find the fourth taylor polynomial of f(x)=1/x at x=2
To find the fourth Taylor polynomial of \( f(x) = \frac{1}{x} \) at \( x = 2 \), we can utilize the formula for the Taylor polynomial centered at a point \( a \):
\[ P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots + \frac{f^n(a)}{n!}(x - a)^n \]
The goal is to find the \( n \)th derivative, \( f^n(a) \), evaluated at \( a = 2 \).
First, let's find the first, second, third, and fourth derivatives of \( f(x) = \frac{1}{x} \):
\( f'(x) = -\frac{1}{x^2} \)
\( f''(x) = \frac{2}{x^3} \)
\( f'''(x) = -\frac{6}{x^4} \)
\( f^{(4)}(x) = \frac{24}{x^5} \)
Now, we can evaluate each of these derivatives at \( a = 2 \):
\( f'(2) = -\frac{1}{2^2} = -\frac{1}{4} \)
\( f''(2) = \frac{2}{2^3} = \frac{1}{2} \)
\( f'''(2) = -\frac{6}{2^4} = -\frac{3}{4} \)
\( f^{(4)}(2) = \frac{24}{2^5} = \frac{3}{2} \)
Substituting these values into the formula for the fourth Taylor polynomial, we get:
\[ P_4(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f'''(2)}{3!}(x - 2)^3 + \frac{f^{(4)}(2)}{4!}(x - 2)^4 \]
\[ P_4(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{2}\left(\frac{1}{2}\right)(x - 2)^2 - \frac{1}{4}\left(\frac{3}{4}\right)(x - 2)^3 + \frac{1}{24}\left(\frac{3}{2}\right)(x - 2)^4 \]
Simplifying further, we have:
\[ P_4(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 - \frac{3}{32}(x - 2)^3 + \frac{1}{16}(x - 2)^4 \]
Thus, the fourth Taylor polynomial of \( f(x) = \frac{1}{x} \) at \( x = 2 \) is \( P_4(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 - \frac{3}{32}(x - 2)^3 + \frac{1}{16}(x - 2)^4 \).
f(x) = 1/x
f(2) = 1/2
f'(2) = -1/4
f''(2) = 2/8
f'''(2) = -6/16
. . .
f(x) = 1/0! f(2) + 1/1! f'(2) (x-2) + 1/2! f''(2) (x-2)^2 + ...
f(x) = ∑(-1)^k/2^(k+1) (x-2)^k