Solutions - 0240-03

Calculate derivatives. Use $a_n={\displaystyle \frac{f^{(n)}(1)}{n!}}$.

$n$	$f^{(n)}(x)$	$f^{(n)}(1)$	$a_n$
0	$+x^{-1}$	$+1$	$+1$
1	$-x^{-2}$	$-1$	$-1$
2	$+2\cdot x^{-3}$	$+2$	$+1$
3	$-3\cdot 2\cdot x^{-4}$	$-3!$	$-1$
4	$+4\cdots 2\cdot x^{-5}$	$+4!$	$+1$
5	$-5\cdots 2\cdot x^{-6}$	$-5!$	$-1$
6	$+6\cdots 2\cdot x^{-7}$	$+6!$	$+1$

Each new derivative takes down the next power as a factor, and switches the sign. The accumulation of powers follows a factorial pattern, and these factorials cancel those added to the denominator to make $a_n$.

So we have:

$$\begin{array}{c}f(x)=\sum _{n=0}^{\infty }{(-1)}^n{(x-1)}^n\\ \\ =1-(x-1)+{(x-1)}^2-{(x-1)}^3+{(x-1)}^4-\cdots \end{array}$$

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Alternate method:

We can derive this Taylor series using some algebraic tricks with the standard geometric series:

$$\begin{array}{c}\frac{1}{x}=\frac{1}{1-(-(x-1))}\gg \gg \sum _{n=0}^{\infty }(-(x-1){)}^n\\ \\ \gg \gg \sum _{n=0}^{\infty }(-1{)}^n(x-1{)}^n\end{array}$$