Question

How do you find the third degree Taylor Polynomial of `f(x)=ln(x^2)` at x=1?

Answer 1

`ln(x^2)approx2(x-1)-(x-1)^2+2/3(x-1)^3`

Explanation:

The Taylor polynomial for `f` centered at `x=c` is given by

`T(x)=sum_(n=0)^oo(f^((n))(c)(x-c)^n)/(n!)`

Since we want a third degree polynomial, we will extend to the `(x-c)^3` term (we need to find three derivatives):

`f(x)=ln(x^2)=2ln(x)" "=>" "f(1)=2ln(1)=0`

`f'(x)=f^((1))(x)=2/x" "=>" "f^((1))(1)=2`

`f^((2))(x)=-2/x^2" "=>" "f^((2))(1)=-2`

`f^((3))(x)=4/x^3" "=>" "f^((3))(1)=4`

Then:

`ln(x^2)approxf(c)+(f^((1))(1)(x-c))/(1!)+(f^((2))(1)(x-c)^2)/(2!)+(f^((3))(1)(x-c)^3)/(3!)`

`ln(x^2)approx0+(2(x-1))/1+(-2(x-1)^2)/2+(4(x-1)^3)/(3!)`

`ln(x^2)approx2(x-1)-(x-1)^2+2/3(x-1)^3`