A curious connection : What's the function $f(x)$?

Solution 1:

Not a full answer, but a set of equations to exploit.

Let $$ g(x) = f'(x)=\sum_{k=1}^{\infty}\frac{k f'(k)}{k!}(-1)^k x^{k-1}= \sum_{k=0}^{\infty}\frac{g(k+1)}{k!}(-1)^{k+1} x^{k} $$ But also, via Taylor expansion $$ g(x) = \sum_{k=0}^{\infty}\frac{g^{(k)}(0)}{k!} x^{k} $$ So we obtain the set of equations, for all $k=0 \cdots \infty$: $$ g^{(k)}(0) = {g(k+1)}(-1)^{k+1} $$

The first few are: $$ g(0) = - g(1)\\ g'(0) = g(2)\\ g''(0) = - g(3)\\ g^{(3)}(0) = g(4)\\ \cdots $$

Can we proceed from here?

Solution 2:

Hint: There is a relationship of $f(x)$ with the Stirling numbers of the second kind.

Using the ansatz $f(x)=\sum_{j=0}^{\infty}a_j\frac{x^j}{j!}$ we obtain \begin{align*} \color{blue}{f(x)}&=\sum_{k=0}^\infty \frac{f^\prime(k)}{k!}(-x)^k\\ &=\sum_{k=0}^\infty\frac{d}{du}\left.\left(\sum_{j=0}^\infty a_j\frac{u^j}{j!}\right)\right|_{u=k}\frac{(-x)^k}{k!}\\ &=\sum_{k=0}^\infty\left.\left(\sum_{j=1}^\infty a_j\frac{u^{j-1}}{(j-1)!}\right)\right|_{u=k}\frac{(-x)^k}{k!}\\ &=\sum_{k=0}^\infty\left(\sum_{j=0}^\infty a_{j+1}\frac{k^j}{j!}\right)\frac{(-x)^k}{k!}\\ &=\sum_{j=0}^\infty\frac{a_{j+1}}{j!}\sum_{k=0}^\infty\frac{k^j}{k!}(-x)^k\\ &=\sum_{j=0}^\infty\frac{a_{j+1}}{j!}e^{-x}\sum_{k=0}^j{j\brace k}(-x)^k\tag{1}\\ &=\sum_{j=0}^\infty \frac{a_{j+1}}{j!}[t^j]e^{-xe^t}\tag{2}\\ &=[t^0]e^{-xe^t}\sum_{j=0}^\infty a_{j+1}\frac{t^{-j}}{j!}\\ &\,\,\color{blue}{=[t^0]e^{-xe^t}\left(\left.f^{\prime}(x)\right|_{x=\frac{1}{t}}\right)}\tag{3} \end{align*}

According to (3) the bivariate formal Laurent series $F(x,t)=e^{-xe^t}\sum_{j=0}^\infty a_{j+1}\frac{t^{-j}}{j!}$ might be useful to find a representation of $f(x)$.

Comment:

• In (1) we note that $\sum_{k=0}^\infty \frac{k^j}{k!}x^k$ admits a representation via Stirling numbers of the second kind times $e^x$.

• In (2) we use the bivariate generating function of the Stirling numbers of the second kind \begin{align*} e^{x(e^t-1)}&=\sum_{k=0}^\infty \frac{(e^t-1)^kx^k}{k!}\\ &=\sum_{k=0}^\infty\left(\sum_{j=k}^\infty {j\brace k} \frac{t^j}{j!}\right)x^k\\ &=\sum_{j=0}^\infty\left(\sum_{k=0}^j{j\brace k}x^k\right)\frac{t^j}{j!}\\ \end{align*} from which we get by using $-x$ instead of $x$ and selecting the coefficient of $t^j$: \begin{align*} \frac{1}{j!}e^{-x}\sum_{k=0}^j{j\brace k}(-x)^k=[t^j]e^{-xe^t} \end{align*}

Solution 3:

Considering the operator

$$ \cal{L}(f) = \sum_{k=0}^{\infty}\frac{f'(k)}{k!}(-x)^k $$

and choosing

$$ f(x) = e^{-W(1) x} $$

with $W(\cdot)$ the Lambert function, we have

$$ \cal{L}(f) =\mbox{ $-\frac{1}{W(1)}f$} $$

because

$$ f(x) = 1-x W(1)+\frac 12 W(1)^2 x^2-\frac 16 W(1)^3 x^3+\cdots + \frac{W(1)^k}{k!}(-x)^k + \cdots\\ \frac{f(x)}{W(1)} = -1 + e^{-W(1)}x-\frac 12e^{-2W(1)}x^2+\frac 16 e^{-3W(1)}x^3+\cdots + \frac{e^{-kW(1)}(-x)^k}{k!}+\cdots $$

with

$$ W(1)^k = e^{-k W(1)} $$

I hope it helps.