It's been a while since I posted about linear recurrence relations. The only two posts were:
- Solving Linear Recurrence on the 16th of June 2020
- More on Linear Recurrence on the 26th December 2020
Today I turned \(28073\) days old and one of the properties of this number is that it's a member of OEIS A105578:$$ \begin{align} &\text{a}(n+3) = 2\text{a}(n+2) - 3\text{a}(n+1) + 2\text{a}(n)\\ &a(0) = 1, a(1) = 1, a(2) = 0 \end{align}$$This recurrence relation corresponds to the polynomial:$$ \begin{align} &x^3 = 2x^2 -3x +1 \\ &x^3-2x^2+3x-1 = 0 \end{align}$$The polynomial \( \text{P}(x) = x^3-2x^2+3x-1 \) has three real roots. See Figure 1.
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Figure 1 |
- \( x \approx -1.34292 \)
- \( x \approx 0.52932 \)
- \( x \approx 2.8136 \)
However the form of the exact roots is interesting:$$x = \frac{2}{3} - \frac{13(1 + i \sqrt{3})}{6 \sqrt[3]{8 + 3i \sqrt{237}}} - \frac{1}{6} (1 - i \sqrt{3}) \sqrt[3]{8 + 3i \sqrt{237}}$$$$x = \frac{2}{3} - \frac{13(1 - i \sqrt{3})}{6 \sqrt[3]{8 + 3i \sqrt{237}}} - \frac{1}{6} (1 + i \sqrt{3}) \sqrt[3]{8 + 3i \sqrt{237}}$$$$x = \frac{1}{3} \left( 2 + \frac{13}{\sqrt[3]{8 + 3i \sqrt{237}}} + \sqrt[3]{8 + 3i \sqrt{237}} \right)$$Even though the formula contains \( i \)'s, they always cancel out to produce real numbers.
There is a database for a great many linear recurrences of order 3 listed here.
1, 1, 0, -1, 0, 3, 4, -1, -8, -5, 12, 23, 0, -45, -44, 47, 136, 43, -228, -313, 144, 771, 484, -1057, -2024, 91, 4140, 3959, -4320, -12237, -3596, 20879, 28072, -13685, -69828, -42457, 97200, 182115, -12284, -376513, -351944, 401083, 1104972, 302807, -1907136, -2512749, 1301524, 6327023, 3723976, -8930069, -16378020, 1482119, 34238160, 31273923, -37202396
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| Figure 2: permalink |
It can be seen that after a while the values between to fluctuate wildly between larger and larger positive and negative values.
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