Partial Fractions

• Jun 23rd 2008, 10:36 AM
JonathanEyoon
Partial Fractions
Need some insight into how to determine A, C, and D

(x^2 - 2x - 1) /(((x - 1)^2)(x^2 + 1)

A / ( x - 1) + B / (x - 1)² + (Cx +D) / (x² + 1)

x² - 2x - 1 = A(x - 1 )(x² + 1) - (x² +1) + (Cx + D)(x - 1)²

Now I know that B = -1. What i'm having trouble is figuring out the others. I've tried foiling out everything but then when I equate coefficients, i'll have two variables equal to each other. Any help is appreciated (Worried)
• Jun 23rd 2008, 11:01 AM
Moo
Hello :)

Quote:

Originally Posted by JonathanEyoon
Need some insight into how to determine A, C, and D

(x^2 - 2x - 1) /(((x - 1)^2)(x^2 + 1)

A / ( x - 1) + B / (x - 1)² + (Cx +D) / (x² + 1)

x² - 2x - 1 = A(x - 1 )(x² + 1) - (x² +1) + (Cx + D)(x - 1)²

Now I know that B = -1. What i'm having trouble is figuring out the others. I've tried foiling out everything but then when I equate coefficients, i'll have two variables equal to each other. Any help is appreciated (Worried)

$\displaystyle x^2- 2x - 1 = A(x - 1 )(x^2 + 1) + B(x^2 +1) + (Cx + D)(x - 1)^2$

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Using rough expanding :

$\displaystyle {\color{red}x^2}-2{\color{green}x}-1=A{\color{magenta}x^3}-A{\color{red}x^2}+A{\color{green}x}-A+B{\color{red}x^2}+B+C{\color{magenta}x^3}+D{\col or{red}x^2}-2C{\color{red}x^2}-2D{\color{green}x}+C{\color{green}x}+D$

x^3 coefficients :
$\displaystyle A+C=0$

x² coefficients :
$\displaystyle 1=-A+B+D-2C$

x coefficients :
$\displaystyle -2=A-2D+C$

constant coefficients :
$\displaystyle -1=-A+B+D$

You're right, you can easily get B.
$\displaystyle \boxed{B=-1}$

Substituting in the previous equations :

$\displaystyle \left\{\begin{array}{llll} 0&=&A+C \\ 2&=&-A-2C+D \\ -2&=&A+C-2D \\ 0&=&-A+D \end{array} \right.$

From the first equation, we have $\displaystyle C=-A$ and from the fourth, $\displaystyle A=D$

Substituting again :

$\displaystyle \left\{\begin{array}{ccc} 2&=&-A+2A+A \\ -2&=&A-A-2A \end{array} \right.$

$\displaystyle \implies \boxed{A=1}$

Therefore $\displaystyle \boxed{D=1}$ and $\displaystyle \boxed{C=-1}$

(Whew)

---------------------------

There is a faster way, in noticing that $\displaystyle x^2+1=(x-i)(x+i)$

But if you didn't study the complex numbers, it won't be easy for you to grasp it immediately :p

---------------------------

There is another way, involving differentiation (you do the same trick as you did for B, after you have differentiated (Wink))
• Jun 23rd 2008, 11:07 AM
Krizalid
Note that

$\displaystyle \frac{x^{2}-2x-1}{(x-1)^{2}\left( x^{2}+1 \right)}=\frac{(x-1)^{2}-2}{(x-1)^{2}\left( x^{2}+1 \right)}=\frac{1}{x^{2}+1}-\frac{2}{(x-1)^{2}\left( x^{2}+1 \right)}.$

So the remaining challenge is to descompose $\displaystyle \frac{2}{(x-1)^{2}\left( x^{2}+1 \right)}.$
• Jun 23rd 2008, 11:33 AM
JonathanEyoon
Moo

That was pretty confusing but after studying it, it makes perfect sense. I appreciate it

Question. You foiled everything out but is there a simpler method to do this without expanding everything? I'm asking because the professor told us we can use three methods : Finger method, Equating Coefficients, and plugging an arbitrary value for X to solve for A, B, C, and D.

I was wondering if you could show me using the last method perhaps?
• Jun 23rd 2008, 11:52 AM
Moo
Quote:

Originally Posted by JonathanEyoon
Moo

That was pretty confusing but after studying it, it makes perfect sense. I appreciate it

Question. You foiled everything out but is there a simpler method to do this without expanding everything? I'm asking because the professor told us we can use three methods : Finger method, Equating Coefficients, and plugging an arbitrary value for X to solve for A, B, C, and D.

I was wondering if you could show me using the last method perhaps?

Uuum, yeah I'm sorry sorry, I have the bad habit to skip some steps in calculations...

For the last method... Ok.

------------------------------
So you know that $\displaystyle B=-1$, by taking $\displaystyle x=1$.

The equation is now :

$\displaystyle x^2- 2x - 1 = A(x - 1 )(x^2 + 1) -(x^2 +1) + (Cx + D)(x - 1)^2$

Take for example $\displaystyle x=2$. This will simplify calculations because $\displaystyle 2-1=1$.

$\displaystyle (2)^2-2(2)-1=A(1)(5)-(5)+(2C+D)(1)$

$\displaystyle -1=5A-5+2C+D \implies 4=5A+2C+D \ {\color{red}(1)}$

Take for example $\displaystyle x=0$

$\displaystyle -1=A(-1)(1)-(1)+D(0-1)^2$

$\displaystyle -1=-A+D-1 \implies A-D=0 \ {\color{red}(2)}$

Take for example $\displaystyle x=-1$

$\displaystyle 1-2(-1)-1=A(-2)(2)-2+(-C+D)(-2)^2$

$\displaystyle 2=-4A-2-4C+4D \implies A+C+1=D \ {\color{red}(3)}$

Use basic substitutions in (1), (2) and (3) (Tongueout)
• Jun 23rd 2008, 12:07 PM
JonathanEyoon
In your first line, I thought since B = -1 it's supposed to be - (x^2 + 1)? Or is there a rule where we don't put in the -1?

Edit : I'm even more lost by that last method. If you have three variables in one equation, how are you supposed to know each one?

you have -6 = 5A + 2C + D. How am I supposed to get those variables from here? Thanks and sorry for the many questions.
• Jun 23rd 2008, 12:12 PM
Moo
Quote:

Originally Posted by JonathanEyoon
In your first line, I thought since B = -1 it's supposed to be - (x^2 + 1)? Or is there a rule where we don't put in the -1?

Edit : I'm even more lost by that last method. If you have three variables in one equation, how are you supposed to know each one?

you have -6 = 5A + 2C + D. How am I supposed to get those variables from here? Thanks and sorry for the many questions.

I made some typos... I edited it (Wink)
I'm sincerely sorry for the confusion ! (Bow)

For example, from (2), you have A=D.
Then, substitute in (1) and (3), etc...

You have to solve this system :

$\displaystyle \left\{\begin{array}{lll} {\color{red}(1)} \\ {\color{red}(2)} \\ {\color{red}(3)} \end{array} \right.$

These (1), (2) & (3) label equations :p
• Jun 23rd 2008, 12:38 PM
JonathanEyoon
Ah ic what you mean now. I'll give it a try!