Problem Sequence:

Calculating a scaling limit for partial sums of the exponential series

On this page are 21 problems. By solving these problems, the reader will be guided through a fairly technical calculation involving the partial sums of the exponential series.

Starting from scratch, we will derive a scaling limit which gives precise information about the locations of the zeros of the partial sums. Along the way the reader will perform a complete application of the saddle point method and derive the equation of the famous Szegő curve.

Our main result here originally appeared (without proof) as Lemma 2 in my last post. The purpose of this problem sequence is to outline its proof.

A blue superscript

1 Definitions

2 The saddle point method

3 Choosing the right contour

4 Applying the method

4.1 Some notation

4.2 Estimating the parts of the integral

5 The zeros of $p_{n-1}(nz)$ and the Szegő curve

6 How quickly do the zeros approach the Szegő curve?

7 The scaling limit

Appendix: Using the scaling limit (Why uniformity?)

Appendix: Further reading

1 Definitions

Let

\[ p_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}. \tag{1.1} \]

This polynomial $p_n(z)$ is commonly referred to as the nProblem 1.
Let $\gamma$ be a simple closed loop around the origin with
positive orientation. Show that \[ \frac{e^z -
p_{n-1}(z)}{z^n} = \frac{1}{2\pi i} \int_\gamma \zeta^{-n}
e^\zeta \,\frac{d\zeta}{\zeta-z} \] for all nonzero $z$ inside
$\gamma$. Then make a change of variables to show that
^{1}

Hint: Show that \[ \int_{\gamma} \zeta^{-m} \,\frac{d\zeta}{\zeta - z} = 0 \] for all integers $m \geq 1$.

\[ \frac{e^{nz} - p_{n-1}(nz)}{(ez)^n} =
\frac{1}{2\pi i} \int_\gamma (es)^{-n} e^{ns}
\,\frac{ds}{s-z} \tag{1.2} \]

for all nonzero $z$ inside $\gamma$.Hint: Show that \[ \int_{\gamma} \zeta^{-m} \,\frac{d\zeta}{\zeta - z} = 0 \] for all integers $m \geq 1$.

The precise nature of the contour $\gamma$ will be determined in Section 3.

Define $\varphi(s) = s-1-\log s$, taking the principal branch of the natural logarithm, so that \[ (es)^{-n} e^{ns} = \exp\{n(s-1-\log s)\} = \exp\{n\varphi(s)\}. \]

2 The saddle point method

The saddle point method (also known as the method of steepest descent)

somethinghere is an integral, and its size will be roughly equal to the integral over neighborhoods of the maxima of its integrand. Of course we are interested in contour integrals in the complex plane, so we need to define what we mean by

maximain this context.

Suppose we have an integral of the form \[ F(n) = \int_\gamma \exp\{n f(s)\}\,ds = \int_\gamma \exp\{n \operatorname{Re} f(s)\} \exp\{in \operatorname{Im} f(s)\}\,ds, \qquad n > 0. \] The size of the integrand is \[ \lvert \exp\{n f(s)\} \rvert = \exp\{n \operatorname{Re} f(s)\}, \] and this is largest where $\operatorname{Re} f(s)$ is largest. However, because $F(n)$ is an integral of a complex-valued function, if we integrate over a generic contour $\gamma$ then the final size of the integral may be sensitive to cancellation or resonance coming from the oscillatory factor $\exp\{in \operatorname{Im} f(s)\}$. The main plan, then, is to deform the contour in such a way that the integral remains unchanged and that these oscillations are easy to handle.

Problem 2.
Let $f$ be a function which is holomorphic at some point $z_0$
with $f'(z_0) = 0$ and $f''(z_0) \neq 0$.

First show that $\operatorname{Re} f$ has a saddle point at $z_0$.

Then show that there are two smooth curves $\Gamma_1$ and $\Gamma_2$ passing through $z_0$ such that

In other words, the two curves passing through a simple critical point $z_0$ of $f$ on which $\operatorname{Im} f$ is constant are the steepest ascent and descent paths through the saddle at $z_0$ on the graph of $\operatorname{Re} f$.

Note: This is Exercise 2 in Chapter 8 of Stein & Shakarchi's Complex Analysis.

First show that $\operatorname{Re} f$ has a saddle point at $z_0$.

Then show that there are two smooth curves $\Gamma_1$ and $\Gamma_2$ passing through $z_0$ such that

- $\Gamma_1$ and $\Gamma_2$ are orthogonal at $z_0$,
- $\operatorname{Im} f$ is constant on $\Gamma_1$ and $\Gamma_2$, and
- $\operatorname{Re} f$ has a maximum at $z_0$ when
restricted to $\Gamma_1$ and a minimum at $z_0$ when
restricted to $\Gamma_2$.

In other words, the two curves passing through a simple critical point $z_0$ of $f$ on which $\operatorname{Im} f$ is constant are the steepest ascent and descent paths through the saddle at $z_0$ on the graph of $\operatorname{Re} f$.

Note: This is Exercise 2 in Chapter 8 of Stein & Shakarchi's Complex Analysis.

Definition 1.
The *axis* of a saddle point $z_0$ of a function $f$ is defined to be the
straight line in the $z$-plane defined by \[
f''(z_0)(z-z_0)^2 \leq 0. \] One can check that the
steepest descent path $\Gamma_1$ from Problem 2 is tangent to the axis of $z_0$
at $z_0$.

Figure 1. Two perspectives of the saddle point at $z=1$ of the function $\varphi(z)$ defined at the end of Section 1. Over the surface $\operatorname{Re} \varphi(z)$ we see the

So, if we can deform $\gamma$ so that it passes through a critical point of $f$ along a path of steepest descent on the surface $\operatorname{Re} f$, then $\operatorname{Im} f$ will be constant on $\gamma$, say $\operatorname{Im} f =: I_0$, and we will have \[ F(n) = \int_\gamma \exp\{n \operatorname{Re} f(s)\} \exp\{in I_0\}\,ds = \exp\{in I_0\} \int_\gamma \exp\{n \operatorname{Re} f(s)\} \,ds. \] What remains is a real integral. The new integrand $\exp\{n \operatorname{Re} f(s)\}$ will have a maximum at the critical point, and we can apply the basic principle discussed above: the integral will be roughly equal to the integral over a neighborhood of the maximum of its integrand.

These are the broad strokes of the method we will apply to this calculation.

3 Choosing the right contour

In this section we will perform a series of calculations to determine which qualities we need the contour $\gamma$ to have.

The contour integral formula $(1.2)$ we derived in Problem 1 is only valid for $z$ inside $\gamma$, so we need to make sure that $\gamma$ is large enough so that it encompasses all zeros of $p_{n-1}(nz)$. To determine a rough bound for the zeros we will use the following classical result.

Eneström-Kakeya
Theorem. If $q(z) = \sum_{k=0}^{m} a_k z^k$ is a
polynomial with \[ 0 \leq a_0 \leq a_1 \leq \cdots \leq a_m,
\] then all zeros of $q$ lie in the closed disk $|z| \leq 1$.

Problem 3.
Use the Eneström-Kakeya theorem to show that all zeros of
$p_{n-1}(nz)$ lie in the open disk $|z| < 1$.

**Bonus:** Prove the Eneström-Kakeya theorem.

Figure 2. The 29 zeros of the polynomial $p_{29}(30z)$ are shown as white dots. As predicted by Problem 3, these zeros lie inside the unit circle (shown in blue).

From this calculation, we know that if we take $\gamma$ to lie on or outside the unit circle, it will contain all zeros of $p_{n-1}(nz)$.

Problem 4.
Show that the only critical point of $\varphi(s)$ is located
at $s=1$ and that \[ \varphi(s) = \tfrac{1}{2} (s-1)^2 +
\text{higher order terms} \] near there.

Further, show that the axis of this critical point is the vertical line $\operatorname{Re} s = 1$.

Further, show that the axis of this critical point is the vertical line $\operatorname{Re} s = 1$.

At the critical point we have $\operatorname{Re} \varphi(1) = 0$. Based on the discussion in the previous section, we want this to be the highest point of $\gamma$ on the graph of $\operatorname{Re} \varphi(s)$.

If we travel away from $s=1$ upwards along the path of steepest descent, we will end up in the region $\operatorname{Re} \varphi(s) < 0$. We should then choose the contour so that it remains in this region $\operatorname{Re} \varphi(s) < 0$, winds counterclockwise around the origin, then returns to the critical point. But is this possible?

Problem 5.
Show that if $\operatorname{Re} \varphi(s) = 0$ then either
$s=1$, $\operatorname{Re} s > 1$, or $|s| < 1$.

So, for example, the circle $|s| = 1$ passes through the critical point of $\varphi$ tangent to the path of steepest descent and lies entirely in the region $\operatorname{Re} \varphi < 0$ except at the point $s=1$.

Problems 3, 4, and 5 motivate the following choice for our contour $\gamma$:

The contour
$\gamma$. Let $\gamma$ be a simple closed loop around
the origin which passes through the point $s=1$ and lies
entirely in the regions $|s| > 1$ and $\operatorname{Re}
\varphi(s) < 0$ except at $s=1$. In a small neighborhood
$U_\gamma$ of $s=1$, let $\gamma$ coincide with the path of
steepest descent through $s=1$ on the graph of
$\operatorname{Re} \varphi(s)$.

Figure 3. A plot of the contour $\gamma$ in white. Note how it avoids the region $\operatorname{Re} \varphi(s) > 0$ (shown in red) except at the saddle point $s=1$. In the region $U_\gamma$ (shown in yellow), the contour coincides with the steepest descent path through the saddle (shown in blue).

From Problem 2 we know that $\operatorname{Im} \varphi(s)$ is constant on $\gamma \cap U_\gamma$. In fact, $\operatorname{Im} \varphi(s) = 0$ there.

These conditions ensure that all zeros of $p_{n-1}(nz)$ lie inside the contour, that $s=1$ is the highest point of the contour on the surface $\operatorname{Re} \varphi(s)$, and that $\gamma$ coincides with the path of steepest descent near that point.

4 Applying the method

In this section we will work through the technical details of the saddle point method. We will carefully break the integral in $(1.2)$ into pieces then estimate the size of each piece. After whittling away the insignificant parts we will be left with a (relatively) simple expression representing the integral's dominant behavior as $n \to \infty$.

4.1 Some notation

In analysis we often only need a rough idea of a thing's size. The point of the notation we'll introduce in this section is to hide irrelevant details when making estimates and to emphasize ballpark comparisons.

The first piece of notation is

big-O notation, written $O(\cdots)$. If $a_n(z)$ and $b_n(z)$ are sequences which depend on a complex parameter $z$, $d_n$ is another sequence which doesn't, and $A$ is some subset of the complex plane, the statement

\[ ``\,\, a_n(z) = b_n(z) + O(d_n) \quad \text{as } n \to \infty \text{ uniformly for } z \in A \," \tag{4.1} \]

means that there are positive constants $C$ and $M$ which do not depend on $z$ such that the inequality \[ \lvert a_n(z) - b_n(z) \rvert \leq C\lvert d_n \rvert \tag{4.2} \]

holds for all $n \geq M$ and all $z \in A$.Qualitatively, writing $a_n(z) = b_n(z) + O(d_n)$ means that the error in estimating $a_n(z)$ by $b_n(z)$ is at most something like $d_n$.

Notice that the constants $C$ and $M$ are not specified by the statement in $(4.1)$. This is a

It's also sometimes useful to consider $O(\cdots)$ as the set all of sequences which are bounded by a constant multiple of $\lvert \cdots \rvert$. We can then compare these sets using $\subset$, $=$, etc. For example, $O(n^{-2}) \subset O(n^{-1})$ as $n \to \infty$. If $a_n$, $b_n$, and $d_n$ are sequences of complex numbers, then the statement $O(b_n) \subset O(d_n)$ is equivalent to \[ a_n = O(b_n) \quad \text{as } n \to \infty \quad \Longrightarrow \quad a_n = O(d_n) \quad \text{as } n \to \infty, \] where both

equalitiesare meant to be interpreted as in $(4.1)$ and $(4.2)$.

The last piece of notation we'll introduce is the $\sim$ notation. The statement \[ ``\,\, a_n(z) \sim b_n(z) \quad \text{as } n \to \infty \text{ uniformly for } z \in A \," \] means that \[ \lim_{n \to \infty} \frac{a_n(z)}{b_n(z)} = 1 \] uniformly for $z \in A$. Here we say that $a_n(z)$ is

4.2 Estimating the parts of the integral

From our discussion in the previous sections, we expect that the main contribution to our integral $(1.2)$ comes from a neighborhood of the point $s=1$. So we break our integral in two,

\[ \int_\gamma = \int_{\gamma \text{ near } 1} +
\int_{\gamma \text{ away from } 1}, \tag{4.3} \]

and expect that the integral over the part of $\gamma$ away from
$s=1$ is much smaller than the part near $s=1$.Problem 6.
Show that there are open sets $U$ and $V$ such that $1 \in U$,
$0 \in V$, and a biholomorphic map $\psi : V \to U$ which
satisfies \[ (\varphi \circ \psi)(x) = x^2 \] for $x \in V$
with $\psi'(0) = \sqrt{2}$.

Note that $\psi(0) = 1$ and that $U$ and $V$ can be made arbitrarily small, so that we may take $U \subset U_\gamma$ (the set $U_\gamma$ is defined above). In fact, we can take $V$ to be an open disk.

Note that $\psi(0) = 1$ and that $U$ and $V$ can be made arbitrarily small, so that we may take $U \subset U_\gamma$ (the set $U_\gamma$ is defined above). In fact, we can take $V$ to be an open disk.

This gives us a neighborhood $U$ of $s=1$ which we can use to make $(4.3)$ precise: let us split the integral in $(1.2)$ as \[ \int_\gamma = \int_{\gamma \cap U} + \int_{\gamma \setminus U}. \] The first thing we do is show that the integral over $\gamma \setminus U$ is exponentially small.

Definition
2. For $\epsilon > 0$, let $N_\epsilon$ be the set of
all points within a distance of $\epsilon$ of the curve
$\gamma$.

Problem 7.
Show that there is a constant $C_1 > 0$ which does not
depend on $z$ such that \[ \left\lvert \int_{\gamma \setminus
U} e^{n\varphi(s)} \,\frac{ds}{s-z} \right\rvert < e^{-C_1
n} \] for all $z \notin N_\epsilon$.

In big-O notation, we have shown that

\[ \int_{\gamma} e^{n\varphi(s)} \,\frac{ds}{s-z} =
\int_{\gamma \cap U} e^{n\varphi(s)} \,\frac{ds}{s-z} +
O\!\left(e^{-C_1 n}\right) \tag{4.4} \]

uniformly for $z \notin N_\epsilon$.Now we turn our attention to the remaining integral over $\gamma \cap U$.

Problem 8.
Show that \[ \int_{\gamma \cap U} e^{n\varphi(s)}
\,\frac{ds}{s-z} = \int_{-\alpha_1}^{\alpha_2} e^{-nt^2}
\frac{i\psi'(it)}{\psi(it)-z}\,dt \] for some positive numbers
$\alpha_1$ and $\alpha_2$.

After this change of variables, the largest contribution to the integral comes from a neighborhood of $t=0$ since the exponential factor eliminates everything else as $n \to \infty$. We thus estimate the subexponential factor at this point.

Problem 9.
Define \[ g_z(t) = \frac{i\psi'(it)}{\psi(it)-z}. \] Show
that, for $z \notin N_\epsilon$, there is a positive constant
$M$ that does not depend on $z$ such that \[ |g_z(t) - g_z(0)|
\leq M|t| \] for $t \in V$.

To apply this approximation, write \[ \begin{align} \int_{-\alpha_1}^{\alpha_2} e^{-nt^2} \frac{i\psi'(it)}{\psi(it)-z}\,dt &= \int_{-\alpha_1}^{\alpha_2} e^{-nt^2} g_z(t)\,dt \\ &= g_z(0)\int_{-\alpha_1}^{\alpha_2} e^{-nt^2}\,dt + \int_{-\alpha_1}^{\alpha_2} e^{-nt^2} (g_z(t) - g_z(0))\,dt. \end{align} \]

Problem 10.
Show that \[ \left\lvert \int_{-\alpha_1}^{\alpha_2} e^{-nt^2}
(g_z(t) - g_z(0))\,dt \right\rvert < \frac{M}{n} \] for $z
\notin N_\epsilon$.

In big-O notation, we have shown that

\[ \int_{-\alpha_1}^{\alpha_2} e^{-nt^2}
\frac{i\psi'(it)}{\psi(it)-z}\,dt =
g_z(0)\int_{-\alpha_1}^{\alpha_2} e^{-nt^2}\,dt +
O\!\left(\frac{1}{n}\right) \tag{4.5} \]

uniformly for $z \notin N_\epsilon$.Problem 11.
Show that there is a positive constant $C_2$ such that \[
\int_{-\alpha_1}^{\alpha_2} e^{-nt^2}\,dt = \sqrt\frac{\pi}{n}
+ O\!\left(e^{-C_2 n}\right), \] and hence that

\[ g_z(0)\int_{-\alpha_1}^{\alpha_2}
e^{-nt^2}\,dt = g_z(0) \sqrt\frac{\pi}{n} + O\!\left(e^{-C_2
n}\right) \tag{4.6} \]

for $z$ bounded away from the point $z=1$. All that remains is to reassemble the various parts.

By combining $(4.5)$ and $(4.6)$ we get the estimate \[ \int_{-\alpha_1}^{\alpha_2} e^{-nt^2} \frac{i\psi'(it)}{\psi(it)-z}\,dt = g_z(0) \sqrt\frac{\pi}{n} + O\!\left(\frac{1}{n}\right) \] uniformly for $z \notin N_\epsilon$ since $O(e^{-C_2 n}) \subset O(1/n)$. Substituting this in $(4.4)$ yields \[ \int_{\gamma} e^{n\varphi(s)} \,\frac{ds}{s-z} = \frac{i}{1-z} \sqrt\frac{2\pi}{n} + O\!\left(\frac{1}{n}\right) \] uniformly for $z \notin N_\epsilon$, where we have used the fact that \[ g_z(0) = \frac{i\sqrt{2}}{1-z}. \] Finally, substituting this in $(1.2)$ yields

\[ \frac{e^{nz} - p_{n-1}(nz)}{(ez)^n} =
\frac{1}{(1-z)\sqrt{2\pi n}} + O\!\left(\frac{1}{n}\right) \tag{4.7}
\]

uniformly for $z$ inside $\gamma$ with $z \neq 0$ and $z \notin
N_\epsilon$.This completes our application of the saddle point method.

5 The zeros of $p_{n-1}(nz)$ and the Szegő curve

Problem 12.

- Show that all zeros $p_{n-1}(nz)$ lie inside $\gamma$ with $z \neq 0$.
- For $\delta > 0$, show that there is an $\epsilon > 0$ such that all zeros satisfying $\lvert z-1 \rvert > \delta$ are not in $N_\epsilon$.

It follows from Problem 12 and equation $(4.7)$ that every zero of $p_{n-1}(nz)$ which is bounded away from $1$ satisfies the equation

\[ \left(z e^{1-z}\right)^{-n} =
\frac{1}{(1-z)\sqrt{2\pi n}} + O\!\left(\frac{1}{n}\right).
\tag{5.1} \]

Problem 13.
Show that, as $n \to \infty$, the zeros of $p_{n-1}(nz)$
converge to the portion of the curve $\left\lvert z e^{1-z}
\right\rvert = 1$ that lies in the disk $\lvert z \rvert \leq
1$.

The curve

\[ S := \left\{ z \in \mathbb C : \left\lvert z
e^{1-z} \right\rvert = 1 \text{ and } \lvert z \rvert \leq 1
\right\} \tag{5.2} \]

is called the Szegő curve.
It is named after Gabor Szegő, who first derived the curve in 1924.Figure 4. The Szegő curve $S$ is shown in red, along with the zeros of $p_{29}(30z)$ in white and the unit circle in blue. Compare with Figure 2.

Note that the condition $\left\lvert z e^{1-z} \right\rvert = 1$ can also be written as $\operatorname{Re} \varphi(z) = 0$.

6 How quickly do the zeros approach the Szegő curve?

Now that we know the global behavior of the zeros of $p_{n-1}(nz)$, we can begin to study their precise locations. In this section we will not try to be totally rigorous, instead focusing on exploratory analysis. Our findings here will allow us to guess the form of the final result. It is not strictly necessary to solve these problems in order to move on to the next section.

If we omit the $O(1/n)$ term, equation $(5.1)$ basically says that the zeros of $p_{n-1}(nz)$ which do not converge to the point $z=1$ satisfy

\[ \exp\{ n \varphi(z) \} \approx \frac{e^{-(\log
n)/2}}{(1-z)\sqrt{2\pi}}. \tag{6.1} \]

Let's assume that $z$ converges to some point $\xi \in S$ with $\xi
\neq 1$, and write $z = \xi + \delta$, where $\delta \to 0$ as $n
\to \infty$.Problem 14.
Show that $1-z \approx 1-\xi$ and that \[\varphi(z) \approx
\varphi(\xi) + \frac{\xi - 1}{\xi}\delta. \]

Our estimate in $(6.1)$ then becomes

\[ \exp\!\left\{ n\varphi(\xi) + \frac{\xi -
1}{\xi}n\delta \right\} \approx \frac{e^{-(\log
n)/2}}{(1-\xi)\sqrt{2\pi}}. \tag{6.2} \]

Problem 15.
Show that $\operatorname{Re} \varphi(\xi) = 0$ and hence that
$\exp\{ n\varphi(\xi)\}$ is bounded.

Deduce that, in order to balance the decay of the factor $e^{-(\log n)/2}$ on the right-hand side of $(6.2)$, we must have \[ \delta \approx \frac{\xi \log n}{2(1-\xi)n}. \]

Deduce that, in order to balance the decay of the factor $e^{-(\log n)/2}$ on the right-hand side of $(6.2)$, we must have \[ \delta \approx \frac{\xi \log n}{2(1-\xi)n}. \]

This estimate for $\delta$ tells us that the zeros in question approach the Szegő curve at a rate proportional to $\log n/n$, but it does not include anything which differentiates between separate zeros of $p_{n-1}(nz)$. We expect that such information will be contained in a subsequent correction term. Perhaps the correct estimate takes the form \[ \delta \approx \frac{\xi \log n}{2(1-\xi)n} + \frac{v_n}{n}, \] where $v_n$ is some bounded sequence. By substituting this into $(6.2)$ we get \[ \exp\!\left\{n \varphi(\xi) + \frac{\xi - 1}{\xi} v_n \right\} \approx \frac{1}{(1-\xi)\sqrt{2\pi}}. \] One thing this estimate tells us is that, as $n \to \infty$, the limit of the left-hand side should exist.

Problem 16.
Let $(v_n)$ be a bounded sequence. Show that, to make \[
\lim_{n \to \infty} \exp\!\left\{n \varphi(\xi) + \frac{\xi -
1}{\xi} v_n \right\} \] exist, we should choose $v_n =
(w_n-i\tau_n)\xi/(1-\xi)$, where $(w_n)$ is a convergent
sequence and \[ \tau_n \equiv n \operatorname{Im} \varphi(\xi)
\pmod{2\pi}. \]

7 The scaling limit

We now have the information necessary to state our main result.

Theorem.
Let $\xi \in S$ with $\xi \neq 1$. Define the sequence
$(\tau_n)$ by \[ \tau_n \equiv n\operatorname{Im}(\xi - 1 -
\log \xi) \pmod{2\pi}, \quad -\pi < \tau_n \leq \pi \] and
set \[ z_n(w) = \xi + \frac{\xi\log n}{2(1-\xi)n} -
\frac{(w-i\tau_n)\xi}{(1-\xi)n}. \] Then

\[ \lim_{n \to
\infty} \frac{p_{n-1}(n z_n(w))}{\exp(n z_n(w))} = 1 -
\frac{e^{-w}}{(1-\xi)\sqrt{2\pi}} \tag{7.1}\]

uniformly on compact
subsets of the $w$-plane.
Figure 5. This animation illustrates how the scaling limit in the theorem tracks the zeros of $p_{n-1}(nz)$ (shown in white) as $n$ increases from 20 to 100. A point $\xi$ (shown in yellow) on the Szegő curve (shown in red) is chosen, and for each $n$ the image of a square in the $w$-plane under the map $z_n(w)$ is shown in blue. This blue square approaches $\xi$ at the same rate that the zeros of $p_{n-1}(nz)$ do. As $n$ increases, the number of zeros in the square becomes fixed and the zeros flatten out into a line parallel with the side of the square. Note that the zeros of the right-hand side of $(7.1)$ also lie on a straight line.

In the following problems, suppose that $w$ is restricted to an arbitrary compact subset of $\mathbb C$.

Problem 17.
Show that $z_n(w)$ is inside $\gamma$ and $z_n(w) \notin N_\epsilon$ for $n$ large enough.

Problem 18.
Use equation $(4.7)$ to show that \[ \frac{\exp(n z_n(w))}{(e z_n(w))^n} \left( \frac{p_{n-1}(n z_n(w))}{\exp(n z_n(w))} - 1 \right) \sim -\frac{1}{(1-\xi)\sqrt{2\pi n}} \] as $n \to \infty$ uniformly with respect to $w$.

Problem 19.
Show that \[ z_n(w)^n \sim \xi^n n^{1/[2(1-\xi)]} \exp\left\{-\frac{w-i\tau_n}{1-\xi}\right\} \] as $n \to \infty$ uniformly with respect to $w$.

Problem 20.
Show that \[ \frac{\exp(n z_n(w))}{(e z_n(w))^n} \sim \frac{e^w}{\sqrt{n}} \] as $n \to \infty$ uniformly with respect to $w$.

Problem 21.
Prove $(7.1)$.

Now that we've obtained this limit we can use it to locate individual zeros of $p_{n-1}(nz)$, as was done in my last post here.

Appendix: Using the scaling limit (Why uniformity?)

The calculations in this problem sequence would be much simpler if we did not require uniform convergence of the limit in $(7.1)$. The main reason we require it is so that we can apply Hurwitz's theorem to connect the zeros of the limit function to the zeros of the polynomials.

Hurwitz's theorem.
Let $f_n(z)$ be a sequence of functions which are analytic in a region $R$ and which converge uniformly to a function $f(z) \not\equiv 0$ in every closed subregion of $R$. Let $\zeta$ be an interior point of $R$. If $\zeta$ is a limit point of the zeros of the $f_n(z)$, then $\zeta$ is a zero of $f(z)$. Conversely, if $\zeta$ is an $m$-fold zero of $f(z)$, then $f_n(z)$ has exactly $m$ zeros (counted with their multiplicities) in any sufficiently small neighborhood of $\zeta$ for all $n$ large enough.

So, the uniformity of the limit in $(7.1)$ allows us to apply Hurwitz's theorem and conclude that if $w = w_0$ is any root of

\[ 1 - \frac{e^{-w}}{(1-\xi)\sqrt{2\pi}} = 0, \tag{A.1} \]

then $p_{n-1}(nz_n(w))$ has a zero of the form \[ w = w_0 + \epsilon_n, \] where $\epsilon_n \to 0$ as $n \to \infty$. Or, stated a bit differently, we conclude that $p_{n-1}(nz)$ has a zero of the form \[ z = z_n(w_0 + \epsilon_n), \] and hence that $z \approx z_n(w_0)$ is a very good approximation when $n$ is large.For example, let's take $\xi = i/e$ (you can check that $i/e \in S$). Equation $(A.1)$ then has a root at \[ w \approx -0.982403 + 0.352513i, \] so we expect that $p_{49}(50z)$ has a zero at approximately \[ \begin{align} z &\approx z_{50}(-0.982403 + 0.352513i) \\ &\approx -0.0221277 + 0.381359i. \end{align} \] Indeed, $p_{49}(50z)$ has a zero at \[ z \approx -0.0227486 + 0.380855i. \]

Appendix: Further reading

Antonio R. Vargas

May 12, 2018