Author Topic: Partial Fractions (Read 1926 times) Tweet Share

cltf · « **on:** October 08, 2010, 09:56:32 pm »

I'm really having trouble understanding partial fractions here are some of the types of questions I'm struggling with:
1. $\frac{x^{2}+1}{x(x+1)^{\left 2 \right}}$
2. $\frac{x^{2}+1}{x(x-1)^{\left 2 \right}}$
3. $\frac{2}{\left ( x+1 \right )\left ( x^{2}+1 \right )}$
4. $\frac{4x+4}{\left ( x+2 \right )\left ( x^{2}-2 \right )}$

thx

brightsky · « **Reply #1 on:** October 08, 2010, 10:10:14 pm »

1. Of the form $\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$ , where A, B, C are all constants. Let the expression equate that and then solve for A, B and C algebraically.

2. Same as 1 except for the last two fractions, x+1 becomes x - 1.

3. Split it up according to the factors. Also, since we have an irreducible quadratic, the partial decomp would be in the form $\frac{A}{x+1} + \frac{Bx + C}{x^2 + 1}$

4. Similar technique as 3.

jasoN- · « **Reply #2 on:** October 08, 2010, 10:22:29 pm »

1. firstly you could split the fraction into partial fractions:
$\frac{x^{2}+1}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
Multiply to get common denominators:
$\frac{x^{2}+1}{x(x+1)^2}=\frac{A(x+1)^2}{x(x+1)^2}+\frac{Bx(x+1)}{x(x+1)^2}+\frac{Cx}{x(x+1)^2}$
Cancel denominators:
$x^2+1=A(x+1)^2+Bx(x+1)+Cx$ (1)
At $x=-1, (-1)^2+1=A(0)^2+B(0)+C(-1)$
$2=-C \Rightarrow C=-2$
At $x=0, C=-2: (0)^2+1=A(1)^2+B(0)-2(0)$
$A=1$
Using all the points solved, sub into the equation 1 and use x=1 (or any value not used for that matter)
$(1)^2+1=(2)^2+B(2)-2(1)$
$\Rightarrow 2=4+2B-2 \Rightarrow B=0$
$\frac{x^{2}+1}{x(x+1)^2}=\frac{1}{x}+\frac{0}{x+1}+\frac{-2}{(x+1)^2}$
$\frac{x^{2}+1}{x(x+1)^2}=\frac{1}{x}-\frac{2}{(x+1)^2}$

~~2. Realise $(x-1)^2$ is an example of 'DOTS' $\Rightarrow (x-1)(x+1)$~~
Sub all values back into original:
$\frac{x^{2}+1}{x(x-1)^2}=\frac{x^{2}+1}{x(x-1)(x+1)}$
$\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x+1)}$
Do similar steps as shown in 1.

3. As $x^2+1$ is irreducible (cannot be factorised), it can be factorised into $}\frac{Bx+C}{x^2+1}$
ie. $\frac{2}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$
again follow the steps shown above to cancel down

4. $x^2-2$ is also irreducible $\Rightarrow$ follow 3.

In general:
Type 1: Linear Factors $\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$
Type 2: Repeated Factor $\frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}$
Type 3: Irreducible quadratic factors $\frac{1}{ax^2+bx+c} = \frac{Ax+B}{ax^2+bx+c}$

m@tty · « **Reply #3 on:** October 08, 2010, 10:26:28 pm »

Quote from: jasoN- on October 08, 2010, 10:22:29 pm

2. Realise $(x-1)^2$ is an example of 'DOTS' $\Rightarrow (x-1)(x+1)$
Sub all values back into original:
$\frac{x^{2}+1}{x(x-1)^2}=\frac{x^{2}+1}{x(x-1)(x+1)}$
$\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x+1)}$
Do similar steps as shown in 1.

With 2:

$(x-1)^2\ne (x+1)(x-1)$

brightsky · « **Reply #4 on:** October 08, 2010, 10:28:03 pm »

Quote from: jasoN- on October 08, 2010, 10:22:29 pm

2. Realise $(x-1)^2$ is an example of 'DOTS' $\Rightarrow (x-1)(x+1)$

$(x-1)^2$ isn't an example of DOPS. :p

EDIT: Beaten.

jasoN- · « **Reply #5 on:** October 08, 2010, 10:34:29 pm »

ah FML! epic fail

m@tty · « **Reply #6 on:** October 08, 2010, 10:44:03 pm »

Also, just being picky, but $\frac{1}{ax^2+bx+c} \ne \frac{Ax+B}{ax^2+bx+c}$ (Well it does, A=0, B=1 ... but that is pointless...)

The $\frac{Ax+B}{ax^2+bx+c}$ term comes only when separating the irreductible quadratic factor from a fraction with another factor on the denominator. ie. When you are splitting into partial fractions.

$\frac{1}{x(ax^2+bx+x)}=\frac{A}{x} + \frac{Bx+C}{ax^2+bx+c}$

cltf · « **Reply #7 on:** October 10, 2010, 09:45:53 pm »

Thank you, I actually get this now

m@tty · « **Reply #8 on:** November 02, 2010, 12:43:41 am »

1st question: It's personal preference really, though in some situations it is easier by a particular method. I tend to sub x-values and solve in a fashion similar to jasoN-.

And you can separate 'irreductible' quadratics into complex partial fractions.

eg.

$\text{Split }\frac{2x+1}{x^2+1} \text{ into partial fractions}$

$x^2+1=x^2-i^2=(x+i)(x-i) \text{ where }i\text{ is the imaginary unit, }i^2=-1$

$\text{Let }\frac{2x+1}{x^2+1}=\frac{A}{x-i}+\frac{B}{x+i}$

$\therefore 2x+1=A(x+i)+B(x-i)$

$\text{Sub }x=i \implies 2i+1=2iA \implies A=1+\frac{1}{2i}=1+\frac{-2i}{4}=1-0.5i$

$\text{Sub }x=-i \implies 1-2i=-2iB \implies B=-\frac{1}{2i}+1=\frac{2i}{4}+1=1+0.5i$

$\therefore \frac{2x+1}{x^2+1}=\frac{1-0.5i}{x-i}+\frac{1+0.5i}{x+i}$

However, I've never seen it done before, and can't think of too many uses (bear in mind that this with my very limited knowledge; there could be many, many applications . . .)

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Author Topic: Partial Fractions (Read 1926 times) Tweet Share

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Partial Fractions

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