Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:53:21Z http://mathoverflow.net/feeds/question/11729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11729/are-there-any-nonlinear-solutions-to-fx1-fx-fx Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? 2010 Joint Meetings 2010-01-14T06:54:48Z 2010-01-15T01:21:35Z <p>(Asked by bcross at math.iuiui.edu on the Q&amp;A board at JMM)</p> <p>Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?</p> <p>[Ed: please retag appropriately]</p> http://mathoverflow.net/questions/11729/are-there-any-nonlinear-solutions-to-fx1-fx-fx/11731#11731 Answer by Mariano Suárez-Alvarez for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? Mariano Suárez-Alvarez 2010-01-14T07:24:40Z 2010-01-14T07:24:40Z <p>Theorem 1 in [Sugiyama, Shohei. On the existence and uniqueness theorems of difference-differential equations. Kōdai Math. Sem. Rep. 12 1960 179--190. <a href="http://www.ams.org/mathscinet-getitem?mr=MR0121552" rel="nofollow">MR0121552</a>] (which you can probably get from <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.kmj/1138844324" rel="nofollow">here</a>) gives an existence and uniqueness theorem which provides non-linear solutions <em>on finite intervals</em>.</p> http://mathoverflow.net/questions/11729/are-there-any-nonlinear-solutions-to-fx1-fx-fx/11733#11733 Answer by Bjorn Poonen for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? Bjorn Poonen 2010-01-14T07:41:44Z 2010-01-14T07:41:44Z <p>Yes, there exist nonlinear solutions.</p> <p>Multiplying by $e^{x+1}$ and setting $g(x):=e^x f(x)$ transforms the question into finding a solution to $g(x+1)=eg'(x)$ not of the form $e^x(ax+b)$.</p> <p>Start with any $C^\infty$ function on $\mathbb{R}$ whose Taylor series centered at $0$ and $1$ are identically $0$, but which is nonzero somewhere inside $(0,1)$. Restrict it to $[0,1]$. Let $g(x)$ on $[0,1]$ be this. Using $g(x+1):=eg'(x)$ for $x \in [0,1]$ extends $g(x)$ to a $C^\infty$ function $g(x)$ on $[0,2]$, which can then be extended to $[0,3]$, and so on. In the other direction, use $g(x) := \int_0^x e^{-1} g(t+1) dt$ to define $g(x)$ for $x \in [-1,0]$, and then for $x \in [-2,-1]$, and so on. These piece together to give a $C^\infty$ function $g(x)$ on all of $\mathbb{R}$. The corresponding $f(x)$ satisfies $f(0)=0$ and $f(1)=0$ but is not identically $0$, so it is not linear.</p> http://mathoverflow.net/questions/11729/are-there-any-nonlinear-solutions-to-fx1-fx-fx/11754#11754 Answer by mnf for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? mnf 2010-01-14T14:43:41Z 2010-01-14T14:43:41Z <p>This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution. </p>