I am trying to approximate $\ln^2(1.2)$ using Taylor series approximation to the 3rd order with $x$ centered at $1$. The value which I get once I am done approximating is $.032$ with an error of less than or equal to about $.00058$. Which to me, does not make much sense considering $\ln^2(1.2)$ is about $.033$ so the error is not correct in this case making it a good approximation. Am I incorrect on my logic?
Here is my work: \begin{align*} f(x) &= \ln^2(x)\\ f'(x) &= 2\ln(x)/x \\ f''(x) &= -2\ln(x)/x^2 + 2/x^2 \\ f'''(x) &= 4\ln(x)/x^3 -6/x^3.
\end{align*}
From there $f(1) = 0, f'(1) = 0, f''(1) = 2, f'''(1) = 6$.
Taylor Series Polynomial expansion would be: $0+0+1(x-1)^2 – 1(x-1)^3$
Error would be less than or equal to $f''''(1.2)/4!(.2)^4$ which is about $.00058$.
Then I plug $1.2$ into the taylor series polynomial: $(.2)^2 – (.2)^3 = .032$.
Best Answer
Let $M$ be the maximum of $|f''''(x)|$ on $[1,1.2]$. Then the error is in your approximation is at most:
$$ \frac{M}{4!} |1 - 1.2|^4. $$
Here we find that $M = 22$. So the error is at most:
$$0.0014\overline{66}.$$
If you use a calculator, you get that $\ln(1.2)^2 \approx 0.0332412$. So the error in your approximation is about:
$$ |0.0332412 - 0.032| = 0.0012412 < 0.0014\overline{66}.$$