How do you find the Taylor polynomial of degree 10 of the function arctan(x^3)arctan(x3) at a = 0?

1 Answer
May 28, 2018

arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)arctanx3=x3x93+x155+o(x15)

Explanation:

Start from the sum of the geometric series:

sum_(n=0)^oo q^n = 1/(1-q)n=0qn=11q

for abs q <1|q|<1.

Let q = -x^2q=x2:

1/(1+x^2) = sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n)11+x2=n=0(x2)n=n=0(1)nx2n

converging for abs x <1|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)dxt0dx1+x2=n=0(1)nt0x2ndx

arctan t = sum_(n=0)^oo (-1)^n t^(2n+1)/(2n+1)arctant=n=0(1)nt2n+12n+1

Let now t=x^3t=x3:

arctan x^3 = sum_(n=0)^oo (-1)^n (x^3)^(2n+1)/(2n+1)arctanx3=n=0(1)n(x3)2n+12n+1

arctan x^3 = sum_(n=0)^oo (-1)^n x^(6n+3)/(2n+1)arctanx3=n=0(1)nx6n+32n+1

and truncating at the third term:

arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)arctanx3=x3x93+x155+o(x15)

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