Taylor series for 1/(1+x) about a=3.

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f(x) =(1+x)^(-1)
f(3) = 4^(-1) = 1/4

f'(x) = (-1)(1+x)^(-2) = -1/(1+x)^2
f'(3) = -1/16

f''(x) = (-1)(-2)(1+x)^(-3) = 2/(1+x)^3
f''(3) = 2/64 = 1/32

f'''(x) =2(-3)(1+x)^(-4) = -6/(1+x)^4
f'''(3) = -6/256 = -3/128

f''''(x) = -6(-4)(1+x)^(-5) = 24/(1+x)^5
f''''(3) = 24/1024 = 3/128

f(x) = f(3) + (x-3) f'(3) /1! + (x-3)^2 f''(3) /2! + (x-3)^3 f'''(3) /3! + (x-3)^4 f''''(3) / 4! -.....
f(x) = 1/4 - (x-3) /16 + (x-3)^2 /64 - (x-3)^3 /256 + (x-3)^4 /1024 + O((x-3)^5)

1/4 - (x-3)/16 + 1/64 (x-3)^2 - 1/256 (x-3)^3 + (x-3)^4/1024

http://www.wolframalpha.com/input/?i=tay…