# How do you find the third degree Taylor polynomial for f(x)= ln x, centered at a=2?

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Daan J. Share
Jul 21, 2015

$\ln \left(2\right) + \frac{1}{2} \left(x - 2\right) - \frac{1}{8} {\left(x - 2\right)}^{2} + \frac{1}{24} {\left(x - 2\right)}^{3}$.

#### Explanation:

The general form of a Taylor expansion centered at $a$ of an analytical function $f$ is f(x)=sum_{n=0}^oof^((n))(a)/(n!)(x-a)^n. Here ${f}^{\left(n\right)}$ is the nth derivative of $f$.

The third degree Taylor polynomial is a polynomial consisting of the first four ($n$ ranging from $0$ to $3$) terms of the full Taylor expansion.

Therefore this polynomial is $f \left(a\right) + f ' \left(a\right) \left(x - a\right) + \frac{f ' ' \left(a\right)}{2} {\left(x - a\right)}^{2} + \frac{f ' ' ' \left(a\right)}{6} {\left(x - a\right)}^{3}$.

$f \left(x\right) = \ln \left(x\right)$, therefore $f ' \left(x\right) = \frac{1}{x}$, $f ' ' \left(x\right) = - \frac{1}{x} ^ 2$, $f ' ' ' \left(x\right) = \frac{2}{x} ^ 3$. So the third degree Taylor polynomial is:
$\ln \left(a\right) + \frac{1}{a} \left(x - a\right) - \frac{1}{2 {a}^{2}} {\left(x - a\right)}^{2} + \frac{1}{3 {a}^{3}} {\left(x - a\right)}^{3}$.

Now we have $a = 2$, so we have the polynomial:
$\ln \left(2\right) + \frac{1}{2} \left(x - 2\right) - \frac{1}{8} {\left(x - 2\right)}^{2} + \frac{1}{24} {\left(x - 2\right)}^{3}$.

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