Find the Taylor Polynomials (Centred at Zero) of Degree (A) $f ( x ) = \sqrt { x + 1 }$

question 116

Multiple Choice

Find the Taylor polynomials (centred at zero) of degree (a) 1, (b) 2, (c) 3, and (d) 4. $f ( x ) = \sqrt { x + 1 }$

• A) \(S _ { 1 } ( x ) = 1 + \frac { x } { 2 } , S _ { 2 } ( x ) = 1 + \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } , S _ { 3 } ( x ) = 1 + \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } , S _ { 4 } ( x ) = 1 + \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } - \frac { x ^ { 4 } } { 128 }\)
• B) \(S _ { 1 } ( x ) = 1 - \frac { x } { 2 } , S _ { 2 } ( x ) = 1 - \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } , S _ { 3 } ( x ) = 1 - \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } - \frac { x ^ { 3 } } { 16 } , S _ { 4 } ( x ) = 1 - \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } - \frac { x ^ { 3 } } { 16 } + \frac { x ^ { 4 } } { 128 }\)
• C) \(S _ { 1 } ( x ) = 1 + \frac { x } { 2 } , S _ { 2 } ( x ) = 1 + \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } , S _ { 3 } ( x ) = 1 + \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } , S _ { 4 } ( x ) = 1 + \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } + \frac { x ^ { 4 } } { 128 }\)
• D) \(S _ { 1 } ( x ) = 1 - \frac { x } { 2 } , S _ { 2 } ( x ) = 1 - \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } , S _ { 3 } ( x ) = 1 - \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } - \frac { x ^ { 3 } } { 16 } , S _ { 4 } ( x ) = 1 - \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } - \frac { x ^ { 3 } } { 16 } - \frac { x ^ { 4 } } { 128 }\)
• E) \(S _ { 1 } ( x ) = 1 - \frac { x } { 2 } , S _ { 2 } ( x ) = 1 + \frac { x } { 2 } - \frac { x ^ { 2 } } { 8 } , S _ { 3 } ( x ) = 1 + \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } , S _ { 4 } ( x ) = 1 + \frac { x } { 2 } + \frac { x ^ { 2 } } { 8 } + \frac { x ^ { 3 } } { 16 } - \frac { x ^ { 4 } } { 128 }\)

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