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If
is a power series, then
the series obtained by differentiating the terms of
is

This is not a power series, but its translate

is.

** 12.15**
**Definition (Formal derivative.)**
If

is a power series, then the

*formal derivative of*
is

I will sometimes write

when I think this will
cause no confusion.

Our fundamental theorem on power series is:

** 12.17**
**Theorem (Differentiation theorem.)***
Let
be a power series. Then
and
have the same radius of convergence. The function associated
with
is differentiable in the disc of convergence, and the function
represented by
agrees with on the disc of convergence.
*
The proof is rather technical, and I will postpone it
until section 12.8. I will derive some consequences of it before
proving it (to convince you that it is worth proving).

** 12.18**
**Example.**
We know that the geometric series

has radius of convergence

and

for

. The differentiation
theorem says

also has radius of convergence

,
and

i.e.,

We can apply the theorem again and get

or

Another differentiation gives us

or

The pattern is clear, and I omit the induction proof that for all

** 12.19**
**Exercise.**
By assuming the differentiation theorem, we've shown that the series

has radius of
convergence

for all

. Verify this directly.

** 12.20**
**Exercise.**
A
Find formulas for

and

that are valid for

. (You may assume the
differentiation theorem.)

** 12.21**
**Example.**
By the differentiation theorem, if

then

and

are differentiable on

and

, and

. (We
saw in earlier examples that both series have radius of convergence

, and
that the formal derivatives satisfy

and

.) Also, clearly

and

are real when

is real. The discussion in example

10.3 then shows
that for real

,

and

agree with the cosine and sine functions you

discussed in your previous calculus course, and in particular that

** Next:** 12.4 The Exponential Function
** Up:** 12. Power Series
** Previous:** 12.2 Radius of Convergence
** Index**