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Analytic Combinatorics/Darboux Method

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Introduction

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Darboux's method is one way of estimating the coefficients of generating functions involving roots.

It is easier than Singularity Analysis, but it applies to a smaller set of functions.

Theorem

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Theorem due to Wilf[1].

If we have a function (1−z)βf(z) where β∉{0,1,2,…} where f(z) has a radius of convergence greater than 1 and an expansion near 1 of ∑j≥0fj(1−z)j, then:

[zn](1−z)βf(z)=∑j=0mfjn−β−j−1Γ(−β−j)+O(n−m−β−2)

Example

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The theorem is a bit abstract, so I will show an example of how you might use it before going into the proof.

Taking an example from Wilf[2]:

e−z/2−z2/41−z=(1−z)−12e−z/2−z2/4

e−z/2−z2/4 is a complete function, so its radius of convergence is greater than 1.

Near 1 it can be expanded using the Taylor series:

e−z/2−z2/4=e−3/4+e−3/4(1−z)+⋯

Therefore, for m=0:

[zn]e−z/2−z2/41−z=e−3/4n−12Γ(12)+O(n−32)

Or, if we want more precision we can set m=1:

[zn]e−z/2−z2/41−z=e−3/4n−12Γ(12)+e−3/4n−32Γ(−12)+O(n−52)

and so on.

Proof

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Proof due to Wilf[3].

We have:

(1−z)βf(z)=∑j≥0fj(1−z)β+j

and:

∑j≥0fj(1−z)β+j=∑j=0mfj(1−z)β+j+∑j>mfj(1−z)β+j

By factoring out (1−z)β+m+1 from the last sum:

∑j>mfj(1−z)β+j=(1−z)β+m+1g(z)

Therefore:

(1−z)βf(z)=∑j=0mfj(1−z)β+j+(1−z)β+m+1g(z)

We have to prove that:

  1. [zn]∑j=0mfj(1−z)β+j=∑j=0mfjn−β−j−1Γ(−β−j)
  2. [zn](1−z)β+m+1g(z)=O(n−m−β−2)

Proof of 1

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By applying #Lemma 1:

[zn]∑j=0mfj(1−z)β+j=∑j=0mfj[zn](1−z)β+j∼∑j=0mfjn−β−j−1Γ(−β−j)

Proof of 2

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[zn](1−z)β+m+1=O(n−m−β−2) (by #Lemma 1)
[zn]g(z)=O(θn)(0<θ<1) (because, by assumption in the theorem, the radius of convergence R is greater than 1 and Cauchy's inequality tells us that [zn]g(z)=O(1Rn) and 1Rn<1)
|∑0≤k≤n2hkgn−k|≤{max0≤k≤n2|hk|}{∑0≤k≤n2Aθn−k}=max⁡{B,Bn−m−β−2}Cθn2≤Dθn2=O(θn2) (for A,B,C,D constants and assuming that −m−β−2<0).
|∑n2≤k≤nhkgn−k|≤{maxn2≤k≤n|hk|}{∑n2≤k≤nAθn−k}≤Bn−m−β−2=O(n−m−β−2)

because {∑n2≤k≤nAθn−k}≤Cmaxn2≤k≤nθn−k=C because θ2≤θ1≤θ0=1.

Putting it all together:

[zn](1−z)β+m+1g(z)=O(θn2)+O(n−m−β−2)=O(n−m−β−2)

because Cθn2=o(n−m−β−2)[4] because θ<1[5].

Lemma 1

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[zn](1−z)β∼n−β−1Γ(−β)

Proof:

[zn](1−z)β=(βn)(−1)n=(n−β−1n)=Γ(n−β)Γ(−β)Γ(n+1)=1Γ(−β)(n−β)(β+1)∼n−β−1Γ(−β)[6]

where x(n) is the rising factorial.

Szegő's theorem

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We can apply a similar theorem to functions with multiple singularities. From Wilf[7] and Szegő[8].

If f(z) is analytic in |z|<1, has a finite number of singularities eiϕ1,eiϕ2,⋯,eiϕr on the unit circle |z|=1 and in the neighbourhood of each singularity has the expansion

f(z)=∑v≥0cv(k)(1−ze−iϕk)αk+vβk(k=1,2,⋯,r and βk>0)

Then we have the asymptotic series

[zn]f(z)=∑v≥0∑k=1rcv(k)(αk+vβkn)(−eiϕk)n

Notes

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  1. ↑ Wilf 2006, pp. 194.
  2. ↑ Wilf 2006, pp. 195.
  3. ↑ Wilf 2006, pp. 193-195.
  4. ↑ w:Big_O_notation#Little-o_notation.
  5. ↑ w:Big_O_notation#Orders_of_common_functions.
  6. ↑ w:Falling_and_rising_factorials#Properties.
  7. ↑ Wilf 2006, pp. 196.
  8. ↑ Szegő 1975, pp. 207-208.

References

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  • Szegő, Gabor (1975). Orthogonal Polynomials (4th ed.). American Mathematical Society.
  • Wilf, Herbert S. (2006). Generatingfunctionology (PDF) (3rd ed.). A K Peters, Ltd.