General solution of cyclic probability expectation

For some problems that need to solve equations but cannot use Gaussian elimination, generally speaking, the equations are solved within the ring, and recursion is performed between the rings. The solution equation in the ring is generally a recursive sequence of numbers connected at the beginning and end by adding an equation.

• Suppose there is a recurrence relationship $f(1)\to f(2)\to f(3)...\to f\left(n\right)$ , in fact, it can also be regarded as an equation in essence, telling you the initial value$f\left(1\right)$ , the equation can be solved by recurrence relation.
  So if you don’t tellf(1) f(1)The value of$f$$($$1$$)$ , but add another recurrence relation$f(n)\to f\left(1\right)$ , forming a ring. Let's ignore$f(n)\to f(1)$。根据$f(1)\to f(2)\to f(3)...\to f\left(n\right)$ , regard it as the initial value$f\left(1\right)$ , then we can obtainf (1) → f (n) f(1)\rightarrow f(n)by solving the general term formula of the sequence$f\left(1\right)\to f\left(n\right)$ , just usef (1) f(1)directly$f\left(1\right)$ means$f\left(n\right)$ , then according to the equation$f(n)\to f\left(1\right)$ can be solved without Gaussian elimination.