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

for

Note that the sum of a geometric series is:

Write then the function as:

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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} ).

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