Determining power series for 3 x 2 − 4 x + 9 ( x − 1 ) 2 ( x + 3 )

Question

Determining power series for             3              x                  2                    −      4      x      +      9              (      x      −      1              )        2            (      x      +      3      )

cesantedz · Accepted Answer

You can use the partial fraction decomposition:

\frac{3 x^{2} - 4 x + 9}{(x - 1)^{2} (x + 3)} = \frac{A}{1 - x} + \frac{B}{(1 - x)^{2}} + \frac{C}{1 + \frac{1}{3} x}

and sum up the series you get, which are known.
If you do the computation, you find A=0, B=2 and C=1, so

\frac{3 x^{2} - 4 x + 9}{(x - 1)^{2} (x + 3)} = \frac{2}{(1 - x)^{2}} + \frac{1}{1 + \frac{1}{3} x}

The development of

(1 - x)^{- 2}

can be deduced from the fact that

\frac{1}{1 - x} = \sum_{n \geq 0} x^{n}

so, by deriving, we get

\frac{1}{(1 - x)^{2}} = \sum_{n \geq 1} n x^{n - 1} = \sum_{n \geq 0} (n + 1) x^{n}

The power series for the other term is again easy:

\frac{1}{1 + \frac{1}{3} x} = \sum_{n \geq 0} \frac{(- 1)^{n}}{3^{n}} x^{n}

so your power series development is

\frac{3 x^{2} - 4 x + 9}{(x - 1)^{2} (x + 3)} = \sum_{n \geq 0} (2 n + 2 + \frac{(- 1)^{n}}{3^{n}}) x^{n}

Determining power series for (3x^2−4x+9)/((x−1)^2(x+3))

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