Find the taylor series of arctan(4x^3) at x = 0?

Answer 1

arctanx = ∫₀^x1/(1 + t²)dt = ∫₀^x(1 - t² + t⁴-...+ (t^2n-2)(-1) ^n-1_{+ ...}= x - x³/3 + x⁵/5 -...+(-1)^n-1x^2n-1/(2n- 1) + ...

Now we replace x by 4x³ and we get:

arctan(4x³) = ∑₁^∞(-1)^{n -1}(4x³)^2n-1/(2n-1);