How do you find the Maclaurin series for #arctan x# centered at x=0?

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Answer 1 · 2017-02-03T23:40:28

#arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)#

Start from the sum of a geometric series, which is:

#sum_(n=0)^oo xi^n = 1/(1-xi)#

now substitute: #xi = -t^2# and we have:

#sum_(n=0)^oo (-t^2)^n = 1/(1+t^2)#

or:

#sum_(n=0)^oo (-1)^nt^(2n) = 1/(1+t^2)#

If we integrate now the series term by term we have:

#sum_(n=0)^oo int_0^x (-1)^nt^(2n)dt = int_0^x (dt)/(1+t^2)#

At the second member we have a standard integral:

#int_0^x (dt)/(1+t^2) = arctanx#

so we have:

#arctanx = sum_(n=0)^oo int_0^x (-1)^nt^(2n)dt = sum_(n=0)^oo (-1)^n [t^(2n+1)/(2n+1)]_0^x#

and finally:

#arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)#