1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Taylor inequality
  1. Taylor Approximation (Replies: 5)

  2. Taylor series (Replies: 15)

  3. Taylor coefficient (Replies: 3)