My friend and I are trying to tackle the following problem:
Write the Taylor series around a = 1 for $xe^{x}$
I approached the problem the following way:
We know that Taylor series for $e^x$ around 1 are:
$f(x) = e^{x} = \Sigma \frac{e^{1}}{n!}(x-1)^{n}$
so if we do g(x) = $xe^{x}$, $g(x) = x(f(x))$ and
$ g(x) = x(\Sigma \frac{e}{n!}(x-1)^{n}) = \Sigma x\frac{e}{n!}(x-1)^{n}$ from n = 0 to infinty
Which, according to WolframAlpha, leads back to $xe^{x}$.
Now, my friend did it the following way: he found f(x) of $xe^{x}$, and then f'(x)=$xe^{x}+e^{x}$, and then f''(x) = $xe^{x} + e^{x} + e^{x}$, and so forth (each iteration added another $e^{x}$). He came up with the following expression:
$\Sigma (n+1)\frac{e(x-1)^{n}}{n!}$ from n=0 to infinity.
I put both equations on a graph and they both seem to approximate $xe^{x}$ fairly well around a=1. So my question is, who has the most acceptable Taylor Series? Is the other answer wrong?
As mentioned in the comments, Taylor series are a special case of power series, which are expressions of the form $$\sum_{n=0}^\infty a_n(x-x_0)^n=a_0+a_1(x-x_0)+\frac12a_2(x-x_0)^2+\dots.$$ What this means is that all the terms in the sum must look like a constant multiplied by a power of $(x-x_0)$. The expression $$\sum_{n=0}^\infty\frac{e}{n!}x(x-1)^n$$ does indeed converge to $xe^x$ (you are correct about this), and it is a series representation of $xe^x$, but it is not the Taylor series expansion---as the terms don't look like $(x-1)^n$. The extra factor of $x$ makes this into not-a-Taylor-series. One way to work around this is to write $$\begin{split}xe^x&=\sum_{n=0}^\infty\frac{e}{n!}(x-1+1)(x-1)^n\\&=e\sum_{n=0}^\infty\frac1{n!}(x-1)^{n+1}+e\sum_{n=0}^\infty\frac1{n!}(x-1)^n\\&=e\sum_{n=1}^\infty\frac{1}{(n-1)!}(x-1)^n+e\sum_{n=0}^\infty\frac1{n!}(x-1)^n\\ &=e+e\sum_{n=1}^\infty\frac{n+1}{n!}(x-1)^n\\&=e\sum_{n=0}^\infty\frac{n+1}{n!}(x-1)^n.\end{split}$$ This is the correct Taylor series representation of $xe^x$.