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Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ s.t. $f(x+1)-2f(x)+f(x-1)=x+1$, $f(0)=0$, and $f(1)=1$.

The problem is stated above. My attempt to solve this question is to plug in $x=0$, and $x=1$. That gives us $f(-1)=0$ and $f(2)=4$. Plugging in $x=2$ can give us $f(3)=10$. After that we can also get $f(4)=20$, $f(5)=35$...etc. I observed the sequence $0,1,4,10,20,35,...$ and found that these numbers are tetrahedral numbers, so the function we want may look like this:

$f(x)=\frac{x(x+1)(x+2)}{6}$

However, by plugging in different integer values, and using mathematical induction can only confirm that the function we want is valid within $\mathbb{Z}$. I don't know if there's a way to expand the domain from $\mathbb{Z}$ to $\mathbb{R}$.

E. Huang
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1 Answers1

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The functional equation tells us only about values at points $x-1,x,x+1$, i.e., point that are either all $\in \Bbb Z$ or none of them is. More generally, these points are always in the same remainder class of $\Bbb R/\Bbb Z$. It follows that the initial values $f(0)$, $f(1)$ cannot restrict what the function looks like on $\Bbb Z+\frac12$ or on $\Bbb Z+\pi$, say. What you can prove for any set $A$ that looks like the translated integers 8i.e., $A=\Bbb Z+a$ for some $a\in\Bbb R$ is that $f(x)=\frac{x(x+1)(x+2)}6+\alpha\cdot x+\beta$, but $\alpha,\beta$ may depend on $a$. It is only by the given values of $f(0)$, $f(1)$ that we can infer $\alpha=\beta=0$ when $a=0$.

So the general solution is (and you can verify this): Let $\alpha,\beta\colon[0,1)\to\Bbb R$ be arbitrary functions with $\alpha(0)=\beta(0)=0$. Then $$ f(x)=\frac{x(x+1)(x+2)}6+\alpha(x-\lfloor x\rfloor)\cdot x+\beta(x-\lfloor x\rfloor)$$ is a solution of the functional equation, and every solution has this form.

Hagen von Eitzen
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