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Francneto
Junior Member

Angola
3 Posts

Posted - 09/13/2007 :  08:05:34  Show Profile
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) = 1

f"(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) = -1/2

That'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.
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skeeter
Advanced Member

USA
5634 Posts

Posted - 09/13/2007 :  13:39:30  Show Profile
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)4/24

f(x) = arctan(x) ... f(1) = /4
f'(x) = 1/(1 + x2) ... f'(1) = 1/2
f''(x) = -2x/(1 + x2)2 ... f''(1) = -1/2
f'''(x) = (6x2 - 2)/(1 + x2)3 ... f'''(1) = 1/2

Edited by - skeeter on 09/13/2007 13:40:35
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