# How do you find the remainder term #R_3(x;1)# for #f(x)=sin(2x)#?

Final Term of the Taylor Series

Through the use of derivatives,

Thus, we have

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To find the remainder term ( R_3(x;1) ) for ( f(x) = \sin(2x) ) using the Taylor series expansion centered at ( a = 1 ), follow these steps:

- Find the first few derivatives of ( f(x) = \sin(2x) ) evaluated at ( a = 1 ):

[ f(x) = \sin(2x) ] [ f'(x) = 2\cos(2x) ] [ f''(x) = -4\sin(2x) ] [ f'''(x) = -8\cos(2x) ]

- Evaluate these derivatives at ( a = 1 ):

[ f(1) = \sin(2) ] [ f'(1) = 2\cos(2) ] [ f''(1) = -4\sin(2) ] [ f'''(1) = -8\cos(2) ]

- Use the Taylor series formula to find the remainder term ( R_3(x;1) ):

[ R_3(x;1) = \frac{f'''(c)}{3!} \cdot (x - 1)^3 ]

where ( c ) is some value between ( 1 ) and ( x ).

- Substitute the values of the derivatives and ( x ):

[ R_3(x;1) = \frac{-8\cos(2)}{6} \cdot (x - 1)^3 ]

Simplify:

[ R_3(x;1) = \frac{-4\cos(2)}{3} \cdot (x - 1)^3 ]

This is the remainder term ( R_3(x;1) ) for ( f(x) = \sin(2x) ) centered at ( a = 1 ).

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To find the remainder term ( R_3(x;1) ) for the function ( f(x) = \sin(2x) ), we use Taylor's theorem with remainder. In this case, we use the third-order Taylor polynomial, which is ( P_3(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 ), where ( f(1) = \sin(2) ), ( f'(1) = 2\cos(2) ), ( f''(1) = -4\sin(2) ), and ( f'''(1) = -8\cos(2) ). Then, the remainder term ( R_3(x;1) ) is given by ( R_3(x;1) = f(x) - P_3(x) ).

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