Solving a $x^2+\varepsilon x=1$ with perturbation theory for two regimes
27 Nov 2022
 Solving the equation directly
 Pertrubation theory for small $\varepsilon$
 Pertrubation theory for large $\varepsilon$
 The strange case of $\varepsilon\approx 1$
 Is this useful?
Solving the equation directly
Let’s look at the equation \(x^2+\varepsilon x=1\) We can rewrite it as \(x^2+\varepsilon x1=0\) for convenience. As with all quadratic equations, this equation has two solutions. Luckily both solutions of our equation are real numbers for any $\varepsilon$.
In this article, we will assume $\varepsilon > 0$ and we will only look at the bigger solution. So we obviously get \(x(\varepsilon)=\frac{1}{2}\big(\sqrt{\varepsilon^2+4}\varepsilon\big)\).
Limits for $0$ and $+\infty$
Let’s look at some limits of the function. At $0$ we trivially get \(\lim_{\varepsilon\rightarrow\infty}x(\varepsilon)=\frac{1}{2}\big(\sqrt{0^2+4}0\big)=1\).
$\lim_{\varepsilon\rightarrow\infty}x(\varepsilon)=0$ is more tricky to proof. The definition for this is^{1}
\[(\forall C > 0) (\exists \varepsilon > 0)(\forall s>\varepsilon) x(s) < C\]Because $x(\varepsilon)$ is strictly monotonically decreasing^{2} and bigger then $0$ this all comes down to finding an $\varepsilon$ so that $x(\varepsilon) < C$ for any $C > 0$. So let’s check:
\[\begin{aligned} &\frac{1}{2}(\sqrt{\varepsilon^2+4}\varepsilon) < C\\ \Leftrightarrow &\sqrt{\varepsilon^2+4} < 2C+\varepsilon\\ \Leftrightarrow &\varepsilon^2+4 < (2C+\varepsilon)^2 = 4C^2+4C\varepsilon+\varepsilon^2 \\ \Leftrightarrow & 4 < 4C^2+4C\varepsilon\\ \Leftrightarrow &\frac{1C^2}{C} < \varepsilon \end{aligned}\]QED
Pertrubation theory for small $\varepsilon$
So let’s assume $\varepsilon$ is small. If we assume $\varepsilon = 0$ we can solve the resulting equation $x^2  1 = 0$ instantly. This is nice and we can formalize this by using a power series for $x$ as
\[x = \sum_{n=0}^\infty x_n\varepsilon^n = x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + \varepsilon^3 x_3 + \varepsilon^4 x_4+\dots\]with unknown coefficients $x_n$.
So we can plug this in our equation and get \(\begin{aligned}0&=x^2+\varepsilon x1\\ &=(x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + \varepsilon^3 x_3 + \varepsilon^4 x_4+\dots)(x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + \varepsilon^3 x_3 + \varepsilon^4 x_4+\dots)\\ &\quad+\varepsilon(x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + \varepsilon^3 x_3 + \varepsilon^4 x_4+\dots)1\end{aligned}\)
We can multiply this out \(\begin{aligned}0=&x_1^2+2\varepsilon x_0x_1+2\varepsilon^2 x_0x_2+\varepsilon^2 x_1^2+2\varepsilon^3x_0x_3+2\varepsilon^3x_1x_2+\\ &2\varepsilon^4x_0x_4+2\varepsilon^4x_1x_3+\varepsilon^4x_2^2+\\ &\varepsilon x_0+\varepsilon^2 x_1+\varepsilon^3 x_2+\varepsilon^4 x_31+\mathcal{O}(\varepsilon^5) \end{aligned}\)
Grouping together by powers of $\varepsilon$ we get \(\begin{aligned} 0=&(x_1^21)+\varepsilon(2x_0x_1+x_0)+\varepsilon^2(2x_0x_2+x_1^2+x_1)\\ &\varepsilon^3(2x_0x_3+2x_1x_2+x_2)+\varepsilon^4(2x_0x_4+2x_1x_3+x_2^2+x_3)+\mathcal O(\varepsilon^5) \end{aligned}\)
This equation needs to be zero for an arbitrary $\varepsilon$. This is only possible if each factor is equal to $0$. So we will look at each factor.
$\varepsilon^0=1$
We get $x_0^21=0$. Again we only want the larger solution so we take $x_0=1$. After this the higher orders are simpler (but mostly not easier), due to only being linear equations.
$\varepsilon$
We have $2x_0x_1+x_0=0$. Solving for $x_1$ we get \(x_1 = \frac{1}{2x_0}x_0=\frac{1}{2}\).
$\varepsilon^2$
We have $2x_0x_2+x_1^2+x_1$ and solving for $x_2$ we get \(x_2 = \frac{1}{2x_0}(x_1^2+x_1)\).
Now we have a small problem. There are still variables in our formula for $x_2$. Luckily (and by construction) we already know $x_0$ and $x_1$ and can plug them in. So with easy calculation $x_2 = \frac{1}{8}$.
$\varepsilon^3$
\(2x_0x_3+2x_1x_2+x_2 \Rightarrow x_3 = \frac{1}{2x_0}(2x_1x_2+x_2)=\frac{1}{32}\)
Higher orders
We can continue this schema as long as we want. The equations get increasingly longer but stay in principle simple.
Summing up the series
So I calculated two more orders before I got bored. Let’s plug them into our definition of the power series again:
\[x = 1\frac{1}{2}\varepsilon+\frac{1}{8}\varepsilon^2\frac{1}{32}\varepsilon^3 +0\varepsilon^4+\frac{1}{512}\varepsilon^5+\mathcal O (\varepsilon^6)\]To get an approximation we just ignore all orders we did not calculate:
\[x \approx 1\frac{1}{2}\varepsilon+\frac{1}{8}\varepsilon^2\frac{1}{32}\varepsilon^3 +0\varepsilon^4+\frac{1}{512}\varepsilon^5\]So for $\varepsilon$ smaller $1$ our approximation works out well and gets worse for larger $\varepsilon$. We assumed $\varepsilon$ is small at the beginning so all this is as expected.
If we look at the limits, at $\varepsilon\rightarrow0$ everything fits, and at $\varepsilon\rightarrow\infty$ our series diverges.
This is all rather wellknown. If you have some formal training in quantum physics or advanced ordinary differential equations you will have heard this before. In the next part, I will show you something quite similar, but it looks funny and it is a little surprising that it works.
Pertrubation theory for large $\varepsilon$
In the last step, we looked at a power series that works well for $\varepsilon$ small. Now let us construct a power series that works well when $\varepsilon$ becomes very large. In this case, $\varepsilon^{1}=\frac{1}{\varepsilon}$ becomes very small, so let’s use it for a series: \(x=\sum_{n=0}^{\infty}x_n\varepsilon^{n}\)
We can again plug this into our quadratic equation and look at the resulting powers of $\varepsilon$.
$\varepsilon$
Somewhat suprisingly we still have a positive power of $\varepsilon$ (from the term with $\varepsilon$ in the equation): \(x_0=0\)
Nothing to do here. Turns out we could have started our series at $n=1$.
$\varepsilon^0=1$
We get $x_0^2+x_11=0$, so $x_1=1$.
Another surprise is that we don’t get to choose between two solutions this time. I think this could be because the smaller solution diverges at infinity. If you work it out tell me what you find.
$\varepsilon^{1}$
We get $2x_0x_1+x_2=0$ and so $x_2=0$. It turns out all even order terms in the series are zero. If you don’t believe me, try to prove it by induction. It is not that hard, but messy to write down correctly.
$\varepsilon^{2}$
Now we get to the first order without a surprise. The equation is $2x_0x_2+x_1^2+x_3=0 \Rightarrow x_3 = x_1^2 = 1$. Now we can go on for the higher(lower) orders.
Summing up
With some more orders, I got
\(x = \varepsilon^{1}\varepsilon^{3}+3\varepsilon^{5}\).
In most cases, it is a very bad omen, when the coefficients in a power series increase, but it is not that bad in this case. For $\epsilon \rightarrow 0$ this diverges badly anyway.
Due to $x_0=0$, this converges at $\varepsilon\rightarrow\infty$ to $0$ as it should.
If we plot this function, as expected it works well for large $\varepsilon$ (in this case larger than $3$) and does not work for small values.
The strange case of $\varepsilon\approx 1$
So we found a good approximation for small and large $\varepsilon$ with simple calculations, but they both break down at $\varepsilon=1$. That seems strange to me. After some months of thought, I have not found something similar that works there.
Of cause, you can calculate more orders and the approximation at $1$ will become good enough for practical applications.
You might try a series of the form \(x = \sum_{n=0}^\infty x_n(\varepsilon1)^n\), but this looks so ugly to me, that I did not even want to work it out for fun.^{3}
Is this useful?
Not really for quadratic (or other algebraic) equations.
The principle for small $\varepsilon$ is used in many calculations in high energy physics. You can think of this quadratic equation as one of the simplest possible model systems for the method.
If one could work out the the method for large $\varepsilon$ for quantum mechanics, it could possibly help for some quantumsystems with strong coupling. Maybe.

Don’t be confused: the names of the variables look kind of strange because we already used $x$ and $\varepsilon$ ↩

For $\varepsilon \ge 0$
\[\begin{aligned} &x'(\varepsilon) = \frac{1}{2}\big(\frac{\varepsilon}{\sqrt{\varepsilon^2+4}}1\big) < 0\\ \Leftrightarrow & \frac{\varepsilon}{\sqrt{\varepsilon^2+4}} < 1\\ \Leftrightarrow & \varepsilon < \sqrt{\varepsilon^2+4} \end{aligned}\]QED ↩

I think it would be beautiful if you can make the combined series \(\sum_{n=\infty}^{\infty}a_n\varepsilon^n\) work, but I just get an infinite number of terms for each order. ↩