(a)Expand each of the following functions as a series in ascending power of x up to and including the term in x^3 .
(1) $\displaystyle \frac{x-1}{x+1}$
(2) $\displaystyle \frac{1}{(x-2)(1+2x)}$
(a)Expand each of the following functions as a series in ascending power of x up to and including the term in x^3 .
(1) $\displaystyle \frac{x-1}{x+1}$
(2) $\displaystyle \frac{1}{(x-2)(1+2x)}$
Hello, thereddevils!
Have you tried Long Division?(a) Expand each of the following functions as a series
in ascending power of $\displaystyle x$ up to and including the term in $\displaystyle x^3$
$\displaystyle (1)\;\;\frac{x-1}{x+1}$
We have: .$\displaystyle \frac{x-1}{x+1} \;=\;\frac{-1+x}{1+x}$
. . $\displaystyle \begin{array}{ccccccccc} & & & -1 & +2x &- 2x^2 & +2x^3 & + \hdots \\ & & -- & -- & -- & -- & -- & -- \\ 1+x & ) & -1 & +x \\ & & -1 & -x \\ & & -- & -- \\ & & & 2x \\ & & & 2x & +2x^2 \\ & & & -- & -- \\ & & & & -2x^2 \\ & & & & -2x^2 & -2x^3 \\ \end{array}$
. . . . . . . . . . . . . . . $\displaystyle \begin{array}{ccccc}
-- & \;\;-- \\ & \;\; 2x^3 \\ &\;\; 2x^3 &\;\; + 2x^4 \\ &\;\; -- & \;\;-- \\
& & \;\;\hdots \end{array}$
Therefore: .$\displaystyle \frac{x-1}{x+1} \;=\;-1 + 2x - 2x^2 + 2x^3 - \hdots$