A doubt in Partial fraction decomposition

[Physics Slayer]

Homework Statement

Don't understand one of the steps in partial fraction decomposition.

Relevant Equations

-

Say you want to find the following Integrals
$$\int \frac{1}{(x-1)(x+2)} (dx)$$
$$\int \frac{1}{(x-1)(x^2 + 2)} (dx)$$
The easiest way to solve them will be by using partial fraction decomposition on both the given functions.
Decomposing the first function,
$$\frac{1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}$$
And then decomposing the second function
$$ \frac{1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+2} $$
We can then solve for the constants($A,B,C$) by setting convenient values of $x$.

So my doubt is why don't we set the numerator for the first function as $(Ax+B)$ our teacher told us that as a rule of thumb whenever you are using partial fraction decomposition first make sure you are dealing with a proper fraction and then set the degree of x in the numerator one less than the degree of x in the denominator, what I think he meant by this is,

$$\frac{f(x)}{(x^n + a)(x^m + b)} = \frac{k_{n}x^{n-1}+k_{n-1}x^{n-2}+ ... +k_{2}x + k_1}{x^n + a}+\frac{r_{m}x^{m-1}+r_{m-1}x^{m-2}+...+r_{2}x + r_1}{x^m + b}$$
Where $f(x)$ can be any arbitrary polynomial function, we can solve for the constants $k_n,k_{n-1},...,k_2, k_1$ and $r_m, r_{m-1},...r_2,r_1$by setting different continent values of $x$.

He didn't give any reasoning to the part in bold, I can see that this method works and I have bashed many integrals using the partial fraction decomposition method but I still don't understand the logic behind it.

 

[anuttarasammyak]

$$\frac{Bx+C}{x^2+2}=\frac{D}{x+\sqrt{2}i}+\frac{E}{x-\sqrt{2}i}$$
may give us the answer more easily. In general I think we can expect to express
$$\prod_i^n (x-a_i)^{-1}=\sum_i^n b_i(x-a_i)^{-1}$$
with a_i and b_i constants including complex numbers in case all a_i’s are different.

 

[Physics Slayer]

$$\frac{Bx+C}{x^2+2}=\frac{D}{x+\sqrt{2}i}+\frac{E}{x-\sqrt{2}i}$$
will give us the answer more easily.

We usually don't want $i$ showing up in our final answer.
Also did u simply apply partial fraction decomposition on $$\frac{Bx+C}{x^2+2} = \frac{Bx+C}{(x + i\sqrt{2})(x - i\sqrt{2})}$$?
Or did u directly go from
$$\frac{1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{D}{x+i\sqrt{2}}+\frac{E}{x-i\sqrt{2}}$$

 

[nuuskur]

Partial fraction decomposition is valid for rational functions that dominate in the denominator. That is, given a rational function $\frac{p}{q}$, if $\deg(p)<\deg (q)$, then one might call this a "proper fraction". It is known that any rational function can be written uniquely as a sum of a polynomial and a proper fraction.

 

[PeroK]

He didn't give any reasoning to the part in bold, I can see that this method works and I have bashed many integrals using the partial fraction decomposition method but I still don't understand the logic behind it.

You can simply try it and see. In your first example, instead of having $A$ and $B$ you could have $A_0 + A_1x + A_2x^2 + \dots$ and similarly for $B$.

You should find that $A_1 = A_2 = 0$ etc.

 

[PeroK]

You can simply try it and see. In your first example, instead of having $A$ and $B$ you could have $A_0 + A_1x + A_2x^2 + \dots$ and similarly for $B$.

You should find that $A_1 = A_2 = 0$ etc.

Actually, we can find additional solutions with higher powers. E.g.
$$\frac{1}{(x+1)(x+2)} = \frac{x+2}{x+1} - \frac{x+3}{x+2} = \frac{1}{x+1} - \frac{1}{x + 2}$$So, the logic is that we can find a solution with constant numerators in this case, and that's the simplest form we want.

The next step would be to show that we need more than a constant (i.e. a linear expression) numerator when one of the terms in the denominator is quadratic. That's the exercise you ought to carry out for yourself. E.g. show that:
$$\frac{1}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{(x+2)^2}$$has no solutions for constants $A, B$.

 

[anuttarasammyak]

Also did u simply apply partial fraction decomposition on Bx+Cx2+2=Bx+C(x+i2)(x−i2)?
Or did u directly go from

In both ways you can easily derive these constants be
$$D=\frac{B}{2}+\frac{C}{2\sqrt{2}}i$$
$$E=\frac{B}{2}-\frac{C}{2\sqrt{2}}i$$

 

[Mark44]

Homework Statement:: Don't understand one of the steps in partial fraction decomposition.
Relevant Equations:: -

So my doubt is why don't we set the numerator for the first function as (Ax+B) our teacher told us that as a rule of thumb whenever you are using partial fraction decomposition first make sure you are dealing with a proper fraction and then set the degree of x in the numerator one less than the degree of x in the denominator, what I think he meant by this is,
$$\frac{f(x)}{(x^n + a)(x^m + b)} = \frac{k_{n}x^{n-1}+k_{n-1}x^{n-2}+ ... +k_{2}x + k_1}{x^n + a}+\frac{r_{m}x^{m-1}+r_{m-1}x^{m-2}+...+r_{2}x + r_1}{x^m + b}$$
Where $f(x)$ can be any arbitrary polynomial function, we can solve for the constants $k_n,k_{n-1},...,k_2, k_1$ and $r_m, r_{m-1},...r_2,r_1$by setting different continent values of $x$.

He didn't give any reasoning to the part in bold, I can see that this method works and I have bashed many integrals using the partial fraction decomposition method but I still don't understand the logic behind it.

Your interpretation is fine, in theory, but in actual practice you're not often going to come across any denominator polynomials that are higher than third degree, unless they can be conveniently factored into a product of linear or irreducible quadratic polynomials, or powers of these. Furthermore, quadratics of degree 5 or higher can't generally be factored, something proven by Evariste Galois, a mathematician who died at the age of 20 after being shot in a duel.

 

[MidgetDwarf]

In general, to apply the method of partial fraction decomposition their are 3 things to consider.

1) A real "root" is not repeated.

2) Repeated real "roots".

3)imaginary "roots".

You fixed the powers of x of the "unknown" constants depending on the 3 cases.