Partial Fractions (1 Viewer)

[Trebla]

Hi there.....

I'm having trouble understanding a particular method of decomposition into partial fractions. (NB: _ means subscript cos I don't know how to make subscripts here)
"One way of carrying out a decomposition by using the fact that c_i = R(a_i)/B'(a_i). This may be derived by noting that if
R(x)/B(x) = c_1 / (x - a_1) +...........+ c_n/ (x - a_n)
where B(x) = (x - a_1).............(x - a_n)
.: B(a_1) = 0
then multiply both sides by (x - a_1)
.: R(x).(x - a_1)/[B(x) - B(a_1)] = c_1 + (x - a_1)c_2/(x - a_2) +......+ (x - a_1)c_n/(x - a_n)
As x --> a_1, then LHS --> R(a_1)/B'(a_1) and RHS --> c_1
.: c_1 = R(a_1)/B'(a_1)"

The part in bold is what I don't understand. Where did the derivative come from?

 

[sasquatch]

Hmm i was just mucking around and i did this:

B(x) = (x -a₁) * P(x), where

P(x) = (x - a₂)....(x - a_n)

B'(x) = (x-a₁) * P'(x) + P(x)
B'(a₁) = P(a₁)

so

R(x) = B(x) [c₁/(x-a₁) + ... + c_n/(x-a_n]
= (x -a₁) P(x) [c₁/(x-a₁) + ... + c_n/(x-a_n)
= P(x) [c₁ + c₂(x-a₁)/(x-a₂) + ... + c_n(x-a₁)/(x-a_n)

So R(a₁) = P(a₁) [ c₁]

B'(a₁) = P(a₁)

R(a₁)/ B'(a₁) = [P(a₁)c₁] / P(a₁)
= c₁

But i dunno it works...

 

[SeDaTeD]

How nifty.

Well, by first principles, B'(a_1) = lim x -> a_1 [B(x) - B(a_1)]/(x-a_1), and R(x) -> R(a_1) as x -> a_1. So, as x -> a_1, R(x)(x - a_1)/[B(x) - B(a_1)] -> R(x)/B'(x).

 

[sasquatch]

 

[Trebla]

OK thanks guys I get it now...but just another thing, when using this method do we have to do all that deriving or can we assume the formula as it is quoted in the syllabus?

 

[sasquatch]

Ive never seen that before... where in the syllabus is it?

 

[SeDaTeD]

Well, it may look simple to get the values you want, but you must note that it involves differentiating B(x), which may get a bit difficult if it had a lot of factors in it, unless you expand the whole thing. Then you have to sub it in for each c_1, which may be quite time consuming also.

Is it actually quoted in the syllabus? If so then you should be able to use it.

 

[Riviet]

Whether it's quoted in the syllabus or not, I would use the more straight forward methods of splitting up a fraction, since you will be racing against time in your trials and final exams.

 

[Trebla]

The method is quoted under section 7.6 under the Applications, Implications and Considerations heading. It's hardly ever used in the course, my teacher mentioned it briefly and my tutors never taught this method.
I was just a bit curious about this method, because this is the first time I've seen it. Now I know 3 methods of finding partial fractions with linear factors in the denominator....lol

e.g. (3x - 2)/(x - 1)(x - 2) ≡ a/(x - 1) + b/(x - 2)

Method 1: (equating coefficients)
(3x - 2) ≡ a(x - 2) + b(x - 1)
(3x - 2) ≡ ax - 2a + bx - b
(3x - 2) ≡ x(a + b) - 2a - b
a + b = 3 ==> (1)
- 2a - b = - 2 => (2)
(1) + (2)
- a = 1
.: a = - 1
Sub into (1)
.: b = 4

Method 2: (substituting selected values of x)
(3x - 2) ≡ a(x - 2) + b(x - 1)
When x = 1
- a = 1
.: a = - 1
When x = 2
.: b = 4

OR use a combination of both of the above...

Method 3: (using a derivative substitution)
(3x - 2)/(x - 1)(x - 2) = (3x - 2)/(x² - 3x + 2)
R(x)/B'(x) = (3x - 2)/(2x - 3)
a = (3(1) - 2)/(2(1) - 3)
.: a = - 1
b = (3(2) - 2)/(2(2) - 3)
.: b = 4

 

[DraconisV]

Guys is this the polynomials topic you are talking about here.
Coz i havent started that and what your typing looks like japanese to me at the moment.

I hope its 4unit polynomials and not something ive done and should know.

 

[pLuvia]

Guys is this the polynomials topic you are talking about here.
Coz i havent started that and what your typing looks like japanese to me at the moment.

I hope its 4unit polynomials and not something ive done and should know.

It is 4 unit polynomials just partial fractions, the method is to split one fraction into two, usually implemented in integration which makes it much easier to integrate