How to take the square root of a dual number ($\sqrt{a + b \cdot \varepsilon}$ with $a, ~b \in \mathbb{R}$ and $\varepsilon^{2} = 0$)?

Question

How to take the square root of a dual number:

$\sqrt{\Xi}$ with $\begin{align*} a, ~b &\in \mathbb{R}\\ \varepsilon^{2} &= 0\\ \varepsilon &\ne 0\\ \\ \Xi &:= a + b \cdot \varepsilon\\ \end{align*}$

My first attempt worked (at least I think so), but the second one came to nothing, but it seems to me that taking the square root of binary numbers can also be derived in a different way, which is why I am here under the question in the question and the answers to show all possibilities to the question. (I wrote my first attempt as an answer under the question.)

Surely the question sounds strange, because why should one calculate that at all, but let's just ask ourselves how that would work?

My attempt $2$

My attempt $2$ was "try and hope that it works": $$ \begin{align*} \Xi^{2} &= \left( a + b \cdot \varepsilon \right)^{2}\\ \Xi^{2} &= a^{2} + 2 \cdot a \cdot + b \cdot \varepsilon + \left( b \cdot \varepsilon \right)^{2}\\ \Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot \varepsilon^{2}\\ \Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot 0\\ \Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + 0\\ \Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon \\ \Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\ a + b \cdot \varepsilon &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\ c := a^{2} &\wedge d := 2 \cdot a \cdot b \cdot \varepsilon\\ a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot a} \cdot \varepsilon^{-1}\\ a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\ \\ \Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\ a + b \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon} \quad\mid\quad a = \sqrt{c} \wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\ \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1} \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon}\\ \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot 1 &= \sqrt{c + d \cdot \varepsilon}\\ \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} &= \sqrt{c + d \cdot \varepsilon}\\ \sqrt{c + d \cdot \varepsilon} &= \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}}\\ \sqrt{a + b \cdot \varepsilon} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \quad\mid\quad \left( ~~ \right)^{2}\\ a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \right)^{2}\\ a + b \cdot \varepsilon &= a + 2 \cdot \sqrt{a} \cdot \frac{b}{2 \cdot \sqrt{a}} + \frac{b^{2}}{4 \cdot a}\\ a + b \cdot \varepsilon &= a + b + \frac{b^{2}}{4 \cdot a}\\ b \cdot \varepsilon &= b + \frac{b^{2}}{4 \cdot a}\\ \end{align*} $$

But that makes no sense...

Answer 1

From calculus, $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$.

With the obvious caveats on the sign of $a$, we have $$ \sqrt{a + b\epsilon} = \pm \sqrt{a} \sqrt{1 + \frac{b\epsilon}{a}} = \pm \sqrt{a} (1 + \frac{b\epsilon}{2a} + O(\epsilon^2)). $$

Answer 2

My attempt $1$:

I wanted to try to derive the method of taking the square root of a dual number analogously to taking the square root of complex numbers. So I started:

** Derivation **

$$ \begin{align*} 0 &\geq k \in \mathbb{Z}\\ \varepsilon &\ne 0\\ \varepsilon^{2} &= 0\\ \varepsilon^{3} &= \varepsilon^{2} \cdot \varepsilon = 0 \cdot \varepsilon = 0\\ \varepsilon^{4} &= \left( \varepsilon^{2} \right)^{2} \cdot \varepsilon = 0^{2}= 0\\ \varepsilon^{k} &= \varepsilon^{2 + k - 2} = \varepsilon^{2} \cdot \varepsilon^{k - 2} = 0 \cdot \varepsilon^{k - 2} = 0\\ \\ x &\in \mathbb{R}\\ \\ e^{x \cdot \varepsilon} &= \exp\left( x \cdot \varepsilon \right)\\ e^{x \cdot \varepsilon} &= \sum_{n = 0}^{\infty} \frac{\left( x \cdot \varepsilon \right)^{n}}{n!}\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \frac{\left( x \cdot \varepsilon \right)^{2}}{2!} + \frac{\left( x \cdot \varepsilon \right)^{3}}{3!} + \frac{\left( x \cdot \varepsilon \right)^{4}}{4!} + \cdots\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \frac{x^{2} \cdot \varepsilon^{2}}{2!} + \frac{x^{3} \cdot \varepsilon^{3}}{3!} + \frac{x^{4} \cdot \varepsilon^{4}}{4!} + \cdots\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \varepsilon^{2} \cdot \left( \frac{x^{2}}{2!} + \frac{x^{3} \cdot \varepsilon}{3!} + \frac{x^{4} \cdot \varepsilon^{2}}{4!} + \cdots \right)\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + 0 \cdot \left( \frac{x^{2}}{2!} + \frac{x^{3} \cdot \varepsilon}{3!} + \frac{x^{4} \cdot \varepsilon^{2}}{4!} + \cdots \right)\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + 0\\ e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon\\ \\ \Xi &= y \cdot \left( 1 + x \cdot \varepsilon \right)\\ \Xi &= y + y \cdot x \cdot \varepsilon \quad\mid\quad \Xi = a + b \cdot \varepsilon\\ a &= y \wedge b = y \cdot x\\ a &= y \wedge \frac{b}{y} = x\\ a &= y \wedge \frac{b}{a} = x\\ \\ \Xi &= y \cdot e^{x \cdot \varepsilon} \quad\mid\quad \sqrt{~~}\\ \sqrt{\Xi} &= \sqrt{y \cdot e^{x \cdot \varepsilon}}\\ \sqrt{\Xi} &= \sqrt{y} \cdot \sqrt{e^{x \cdot \varepsilon}}\\ \sqrt{\Xi} &= \sqrt{y} \cdot \left( e^{x \cdot \varepsilon} \right)^{\frac{1}{2}}\\ \sqrt{\Xi} &= \sqrt{y} \cdot e^{\frac{x}{2} \cdot \varepsilon} \quad\mid\quad e^{x \cdot \varepsilon} = 1 + x \cdot \varepsilon\\ \sqrt{\Xi} &= \sqrt{y} \cdot \left( 1 + \frac{x}{2} \cdot \varepsilon \right) \quad\mid\quad a = y \wedge \frac{b}{a} = x\\ \sqrt{\Xi} &= \sqrt{a} \cdot \left( 1 + \frac{\frac{b}{a}}{2} \cdot \varepsilon \right)\\ \sqrt{\Xi} &= \sqrt{a} \cdot \left( 1 + \frac{b}{2 \cdot a} \cdot \varepsilon \right)\\ \sqrt{\Xi} &= \sqrt{a} \cdot 1 + \sqrt{a} \cdot \frac{b}{2 \cdot a} \cdot \varepsilon\\ \sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\ \end{align*} $$

** Proof **

$$ \begin{align*} \sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\ \sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\ \sqrt{a + b \cdot \varepsilon} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \quad\mid\quad \left( ~~ \right)^{2}\\ a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\ a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\ a + b \cdot \varepsilon &= a + 2 \cdot \sqrt{a} \cdot \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\ a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\ a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \right)^{2} \cdot \varepsilon^{2}\\ a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \right)^{2} \cdot 0\\ a + b \cdot \varepsilon &= a + b \cdot \varepsilon \quad \square\\ \end{align*} $$

Aka the solution that I find and proved is $\sqrt{a + b \cdot \varepsilon} = \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon$.

Answer 3

You are making things too complicated. Let $w=a+b\epsilon$ be a dual number (with $a,b$ real). We want to find the dual numbers $z$ such that $z^2=w$.

We write $z=x+y\epsilon$, with $x$ and $y$ real numbers. A short computation shows that $z^2 = x^2 + 2xy\epsilon$. Since $(1,\epsilon)$ is a base of dual numbers (as a vector space over the real numbers), we get the system

$$ \left\{ \begin{array}{lcl} x^2 & = & a \\ 2xy & = & b\\ \end{array} \right. $$

1. The first thing to remark is that, if $a<0$, then there is no solution.
2. If $a=0$ but $b\neq 0$, then the first equation implies that $x=0$, so the second equation becomes $0=b$, so no solution.
3. If $a=b=0$, then $x=0$ and $y$ can be any real number. So $0=0+0\epsilon$ has infinitely many square roots in the dual numbers.
4. Finally, if $a>0$, then $x=\pm \sqrt{a}$ and $y=\frac{b}{2x}=\pm \frac{b}{2\sqrt{a}}$, so $\omega=a+b\epsilon$ has two opposite square roots: $\pm(\sqrt{a}+\frac{b}{2\sqrt{a}}\epsilon)$