find t_n centered at x = 0 for all n for the function f(x) = 1/(1 + x). (use symbolic notation and fractions…

Explanation:

Step1: Recall the formula for Taylor series

The Taylor series of a function $f(x)$ centered at $x = a$ is given by $T_n(x)=\sum_{i = 0}^{n}\frac{f^{(i)}(a)}{i!}(x - a)^i$. Here $a = 0$. First, find the $i$-th derivative of $f(x)=\frac{1}{1 + x}=(1 + x)^{-1}$.

Step2: Find the $i$-th derivative of $f(x)$

Using the power - rule for differentiation $(x^n)'=nx^{n - 1}$, we have $f'(x)=(-1)(1 + x)^{-2}$, $f''(x)=(-1)(-2)(1 + x)^{-3}$, $f^{(i)}(x)=(-1)^i i!(1 + x)^{-(i + 1)}$. Evaluating at $x = 0$, we get $f^{(i)}(0)=(-1)^i i!$.

Step3: Substitute into the Taylor - series formula

Substitute $f^{(i)}(0)$ into $T_n(x)=\sum_{i = 0}^{n}\frac{f^{(i)}(0)}{i!}(x - 0)^i$. We have $\frac{f^{(i)}(0)}{i!}=\frac{(-1)^i i!}{i!}=(-1)^i$. So $T_n(x)=\sum_{i = 0}^{n}(-1)^i x^i$.

Answer:

$(-1)^i x^i$