How do you find the Taylor polynomial of degree 10 of the function #arctan(x^3)# at a = 0?

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Answer 1 · 2018-05-28T14:59:50

#arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)#

Start from the sum of the geometric series:

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

for #abs q <1#.

Let #q = -x^2#:

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

converging for #abs x <1#.

In the interval of convergence we can integrate term by term:

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

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

Let now #t=x^3#:

#arctan x^3 = sum_(n=0)^oo (-1)^n (x^3)^(2n+1)/(2n+1)#

#arctan x^3 = sum_(n=0)^oo (-1)^n x^(6n+3)/(2n+1)#

and truncating at the third term:

#arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)#

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