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For Example:

$$\frac{ax^2 + bx+c}{(dx+e)(fx^2+g)}\equiv\frac{A}{dx+e}+\frac{Bx+C}{fx^2+g}$$

and

$$\frac{ax^4 + bx^3+cx^2+dx+e}{(x+f)(x^2+g)}\equiv Ax+B+\frac{C}{x+f}+\frac{Dx+E}{x^2+g}$$

How do you know how to format the right hand side, in the equation below in partial fractions? (what is the 'pattern'?)

$$\frac{2ax}{(x-2a)(x^2+a^2)}$$ where a is a non zero constant

Could you look at it as the top of the fraction being a simple derivative of the bottom? If so how would you know to add the Ax+B in the latter example?

maxmitch
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  • Not quite a duplicate. That question is on derivation of the general form, this question is on the actual method itself - as in, how to express the general form, rather than the mathematical derivation of it. – Glen O May 19 '13 at 15:16
  • @GlenO I'm not sure I agree, but assuming you're right, it's not a duplicate in the usual sense of the word, but the other question answers this one, so it is a MSE-duplicate. – Git Gud May 19 '13 at 15:21

2 Answers2

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Step 1: Look at the orders of the numerator and the denominator. For instance, in the second example, the order of the numerator is 4 ($x^4$ is the largest power) and the order of the denominator is 3 ($x\cdot x^2=x^3$ is the largest power). If $n\geq m$, where $n$ is the order of the numerator and $m$ is the order of the denominator, then you need to add linear terms up to order $n-m$. In this case, up to order $1$, so you have $$ Ax+B+\frac??+\frac?? $$

Step 2: Separate out the factors in the denominator into separate fractions. So in your question case, you have $$ \frac{2ax}{(x-2a)(x^2+a^2)} = \frac?{x-2a}+\frac?{x^2+a^2} $$

Step 3: Give each numerator powers up to, but not including the order of the denominator. So the denominator of the first fraction is order 1, so you have up to order 0, meaning just a constant, $A$. The denominator of the second fraction is order 2, so you have up to order 1, so it's $Bx+C$, giving $$ \frac{2ax}{(x-2a)(x^2+a^2)} = \frac{A}{x-2a}+\frac{Bx+C}{x^2+a^2} $$ If a denominator had a cubic, order 3, then the numerator would be up to order 2, so $Ax^2+Bx+C$.

EDIT: Note that you may include a fourth step if you have a repeated factor. If you have $(x-a)^2$ as a factor of the denominator, then you could write $$ \frac{Ax+B}{(x-a)^2} $$ But you could also write it as $$ \frac{A}{x-a}+\frac{B}{(x-a)^2} $$ which is equivalent for different values of $A$ and $B$. It is often viewed as better to use the latter expression.

Glen O
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Using Partial Fraction Decomposition, $$\frac{2ax}{(x-2a)(x^2+a^2)}=\frac A{(x-2a)}+\frac {Bx+C}{(x^2+a^2)} $$

lab bhattacharjee
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  • I notice you made a change from it being B at the top of the right fraction to Bx+c can you explain to me how you knew to change it? – maxmitch May 19 '13 at 15:07
  • @maxmitch, have you followed the link? For irreducible polynomial $$\frac1 {\sum_{0\le r\le n}a_rx^r},$$ it should be $$\frac{{\sum_{0\le r\le n-1}}b_rx^r}{{\sum_{0\le r\le n}}a_rx^r}$$ – lab bhattacharjee May 19 '13 at 15:09
  • @maxmitch: Notice that we need a degree two polynomial to cancel out the other degree two polynomial. – Inceptio May 19 '13 at 15:09