Estimating Accuracy of Taylor Polynomial w/ Taylor Inequality

[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?

 

[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.

 

[frasifrasi]

so, what part does x = 0 play ?

 

[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.