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.

  • 8
    $\begingroup$ 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 $y=0$. $\endgroup$ – egreg Jul 21 '14 at 22:50
  • $\begingroup$ @egreg : Maybe your comment should be an answer. $\endgroup$ – Michael Hardy Jul 21 '14 at 22:52
  • 3
    $\begingroup$ How about $\sum_{n=0}^\infty y^{(n)}(x)=\text{some nonzero function}$? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 21 '14 at 22:52
  • 2
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Jul 21 '14 at 22:55
  • 1
    $\begingroup$ @WillO $y+y'+\dotsb=0$; differentiate: $y'+y''+\dotsb=0$; substitute: $y=0$. $\endgroup$ – 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 $n\ge 0$.

  • $\begingroup$ This does not preclude the existence of smooth solutions, though. $\endgroup$ – 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 $\alpha\in \mathbb{K}$

  • $\begingroup$ 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 $x$ $\endgroup$ – CameronJWhitehead Jul 21 '14 at 23:14
  • $\begingroup$ I confused this equation with the equation $1+x+x^2+\ldots+x^n=0$ $\endgroup$ – 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.