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I am (as here) working on Machin formula and with some works I'd like to explore the following polynomial recurrence relation :

$$P_0 = 1, P_1 =b,\quad P_{n+2} = XP_{n+1} - P_n$$

It seems relation to Chebyshev_polynomials but I am unfamiliar with these polynomials.

Is it possible to have an explicit expression for $P_n$?

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  • Use linear algebra: put it in a matrix and then diagonalize it. – Lolman Jul 27 '14 at 13:27
  • Is $b$ any number ? Chebyshev uses $x$ for $P_1$, and the nice thing is that the roots can be expressed in terms of $\cos \frac{(2k-1)\pi}{2n}$ – Matt B. Jul 27 '14 at 13:29
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    I'd expect that your $P_n$ is just a linear combination of the Chebyshev polynomials of the first and second kind. (One takes $X=2b$, the other $X=b$. So both together presumably allow any $X$.) – Semiclassical Jul 27 '14 at 13:31
  • Not clear what you mean by "allow any $X$." Certainly, knowing polynomials at two values in general doesn't let us know the polynomial. @Semiclassical – Thomas Andrews Jul 27 '14 at 13:59
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    In particular, it will be a "linear" combination of polynomials $T_n(X/2)$ and $U_n(X/2)$, but the coefficients will be rational functions of $X$, not real numbers. – Thomas Andrews Jul 27 '14 at 14:06
  • @ThomasAndrews: I was speaking rather too loosely I suppose. In any case, your answer quite captures the spirit of what I had in mind. – Semiclassical Jul 27 '14 at 14:36

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Let $a_n(X) = T_n(\frac{X}{2})$ and $b_n(X)= U_n(\frac{X}{2})$, where $T_n,U_n$ are the Chebyshev polynomials. Then satisfy our recurrence $a_{n+2}(X)=xa_{n+1}(X)-a_n(X)$ and the same for $b_n(X)$.

Now find rational expressions in $X$, $p(X),q(X)$ so that:

$$p(X)a_0(X)+q(X)a_1(X)=p(X)+q(X)=1\\p(X)a_1(X)+q(X)b_1(X)=p(X)\frac{X}{2}+q(X)X = b$$

The second can be rewritten as:

$$p(X) + 2q(X) =\frac{2b}{X}$$

So $q(X) =\frac{2b-X}{X}$ and $p(X)=\frac{2X-2b}{X}$.

So the sequence of rational functions:

$$Z_n(X) = \frac{2X-2b}{X} T_n\left(\frac{X}{2}\right) + \frac{2b-X}{X}U_n\left(\frac{X}{2}\right)$$

satisfy the recurrence, $Z_0(X) = 1$, and $Z_1(X)=b$. So the $Z_n(X)$ are actually polynomials, and $Z_n(X)=P_n(X)$.

So $P_n(X)=Z_n(X)$, and you are done.

The more general solution would be to use the general solution to linear recurrences to write:

$$P_n(X) = c_1(X)\left(\frac{X+\sqrt{X^2-4}}{2}\right)^n + c_2(X)\left(\frac{X-\sqrt{X^2-4}}{2}\right)^n$$

and solving for $c_1(X)$ and $c_2(X)$ to solve the initial condition $P_0(X)=1$ and $P_1(X)=b$. This will yield $c_1(X),c_2(X)$ that are not even rational functions of $X$, but rational functions of $X$ and $\sqrt{X^2-4}$, but the end expression will represent a polynomial.

Thomas Andrews
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