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Solve the following homogeneous recurrence relation \begin{equation} c_{n} - 2c_{n-1} - 5c_{n-2} - c_{n-3} = 0 \quad\quad n \geq 3 \end{equation} with initial conditions $c_{0} = 0, c_{1} = 1$ and $c_{2} = 14$.

I intended to solve this the usual way; by finding the characteristic polynomial, an expression for $c_{n}$, and the initial conditions. But, it seems that the characteristic polynomial has no roots in $\mathbb{Z}$.

Does anybody have any suggestions on how I should proceed?

RobPratt
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JLGL
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    Do you mean $c_n=2c_{n-1}-5c_{n-2}-c_{n-3}$? – Comma May 24 '24 at 13:47
  • Sadly no, this is a problem in my syllabus and it seems to have some formulation mistake. I am asking whether this is indeed a mistake, or if I'm missing something regarding how to approach it. – JLGL May 24 '24 at 13:51
  • As it stands, the expression isn't a recurrence relation, it's just some value. Surely there must be some equality somewhere? Under this assumption I do agree that the polynomial you get is somewhat nasty and has some fairly bad roots (though that is fine, roots do not need to be in Z. – Comma May 24 '24 at 13:53
  • @Comma My mistake, I did make a typo and have edited the post now. But my question still stands. – JLGL May 24 '24 at 13:59

3 Answers3

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For $x^3-2x^2-5x-1$, the discriminant is $361 = 19^2,$ and the roots are \begin{align*} \small r_1 = 1 + 2 \cos \left( \frac{8 \pi}{19} \right) + 2 \cos \left( \frac{18 \pi}{19} \right) + 2 \cos \left( \frac{26 \pi}{19} \right) \approx -1.285142481829 \\ \small r_2 = 1 + 2 \cos \left( \frac{2 \pi}{19} \right) + 2 \cos \left( \frac{14 \pi}{19} \right) + 2 \cos \left( \frac{16 \pi}{19} \right) \approx -0.221876162263 \\ \small r_3 = 1 + 2 \cos \left( \frac{4 \pi}{19} \right) + 2 \cos \left( \frac{28 \pi}{19} \right) + 2 \cos \left( \frac{32 \pi}{19} \right) \approx 3.5070186440929 \\ \end{align*}

The coefficients of the linear recurrence should be equal to certain rational polynomials in $r_1, r_2, r_3,$ these still to be found.

I got interested in this material from some questions by Tito, On the trigonometric roots of a cubic A standardized version of this cubic (replace $x$ by $x+1$) is in Reuschle page 26 A modern presentation of Gauss's method is in Cox Galois Theory, very nice but he does just a few examples ( which is where Reuschle comes in).

Will Jagy
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The characteristic polynomial approach does indeed work. With polynomial $x^3-2x^2-5x-1$, you have roots \begin{align*} r_1 &\approx -1.28514248182979 \\ r_2 &\approx -0.221876162263191 \\ r_3 &\approx 3.50701864409298 \end{align*} Then we expect a solution of the form $Ar_1^n+Br_2^n+Cr_3^n$, so by subbing in our initial values, we get a system of 3 equations that can be represented by the matrix equation $$ \left[ {\begin{array}{ccc} 1 & 1 & 1\\ r_1 & r_2 & r_3\\ r_1^2 & r_2^2 & r_3^2\\ \end{array} } \right] \left[ {\begin{array}{c} A\\ B\\ C\\ \end{array} } \right] = \left[ {\begin{array}{c} 0\\ 1\\ 14\\ \end{array} } \right] $$ Multiply by the inverse of the matrix gives us values for $A,B,C$. \begin{align*} A &\approx 2.10287 \\ B &\approx -2.97066 \\ C &\approx 0.86779 \end{align*} Putting it all together we get $$ c_n \approx 2.10287*(-1.28514248182979)^n + -2.97066 * (-0.221876162263191)^n + 0.86779 * 3.50701864409298^n $$

Comma
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We can determine $(c_n)_{n\geq 0}$ by deriving a generating function \begin{align*} C(z)=\sum_{n=0}^{\infty}c_nz^n \end{align*}

$C(n)$ is a rational function and we can derive the denominator of $C(z)$ directly from the recurrence relation \begin{align*} c_{n}& - 2c_{n-1} - 5c_{n-2} - c_{n-3} = 0 \quad\quad n \geq 3\\ c_0&=0, c_1=1, c_2=14\tag{1} \end{align*} as \begin{align*} \color{blue}{C(z)=\frac{a_0+a_1z+a_2z^2}{1-2z-5z^2-z^3}} \tag{2} \end{align*} A theorem asserting this relationship is stated here. We calculate the numerator of $C(z)$ by multiplication of (2) with the denominator. We obtain by comparing coefficients: \begin{align*} C(z)&\left(1-2z-5z^2-z^3\right)=a_0+a_1z+a_2z^2\\ a_0&=c_0\color{blue}{=0}\\ a_1&=-2c_0+c_1\color{blue}{=1}\\ a_2&=-5c_0-2c^1+c_2=-2+14\color{blue}{=12}\\ \end{align*} and the generating function $C(z)$ is

\begin{align*} \color{blue}{C(z)}&\color{blue}{=\frac{z+12z^2}{1-2z-5z^2-z^3}}\tag{3}\\ &\color{blue}{=z+14z^2+33z^3+137z^4+453z^5+\cdots} \end{align*} where the last line was calculated with some help of Wolfram Alpha.

Coefficients of $C(z)$: We calculate the coefficients of $C(z)$. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.

We obtain for $n>2$: \begin{align*} \color{blue}{c_n}&=[z^n]C(z)\\ &=\left([z^{n-1}]+12[z^{n-2}]\right)\sum_{q=0}^{\infty}z^q\left(2+5z+z^2\right)^q\tag{4}\\ &=\sum_{q=0}^{n-1}[z^{n-1-q}]\left(2+5z+z^2\right)^q\\ &\qquad+12\sum_{q=0}^{n-2}[z^{n-2-q}]\left(2+5z+z^2\right)^q\tag{5}\\ &\color{blue}{=\sum_{q=0}^{n-1}\sum_{{b+2c=n-1-q}\atop{a+b+c=q}} \binom{n-1-q}{a,b,c}2^a5^b}\\ &\qquad\color{blue}{+12\sum_{q=0}^{n-2}\sum_{{b+2c=n-2-q}\atop{a+b+c=q}} \binom{n-2-q}{a,b,c}2^a5^b}\tag{6}\\ \end{align*}

Comment:

  • In (4) we perform a geometric series expansion of $C(z)$ and use $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

  • In (5) we use again the coefficient of operator rule as in (4).

  • In (6) we do an expansion using trinomial coefficients.

Calculation of $c_5$: We make a small plausibility check and calculate $c_5$. We obtain from (6) \begin{align*} \color{blue}{c_5}&=\left(2\binom{2}{1,0,1}+5^2\binom{2}{0,2,0} +2^2\cdot 5\binom{3}{2,1,0}+4^2\binom{4}{4,0,0}\right)\\ &\qquad\quad+12\left(\binom{1}{0,0,1}+2\cdot 5\binom{2}{1,1,0}+2^3\binom{3}{3,0,0}\right)\\ &=2\frac{2!}{1!0!1!}+25\frac{2!}{1!1!0!}+20\frac{3!}{2!1!0!}+16\frac{4!}{4!0!0!}\\ &\qquad\quad+12\left(\frac{1!}{0!0!1!}+10\frac{2!}{1!1!0!}+8\frac{3!}{3!0!0!}\right)\\ &=4+25+60+16+12\left(1+20+8\right)\\ &\,\,\color{blue}{=453} \end{align*} in accordance with the coefficient of $z^5$ in (3).

Markus Scheuer
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