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

##### 1 Answer

Mar 11, 2017

#### Explanation:

The Taylor polynomial for

#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

#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#