I wonder how we can find all $f(x)$ that satisfies$$\sum_{k=0}^{\infty}f^k\left(x\right)=\sum_{k=0}^{\infty}f^{\left(k\right)}\left(x\right)\tag{1}$$Here $f^{(k)}(x)$ means the $k$th derivative of $f(x)$.
We could tell that this DE is not a linear differential equation, let alone has a finite order. For the LHS to converge, $|f(x)|<1$. So we could use geometric series to write this as $$\frac1{1-f(x)}=\sum_{k=0}^{\infty}f^{\left(k\right)}\left(x\right)$$Replacing $f(x)=y$ and integrating with respect to $y$, we get $$-\ln|1-y|=\frac{y^2}2+\int\sum_{k=1}^\infty y^{(k)}dy+C$$I was thinking of substituting the LHS of $(1)$ into the integrand, but that just gives $-\ln|1-y|=-\ln|1-y|$. How can we solve this differential equation? If we don't have a closed form, can we write the equation differently (not something obvious like adding and subtracting the same number), obtain asymptotics for $f(x)$, or find approximations?
Formally, we can reduce this to a first order separable ODE for $y$. Let $D = \frac{d}{dx}$, then our equation reads $$\sum_{k=0}^\infty y^k = \left(\sum_{k=0}^\infty D^k\right)y.$$ Again operating formally, we now sum both geometric series to obtain $$\frac{1}{1-y} = (I - D)^{-1}y,$$ where $(I-D)^{-1}$ denotes the inverse of the operator $I - D$ whose action on a function $f$ is $f - f'$. Left multiplying by $I-D$, we obtain $$(I-D)\left(\frac{1}{1-y}\right) = \frac{1}{1-y} - \frac{y'}{(1-y)^2} = y.$$ We can then rearrange into standard form to obtain $$y' = -y^3 + 2y^2 - 2y + 1.$$ Such an equation can be solved locally for most initial conditions and WolframAlpha gives some implicit equations for $y$, although it isn't clear which, if any, of these solutions satisfy the necessary conditions to justify the convergence of the original infinite series.