Final answer:
The Taylor polynomial approximation of f(x) = 1 / (1 + x) around the point 0 results in a sum of terms with alternating signs, consistent with a geometric series expansion involving negative exponents.
Step-by-step explanation:
The student has asked to show that for a given function f(x) = 1 / (1 + x), the Taylor polynomial approximation around 0, denoted by T_{n}(f, 0, x), is equal to \(\sum_{k=0}^{n}(-1)^{k} x^{k} \).
The question involves using a series expansion known as the Taylor series or Maclaurin series when the expansion is around 0, which is specific to a function of x.
The binomial theorem or geometric series are typically used when expressing such approximations. In this case, the function is a geometric series where the base is -x. Hence, expanding it we get the mentioned series where the terms alternate between positive and negative corresponding to the power of x being even or odd.
Step-by-step, the solution involves expanding f(x) in terms of x, starting from exponents of 0 up to n, with each term being negated with respect to the previous one, which aligns with the negative exponents in the original function suggesting division.