Given the recurrence relation ππ =β2ππβ1 +15ππβ2 with the initial conditions π0=1 and π1=7.
(a) Write the characteristic equation.
(b) Solve the recurrence relation.
an=β2anβ1+15anβ2a:Ξ»2=β2Ξ»+15Ξ»2+2Ξ»β15=0Ξ»β{β5,3}b:an=C1(β5)n+C23n{a0=1a1=7β{C1+C2=1β5C1+3C2=7β{C1=β0.5C2=1.5an=β0.5(β5)n+1.5β 3na_n=-2a_{n-1}+15a_{n-2}\\a:\\\lambda ^2=-2\lambda +15\\\lambda ^2+2\lambda -15=0\\\lambda \in \left\{ -5,3 \right\} \\b:\\a_n=C_1\left( -5 \right) ^n+C_23^n\\\left\{ \begin{array}{c} a_0=1\\ a_1=7\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_1+C_2=1\\ -5C_1+3C_2=7\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_1=-0.5\\ C_2=1.5\\\end{array} \right. \\a_n=-0.5\left( -5 \right) ^n+1.5\cdot 3^nanβ=β2anβ1β+15anβ2βa:Ξ»2=β2Ξ»+15Ξ»2+2Ξ»β15=0Ξ»β{β5,3}b:anβ=C1β(β5)n+C2β3n{a0β=1a1β=7ββ{C1β+C2β=1β5C1β+3C2β=7ββ{C1β=β0.5C2β=1.5βanβ=β0.5(β5)n+1.5β 3n
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