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Find the explicit sequence that satisfy:

$$ x_{n+2}-3x_{n+1}+2x_n= \cos^2(\frac{\pi}{2}n)\sin(\frac{\pi}{6}n)$$

and the initial condition $$x_0=x_1=0$$

My first attempt was to compute some term and guess the law: $x_2=x_3=0$

$x_4=\frac{\sqrt3}{2}$

$x_5=3\frac{\sqrt3}{2}$

$x_6=4\sqrt3$

$x_7=9\sqrt3$

$x_8=19\sqrt3$

$x_9=39\sqrt3$

$x_{10}=\frac{391}{2}\sqrt3$

then i try to follow the wikipedia article https://en.wikipedia.org/wiki/Recurrence_relation#Solving_non-homogeneous_linear_recurrence_relations_with_constant_coefficients

but i don't know the form of the inhomogeneous solution.

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    $\begingroup$ What have you tried? $\endgroup$ – Michael Burr Oct 8 '18 at 11:44
  • $\begingroup$ i compute some of the first term to find a law, but nothing... then i tried variation of parameter but the integral becomes ugly (there must be another easiest method) $\endgroup$ – polbos Oct 8 '18 at 11:48
  • $\begingroup$ Please include your attempt in your question. Otherwise, your question is in danger of being closed due to missing context. $\endgroup$ – Michael Burr Oct 8 '18 at 11:57
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    $\begingroup$ Do you know Z transforms? $\endgroup$ – Paul Oct 8 '18 at 12:01
  • $\begingroup$ no but thank you for the advice $\endgroup$ – polbos Oct 8 '18 at 12:30
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This is a linear difference equation so the solution can be written as

$$ x_n = x^h_n + x^p_n \\ x^h_{n+2}-3x^h_{n+1}+2x^h_n= 0\\ x^p_{n+2}-3x^p_{n+1}+2x^p_n= \cos^2(\frac{\pi}{2}n)\sin(\frac{\pi}{6}n) $$

the difficult part is to obtain $x^p_n$ so we will focus that. Making

$$ x^p_n = a e^{\frac{i \pi n}{3}}+b e^{-\frac{1}{3} i \pi n}+c e^{\frac{2 i \pi n}{3}}+d e^{-\frac{2}{3} i \pi n} $$

after substituting and grouping terms we get at

$$ \frac{1}{4} e^{-\frac{2}{3} i \pi n} \left(-i \left(4 \sqrt{3} a+1\right) e^{i \pi n}+i \left(4 \sqrt{3} b+1\right) e^{\frac{i \pi n}{3}}+\left(-8 i \sqrt{3} c+12 c+i\right) e^{\frac{4 i \pi n}{3}}+4 \left(3+2 i \sqrt{3}\right) d-i\right) = 0 $$

and then solving

$$ \left\{ \begin{array}{rcl} 4 \left(3+2 i \sqrt{3}\right) d-i&=&0 \\ 4 \sqrt{3} b+1&=&0 \\ 4 \sqrt{3} a+1&=&0 \\ -8 i \sqrt{3} c+12 c+i&=&0 \\ \end{array} \right. $$

we obtain

$$ \left\{ \begin{array}{rcl} a&=&-\frac{1}{4 \sqrt{3}} \\ b&=&-\frac{1}{4 \sqrt{3}} \\ c&=&\frac{1}{336} \left(-12 i+8 \sqrt{3}\right) \\ d&=&\frac{1}{336} \left(12 i+8 \sqrt{3}\right) \\ \end{array} \right. $$

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  • $\begingroup$ @polbos Here $x^p_n$ represents a particular solution to the complete recurrence equation. $\endgroup$ – Cesareo Oct 8 '18 at 16:18

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