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I'm an engineer, not a math guy. Please use small words if possible. ;-)

I am going through this neat paper on a method of partial fraction decomposition by repeated synthetic division. On page 157 it is covering case with repeated irreducible quadratics in denominator - and it says this:

"Multiply both sides by the denominator...then We reduce the right hand side modulo $x^2+bx+c $ by sending it to the field $\mathbb R[x]/(x^2+bx+c) = $ {$px+q|p,q\in \mathbb{R},x^2+bx+c=0$}. Modulo $x^2+bx+c$....

I have no clue how to "reduce the r.h.s. modulo by sending it to the field". Can someone give me example of how to do such? Preferably with real equations i can ponder like an engineer.

UPDATE 1: Adding real example we can use. Let, $$\frac{N(s)}{D(s)}=\frac{10(s+1)}{(s+1)^3(s^2+5s+6)^3}=\frac{K_1}{s+1}+\frac{K_2}{(s+1)^2}+\frac{K_3}{(s+1)^3}+\frac{M_1s+C_1}{s^2+5s+6}+\frac{M_2s+C_2}{(s^2+5s+6)^2}+\frac{M_3s+C_3}{(s^2+5s+6)^3}$$

Multiplying both sides by D(s) I get, $$N(s) = 10(s+3) = K_1(s+1)^2(s^2+5s+6)^3+K_2(s+1)(s^2+5s+6)^3+K_3(s^2+5s+6)^3+(M_1s+C_1)(s+1)^3(s^2+5s+6)^2+(M_2s+C_2)(s+1)^3(s^2+5s+6)+(M_3s+C_3)(s+1)^3$$

How do i proceed now?

relayman357
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  • Please give more context: which of these topics do you know: congruences / modular arithmetic, rings, fields, ideals, equivalence relation, quotient rings, – Bill Dubuque Nov 05 '20 at 02:21
  • You may find of interest this related post where I show how to extend to nonlinear denominators the Heaviside cover-up method for partial fraction decomposition, by using modular polynomial arithmetic. – Bill Dubuque Nov 05 '20 at 02:26
  • Hi Bill, I understand modular arithmetic and congruence, but that’s about all from your list. – relayman357 Nov 05 '20 at 03:13
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    It just means to divide the polynomial by $x^2+bx +c$ and consider only the remainder, which will be a polynomial of degree at most $1$. It's entirely analogous to modular arithmetic. – saulspatz Nov 05 '20 at 04:03
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    i.e. use congruences modulo $,x^2+bx+c,,$ where $,x^2 \equiv -bx -c,$. Such congruences work the same was as for integers, i.e. we have the congruence sum and product rules, etc. – Bill Dubuque Nov 05 '20 at 04:26
  • @BillDubuque i have added an UPDATE 1 with real equations we can work with. I don't know how to proceed as you suggest in your comments. Can you help? – relayman357 Nov 06 '20 at 18:51
  • @saulspatz - see my UPDATE 1 above and my comment to Bill. – relayman357 Nov 06 '20 at 18:51
  • I don't know if this matters, but your example doesn't have an irreducible quadratic in the denominator. $s^2+5s+6=(s+2)(s+3)$ – saulspatz Nov 06 '20 at 19:11
  • @saulspatz I know, just trying to find example to learn the method....the actual mechanics of the process. Can you help? – relayman357 Nov 06 '20 at 20:26
  • I'm sorry, but I don't know anything about this, and I'm really not interested enough to read the paper. I took a quick look at it, but I didn't understand the sentence you're asking about, either. I'd need to read the paper to figure out the context. – saulspatz Nov 06 '20 at 20:55

1 Answers1

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$$\begin{align} 10(s+3) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &+ K_3(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$

Let $s = -1$ and you get $20 = 8K_3$. So $K_3 = \dfrac 52$

$$\begin{align} 10(s+3) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &-\dfrac 52(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$

$$\begin{align} -\dfrac 52 (s + 3) (s + 1) (s^4 + 11 s^3 + 46 s^2 + 88 s + 68) &= K_1(s+1)^2(s^2+5s+6)^3 \\ & +K_2(s+1)(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^3 \end{align}$$

$$\begin{align} -\dfrac 52 (s + 3)(s^4 + 11 s^3 + 46 s^2 + 88 s + 68) &= K_1(s+1)(s^2+5s+6)^3 \\ & +K_2(s^2+5s+6)^3 \\ &+ (M_1s+C_1)(s+1)^2(s^2+5s+6)^2 \\ &+ (M_2s+C_2)(s+1)^2(s^2+5s+6) \\ &+ (M_3s+C_3)(s+1)^2 \end{align}$$

Let $s=-1$ again and you get...

Added 11/22/2020

Your original equation is wrong. It should look like

\begin{align} \frac{10}{(s+1)^2 (s+2)^3 (s+3)^3} &=\\ &\frac{K_1}{(s+1)} + \frac{K_2}{(s+1)^2} + \\ &\frac{L_1}{(s+2)} + \frac{L_2}{(s+2)^2} + \frac{L_3}{(s+2)^3} + \\ &\frac{M_1}{(s+3)} + \frac{M_2}{(s+3)^2} + \frac{M_3}{(s+3)^3} \end{align}

You then get the equation

\begin{align} 10 &= K_1(s+1)(s+2)^2 (s+3)^3 \\ &+ K_2(s+2)^3 (s+3)^3 + \\ &+ L_1(s+1)^2 (s+2)^2 (s+3)^3 \\ &+ L_2(s+1)^2 (s+2) (s+3)^3 \\ &+ L_3(s+1)^2 (s+3)^3 \\ &+ M_1 (s+1)^2 (s+2)^3 (s+3)^2 \\ &+ M_2 (s+1)^2 (s+2)^3 (s+3) \\ &+ M_3 (s+1)^2 (s+2)^3 \\ \end{align}

Now let $s=-1$. You get $10=8K_2 \implies K_2 = \dfrac 54$. Simplifying, we get

\begin{align} \dfrac 58 (s + 1) (9 s^5 + 124 s^4 + 685 s^3 + 1902 s^2 + 2668 s + 1528) &= K_1(s+2)^3 (s+3)^3 \\ &+ L_1(s+1) (s+2)^2 (s+3)^3 \\ &+ L_2(s+1) (s+2) (s+3)^3 \\ &+ L_3(s+1) (s+3)^3 \\ &+ M_1 (s+1) (s+2)^3 (s+3)^2 \\ &+ M_2 (s+1) (s+2)^3 (s+3) \\ &+ M_3 (s+1) (s+2)^3 \\ \end{align}

Again let $s=-1$. You get $-45 = 8K_1 \implies K_1 = -\dfrac{45}{8}$. Simplifying, we get

\begin{align} -\dfrac 54 (4 + s)(52 + 70 s + 39 s^2 + 10 s^3 + s^4) &= L_1 (s+2)^2 (s+3)^3 \\ &+ L_2 (s+2) (s+3)^3 \\ &+ L_3 (s+3)^3 \\ &+ M_1 (s+2)^3 (s+3)^2 \\ &+ M_2 (s+2)^3 (s+3) \\ &+ M_3 (s+2)^3 \\ \end{align}

Steven Alexis Gregory
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