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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: $(.2)^2 - (.2)^3 = .032$.

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  • $\begingroup$ Pulse, Check my edit if every thing you want to ask is ok. $\endgroup$ – Jaideep Khare Apr 24 '17 at 23:37
  • $\begingroup$ It is impossible to say whether your logic is correct when you don't tell us what your logic is! What did you get as a Taylor's series for (ln(x))^2 about x= 1? $\endgroup$ – user247327 Apr 24 '17 at 23:39
  • $\begingroup$ Can you show how you got $0.032$? $\endgroup$ – Paul Apr 24 '17 at 23:39
  • $\begingroup$ @Paul sorry about that, I fixed it. $\endgroup$ – Pulse Apr 24 '17 at 23:56
  • $\begingroup$ @user247327 Does this edit make more sense? $\endgroup$ – Pulse Apr 25 '17 at 1:06
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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:

$$ |0.0332412 - 0.032| = 0.0012412 < 0.0014\overline{66}.$$

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  • $\begingroup$ For "M" would you be plugging in the value which gives it the maximum on that interval? So in this case, for the 4th derivative you would be plugging in (1.2) for x? $\endgroup$ – Pulse Apr 25 '17 at 4:47
  • $\begingroup$ For $M$ you plug in the value which maximizes $|f''''(x)|$ on $[1,1.2]$ which is $1$. $\endgroup$ – Ken Duna Apr 25 '17 at 12:47

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