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I have to calculate an approximation for $\ln(1.3)$ using degree $2$ expansion for Taylor polynomial:

$$P_2(x) = f(x_0) + f'(x_0)(x-x_0) + f''(x_0)(x-x_0)^2$$

So I can take $x_0 = 1$ and $x = 1.3$ right?

Then I get

$$P_2(1.3) = \ln(1) + \frac{1}{1}(1.3-1) - \frac{1}{1^2}(1.3-1)^2$$

which gives me $0.21$, but my book says $2.55$. What am I doing wrong?

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    $\begingroup$ That should be $\ln(1)$. Also, the Taylor approximation should be $P_2(x)=f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{f^{\prime\prime}(x_0)}{2}(x-x_0)^2$ $\endgroup$ – Miel Sharf Dec 15 '14 at 15:55
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    $\begingroup$ Your solution is more reasonable - 1.3 is smaller than $e$, so $\ln(1.3)<\ln(e)=1$ $\endgroup$ – Miel Sharf Dec 15 '14 at 15:58
  • $\begingroup$ Well, elementary use of a calculator shows ln(1.3) is roughly .26, so I suspect your book is wrong, unless there's an order of magnitude off...plus it has to be much less than 1, due to Miel's comment $\endgroup$ – Alan Dec 15 '14 at 15:58
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    $\begingroup$ Your book probably should say $0.255$. $\endgroup$ – Michael Lugo Dec 15 '14 at 16:01
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In fact, $\ln 1.3 \approx 0.262$ so you are much closer than the book. The coefficient on the second term should be $-\frac 12$ because you lost the $\frac 1{2!}$ in the Taylor series. You book is badly wrong.

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The 2nd degree Taylor polynomial for $f(x)=\ln x$ at $x=1$ is $$ T_2(x)=0+(x-1)-{1\over 2}(x-1)^2, $$ and $$ T_2(x)\approx \ln x\text{ when }x\approx 1 $$ yields the approximation $$ \ln(1.3)\approx T_2(1.3)=(1.3-1)-{1\over 2}(1.3-1)^2=\color{blue}{0.255}. $$ Looks like your book just put the decimal in the wrong place.

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