# If f(x) = ln(1+2x), where a = 2 and n = 3 how do you approximate f by a Taylor polynomial with degree n at the number a?

Feb 21, 2017

$= \ln 5 + \frac{2}{5} \left(x - 2\right) - \frac{2}{25} {\left(x - 2\right)}^{2} + \frac{8}{375} {\left(x - 2\right)}^{3}$

#### Explanation:

The value of $f \left(a\right)$ is straightforward ... it's $\ln \left(1 + 2 a\right)$.

So maybe you want to approximate the behaviour of the function around $x = a$?

If that is the case, you can apply the Taylor Series formulation to third degree:

f(x-a) = f(a) + ((x-a) f'(a))/(1!) + ((x-a)^2 f''(a))/(2!) + ((x-a)^3 f'''(a))/(3!)

And so we need these:

$f = \ln \left(1 + 2 x\right)$

$f ' = \frac{2}{1 + 2 x}$

$f ' ' = - \frac{4}{1 + 2 x} ^ 2$

$f ' ' ' = \frac{16}{1 + 2 x} ^ 3$

We can then say that:

f(x-2) = ln 5 + ((x-2) (2/5))/(1!) + ((x-2)^2 (- 4/25))/(2!) + ((x-2)^3 (16/125))/(3!)

$= \ln 5 + \frac{2}{5} \left(x - 2\right) - \frac{2}{25} {\left(x - 2\right)}^{2} + \frac{8}{375} {\left(x - 2\right)}^{3}$