Partial Fraction Decomposition - Building Up Powers

WakelessFoil

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Apr 7, 2021
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I cannot seem to wrap my head around the idea that when doing partial fraction decomposition you "build up powers". My professor gave me a simplified explanation (lower left corner), but it doesn't make sense to me that the denominator can be split into several with different powers.
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nasi112

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Aug 23, 2020
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In the beginning, you start by memorizing the rules of partial fraction decomposition, then by solving some problems, you will get the idea.

I will give you some examples

\(\displaystyle \frac{1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x - 2}\)

\(\displaystyle \frac{1}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 2}\)

\(\displaystyle \frac{1}{(x+1)^3(x-2)} = \frac{A}{x+1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3} + \frac{D}{x - 2}\)

\(\displaystyle \frac{1}{(x+1)^3(x-2)^2} = \frac{A}{x+1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3} + \frac{D}{x - 2} + \frac{E}{(x - 2)^2}\)

Did you get the idea?

Things will get interesting when we have \(\displaystyle x^2\) inside the brackets

\(\displaystyle \frac{1}{(x^2+1)(x-2)} = \frac{A + Bx}{x^2+1} + \frac{C}{x - 2}\)

\(\displaystyle \frac{1}{(x^2+1)^2(x-2)} = \frac{A + Bx}{x^2+1} + \frac{C + Dx}{(x^2 + 1)^2} + \frac{E}{x - 2}\)

\(\displaystyle \frac{1}{(x^2+1)^2(x-2)^2} = \frac{A + Bx}{x^2+1} + \frac{C + Dx}{(x^2 + 1)^2} + \frac{E}{x - 2} + \frac{F}{(x - 2)^2}\)

And so on......
 

lookagain

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Aug 22, 2010
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\(\displaystyle \frac{1}{(x^2+1)(x-2)} = \frac{A + Bx}{x^2+1} + \frac{C}{x - 2}\)

\(\displaystyle \frac{1}{(x^2+1)^2(x-2)} = \frac{A + Bx}{x^2+1} + \frac{C + Dx}{(x^2 + 1)^2} + \frac{E}{x - 2}\)

\(\displaystyle \frac{1}{(x^2+1)^2(x-2)^2} = \frac{A + Bx}{x^2+1} + \frac{C + Dx}{(x^2 + 1)^2} + \frac{E}{x - 2} + \frac{F}{(x - 2)^2}\)

nasi112, it is standard for these numerators to be written as Ax + B, Cx + D, etc.,
respectively.
 

apple2357

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Mar 9, 2018
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436
I cannot seem to wrap my head around the idea that when doing partial fraction decomposition you "build up powers". My professor gave me a simplified explanation (lower left corner), but it doesn't make sense to me that the denominator can be split into several with different powers.
View attachment 26273
Good question. I think the problem arises because when you are looking at partial fractions like
1/(x-3)(x-4) say it feels natural to break this up as A/(x-3) + B/(x-4)

So when you come across 1/ (x-3)^2 , the tendency is to think A/(x-3) + B/(x-3) ???
I would invite you to try finding A & B and see what happens?

In essence you need to think about factors of the denominator when breaking up so it would be A/(x-3) + B/(x-3)^2 which looks odd!

Alternatively how would you simplify:

2/(x-3)+ 5/(x-3)^2

What would be the lowest common denominator?

Play with the 'doing' before thinking about the 'undoing' and that might help too.
Its tricky and you are not the first to ask this question, its a good one.
 

lookagain

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Aug 22, 2010
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I think the problem arises because when you are looking at partial fractions like
1/(x-3)(x-4) say it feels natural to break this up as A/(x-3) + B/(x-4)

Don't leave out the grouping symbols:

1/[(x - 3)(x - 4)]
 
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