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The Problem

I have two recursively defined polynomials (skip to the bottom for background and motivation if you care about that) that represent the numerator and denominator of a factor and I want to find the limit of that factor as n goes to infinity.

$$n_0 = d_0 = 1$$ $$n_n = d_{n-1}x - n_{n-1}$$ $$d_n = d_{n-1}(x-1)-n_{n-1}$$

I was able represent this recursive relationship as a matrix and use eigenvalue matrix decomposition to find a closed form for $d_n$ and $n_n$. They are not very pretty:

$$d_n = \frac{2^{-n-1}}{\sqrt{x-4}} \left(\left(\sqrt{x-4}+\sqrt{x}\right) \left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(\sqrt{x-4}-\sqrt{x}\right) \left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{2 \left(\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n\right)}{\sqrt{x}}\right)$$ $$ n_n = 2^{-n-1} \left(\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n}{\sqrt{\frac{x-4}{x}}}\right)$$

The first few terms of the ratio ($r_n = n_n/d_n$) are:

$$r_1 = \frac{x-1}{x-2}$$ $$r_2 = \frac{x^2-3 x+1}{x^2-4 x+3}$$ $$r_3 = \frac{x^3-5 x^2+6 x-1}{x^3-6 x^2+10 x-4}$$ $$r_4 = \frac{x^4-7 x^3+15 x^2-10 x+1}{x^4-8 x^3+21 x^2-20 x+5}$$

And so on.

All of these equations seem to have poles located solely at $0\le x\le 4$, moreover the denominator and numerator seem to never have imaginary roots, but I haven't proven that.

$x$ is assumed to be a positive real number, and because there are no poles $x > 4$ the limit seems to be well behaved and equal to 1(?) outside of this range. But within this range, I am extremely curious about what the limit is, if it even exists or is possible to evaluate.

Can someone let me know if it's possible to evaluate this limit in this range? And if so can you let me know how to go about it?

Background and Motivation

An ideal transmission line can be modeled as an inductor and capacitor, the inductor is in series with the load and the capacitor is in parallel.

The impedance of an inductor is given by $Z_L = i \omega L$ and the impedance of a capacitor is $Z_C = \frac{1}{i \omega C}$.

If we string several transmission lines together and end in an open circuit then we can start evaluating from the end using the impedance addition laws. We first add $Z_L + Z_C$ and then to combine with the second to last capacitor we must $\frac{1}{\frac{1}{Z_L+Z_C}+\frac{1}{Z_C}}$.

When we do that we find, interestingly that the ratio between $Z_C$ and this value is our real valued ratio $r_1$. We can then solve for $r_n$ by breaking up the ratio into a numerator and denominator and starting from the (n-1)th iteration, and find the relationship given at the top of the question, with $x = \omega ^2 L C$.

Because a series of short transmission lines strung together is the same thing as a long transmission line, I expected the limit of this ratio to be equal to 1. However, when attempting to evaluate this limit it seems that it is not that simple.

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  • $\begingroup$ It appears to me that your limit is $1$ if $x>1$ and $0$ if $0$ if $0\leq x<1$, whenever $x$ is not a pole. Can you check that (at least) numerically? One has to be careful at $x=1$, it can't be done by quick inspection. $\endgroup$ – T. Amdeberhan Sep 10 '16 at 19:58
  • $\begingroup$ Because there are an infinite number of poles as you take the limit, I tried some transcendental numbers. I was unable to coax Mathematica into giving me a limit as n->infinity with x = Pi/2 or Pi/4. Checking n in the neighborhood of 10^10 showed the value jumping around quite a bit. $\endgroup$ – OmnipotentEntity Sep 10 '16 at 20:26
  • $\begingroup$ Oddly, when I use a decimal value (eg, Pi*0.5) Mathematica shows the value as near 0, but when I use a fractional one (eg, Pi/2) it's jumpy. Wat. $\endgroup$ – OmnipotentEntity Sep 10 '16 at 20:30
  • $\begingroup$ The behavior I described above seems to be related to floating point error. $\endgroup$ – OmnipotentEntity Sep 10 '16 at 21:09
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Here is an explicit formula for your ratio $r_n=\frac{n_n}{d_n}$: $$r_n= \frac{\sum_{k=0}^n\binom{n+k}{2k}(-x)^k} {\sum_{k=0}^n\binom{n+k+1}{2k+1}(-x)^k}.$$ Let $P_n(x)$ and $Q_n(x)$ be the numerator and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

In fact (Fedor!), $$r_n=\frac{P_n(x)}{Q_n(x)}=\sqrt{x}\,\frac{U_{2n}(\sqrt{x}/2)}{U_{2n+1}(\sqrt{x}/2)}$$ where $U_n(y)$ are Chebyshev polynomials of the 2nd kind, expressible as $$U_n(y)=\frac{(y+\sqrt{y^2-1})^n-(y-\sqrt{y^2-1})^n}{2\sqrt{y^2-1}}.$$ If $y\geq1$, or equivalently $z=y+\sqrt{y^2-1}\geq1$, then $$\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}= \lim_{n\rightarrow\infty}\frac{z^n-z^{-n}}{z^{n+1}-z^{-n-1}}=\frac1z=y-\sqrt{y^2-1}.$$ If $0<y<1$ then the complex modulus $\vert z\vert=1$ and hence $\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}$ fails to exist.

If $y=0$ then apparently the limit is $0$.

When $y=\frac{\sqrt{x}}2$, the conditions become $$\lim_{n\rightarrow\infty}r_n(x)=\frac{x-\sqrt{x^2-4x}}2$$ if $x\geq4$ or $x=0$. Otherwise (if $0<x<4$) this limit does not exist for being oscillatory!

Finally, since the roots of Chebyshev polynomials $U_n(y)$ lie in $[-1,1]$ it follows that the roots of $U_n(\sqrt{x}/2)$ must be limited in the range $[0,4]$.

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  • $\begingroup$ That is a much nicer simplification as compared to my approach. How did you arrive at this? $\endgroup$ – OmnipotentEntity Sep 10 '16 at 20:53
  • $\begingroup$ Similar to "decoupling" of system of ODEs, whenever possible. $\endgroup$ – T. Amdeberhan Sep 10 '16 at 21:19
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    $\begingroup$ Coefficients look very much like these of Chebyshev polynomials. But well, the explicit formulae also. And they should help to find the limit, why they do not? $\endgroup$ – Fedor Petrov Sep 10 '16 at 23:13
  • $\begingroup$ Thank you so much! I suppose the reason why this works for real transmission lines is the resistance dampens oscillations. $\endgroup$ – OmnipotentEntity Sep 11 '16 at 1:39
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The explanation for your observations is that you are dealing with a self-adjoint difference equation in disguise. Let me make a transformation along these line, though I won't analyze it through to the end.

If we let $Y_n=(n_n,d_n)^t$, then this satisfies $$ Y_n = \begin{pmatrix} -1 & x \\ -1 & x-1 \end{pmatrix} Y_{n-1} . $$ Now write $Y_n=A^n W_n$, with $A=\begin{pmatrix} -1 & 0\\ -1&-1\end{pmatrix}$, so $$ A^n = (-1)^n \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} $$ (this is variation of constants, relative to $x=0$). By a calculation, $$ J(W_n-W_{n-1}) = -xH_n W_n, \quad\quad H_n=-JA^{-n}\begin{pmatrix} 0&1\\ 0&1\end{pmatrix}A^{n-1},\quad J=\begin{pmatrix} 0&-1\\ 1&0\end{pmatrix}. $$ An equation of this form is called a (discrete) canonical system, and it corresponds to a symmetric difference expression in the space $\ell^2_H$ if $H_n$ is a positive definite matrix.

Our luck holds here because a calculation gives that $$ H_n = \begin{pmatrix} (n-1)^2 & n-1 \\ n-1 & 1 \end{pmatrix} \ge 0 , $$ as required.

This explains why the zeros are real, though it doesn't show why they are in $[0,4]$. For this, one would have to analyze the spectrum of this system more carefully (you could rewrite it as a Jacobi difference equation).

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