Differentiability of Power Series
Consider a power series with radius of convergence R > 0).
f(z) =
∞ ∑ n=0
anzn.
Then f(z) is differentiable at z=0 and f′(0) = a1:
f(z) − f(0)
=
∞ ∑ n=1
an zn
= a1 z +
∞ ∑ n=2
an zn
= a1 z + o(z).
If |z0| < R, and z is near1 z0, the power series for
f(z) converges absolutely so that
f(z)
=
∞ ∑ n=0
an zn
=
∞ ∑ n=0
an ((z − z0) + z0)n
=
∞ ∑ n=0
an
n ∑ k=0
⎛ ⎝
n
k
⎞ ⎠
(z − z0)k z0n−k
=
∞ ∑ k=0
(z − z0)k
∞ ∑ n=k
⎛ ⎝
n
k
⎞ ⎠
an z0n−k
=
∞ ∑ k=0
dk (z − z0)k,
where
dk
=
∞ ∑ n=k
⎛ ⎝
n
k
⎞ ⎠
an z0n−k
=
∞ ∑ n=0
⎛ ⎝
n + k
k
⎞ ⎠
an+k z0n
The argument for z=0 shows that f(z) is differentiable at
z = z0, with f′(z0) = d1, so that
f′(z0)
=
∞ ∑ n=0
⎛ ⎝
n
1
⎞ ⎠
an+1 z0n
=
∞ ∑ n=1
n an+1 z0n
.
We have shown
•
If f(z) is represented by a convergent power series for
|z| < R, then f(z) is an analytic function in the
region |z| < R and its derivative is represented by the
convergent series ∑n=1∞ n an zn−1, |z| < R.
All Analytic Functions Can Be Represented by Power Series
With a great deal more work, we will show that every analytic
function can be represented locally as a convergent power series:
•
If f(z) is an analytic function in the
region |z| < R, then f(z) is represented by a convergent power series for
|z| < R. Moreover, the derivatives of all orders exist and can
be represented by the formally differentiated series for
|z| < R.
An Exercise Using Absolute Convergence of Power Series
The power series for the exponential function is
exp(z) =
∞ ∑ n=0
1
n!
zn, |z| < ∞.
Use absolute convergence of the series and changing the order of summation to show that