How do you find the Taylor polynomial for #1/(2-x)#?

Question

Isaiah Allen · Accepted Answer

#1/(2-x) =sum_(n=0)^oo x^n/2^(n+1)#

for #x in (-2,2)#.

Note that the sum of a geometric series is: #sum_(n=0)^oo q^n = 1/(1-q)# for #q in (-1,1)# Write then the function as: #1/(2-x) = 1/2 1/(1-x/2)# Posing: #q=x/2#, we have: #1/(2-x) = 1/2 sum_(n=0)^oo (x/2)^n = sum_(n=0)^oo x^n/2^(n+1)# for #x in (-2,2)#.

Christopher Agresta · Answer

undefined To find the Taylor polynomial for $ \frac{1}{2-x} $, you need to compute its Taylor series expansion centered at the point where you want to approximate the function. In this case, we'll center the expansion at $ x = 0 $. The general formula for the Taylor series expansion of a function $ f(x) $ centered at $ x = a $ is:

$$ T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x - a)^n $$

First, let's find the derivatives of $ \frac{1}{2-x} $:

$$ f(x) = \frac{1}{2-x} $$

$$ f'(x) = \frac{1}{(2-x)^2} $$

$$ f''(x) = \frac{2}{(2-x)^3} $$

$$ f'''(x) = \frac{6}{(2-x)^4} $$

$$ f^{(4)}(x) = \frac{24}{(2-x)^5} $$

Now, evaluate these derivatives at $ x = 0 $ since we're centering the expansion there:

$$ f(0) = \frac{1}{2} $$

$$ f'(0) = 1 $$

$$ f''(0) = 2 $$

$$ f'''(0) = 6 $$

$$ f^{(4)}(0) = 24 $$

Now, plug these values into the formula for the Taylor polynomial:

$$ T_n(x) = \frac{1}{2} + 1x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 $$

Simplify this expression to get the Taylor polynomial for $ \frac{1}{2-x} $.