Finding Maclaurin Error < 0.0001 for f(x)=cos(2x) at x=0.6

[kenny87]

Here's the problem:

Determine the degree of Maclaurin polynomial required for the error to be less than .0001 if f(x)=cos(2x) and you are approximating f(0.6)

I really don't know what I am doing. Here's what I've tried to do:

Rn(.6) = ( (f^(n+1)(z)) / (n+1)! ) (.6)^(n+1)

I don't know where to go from here...

 

[pocoman]

You must find the form of $$f^{n+1}(0)$$ first, then substitute it to Rn(0.6).
Solve the inequation : Rn(0.6) < 0.001 for n.
n is the required degree of Maclaurin polynomial for the given error.

 

[e(ho0n3]

Where did this come from: Rn(.6) = ( (f^(n+1)(z)) / (n+1)! ) (.6)^(n+1) ?

I would do this by trial and error, i.e calculating a larger and larger Maclaurin series expansion of cos (2x) until f(0.6) is where you need it to be.