Given a function $\sigma: \mathbb{R}\rightarrow\mathbb{R}$ which satisfies the following properties$(x\in\mathbb{R}, n\in \mathbb{N}):$
$$\sigma(n)=0, \sigma(x+1)=\sigma(x), \sigma(n+\frac{1}{2})=1$$ Find two functions $f,g:\mathbb{R}\rightarrow \mathbb{R}$ with $g(x)\neq0$ for all $x$ such that $\sigma(x)=g(x)(f(x)+1)$.
I have tried:
We have $\sigma(n)=0, \forall n\in\mathbb{N}$ $$\Rightarrow \sigma(1)=\sigma(2)=\cdots=0$$ And $\sigma(x+1)=\sigma(x)\Rightarrow \sigma(x)$ is periodic function with period $=1$.
$\sigma(n+\frac{1}{2})\Rightarrow \sigma(\frac{3}{2})=\sigma(\frac{5}{3})=\cdots=1 $
$\sigma(x)=g(x)(f(x)+1)$ with $g(x)\neq 0$ Since, we have periodic function, so we will use trigonometric function for $f(x)$ and $g(x)$
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Please kindly help me on this.
This diagram shows the set of points that are defined by the description you have given : zero for integer values and 1 at the mid-points.
You need a curve that will pass through these points, and you are right to think that something trigonometric will do the task.
You should be aware that one obvious candidates is $\sigma(x)=\sin^2 (\pi x)$
Let $g(x)=1$ as suggested in the comments and let $f(x)=\sin^2 (\pi x)-1$ and then you have $\sigma(x)=g(x)(f(x)+1)$
Pick any periodic function. $g(x)=0.4 \sin^2 (\pi x) + 0.1$ has the characteristics you require since $0.1 \le g(x) \le 0.5$. You will then need to change $f(x)$ to $f(x)=2 \sin^2 (\pi x)-1$ to keep your peaks at $1$.