Partial Fractions

[jonnburton]

Hi all,

There's a worked example on partial fractions in my textbook and I have a couple of questions which are bothering me. I was wondering whether anyone could help me with this?

Express \(\displaystyle \frac {1}{x^2(x-1)}\) in the form \(\displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\)


Solution:
\(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\)

The first thing I can't see is how the denominator can be split up in this way. When you 'split up' a denominator, the product of the parts should be the original denominator. But That's not what I get here: \(\displaystyle x * x^2 * (x-1) = x^3(x-1)\).


The other thing which doesn't make sense to me is the next stage, which goes on to say:

\(\displaystyle \frac {1}{x^2(x-1)} = \frac {Ax(x-1) + B(x-1) +Cx^2}{x^2(x-1)}\)

Given \(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\), I would have written this part as \(\displaystyle \frac {Ax^2(x-1) + Bx(x-1) +Cx^3}{x^2(x-1)}\)

As ever, I'd be very grateful for any help with this because it is not making very much sense at the moment!

 

[mathmari]

The first thing I can't see is how the denominator can be split up in this way. When you 'split up' a denominator, the product of the parts should be the original denominator. But That's not what I get here: \(\displaystyle x * x^2 * (x-1) = x^3(x-1)\).

If the denominator is raised to a power r, in your case r=2, then when you split it up you have to involve all the powers from 1 to r,
\(\displaystyle \frac {1} {x^2}= \frac {A} {x^1} + \frac {B} {x^2} \)

The other thing which doesn't make sense to me is the next stage, which goes on to say:

\(\displaystyle \frac {1}{x^2(x-1)} = \frac {Ax(x-1) + B(x-1) +Cx^2}{x^2(x-1)}\)

Given \(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\), I would have written this part as \(\displaystyle \frac {Ax^2(x-1) + Bx(x-1) +Cx^3}{x^2(x-1)}\)

\(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\)
At the second part, you want that all the terms have as denominator \(\displaystyle x^2(x-1) \), so the term \(\displaystyle \frac{A}{x} \) has already the term x at the denominator so you need only to multiply \(\displaystyle x(x-1) \) both at the numerator and the denominator, the term \(\displaystyle \frac{B}{x^2} \) has already \(\displaystyle x^2 \) so you need only to multiply \(\displaystyle x-1 \) and the term \(\displaystyle \frac{C}{x-1} \) has already \(\displaystyle x-1 \) so you need to multiply this term by \(\displaystyle x^2 \) .

 

[Deleted member 4993]

Hi all,

There's a worked example on partial fractions in my textbook and I have a couple of questions which are bothering me. I was wondering whether anyone could help me with this?

Express \(\displaystyle \frac {1}{x^2(x-1)}\) in the form \(\displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\)


Solution:
\(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\)

The first thing I can't see is how the denominator can be split up in this way. When you 'split up' a denominator, the product of the parts should be the original denominator (No ... the LCM of the individual denominator should be the original denominator). But That's not what I get here: \(\displaystyle x * x^2 * (x-1) = x^3(x-1)\).


The other thing which doesn't make sense to me is the next stage, which goes on to say:

\(\displaystyle \frac {1}{x^2(x-1)} = \frac {Ax(x-1) + B(x-1) +Cx^2}{x^2(x-1)}\)

Given \(\displaystyle \frac {1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\),

I would have written this part as

\(\displaystyle \frac {Ax^2(x-1) + Bx(x-1) +Cx^3}{x^2(x-1)}\) ..... How did you get that!!

That simplifies to:

\(\displaystyle \displaystyle A \ + \ \frac{B}{x} \ + \ \frac{C*x}{(x-1)}\)

As ever, I'd be very grateful for any help with this because it is not making very much sense at the moment!

The LCM of x, x² & (x-1) is x² * (x-1)

Just like LCM of 3, 9 and 2 is (9 * 2 =) 18 (and not 54)

Can you add \(\displaystyle \displaystyle \frac{2}{3} \ + \ \frac{5}{9} \ + \ \frac{1}{2}\) - do it and check the answer with calculator.

 

[lookagain]

If the denominator is raised to a power r, in your case r=2, then when you split it up you have to involve all the powers from 1 to r,
> > \(\displaystyle \frac {1} {x^2}= \frac {1} {x^1} + \frac {1} {x^2} \) < <

mathmari, that highlighted part is not true.

 

[mathmari]

mathmari, that highlighted part is not true.

Oh, sorry! It's \(\displaystyle \frac {1} {x^2}= \frac {A} {x} + \frac {B} {x^2} \) .

 

[jonnburton]

The LCM of x, x² & (x-1) is x² * (x-1)

Just like LCM of 3, 9 and 2 is (9 * 2 =) 18 (and not 54)

Can you add \(\displaystyle \displaystyle \frac{2}{3} \ + \ \frac{5}{9} \ + \ \frac{1}{2}\) - do it and check the answer with calculator.

Thank you for the posts...

I think my problem is I find these things much easier to visualise when using numbers: I managed to add those fractions correctly, after checking with a calculator.

Having said that, I think I am starting to understand this now!

 

[jonnburton]

Oh, sorry! It's \(\displaystyle \frac {1} {x^2}= \frac {A} {x} + \frac {B} {x^2} \) .

Thanks! I had to write that down and work it out but I get that now. These things are never immediatly obvious to me...

 

[HallsofIvy]

Oh, sorry! It's \(\displaystyle \frac {1} {x^2}= \frac {A} {x} + \frac {B} {x^2} \) .

That is trivially true with A= 0, B= 1. I think what you intended was
\(\displaystyle \frac{ax+ b}{x^2}= \frac{a}{x}+ \frac{b}{x^2}\) which is still pretty trivial. Of more interest is
\(\displaystyle \frac{ax+ b}{x^2(x- c)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x- c}\)
(Of course, A, B, and C are not necessarily equal to a, b, and c.)