Reindexing Exponential Generating Function

Question

I have an exponential generating function, and I need to double check what the teacher said, because I'm having trouble coming to the same result. Also, I need to verify what I am coming up with, and the reasoning behind it.

---

My generating function is: $$xe^{2x} = x\sum_{n=0}^{\infty}{2^n \frac{x^n}{n!}}$$

Next, I multiply through with the x:

$$ \begin{align} &= \sum{2^n \frac{x^{n+1}}{n!}}\\ &=\frac{1}{2}\sum{(n+1)2^{n+1}\frac{x^{n+1}}{(n+1)!}} \end{align} $$

---

And this is where I am stuck. First, how do I index the last sum? Does it go from $n=-1,\text{ or }0$?

And, for the sequence $a(n)$ such that the final sum is $\sum{a(n)\frac{x^n}{n!}}$, what do I write?

---

Right now, my guess is:

$$ \begin{align} xe^{2x} &= \sum_{n=1}^{\infty}{n2^{n-1}\frac{x^n}{n!}}\\ \end{align} $$

Where you can only get away with writing:

$$\sum_{n=0}^{\infty}{...}$$

because the modifier, $n2^{n-1}$ will be zero when $n=0$ anyways...then you can write:

$$a(n) = n2^{n-1}$$

but I think logically, the sum should start from $n=1$ once you multiply the $x$ in there, so does there need to be the step where you handle the a(0) case?

---

Anyhow, my teacher, in an example, wrote: $a(n) = n2^n$ on the board, skipping steps, and I can't get there without adding an extra $2$...and my sum doesn't start from zero unless I do the "it'll be zero anyway" part.

---

update

I went here first; then I remembered Wolfram Alpha. Anyhow, I confirmed that the sequence $a(n)$ is $n2^{n-1}$ with the following two searches:

taylor series for xe^(2x)

n(2^(n-1))/(n!) for n = {0,1,2,3,4,5}

So I guess what I am looking for is a rationale for how to index the sum at zero rather than 1 after multiplying the $x$ through...because I think that the mechanics of doing that are important for being able to define $a(n)$ for all values $n \geq 0$, and that this answer only works out nicely because $x^n$ is multiplied by $n$ (the zero thing I mentioned earlier)...

Answer 1

Here is a slightly different answer, which might also be helpful for this and similar tasks.

Situation: The generating function for the exponential series is already known. \begin{align*} e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} \end{align*} We want to find the exponential series of $xe^{2x}$, i.e. the coefficients $a_n$ of \begin{align*} xe^{2x}=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}\tag{1} \end{align*}

We obtain \begin{align*} xe^{2x}&=x\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}\tag{2}\\ &=x\sum_{n=0}^{\infty}2^n\frac{x^n}{n!}\tag{3}\\ &=\sum_{n=0}^{\infty}2^n\frac{x^{n+1}}{n!}\tag{4}\\ &=\sum_{n=1}^{\infty}2^{n-1}\frac{x^{n}}{(n-1)!}\tag{5}\\ &=\sum_{n=1}^{\infty}n2^{n-1}\frac{x^{n}}{n!}\tag{6}\\ \end{align*} and have finally reached in (6) a represenation of the form \begin{align*} \sum_{n=0}^{\infty}a_n\frac{x^{n}}{n!}\qquad\text{with}\qquad a_n=n2^{n-1} \end{align*} Note, that the index in (6) could also start from $0$, since $a_0=0\cdot2^{-1}=0$ .

Coment:

• In (2) we use the generating function of $\exp$ from (1)

• In (3) we do some rearrangement to isolate $x^n$

• In (4) we multiply $x^n$ with the factor $x$

• In (5) we shift the index by one to obtain $x^n$ again

• In (6) we multiply with $\frac{n}{n}$ since we want $\frac{x^n}{n!}$

Answer 2

I don’t know if this exactly answers your question, but yes, you do sort of have to see that you can add $n=0$ to the summation and not change anything because of the factor of $n$.

$$xe^{2x} = x\sum\limits_{n=0}^{\infty}{2^n \dfrac{x^n}{n!}}=x\left(2^0\frac{x^0}{0!}+2^1\frac{x^1}{1!}+\cdots\right)=\left(2^0\frac{x^1}{0!}+2^1\frac{x^2}{1!}+\cdots\right)$$

$$=\frac{1}{2}\left(2^1\frac{x^1}{0!}+2^2\frac{x^2}{1!}\right)=\frac{1}{2}\left(2^1x^1\frac{1}{1!}+2^2x^2\frac{2}{2!}\right)=\frac{1}{2}\left(2^0x^0\frac{0}{0!}+2^1x^1\frac{1}{1!}+2^2x^2\frac{2}{2!}\right)$$

$$=\frac{1}{2}\left(2^0\cdot0\cdot\frac{x^0}{0!}+2^1\cdot1\cdot\frac{x^1}{1!}+2^2\cdot2\cdot\frac{x^2}{2!}\right)=\left(2^{-1}\cdot0\cdot\frac{x^0}{0!}+2^0\cdot1\cdot\frac{x^1}{1!}+2^1\cdot2\cdot\frac{x^2}{2!}\right)$$

$$=\sum\limits_{n=0}^{\infty}2^{n-1}\cdot n\cdot\frac{x^n}{n!}=\sum\limits_{n=0}^{\infty}n\cdot2^{n-1}\frac{x^n}{n!}$$