Solve recurrence equations.
与 .wl 参考的差异： `RSolve` first-order linear inhomogeneous
可独立核查：此函数的答案会通过另一条途径重新推导并比较——工作台和 verify 工具会自动完成，因此错误答案会被发现而不是被采信。 历史 →
RSolve[eqn, a[n], n]
RSolve[{eqn1, eqn2, …}, {a1[n], a2[n], …}, n]
RSolve[eqn, a[n1, n2, …], {n1, n2, …}]
RSolve[a[n] == 2 a[n-1], a[n], n]
→ {{a[n] -> 2^(-1 + n)*C[1]}}RSolve[{a[n + 1] == 2 a[n], a[0] == 1}, a, n]
→ {{a -> Function[{n}, 2^n]}}RSolve[{a[n] == 2 a[n-1], a[2] == 5}, a[n], n]
→ {{a[n] -> 5*2^(-2 + n)}}RSolve[a[n] == 4 a[n-1] - 4 a[n-2], a[n], n]
→ {{a[n] -> 2^n*C[1] + 2^n*n*C[2]}}RSolve[{a[n] == a[n-1] + a[n-2], a[0] == 0, a[1] == 1}, a[n], n]
→ {{a[n] -> Fibonacci[n]}}RSolve[a[n] == a[n-1] + a[n-2], a[n], n]
→ {{a[n] -> C[1]*Fibonacci[n] + C[2]*LucasL[n]}}RSolve[{x[n + 1] == 4 x[n] (1 - x[n]), x[0] == 1/10}, x, n]
→ .* (regex*)
{{x -> Function[{n}, (1 - Cos[2^n*ArcCos[4/5]])/2]}}RSolve[{x[n + 1] == 4 x[n] (1 - x[n])}, x, n]
→ {{x -> Function[{n}, 1/2 - Cos[2^n*C[1]]/2]}}