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So I'm confused about the partial fraction decomposition of fractions that involved repeated factors in the denominator. From what I've been taught, when you are breaking your fraction into partial fractions, the numerator of each fraction has a degree of at least one less than the denominator. However, from what I've seen in textbooks and online the partial fraction decomposition of fraction involving repeated factors in denominators often looks like this for linear factors. The degree of the numerator is only one less than the degree of the repeated factor which is $(x-a)$ rather than than the degree of the entire denominator

$\frac{P(x)}{(x - a)^n} = \frac{A}{(x - a)} + \frac{B}{(x -a)^2} + \frac{C}{(x -a)^3} + \dots$

But if I consider the general rule I was taught. Then shouldn't each partial fraction look like this.

$\frac{P(x)}{(x - a)^n} = \frac{A}{(x - a)} + \frac{Bx+C}{(x -a)^2} + \frac{Dx^2+Ex+F}{(x -a)^3} + \dots$

Why is this so? Is there a proof or explanation for this? because I can't seem to find one online. I've found one proof from Cambridge Mathematics 4 unit Year 12, but I don't understand the proof. So could you explain their proof as well, https://i.gyazo.com/ad52aefdf68e4728985498b48262a1ed.png (the proof) especially the part where it express $P(x)$ as a polynomial in $(x-a)$, what does that mean? and what are they doing when they say let $G(x) = P(x+a)= b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + \dots$

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With repeated factors your numerator should one less than the basic factor. Like you wrote:

$$\frac{P(x)}{(x - a)^n} = \frac{A}{(x - a)} + \frac{B}{(x -a)^2} + \frac{C}{(x -a)^3} + \dots$$

This is because the fraction $\frac{Dx^2+Ex+F}{(x-a)^3}$ can be rewritten as:

$$=\frac{Dx^2-2Dax+a^2+2Dax+Ex-Ea-2Da^2+F-a^2+2Da^2+Ea}{(x-a)^3}$$

$$=\frac{Dx^2-2Dax+a^2+2Dax+Ex-Ea-2Da^2+F-a^2+2Da^2+Ea}{(x-a)^3}$$

$$=\frac{D(x-a)^2}{(x-a)^3}+\frac{(2Da+E)(x-a)}{(x-a)^3}+\frac{Ea+F-a^2}{(x-a)^3}$$

$$=\frac{D}{x-a}+\frac{2Da+E}{(x-a)^2}+\frac{F-a^2+2Da^2+Ea}{(x-a)^3}$$

$$=\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}$$

This will work for any degree of polynomials involved and this final form is preferred as it straight forward to integrate.

Update in response to comment:

That is what your textbook is doing when it says let $G(x)=P(x+a)$ - i.e we take your polynomial and replace $x$ with $x+a$ and expand it. This new polynomial will have particular coefficients labelled $b_i$. I.e. you now have a polynomial like:

$$G(x) = P(x+a)= b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + \dots + b_1x + b_0$$

If we then consider your original polynomial $P(x)$ you have:

$$\frac{P(x)}{(x-a)^n}=\frac{G(x-a)}{(x-a)^n}=\frac{b_{n-1}(x-a)^{n-1} + b_{n-2}(x-a)^{n-2} + \dots +b_1(x-a) +b_0}{(x-a)^n}$$

$$=\frac{b_{n-1}(x-a)^{n-1}}{(x-a)^n} + \frac{b_{n-2}(x-a)^{n-2}}{(x-a)^n} + \cdots + \frac{b_1(x-a)}{(x-a)^n} + \frac{b_0}{(x-a)^n}$$

$$=\frac{b_{n-1}}{(x-a)} + \frac{b_{n-2}}{(x-a)^2} + \cdots + \frac{b_1}{(x-a)^{n-1}} + \frac{b_0}{(x-a)^n}$$

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  • $\begingroup$ Ah I see. But how do you know it ill work for all degrees can you explain or proof that, maybe using a repeated factor with the degree n? $\endgroup$ – D.Pham Apr 22 '17 at 15:10

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