Manipulation of Taylor expansion of $e^x$ I have an expression:
$$ f(x) = \sum\limits_{N=0}^\infty (N-x)^2 \frac{x^N}{N!}$$
I want to figure out whose Taylor Expansion this is.
I've found that I can separate out the above expression, and get the following:
$$f(x) = \sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} - 2 \sum\limits_{N=0}^\infty xN \frac{x^N}{N!} + \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!}$$
I figured out this much:
$$x^2 e^x = \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!} = \sum\limits_{N=0}^\infty xN \frac{x^N}{N!}$$
$$xe^x = \sum\limits_{N=0}^\infty N \frac{x^N}{N!}$$
But I cannot figure out who the first mysterious sum belongs to!

Let's look at the mysterious sum. First, let's just write out some of the terms
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = 0 + x + \frac{4 x^2}{2!} + \cdots$$
so the first term is zero, let's drop it
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} =\sum\limits_{N=1}^\infty N^2 \frac{x^N}{N!}$$
and then shift the index by saying $N=n+1$
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty (n+1)^2 \frac{x^{n+1}}{(n+1)!}$$
now let's cancel out one of the $n+1$ in the numerator with its match in the denominator
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty (n+1) \frac{xx^n}{n!}$$
And split the sum
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty x n \frac{x^n}{n!} + \sum\limits_{n=0}^\infty x \frac{x^n}{n!}$$
The first term is the same as what you identified for $x^2 e^x$ and the second term is $x$ times the series for $e^x$. Therefore
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = x^2 e^x + x e^x$$
