Solving a Cubic Equation by Perturbation Theory

You probably do not know how to solve the equation

x^{3} + \frac{1}{10} x + 8 = 0

exactly. Except in those few cases where one root is obvious (and hence the cubic can be reduced to a quadratic), cubic equations are nearly always solved in practice by a numerical or approximate method.

In this case, the coefficient \frac{1}{10} is smaller than the others. This suggests that we study the equation

x^{3} + ϵx + 8 = 0,

find an approximation to the solutions that’s accurate when ϵ is small, and set ϵ equal to \frac{1}{10} at the end. Let’s assume that

x ≈ x_{0} + ϵx_{1}

and find the numbers x_{0} and x_{1} . The idea is that if ϵ is small, then ϵ^{2} is even smaller, and terms in the Taylor series involving ϵ^{2} or higher powers can probably be ignored.

We calculate

x^{3} ≈ x_{0}^{3} + 3ϵx_{0}^{2}x_{1} + 3ϵ^{2}x_{0}x_{1}^{2} + ϵ^{3}x_{1}^{3}.

Only the first two terms of this formula are “significant”, because a term 3ϵ^{2}x_{0}^{2}x_{2} has been neglected already in our approximation. So we will throw away all terms that involve power of ϵ higher than the first.

Now the equation becomes

0 = x^{3} + ϵx + 8
≈ x_{0}^{3} + 3ϵx_{0}^{2}x_{1}

+ ϵx_{0}

+ 8.

The general strategy in perturbative calculations is to make the coefficient of each power of ϵ separately equal to 0, so that the equation is satisfied for all values of ϵ.

The lowest-order equation is

0 = x_{0}^{3} + 8.  (ϵ^{0})

Its principal solution is x_{0} = −2. (The equation also has two complex roots, but we will ignore them today.)

Substitute this result into the next equation:

0 = 3x_{0}^{2}x_{1} + x_{0} = 12x_{1} − 2.  (ϵ^{1})

Thus x_{1} = \frac{1}{6} .

So we have found a first-order perturbative solution,

x ≈ −2 + \frac{ϵ}{6} ,

which is actually the Taylor polynomial T_{1}(ϵ) of the exact solution.

Let us check this solution by substituting it into the original cubic equation. After working out the algebra we get

x^{3} + ϵx + 8 = \frac{ϵ^{3}}{216} .

The right-hand side (called the residual) is not exactly zero, but it is small compared to ϵ if ϵ itself is small.*

If ϵ = \frac{1}{10} , our approximation is x ≈ −2.01666 . . .. Compare this with the “exact” answer calculated by Maple.

* We had no right to expect the residual to be smaller than the order ϵ^{2}, but by accident it is of order ϵ^{3}. If you go back and put terms ϵ^{2}x_{2} + ϵ^{3}x_{3} into the assumed form of the answer, you will find that x_{2} = 0 but x_{3} is not zero.