A Simple Perturbation Theory Example: Solving the Quintic with Power Series.

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Perturbation theory can be applied to many systems, including certain polynomial solutions. Below, we discuss a specific application to a certain quintic polynomial.

Perturbation theory often takes on a rather scary reputation, since it's generally only called out by name late in one's academic career, and has a somewhat strong association with particle physics (see Feynman diagrams and Dyson series). However, the essential concepts are ubiquitous and it's applications are more broad than they may first seem.

Quintic polynomials, famously, have no general formula for their solution. Below, as an example of the simple and powerful concepts of perturbation theory, we solve a specific quintic polynomial, giving a Taylor series expansion that approximates the solution after truncation.

The Approach

Suppose we'd like to solve the polynomial below for \(x\).

\[ x^5+x-1=0 \]
Of course, we can't apply a general formula for x since there is no quintic equation. However, we may effectively solve for an expansion of x in terms of some parameter. But what parameter?
\[ x^5+ \epsilon x-1=0 \]
It turns out, our polynomial just so happens to coincide with the above equation when \(\epsilon = 1\). And, since the value of \(x\) itself now depends on \(\epsilon\), we can assume it has a Taylor expansion in \(\epsilon\) of the form:
\[ x(\epsilon) = x_0 + \frac{\partial x}{\partial \epsilon}\epsilon + \frac{1}{2} \frac{\partial^2 x}{\partial \epsilon^2}\epsilon^2 + ... = \sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^n x}{\partial \epsilon^n}\epsilon^n \]

We can then substitute this form of \(x(\epsilon)\) into the polynomial (with the parameter), and collect terms according to order in \(\epsilon\), as below.

\begin{align} \left[x(\epsilon)\right]^5+ \epsilon x(\epsilon)-1&=\left[\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^n x}{\partial \epsilon^n}\epsilon^n\right]^5+\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^n x}{\partial \epsilon^n}\epsilon^{n+1}-1\\ \\ & = \left[\sum_{n=0}^{\infty}x_n\epsilon^n\right]^5+\sum_{n=0}^{\infty}x_n\epsilon^{n+1}-1\\ \\ & = \big(x_0^5 + 5x_0^4x_1\epsilon + (10x_0^3x_1^2+5x_0^4x_2)\epsilon^2+...\big)\\ &+\epsilon\big( x_0 + x_1 \epsilon + x_2 \epsilon^2 + \quad x_3 \epsilon^3 \quad +\ ...\big)-1\\ \\ & = (x_0^5 - 1)\epsilon^0 + (5x_0^4x_1+x_0)\epsilon^1 + (10x_0^3x_1^2+5x_0^4x_2 +x_1) \epsilon^2 + ...\\ \\ & = 0\\ \end{align}
Now, since we expect the above to be satisfied regardless of the value of \(\epsilon\), the series above must vanish term by term. That is, each coefficient of the expansion in \(\epsilon\) above must vanish independently, and hence, we may solve for the expansion of \(x\), term by term.
\begin{align} (x_0^5 - 1)=0 &\ \ \quad\Rightarrow x_0 =1 \\ (5x_0^4x_1+x_0)=0&\ \ \quad\Rightarrow x_1 =-1/5 \ \\ (10x_0^3x_1^2+5x_0^4x_2 +x_1)=0&\ \ \quad\Rightarrow x_2 =-1/25 \\ \quad\vdots\quad\quad\quad \\ \end{align}
So, to second order in \(\epsilon\), \(x\) has the following expansion:
\begin{align} x(\epsilon) & = x_0 + x_1\epsilon + x_2\epsilon^2+...\\ & = 1\ +\ -\frac{1}{5}\epsilon \ +\ -\frac{1}{25}\epsilon^2+...\\ \end{align}
And so, at our point of interest, when \(\epsilon =0\), we then have the approximate solution for \(x\) below:
\[ x(1)\approx 1-\frac{1}{5}-\frac{1}{25} = .76 \]
Wolfram|Alpha gives a (real) solution of approximately \(~.788\), so we're not too far off!

Further Resources

This example was meant to be short but informative enough to show the general power of perturbative approaches. Many details have been neglected here such as the choice of placement of the parameter \(\epsilon\), and the assumption that the power series of \(x\) in \(\epsilon\) converges.

The same proble is given as an introduction in Carl Bender's wonderful lecture series. It's also used in his book, Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory. For more detail, see these resources.