1798 CHAPTER 57. INFINITE PRODUCTS

where cmk+1 ̸= 0. You choose b1,b2, · · · ,bmk+1 such that

f (z)

(mk+1

∑l=1

bl

(z− zk)k

)=

mk

∑j=0

akj (z− zk)

j +∞

∑k=mk+1

ckj (z− zk)

j .

Thus you need

mk+1

∑l=1

∞

∑j=mk+1

c jbl (z− zk)j−l =

mk

∑r=0

akr (z− zk)

r +Higher order terms.

It follows you need to solve the following system of equations for b1, · · · ,bmk+1.

cmk+1bmk+1 = ak0

cmk+2bmk+1 + cmk+1bmk = ak1

cmk+3bmk+1 + cmk+2bmk + cmk+1bmk−1 = ak2

...cmk+mk+1bmk+1 + cmk+mk bmk + · · ·+ cmk+1b1 = ak

mk

Since cmk+1 ̸= 0, it follows there exists a unique solution to the above system. You firstsolve for bmk+1 in the top. Then, having found it, you go to the next and use cmk+1 ̸= 0again to find bmk and continue in this manner. Let Sk (z) be determined in this manner foreach zk. By the Mittag-Leffler theorem, there exists a Meromorphic function, g such thatg has exactly the singularities, Sk (z) . Therefore, f (z)g(z) has removable singularities ateach zk and for z near zk, the first mk terms of f g are as prescribed. This proves the theorem.

Corollary 57.3.2 Let P ≡ {zk}∞

k=1 be a set of points in Ω, an open set such that P has nolimit points in Ω. For each zk, consider

mk

∑j=0

akj (z− zk)

j . (57.3.19)

Then there exists an analytic function defined on Ω such that the Taylor series of f at zkhas the first mk terms given by 57.3.19.

Proof: The proof is identical to the above except you use the versions of the Mittag-Leffler theorem and Weierstrass product which pertain to open sets.

Definition 57.3.3 Denote by H (Ω) the analytic functions defined on Ω, an open subsetof C. Then H (Ω) is a commutative ring2 with the usual operations of addition and mul-tiplication. A set, I ⊆ H (Ω) is called a finitely generated ideal of the ring if I is of theform {

n

∑k=1

gk fk : fk ∈ H (Ω) for k = 1,2, · · · ,n

}2It is not a field because you can’t divide two analytic functions and get another one.