Finding Taylor Polynomials and Error for f(x)=3x+cos(3x) about X0=0

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Homework Statement



Find the Taylor Polynomial T2(x) (degree 2) for f(x) expanded about X0.

f(x)=3x + cos(3x)

X0= 0

Find the error formula and then find the actual (absolute) error using T2(0.6) to approx. f(0.6).


The Attempt at a Solution



As I've said on this forum before, the text isn't very clear on Taylor Polynomials. They only give me one example and all I could think of doing here was following along with that. I also used Maple, which I've had trouble with before when it comes to trig. functions.

I came up with 1+3x-9/2 X2 Is that the answer to the first part?

And then the last part (actual absolute error) would be to sub in 0.6 for x? They never once mentioned error formula in my text and my teacher is mia.

Thanks for any help!
 
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Oop... edited question to make more sense. :)
 
You need to find f'(x) and f''(x), then f(0), f'(0) and f''(0).. sub these into the Taylor polynomial general formula (should be able to find it online).

I'm presuming error formula is just what you do to find the error- ie its the absolute value of f(x) - T2(x)
(can't do subscript..)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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