Dual number $(a+b\varepsilon)$ raised to a dual power, e.g. $(a+b\varepsilon)^{(c+d\varepsilon)}$

Question

I'm working on some code which utilizes Newton's method, and I would like to take advantage of dual numbers to simplify taking the derivative. I've worked out a class definition Dual which works great for polynomials, pow (where either base or exponent is real) exp, and log.

However, I am a bit stymied at raising a dual number to the power of another dual number, e.g.

$(a+b\varepsilon)^{(c+d\varepsilon)}$

I used the general form derived from the Taylor series $f(a+b\varepsilon) = f(a) + bf'(a)\varepsilon$ to derive the rules for $x^n$ and $n^x$ where $n$ is real and $x$ is dual. For those, I got:

$x^n = a^n + bna^{n-1}$ , where $ x = a+b\varepsilon$

$n^y = n^c + d\ln(n)n^{c}$ , where $ y = c+d\varepsilon$

Substituting and simplifying, I end up with:

$x^y = a^c + (d\ln(a)a^c + bca^{c-1})\varepsilon$

This seems to work right in my code, but I do not know if this is actually correct. The implementations of exp and pow work properly with this more general code, and $\varepsilon^\varepsilon = 1 + NaN\varepsilon$, which is a good-ish sign (if I got something well-formed, that would be more troubling).

Is this formula valid?

Answer 1

$\def\e{\varepsilon}$\begin{align*} (a+b\e)^{c+d\e} &= \exp((c+d\e)\ln(a+b\e)) & x=e^{\ln x} \\ &=\exp((c+d\e)(\ln a+\ln(1+b\e/a))) \\ &=\exp((c+d\e)(\ln a+b\e/a)) & \ln(1+x)=x+\mathrm{h.o.} \\ &=\exp(c\ln a+\e(d\ln a+bc/a))\\ &=a^c e^{\e(d\ln a+bc/a)} \\ &=a^c (1+\e(d\ln a+bc/a)). & e^x=1+x+\mathrm{h.o.} \end{align*}

Answer 2

The other answer is not very consistent with the set of Dual Numbers. You don't need any approximation to prove this. First suppose that $a\neq 0\\$. The second step is to prove that for a generic Dual Number $z=a+\epsilon b\,\,\text{with } a\neq 0$, holds : \begin{align} z=e^{\log(z)} \end{align} and holds since for the Dual Numbers set we have the following: \begin{align*} \log(z)= \log(a+\epsilon b)=\log(a)+\frac{b}{a}\epsilon \end{align*} \begin{align*} e^{z}= e^{a+\epsilon b}=e^a+e^ab\epsilon \end{align*} Combining them we have \begin{align*} e^{\log(z)}= e^{\log(a)+\epsilon \frac{b}{a} }=e^{\log(a)}+e^{\log(a)}\frac{b}{a}\epsilon=a+\epsilon b=z \end{align*} Coming back to the original problem we have: \begin{align*} (a+b\epsilon)^{c+d\epsilon} &= e^{ (c+d\epsilon)\log(a+b\epsilon) } & z=e^{\log z} \\ &=e^{ (c+d\epsilon)(\log a+b\epsilon/a) } & \log(a+\epsilon b)=\log(a)+\frac{b}{a}\epsilon\\ &=e^{ c\log a+\epsilon(d\log a+b c/a) }& \epsilon^2=0 \\ &=e^{c\log a}\,e^{ \epsilon\,d\log a }\,e^{ \epsilon \frac{b\,c}{a} }\\ &=a^{c+\epsilon d}\,e^{\epsilon \frac{b\,c}{a}} \end{align*} or, if you prefer \begin{align*} (a+b\epsilon)^{c+d\epsilon} &=a^{c+\epsilon d}\,e^{\epsilon \frac{b\,c}{a}}=a^{c+\epsilon d}\,(1+\epsilon \frac{b\,c}{a}) \end{align*}

Answer 3

In Mathematica use this code:

$Pre = MatrixFunction[
      Function[ε, #], {{0, 0}, {1, 0}}] /. {{a_, 
        b_}, {0, a_}} -> a + ε b /. {{a_, 0}, {b_, a_}} -> a + ε b &;

(a + b ε)^(c + d ε)

The result:

a^c + a^(-1 + c) (b c + a d Log[a]) ε