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Past the Black's question at Yahoo! Answers regarding partial fractions and Maclaurin series

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MarkFL

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Feb 24, 2012
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Here is the question:

PARTIAL FRACTIONS: How to solve this?


Let f(x) =

(x^2 + 7x − 6)/((x − 1)(x − 2)(x + 1))
.
(i) Express f(x) in partial fractions.

(ii) Show that, when x is sufficiently small for x4 and higher powers to be neglected,

f(x) = −3 + 2x − (3/2)x^2 + (11/4)x^3.

Please work it out step by step so I can figure out how to work out similar questions on my own.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Re: Past the Black's question at Yahoo Answers regarding partial fractions and Maclaurin series

Hello Past the Black,

We are given:

\(\displaystyle f(x)=\frac{x^2+7x-6}{(x-1)(x-2)(x+1)}\)

i) We may assume the partial fraction decomposition of $f$ will take the form:

\(\displaystyle \frac{x^2-7x-6}{(x-1)(x-2)(x+1)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+1}\)

Using the Heaviside cover-up method, we find:

\(\displaystyle A=\frac{1^2+7\cdot1-6}{(1-2)(1+1)}=\frac{2}{(-1)(2)}=-1\)

\(\displaystyle B=\frac{2^2+7\cdot2-6}{(2-1)(2+1)}=\frac{12}{(1)(3)}=4\)

\(\displaystyle C=\frac{(-1)^2+7\cdot(-1)-6}{(-1-1)(-1-2)}=\frac{-12}{(-2)(-3)}=-2\)

Hence, we find:

\(\displaystyle f(x)=\frac{x^2-7x-6}{(x-1)(x-2)(x+1)}=-\frac{1}{x-1}+\frac{4}{x-2}-\frac{2}{x+1}\)

ii) The Maclaurin expansion is given by:

\(\displaystyle f(x)=\sum_{k=0}^{\infty}\left[\frac{f^{(k)}(0)}{k!}x^k \right]\)

Using the partial fraction decomposition written in the following form:

\(\displaystyle f(x)=-(x-1)^{-1}+4(x-2)^{-1}-2(x+1)^{-1}\)

We may state:

\(\displaystyle f^{(k)}(0)=(-1)^kk!\left(-(-1)^{-(k+1)}+4(-2)^{-(k+1)}-2 \right)\)

And so the series expansion becomes:

\(\displaystyle f(x)=\sum_{k=0}^{\infty}\left[(-1)^k\left(-(-1)^{-(k+1)}+4(-2)^{-(k+1)}-2 \right)x^k \right]\)

And so the first 4 terms are:

\(\displaystyle f(x)\approx-3+2x-\frac{3}{2}x^2+\frac{11}{4}x^3\)