# Confused about taylor series first term.

So I ama trying to find the MacLaurin series of $$xe^{-x}$$ and since I know that

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

so $$e^{-x} =$$

$$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}$$

so $$xe^{-x} =$$

$$xe^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n!}$$

So I know this: But when $$n = 0$$, the right side is = x but the f(x) is equal to 0? What is f(a) here? I'm trying to determine $$T_0$$ and $$T_1$$ and I think I'm a bit connfused here.

When figure out $$T_0$$, I just have to plug n = 0 into the equation I got right?

• $a=0$ here..... – amsmath Apr 17 at 16:12

## 1 Answer

Yes that is correct, $$T_0$$ would be zero. This is because $$xe^x$$ evaluated at $$0$$ is $$0$$.

You'll need more terms in the series before it becomes a good approximation.

• So I can't just olug in 0 into the sigma equation to find t0? – Jwan622 Apr 17 at 16:15
• Let $f(x) = xe^x$, $a = 0$, $T_0 = 0*e^0$ ,$f'(x) = (xe^x)' = e^x+xe^x$ , $T_1(x) = (x-0)f'(0) = x$ – George Dewhirst Apr 17 at 16:19
• Oh wait 0^1 is 0 right? Is 0^0 = 0 as well? – Jwan622 Apr 17 at 16:44
• Usually $0^0 = 1$, but I can't see how we use it here! – George Dewhirst Apr 17 at 16:44
• I am trying to find t0 so n would equal 0 right? – Jwan622 Apr 17 at 19:13