Taylor inequality

1. Dec 5, 2007

frasifrasi

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. Dec 5, 2007

Dick

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.

3. Dec 5, 2007

frasifrasi

so, what part does x = 0 play ?

4. Dec 5, 2007

Dick

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