Differentiability of Power Series
Consider a power series with radius of convergence R > 0).
f(z) =
¥ å
n=0
anzn.
Then f(z) is differentiable at z=0 and f¢(0) = a1:
f(z) - f(0)
=
¥ å
n=1
an zn
= a1 z +
¥ å
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)
=
¥ å
n=0
an zn
=
¥ å
n=0
an ((z - z0) + z0)n
=
¥ å
n=0
an
n å
k=0
æ è
n
k
ö ø
(z - z0)k z0n-k
=
¥ å
k=0
(z - z0)k
¥ å
n=k
æ è
n
k
ö ø
an z0n-k
=
¥ å
k=0
dk (z - z0)k,
where
dk
=
¥ å
n=k
æ è
n
k
ö ø
an z0n-k
=
¥ å
n=0
æ è
n + k
k
ö ø
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)
=
¥ å
n=0
æ è
n
1
ö ø
an+1 z0n
=
¥ å
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 ån=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) =
¥ å
n=0
1
n!
zn, |z| < ¥.
Use absolute convergence of the series and changing the order of summation to show that