# Solution to $y(x) + y'(x) + y''(x) + y'''(x) + \cdots = 0$

Is there a non-trivial solution to the following differential equation?

$$y(x) + y'(x) + y''(x) + y'''(x) + \cdots= 0$$

That is, is there a smooth function $y : \mathbb{R} \to \mathbb{R}$ such that for each $x$, the series $$\sum_{n = 0}^{\infty}\frac{d^n y}{dx^n}(x)$$ converges to zero.

• Suppose you can give a meaning to the infinite sum and a solution exists; if differentiating term by term is possible, then $(y+y'+\dotsb)'=0$, so $y=0$. – egreg Jul 21 '14 at 22:50
• @egreg : Maybe your comment should be an answer. – Michael Hardy Jul 21 '14 at 22:52
• How about $\sum_{n=0}^\infty y^{(n)}(x)=\text{some nonzero function}$? ${}\qquad{}$ – Michael Hardy Jul 21 '14 at 22:52
• Maybe the question of how to make it precise should come later. Dirac didn't make his delta function and its derivatives precise. Cardano didn't make imaginary numbers precise. – Michael Hardy Jul 21 '14 at 22:55
• @WillO $y+y'+\dotsb=0$; differentiate: $y'+y''+\dotsb=0$; substitute: $y=0$. – egreg Jul 21 '14 at 23:43

There is no non-trivial real analytic solution.

Indeed, if $$y(x)=\sum_{n=0}^\infty \frac{a_n x^n}{n!},$$ then $\sum_{n=0}^\infty y^{(n)}(x)=0$, implies that $\sum_{n\ge k}a_n=0$, for all $k\ge 0$, and hence $a_n=0$, for all $n\ge 0$.

• This does not preclude the existence of smooth solutions, though. – Andrés E. Caicedo Jul 29 '14 at 22:30

The previous solution is the solution for the equation $y+y'+\ldots+y^{(n)}=0$. But for this equation you can compose with the operator $I-D$ where $D$ is the operator $D(y)=y'$, your equation write $\displaystyle\sum_{n=0}^{+\infty}D^{(n)}(y)=0$, then $(I-D)\circ (\displaystyle\sum_{n=0}^{+\infty}D^{(n)})(y)=I(y)=y=0$

$t\mapsto \sum_{\lambda\in \Gamma}\alpha e^{\lambda t}$ where $\Gamma$ is the set of solutions of the characteristic equation $1+x+x^2+x^2+\ldots=0$, and $\alpha\in \mathbb{K}$

• There are no solutions to this equation though, $1+x + x^2 + \ldots$ converges only if $\| x \| < 1$ in which case the sum equals $\frac{1}{1-x} \neq 0$ for any $x$ – CameronJWhitehead Jul 21 '14 at 23:14
• I confused this equation with the equation $1+x+x^2+\ldots+x^n=0$ – Hamou Jul 21 '14 at 23:18

As others have pointed out, there is no solution to your equation.

There are, however, solutions to the closely related equation

$$(y(x)+y'(x))+(y''(x)+y'''(x))+\ldots =0$$

Namely: $y=Ae^{-x}$ ($A$ constant).