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I was just trying to solve a problem that had the following item which I needed to split into separate generating functions:

$$\frac{x}{(1-2x)^2(1-5x)}$$

I had assumed I needed to split it into:

$$\frac{A}{1-2x} + \frac{B}{1-2x} + \frac{C}{1-5x}$$

But according to Wolfram Alpha it appears I had to split it into:

$$\frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{1-5x}$$

Can anyone explain the intuition behind this? Is this a general rule that when you have a repeated factor in the denominator, you split it into all powers of that factor?

AJJ
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  • Yes, it is a general rule. With your splitting, the LCM wouldn't be $(1-2x)^2(1-5x)$. – Vincenzo Tibullo Jan 07 '16 at 17:20
  • Check out the answer to this question. –  Jan 07 '16 at 17:23
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    Imagine you tried to do partial fractions on $\dfrac{1+x}{x^2}$. You could try $\dfrac{1+x}{x^2}=\dfrac Ax+\dfrac Bx$, but that's just $\dfrac{A+B}x$ on the right so it won't work. However, $\dfrac{1+x}{x^2}=\dfrac1{x^2}+\dfrac x{x^2}$, which simplifies into $\dfrac1{x^2}+\dfrac1x$, which is the best we can do. – Akiva Weinberger Jan 07 '16 at 17:44
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    What if you had $\dfrac{1+x}{(x)(x+0.001)}$? Then you'd get $\dfrac{1000}x-\dfrac{999}{x+0.001}$. The closer the factors in the denominator get to each other, the higher the coefficients need to be. (Also, notice the minus sign.) – Akiva Weinberger Jan 07 '16 at 17:51
  • @AkivaWeinberger Great point – AJJ Jan 07 '16 at 17:53

3 Answers3

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Consider the simplest of cases. $$\frac{\xi}{x^3}$$ Assuming $\xi$ is some polynomial, one could carry on long division to determine the quotient and the remainder. Thus one could get $$\frac{\xi}{x^3}=q(x) + \frac{r(x)}{x^3}$$ Now what can we say for certain about $r(x)$ ? We can say that it is the remainder thus of smaller degree than $x^3$. The most general possible polynomial of degree 2 or smaller is $A+Bx+Cx^2$. Then $$\frac{r(x)}{x^3}=\frac{A+Bx+Cx^2}{x^3}$$ Further more, we can split the right side $$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{Bx}{x^3}+\frac{Cx^2}{x^3}$$ which becomes $$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{B}{x^2}+\frac{C}{x}$$ Thus the above would be a sensible what to try to split any proper fraction $\frac{r(x)}{x^3}$ Moreover, the similar idea would hold for $\frac{r(x)}{(x+a)^3}$, meaning a sensible way to split it would be $$\frac{r(x)}{(x+a)^3}=\frac{A}{(x+a)^3} +\frac{B}{(x+a)^2}+\frac{C}{(x+a)} $$ hope that helps..

userX
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Yep - if there's a factor of $(x-a)^k$ in the denominator, you need to include $\frac{1}{x-a},\frac{1}{(x-a)^2},...,\frac{1}{(x-a)^k}$ terms in your expansion.

Note that in your attempt, you could combine the first two terms to $\frac{A+B}{1-2x}$, making one of those constants redundant.

πr8
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Note that if $A$ and $B$ are constants then

$$\frac{A}{1-2x}+\frac{B}{1-2x}=\frac{A+B}{1-2x},$$

so that really these two terms are only one.

JP McCarthy
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