2
$\begingroup$

Let $p_n(z)$ be the $n^\text{th}$ partial sum of the Maclaurin series for $f(z) = (1+z)/(1-z)$. For large $n$ the zeros of $p_n$ appear to avoid the point $z=-1$:

enter image description here

Figure: Zeros of $p_{40}$ and the unit circle.

To calculate the asymptotic behavior of zeros like this one can sometimes calculate a scaling limit of the polynomials. For example, consider instead

$$ g(z) = \frac{a+z}{1-z}. $$

The partial sums of $g$, which we'll call $q_n$, are given by

$$ q_n(z) = a + (a+1) z\cdot \frac{1-z^n}{1-z}. $$

Using this expression we can obtain the following formulas:

$$ \lim_{\substack{n \to \infty \\ n\text{ even}}} q_n\!\left(-1+\frac{w}{n}\right) = \frac{a-1}{2} + \frac{a+1}{2} e^{-w}, \tag{1} $$

$$ \lim_{\substack{n \to \infty \\ n\text{ odd}}} q_n\!\left(-1+\frac{w}{n}\right) = \frac{a-1}{2} - \frac{a+1}{2} e^{-w}. \tag{2} $$

So when $a \neq \pm 1$ we can deduce, for example, that the limit points of the zeros of $q_n(-1+w/n)$ for $n$ odd are

$$ w_k := \log\left|\frac{a+1}{a-1}\right| + i\left(\arg \frac{a+1}{a-1} + 2\pi k\right), \qquad k \in \mathbb Z, $$

and hence that the zeros $z_n$ of $q_n(z)$ for $n$ odd which tend to $z=-1$ have the form

$$ z_n = -1 + \frac{w_k}{n} + o(n^{-1}), \qquad k \in \mathbb Z. $$

However, if $a=1$ then these scaling limits $(1)$ and $(2)$ have no zeros, which tells us that the scale $1/n$ is incorrect, and that we should choose another scaling which is $\gg 1/n$. But, carrying out the calculations, it doesn't seem possible to get a limit function with zeros at all.

Is there another approach we could use to find the asymptotics of these zeros near $z=-1$?

$\endgroup$
1
  • 5
    $\begingroup$ The roots of $p_n(z)$ lie on the curve given in polar coordinates by $\cos\varphi=(4r^{2n+2}-r^2-1)/2r$, the intersection of this curve with the negative part of real axis is approximately $-1+\ln(n)/n$. Hope that helps. $\endgroup$ Nov 5 '15 at 2:02
2
$\begingroup$

The partial sums are $p_n=1+2z+2z^2+...+2z^{n}=-1+2\frac{1-z^{n+1}}{1-z}$ so $p_n(-1)=(-1)^n$ and it is no surprise that the roots are pushed away from $z=-1$.

I haven't checked numerically how well it works, but one could

  1. Estimate values $(-1+\varepsilon)^m$ as $(-1)^m(1-m\varepsilon)$
  2. Use that to estimate the location and value of the local minimum of $p_n(z)$ on the real line near $z=-1$.
  3. Try to go from there to the nearby pair of complex conjugate roots (with about that real part?).

    Or perhaps a quadratic approximation would be worth the extra work.

$\endgroup$
1
$\begingroup$

I was hoping to apply the general theory here about sequences of polynomials satisfying linear recurrences.

The partial sums $p_n(z)$ satisfy the linear recurrence $p_{n} = z p_{n-1} + p_{n-2} - z p_{n-3}=0$.

We construct the symbol of this recurrence, $t^3 - zt^2 + t-z = (t-1)(t+1)(t-z)$.

However, we cannot apply the result in the reference, since two of our roots differ by a constant multiple of a complex number on the unit circle. This is another indication that your example is degenerate in some sense.

$\endgroup$

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.