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Hello I was wondering what was the function $f$ defines like this :

Let $f(x)$ a continuous and differentiable function such that : $$f(x)=\sum_{k=0}^{\infty}\frac{f'(k)}{k!}(-x)^k$$

In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain :

We have :

$$\int_{0}^{\infty}x^{-s-1}f(x)dx=\Gamma{(-s)}f'(s)$$

Or : $$\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dx=f'(s)$$

Now we use the Fundamental theorem of calculus to get : $$\int_{0}^{s}\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dxds=f(s)-f(0)$$

Now we take the limit to get : $$\lim_{s\to\infty}\int_{0}^{s}\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dxds=f(\infty)-f(0)$$

Wich is equal to:

$$\int_{0}^{\infty}\frac{f(ax)-f(bx)}{ln(\frac{a}{b})x}$$

So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ?

Thanks

Ps:I know it's not very rigorous but I think it's interesting

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    $\begingroup$ Just a quick observation, but $f(x)=0$ satisfies the condition. $\endgroup$ – Karn Watcharasupat Aug 5 '18 at 16:35
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    $\begingroup$ @KarnWatcharasupat The set of solutions form a vector space. $\endgroup$ – saulspatz Aug 5 '18 at 16:37
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    $\begingroup$ If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^{-x}$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something). $\endgroup$ – Dog_69 Aug 5 '18 at 16:40
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    $\begingroup$ @Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately $\endgroup$ – user88595 Aug 5 '18 at 17:04
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    $\begingroup$ I am right. I was wrong after all. $f'$ gives the sum $$ -\sum_{k=0}^\infty \frac{f'(k+1)}k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true. $\endgroup$ – Dog_69 Aug 5 '18 at 17:10
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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*}

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  • $\begingroup$ Can you show at least one real sample of $f(x)\not= const$? $\endgroup$ – Yuri Negometyanov Aug 17 '18 at 12:39
  • $\begingroup$ @FatsWallers: Many thanks for accepting my answer and granting the bounty. $\endgroup$ – Markus Scheuer Aug 17 '18 at 13:23
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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?

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  • $\begingroup$ Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^{(n)}(0)$? $\endgroup$ – Simply Beautiful Art Aug 5 '18 at 19:38
  • $\begingroup$ @SimplyBeautifulArt Using your suggestion, one gets $g^{(n)}(0) = f'(n+1) = - g(-(n+1))$. $\endgroup$ – Andreas Aug 5 '18 at 20:50
  • $\begingroup$ Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe. $\endgroup$ – Simply Beautiful Art Aug 6 '18 at 0:28
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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.

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    $\begingroup$ If $f(x)=e^{-ex}$, then $e^k \neq f'(k) =-ee^{-ek}$. $\endgroup$ – Rócherz Aug 5 '18 at 17:26
  • $\begingroup$ What you have established, using your proposed $f(x)$, is that $ \frac{-1}{W(1)}\sum_{k=0}^{\infty}\frac{f'(k)}{k!}(-x)^k = 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 = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^{-k W(1)}$. Now, the prefactor $\frac{-1}{W(1)}$ unfortunately is NOT given in the original task. $\endgroup$ – Andreas Aug 5 '18 at 21:10
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Let us try to find the sequence $$d_n = \frac1{(n-1)!}f'(n),\quad n=1,2\cdots\tag1$$ from the equations $$f'(x)= -\sum_{k=1}^\infty\frac{(-x)^{k-1}}{(k-1)!}f'(k).\tag2$$ Then $$(n-1)!d_n= -\sum_{k=0}^\infty(-n)^kd_{k+1},\quad n= 1, 2\dots,\tag3$$ $$\begin{cases} 0!d_1 = -d_1 + d_2 - d_3 + d_4 - d_5+\dots\\[4pt] 1!d_2 = -d_1 + 2d_2 - 2^2d_3 + 2^3d_4 - 2^4d_5+\dots\\[4pt] 2!d_3 = -d_1 + 3d_2 - 3^2d_3 + 3^3d_4 - 3^4d_5+\dots\\[4pt] 3!d_4 = -d_1 + 4d_2 - 4^2d_3 + 4^3d_4 - 4^4d_5+\dots\\[4pt] \hspace{100pt}\dots \end{cases}\tag4$$ The linear system $(2)$ is homogenius.
Let us consider the sequence of determinants $${\small D_n= \begin{vmatrix} -2&1&-1&1&\dots&-(-1)^{n-2}&-(-1)^{n-1}\\[4pt] -1&1&-2^2&2^3&\dots&-(-2)^{n-2}&-(-2)^{n-1}\\[4pt] -1&3&-3^2-2!&3^3&\dots&-(-3)^{n-2}&-(-3)^{n-1}\\[4pt] -1&4&-4^2&4^3-3!&\dots&-(-4)^{n-2}&-(-4)^{n-1}\\[4pt] &&\dots&&\dots&&\dots\\[4pt] -1&n-1&-(n-1)^2&(n-1)^3&\dots &\large{\genfrac{}{}{0}{}{-(1-n)^{n-2}}{-(n-2)!}} &-(1-n)^{n-1}\\[4pt] -1&n&-n^2&n^3&\dots&-(-n)^{n-2} &\large{\genfrac{}{}{0}{}{-(-n)^{n-1}}{-(n-1)!}}\\ \end{vmatrix}}.\tag5$$ Calculations give: $$ |D_1|=|-2|\ge 1,\\ |D_2|=|-1|\ge 1,\\ |D_3|=|-7|\ge 2^2 = 4,\\ |D_4|=|-134|\ge 2^23^3=108,\\ |D_5|=|242040|\ge 2^23^34^4=27648,\\ |D_6|=|4046'51376|\ge 2^23^34^45^5=864'00000\dots\tag6 $$

At the same time, if $\lim\limits_{n\to\infty} |D_n| > 0,$ then the system $(3)$ has only zero solution.

Therefore I think, that $\color{brown}{\textbf{a unique solution f(x) of the given condition is a constant.}}$

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