Error for Taylor Polynomial for sin^2(0.2) w/ 4th-deg poly

[thatguy47]

Please help! I've been trying do this problem for over and hour and I'm lost. I think I'm doing it completely wrong. Here's the problem:

I was doing: f(1) + f'(1)(x-1) + [f''(1)(x-1)^2]/2 + [f'''(x)(x-1)^3]/6 + [f''''(x)(x-1)^4]/24 to solve for the error
When I did that I got 0.08173...
I then did on my calcular 0.08173 - (sin(.2))^2 and got -0.042... for the error which is wrong.
The answer is

Does anyone know how you get that? I'm really lost. Please help me :? :? :? .

edit: the more I work on this problem, I find different ways to do it. I was doing some of the above calculations wrong. When I did it again I didn't get 0.08173. I instead got .5034 and then subtracted (sin(.2))^2) and got .4639... which is still wrong. I'm so confused...

 

[daon]

Why are your calculations using the center as 1 if you're told its at 0.

 

[thatguy47]

Ya, I messed up the first time. Even when I use 0 I still cant get the answer.

 

[daon]

Ya, I messed up the first time. Even when I use 0 I still cant get the answer.

are you plugging in 1 like you typed or 0.2?

 

[thatguy47]

I got it. Instead of doing the 5th degree which is zero I did the sixth degree and got 36(x)^6 divided by 6! and got 2.84 x 10^-6. I was doing a lot wrong earlier. I'm just glad that its starting to make sense now. Thanks.