Partial Fraction Integration: Proving the Linear Factor Rule

  • Context: Undergrad 
  • Thread starter Thread starter The Anomaly
  • Start date Start date
  • Tags Tags
    Linear
Click For Summary
SUMMARY

The discussion centers on the Partial Fraction method of integration, specifically the Linear Factor Rule, which states that for each factor of the form (ax+b)m, the decomposition includes a sum of m partial fractions: A1/(ax+b) + A2/(ax+b)2 + ... + Am/(ax+b)m. Participants clarify that while the proof of the existence of these coefficients is typically covered in advanced courses, it is not necessary for performing integration. Understanding this rule allows students to proceed with integration without needing to validate the coefficients' existence.

PREREQUISITES
  • Understanding of Partial Fraction Decomposition
  • Familiarity with Linear Factor Rule
  • Basic knowledge of integration techniques
  • Concept of polynomial factors in algebra
NEXT STEPS
  • Study the proof of the Linear Factor Rule in advanced algebra contexts
  • Explore advanced integration techniques beyond Partial Fractions
  • Learn about polynomial long division as a precursor to Partial Fraction Decomposition
  • Review prerequisites in Precalculus algebra for a solid foundation
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking to clarify the Linear Factor Rule and its applications in Partial Fraction Decomposition.

The Anomaly
Messages
8
Reaction score
0
I'm studying the Partial Fraction method of integration, and I believe I understand the fundamental idea of it. However, much of it is based on a rule that the book calls the Linear Factor Rule. It is the following:

For each factor of the form (ax+b)m the partial fraction decomposition contains the following sum of m partial fractions:

\frac{A_1}{(ax+b)} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_m}{(ax+b)^m}

I'm assuming that the proof of this is either assumed, or was done in a Precalculus course or something. But could you help me out with proving it? It just doesn't make much sense at this point.
 
Physics news on Phys.org
The Anomaly said:
I'm studying the Partial Fraction method of integration, and I believe I understand the fundamental idea of it. However, much of it is based on a rule that the book calls the Linear Factor Rule. It is the following:

For each factor of the form (ax+b)m the partial fraction decomposition contains the following sum of m partial fractions:

\frac{A_1}{(ax+b)} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_m}{(ax+b)^m}

I'm assuming that the proof of this is either assumed, or was done in a Precalculus course or something. But could you help me out with proving it? It just doesn't make much sense at this point.

No, the proof that such numerators exist is found only in more advanced courses. But to proceed in integration you do not need that proof. Even if you have not proved that such coefficients exist in all cases, if such coefficients exist (as they always do) in the case you are doing, then you can go ahead with the integration.
 
g_edgar said:
No, the proof that such numerators exist is found only in more advanced courses. But to proceed in integration you do not need that proof. Even if you have not proved that such coefficients exist in all cases, if such coefficients exist (as they always do) in the case you are doing, then you can go ahead with the integration.

Alright, that makes sense. I just wanted to make sure that I wasn't missing some obvious proof from precalculus.

Thanks for the reply!

EDIT: And for curiosity, when my Calculus book refers to advanced algebra (Not exactly for this example, but for others) are they referring to Precalculus algebra? Or way more advanced stuff that I have not approached yet?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K