1
$\begingroup$

I would need some guidance in the below. I am more or less confident about what I am doing but I still have a couple of questions that I cannot confidently answer.

Consider a sequence $x_n$ of real numbers for $n=0,1,2,\dots$ (causal sequence). In my problem $x_n$ are defined by a non-linear recursion equation which can be Z-transformed yielding an exact solution for the transform $$ X(s)\equiv\sum_{n=0}^{\infty}x_ns^{-n}= \sqrt{\frac{s^2}{(s-e^{i\phi})(s-e^{-i\phi})}}, $$ where $s$ is a complex variable and $\phi$ is real. From this we obtain the terms of the sequence as $$ x_n=\oint_C\frac{ds}{2\pi i}s^{n-1} \sqrt{\frac{s^2}{(s-e^{i\phi})(s-e^{-i\phi})}},\qquad n=0,1,\dots, $$ where $C$ is a closed contour that (i) goes around the origin in a counterclockwise sense, (ii) is large enough to be fully contained in the region of convergence (ROC) of $X(s)$, and (iii) should contain all the poles of $X(s)$ (and perhaps all other non-analytic/singular points?).

Questions:

  • from complex integration theory and the definition/representation of $X$ as a power series in $1/s$ it is clear that $C$ should contain all the poles. But what if - as in my case - there are other singular points or cuts in the finite complex plane? Does the inverse Z-transform apply here too?
  • $X(s)$ has two branch points $s=e^{\pm i\phi}$ with $1/\sqrt{s}$ (integrable) singularity so the integral above is surely convergent for all $n$. (Note that $z=\infty$ is not a branch point, neither is $s=0$.) If I take the principal branch of the square root (with its cut on the negative real line) I get the correct results for $x_n$ (checked with Mathematica and numerical integration). But can I take a different branch of the square root?
  • Suppose I take the other commonly used branch of the square root which has a cut on the positive reals. Then the cut of $X(s)$ would extend to infinity (although $s=\infty$ is not a branch point) and hence no matter how $C$ is chosen it would not be fully contained in the region of analyticity. How can we resolve this?
  • Can in general $x_n$ depend on the choice of the branch of the square root? (It should not; the recursion gives unique solution...)

Many thanks

$\endgroup$
0
$\begingroup$

$\frac{s^2}{(s-e^{i \phi})(s-e^{-i \phi})}$ has two zeros/poles of odd order : at $s = e^{\pm i \phi}$. Thus $X(s) = (\frac{s^2}{(s-e^{i \phi})(s-e^{-i \phi})})^{1/2}$ has two branch points at $s = e^{\pm i \phi}$. If you put a branch cut on the line segment $[e^{i \phi} , e^{-i\phi}]$ then it is analytic on $\mathbb{C} \setminus [e^{i \phi}, e^{-i\phi}]$. Therefore, is has a Laurent series $$X(s) = \sum_{n=\color{red}{-\infty}}^\infty x_n s^n \quad ( |s| > 1), \qquad x_n=\frac{1}{n! 2\pi i }\int_{|s| = r} \frac{X(s)}{s^{n+1}}ds\quad (r > 1)$$

Also, since $X(1/s)$ is analytic at $s=0$, then $x_n = 0$ for $n > 0$, and hence $$X(s) = \sum_{n=-\infty}^0 x_n s^n \quad ( |s| > 1)$$

Finally for $Re(s) > 0,\ \displaystyle\int_0^\infty t^{n-1} e^{-st}dt = s^{-n} (n-1)!$ which means that $$X(s)-x_0 =\int_0^\infty f(t) e^{-st}dt \quad ( Re(s) > 1), \qquad f(t) = \sum_{n=1}^\infty \frac{x_n}{(n-1)!} t^{n-1}$$ where the series for $f(t)$ converges because..

$\endgroup$
  • $\begingroup$ I agree with what you write (except that the $n!$ should not be there in your formula for $x_n$) but where are you trying to get at? $\endgroup$ – Andras Vanyolos Apr 21 '17 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.