# Am I calculating my partial sums correctly? Taylor series.

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:

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?

• A zeroth degree polynomial is constant. Is $x$ constant? – Peter Foreman Apr 17 at 20:25

$$T_i$$ 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.

• 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? – Jwan622 Apr 17 at 20:47
• 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? – Jwan622 Apr 17 at 20:54
• 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 $n = 0$. – mihaild Apr 17 at 20:54
• Oh I think that's my problem. That's why I'm getting $T_0$ = x and $T_1 = x - x^2$ right? – Jwan622 Apr 17 at 20:55
• Yes. For the first term to be $T_0$ you need to start series from $x^0$. – mihaild Apr 17 at 20:58

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: $$T_2 = 0 + \frac{(-1)^{0} x}{0!} + \frac{(-1)^{1} x^2}{1!} = x - x^2$$

• When finding the 0 term, I can't just plug in n=0 into the general termformula? – Jwan622 Apr 17 at 20:29
• So teh sigma part doesn't emcompass the first term of hte series which is 0? – Jwan622 Apr 17 at 20:30
• 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? – Jwan622 Apr 17 at 20:39
• Why did I have to rewrite the series? – Jwan622 Apr 17 at 20:54