a) Consider the function arctan (x^2). Write a partial sum for the power series

Question

a) Consider the function arctan (x^2). Write a partial sum for the power series which represents this function consisting of the nonzero terms. For example, if the series were Sigma^infinity_n=0 3^n x^2n, you would write 1 + 3x^2 + 3^2 x^4 + 3^3 x^6. Also indicate the radius of convergence. Partial Sum: Radius of Convergence: b) Use part a) to write the partial sum for the power series which represents integral arctan (x^2) dx. Write the first 4 nonzero terms. Also indicate the radius of convergence. Partial Sum: Radius of Convergence: c) Use part b)to approximate the integral integral^0.6_0 arctan (x^2) dx.

Explanation / Answer

a)

1/(1-x) =[n=0 to ] xn

1/(1+x2) =1/(1-(-x2)) =[n=0 to ] (-x2)n

1/(1+x2) =1/(1-(-x2)) =[n=0 to ] (-1)nx2n

[1/(1+x2)]dx=[n=0 to ] (-1)nx2n dx

arctan(x)=[n=0 to ] (-1)n(1/(2n+1))x2n+1

arctan(x2)=[n=0 to ] (-1)n(1/(2n+1))(x2)2n+1

arctan(x2)=[n=0 to ] (-1)n(1/(2n+1))x4n+2

arctan(x2)=[(-1)0(1/(0+1))x0+2]+[(-1)1(1/(2+1))x4+2]+[(-1)2(1/(4+1))x8+2]+[(-1)3(1/(6+1))x12+2]

arctan(x2)=x2-(1/3)x6+(1/5)x10-(1/7)x14

an=(-1)n(1/(2n+1))x4n+2

an+1=(-1)n+1(1/(2n+3))x4n+6

ratio test for convergence:

limn->|an+1/an|<1

limn->|[(-1)n+1(1/(2n+3))x4n+6]/[(-1)n(1/(2n+1))x4n+2]|<1

|x4|<1

|x|<1

radius of convergence =1

b)

arctan(x2) dx=[x2-(1/3)x6+(1/5)x10-(1/7)x14]dx

arctan(x2) dx=(1/3)x3-(1/3*7)x7+(1/5*11)x11-(1/7*15)x15

arctan(x2) dx=(1/3)x3-(1/21)x7+(1/55)x11-(1/105)x15

radius of convergence = 1

c)

[0 to 0.6] arctan(x2) dx=[(1/3)0.63-(1/21)0.67+(1/55)0.611-(1/105)0.615]-[(1/3)03-(1/21)07+(1/55)011-(1/105)015]

[0 to 0.6] arctan(x2) dx=[0.0707285]-0

[0 to 0.6] arctan(x2) dx=0.0707285

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