Is there an easier way to calculate the fifth degree Taylor polynomial?

Question

Here's the problem:

Calculate the fifth degree taylor polynomial at x = 0 for the following function:

$f(x) = (3x^2+4x-2)e^\frac{(-x^2)}{3}$

Now I know that I'm supposed to take the derivative 4 times and plug in 0 in order to get the terms, but I get a really large equation after doing the product rule just once. Is there any way to simplify this to make it easier for myself so that I don't have to do a grueling amount of work?

Answer 1

Insert the Taylor polynomial for $e^{-x^2/3}$ (which is just the Taylor polynomial for $e^x$, only you swap $x$ for $\frac{-x^2}3$), and multiply: $$ f(x)=(3x^2+4x+2)\left(1+\frac{-x^2}{3}+\frac12\left(\frac{-x^2}3\right)^2\right) $$ You don't need higher order terms for $e^{-x^2/3}$, since you're not going higher than degree $5$.

Answer 2

You can replace $e^{-\frac{x^2}{3}}$ by its $5$th degree polynomial (which you get by truncating its Taylor series), and multiply it to $3x^2 + 4x - 2$. You should obtain the correct polynomial, as all higher order terms in the Taylor series of $e^{-\frac{x^2}{3}}$ will multiply to give terms of at least degree $6$. Then, truncate any higher order terms.

Answer 3

Via a back-of-the-napkin calculation, $$\begin{align} f(x) &= (3x^2+4x-2)\exp(-x^2/3)\\ &= 3x^2\exp(-x^2/3)+4x\exp(-x^2/3)-2\exp(-x^2/3)\\ &= 3x^2\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k}}{k!}+4x\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k}}{k!}-2\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k}}{k!}\\ &= 3\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k+2}}{k!}+4\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k+1}}{k!}-2\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k}}{k!}\\ &= 3\sum_{k=1}^\infty\frac{(-3)^{-k-1}x^{2k}}{(k-1)!}+4\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k+1}}{k!}-2\sum_{k=0}^\infty\frac{(-3)^{-k}x^{2k}}{k!}\\ &= -2-\sum_{k=1}^\infty\left(\frac{3\cdot(-3)^{-k-1}}{(k-1)!}-\frac{2\cdot(-3)^{-k}}{k!}\right)x^{2k}+\sum_{k=0}^\infty\left(\frac{4\cdot(-3)^{-k}}{k!}\right)x^{2k+1}\\ &= -2-\sum_{k=1}^\infty a_k x^{2k}+\sum_{k=0}^\infty b_k x^{2k+1}\\ &= -a_0+b_0x-a_1x^2+b_1x^3-a_2x^4+\cdots\qquad{\text{(where we let $a_0=2$)}} \end{align}$$ Even if I am slightly off at some or multiple parts, the idea should be the same.