For the following function, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series and the open interval of convergence. f(x)=arctan(x/9)

Answer 1

This problem is similar to the last one. First, observe the derivative of arctan(9/x) is -9/(x²+81) which can be rewritten as (-1/9)(1/(1+(x²/81)). Let u=x²/81 and consider

-(1/9)(1/(1+u))

we know 1/1+u = ∑_n=0^∞ (-1)ⁿuⁿ, and using our substitution for u, we have

(-1/9) ∑_n=0^∞ ((-1)ⁿx²ⁿ)/9²ⁿ = ∑_n=0^∞ ((-1)ⁿ⁺¹x²ⁿ)/9²ⁿ⁺¹

By the Ratio Test, this power series converges for |x| < 9. So as long as x is in (-9,9), we can integrate this series to get a power series for arctan(9/x). ie,

arctan(9/x) = ∫₀^x ∑_n=0^∞ ((-1)ⁿ⁺¹t²ⁿ)/9²ⁿ⁺¹ dt = ∑_n=0^∞ ((-1)ⁿ⁺¹x²ⁿ⁺¹)/((2n+1)9²ⁿ⁺¹ )