I have a recurrence equation tnt_n which is defined as

tn=2.5tn−1−1.5tn−4,for n≥5

t_n=2.5t_{n-1}-1.5t_{n-4},\qquad \text{for }n\geq5

Where

t1=5t2=10.5t3=26.25t4=62.625

\begin{align}

t_1&=5\\

t_2&=10.5\\

t_3&=26.25\\

t_4&=62.625

\end{align}

I need to solve my recurrence equation for nn when tn=13×1010t_n=13\times10^{10}. How can I achieve this in mathematica?

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Feb 12 ’15 at 2:50

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2 Answers

2

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RSolve can provide an exact expression for t[n]

Clear[t];

t[n_] = t[n] /. RSolve[{t[1] == 5, t[2] == 21/2,

t[3] == 105/4, t[4] == 501/8,

t[n] == (5 t[n – 1] – 3 t[n – 4])/2},

t[n], n][[1]] // ToRadicals // Simplify

(1/219)(2^(1 – n)

(1 + (10 – Sqrt[73])^(1/3) +

(10 + Sqrt[73])^(1/3))^n*

(73 + (73*(5767 – 520*Sqrt[73]))^(1/3) +

(73*(5767 + 520*Sqrt[73]))^(1/3)) +

((1/4)(2 + (-1 – ISqrt[3])*

(10 – Sqrt[73])^(1/3) +

I*(I + Sqrt[3])(10 + Sqrt[73])^

(1/3)))^n(146 + I*(I + Sqrt[3])*

(73*(5767 – 520*Sqrt[73]))^(1/3) +

(-1 – ISqrt[3])

(73*(5767 + 520*Sqrt[73]))^(1/3)) +

((1/4)(2 + I(I + Sqrt[3])*

(10 – Sqrt[73])^(1/3) +

(-1 – ISqrt[3])(10 + Sqrt[73])^

(1/3)))^n*(146 + (-1 – ISqrt[3])

(73*(5767 – 520*Sqrt[73]))^(1/3) +

I*(I + Sqrt[3])*

(73*(5767 + 520*Sqrt[73]))^(1/3)))

Table[t[n], {n, 0, 5}] // N // Rationalize

{2, 5, 21/2, 105/4, 501/8, 2385/16}

FindRoot[t[n] == 13*^10, {n, 25}][[1]] // Chop // Quiet

n -> 28.6241

(t[#] // Chop) & /@ {28., 29.}

{7.54702*10^10, 1.80385*10^11}

LogPlot[{t[n] // Chop, 13*^10}, {n, 28, 29}]

t[1] = 5;

t[2] = 10.5;

t[3] = 26.25;

t[4] = 62.625;

t[n_Integer] := 2.5 t[n – 1] – 1.5 t[n – 4];

n = 5; While[t[n] < 13 10^10, Null; n++]; n (* 29 *)