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

Mar 11, 2017

$\ln \left({x}^{2}\right) \approx 2 \left(x - 1\right) - {\left(x - 1\right)}^{2} + \frac{2}{3} {\left(x - 1\right)}^{3}$

#### 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 ${\left(x - c\right)}^{3}$ term (we need to find three derivatives):

$f \left(x\right) = \ln \left({x}^{2}\right) = 2 \ln \left(x\right) \text{ "=>" } f \left(1\right) = 2 \ln \left(1\right) = 0$

$f ' \left(x\right) = {f}^{\left(1\right)} \left(x\right) = \frac{2}{x} \text{ "=>" } {f}^{\left(1\right)} \left(1\right) = 2$

${f}^{\left(2\right)} \left(x\right) = - \frac{2}{x} ^ 2 \text{ "=>" } {f}^{\left(2\right)} \left(1\right) = - 2$

${f}^{\left(3\right)} \left(x\right) = \frac{4}{x} ^ 3 \text{ "=>" } {f}^{\left(3\right)} \left(1\right) = 4$

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!)

$\ln \left({x}^{2}\right) \approx 2 \left(x - 1\right) - {\left(x - 1\right)}^{2} + \frac{2}{3} {\left(x - 1\right)}^{3}$