Partial fractions nth degree factors in the denominator

[nobahar]

Hello!
Quick question reagrding partial fractions.
When there is a factor such as (x+2)³ in the denominator, then the fraction is separated into the components (x+2)¹+...+(x+2)³.
I am not convinced I understand quite why this is so. Partial fractions guides all offer something on the theme of: "Here's a hint..." but it doens't help an awful lot.
When decomposing into partial fractions, will one always arrive at a set of fractions with the numerators consisting of powers one degree less than in the denominator? Which seems to be what it hints at when the farctions are separated as above.
Many thanks.

 

[jtyler05si]

Hello!
Quick question reagrding partial fractions.
When there is a factor such as (x+2)³ in the denominator, then the fraction is separated into the components (x+2)¹+...+(x+2)³.
I am not convinced I understand quite why this is so. Partial fractions guides all offer something on the theme of: "Here's a hint..." but it doens't help an awful lot.
When decomposing into partial fractions, will one always arrive at a set of fractions with the numerators consisting of powers one degree less than in the denominator? Which seems to be what it hints at when the farctions are separated as above.
Many thanks.

Not necessarily. It is true until you get something with a multiplicity greater than one; as in your case. Normally, yes you would break everything into separate pieces, and if they are all different then the numerator will follow that pattern. Special cases arise when you have more than one of something in the denominator.

 

[jtyler05si]

Oh and as for the reasoning behind having (x+2)¹+...+(x+2)³ in the denominator, as far as I understand it is that we don't know what contributed to get the end product of (x+2)³ so you have to include every possible way (different combination of those three above.) The problem works itself out in the end. If you only tried (x+2)³, you would only be offering one way to get to the original problem, which could easily be wrong.

Not sure if I am very clear when explaining this, I hope it helps.

 

[Mark44]

When you decompose a rational expression such as $$\frac{1}{(x + 2)^3}$$
where there are repeated linear factors, the only way you can get an equation that is identically true (i.e. for all values of the variable for which all expressions are defined, is like this:
$$\frac{1}{(x + 2)^3} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{(x + 2)^3}$$

If one factor in the rational expression is an irreducible quadratic, the numerator takes a different form:

$$\frac{1}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}$$

If there is a repeated irreducible quadratic, the decomposition looks like this:
$$\frac{1}{(x^2 + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}$$
In all cases, the numerator will be of one degree less than the degree of the factor that is repeated. IOW, if there is a linear factor in the denominator, whether repeated or not, the numerator will be a constant. If there is an irreducible quadratic factor in the denominator, whether repeated or not, the numerator will be of degree one; e.g., Ax + B.

If a linear factor is repeated n times, there will be n terms in the decomposition, all of the form a_p/(x - c)^p, where p ranges from 1 through n.

If an irreducible quadratic factor is repeated n times, there will be n terms in the decomposition, all of the form (a_px + b_p)/(x² + cx + d)^p, where p ranges from 1 through n.

I hope that's clear.

 

[The Chaz]

Mark is the man.

 

[Mark44]

(Blush...)

 

[nobahar]

Many thanks jtyler05si and Mark 44 for the responses.
If I have understood both correctly, there is disagreement on the issue of degree in the numerator and denominator. jtyler05si suggests that the numerator will not necessarily be one degree less. Mark 44 suggests that the numerator will be one degree less than the denominator.
Any further clarification appreciated.

 

[The Chaz]

The degree will be one degree less than the IRREDUCIBLE polynomial in the bottom, regardless of its multiplicity. The number of separate fractions depends on the number of factors and their multiplicity.
x^2 + 1 is irreducible; x^2 - 1 is not.
x^2 + 1 would have a numerator of Ax + B. x^2 - 1 would be split into (x+1)(x - 1)

 

[Mark44]

Many thanks jtyler05si and Mark 44 for the responses.
If I have understood both correctly, there is disagreement on the issue of degree in the numerator and denominator. jtyler05si suggests that the numerator will not necessarily be one degree less. Mark 44 suggests that the numerator will be one degree less than the denominator.
Any further clarification appreciated.

I don't think I said it as clearly as I might have. The degree of the numerator is always one less than the degree of the linear factor or irreducible quadratic factor (not counting exponents used for repeated factors).

For example, if one term in the decomposition is
$$\frac{C}{(x + 2)^3}$$,
I'm considering only the degree of the linear polynomial in parentheses in the denominator - I'm not counting the exponent 3. Here the degree of the numerator is 0 and the degree of the linear factor is 1.

I don't know a better way to say this.

 

[jtyler05si]

I don't think I said it as clearly as I might have. The degree of the numerator is always one less than the degree of the linear factor or irreducible quadratic factor (not counting exponents used for repeated factors).

For example, if one term in the decomposition is
$$\frac{C}{(x + 2)^3}$$,
I'm considering only the degree of the linear polynomial in parentheses in the denominator - I'm not counting the exponent 3. Here the degree of the numerator is 0 and the degree of the linear factor is 1.

I don't know a better way to say this.

haha I agree.

My earlier statement about the degree not always being one less is relative to that exact term. The way Mark is explaining is probably more correct in that the degree of the top is always one less than that of the linear polynomial in this case.
My explanation was probably confusing, but I was just pointing out that having $$\frac{C}{(x + 2)^3}$$ in this special case is okay and still follows the rules. Sorry for the confusion, ha.

 

[nobahar]

Many thanks again for the responses, Mark 44, The Chaz and jtyler05si.
The explanations were clear, it was a misunderstanding on my part; I was always referring to the degree inside the parentheses, but I misunderstood what jtyler05si was saying.
I understand now (or at least, I think I do!).
Thanks Mark 44, jtyler05si and The Chaz for all your help, it is much appreciated.

 

[nobahar]

Hello!
I have a further question regarding partial fractions and I htought I might as well post it here since it is somewhat of continuation of my previous question.
If you have a set of partial fractions such as:
$$\frac{A}{x-2} + \left \left \frac{B}{(x-2)^2} + \left \left \frac{C}{(x-2)^3} + \left \left \frac{D}{(x-2)^4} + \left \left \frac{E}{x+3}$$
Then is it fair to say that when multiplied through:
A and will be a coefficient of the highest degree in the numerator; B will be the coefficient for the second highest; C for the third highest; and D for the fourth.
For example, A will be multipled by (x-2)³(x+3), E by (x-2)⁴ and D by (x+3), therefore you will have AX⁴, Ex⁴ and Dx as the highest powers for each coefficient when expanded out the brackets. This may sound obvious to some (if it's even correct, that is), but it means that the highest power factors in the denominator in "partial fraction form" of repeated factors will contirbute the 'least highest powers'. I say 'least highest' because yesterday, when I was deliberating it, I'm sure I came to the conclusion that it should be so, I think it has something to do with not necessarily contributing to, either at all or solely, the lowest degree(s) (e.g. x¹ or x⁰).
I apologise if these questions are tiresome, but I'm excogitating partial fractions as I feel they are important and that I may have overlooked a fair amount.

 

[Mark44]

Yes, what you say mostly makes sense. A and E will be the coefficients of 4th powers of x and lower powers, B will be the coefficient of a 3rd power of x and lower powers, C the coefficient of x squared and lower powers, and D the coefficient of x and a constant.

 

[nobahar]

Many thanks Mark 44.