I am asked to find the Taylor (Maclaurin) series for $9xe^x$ at $x=0$.
I did the following:
$f(x)=9xe^x \implies f'(x)=9e^x(1+x) \implies f''(x)=9e^x(2+x)$ et cetera. This yields:
$P_0(x)=0, P_1(x)=\frac{9}{1!}x, P_2(x)=\frac{2*9}{2!}x^2$ et cetera. So then I made the series:
$\large \Sigma_{n=0}^\infty \frac{(9n)x^n}{n!}$. What am I doing wrong? Where is my logic failing?
I know what the solution is and the answer, but I would like to know why my logic is not correct. Even my terms are wrong.
The solution:
$e^x=\Sigma_{n=0}^\infty \frac{x^n}{n!} \implies 9xe^x=9\Sigma_{n=0}^\infty \frac{x^{n+1}}{n!}$
Nothing, really. You're not using the fastest, least-work, solution, but that doesn't make yours wrong in any way.
Well, provided your "et cetera" contains a proof, by induction or otherwise, of
$$f^{(n)}(x) = 9e^x(n+x).$$
If you merely spotted a pattern and concluded that it will continue without being able to say why it continues [and a correct explanation of that is easily transformable into a proof, if it is not yet one], that would be a mistake.
No, they are correct, just not simplified as much as one could. You have
$$9xe^x = \sum_{n=0}^\infty \frac{(9n)x^n}{n!}.$$
We can pull the constant factor $9$ of each term out of the series to obtain
$$9\sum_{n=0}^\infty \frac{nx^n}{n!}.$$
Next we observe that the term for $n = 0$ is $0$, so we can drop that from the series and get
$$9\sum_{n=1}^\infty \frac{nx^n}{n!}.$$
Then we recall that for $n \geqslant 1$ we have $n! = n\cdot (n-1)!$, and hence
$$9xe^x = 9\sum_{n=1}^\infty \frac{nx^n}{n\cdot (n-1)!} = 9\sum_{n=1}^\infty \frac{x^n}{(n-1)!}.$$
Finally, we can shift the index of summation, call $n = m+1$, to get
$$9xe^x = 9\sum_{m=0}^\infty \frac{x^{m+1}}{m!},$$
rename the summation index from $m$ to $n$, and lo-and-behold:
$$9xe^x = 9\sum_{n=0}^\infty \frac{x^{n+1}}{n!}.$$