Find the Taylor Polynomial $T_{3}$ for the Function $f(x) = \frac{5x}{2+4x}$

Question

Find the Taylor Polynomial $T_{3}$ for the Function $f(x) = \frac{5x}{2+4x}$

So I have this problem and I'm struggling, but below is what I am attempting to do:

Plan: Attempt to translate series into $\frac{1}{1-x}$ form and convert to series $\sum_{n=0}^{\infty}x^n$ then plug in $x^3$ to get the Taylor polynomial.

So here is my attempt: $$f(x) = \frac{5x}{2+4x} = \frac{5x}{2} \frac{1}{1-(-2x)} = \frac{5x}{2}\sum_{n=0}^{\infty}(-1)^n 2x^n$$ Then, I list until I get something with $x^3$: $$5x -\frac{10x^3}{2}$$

However, I don't believe that answer to be right, so what did I do wrong/what can I do to improve?

Answer 1

You're forgetting the term for $n=1$ and you should write $2^nx^n$, not $2x^n$: $$ \frac{5x}{2}\sum_{n=0}^{\infty}(-2x)^n=\frac{5x}{2}\sum_{n=0}^{\infty}(-1)^n2^nx^n $$

You can also easily compute the derivatives: $$ f(x)=\frac{5}{4}\frac{2x}{1+2x}=\frac{5}{4}\left(1-\frac{1}{1+2x}\right) $$ and $f(0)=0$. Therefore \begin{align} f'(x)&=\frac{5}{2}(1+2x)^{-2} & f'(0)&=\frac{5}{2} \\[2px] f''(x)&=-10(1+2x)^{-3} & f''(0)&=-10 \\[6px] f'''(x)&=60(1+2x)^{-4} & f'''(0)&=60 \end{align} Thus the Taylor polynomial of degree $3$ is $$ f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f''(x)}{6}x^3=\frac{5}{2}x-5x^2+10x^3 $$

Answer 2

$$f(x) = \frac{5x}{2+4x} = \frac{5x}{2} \frac{1}{1-(-2x)} = \frac{5x}{2}\sum_{n=0}^{\infty}(-1)^n (2x)^n$$

$$=\frac{5x}2(1-2x+4x^2...)=\frac{5x}2-5x^2+10x^3...$$