Why Can't x/(x+1)^2 Be Split Like x/((x+1)(x+2))?

In summary, partial fractions involve decomposing a fraction with a polynomial in the denominator into simpler fractions. However, when there is a squared term in the denominator, a different procedure is used because the original method does not work. This is due to the fact that in this case, the factors are not coprime and therefore cannot be split into separate fractions.
  • #1
FS98
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im a bit confused about partial fractions

If we have something like x/((x+1)(x+2)) we could decompose it into a/(x+1) +b/(x+2)

If we had something like x/(x+1)^2 we could decompose it into a/(x+1) + b/(x+1)^2

We use a different procedure when there is a square in part of the polynomial in the denominator.

What I don’t understand is why in the second case it wouldn’t be split into a/(x+1) + b/(x+1).

x/(x+1)^2 can be written as x/(x+1)(x+1). Then couldn’t we just use the procedure used in the first case?
 
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  • #2
In the decomposition to partial fractions, we use the fact that factors are coprime. For coprime elements ##n,m## we can find other elements ##a,b## such that ##1=an+bm## by the Euclidean algorithm and the ##1## can be extended to what we need.

Say we want to solve ##\dfrac{z}{n\cdot m} = \dfrac{q}{n}+\dfrac{p}{m}## with coprime ##n,m## and we found ##a,b## such that ##1=an+bm##. Then we have ##z= qm+pn=z\cdot 1 = zan+zbm## and thus ##q=zb## and ##p=za## or ##\dfrac{z}{n\cdot m}=\dfrac{zb}{n}+\dfrac{za}{m}##. All this because of Bézout's identity, resp. the fact that ##n,m## are coprime.
 
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  • #3
FS98 said:
What I don’t understand is why in the second case it wouldn’t be split into a/(x+1) + b/(x+1).
The simple answer is because it doesn't work. Notice that ##\frac a {x + 1} + \frac b {x + 1}## is really the same as ##\frac c {x + 1}##, so you won't get that second term with the squared factor in the denominator.
 
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Related to Why Can't x/(x+1)^2 Be Split Like x/((x+1)(x+2))?

What are partial fractions?

Partial fractions are a mathematical method used to simplify and decompose a rational function into smaller, simpler fractions. This allows for easier integration and solving of equations involving rational functions.

Why do we use partial fractions?

Partial fractions are used to break down more complex rational functions into simpler ones, making it easier to solve equations involving these functions. They also allow for more efficient integration and manipulation of rational functions.

How do we find the partial fraction decomposition for a given rational function?

The partial fraction decomposition involves finding the individual fractions that, when added together, form the original rational function. This is done by setting up a system of equations, equating the numerators and denominators of the individual fractions, and solving for the unknown coefficients.

What are the different types of partial fraction decomposition?

The two main types of partial fraction decompositions are known as "proper" and "improper" fractions. Proper fractions have a degree of the numerator that is less than the degree of the denominator, while improper fractions have a numerator with a degree greater than or equal to the denominator.

Can we always use partial fractions to solve rational functions?

No, not all rational functions can be broken down into simpler partial fractions. Some may have complex roots, repeated roots, or other special cases that require different methods for solving.

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