Partial Fractions: Solving s/(s^2 + 4)(s^2 + 9)

AI Thread Summary
The discussion focuses on the correct approach to decomposing the expression s/[(s^2 + 4)(s^2 + 9)] into partial fractions. Participants clarify that the initial assumption of using constants A and B alone is incorrect, as the irreducible quadratic factors require a more complex decomposition involving additional constants. The correct form includes both linear terms for each quadratic, leading to four unknowns to solve for. A method is suggested to derive equations by substituting specific values for s, but it is emphasized that A and B must be constants, not functions of s. The conversation ultimately highlights the importance of ensuring that the equations derived from the decomposition are identically true for all values of s.
a.mlw.walker
Messages
144
Reaction score
0
Hi, had to learn partial fractions last year for laplace transforms, but have forgotten the general rules, and now i can't work out how to turn this into partial fractions:

\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}

\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}=\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}
so

s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)

i understand that i am supposed to strategically choose s to make A or B zero, but sith the s^2 I'm not sure what to do

thanks
 
Mathematics news on Phys.org
make the coefficient 0
 
by setting s to what?
 
a.mlw.walker said:
Hi, had to learn partial fractions last year for laplace transforms, but have forgotten the general rules, and now i can't work out how to turn this into partial fractions:

\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}

\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}=\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}
This is not the way to decompose irreducible quadratic factors. Here's what you should have:
\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)}=\frac{As + B}{s^{2}+4}+\frac{Cs + D}{s^{2}+9}

a.mlw.walker said:
so

s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)

i understand that i am supposed to strategically choose s to make A or B zero, but sith the s^2 I'm not sure what to do

thanks
 
ok, yeah i wondered, but mupad's soln looked so simple i didnt think there could be 4 unkowns. So can i now multiply everything by the denomiator of s on the LHS.

What i don't understand is usually you would choose values for s to make most unkowns 0, then calculate for each unkown. but here, there is no s that can make
S^2 + 4 =0 without imaginary numbers?
 
a.mlw.walker said:
ok, yeah i wondered, but mupad's soln looked so simple i didnt think there could be 4 unkowns. So can i now multiply everything by the denomiator of s on the LHS.

What i don't understand is usually you would choose values for s to make most unkowns 0, then calculate for each unkown. but here, there is no s that can make
S^2 + 4 =0 without imaginary numbers?

You should then write the partial fraction expansion as follows:

s/[(s^2+4)(s^2+9)] = A/(s+2i) + A*/(s-2i) + B/(s+3i) + B*(s-3i)

Where the star denotes complex conjugation. The "strategic value" method now becomes trivial, you don't have to actually write down any equations. If you know about complex analysis, you'll recognize the coefficients as the residues of the function at the poles and compute them directly. If you don't know this, you can easily see that e.g. multiplying boith sides by s + 2i and taking the limit s to -2i will yield A.
 
An obvious way to simplify
\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{As+B}{s^{2}+4}+\frac{Cs+D}{s^{2}+9}
is to multiply both sides by (s^2+ 4)(s^2+ 9) to get
s= (As+B)(s^2+9)+ (Cs+D)(s^2+4).

Because those denominators were "irreducible" you cannot "make a coefficient 0" but you still need only 4 equations to solve for the 4 constants.

One way would be to take s to be 4 simple numbers, say s= 0, s= 1, s= -1, and s= 2.

If n= 0, your equation is 0= 9B+ 4D. If s= 1, it is 1= 10A+ 10B +5s+ 5D, etc.

Another way to get 4 equations is to multiply out that right side:

s= (As+B)(s^2+9)+ (Cs+D)(s^2+9)= As^3+ Bs^2+ As+ 9B+ Cs^3+ Ds^2+ Cs+ D.
0s^3+ 0s^2+ s+ 0= (A+ C)s^3+ (B+D)s^2+ (A+C)+ B+D.

In order for those to be equal for all s, the "corresponding coefficients" must be the same: we must have A+ C= 0, B+ D= 0, A+ C= 1, and B+ D= 0.
 
Why can't this be done:
As^2 +9A +Bs^2 +4B=0
A+B=0
9A+4B=s
A=s/5, B=-s/5
 
vin300 said:
Why can't this be done:
As^2 +9A +Bs^2 +4B=0
A+B=0
9A+4B=s
A=s/5, B=-s/5
Mainly because it isn't correct. It took me some time to figure out what you had done, but going back to the OP, I was able to follow your reasoning. In the first post in this thread, a.mlw.walker thought that (incorrectly) that this would be the way to go:
\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}
And this leads to:
s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)

Your first equation is As^2 +9A +Bs^2 +4B=0, but where you have zero on the right side, you should have s, but I think that was a typo on your part. You are reasoning that since there is no s^2 term on the right side, A + B (the coefficient of s^2) has to be zero, and that 9A + 4B has to be s.

Your values for A and B satisfy the equation A + B = 0, but they don't satisfy 9A + 4B = s.
 
  • #10
Why not? What is 9s/5 -4s/5 ?
 
  • #11
Solved it by doing what Mark44 said.
 
  • #12
vin300 said:
Why not? What is 9s/5 -4s/5 ?
9s/5 - 4s/5 = s

The main problem with your technique is the difference between equations that are identically true (true for all values of the variable) and those that are conditionally true (true only for certain values of the variable). In partial fractions decomposition, the idea is to rewrite the product on the left side below as a sum in such a way that the equation is identically true. A and B on the right side have to be constants.
\frac{s}{\left(s^{2}+4\right)\left(s^{2}+9\right)} =\frac{A}{s^{2}+4}+\frac{B}{s^{2}+9}

By multiplying both sides by (s2 + 4)(s2 + 9), you get the next equation, which also has to be identically true.
s=A\left(s^{2}+9\right)+B\left(s^{2}+4\right)

If we group like terms together on the right side, we get
s = (A + B)s2 + (9A + 4B)

This is the same as 0s2 + 1s + 0 = (A + B)s2 + (9A + 4B)

For this equation to be identically true, we have to have A + B = 0 and 9A + 4B = 0 so that the coefficients of s2, s, and the constant term are equal on both sides of the equation. The only way two polynomials of the same degree can be identically equal is for their corresponding coefficients to be equal.

Using you technique, we have A + B = 0 and 9A + 4B = 0, which gives A = B = 0, and we also have an s term on the left side that is not matched on the right side.

As it turns out in this problem, your values of A = s/5 and B = -s/5 happen to work out, but I'm reasonably sure this is only an accident.
 
  • #13
you were talking to vin300 not me right?
 
  • #14
vin300 said:
Why can't this be done:
As^2 +9A +Bs^2 +4B=0
A+B=0
9A+4B=s
A=s/5, B=-s/5
The whole point of "partial fractions" is that A and B must be constants, not functions of s.
 
Back
Top