Question

Partial Fractions. Could someone please explain how I could use elimination to find the values A,B,C,D? Attached problem and solution...

Answer 1

I don't like the substitution method because it seems like they just guess the values for A & B.

Answer 2

you mean the Heaviside method?

Answer 3

They randomnly chosen x values to solve for A,B,C, and D.
say you have
5+2x=A(x+1)+B(x-1)
You can solve this for A and B...by choosing values of x such that the equation is easier to solve for A and B like choosing x=1 and x=-1

like since
5+2x=A(x+1)+B(x-1) is an equality it must hold for all x values...
So choosing x=-1 and x=1 is fine to solve for A and B.
input x=-1:
5+2(-1)=A(-1+1)+B(-1-1)
5-2=A(0)+B(-2)
3=B(-2)
B=-3/2

And
input x=1: 5+2x=A(x+1)-3/2(x-1)
5+2=A(1+1)-3/2(1-1)
7=A(2)
A=7/2

so
we have
5+2x=7/2(x+1)-3/2(x-1)
you can expand the right hand side to test to see if this is indeed true
7/2(x+1)-3/2(x-1)
=
7/2 x+7/2-3/2 x +3/2
4/2x+10/2
2x+5
and yes 2x+5 is the same as 5+2x

Answer 4

however if you don't like Heaviside method you could just solve a system of linear equations but sometimes you have to do this even with Heaviside method.

Answer 5

okay, so looking at the problem I gave how did they choose the values 1/4 & -1/4 for A and B? Why not 1 and -1 that would also equal 0...

Answer 6

\[A(x+1)(x-1)^2+B(x-1)(x+1)^2+C(x-1)^2+D(x+1)=1 \\ A(x^2-1)(x-1)+B(x^2-1)(x+1)+C(x^2-2x+1)+D(x+1)=1 \\ A(x^3-x^2-x+1)+B(x^3+x^2-x-1)+C(x^2-2x+1)+D(x+1)=1 \\ x^3(A+B)+x^2(-A+B+C)+x(-A-B-2C+D)+(A-B+C+D)=1 \\ \text{ so you have the follow system \to solve } \\ A+B=0 \\ -A+B+C=0 \\ -A-B-2C+D=0 \\ A-B+C+D=1\]

Answer 7

so you want to do heaviside method?

Answer 8

If it will make more sense, lets do it.

Answer 9

I don't know which makes more sense to you...They both make sense to me.
\[A(x+1)(x-1)^2+B(x-1)(x+1)^2+C(x-1)^2+D(x+1)^2=1\]
notice some of these would be 0 if you choose x=1 or x=-1
So choosing input x=1: do you see how we get D(1+1)^2=1?

Answer 10

if you don't see how, please let me know

Answer 11

Gotcha, I see what you did there, so the input values don't matter only the output. Like "D" would equal 1/4. And I would continue putting in values for x until I have only one variable left and solve for it. Big thank you!!

Answer 12

yeah starting with x=1 and x=-1 makes things easier
since most terms will be 0
So the solving for A and B part are a little more difficult but yep you can choose 2 more x values to get a system of 2 linear equations.
\[A(x+1)(x-1)^2+B(x-1)(x+1)^2+\frac{1}{4}(x-1)^2+\frac{1}{4}(x+1)^2=1 \\ \text{ so choose } x=2 \\ A(3)(1)^2+B(1)(3)^2+\frac{1}{4}(1)^2+\frac{1}{4}(2)^2=1\]
you will choose another x value and you will get another equation in terms of A and B
And then solve the system .

Answer 13

now that isn't exactly what they did

Answer 14

it looks like they expanded everything and compared both sides of the equations

Answer 15

but I think the way I'm suggesting might be less work

Answer 16

Great thanks for clarifying.

Answer 17

ok you try to see if you can get their A and B
if you cannot holler at me

Answer 18

Going back to the post where you said make x=2, leftover I would have A and B with constants. So how would I isolate each of them to find individual values?

Answer 19

you need to choose another x value to plug in so you have two linear equations involving 2 different constant variables

Answer 20

like choose x=-2

Answer 21

or any x that we haven't already used

Answer 22

\[3A+9B+\frac{1}{4}+\frac{1}{4}(9)=1 \\ -9A-3B+\frac{1}{4}(9)+\frac{1}{4}=1 \\ \text{ simplifying both equations } \\ 3A+9B=1-\frac{10}{4} =\frac{-6}{4}=\frac{-3}{2} \\ -9A-3B=1-\frac{10}{4}=\frac{-3}{2}\]

so we have the following system to solve:
\[3A+9B=\frac{-3}{2} \\ -9A-3B=\frac{-3}{2} \\ \text{ we can solve by elimination } \\ \text{ multiply first equation by 3 } \\ 9A+27B=\frac{-9}{2} \\ -9A-3B=\frac{-3}{2} \\ \text{ now add equations } \\ 24B=\frac{-12}{2} \\ 24B=-6 \\ B=\frac{-6}{24} =\frac{-1}{4} \\ \text{ now you can plug into either of the two equations here to find } A\]

Answer 23

Oh yeah, I feel stupid for forgetting middle school maths. Thanks I appreciate it.

Answer 24

np

Answer 25

And you aren't stupid. :p