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Francneto
Junior Member

Angola
3 Posts

 Posted - 09/13/2007 :  08:05:34 I still need help to splve the following problems:1. Let f(x)=tan^-1(x) a) Obtain the 3rd degree Taylor polynomial P3(x) and the Taylor remainder R3 (x) about c=1 for the given function.this is what I have tried:The first derivative is f'(x) = 1/(x^2 + 1)factoring(1 + x^2) within the complex numbers: (x^2 + 1) = (x + i)(x - i)So f'(x) = 1/(x+i)(x-1)f'(x) = (i/2)/(x+i) - (i/2)/(x-i) f'(0) = 1f"(x) = -(i/2)/(x+i)^2 + (i/2)/(x-i)^2 f"(0) = 0 f'"(x) = i/(x+i)^3 - i/(x-i)^3 f"'(0) = -1/2That's all I could do. Am I on the right track? Could you please help to find: 1. the remainder R3(X) R3 (x) about c=1 for the given function?2. Use the Use P3(x) to estimate tan^-1(0.7) and use R3(x) to estimate the error in the approximation.

skeeter
Advanced Member

USA
5634 Posts

 Posted - 09/13/2007 :  13:39:30 f(x) = arctan(x), centered at c = 1 ...P3(x) = f(1) + f'(1)(x-1) + f"(1)(x-1)2/2 + f'''(x)(x-1)3/6 + R3(x)R3(x) = f4(z)(x - 1)4/24f(x) = arctan(x) ... f(1) = /4f'(x) = 1/(1 + x2) ... f'(1) = 1/2f''(x) = -2x/(1 + x2)2 ... f''(1) = -1/2f'''(x) = (6x2 - 2)/(1 + x2)3 ... f'''(1) = 1/2 Edited by - skeeter on 09/13/2007 13:40:35
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