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Summation of Series
Adrian Down
November 22, 2005
1 f (z) with 1
z^2
dependence
1.1 Statement
Derivate an identity for the sum of some infinite series by integrating the function
f (z) = z−^2 csc(πz)
over a big contour.
Note. • This is the reverse of the usual statement of problems of this type. Usually, we are given a sum for which we must choose a function. Experience with these problems indicates a good choice of function.
- We will let the size of the contour tend to ∞. Letting a contour “tend to ∞” is not rigorously defined. The operation simply means we take all parts of the contour farther and farther from the origin.
1.2 Locate the singularities
f is meromorphic on C. f has poles at all points such that
sin(πz) = 0 ⇔ z ∈ Z
The poles are located at the integers on the real line.
Note. The pole at z = 0 is unique in that (^) z^12 is also singular at this point.
Our goal is to utilize Cauchy’s Residue Theorem,
1 2 πı
ΓN
f =
z 0 ∈ΓN
res (f, zj )
We would like to do two things:
- Show that the integral of f over ΓN tends to 0 as N → ∞.
- Find an expression for the value of the residue at each singularity as a function of N.
If these conditions are satisfied, Cauchy’s Residue Theorem gives a sum of all of the residues equalling 0. Usually there are several unique residues which can then be isolated and set equal to an infinite sum, thus giving the desired identity.
Note. Sometimes, some algebra must be done to combine residues, such as adding residues at +n and −n to get an identity for a sum over the positive integers.
1.3 Choose a contour
In choosing a contour, our goals are:
- To enclose all of the poles of f (z).
- To use the Estimation Lemma to bound f (z) on the contour.
Note. The trick of obtaining the same integral on one side of the contour as that on the other side of the contour almost never works with this method.
With these considerations, we choose a contour that satisfies the follow- ing:
- Always pick a simple loop. Often, the best path to use is a square with corners located at ±(N + 12 ) ± ı(N + 12 ).
- Stay away from the poles as much as possible. We have shown that f (z) tends to ∞ in the neighborhood of any pole, so it will be more difficult to bound f near a singularity.
Since we seek an upper bound on (^) sin(^1 πz) , we seek a lower bound on sin(πz). Examine the function directly,
2 ı sin
π
N +
= eıπ(N^ +^
(^12) +ıt) − e−ıπ(N^ +^
(^12) +ıt)
Note. It is often easier to convert trigonometric functions to exponentials, since the latter are easier to manipulate.
Taking the absolute value, ∣ ∣ ∣ ∣ ∣ e ︸ ︷︷ ︸^2 ıπN 1
eı^
π ︸︷︷︸^2 ı
e−πt^ − e ︸ ︷︷ ︸^2 ıπN 1
e−ı^
N ︸ ︷︷ ︸^2 −ı
eπt
∣ıe−πt^ + ıeπt
∣ (^) = eπt^ + e−πt^ ≥ 1
Thus ∀z ∈ Γright,N ,
2 |sin(πz)| ≥ 1
⇒
|sin(πz)|
1 2
We have obtained the desired upper bound.
z ∈ Γright ⇒ |f (z)| ≤ 2 |z|−^2 ≤ 2 N −^2
By the arguments above using the Estimation Lemma, the integral of f along ΓN goes to 0 as the contour is made very large.
1.5 Calculate residues
From Cauchy’s Residue Theorem,
2 πı
ΓN
f = res (f, 0) +
1 ≤|n|≤N
res (f, n))
= res (f, 0) +
n 6 = 0 n ∈ Z
res (f, n)
The residues are easily calculated using the derivative trick.
f (z) =
z−^2 sin(πz)
For N 6 = 0,
res (f, n) =
n−^2 π cos(πn)
=
πn^2
(−1)n πn^2 Now consider z = 0. f has a pole of order 3 at z = 0.
Note. The higher the order of the pole, the harder it is to calculate the residue, since the coefficient of (z − z 0 )−^1 is “buried” more deeply in the expansion.
It is often useful to do a back-of-the-envelope type of calculation by cal- culating only the part of the series expansion for f necessary to find the coefficient of (z − z 0 )−^1. To do so, write f (z) as,
f (z) = z−^3
sin(πz) z
This is useful because sin( zπz ) has a removable singularity at z = 0, and thus has a Taylor series near z = 0.
sin(πz) z
z
πz −
(πz)^3 6
= π
π^2 6
z^2 +...
We can then write f as
f =
z−^3 π
1 − π 2 6 z
2 +... =^
z−^3 π
π^2 6
z^2 +...
πz^3
π^2 6 π
z^2 z^3
Note. To take the reciprocal of the series, make use of the approximation,
1 1 − x
≈ 1 + x
Thus the residue of f at 0 is
res (f, 0) =
π 6
With this choice,
res(f, w) = cot (π, w)
We can find that
max z∈Γn | cot(πz)| ≤ C < ∞
where the bound is independent of N. Thus the bound on |f | goes as (^) N^1 , whereas the path length goes as N. We cannot conclude from the Estimation Lemma that the integral of f around ΓN goes to 0 as N is made large.
Note. The integral does tend to 0, but it does not follow from anything that we have done thus far.
Nothing that we have learned thus far can resolve our dilemma. Abandon the first choice of f , and as another try, consider the function
f (z) =
cot(πz) (z − w)z
We can then bound |f |
max z∈ΓN
|(z − w)z|
N −^2
when N ≥ 2 |w|. Since cot(πz) is uniformly bounded by the arguments above, the Estima- tion Lemma shows that the integral of f around ΓN goes to 0 as N → ∞.
2.3 Derive an identity
By considering the integral of f along a path γN , we have shown thus far that the sum of the residues of f inside of ΓN is 0. We hope that the residues of the new function are the same as those of the initial guess for f. It turns out that the residues are the same, albeit with unimportant factors of π and w.
3 Counting zeros
3.1 Setup
Consider
- D a domain
- γ a simple loop in D
- f meromorphic it D
- f has no zeroes or poles on γ
We would like to know how many zeroes and poles f has inside γ.
Note. Inside is defined in the technical sense done previously.
3.2 Number of zeros and poles
Lemma.
1 2 πı
γ
f ′ f
= # zeros inside γ − # poles inside γ
Note. • This lemma gives no information N or P individually. However, it can be useful if we know for some other reason that N or P is 0. We can then derive information about the other.
- The poles are counted according to multiplicity, so that a pole of order n is counted n times in P.
Example.
f (z) =
z^3 (z − 1)^2 (z − 2)
The numerator gives 3 zeroes. The denominator gives 3 poles. Thus the lemma would tell us 3 − 3 = 0.
