이원수 환 위에서 정의되는 미분가능한 실함수 📂Abstract Algebra

이원수 환 위에서 정의되는 미분가능한 실함수

Build-up¹

Assume the given smooth function $f : \mathbb{R} \to \mathbb{R}$. The Taylor series of $f$ at $a$ is as follows:

$$ \begin{align*} f(x) &= f(a) + f^{\prime}(a)(x - a) + \dfrac{f^{\prime \prime}(a)}{2!}(x - a)^{2} + \cdots \\ &= f(a) + f^{\prime}(a)(x - a) + \sum_{n=2}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x - a)^{n} \end{align*} $$

Although the above equation is obtained for a function defined on the real space, let’s substitute the dual number $a + b\epsilon = (a, b)$ instead of $x$.

$$ \begin{align*} f(a + b\epsilon) &= f(a) + f^{\prime}(a)(a + b\epsilon - a) + \sum_{n=2}^{\infty} \dfrac{f^{(n)}(a)}{n!}(a + b\epsilon - a)^{n} \\ &= f(a) + f^{\prime}(a)b\epsilon + \sum_{n=2}^{\infty} \dfrac{f^{(n)}(a)}{n!}b^{n}\epsilon^{n} \\ &= f(a) + f^{\prime}(a)b\epsilon \\ &= \big( f(a), b f^{\prime}(a) \big) \end{align*} $$

Since $\epsilon^{2} = 0$, all terms from the third one onward are $0$.

Definition

Assume the given differentiable function $f : \mathbb{R} \to \mathbb{R}$. For the dual number $a + b\epsilon$, we define $f(a + b\epsilon)$ as follows:

$$ f(a + b\epsilon) := \big( f(a), b f^{\prime}(a) \big) \tag{1} $$

Explanation

Note that notation was abused from $(1)$. The $f$ on the left side of $(1)$ is actually a function defined as follows for a fixed $f$.

$$ \begin{align*} F_{f} : \left\{ a + b\epsilon : a, b \in \mathbb{R} \right\} &\to \left\{ a + b\epsilon : a, b \in \mathbb{R} \right\} \\ a + b\epsilon &\mapsto f(a) + b f^{\prime}(a)\epsilon = \big( f(a), b f^{\prime}(a) \big) \end{align*} $$

However, considering the case $b = 0$, since it is $F_{f}(a,0) = (f(a), 0)$, it can be seen that $f$ extends naturally. Therefore, note that it has been denoted for convenience as $f \equiv F_{f}$.

When extending the real number $x$ to the dual number $(x, 1)$ and performing addition or multiplication, the differential is preserved in the second component. Hence, it can be seen that the above definition is very naturally defined from the perspective of preserving the differential.

Composite Function

For the composite function $f \circ g$, we define $f \circ g (a + b\epsilon)$ as follows:

$$ f \circ g (a + b\epsilon) := f(g(a)) + f^{\prime}(g(a)) g^{\prime}(a)b\epsilon = \big( f(g(a)), b f^{\prime}(g(a)) g^{\prime}(a) \big) $$

The first component is the value of the function $f \circ g$ at $a$, and the second component is the differential.

Derivation

It is obtained directly by calculating as defined in $(1)$.

$$ \begin{align*} f \circ g (a + b\epsilon) &= f(g(a + b\epsilon)) \\ &= f(g(a) + g^{\prime}(a)b\epsilon) \\ &= f(g(a)) + f^{\prime}(g(a)) g^{\prime}(a)b\epsilon \\ &= \big( f(g(a)), b f^{\prime}(g(a)) g^{\prime}(a) \big) \end{align*} $$