Complete the proof of Theorem 4.38(a) by showing that if the recurrence relation x n = ax
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Complete the proof of Theorem 4.38(a) by showing that if the recurrence relation xn = axn-1 + bxn-2 has distinct eigenvalues λ1 ≠ λ2, then the solution will be of the form
xn = c1λn1 + c2λn2
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Since A is diagonalizable we have Suppos...View the full answer
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