its derivative has the Taylor expansion
The last step follows from the fact that
The hyperbolic functions differ from their trigonometric counterparts in that they do not sport alternating signs.
Since integration is the inverse operation of differentiation, you should expect that it is also possible to integrate Taylor series term-by-term.
then its indefinite integral has the Taylor expansion
Once again, this can be exploited to find Taylor series. Consider for instance the inverse tangent function . Its derivative is , which has the Taylor series expansion
valid for -1<x<1. (This expansion can be found by using the geometric series!)
Consequently its integral has the Taylor series expansion
What is the appropriate choice for C? Plug in the center of the power series, in our case , on both sides of the equation:
so in this example C=0.
Integrating does not change the radius of convergence of a power series, so we expect the Taylor series expansion for the inverse tangent function to be valid for all -1<x<1.
has radius of convergence r. Show that its integral has radius of convergence r.
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