Question

D) Find the 5th degree Taylor polynomial centered at x = 0 for the function y = \tiny \frac{x}{1+x}

Answer 1

Answer:

$P(x) = x - x^2 + x^3 - x^4+x^5$

Step-by-step explanation:

Let us first remember how a Taylor polynomial looks like:

Given a differentiable function $f$ then we can find its Taylor series to the $nth$ degree as follows:

$P(x) = f(x_{0}) + f'(x_{0}).(x-x_{0}) + \frac{f''(x_{0})}{2!}.(x-x_{0})^2+.....+\frac{f^n(x_{0})}{n!}.(x-x_{0})^n + R_{n}(x).(x-x_{0})^n$

Where $R_{n}(x)$ represents the Remainder and $f^n(x)$ is the $nth$ derivative of $f$.

So let us find those derivatives.

$f(x) = \frac{x}{1+x}\\f'(x) = \frac{1}{(1+x)^2}\\f''(x) = \frac{-2}{(1+x)^3}\\f'''(x) = \frac{6}{(1+x)^4}\\f''''(x) = \frac{-24}{(1+x)^5}\\f'''''(x) = \frac{120}{(1+x)^6}$

The only trick for this derivatives is for the very first one:

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

Then it's only matter of replacing on the Taylor Series and replacing $x_{0}=0$