Find the Taylor polynomials (centred at zero) of degree (a) 1, (b) 2, (c) 3, and (d) 4. $f ( x ) = \frac { x } { x + 1 }$
A) $S _ { 1 } ( x ) = x , S _ { 2 } ( x ) = x - x ^ { 2 } , S _ { 3 } ( x ) = x - x ^ { 2 } + x ^ { 3 } , S _ { 4 } ( x ) = x - x ^ { 2 } + x ^ { 3 } - x ^ { 4 }$
B) $S _ { 1 } ( x ) = x , S _ { 2 } ( x ) = x + x ^ { 2 } , S _ { 3 } ( x ) = x + x ^ { 2 } + x ^ { 3 } , S _ { 4 } ( x ) = x + x ^ { 2 } + x ^ { 3 } + x ^ { 4 }$
C) $S _ { 1 } ( x ) = x , S _ { 2 } ( x ) = x - x ^ { 2 } , S _ { 3 } ( x ) = x - x ^ { 2 } - x ^ { 3 } , S _ { 4 } ( x ) = x - x ^ { 2 } - x ^ { 3 } - x ^ { 4 }$
D) $S _ { 1 } ( x ) = x , S _ { 2 } ( x ) = x + x ^ { 2 } , S _ { 3 } ( x ) = x + x ^ { 2 } - x ^ { 3 } , S _ { 4 } ( x ) = x + x ^ { 2 } - x ^ { 3 } + x ^ { 4 }$
E) $S _ { 1 } ( x ) = x , S _ { 2 } ( x ) = x + x ^ { 2 } , S _ { 3 } ( x ) = x + x ^ { 2 } - x ^ { 3 } , S _ { 4 } ( x ) = x + x ^ { 2 } + x ^ { 3 } - x ^ { 4 }$

Question

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The function $f(x) = \frac{x}{x+1}$ can be rewritten as $f(x) = x(1+x)^{-1}$, and its Taylor series expansion is $x - x^2 + x^3 - x^4 + \cdots$, which matches the polynomials given in option A for degrees 1 through 4.

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

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