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:
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,$$T_3(x)=x-2x^2+\frac{4x^3}{2!}.$$