x^5+x = 1

$x^5 + x=1$

Solve this equation perturbatively.

Replace the equation by

$ x^5 + \epsilon x=1  $

Take the series solution in ε to be

$x = \Sigma_{n=0}^{\infty}a_n\epsilon^n$

Then the equation becomes

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

When

$\epsilon = 0\\ a_0 = 1$

For ε.

$\begin{align}\epsilon &\Rightarrow \frac{5!}{4!}a_0^4a_1 + a_0 = 0\\&\rightarrow 5a_0^4a_1 + a_0 = 0 \\&\therefore a_0=-\frac{1}{5}\end{align}$

For ε²,

$\begin{align}\epsilon^2 &\Rightarrow \frac{5!}{3!2!}a_0^3a_1^2+\frac{5!}{4!}a_0^4a_2+a_1 = 0  \\ &\rightarrow10a_0^3a_1^2+5a_0^4a_2+a_1=0\\&\rightarrow\frac{1}{5}+5a_2 = 0\\&\therefore a_2 = -\frac{1}{25}\end{align}$

For ε³,

$\begin{align}\epsilon^3 &\Rightarrow \frac{5!}{4!}a_0^4a_3 + \frac{5!}{3!}a_0^3a_1a_2 + \frac{5!}{2!3!}a_0^2a^3_1 + a_2 = 0 \\&\rightarrow 5a_0^4a_3 + 20a_0^3a_1a_2 + 10a_0^2a^3_1 + a_2 = 0 \\&\rightarrow 5a_3 + \frac{20}{125}- \frac{10}{125}- \frac{5}{125}=0\\&\therefore a_3 = -\frac{1}{125}\end{align}$

If we choose the ε = 1, the solution is 

$\begin{align}x &= a_0 + a_1\epsilon+a_2\epsilon^2+a_3\epsilon^3\\&= a_0 +a_1+a_2+a_3\\&= a-\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\\\therefore x &= 0.752\end{align}$

and the numerical solution is ≃0.755