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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# will indicate that there are relevant papers or books the interested reader might like to see.


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 pn1(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
pn(z)=nk=0zkk!.
This polynomial pn(z) is commonly referred to as the nth partial sum of the exponential series or the nth Taylor polynomial of the exponential function.


Problem 1. Let γ be a simple closed loop around the origin with positive orientation. Show that ezpn1(z)zn=12πiγζneζdζζz for all nonzero z inside γ. Then make a change of variables to show that
enzpn1(nz)(ez)n=12πiγ(es)nensdssz
for all nonzero z inside γ.1

Hint: Show that γζmdζζz=0 for all integers m1.


The precise nature of the contour γ will be determined in Section 3.

Define φ(s)=s1logs, taking the principal branch of the natural logarithm, so that (es)nens=exp{n(s1logs)}=exp{nφ(s)}.

2 The saddle point method


The saddle point method (also known as the method of steepest descent)2 is a rigorous application of a simple idea: the size of something is roughly equal to the size of its biggest parts. The something here 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 maxima in this context.

Suppose we have an integral of the form F(n)=γexp{nf(s)}ds=γexp{nRef(s)}exp{inImf(s)}ds,n>0. The size of the integrand is |exp{nf(s)}|=exp{nRef(s)}, and this is largest where Ref(s) is largest. However, because F(n) is an integral of a complex-valued function, if we integrate over a generic contour γ then the final size of the integral may be sensitive to cancellation or resonance coming from the oscillatory factor exp{inImf(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 z0 with f(z0)=0 and f(z0)0.

First show that Ref has a saddle point at z0.

Then show that there are two smooth curves Γ1 and Γ2 passing through z0 such that
  1. Γ1 and Γ2 are orthogonal at z0,
  2. Imf is constant on Γ1 and Γ2, and
  3. Ref has a maximum at z0 when restricted to Γ1 and a minimum at z0 when restricted to Γ2.
We will call Γ1 the steepest descent path and Γ2 the steepest ascent path.

In other words, the two curves passing through a simple critical point z0 of f on which Imf is constant are the steepest ascent and descent paths through the saddle at z0 on the graph of Ref.

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


Definition 1. The axis of a saddle point z0 of a function f is defined to be the straight line in the z-plane defined by f(z0)(zz0)20. One can check that the steepest descent path Γ1 from Problem 2 is tangent to the axis of z0 at z0.




Figure 1. Two perspectives of the saddle point at z=1 of the function φ(z) defined at the end of Section 1. Over the surface Reφ(z) we see the steepest descent path in blue, the steepest ascent path in red, and the axis of the saddle point in white.


So, if we can deform γ so that it passes through a critical point of f along a path of steepest descent on the surface Ref, then Imf will be constant on γ, say Imf=:I0, and we will have F(n)=γexp{nRef(s)}exp{inI0}ds=exp{inI0}γexp{nRef(s)}ds. What remains is a real integral. The new integrand exp{nRef(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 γ to have.

The contour integral formula (1.2) we derived in Problem 1 is only valid for z inside γ, so we need to make sure that γ is large enough so that it encompasses all zeros of pn1(nz). To determine a rough bound for the zeros we will use the following classical result.3


Eneström-Kakeya Theorem. If q(z)=mk=0akzk is a polynomial with 0a0a1am, then all zeros of q lie in the closed disk |z|1.


Problem 3. Use the Eneström-Kakeya theorem to show that all zeros of pn1(nz) lie in the open disk |z|<1.

Bonus: Prove the Eneström-Kakeya theorem.



Figure 2. The 29 zeros of the polynomial p29(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 γ to lie on or outside the unit circle, it will contain all zeros of pn1(nz).


Problem 4. Show that the only critical point of φ(s) is located at s=1 and that φ(s)=12(s1)2+higher order terms near there.

Further, show that the axis of this critical point is the vertical line Res=1.


At the critical point we have Reφ(1)=0. Based on the discussion in the previous section, we want this to be the highest point of γ on the graph of Reφ(s).

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


Problem 5. Show that if Reφ(s)=0 then either s=1, Res>1, or |s|<1.


So, for example, the circle |s|=1 passes through the critical point of φ tangent to the path of steepest descent and lies entirely in the region Reφ<0 except at the point s=1.

Problems 3, 4, and 5 motivate the following choice for our contour γ:


The contour γ. Let γ be a simple closed loop around the origin which passes through the point s=1 and lies entirely in the regions |s|>1 and Reφ(s)<0 except at s=1. In a small neighborhood Uγ of s=1, let γ coincide with the path of steepest descent through s=1 on the graph of Reφ(s).



Figure 3. A plot of the contour γ in white. Note how it avoids the region Reφ(s)>0 (shown in red) except at the saddle point s=1. In the region Uγ (shown in yellow), the contour coincides with the steepest descent path through the saddle (shown in blue).


From Problem 2 we know that Imφ(s) is constant on γUγ. In fact, Imφ(s)=0 there.

These conditions ensure that all zeros of pn1(nz) lie inside the contour, that s=1 is the highest point of the contour on the surface Reφ(s), and that γ 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.



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(). If an(z) and bn(z) are sequences which depend on a complex parameter z, dn is another sequence which doesn't, and A is some subset of the complex plane, the statement
an(z)=bn(z)+O(dn)as n uniformly for zA"
means that there are positive constants C and M which do not depend on z such that the inequality
|an(z)bn(z)|C|dn|
holds for all nM and all zA.

Qualitatively, writing an(z)=bn(z)+O(dn) means that the error in estimating an(z) by bn(z) is at most something like dn.

Notice that the constants C and M are not specified by the statement in (4.1). This is a feature, not a bug! We won't care what those constants are, only that they exist. The beauty of big-O notation is that we can hide these constants and rewrite the inequality (4.2) as a sort of faux equality; algebraically, (4.1) behaves pretty much like a normal equation.

It's also sometimes useful to consider O() as the set all of sequences which are bounded by a constant multiple of ||. We can then compare these sets using , =, etc. For example, O(n2)O(n1) as n. If an, bn, and dn are sequences of complex numbers, then the statement O(bn)O(dn) is equivalent to an=O(bn)as nan=O(dn)as n, where both equalities are meant to be interpreted as in (4.1) and (4.2).

The last piece of notation we'll introduce is the notation. The statement an(z)bn(z)as n uniformly for zA" means that limnan(z)bn(z)=1 uniformly for zA. Here we say that an(z) is asymptotically equivalent to bn(z).



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,
γ=γ near 1+γ away from 1,
and expect that the integral over the part of γ 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 1U, 0V, and a biholomorphic map ψ:VU which satisfies (φψ)(x)=x2 for xV with ψ(0)=2.

Note that ψ(0)=1 and that U and V can be made arbitrarily small, so that we may take UUγ (the set Uγ 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 γ=γU+γU. The first thing we do is show that the integral over γU is exponentially small.


Definition 2. For ϵ>0, let Nϵ be the set of all points within a distance of ϵ of the curve γ.


Problem 7. Show that there is a constant C1>0 which does not depend on z such that |γUenφ(s)dssz|<eC1n for all zNϵ.


In big-O notation, we have shown that
γenφ(s)dssz=γUenφ(s)dssz+O(eC1n)
uniformly for zNϵ.

Now we turn our attention to the remaining integral over γU.


Problem 8. Show that γUenφ(s)dssz=α2α1ent2iψ(it)ψ(it)zdt for some positive numbers α1 and α2.


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. We thus estimate the subexponential factor at this point.


Problem 9. Define gz(t)=iψ(it)ψ(it)z. Show that, for zNϵ, there is a positive constant M that does not depend on z such that |gz(t)gz(0)|M|t| for tV.


To apply this approximation, write α2α1ent2iψ(it)ψ(it)zdt=α2α1ent2gz(t)dt=gz(0)α2α1ent2dt+α2α1ent2(gz(t)gz(0))dt.

Problem 10. Show that |α2α1ent2(gz(t)gz(0))dt|<Mn for zNϵ.


In big-O notation, we have shown that
α2α1ent2iψ(it)ψ(it)zdt=gz(0)α2α1ent2dt+O(1n)
uniformly for zNϵ.


Problem 11. Show that there is a positive constant C2 such that α2α1ent2dt=πn+O(eC2n), and hence that
gz(0)α2α1ent2dt=gz(0)πn+O(eC2n)
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 α2α1ent2iψ(it)ψ(it)zdt=gz(0)πn+O(1n) uniformly for zNϵ since O(eC2n)O(1/n). Substituting this in (4.4) yields γenφ(s)dssz=i1z2πn+O(1n) uniformly for zNϵ, where we have used the fact that gz(0)=i21z. Finally, substituting this in (1.2) yields
enzpn1(nz)(ez)n=1(1z)2πn+O(1n)
uniformly for z inside γ with z0 and zNϵ.

This completes our application of the saddle point method.



5 The zeros of pn1(nz) and the Szegő curve


Problem 12.
  1. Show that all zeros pn1(nz) lie inside γ with z0.
  2. For δ>0, show that there is an ϵ>0 such that all zeros satisfying |z1|>δ are not in Nϵ.
Hint: Use Problem 3.


It follows from Problem 12 and equation (4.7) that every zero of pn1(nz) which is bounded away from 1 satisfies the equation4
(ze1z)n=1(1z)2πn+O(1n).


Problem 13. Show that, as n, the zeros of pn1(nz) converge to the portion of the curve |ze1z|=1 that lies in the disk |z|1.


The curve
S:={zC:|ze1z|=1 and |z|1}
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 p29(30z) in white and the unit circle in blue. Compare with Figure 2.


Note that the condition |ze1z|=1 can also be written as Reφ(z)=0.



6 How quickly do the zeros approach the Szegő curve?


Now that we know the global behavior of the zeros of pn1(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 pn1(nz) which do not converge to the point z=1 satisfy
exp{nφ(z)}e(logn)/2(1z)2π.
Let's assume that z converges to some point ξS with ξ1, and write z=ξ+δ, where δ0 as n.


Problem 14. Show that 1z1ξ and that φ(z)φ(ξ)+ξ1ξδ.


Our estimate in (6.1) then becomes
exp{nφ(ξ)+ξ1ξnδ}e(logn)/2(1ξ)2π.


Problem 15. Show that Reφ(ξ)=0 and hence that exp{nφ(ξ)} is bounded.

Deduce that, in order to balance the decay of the factor e(logn)/2 on the right-hand side of (6.2), we must have δξlogn2(1ξ)n.


This estimate for δ tells us that the zeros in question approach the Szegő curve at a rate proportional to logn/n, but it does not include anything which differentiates between separate zeros of pn1(nz). We expect that such information will be contained in a subsequent correction term. Perhaps the correct estimate takes the form δξlogn2(1ξ)n+vnn, where vn is some bounded sequence. By substituting this into (6.2) we get exp{nφ(ξ)+ξ1ξvn}1(1ξ)2π. One thing this estimate tells us is that, as n, the limit of the left-hand side should exist.


Problem 16. Let (vn) be a bounded sequence. Show that, to make limnexp{nφ(ξ)+ξ1ξvn} exist, we should choose vn=(wniτn)ξ/(1ξ), where (wn) 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.5 Its proof is broken up into the five problems which follow it.


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.




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


1 The integral formula in Problem 1 is a special case of the one given on page 436 of the paper The Padé table of functions having a finite number of essential singularities by A. Edrei. [pdf]

2 Excellent references for the saddle point method / the method of steepest descent are Miller's Applied Asymptotic Analysis and de Bruijn's Asymptotic Methods in Analysis. Each gives a different proof of the method and both are worth reading for their complementary perspectives.

3 The paper Eneström–Kakeya Theorem and Some of Its Generalizations by R. B. Gardner and N. K. Govil contains some interesting historical information about this result. [pdf]

4 The asymptotic formula (5.1) was essentially originally derived by Szegő in his 1924 paper Über eine Eigenschaft der Exponentialreihe. An alternative derivation can be found in the paper Asymptotics for the zeros of the partial sums of e^z. I. by A. J. Carpenter, R. S. Varga, and J. Waldvogel. [pdf]

5 The main result of this problem sequence was essentially originally obtained by A. Edrei, E. B. Saff, and R. S. Varga in their monograph Zeros of Sections of Power Series. I derived a generalization of it in my paper The Saff-Varga Width Conjecture and entire functions with simple exponential growth. [pdf].




Antonio R. Vargas
May 12, 2018