, n=4 and the interval is 0<= x <= 2pi/3

    the fifth derivative is -sin x

    to get the M in taylor's inequality, wouldn't we have to plug 0 into |-sin x|?

    Why does the book say |-sin x|<= 1 = M?

    Does it work differently with trig functions?
  2. jcsd
  3. Dec 5, 2007 #2


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    A Taylor series remainder term contains a derivative which is evaluated at some point between x=0 (the point you are expanding around) and x=a. It doesn't say at which point. So the only thing you can say about -sin(x) in that interval is that |-sin(x)|<=1.
  4. Dec 5, 2007 #3
    so, what part does x = 0 play ?
  5. Dec 5, 2007 #4


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    You are expanding around a=pi/3. The point at which you want the approximation is in [0,2pi/3]. The x in the derivative part of the taylor error term is ANOTHER point in that interval, you don't know which one. Look up a discussion like http://en.wikipedia.org/wiki/Taylor's_theorem. [Broken]
    Last edited by a moderator: May 3, 2017
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