How do you write the first four terms and the general term of the Taylor series expansion of f(x) = 1/(x-1) about x = 2?

May 5, 2015

The Taylor series for $f \left(x\right)$ centered at $a$ is:

f(x) = f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * *

For this question: $f \left(x\right) = \frac{1}{x + 1}$ and $a = 2$

$f \left(x\right) = \frac{1}{x + 1}$ $\textcolor{w h i t e}{\text{sssssssssssss}}$ so $f \left(a\right) = f \left(2\right) = \frac{1}{3}$

$f ' \left(x\right) = - \frac{1}{x + 1} ^ 2$ $\textcolor{w h i t e}{\text{ssss-s}}$ so $f ' \left(a\right) = - \frac{1}{3} ^ 2$

$f ' ' \left(x\right) = \frac{2}{x + 1} ^ 3$ $\textcolor{w h i t e}{\text{sssss-ss}}$ so $f ' ' \left(a\right) = \frac{2}{3} ^ 3$

$f ' ' ' \left(x\right) = - \frac{6}{x + 1} ^ 4$ $\textcolor{w h i t e}{\text{ssss}}$ so $f ' ' ' \left(a\right) = - \frac{6}{3} ^ 4$

The general $n - {1}^{t h}$ derivative will have

f^(n-1)(x)=(-1)^(n-1)((n-1)!)/(x+1)^n

so f^(n-1)(a)= (-1)^(n-1)((n-1)!)/3^n

Notice that in each term of the Taylor expansion for this function, I get: (n-1)! in both the numerator (from the derivative) and the denominator (from the formula for the expansion). Obviously, these cancel, leaving:

$\frac{1}{3} - \frac{1}{3} ^ 2 \left(x - 2\right) + \frac{1}{3} ^ 3 {\left(x - 2\right)}^{2} - \frac{1}{3} ^ 4 {\left(x - 2\right)}^{3}$ for the first four terms, and

${\left(- 1\right)}^{n - 1} \frac{1}{3} ^ n {\left(x - 2\right)}^{n - 1}$ for the ${n}^{t h}$ term.