# 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