Re: Past the Black's question at Yahoo Answers regarding partial fractions and Maclaurin series
Hello Past the Black,
We are given:
$$f(x)=\frac{x^2+7x-6}{(x-1)(x-2)(x+1)}$$
i) We may assume the partial fraction decomposition of $f$ will take the form:
$$\frac{x^2-7x-6}{(x-1)(x-2)(x+1)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+1}$$
Using the
Heaviside cover-up method, we find:
$$A=\frac{1^2+7\cdot1-6}{(1-2)(1+1)}=\frac{2}{(-1)(2)}=-1$$
$$B=\frac{2^2+7\cdot2-6}{(2-1)(2+1)}=\frac{12}{(1)(3)}=4$$
$$C=\frac{(-1)^2+7\cdot(-1)-6}{(-1-1)(-1-2)}=\frac{-12}{(-2)(-3)}=-2$$
Hence, we find:
$$f(x)=\frac{x^2-7x-6}{(x-1)(x-2)(x+1)}=-\frac{1}{x-1}+\frac{4}{x-2}-\frac{2}{x+1}$$
ii) The Maclaurin expansion is given by:
$$f(x)=\sum_{k=0}^{\infty}\left[\frac{f^{(k)}(0)}{k!}x^k \right]$$
Using the partial fraction decomposition written in the following form:
$$f(x)=-(x-1)^{-1}+4(x-2)^{-1}-2(x+1)^{-1}$$
We may state:
$$f^{(k)}(0)=(-1)^kk!\left(-(-1)^{-(k+1)}+4(-2)^{-(k+1)}-2 \right)$$
And so the series expansion becomes:
$$f(x)=\sum_{k=0}^{\infty}\left[(-1)^k\left(-(-1)^{-(k+1)}+4(-2)^{-(k+1)}-2 \right)x^k \right]$$
And so the first 4 terms are:
$$f(x)\approx-3+2x-\frac{3}{2}x^2+\frac{11}{4}x^3$$