Real Rational Functions#

Definition: Real Rational Functions

A real rational function is a real function \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) for which there exist real polynomials \(P\) and \(Q\) such that

\[ f(x) = \frac{P(x)}{Q(x)} \]

for all \(x \in \mathcal{D}\).

Theorem: Limits at Infinity

Let \(f\) be a real rational function given by

\[ f(x) = \frac{P(x)}{Q(x)} = \frac{\displaystyle{\sum_{k=0}^m p_k x^k}}{\displaystyle{\sum_{k=0}^n q_k x^k}}. \]

If \(m \lt n\), then \(f\) has a [horizontal asymptote](./Limits%20(Real%20Functions.md) at \(y = 0\):

\[ m \lt n \implies \lim_{x \to \pm \infty} f(x) = 0 \]

If \(m = n\), then \(f\) has a [horizontal asymptote](./Limits%20(Real%20Functions.md) at \(y = \frac{p_m}{q_n}\):

\[ m = n \implies \lim_{x \to \pm \infty} f(x) = \frac{p_m}{q_n} \]

If \(m \gt n\), then the [limits](./Limits%20(Real%20Functions.md) of \(f\) at \(\pm_{\infty}\) depend on \((n-m)\) and \(\frac{p_m}{q_n}\):

\[ \begin{aligned} &\lim_{x \to +\infty} f(x) = \begin{cases} +\infty \qquad \text{when } \frac{p_m}{q_n} \gt 0 \\ -\infty \qquad \text{when } \frac{p_m}{q_n} \lt 0 \end{cases} \\ &\lim_{x \to -\infty} f(x) = \begin{cases} +\infty \qquad \text{when } (n-m) \text{ is even and } \frac{p_m}{q_n} \gt 0 \text{ or } (n - m) \text{ is odd and } \frac{p_m}{q_n} \lt 0 \\ -\infty \qquad \text{when } (n-m) \text{ is even and } \frac{p_m}{q_n} \lt 0 \text{ or } (n - m) \text{ is odd and } \frac{p_m}{q_n} \gt 0 \end{cases} \end{aligned} \]

Proof

TODO

Theorem: Vertical Asymptotes

Let \(f\) be a real rational function given by

\[ f(x) = \frac{P(x)}{Q(x)} = \frac{\displaystyle{\sum_{k=0}^m p_k x^k}}{\displaystyle{\sum_{k=0}^n q_k x^k}}. \]

If \(r \in \mathbb{R}\) is a root of \(Q\) but not of \(P\), then \(f\) has a [vertical asymptote](./Limits%20(Real%20Functions.md#Asymptotes) at \(x = r\).

If \(r \in \mathbb{R}\) is a root of both \(P\) and \(Q\) but its multiplicity for \(P\) is lower than its multiplicity for \(Q\), then \(f\) again has a [vertical asymptote](./Limits%20(Real%20Functions.md#Asymptotes) at \(x = r\).

Proof

TODO

Theorem: Partial Fraction Decomposition

Every real rational function \(f(x) = \frac{P(x)}{Q(x)}\) with \(\deg P \lt \deg Q\) can be represented as a sum of real rational functions

\[f(x) = \sum_{i = 1}\sum_{j = 1}^{m_{l_i}} \frac{A_{ji}}{l_i(x)^j} + \sum_{i=1}\sum_{j = 1}^{m_{q_i}} \frac{B_{ji}x + C_{ji}}{q_i(x)^j},\]

where:

\(l_i(x)\) are the distinct linear factors of \(Q(x)\) and \(m_{l_i}\) are their respective multiplicities;
\(q_i(x)\) are the distinct irreducible quadratic factors of \(Q(x)\) and \(m_{q_i}\) are their respective multiplicities;
\(A_{ji}, B_{ji}, C_{ji} \in \mathbb{R}\).

Example: \(\frac{x^4 + 3x^2 - 2x}{x^6 + x^4 -x^2 -1}\)

We want to find the partial fraction decomposition of the following rational function:

\[f(x) = \frac{x^4 + 3x^2 - 2x}{x^6 + x^4 -x^2 - 1}\]

The denominator can be factored as \(x^6 + x^4 -x^2 -1 = (x-1)(x+1)(x^2+1)^2\):

\[\frac{x^4 + 3x^2 - 2x}{x^6 + x^4 -x^2 -1} = \frac{x^4 + 3x^2 - 2x}{(x-1)(x+1)(x^2+1)^2}\]

The theorem tells us that

\[\frac{x^4 + 3x^2 - 2x}{x^6 + x^4 -x^2 -1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1} + \frac{Ex + F}{(x^2 +1)^2}\]

for some \(A, B, C, D, E, F \in \mathbb{R}\). We bring the right-hand side to a single denominator:

\[\begin{aligned}& \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1} + \frac{Ex + F}{x^2 +1} = \\ & = \frac{A(x+1)(x^2+1)^2 + B(x-1)(x^2+1)^2 + (Cx+D)(x-1)(x+1)(x^2+1) + (Ex+F)(x-1)(x+1)}{(x-1)(x+1)(x^2+1)^2}\end{aligned}\]

We multiply the numerator out and group according to the powers of \(x\):

\[(A + B + C)x^5 + (A - B + D)x^4 + (2A + 2B + E)x^3 + (2A - 2B + F)x^2 + (A + B - C - E)x + (A - B - D - F)\]

We want this to be equal to the numerator of the original expression. By comparing the coefficients in front of the respective powers of \(x\), we obtain the following system of linear equations:

\[\left\vert\begin{array}{cccccccc}+A & +B & +C & & & & = & 0 \\ +A & -B & & +D & & & = & 1 \\ +2A & +2B & & & +E & & = & 0 \\ +2A & -2B & & & & +F & = & 3 \\ +A & +B & -C & & -E & & = & -2 \\ +A & -B & & -D & & -F & = & 0\end{array}\right.\]

Solving it, we obtain:

\[\left\vert\begin{aligned}A & = \frac{1}{4} \\ B & = -\frac{3}{4} \\ C & = \frac{1}{2} \\ D & = 0 \\ E & = 1 \\ F & = 1\end{aligned}\right.\]

We can now substitute these values back to obtain the partial fraction decomposition:

\[\begin{aligned}\frac{x^4 + 3x^2 - 2x}{x^6 + x^4 -x^2 - 1} & = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1} + \frac{Ex + F}{(x^2 +1)^2} \\ & = \frac{1}{4(x-1)}-\frac{3}{4(x+1)}+\frac{x}{2(x^2+1)}+\frac{x+1}{(x^2+1)^2}\end{aligned}\]

Proof

TODO

Theorem: Antidifferentiation of Rational Functions

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(I \subseteq \mathcal{D}\) be an interval.

If \(f\) is rational on \(I\) with \(f(x) = \frac{1}{(x-a)^m}\), where \(m \in \mathbb{N}\), then \(f\) is antidifferentiable on \(I\):

\[\int \frac{\mathrm{d}x}{(x-a)^m} = \begin{cases} \ln |x - a| + C & \text{if } m = 1 \\ \\ -\frac{1}{(m-1)(x-a)^{m-1}} + C & \text{if } m \ge 2 \end{cases}\]

If \(f\) is rational on \(I\) with \(f(x) = \frac{1}{(x-a)^2+b^2}\), where \(b \gt 0\), then \(f\) is antidifferentiable on \(I\):

\[\int \frac{1}{(x-a)^2+b^2} \,\mathrm{d}x = \frac{1}{b} \arctan \frac{x-a}{b} + C, \qquad b \gt 0\]

If \(f\) is rational on \(I\) with \(f(x) = \frac{x-a}{(x-a)^2+b^2}\), where \(b \gt 0\), then \(f\) is antidifferentiable on \(I\):

\[\int \frac{x-a}{(x-a)^2+b^2} \,\mathrm{d}x = \frac{1}{2}\ln((x-a)^2 + b^2)+C\]

If \(f\) is rational on \(I\) with \(f(x) = \frac{1}{(x^2 + px + q)^m}\), where \(m \in \mathbb{N}\) and \(p, q \in \mathbb{R}\) with \(p^2 - 4q \lt 0\), then \(f\) is antidifferentiable on \(I\):

\[\int \frac{\mathrm{d}x}{(x^2+px+q)^m} = \begin{cases} \frac{2}{\sqrt{4q-p^2}} \arctan \left(\frac{2x+p}{\sqrt{4q-p^2}}\right) + C & \text{if } m = 1 \\ \frac{2x+p}{(m-1)(4q-p^2)(x^2+px+q)^{m-1}} + \frac{4m-6}{(m-1)(4q-p^2)}\int \frac{1}{(x^2+px+q)^{m-1}} + C & \text{if } m \ge 2\end{cases}\]

If \(f\) is rational on \(I\) with \(f(x) = \frac{Ax + B}{(x^2 + px + q)^m}\), where \(m \in \mathbb{N}\) and \(A, B, p, q, \in \mathbb{R}\) with \(p^2 - 4q \lt 0\), then \(f\) is antidifferentiable on \(I\):

\[\int \frac{Ax + B}{(x^2+px+q)^m} \mathop{\mathrm{d}x} = \begin{cases}\frac{A}{2}\ln (x^2+px+q) + \left(B - \frac{Ap}{2}\right)\int \frac{\mathop{\mathrm{d}x}}{x^2+px+q} + C & \text{if } m = 1 \\ -\frac{A}{2(m-1)(x^2+px+q)} + \left(B - \frac{Ap}{2}\right)\int \frac{\mathop{\mathrm{d}x}}{(x^2+px+q)^{m-1}} + C & \text{if } m \ge 2 \end{cases}\]

Proof

TODO

Algorithm: Antidifferentiation of Rational Functions

We are given the following real rational function:

\[f(x) = \frac{N(x)}{D(x)}\]

If \(\deg N \lt \deg D\), then find the partial fraction decomposition of \(f\) and use the antiderivatives of the elementary rational functions:

\[\int f(x) \mathop{\mathrm{d}x} = \sum_{i}\sum_{j = 1}^{m_{l_i}} \int \frac{A_{ji}}{l_i(x)^j} \mathop{\mathrm{d}x} + \sum_{i}\sum_{j = 1}^{m_{q_i}}\int \frac{B_{ji}x + C_{ji}}{q_i(x)^j} \mathop{\mathrm{d}x}\]

If \(\deg N \ge \deg D\), then divide \(N\) by \(D\) to find \(f(x) = Q(x) + \frac{R(x)}{D(x)}\). Find the partial fraction decomposition of \(\frac{R(x)}{D(x)}\) and use the antiderivatives:

\[\int f(x) \mathop{\mathrm{d}x} = \int Q(x) + \frac{R(x)}{D(x)} \mathop{\mathrm{d}x} = \int Q(x) \mathop{\mathrm{d}x} + \int \frac{R(x)}{D(x)} \mathop{\mathrm{d}x}\]

Example: \(\int \frac{x^4+3x^2-2x}{x^4-1} \,\mathrm{d}x\)

We want to determine the antiderivatives of

\[f(x) = \frac{x^4+3x^2-2x}{x^4-1}.\]

Polynomial division gives us the following:

\[f(x) = \frac{3x^2-2x+1}{x^4 - 1} + 1\]

We have:

\[\begin{aligned}\int f(x) \,\mathrm{d}x & = \int \frac{3x^2-2x+1}{x^4 - 1} + 1\,\mathrm{d}x \\ & = \int 1 \,\mathrm{d}x + \int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x \\ & = x + \int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x\end{aligned}\]

To find \(\int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x\), we need to find the partial fraction decomposition of \(\frac{3x^2-2x+1}{x^4 - 1}\):

\[\frac{3x^2-2x+1}{x^4 - 1} = \frac{1}{2(x-1)}-\frac{3}{2(x+1)}+\frac{x+1}{x^2+1}\]

We have:

\[\begin{aligned}\int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x & = \int \frac{1}{2(x-1)}-\frac{3}{2(x+1)}+\frac{x+1}{x^2+1}\,\mathrm{d}x \\ & = \int \frac{1}{2(x-1)} \,\mathrm{d}x + \int -\frac{3}{2(x+1)} \,\mathrm{d}x + \int \frac{x+1}{x^2+1} \,\mathrm{d}x \\ & = \frac{1}{2} \int \frac{1}{x-1} \,\mathrm{d}x - \frac{3}{2}\int \frac{1}{x+1}\,\mathrm{d}x + \int \frac{x+1}{x^2+1} \,\mathrm{d}x \\ & = \frac{1}{2} \int \frac{1}{x-1} \,\mathrm{d}x - \frac{3}{2}\int \frac{1}{x+1}\,\mathrm{d}x + \int \frac{x}{x^2 + 1} \,\mathrm{d}x + \int \frac{1}{x^2 + 1} \,\mathrm{d}x\end{aligned}\]

By using the antiderivatives of the elementary rational functions:

\[\begin{aligned}\int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x & = \frac{1}{2} \int \frac{1}{x-1} \,\mathrm{d}x - \frac{3}{2}\int \frac{1}{x+1}\,\mathrm{d}x + \int \frac{x}{x^2 + 1} \,\mathrm{d}x + \int \frac{1}{x^2 + 1} \,\mathrm{d}x \\ & = \frac{1}{2}\ln|x-1| - \frac{3}{2}\ln|x+1| + \frac{1}{2}\ln(x^2 + 1) + \arctan x + C\end{aligned}\]

Finally:

\[\begin{aligned}\int f(x) \,\mathrm{d}x & = x + \int \frac{3x^2-2x+1}{x^4 - 1} \,\mathrm{d}x \\ & = x + \frac{1}{2}\ln|x-1| - \frac{3}{2}\ln|x+1| + \frac{1}{2}\ln(x^2 + 1) + \arctan x + C\end{aligned}\]