0
$\begingroup$

So i have a function: $$f: \mathbb{R} \to \mathbb{R}$$ $$f(x)=(1-4x^2)e^{2x}$$

I have to formulate it's Taylor Series around x=0 (so i think it is a Maclaurin series if i recall correctly). And then i have to use it as help for calculating sum : $$ \sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!}$$

So i know i will have to use the sum of $e^x$ that is a known Maclaurin series. Butother than that, help would be appreciated.

$\endgroup$
1
$\begingroup$

Since the taylor series for $f(x)=(1-4x^2)e^{2x}$ is $$\sum_{n=0}^{1} \frac{2^n}{n!} x^n- \sum_{n=2}^{\infty} \frac{(2)^n(n^2 -n-1)}{n!}x^n$$

Let $x = -1$. $$\sum_{n=0}^{1} \frac{2^n}{n!} (-1)^n- \sum_{n=2}^{\infty} \frac{(2)^n(n^2 -n-1)}{n!}(-1)^n = -1- \sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!} = -3/e^2 $$

Then $$\sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!} = 3/e^2 -1$$

$\endgroup$
  • $\begingroup$ How do you get the Taylor series ? $\endgroup$ – Yves Daoust Aug 1 '16 at 7:43
  • $\begingroup$ @MathIsTheWayOfLife you are welcome. $\endgroup$ – Zack Ni Aug 1 '16 at 7:45
2
$\begingroup$

Notice that for $n\ge2$,

$$\frac{n^2-n-1}{n!}=\frac1{(n-2)!}-\frac1{n!}$$

so that

$$\sum_{n=2}^\infty\frac{n^2-n-1}{n!}(2x)^n=\sum_{n=2}^\infty(2x)^2\frac{(2x)^{n-2}}{(n-2)!}-\sum_{n=2}^\infty\frac{(2x)^n}{n!}=(2x)^2e^{2x}-\left(e^{2x}-1-2x\right)\\ =(4x^2-1)e^{2x}+1+2x.$$

Then with $x=-1$,

$$\sum_{n=2}^\infty\frac{n^2-n-1}{n!}(-2)^n=3e^{-2}-1.$$

$\endgroup$

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.