Formal Power Series as Linear Operators

Let $t^k$ act as the $k$-th derivative operator on the set of polynomials. So

$$t^k(x^n)=t^k x^n=(n)_kx^{n-k}$$

where $(n)_k=n(n-1)(n-2)...(n-k+1)$ is the falling factorial. Then with a formal power series, $f(t)=\sum_{k\ge 0}a_k\frac{t^k}{k!}$, the linear operator $f(t)$ acts as such that

$$f(t)(x^n)=f(t)x^n=\sum_{k=0}^n\binom{n}{k}a_k x^{n-k}$$

Therefore, depending on the coefficients of the power series, we can get some interesting binomial identites. For example, if $f(t)=e^{yt}$, since the coefficients $a_n=y^n$, we get

$$e^{yt}x^n=\sum_{k=0}^n\binom{n}{k}y^k x^{n-k}=(x+y)^n$$

by linearity,

$$(e^{yt}-1)x^n=(x+y)^n-x^n=\sum_{k=1}^{n}\binom{n}{k}y^k x^{n-k}$$

and perhaps not as obvious

$$\left(\frac{e^{yt}-1}{t}\right)x^n=\int_{x}^{x+y}u^ndu$$

Now suppose that $f(t)=e^{yt}-1-yt$. Then

$$(e^{yt}-1-yt)x^n=(x+y)^n-x^n-ynx^{n-1}=\sum_{k=2}^{n}\binom{n}{k}y^k x^{n-k}$$

Obviously there is a nice formed forward difference equation in the previous case that is not happening here. But there is a relationship with subtracted terms of the binomial expansion. What i would really like help understanding is whether or not a possible analogous integral representation exists for the following operator:

$$\left(\frac{e^{yt}-1-yt}{t^2}\right)x^n=\left(\sum_{k=0}^\infty\frac{y^{k+2}}{(k+2)(k+1)}\frac{t^k}{k!}\right)x^n=\sum_{k=0}^n\binom{n}{k}\frac{y^{k+2}}{(k+1)(k+2)}x^{n-k}$$

$$=\sum_{k=0}^n\binom{n+2}{k+2}\frac{y^{k+2}}{(n+1)(n+2)}x^{n-k}=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$

It is not as simple. Clearly $\frac{d^2}{dx^2}\frac{x^{n+2}}{((n+2)(n+1)}$. If I integrated below I think the math is correct

$$\int_x^{x+y}{\frac{u^{n+1}}{n+1}}du=\frac{1}{(n+1)(n+2)}\sum_{k=1}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$

Which is really close, but the lower bound on the summation is $1$, not $2$. Does any one have any insight in how i can fix this, if possible?

• did you try rewriting everything in term of convolution products ? $D g = \delta' \ast g$ and $D^ {-1} g = 1_{x > 0} \ast g$ – reuns Oct 7 '16 at 20:50
• The reason it is in $t$ is that the above comes from Umbral Calculus. Here, $\mathcal{F}$ is the set of all formal power series in $t$ over an arbitrary field $F$ with characteristic 0. That is why it is in $t$. – Eleven-Eleven Oct 7 '16 at 20:51
• The usual notation is $f(t) = \sum_{k\ge 0} c_k t^k\implies f(D) g =(\sum_{k\ge 0} c_k D^k) g$, compatible with the holomorphic functional calculus and things like this – reuns Oct 7 '16 at 20:53
• Can you elaborate more with references as well? – Eleven-Eleven Oct 7 '16 at 21:01
• That's the impression I've got from trying to understand the question. But upon reading the Wikipedia reference, I now see that Umbral calculus is quite a different kind of thing. – Han de Bruijn Oct 17 '16 at 15:02

Note: OPs calculations are quite ok and it shows the operators are closely related, but different. I don't think there is a necessity to fix anything.

I skimmed through the classic The Umbral Calculus by Steven Roman, but there was no indication that something more is going on regarding OPs question. Another source I've checked without success was The Calculus of Finite Differences by C. Jordan.

It might be helpful to list a few higher powers of the operators under consideration.

In the following I use OPs notation which is precisely the same used by Steven Roman.

Translation operator: $e^{yt}$

Since this operator satisfies: \begin{align*} e^{yt}x^n&=\sum_{k=0}^\infty \frac{y^k}{k!}t^kx^n =\sum_{k=0}^n \frac{y^k}{k!}(n)_kx^{n-k}=\sum_{k=0}^n\binom{n}{k}y^kx^{n-k}\\ &=(x+y)^n \end{align*} we obtain \begin{align*} \left(e^{yt}\right)^2 x^n=e^{yt}(x+y)^n=(x+2y)^n \end{align*} and in general for $j\geq 1$ \begin{align*} e^{jyt}x^n=(x+jy)^n \end{align*}

Since $x^n, n\geq 0$ form a basis of the vector space of all polynomials $p$ in a single variable $x$ and the translation operator is linear, we obtain \begin{align*} e^{jyt}p(x)=p(x+jy)\tag{1} \end{align*}

hence the name translation operator.

Forward difference operator: $e^{yt}-1$

Here we obtain for polynomials $p$ using (1)

\begin{align*} \left(e^{yt}-1\right)p(x) = p(x+y)-p(x) \end{align*}

The next one is

Operator: $\frac{\exp(yt)-1}{t}$

We obtain \begin{align*} \left(\frac{e^{yt}-1}{t}\right)x^n&=\sum_{k=1}^\infty \frac{y^k}{k!}t^{k-1}x^n\\ &=\sum_{k=1}^{n+1}\frac{y^k}{k!}(n)_{k-1}x^{n-(k-1)}\\ &=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}y^kx^{n+1-k}\\ &=\frac{1}{n+1}\left((x+y)^{n+1}-x^{n+1}\right)\tag{2}\\ &=\frac{1}{n+1}\int_x^{x+y}u^n\,du \end{align*}

Similarly we obtain from (2) by linearity \begin{align*} \left(\frac{e^{yt}-1}{t}\right)^2x^n &=\left(\frac{e^{yt}-1}{t}\right)\frac{1}{n+1}\left((x+y)^{n+1}-x^{n+1}\right)\\ &=\frac{1}{n+1}\left[\frac{1}{n+2}\left((x+2y)^{n+2}-(x+y)^{n+2}\right)\right.\\ &\qquad\qquad\quad\left.-\frac{1}{n+2}\left((x+y)^{n+2}-x^{n+2}\right)\right]\\ &=\frac{1}{(n+1)(n+2)}\left((x+2y)^{n+2}-2(x+y)^{n+1}+x^{n+2}\right)\\ &=\frac{1}{(n+2)_2}\left(\int_{x+y}^{x+2y}u\, du-\int_x^{x+y}u\,du\right)\tag{3} \end{align*}

Here at (2) and (3) we can see quite nicely how the operator $\frac{\exp(yt)-1}{t}$ is connected with the integral operator. It can be extended to higher powers without too much effort and the relationship with the integral operator looks plausible.

Operator: $\frac{\exp(yt)-1-t}{t^2}$

Now we take a look at the operator which is on the focus of OP and its generalisation.

\begin{align*} \left(\frac{e^{yt}-1-t}{t^{2}}\right)x^n &=\sum_{k=2}^\infty\frac{y^k}{k!}t^{k-2}x^n\\ &=\sum_{k=2}^{n}\frac{y^{k}}{k!}(n)_{k-2}x^{n-(k-2)}\\ &=\frac{1}{(n+2)_2}\sum_{k=2}^n\binom{n+2}{k}y^kx^{n+2-k}\\ &=\frac{1}{(n+2)_2}\left((x+y)^{n+2}-x^{n+2}-nyx^{n+1}\right) \end{align*}

Comparing the final expression with (3) we do not see a plausible representation via integrals since the term $nyx^{n+1}$ don't provide anything nicely of the form \begin{align*} \text{integrated expression (end point) - integrated expression (starting point)} \end{align*}

This impression becomes more strongly when looking at the general case. We obtain for $j\geq 1$ \begin{align*} \left(\frac{e^{yt}-1-\frac{t^2}{2}-\cdots-\frac{t^{j-1}}{(j-1)!}}{t^j}\right)x^n &=\sum_{k=j}^\infty\frac{y^k}{k!}t^{k-j}x^n\\ &=\frac{1}{(n+j)_j}\sum_{k=j}^\infty\binom{n+j}{k}y^kx^{n+j-k}\\ &=\frac{1}{(n+j)_j}\left((x+y)^{n+j}-\sum_{k=0}^{j-1}\binom{n+j}{k}y^kx^{n+j-k}\right) \end{align*}

• I upvoted... I just haven't voted for the 50 yet... – Eleven-Eleven Oct 16 '16 at 0:28
• @Eleven-Eleven: Many thanks for your friendly comment! Regards, – Markus Scheuer Oct 16 '16 at 7:09
• @Eleven-Eleven: Thanks a lot for accepting my answer and granting the bounty! :-) – Markus Scheuer Oct 16 '16 at 19:16
• No problem. I was glad to see your answer and I have been looking at general cases as you did above. I had the feeling the analogous integral expression may have been a far stretch, but there is still a nice relationship regardless. – Eleven-Eleven Oct 16 '16 at 19:48
• @Eleven-Eleven: Thanks, you're welcome and I agree with your impression. I strongly recommend to go through Steven Romans small booklet umbral calculus as it provides a lot of related interesting information. – Markus Scheuer Oct 16 '16 at 19:55

Introductionary reference:

So why not replace $\,t\,$ by the differential operator $\,d/dx\,$ everywhere it occurs? Next reference:

Therefore we have in general and especially: $$e^{y\,d/dx} f(x) = f(x+y) \quad \Longrightarrow \quad e^{y\,d/dx} x^n = (x+y)^n$$ Indeed the operator $1/(d/dx)$ is the inverse of differentiation, which is integration. So let's take a look at your perhaps not as obvious expression: $$\left(\frac{e^{y\,d/dx}-1}{d/dx}\right)x^n=\frac{1}{d/dx}\left(e^{y\,d/dx}-1\right)x^n\\ =\int\left[(x+y)^n - x^n\right] dx = \frac{(x+y)^{n+1}-x^{n+1}}{n+1} + C$$ with $C$ an arbitrary constant. Evaluation is not ambiguous, because the operators $1/(d/dx)$ and $\left(e^{y\,d/dx}-1\right)$ do indeed commute. Which means that we can do algebra with these operators as with common numbers. The same in, last but not least: $$\left(\frac{e^{y\,d/dx}-1-y\,d/dx}{(d/dx)^2}\right)x^n=\int\left\{\int\left[(x+y)^n-x^n\right]dx\right\}dx-\int y\,x^n\,dx \\=\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + Cx + D$$ with $C$ and $D$ arbitrary constants. This result is deviant from your last formula. It is noticed that: $$\frac{d^2}{dx^2} \left[\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + C\,x + D\right] = (x+y)^n - x^n - y\frac{dx^n}{dx}$$ EDIT. I almost forgot to mention the stuff where it's all about: the formal power series (Wikipedia). $$(y+x)^{n}=\sum _{k=0}^{n}{n \choose k}y^{n-k}x^{k} \quad \Longrightarrow \\ \frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} =\\ \frac{1}{(n+1)(n+2)}\left[\sum _{k=0}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k} - x^{n+2} - (n+2)y\,x^{n+1}\right]=\\ \frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k} + {n+2 \choose 0}y^0 x^{n+2} - x^{n+2} + {n+2 \choose 1}y^1 x^{n+1} - (n+2)y\,x^{n+1} \right]\\ =\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}\right]$$ So here is your fix on the lower bound of the summation.