What is this change of variable in a polynomial?

Question

The autocovariance generating function, $\gamma(z)$, is defined as the $z$-transform of the autocovariance function, $\gamma_\tau$: $$ \gamma(z) = \gamma_0 + \gamma_1(z+z^{-1}) + \gamma_2(z^2+z^{-2}) + \gamma_3(z^3+z^{-3}) + \cdots = \displaystyle\sum_{\tau=-\infty}^\infty \gamma_\tau z^\tau \,, \qquad\quad (1) $$ where $z$ is a complex variable.

I am studying the code in this software package [1] and have found a function that performs the following operation on the equation given above for $\gamma(z)$. According to the documentation:

This function transforms a polynomial in the variables S(n)=z^n + z^(-n) into a polynomial in the variable U=z + z^(-1).

To do so, the following recursions are implemented in the code: $$ \begin{eqnarray} \begin{array}{ll} S(0) = 2, \quad S(1) = 1, \quad \hbox{initialisation} \\ S(n) = U \times S(n-1) - S(n-2), \quad n >= 2 \end{array} \end{eqnarray} $$ where $S(n)$ is the input polynomial of the form: $$ c(S) = c_1 + c_2 S(1) + \dots + c_{n-1} S(n-2) + c_n S(n-1) $$
and the output contains the coefficients of the polynomial: $$ p(U) = p_1 U(n-1) + p_2 U(n-2) + \dots + p_{n-1}U + p_n \,. $$

What may be the purpose of this transformation? For what purposes can this transformation be convenient?

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Is it equivalent to work with the original polynomial or the other? One of the reasons why I don't understand this transformation is the following. If $z$ is replaced in equation (1) by $e^{-iw}$, with $\omega \in[0, \pi]$, then the Fourier transform of the autocovariances $\gamma_\tau$ is obtained (multiplied by $2\pi$). I think there should be a mapping for the variable $U$ that also returns the Fourier transform of $\gamma_\tau$, but I don't know which values should $U$ take in order to get it.

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Hints: Maybe the algorithms employed in the code require the polynomials in that format, so here is some context:

The transformation of the polynomial is performed before running an algorithm that computes the greatest common divisor of polynomials and partial fraction decomposition of a polynomial.

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The code is available in the file SSMMATLAB/sn2u.m of the source code.

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[1] SSMMATLAB, a Set of MATLAB Programs for the Statistical Analysis of State Space Models by Víctor Gómez.

Answer 1

I have received valuable feedback from other sources and have come up with an understanding of the transformation. For $e^{-iw}$ we have that the reciprocal $1/z$ is the same as the conjugate of $z$, so that $(e^{-iw})^n + (e^{iw})^{-n} = \cos(nw) - i\sin(nw) + \cos(nw) + i\sin(nw) = 2\cos(n w)$.

Thus, for a given frequency $w$ the original polynomial is defined as: $$ A(z) = a_0 + a_1 (2\cos(w)) + a_2 (2\cos(2w)) + a_3 (2\cos(3w)) + ... $$ The function sn2u returns the coefficients $b_i$ of the following polynomial: $$ B(u) = b_0 + b_1 (2cos(w)) + b_2 (2cos(w))^2 + b_3 (2cos(w))^3 + ... $$ rewriting $x = 2cos(w)$: $$ B(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ... $$

The last representation is the usual expression of a polynomial. This makes the quotient of these kinds of polynomials liable to be decomposed according to standard techniques for partial fraction decomposition.

Answer 2

If $f(z)$ is an analytic function, then

$$f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}2(z-z_0)^2+\dots$$

Set $z_0=0:$

$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n$$

Set $z\to\frac1z:$

$$f(1/z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^{-n}$$

Add the two together and you get

$$f(z)+f(1/z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(z^n+z^{-n})$$

So that $\gamma(z)=f(z)+f(1/z)$ and $\gamma_\tau=\frac{f^{(\tau)}(0)}{\tau!}$. One can easily calculate $\gamma(z)$ from $f(z)$ and the series expansion, or given the series expansion, calculate $\gamma(z)$ from $f(z)$.