It is stated, that given a factorization $$g(x) = (x - a)^m(bx^2 + cx + d)^n$$ we can create partial fractions of the type:

$$\int f(x)/g(x) dx = \int \sum_{i = 0}^m \frac{A_i}{(x - a)^i} + \sum_{j = 0}^n\frac{B_jx + C_j}{(bx^2 + cx + d)^j} dx$$

Can someone explain to me, why we have to repeat fractions like that? Apostol himself states that this is required to equate the number of equations and coefficients when solving for the nominator.

1) However, if the nominator is of degree $d = 3$, and the denominator is degree $10$, then after the factors multiply nominator coefficients (to solve the equations), the possible degree of the nominator becomes $d = 9$. Isn't it an incorrect strategy somewhat?

2) Sometimes, when multiplying nominator coefficients by factors which correspond to other partial fractions, the resulting degree of this particular partial fraction nominator becomes less than the possible degree of $f(x)$. To illustrate: $g(x) = (x - 2)^2(x^2 + 4)^4$. Let $f(x) = x^5$. We would have the partial fraction $\frac{B_4x + C_4}{(x^2 + 4)^4}$. When we multiply its nominator by $(x - 2)^2$ to solve the equation and obtain coefficients, the resultant $(B_4x + C_4)(x - 2)^2$ has the degree of $3$, while i expected it to possibly have the degree of 5 like the nominator. Why doesn't it break the solution?

In Apostol the partial fractions' method is not proved, and instead the author says there exists some theorem that each rational function can be expressed as a sum of partial fractions as above^. I do not know the proof so cannot justify this method Apostol outlines (which i wrote above^). I am looking for both rigorous and intuitive explanations - whatever helps to understand the technique better.

  • 1
    $\begingroup$ Related (maybe duplicate?): math.stackexchange.com/questions/368665/… and the linked questions there. $\endgroup$ – Hans Lundmark Aug 13 at 21:21
  • $\begingroup$ @HansLundmark, it is similar, but not exactly. 1) The OP does list the case when there is a quadratic power as a factor 2) I would like to know specifically about the case with repeated factors/roots in partial fractions. Maybe the answers there contain the required answer, idk, let's see, cannot read atm. $\endgroup$ – John Aug 13 at 21:34
  • $\begingroup$ It's somewhat unclear what both (1) and (2) are asking for. But maybe it's useful to point out that higher powers of factors in the denominator mean that one gets more terms corresponding to each factor. For example, for $\frac{p(x)}{x (x^2 + 1)^2}$, where $\deg p < \deg [x (x^2 + 1)^2] = 5$, the partial fractions expansion is $$\frac{A}{x} + \frac{B x + C}{x^2 + 1} + \frac{D x + E}{(x^2 + 1)^2} .$$ After clearing denominators, the right-hand side becomes $A (x^2 + 1)^2 + \cdots$ and so can have degree up to $4$, as is necessary. $\endgroup$ – Travis Aug 14 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.