Partial Fraction Decomposition

In summary, partial fraction decomposition is a mathematical method used to simplify rational functions by breaking them down into smaller fractions. It is especially useful in integration and involves factoring the denominator, writing the function as a sum of simpler fractions, and determining unknown coefficients. There are three types of partial fraction decomposition: proper, improper, and mixed. However, it is not possible when the denominator cannot be factored or when the numerator's degree is equal or greater than the denominator's degree.
  • #1
ns5032
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0

Homework Statement


I am just trying to do partial fraction decomposition on an equation. I'm not too good with it, as far as knowing if I need just A or Ax+B, etc.

[e^(-2s) / (s^2+1)(s-1)(s+1)^2]

Homework Equations



The Attempt at a Solution



I'm not quite sure how to work with the e^-2s, but as far as doing the partial fractions, is this right: ?

Ax+B/(s^2+1) + C/(s-1) + D/(s+1) + E/(s+1)

I really just need help with that first step, making sure I set it up right, and also on how to deal with the e^(-2s). Do I take it out and treat it as 1? Do I set what I get from the partial fraction decomposition equal to e^(-2s) or something else??

Thanks!
 
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  • #2
ns5032 said:

Homework Statement


I am just trying to do partial fraction decomposition on an equation. I'm not too good with it, as far as knowing if I need just A or Ax+B, etc.

[e^(-2s) / (s^2+1)(s-1)(s+1)^2]

Homework Equations



The Attempt at a Solution



I'm not quite sure how to work with the e^-2s, but as far as doing the partial fractions, is this right: ?

Ax+B/(s^2+1) + C/(s-1) + D/(s+1) + E/(s+1)
The last one should be E/(s+1)^2.

I really just need help with that first step, making sure I set it up right, and also on how to deal with the e^(-2s). Do I take it out and treat it as 1? Do I set what I get from the partial fraction decomposition equal to e^(-2s) or something else??

Thanks!
No, you do not treat e^(-2s) as 1- it isn't!
You write
[tex]\frac{e^{-2s}}{(s^2+1)(s-1)(s+1)^2}= \frac{As+B}{s^2+1}+ \frac{C}{s-1}+ \frac{D}{s+1}+ E/(s+1)^2[/itex]
for all x and solve for A, B, C, D, E.

Probably the simplest way is to multiply both sides by [itex](s^2+ 1)(s- 1)(s+1)^2[/itex] to get rid of the fractions, then take s equal to whatever 5 numbers you wish so you get 5 equations to solve.
 

1. What is partial fraction decomposition?

Partial fraction decomposition is a mathematical method used to break down a rational function into simpler fractions. It involves expressing a single fraction as a sum of smaller fractions with denominators that are irreducible (cannot be further simplified).

2. Why is partial fraction decomposition useful?

Partial fraction decomposition allows us to simplify complex rational functions and make them easier to work with. It is especially useful in integration, as it can help us solve integrals that would otherwise be difficult or impossible to solve.

3. How do you perform partial fraction decomposition?

The steps for performing partial fraction decomposition are as follows:

  1. Factor the denominator of the rational function into irreducible factors.
  2. Write the rational function as a sum of simpler fractions with these irreducible factors as denominators.
  3. Determine the unknown coefficients in each fraction by equating the numerators of the fractions.
  4. Combine like terms and simplify the resulting fractions.

4. What are the types of partial fraction decomposition?

There are three types of partial fraction decomposition: proper, improper, and mixed. Proper decomposition involves only using proper fractions (where the numerator is smaller than the denominator). Improper decomposition involves using at least one improper fraction. Mixed decomposition involves a combination of proper and improper fractions.

5. When is partial fraction decomposition not possible?

Partial fraction decomposition is not possible when the denominator of the rational function cannot be factored into irreducible factors. It is also not possible when the degree of the numerator is greater than or equal to the degree of the denominator.

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