Taylor Series of $2xe^x$ I have to find the Taylor Series for $2xe^x$ centred at $x=1$. I came up with the following.
$$e^x = e^{x-1} \times e = e \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg)$$
Then consider $2xe^x$.
$$2xe^x = 2xe \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg) = \bigg( \sum_{n=0}^\infty \frac{2xe(x-1)^n}{n!}\bigg) $$
I am wondering whether this is a right Taylor Series centred at $x=1$. I do understand that the $(x -1)^n$ implies that it is centred at $1$. But, I am very not sure with my answer since there are two terms of $x$, that is $x$ and $(x-1)^n$. 
Any clarification and explanation would be highly appreciated.
 A: The expression you have is correct, but I doubt it's what they're looking for. Namely, you'd like to have only terms of $(x-1)^n$ and none of $x$ (since otherwise, it's not really "centered" at $x=1$). It would be a fine solution, as comments suggest, to write $x=(x-1)+1$ and reconstruct the series with that, but an alternate solution which is more calculus-y and makes use of the fact that $e^x$ is a cool function to multiply other things with:
In particular, let $f(x)=2xe^x$. We can easily compute the first few derivatives via the product rule:
$$f'(x)=2xe^x+2e^x$$
$$f''(x)=2xe^x+4e^x$$
$$f'''(x)=2xe^x+6e^x$$
and so on - notice that we always have a term of the form $2xe^x$ since the derivative of $e^x$ is $e^x$. More generally, we could write $f'(x)=f(x)+2e^x$ to capture this recurrence - which allows us to expand, for instance:
$$f^{(n+1)}(x)=f^{(n)}(x)+\frac{d^n}{dx^n}2e^x=f^{(n)}(x)+2e^x$$
meaning that, for each derivative, we just add a new term of the form $2e^x$. So, we get, in closed form, that:
$$f^{(n)}(x)=2xe^x + 2ne^x$$
and this is trivial to evaluate and use to make a Taylor series.
A: First write down Taylor series for $2xe^x$ centered at $x=0$.
$$2xe^x=2x \cdot \sum_{n=0}^{\infty}\frac{x^n}{n!}=\sum_{n=0}^{\infty}\frac{2x^{n+1}}{n!}$$
Now put $x=t-1$, you have:
$$2(t-1)e^{t-1}=\sum_{n=0}^{\infty}\frac{2(t-1)^{n+1}}{n!}$$
But $2(t-1)e^{t-1}=e^{-1}2te^{t}-e^{-1}2e^{t}$ and you have Taylor expansion for $e^{-1}2e^{t}$ at $t=1$,so:
$$e^{-1}2te^{t}-e^{-1}2e^{t}=e^{-1}2e^{t}- \sum_{n=0}^\infty \frac{(t-1)^n}{n!}=\sum_{n=0}^{\infty}\frac{2(t-1)^{n+1}}{n!}=\sum_{n=1}^{\infty}\frac{2(t-1)^{n}}{(n-1)!}$$
Finally:
$$e^{-1}2e^{t}=\sum_{n=0}^\infty \frac{(t-1)^n}{n!}+\sum_{n=1}^{\infty}\frac{2(t-1)^{n}}{(n-1)!}$$
A: Your first insight is valid, you can rewrite the exponential as $e^x=e\cdot e^{x-1}$. But then you should also transform the factor $2x$ using $2x=2(x-1)+2$.
Then,
$$2xe^x=(2(x-1)+2)e\sum_{n=0}^\infty \frac{(x-1)^n}{n!}
\\=2e\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{n!}+2e\sum_{n=0}^\infty \frac{(x-1)^{n}}{n!}
\\=2e\sum_{n=1}^\infty \frac{(x-1)^{n}}{(n-1)!}+2e\sum_{n=0}^\infty \frac{(x-1)^{n}}{n!}
\\=2e+2e\sum_{n=1}^\infty \frac{(n+1)}{n!}(x-1)^{n}.$$
A: Write $2xe^x=2(x-1)e^x+2e^x$ and $e^x=e\cdot e^{x-1}$ to obtain $2xe^x=2e(x-1)e^{x-1}+2e\cdot e^{x-1}$. Since
$$
e^{x-1}=\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k,
$$
we have
\begin{align*}
2xe^x&=2e(x-1)e^{x-1}+2e\cdot e^{x-1} \\ &= 2e(x-1)\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k+2e
\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k \\ &=
2e\left(\sum_{k=1}^\infty\frac{1}{(k-1)!}(x-1)^k+\sum_{k=1}^\infty\frac{1}{k!}(x-1)^k+1\right) \\ &=
2e\left(1+\sum_{k=1}^\infty\left(\frac{1}{(k-1)!}+\frac{1}{k!}\right)(x-1)^k\right) \\ &=
2e\sum_{k=0}^\infty\frac{k+1}{k!}(x-1)^k.
\end{align*}
A: The classical way:
$$\begin{align}&f(x)=2xe^x,&f(1)=2e
\\&f'(x)=2e^x+2xe^x,&f'(1)=4e
\\&f''(x)=2e^x+2e^x+2xe^x=4e^x+2xe^x,&f''(1)=6e
\\&f'''(x)=4e^x+2e^x+2xe^x=6e^x+2xe^x,&f'''(1)=8e\end{align}$$
$$...$$
More generally,
$$f^{(n)}(1)=2(n+1)e,$$
and
$$2xe^x=2e\sum_{n=0}^\infty\frac{n+1}{n!}(x-1)^n.$$
