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Analytic Functions and Taylor Series

Module by: Lawrence Baggett. E-mail the author

Summary: A summary of taylor series functions and their properties, including some practice exercises relating the taylor series to the identity theorem.

Definition 1:

Let SS be a subset of C,C, let f:S→Cf:S→C be a complex-valued function, and let cc be a point of S.S. Then ff is said to be expandable in a Taylor series around c with radius of convergence rr if there exists an r>0r>0 such that Br(c)⊆S,Br(c)⊆S, and f(z)f(z) is given by the formula

f ( z ) = ∑ n = 0 ∞ a n ( z - c ) n f ( z ) = ∑ n = 0 ∞ a n ( z - c ) n
(1)

for all z∈Br(c).z∈Br(c).

Let SS be a subset of R,R, let f:S→Rf:S→R be a real-valued function on S,S, and let cc be a point of S.S. Then ff is said to be expandable in a Taylor series around c with radius of convergence rr if there exists an r>0r>0 such that the interval (c-r,c+r)⊆S,(c-r,c+r)⊆S, and f(x)f(x) is given by the formula

f ( x ) = ∑ n = 0 ∞ a n ( x - c ) n f ( x ) = ∑ n = 0 ∞ a n ( x - c ) n
(2)

for all x∈(c-r,c+r).x∈(c-r,c+r).

Suppose SS is an open subset of C.C. A function f:S→Cf:S→C is called analytic on S if it is expandable in a Taylor series around every point cc of S.S.

Suppose SS is an open subset of R.R. A function f:S→Cf:S→C is called real analytic on S if it is expandable in a Taylor series around every point cc of S.S.

Theorem 1

Suppose SS is a subset of C,C, that f:S→Cf:S→C is a complex-valued function and that cc belongs to S.S. Assume that ff is expandable in a Taylor series around cc with radius of convergence r.r. Then ff is continuous at each z∈Br(c).z∈Br(c).

Suppose SS is a subset of R,R, that f:S→Rf:S→R is a real-valued function and that cc belongs to S.S. Assume that ff is expandable in a Taylor series around cc with radius of convergence r.r. Then ff is continuous at each x∈(c-r,c+r).x∈(c-r,c+r).

Proof

If we let gg be the power series function given by g(z)=∑anzn,g(z)=∑anzn, and TT be the function defined by T(z)=z-c,T(z)=z-c, then f(z)=g(T(z)),f(z)=g(T(z)), and this theorem is a consequence of (Reference) and (Reference).

Exercise 1

Prove that f(z)=1/zf(z)=1/z is analytic on its domain.

HINT: Use r=|c|,r=|c|, and then use the infinite geometric series.

Exercise 2

State and prove an Identity Theorem, analogous to (Reference), for functions that are expandable in a Taylor series around a point c.c.

Exercise 3

  1. Prove that every polynomial is expandable in a Taylor series around every point c.c. HINT: Use the binomial theorem.
  2. Is the exponential function expandable in a Taylor series around the number -1?-1?

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