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RSolve

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Solve recurrence equations.

Boleh disemak secara bebas: jawapan untuk fungsi ini diterbitkan semula melalui laluan lain dan dibandingkan - ruang kerja dan alat verify melakukannya secara automatik, jadi jawapan salah dikesan dan bukannya dipercayai. Sejarah →

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]}}
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Topik

Berkaitan

RSolveValueAsymptoticRSolveValueRecurrenceTableRFixedPointsRStabilityConditionsSumZTransformGeneratingFunctionDifferenceRootDSolveFindSequenceFunctionFindGeneratingFunction

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