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Alternating Function 📂Functions

Alternating Function

Definition

Let set XX be given. A function that satisfies the following is called an alternating function.

ϕ:X×X××XnRϕ(x1,,xi,xi+1,,xn)=ϕ(x1,,xi+1,xi,,xn) \phi : \overbrace{X \times X \times \cdots \times X}^{n} \to \mathbb{R} \\ \phi (x_{1}, \dots, x_{i}, x_{i+1}, \dots, x_{n}) = - \phi (x_{1}, \dots, x_{i+1}, x_{i}, \dots, x_{n})

Explanation

It is a function whose sign changes when two adjacent variables are swapped. Of course, it can also be shown that this holds true for variables that are not adjacent, according to the definition.

If there is at least one duplicate element among the variables, the value of the function is 00.

Properties

ϕ(x1,,xi,,xi+k,,xn)=ϕ(x1,,xi+k,,xi,,xn) \begin{equation} \phi (x_{1}, \dots, x_{i}, \dots, x_{i+k}, \dots, x_{n}) = - \phi (x_{1}, \dots, x_{i+k}, \dots, x_{i}, \dots, x_{n}) \end{equation}

The necessary and sufficient condition for ϕ\phi to be an alternating function is as follows.

ϕ(x1,,xi,xi,,xn)=0 \begin{equation} \phi (x_{1}, \dots, x_{i}, x_{i}, \dots, x_{n}) = 0 \end{equation}

Proof

Proof (1)

ϕ(x1,,xi,,xi+k1,xi+k,,xn)= ϕ(x1,,xi,,xi+k,xi+k1,,xn)= (1)2ϕ(x1,,xi,,xi+k,xi+k2,xi+k1,,xn)= (1)kϕ(x1,,xi+k,xi,,xi+k2,xi+k1,,xn)= (1)k+1ϕ(x1,,xi+k,xi+1,xi,,xi+k2,xi+k1,,xn)= (1)k+(k1)ϕ(x1,,xi+k,,xi+k1,xi,,xn)= ϕ(x1,,xi+k,,xi+k1,xi,,xn) \begin{align*} & \phi (x_{1}, \dots, x_{i}, \dots, x_{i+k-1}, x_{i+k}, \dots, x_{n}) \\ =&\ - \phi (x_{1}, \dots, x_{i}, \dots, x_{i+k}, x_{i+k-1}, \dots, x_{n}) \\ =&\ (-1)^{2} \phi (x_{1}, \dots, x_{i}, \dots, x_{i+k}, x_{i+k-2}, x_{i+k-1}, \dots, x_{n}) \\ \vdots& \\ =&\ (-1)^{k} \phi (x_{1}, \dots, x_{i+k}, x_{i}, \dots, x_{i+k-2}, x_{i+k-1}, \dots, x_{n}) \\ =&\ (-1)^{k+1} \phi (x_{1}, \dots, x_{i+k}, x_{i+1} ,x_{i}, \dots, x_{i+k-2}, x_{i+k-1}, \dots, x_{n}) \\ \vdots& \\ =&\ (-1)^{k+(k-1)} \phi (x_{1}, \dots, x_{i+k}, \dots, x_{i+k-1}, x_{i}, \dots, x_{n}) \\ =&\ - \phi (x_{1}, \dots, x_{i+k}, \dots, x_{i+k-1}, x_{i}, \dots, x_{n}) \end{align*}

Proof (2)

ϕ(x1,,xi,xi,,xn)= ϕ(x1,,xi,xi,,xn)    2ϕ(x1,,xi,xi,,xn)= 0    ϕ(x1,,xi,xi,,xn)= 0 \begin{align*} && \phi (x_{1}, \dots, x_{i}, x_{i}, \dots, x_{n}) =&\ - \phi (x_{1}, \dots, x_{i}, x_{i}, \dots, x_{n}) \\ \iff && 2\phi (x_{1}, \dots, x_{i}, x_{i}, \dots, x_{n}) =&\ 0 \\ \iff && \phi (x_{1}, \dots, x_{i}, x_{i}, \dots, x_{n}) =&\ 0 \end{align*}