Manipulation of Taylor expansion of $e^x$

Question

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!

Answer 1

Note that differentiating and multiplying by $x$ in the power series $\sum_{n=0}^\infty a_nx^n$ gives the power series $(xD)\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty na_nx^n$, where $D$ is the differentiation operator. So in general for any polynomial $p(n)$, we have the identity $$p(xD)\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty p(n)a_nx^n.$$ In this case taking $a_n=1/n!$ gives that the desired expression is $(xD)^2e^x$, or $$ (xD)^2e^x=(xD)xe^x=x(e^x+xe^x)=xe^x+x^2e^x. $$

Answer 2

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

$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = x^2 e^x + x e^x$$

Answer 3

Hint:

$$(x-n)^2=x^2-2nx+n^2=x^2-2nx+n(n-1)+n$$

so that $$\sum_{n=0}^\infty(x-n)^2\frac{x^n}{n!}\\ =\sum_{n=0}^\infty x^2\frac{x^n}{n!}-2\sum_{n=0}^\infty nx\frac{x^n}{n!}+\sum_{n=0}^\infty n(n-1)\frac{x^n}{n!}+\sum_{n=0}^\infty n\frac{x^n}{n!}\\ =x^2\sum_{n=0}^\infty \frac{x^{n}}{n!}-2x\sum_{n=1}^\infty \frac{x^{n}}{(n-1)!}+\sum_{n=2}^\infty\frac{x^n}{(n-2)!}+\sum_{n=1}^\infty\frac{x^n}{(n-1)!}\\ =x^2\sum_{n=0}^\infty \frac{x^{n}}{n!}-2x\sum_{n=0}^\infty \frac{x^{n+1}}{n!}+\sum_{n=0}^\infty\frac{x^{n+2}}{n!}+\sum_{n=0}^\infty\frac{x^{n+1}}{n!}.$$

$$xe^x.$$

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The key to the solution is to decompose $n^2$ in terms of $n$ and $n(n-1)$ and to simplify with the factorials at the denominators. This generalizes to any polynomial in $n$.