How do you find the third degree Taylor polynomial for f(x)= ln x, centered at a=2?

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##ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3##.

The unconcealed fashion of a Taylor exposition centered at ##a## of an analytical operation ##f## is ##f(x)=sum_{n=0}^oof^((n))(a)/(n!)(x-a)^n##. Here ##f^((n))## is the nth derivative of ##f##.

The third step Taylor polynomial is a polynomial consisting of the primeval impure (##n## ranging from ##0## to ##3##) conditions of the liberal Taylor exposition.

Therefore this polynomial is ##f(a)+f'(a)(x-a)+(f''(a))/2(x-a)^2+(f'''(a))/6(x-a)^3##.

##f(x)=ln(x)##, hence ##f'(x)=1/x##, ##f''(x)=-1/x^2##, ##f'''(x)=2/x^3##. So the third step Taylor polynomial is: ##ln(a)+1/a(x-a)-1/(2a^2)(x-a)^2+1/(3a^3)(x-a)^3##.

Now we possess ##a=2##, so we possess the polynomial: ##ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3##.

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