Taylor Series



its derivative tex2html_wrap_inline85 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.

Taylor series and integration

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 tex2html_wrap_inline97 . Its derivative is tex2html_wrap_inline99 , 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 tex2html_wrap_inline105 , 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.

Try it yourself!

Convince yourself that integration does not change the radius of convergence: Suppose the power series


has radius of convergence r. Show that its integral tex2html_wrap_inline115 has radius of convergence r.

Click here for the answer, or to continue.

Helmut Knaust
Tue Jul 16 11:57:13 MDT 1996

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