56.1. RUNGE’S THEOREM 1773

and so, letting Qb (z) = 1(z−b)n ∑

Mk=0 ak

(bn−bz−b

)k,

||R−Qb||K,∞ ≤ ||R−Qbn ||K,∞ + ||Qbn −Qb||K,∞

<ε

2+

ε

2= ε

showing that b ∈ S. Since S is both open and closed in V it follows that, since S ̸= /0, S =V .Otherwise V would fail to be connected.

It remains to consider the case where V is unbounded. Pick b ∈V large enough that∣∣∣ zb

∣∣∣< 12

(56.1.8)

for all z ∈ K. From what was just shown, there exists a rational function, Qb having a poleonly at b such that ||Qb−R||K,∞ < ε

2 . It suffices to assume that Qb is of the form

Qb (z) =p(z)

(z−b)n = p(z)(−1)n 1bn

1(1− z

b

)n

= p(z)(−1)n 1bn

(∞

∑k=0

( zb

)k)n

Then by an application of Corollary 56.1.7 there exists a partial sum of the power series forQb which is uniformly close to Qb on K. Therefore, you can approximate Qb and thereforealso R uniformly on K by a polynomial consisting of a partial sum of the above infinitesum. This proves the theorem.

If f is a polynomial, then f has a pole at ∞. This will be discussed more later.

56.1.4 Runge’s TheoremNow what follows is the first form of Runge’s theorem.

Theorem 56.1.9 Let K be a compact subset of an open set, Ω and let{

b j}

be a set whichconsists of one point from each component of Ĉ\K. Let f be analytic on Ω. Then for eachε > 0, there exists a rational function, Q whose poles are all contained in the set,

{b j}

such that||Q− f ||K,∞ < ε. (56.1.9)

If Ĉ\K has only one component, then Q may be taken to be a polynomial.

Proof: By Lemma 56.1.4 there exists a rational function of the form

R(z) =M

∑k=1

Ak

wk− z

where the wk are elements of components of C\K and Ak are complex numbers such that

||R− f ||K,∞ <ε

2.