0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ should be $(2x)^n$ $\endgroup$ – J. W. Tanner Apr 15 at 19:12
  • $\begingroup$ but the plan was good $\endgroup$ – J. W. Tanner Apr 15 at 21:40
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.