notation for first and second derivatives of a power series

Question

I have a power series $$\sum_{k=0}^\infty\frac{c_k}{k!}x^k$$ where $c_k$ is an arbitrary $k$-th term of some sequence. Then $$\frac{d^2}{dx^2}\left[\sum_{k=0}^\infty\frac{c_k}{k!}x^k\right]=\frac{d}{dx}\left[\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k-1}\right]=\sum_{k=0}^\infty\frac{k(k-1)\cdot c_k}{k!}x^{k-2}$$ Now, is it okay to leave this like this? I know that the first two terms are $0$, which removes the $1/x^2 $ term when $k=0$ and $1/x$ term when $k=1$. I figure this is no problem, but perhaps I am wrong. Should I re-index or can I leave it like this?

EDIT: One of the reasons I'm posting this is I have an expression with a polynomial in front of the power series. For example, $$(x-x^2)\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k-1}$$ In this example, I have not re-indexed as has been mentioned in the previous posts and comments. Now, since I haven't re-indexed, and if I distribute I get $$\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k}+\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k+1}$$ Now, given that there is a polynomial coefficient in front of the derivative of a power series, can I NOT re-index and distribute first?

Answer 1

We should keep in mind the following:

If there is a valid expression then

• re-indexing by adding one or more zeros or skipping one or more zeros does not alter the value of the expression

• index shifting does not alter the value of the expression

Therefore we conclude:

• Re-indexing and index shifting of an expression are only referring to presentational aspects and so they are always only a matter of convenience.

So, it belongs only to your valuation if re-indexing or index-shifting is convenient. But as you already know, a proper presentation is often beneficial since it can be used to

• simplify the expression

• clarify aspects you want to indicate

• make further transformations of the expression easier (or possible :-))

• etc.

So, let's have a look at your first question:

Is it ok to leave the power series like this?

\begin{align*} \sum_{k=0}^\infty\frac{k(k-1)c_k}{k!}x^{k-2} \tag{1} \end{align*}

The answer is: Of course it's ok, since nothing is wrong with it.

BUT, presumably the presentation is not convenient because as you correctly noted, the first two summands are \begin{align*} \frac{0\cdot(-1)\cdot c_0}{0!}x^{-2}=0&\qquad\qquad(k=0)\\ \frac{1\cdot0\cdot c_1}{1!}x^{-1}=0&\qquad\qquad(k=1)\\ \end{align*} Therefore we could simplify the expression (1) by re-indexing, index shifting and cancelling and also indicate thereby which summands are more interesting

\begin{align*} \sum_{k=0}^\infty\frac{k(k-1)c_k}{k!}x^{k-2}&=\sum_{k=2}^\infty\frac{k(k-1)c_k}{k!}x^{k-2}\\ &=\sum_{k=2}^\infty\frac{c_k}{(k-2)!}x^{k-2}\\ &=\sum_{k=0}^\infty\frac{c_{k+2}}{k!}x^{k}\tag{2}\\ \end{align*}

Observe, that the presentation (2) represents exactly the same value as (1) and so they are completely interchangeable.

Now let's go on with your other question regarding:

Can you distribute without re-indexing the following: \begin{align*} (x-x^2)\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k-1}= \sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k}-\sum_{k=0}^\infty\frac{k\cdot c_k}{k!}x^{k+1} \end{align*} The answer is again: Of course you can, since multiplication with $x$ and $x^2$ as well as applying the distributive law are valid transformations regarding this expression.

Any re-indexing or index-shifting first or second, yes or no simply don't matter, since this is only a presentional issue and belongs exclusively to your convenience.

---

(Important) Note: While re-indexing and index-shifting is harmless, we should be careful when cancellation comes into play. Observe: \begin{align*} \sum_{k=0}^\infty\frac{k(k-1)c_k}{k!}x^{k-2} \neq \sum_{k=0}^\infty\frac{c_k}{(k-2)!}x^{k-2}\tag{3}\\ \sum_{k=2}^\infty\frac{k(k-1)c_k}{k!}x^{k-2} = \sum_{k=2}^\infty\frac{c_k}{(k-2)!}x^{k-2}\tag{4} \end{align*} Do you see the crucial difference? While cancellation in (4) causes no problems, since $\frac{k(k-1)}{k!}=\frac{1}{(k-2)!}$ is well defined for $k\geq 2$, the cancellation of the first two summands with $k=0$ and $k=1$ in (3) is not valid! The cancellation transforms these summands into

\begin{align*} \frac{0\cdot(-1)\cdot c_0}{0!}x^{-2}=0&\qquad\longrightarrow\qquad\frac{c_0}{(-2)!}x^{-2}&\qquad(k=0)\\ \frac{1\cdot0\cdot c_1}{1!}x^{-1}=0&\qquad\longrightarrow\qquad\frac{c_1}{(-1)!}x^{-2}&\qquad(k=1)\\ \end{align*}

The crucial fact here is, that cancellation removes the factor $0$ leaving an invalid expression. Using these invalid summmands erroneously in further calculations may lead to strange results!

---

(Harmless) Note: You may expect that we always skip zeros in power series in order to simplify sums. This is not the case. Sometimes it's in fact convenient to add zeros to gain a better presentation. As an example you may have a look at my answer to this question; see number (3).

Answer 2

It is

$$\frac{d^2}{dx^2}\left[\sum_{k=0}^\infty\frac{c_k}{k!}x^k\right]=\frac{d}{dx}\left[\sum_{k=1}^\infty\frac{k\cdot c_k}{k!}x^{k-1}\right]=\sum_{k=2}^\infty\frac{k(k-1)\cdot c_k}{k!}x^{k-2}.$$

Note that

$$\frac{d^2}{dx^2}\left[c_0+c_1x+\sum_{k=2}^\infty\frac{c_k}{k!}x^k\right]=\frac{d}{dx}\left[c_1+\sum_{k=1}^\infty\frac{k\cdot c_k}{k!}x^{k-1}\right]=\sum_{k=2}^\infty\frac{k(k-1)\cdot c_k}{k!}x^{k-2}.$$

So, you must reindex in your solution.