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I am trying to figure out the Taylor polynomial of degree $3$, denoted as $T_3(x)$, for $f(x) = xe^{-2x}$.

I am a bit confused about what form the general term of the series needs to be in for me to determine the 3rd degree partial sum. My algorithm for doing this:

  • find the general term equation for the function;
  • find the series for the function assuming one exists (the remainder term is $0$);
  • plug in $0, 1, 2, 3$ into the general term equation to find $T_0$, $T_1$ and $T_2$ and $T_3$.

So we know that:

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!},\quad e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!},\quad xe^{2x} = \sum_{n=0}^{\infty} \frac{2^nx^{n+1}}{n!},\quad xe^{-2x} = \sum_{n=0}^{\infty} \frac{2^n(-x)^{n+1}}{n!}$$

$$xe^{-2x} = \sum_{n=0}^{\infty} \frac{2^n(-1)(x)^{n+1}}{n!}$$

Am I correct so far?

To get the first term (i.e. $T_0(x)$), can't I just plug in $n = 0$ into the general term? Unfortunately, that just gives me $\frac{2^0(-x)^{1}}{0!} = -x$. But that it does not equal $f(0)$ which is $0$. What am I doing wrong?

I have a similar problem: Am I calculating my partial sums correctly? Taylor series.

and I think I am misunderstanding something very basic.

The following is my book's definition of Taylor/Maclaurin:

enter image description here

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  • $\begingroup$ Why doesn't plugging in n=0 give me the right T0? $\endgroup$ – Jwan622 Apr 18 at 22:04
  • $\begingroup$ You don’t want $T_0$, you want $T_3$. But you’re right, evaluating at $0$ will give $T_0$. $\endgroup$ – Lubin Apr 19 at 16:55
  • $\begingroup$ But when I plug in 0 into my equation... it does not work? $\endgroup$ – Jwan622 Apr 19 at 22:47
  • $\begingroup$ You just need to reindex. $T_0$ has to be the constant term ($0$), and $T_1$ has to be the $x$-term. (And note that you have an error just before “right so far?”: $(-x)^{n+1}\ne-x^{n+1}$.) $\endgroup$ – Lubin Apr 20 at 15:27
  • $\begingroup$ But why do I need to reindex? How do I know if I have to reindex? $\endgroup$ – Jwan622 Apr 21 at 16:40
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Since$$e^{2x}=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\cdots,$$then$$e^{-2x}=1-2x+\frac{4x^2}{2!}-\frac{8x^3}{3!}+\cdots$$and so$$xe^{-2x}=x-2x^2+\frac{4x^3}{2!}-\frac{8x^4}{3!}+\cdots$$Therefore,$$T_3(x)=x-2x^2+\frac{4x^3}{2!}.$$

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  • $\begingroup$ Why doesn't plugging in 0 give me the right T0??? $\endgroup$ – Jwan622 Apr 18 at 22:04
  • $\begingroup$ Why should you get it? You only should get $T_0(x)$ if you had an expression of the type $\sum_{n=0}^\infty a_nx^n$. But in your case, the exponent of $x$ is $n+1$, not $n$. $\endgroup$ – José Carlos Santos Apr 18 at 22:07
  • $\begingroup$ Why does it need to be in the x^n form? My book doesn't specify this does it? $\endgroup$ – Jwan622 Apr 18 at 22:23
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    $\begingroup$ Your book mentions$$\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n,$$doesn't it? $\endgroup$ – José Carlos Santos Apr 18 at 22:30
  • $\begingroup$ so I guess the sigma iterator and the exponent have to be equal? weird.... $\endgroup$ – Jwan622 Apr 18 at 22:35

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