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

1 Answer
Daan J.
Jul 21, 2015

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

Explanation:

The general form of a Taylor expansion centered at #a# of an analytical function #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 degree Taylor polynomial is a polynomial consisting of the first four (#n# ranging from #0# to #3#) terms of the full Taylor expansion.

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

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

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