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the equation $\dfrac {d^xf}{(dx)^x)} = f(x)$ where $ \dfrac{d^xf}{(dx)^x}$ means we are taking the xth derivative of f(x)(using fractional calculus, assuming the Riemann–Liouville fractional derivative/integral). This function would have the property that $f(1) = f'(1), f(2) = f''(2), f(3) = f'''(3)..... f(n) = f^{(n)}(n)$ And this also holds for x. I have tried to solve this like a normal Separable DE and gotten nowhere.(you have to find the integral of (dx)^x, which maybe could be interpreted as d(x^x), but i am not sure if that would work.) But I don't know if you can do that for ODEs of greater then degree 1(this differential equation's degree isn't even well defined) even if you can it requires finding the xth integral with respect to $d^xf$, and I am not even sure if that makes sense.) Albeit there might be some way to use Cauchy's formula for repeated integration. But other then that i am lost

gciriani
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Colonizor48
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1 Answers1

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The solution is the Mittag-Leffler function in which the variable is replaced by $x^\alpha$. Thus: $$ f(x)=E_\alpha(x^\alpha) \tag1$$ since $$ {}^CD^\alpha E_\alpha(x^\alpha)=E_\alpha(x^\alpha) \tag2 $$ It can be demonstrated by using the Laplace transform, as pointed out in a fairly recent paper.[a]

The Mittag-Leffler function is defined as follows: $$ E_\alpha(z)=\sum_{k=0}^{\infty} \frac{z^k}{\Gamma(\alpha k+1)}, \;\alpha>0, \;z\in \mathbb{C} \tag3$$

Please note that since $E_1(x)$ is the exponential function, the special case pointed out by one of the comments, and which you defined as trivial, is part of the solution space.

[a] Mainardi, F. Why the Mittag-Leffler Function Can Be Considered the Queen Function of the Fractional Calculus? Entropy (Basel) 2020, 22 (12), 1359. https://doi.org/10.3390/e22121359.

gciriani
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