As long as we are strictly inside
the interval of convergence, we can take derivatives and integrals of power series
allowing us to get new series.
(Let R be the radius of convergence of a series. x is strictly inside the interval
of convergence of the series when −R<x<R,
so x is not equal to either of the two possible endpoints R or
−R of the interval.)
To see how this works with a series
centered at the origin, first consider that for any
constant cn, ddx(cnxn)=ncnxn−1. Similarly, ∫cnxndx=cnxn+1n+1+C. Now consider the
power series ∞∑n=0c0+c1x+c2x2+c3x3+c4x4+c5x5+⋯. When x is
strictly inside the interval of convergence of this series, we can
differentiate and integrate term by term. A beautiful fact
is that the radius of convergence will
not change when you differentiate or antidifferentiate.
To differentiate, we simply differentiate each term (not worrying
that we have infinitely many terms) and then put the terms back
into summation notation. Notice that in the derivative
series we must change our index to begin
at n=1.(Why?) ddx(∞∑n=0cnxn)=ddx(c0+c1x+c2x2+c3x3+c4x4+c5x5+⋯)=0+c1+2c2x+3c3x2+4c4x3+5c5x4+⋯=∞∑n=1ncnxn−1
Similarly, while we get an infinite integrand, we don't worry and
just antidifferentiate each term, and then add a constant.
We do not need to change our index starting point here. (Why not?)
Succinctly, we get the following for power
series centered at the origin:
Let
∞∑n=0cnxn have radius of
convergence R. As long as x is strictly inside
the interval of convergence of the series, i.e.
−R<x<R, ddx(∞∑n=0cnxn)=∞∑n=1ncnxn−1
∫(∞∑n=0cnxn)dx=(∞∑n=0cnxn+1n+1)+C
and the new series have the same R as the original
series.
The same holds for power series
centered at a.
Let ∞∑n=0cn(x−a)n have radius of convergence R. As long
as x is strictly inside the interval of convergence of the
series, i.e. a−R<x<a+R, ddx(∞∑n=0cn(x−a)n)=∞∑n=1ncn(x−a)n−1
∫(∞∑n=0cn(x−a)n)dx=(∞∑n=0cn(x−a)n+1n+1)+C
and the new series have
the same R as the original series
This video will discuss the derivatives and antiderivatives of power
series, and explain that they have the same radius of convergence as
the original series. Convergence at the endpoints does not
carry through to the derivatives and antiderivatives, where
convergence at the endpoints may be different. The new series must be tested at the endpoints
to determine their convergence.