Consider the function arctan(x2). Write a partial sum for the power series which
ID: 2829832 • Letter: C
Question
Consider the function arctan(x2). Write a partial sum for the power series which represents this function consisting of the first 4 nonzero terms. For example, if the series were , you would write 1 + 3x2 + 32xc4 + 33xc6 . Also indicate the radius of convergence. Partial Sum: Radius of Convergence: Use part a) to write the partial sum for the power series which represents Write the first 4 nonzero terms. Also indicate the radius of convergence. Partial Sum: Radius of Convergence: Use part b) to approximate the integral arctan(x2)dx.Explanation / Answer
The derivative of arctan(x) = 1/(1+x^2) = 1 - x^2 + x^4 - x^6 ....
Then, arctan is the integral of this is x - x^3 + x^5/5 - x^7/7 ... = the sum of (-1)^(n+1)x^(2n+1)/2n+1 for n=0 to infinity
Then, arctan(x^2) = the sum of (-1)^(n+1)*(x^2)^(2n+1)/(2n+1) for n = 0 to infinity =
the sum of (-1)^(n+1)*x^(4n+2)/(2n+1) for n=0 to infinity
Then, the first 4 terms are x^2 - x^6/3 + x^10/5 - x^14/7
The ratio between successive terms is- x^4*(2n+1)/(2n+3)
The limit of this ratio as n goes to infinity is x^4.
Thus, this converges on [-1, 1] (because it is an alternating series, this converges at -1 and 1.
b) The function is the sum for n=0 to infinity of (-1)^(n+1) * x^(4n+2)/(2n+1)
Then, the integral is the sum of (-1)^(n+1) * x^(4n+3)/((2n+1)(4n+3)
The first 4 terms are x^3/3 - x^7/21 + x*11/55 - x^15/105
The ratio is -x^4 *(2n+1)*(4n+3)/((2n+3)(4n+7)
The absolute value of the ratio is less than 1 for -1 <= x <= 1, converging to 1 at -1 and 1; because it is an alternating series, it converges at 1. Thus, it is [-1, 1]
Then, the partial integral of arctan(x^2) from 0 to 6 is
x^3/3 - x^7/21 + x*11/55 - x^15/105 E[0, .6] =
.6^3/3 - .6^7/21 + ,6*11/55 - .6^15/105 - 0 =
0.0707284565774283