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

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SUMMARY

The discussion focuses on finding the Taylor Polynomial T2(x) for the function f(x) = 3x + cos(3x) expanded about X0 = 0. The user derived T2(x) as 1 + 3x - (9/2)x² and sought clarification on the error formula. The error is defined as the absolute value of f(x) - T2(x), and the user is encouraged to compute the actual absolute error by substituting x = 0.6 into the polynomial.

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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!
 
Last edited:
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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..)
 

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