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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:

enter image description here

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?

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

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

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