derive the Maclaurin series by using partial fractions.

Question

derive the Maclaurin series for the function $(x^3+x^2+2x-2)/(x^2-1)$ by using partial fractions and a known Maclaurin series.

question. how can I use partial fractions in this case? Is this case the special one? plz provide me a hint. thanks.

Answer 1

HINT:

By Actual division, $$\frac{(x^3+x^2+2x-2)}{(x^2-1)}=x+1+\frac{3x-1}{x^2-1} $$

Using Partial Fraction Decomposition, $$\frac{3x-1}{x^2-1}=\frac A{x+1}+\frac B{x-1}$$

Answer 2

Using polynomial long division:

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

For partial fraction decomposition, think of it as the inverse of finding a common denominator for two fractions: first factor the denominator: $$x + 1 + \frac{3x-1}{x^2-1} = x + 1 + \frac{3x-1}{(x - 1)(x+ 1)}$$

Then we split the fraction into the sum of two partial fractions, setting up the decomposition to solve for the constants $A$ and $B$: $$x + 1 + \frac{3x-1}{(x - 1)(x+ 1)} = \frac A{x-1} + \frac B{x + 1}$$

Now, we know that $$A(x+1) + B(x - 1) = (A + B)x + (A - B) = 3x - 1 $$ $$\iff A+ B = 3 \;\; \text{and}\;\; A - B = -1$$

So solving for the system of equations in $A, B$: $$A = B - 1 \implies 2B - 1 = 3 \iff B = 2 \implies A = 1$$

Using these values, we have that $$\frac{(x^3+x^2+2x-2)}{(x^2-1)}=x + 1 + \frac 1{x-1} + \frac 2{x + 1}$$

With practice, partial fraction decomposition becomes easier (as with any skill), and there are "short-cuts" you'll pick up along the way that can simplify the process.

Answer 3

Perform polynomial long division

to conclude that

$$\frac{x^{3}+x^{2}+2x-2}{x^{2}-1}=x+1+\frac{3x-1}{x^{2}-1}=x+1+\frac{3x-1}{(x-1)(x+1)}.\tag{1}$$

The theory of partial fractions guarantees that there exist constants $A$ and $B$ such that

$$\frac{3x-1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.\tag{2}$$

You can find the constants in $(2)$ by several methods, one of them is as follows:

(a) Make e.g. $x=0$ on both sides of $(2)$

$$\frac{-1}{-1}=\frac{A}{-1}+\frac{B}{1}\Leftrightarrow B=A+1\tag{3}.$$ Hence $$\frac{3x-1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{A+1}{x+1}.\tag{4}$$ (b) Make e.g. $x=1/2$ on both sides of $(4)$ $$\frac{3/2-1}{(1/2-1)(1/2+1)}=\frac{A}{1/2-1}+\frac{A+1}{1/2+1}\Leftrightarrow A=1.\tag{5}$$ Hence $B=2$ and $$1+x+\frac{3x-1}{(x-1)(x+1)}=1+x+\frac{1}{x-1}+\frac{2}{x+1}.\tag{6}$$

Use the Maclaurin series expansions for $\frac{1}{x-1}$ and $\frac{1}{x+1}$ in $(6)$

$$\frac{1}{x-1}=-\frac{1}{1-x}=-\sum_{n=1}^{\infty }x^{n-1},\quad \frac{1}{x+1}=\frac{1}{1+x}=\sum_{n=1}^{\infty }(-1)^{n-1}x^{x-1},$$ $$\tag{7}$$ both derived from the sum of the geometric series

$$\frac{1}{1-x}=\sum_{n=1}^{\infty }x^{n-1},\qquad (|x|<1).\tag{8}$$