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

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

I know the following:

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I am trying to calculating $T_0$, $T_1$, and $T_2$

So, is this correct?

$$T_0 =\frac{(1)x^1}{1}$$

and at x = 0, both the original function $xe^{-x}$ and the partial sum are $ = 0$ right?

$$T_1 = \frac{(-1)x^2}{1} = -x^2$$

and at x = 1, the original function $xe^{-x} = e^{-1}$ and the partial sum are $ = -1$ right?

$$T_2 = -x^2 + \frac{(1)x^3}{2}$$

and at x = 1, the original function $xe^{-x} = 2*e^{-2}$ and the partial sum are $ = -1 + \frac{1}{2}$ right?

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  • $\begingroup$ A zeroth degree polynomial is constant. Is $x$ constant? $\endgroup$ – Peter Foreman Apr 17 at 20:25
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$T_i$ is a partial sum of original series, not a result of substituting something into it. As $x e^{-x} = 0 + x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + \ldots$, Taylor polynomials at $0$ for $x e^{-x}$ are $T_0(x) = 0$, $T_1(x) = x$, $T_2(x) = x - x^2$ and so on.

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  • $\begingroup$ So I don't get it @mihaild. I can't plug in n = 0 into my summation notation to give me the first term? I thought T_0 is just a partial sum using 1 term so if I plug in n = 0, won't that just give me one term? $\endgroup$ – Jwan622 Apr 17 at 20:47
  • $\begingroup$ So the first term in a Taylor series is f(a) which is f(0) in this case. Using the original function, when x = 0, xe^{-x} = 0. But I Don't get this when I plug x = 0 into the series T_0. Why is this? $\endgroup$ – Jwan622 Apr 17 at 20:54
  • $\begingroup$ If you have your series in form $\sum_{n=0}^\infty a_n x^n$, then yes, $T_0$ is just $a_0$. The series you wrote isn't of this form, it's of form $\sum_{n=0}^\infty a_n x^{n+1}$, so you don't get $T_0$ as just term corresponding to $n = 0$. $\endgroup$ – mihaild Apr 17 at 20:54
  • $\begingroup$ Oh I think that's my problem. That's why I'm getting $T_0$ = x and $T_1 = x - x^2$ right? $\endgroup$ – Jwan622 Apr 17 at 20:55
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    $\begingroup$ Yes. For the first term to be $T_0$ you need to start series from $x^0$. $\endgroup$ – mihaild Apr 17 at 20:58
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No. You already wrote down the series:

$$ x e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n!}$$ You might find it convenient to rewrite the right side in terms of $x^n$ instead of $x^{n+1}$. It helps to use a new name for the index variable. Taking $m = n+1$, we have $$ x e^{-x} = \sum_{m=1}^\infty \frac{(-1)^{m-1} x^m}{(m-1)!} $$

$T_0$ is just the $m=0$ term, which is not there, so $T_0 = 0$.

$T_1$ consists of the $m=0$ and $m=1$ terms: $$T_1 = 0 + \frac{(-1)^{0} x}{0!} = x$$

$T_2$ consists of the $m=0, 1$ and $2$ terms: $$T_2 = 0 + \frac{(-1)^{0} x}{0!} + \frac{(-1)^{1} x^2}{1!} = x - x^2$$

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  • $\begingroup$ When finding the 0 term, I can't just plug in n=0 into the general termformula? $\endgroup$ – Jwan622 Apr 17 at 20:29
  • $\begingroup$ So teh sigma part doesn't emcompass the first term of hte series which is 0? $\endgroup$ – Jwan622 Apr 17 at 20:30
  • $\begingroup$ But isn't what I wrote for the series correct? Doesn't it emcompass the first term i.e. when n = 0? When n = 0, don't we have just x? $\endgroup$ – Jwan622 Apr 17 at 20:39
  • $\begingroup$ Why did I have to rewrite the series? $\endgroup$ – Jwan622 Apr 17 at 20:54

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