Solve linear recurrence relations with characteristic equations. Visualize sequences and compare closed-form solutions.
Enter a linear recurrence with constant coefficients - a(n) = 3a(n-1) - 2a(n-2) - along with starting values, and the tool forms the characteristic equation, factors it, and returns a closed-form solution, a term-by-term table, and a graph of the sequence. Two things surprise people. First, the closed form is fitted to the exact initial values you typed: the same recurrence with different starting values produces a different formula. Second, a repeated root changes the shape of the answer, from A*r1^n + B*r2^n to (A + Bn)*r^n; that is the method working as intended, not an error.
A concrete input and expected output from the current implementation.
Input
a(n) = 3a(n-1) - 2a(n-2), a(0) = 1, a(1) = 2
Expected output
Characteristic equation: x^2 - 3x + 2 = 0, roots 1 and 2. Closed form: a(n) = 2^n. Sequence: 1, 2, 4, 8, 16, 32, 64, ...
The roots 1 and 2 give the general form A*1^n + B*2^n; the initial values force A = 0 and B = 1, so a(n) = 2^n. Direct iteration agrees: a(2) = 3*2 - 2*1 = 4, matching 2^2.
For a(n) = c1*a(n-1) + c2*a(n-2), the tool substitutes the trial solution a(n) = r^n, which turns the relation into the polynomial r^2 - c1*r - c2 = 0. Each root r contributes a candidate r^n; two distinct roots combine as A*r1^n + B*r2^n, while a repeated root replaces the second term with B*n*r^n. The initial values then fix A and B through two linear equations.
Before showing results, the tool evaluates the derived formula against the recurrence itself for the first several indices; a mismatch would betray a sign error in the characteristic polynomial. The output then pairs the closed form with a sequence table and a graph, so you can read growth, decay, or alternation directly instead of trusting the algebra sight unseen.
Two constants A*r^n and B*r^n are the same family when both roots equal r, so they cannot supply the two free constants a second-order recurrence needs. Multiplying one candidate by n produces the missing independent solution, giving (A + Bn)*r^n. For a(n) = 4a(n-1) - 4a(n-2) with a(0) = 1 and a(1) = 4, the result is (1 + n)*2^n, and iterating the recurrence confirms it.
The characteristic-equation method as usually described applies to the homogeneous case, with no extra term. A constant added on the right needs a particular solution on top of the homogeneous form, so a(n) = 2a(n-1) + 3 is treated differently from a(n) = 2a(n-1). Whether this tool accepts that input is worth testing directly on the page, since the method alone does not cover it.
The example and behavioral notes were checked against the browser implementation. Standards and primary references below define the relevant format, formula, or platform behavior.
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