Rational Algebraic Expression

A rational algebraic expression e(x) is a numerator polynomial n(x) divided by a denominator polynomial d(x), namely

e(x) = n(x) / d(x).

Perform the division, to obtain a quotient polynomial q(x) and a remainder polynomial r(x), namely

e(x) = q(x) + r(x) / d(x),

where the degree of of the remainder polynomial r(x) is strictly less that that of d(x).

Sketch of the Expression

We presume that we either know or have found the roots of n(x) and of d(x). The roots of n(x) are the zero's of the expression. The roots of d(x) are the poles of the expression. If q(x) is identically zero, then the expression has neither horizontal nor oblique asymptotes. Else, if the degree if q(x) is one, then it is the horizontal asymptote of the expression. Else, q(x) is the oblique asymptote of the expression.

Partial Fraction Expansion

The r(x) / d(x) may be expanded into a sum of terms of the form of a constant divided by a linear expression, namely

k / (x - p).

This is a first-order pole. The constant k may be evaluated as the limit

k = limit, as x approaches p, of r(x) / d(x) (x - p).

Once such a term is found, the r(x) / d(x) may be reduced by subtraction, namely

r(x) / d(x) - k / (x - p).

The degree of the resulting numerator and denominator will be one less.
Then, it is easier to find the next partial fraction from this reduced expression.

In the foregoing, we have demonstrated only first-order poles. Any second-order pole has to be combined into a single term. Usually it is convenient to combine a pair of complex-conjugate poles into a single term. Sometimes we may combine a pair of irrational-conjugate poles into a single term. In each of these cases, since the denominator is of second degree, the numerator will have to be of first degree.
Namely, for a second-order pole, we have the term

(a (x - p) + b) / (x - p)^2 = b / (x - p)^2 + a / (x - p).

We evaluate the constant b as the limit

b = limit, as x approaches p, of r(x) / d(x) (x - p)^2.

Then, subtract the term

b / (x - p)^2

from the r(x) / d(x), to reduce it. Now, the coefficient a may be evaluated as the k of a first-order pole, in the foregoing.
For a higher-order pole, we first evaluate the term with the highest exponent in the denominator and reduce the expression accordingly. The reduced expression will have the pole of one order less. We proceed with the evaluation and reduction, until we have removed the pole from the expression.
Hopefully, eventually, only one conjugate pair of poles remains. Otherwise, there is no really easy way of evaluating them. One could evaluate one pole of a conjugate pair. The other pole of the pair will be conjugate to it. Then, add the two terms. Finally, reduce the original expression by this conjugate pair of poles. Proceed this way iteratively.

Integration

The primary application of the partial fraction decomposition is to facilitate integration. Usually, it is easier to integrate the partial fractions and add them that to integrate their sum.

Please see the hyperbolic for these two equations -- the first expresses the inverse hyperbolic cotangent in terms of logarithms, while the second provides the derivative of the inverse hyperbolic cotangent.

Arccoth(x) = (1 / 2) ln((x + 1) / (x - 1)).
d Arccoth(x) / dx = - 1 / (x^2 - 1)

Also, please see logarithmic for the integral of the reciprocal of x, as

integral of (1 / x) dx = ln(x) + C.

For example, we integrate the partial fraction expansion

- 2 / (x^2 - 1) = 1 / (x + 1) - 1 / (x - 1)

2 Acoth(x) = ln(x + 1) - ln(x - 1) + C,

which checks from the logarithmic expression of the inverse hyperbolic cotangent function.

For another example, we integrate the partial fraction expansion

2 x / (x^2 - 1) = 1 / (x + 1) + 1 / (x - 1)

ln(x^2 - 1) = ln(x + 1) + ln(x - 1) + C,

which checks from the factorization of (x^2 - 1).

Webmaster@rism.com