Partial Fraction Decomposition

In summary: If you let s = +/-2i then +/-2iC + D = -1/4 (2)where as before I equate coefficients it's obvious to me that +/-2iXC != -1/4 and D = -1/4 so C = 0 If you let s = 1 after you get A = 0 also, my questions are how do I come to the conclusion that C = 0 in (2)? The easiest way to solve for C is to use the quadratic equation: C = (-1)s^2 + 2s^2. You can use this to solve for C in (2) by substituting in the values for s that
  • #1
icesalmon
270
13

Homework Statement


use partial fraction decomposition to re-write 1/(s2(s2+4)




The Attempt at a Solution


I thought it would break down into (A/s) + (B/s2) + ((cx+d)/(s2+4)
but it doesn't.
 
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  • #2
icesalmon said:

Homework Statement


use partial fraction decomposition to re-write 1/(s2(s2+4)

The Attempt at a Solution


I thought it would break down into (A/s) + (B/s2) + ((cx+d)/(s2+4)
but it doesn't.
The only thing wrong is that the last term should be (cs + d)/(s2 + 4). Otherwise, your decomposition is correct.
 
  • #3
Okay, thank you sir.
 
  • #4
okay I had this problem solved, but I went back after changing my variable from x to s
and I get A(s)(s2+4) + B(s2+4) +(cs+d)(s2) = 1 (1)
if I let s = 0 then B(4) = 1 -> B = 1/4
if I let s = +/-2i then +/-2iC + D = -1/4 (2)
where as before I equate coefficients it's obvious to me that +/-2iXC != -1/4 and D = -1/4 so C = 0
Letting s = 1 after I get A = 0 also,
my questions are how do I come to the conclusion that C = 0 in (2)? Should I keep plugging values into (1) and try to create a system? Also, how do I pick values for A, once I've actually calculated D, C, and B. What would cause me to choose 1 for S as opposed to any other value?
 
  • #5
icesalmon said:

Homework Statement


use partial fraction decomposition to re-write 1/(s2(s2+4)




The Attempt at a Solution


I thought it would break down into (A/s) + (B/s2) + ((cx+d)/(s2+4)
but it doesn't.

Just write ##s^2 = x##, so you have
[tex] \frac{1}{x(x+4)}[/tex]
The partial fraction expansion for this is
[tex] \frac{1}{4x} - \frac{1}{4(x+4)},[/tex]
and you can now put back ##x = s^2## to get
[tex] \frac{1}{4 s^2} - \frac{1}{4(s^2+4)}[/tex]
If you want, you can even introduce complex numbers and bread this down further into
[tex] \frac{1}{4s^2} +\frac{1}{16 i (s - 2i)} - \frac{1}{16 i (s + 2i)}[/tex]
where ##i = \sqrt{-1}##.
 
  • #6
wow, very cool. I've never done it like that before. Thanks a lot
 
  • #7
icesalmon said:
okay I had this problem solved, but I went back after changing my variable from x to s
and I get A(s)(s2+4) + B(s2+4) +(cs+d)(s2) = 1 (1)
if I let s = 0 then B(4) = 1 -> B = 1/4
if I let s = +/-2i then +/-2iC + D = -1/4 (2)
where as before I equate coefficients it's obvious to me that +/-2iXC != -1/4 and D = -1/4 so C = 0
Letting s = 1 after I get A = 0 also,
my questions are how do I come to the conclusion that C = 0 in (2)?
Because -1/4 has no imaginary part. You can think of your equation as being written like this:
D + 2Ci = -1/4 + 0i
From this we see that D = -1/4 and C = 0.
icesalmon said:
Should I keep plugging values into (1) and try to create a system? Also, how do I pick values for A, once I've actually calculated D, C, and B. What would cause me to choose 1 for S as opposed to any other value?
It doesn't matter what values you choose for s. The only thing to be concerned with is how convenient a particular value is.

The equation you started with -- A/s + B/s2 + (Cs + D)/(s2 + 4) -- has to be identically true. IOW, it has to be true for all values of s. Any four values you choose will give you four equations for the unknowns A, B, C, and D. The strategy is to pick values so that some of the terms go away, making your task of solving the system easier.
 

FAQ: Partial Fraction Decomposition

What is Partial Fraction Decomposition?

Partial Fraction Decomposition is a mathematical process used to break down a rational expression into simpler fractions. It is often used in integration, solving linear differential equations, and simplifying complex algebraic expressions.

Why is Partial Fraction Decomposition useful?

Partial Fraction Decomposition allows us to solve problems that involve rational expressions more easily. It also helps us to identify and understand the behavior of complex functions.

How do you perform Partial Fraction Decomposition?

To perform Partial Fraction Decomposition, you first need to factor the denominator of the rational expression. Then, you write the partial fraction decomposition in the form A/(x-a) + B/(x-b) + ... where A, B, etc. are constants to be determined. You can use various methods such as equating coefficients or the cover-up method to determine the values of the constants.

What are the different types of Partial Fraction Decomposition?

There are three types of Partial Fraction Decomposition: proper, improper, and mixed. In proper decomposition, the degree of the numerator is less than the degree of the denominator. In improper decomposition, the degree of the numerator is equal to or greater than the degree of the denominator. In mixed decomposition, the rational expression has both proper and improper terms.

What are some applications of Partial Fraction Decomposition?

Partial Fraction Decomposition is used in various fields of science and engineering, including physics, chemistry, and electrical engineering. It is particularly useful in solving differential equations, evaluating integrals, and simplifying complex mathematical expressions.

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