Approximaths

A quintic equation

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So far, we’ve introduced the perturbation technique, described regular and irregular problems and presented solutions to simple problems that we could have solved exactly without using perturbative series. We will now consider a problem that cannot be solved exactly. In this way, we hope to illustrate the power of this method.

We want to find a real zero of the following quintic equation:

\displaystyle x^5 + x = 1

We will proceed as usual (see previous sections):

\displaystyle x^5 + \epsilon x = 1
\displaystyle (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots)^5 + \epsilon (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots) = 1

Solving the unperturbed problem (where \epsilon = 0) we see that a_0 = 1 (a_0 is always the answer to the unperturbed problem). We can use Python to expand the perturbative series and compute the coefficients (see code below). In this example we will calculate up to six coefficients.

from sympy import *

# set variables
a1 = Symbol('a1')
a2 = Symbol('a2')
a3 = Symbol('a3')
a4 = Symbol('a4')
a5 = Symbol('a5')
x = Symbol('x') # epsilon

# from the unperturbed problem we know that a0 = 1

# expand the quintic equation
equ = expand ((1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5)** 5 + x*(1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5))

# get the coefficients separately
equ.coeff(x)

5*a1+1

equ.coeff(x**2)

10*a1**2+a1+5*a2

# solve the system of equations to calculate coefficients
solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3)], (a1, a2, a3))
solution

[(-1/5, -1/25, -1/125)]

solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3),equ.coeff(x**4), equ.coeff(x**5)], (a1, a2, a3, a4, a5))
solution

[(-1/5, -1/25, -1/125, 0, 21/15625)])

Therefore (using the results above calculated using Python):

\displaystyle a_0 = 1
\displaystyle a_1 = -\frac{1}{5}
\displaystyle a_2 = -\frac{1}{25}
\displaystyle a_3 = -\frac{1}{125}
\displaystyle a_4 = 0
\displaystyle a_5 = \frac{21}{15625}

We have:

\displaystyle x(\epsilon) = 1 - \frac{1}{5}\epsilon - \frac{1}{25}\epsilon^2 - \frac{1}{125}\epsilon^3 + 0\cdot\epsilon^4 + \frac{21}{15625}\epsilon^5 + \cdots

An approximation (fifth-order approximation) of the real solution of x^5 + x = 1 is (setting \epsilon = 1):

\displaystyle 1 - \frac{1}{5} - \frac{1}{25} - \frac{1}{125} + 0 + \frac{21}{15625} \approx 0.7533

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