Maclaurin-Series of $xe^x$

Question

I want to find the Maclaurin Series of $f(x) = xe^x$.

So naturally I derived the function a bunch of times and found that:

$f(x) = e^x(x+0)$
$f'(x) = e^x(x+1)$
$f''(x) = e^x(x+2)$
$f'''(x) = e^x(x+3)$
...
$f^{(n)}(x) = e^x(x+n)$

And with $x_0 = 0$ I get $f^{(n)}(x_0) = n$

Now the formula for building this series is $\sum_{k=0}^{\infty}\frac{f^{(k)(x_0)}}{k!}x^k$

So my question is why is this series $\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}$ and not $\sum_{k=0}^{\infty}\frac{k}{k!}x^k$ ?

I'm new to the subject and might have got something totally wrong. In any case, thanks a lot for your answers!

Answer 1

This is just a matter of re-indexation.

$$\sum\limits_{k=0}^\infty \dfrac{kx^k}{k!}=\sum\limits_{k=1}^\infty \dfrac{kx^k}{k!}=\sum\limits_{k=1}^\infty \dfrac{x^k}{(k-1)!}=\sum\limits_{n=0}^\infty \dfrac{x^{n+1}}{n!}=\sum\limits_{k=0}^\infty \dfrac{x^{k+1}}{k!}$$

• first $=$ : the value is $0$ for $k=0$ so we remove this term
• second $=$ : we simplify $k/k!=1/(k-1)!$
• third $=$ : re-indexing $n=k-1$
• fourth $=$ : $n$ being a dummy variable, we can rename it $k$ again.

Answer 2

(i)write down the Maclaurn series for $e^x$. Ifyou don't know it, it can be found esily, since the $n$-th derivative of $e^x$ is $e^x.$ (ii) Multiply by $x$.You now have a power series for $xe^x$. A power series is its own Maclaurin series. so you are finished.