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Find the Taylor polynomial of 1 / (2-x)

by RoRi

Show that

    \[ T_n \left( \frac{1}{2-x} \right) = \sum_{k=0}^n \frac{x^k}{2^{k+1}}. \]

First, we write

    \[ f(x) = \frac{1}{2-x} = \frac{1}{2} \left( \frac{1}{1-\frac{x}{2}} \right). \]

Then, using the algebraic identity \frac{1}{1-x} = 1+x+x^2 + \cdots we have

    \begin{align*} \frac{1}{1-\frac{x}{2}} &= 1 + \left( \frac{x}{2} \right) + \left( \frac{x}{2} \right)^2 + \cdots + \left( \frac{x}{2} \right)^n + \frac{\left(\frac{x}{2} \right)^{n+1}}{1-\frac{x}{2}} \\[9pt] &= P_n (x) + x^n \frac{x}{2^n(2 - x)} \\ &= P_n (x) + x^n g(x) \end{align*}

where

    \[ g(x) = \frac{x}{2^n(2-x)} \quad \implies \quad \lim_{x \to 0} g(x) = 0. \]

Therefore, we may apply Theorem 7.4 to conclude

    \begin{align*} T_n \left( \frac{1}{2-x} \right) &= \frac{1}{2} T_n \left( \frac{1}{1-\frac{x}{2}} \right)\\ &= \frac{1}{2} P_n (x) \\ &= \frac{1}{2} \sum_{k=0}^n \left( \frac{x}{2} \right)^k \\ &= \sum_{k=0}^n \frac{x^k}{2^{k+1}}. \end{align*}

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