Generalization of Gauss Trick(1+2+3+...+100)

The initial idea comes from the proof of (1)

$$\begin{array}{}\text{(1)}& \prod _{d|n}d=n^{\frac{d(n)}{2}}\end{array}$$

As reader will see, it is just a generalization of the trick from Gauss.

Let $(L,T,\le ,\otimes ,\mathrm{\neg })$$(L,T,\le,\otimes,\neg)$ be a finite poset. $T$$T$ is someting else. Here $\mathrm{\neg }$$\neg$ is an anti-isomorphism on $L$$L$. i.e. $a\le b⟺\mathrm{\neg }b\le \mathrm{\neg }a$$a\le b\iff\neg b\le\neg a$.

Here $\otimes$$\otimes$ is a commutative, associate operator ($L$$L$ need not be closed under $\otimes$$\otimes$ ), but $\mathrm{\forall }a\in L,a\otimes \mathrm{\neg }a=T$$\forall a\in L,a\otimes \neg a=T$.

Again, Let $f:L\cup \{T\}\to A$$f:L\cup\{T\}\to A$ be a function that satisfies $f(T)=f(a\otimes \mathrm{\neg }a)=f(a)\ast f(\mathrm{\neg }a)$$f(T)=f(a\otimes \neg a)=f(a)*f(\neg a)$.

Here $(A,\ast )$$(A,*)$ is a commutative semigroup.

Then we have

$$\begin{array}{}\text{(2)}& {\displaystyle \prod _{a\in L}f(a)=f(T{)}^{\frac{|L|}{2}}}\end{array}$$

Proof.

$$\begin{array}{}\text{(3)}& \prod _{a\in L}f(a)\ast \prod _{a\in L}f(\mathrm{\neg }a)={(\prod _{a\in L}f(a))}^2=\prod _{a\in L}f(a)f(\mathrm{\neg }a)=f(T{)}^{|L|}\end{array}$$

Example.

Let $100\downarrow :=\{1,...,100\}$$100\downarrow:=\{1,...,100\}$. Here $F=1,T=101$$F=1,T=101$ , $a\otimes b=a+b,\mathrm{\neg }a=100-a$$a\otimes b=a+b,\neg a=100-a$.

Let the monoid be $(\mathbb{N},+)$$(\mathbb N,+)$, $f(a)=a$$f(a)=a$, then ${\displaystyle \sum _{1\le a\le 100}a=\frac{1}{2}(\sum _{1\le a\le 100}a+\sum _{1\le a\le 100}\mathrm{\neg }a)=\frac{100}{2}\times 101=5050}$$\displaystyle\sum_{1\le a\le 100}a=\frac{1}{2}\left(\sum_{1\le a\le 100}a+\sum_{1\le a\le 100}\neg a\right)=\frac{100}{2}\times101=5050$

Example.

The lattice here is $(n\downarrow ,0,n,|,\times ,\mathrm{\neg })$$(n\downarrow,0,n,|,\times,\neg)$. Here $n\downarrow =\{d\in \mathbb{N},d|n\}$$n\downarrow=\{d\in\mathbb N,d|n\}$ and $\begin{array}{c}\mathrm{\neg }d=\frac{n}{d}\end{array}$$\neg d=\frac{n}{d}\\$.

Let $d(n)=|n\downarrow |$$d(n)=|n\downarrow|$,

Then

$$\begin{array}{}\text{(4)}& \prod _{d|n}d=n^{\frac{d(n)}{2}}\end{array}$$

Example.

Let $(M,\mathrm{\Omega },\mu )$$(M,\Omega,\mu)$ be a measure space, suppose that $\mu (M)$$\mu(M)$ and $\mathrm{\Omega }$$\Omega$ are finite.

Then $\begin{array}{c}\sum _{\omega \in \mathrm{\Omega }}\mu (\omega )=\frac{|\mathrm{\Omega }|}{2}\mu (M)\end{array}$$\sum_{\omega\in\Omega }\mu(\omega)=\frac{|\Omega|}{2}\mu(M)\\$.

One of the key point of this proof is $\begin{array}{c}\prod _{a\in L}f(\mathrm{\neg }a)=\prod _{a\in L}f(a)\end{array}$$\prod_{a\in L}f(\neg a)=\prod_{a\in L}f(a)\\$.

It only depend on $(A,\ast )$$(A,*)$ is commutative semigroup.

Let $S$$S$ be a set, $f:S\to (A,\ast )$$f:S\to (A,*)$ is a function. Then $\begin{array}{c}\prod _{s\in S}f(\sigma s)=\prod _{s\in S}f(s)\end{array}$$\prod_{s\in S}f(\sigma s)=\prod_{s\in S}f(s)\\$

Example.

Consider $U_m$$U_m$, $\mathrm{id}:U_m\to U_m$$\mathrm{id}:U_m\to U_m$, for $a\in U_m$$a\in U_m$, define $\sigma (s)=as$$\sigma(s)=as$.

Then

$$\begin{array}{}\text{(5)}& \prod _{s\in U_m}as=a^{\phi (m)}\prod _{s\in U_m}s=\prod _{s\in U_m}s⟹a^{\phi (m)}=1\end{array}$$