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

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?

• should be $(2x)^n$ – J. W. Tanner Apr 15 at 19:12
• but the plan was good – J. W. Tanner Apr 15 at 21:40

## 2 Answers

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 $$f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f''(x)}{6}x^3=\frac{5}{2}x-5x^2+10x^3$$

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

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