Question

You have given a function λ:R-> R with the following properties (x ∈ R, n∈ N) :
Λ(n) =0, λ(x+1)= λ(x), λ(n+1/2)=1
Find two functions p,q:R-> Rwith q(x) not equal to 0 for all x such that λ(x)= q(x)(p(x)+1)

Answer 1

Let $x$ be a real number.The floor function $\lfloor x \rfloor$ is defined to be the greatest integer less than or equal to the real number $x$. The fractional part function $\{ x \}$  is defined to be $\{x\}= x -\lfloor x \rfloor$. Define two functions $p,q:\mathbb R\to\mathbb R$ with $q(x)$ not equal to 0 for all $x$ in the following way: $p(x)=2\{x\}-1, q(x)=\begin{cases}4\{x\}-1,\ x\ne\frac{1}{4}+n,n\in\mathbb N \\2,\ x=\frac{1}{4}+n,n\in\mathbb N\end{cases}$.

Then the function $λ:\mathbb R\to\mathbb R,$ $λ(x)= q(x)(p(x)+1)=\begin{cases}(4\{x\}-1)(2\{x\}), \ x\ne\frac{1}{4}+n,n\in\mathbb N\\ 1,\ \ \ x=\frac{1}{4}+n,n\in\mathbb N\end{cases}$ has the following properties:

$\lambda(n)=(4\{n\}-1)(2\{n\}) =-1\cdot0=0,\ \ \lambda(x+1)= \lambda(x),$ and

$\lambda(n+\frac{1}{2})=(4\{n+\frac{1}{2}\}-1)(2\{n+\frac{1}{2}\})=(4\cdot\frac{1}{2}-1)(2\cdot\frac{1}{2})=1$

for all $x \in\mathbb R, n\in\mathbb N.$