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 → R with q(x)6=0 for all x such that λ(x) = q(x)(p(x) + 1).​

Answer 1

One possible combination:

`p(x) = \sin(2\, \pi \, x) - 1`.

`\displaystyle q(x) = (1)/(2)`.

`\displaystyle \lambda(x) = (1)/(2)\, \sin(2\, \pi\, x) - (1)/(2) + 1 = (1)/(2)\, \sin(2\, \pi\, x) + (1)/(2)`.

Explanation:

The requirement that
`\lambda(x + 1) = \lambda(x)` for all
`x \in \mathbb{R}` suggests that
`\lambda` should be a cyclic function with a period of
`1`.

One such example is the sine function. Let
`\omega` denote a non-zero real number. Consider the expression
`\sin(\omega\, x)`.

`\begin{aligned}\sin(\omega\, x) &= \sin(\omega\, x + 2\, \pi) \\ &= \sin\left(\omega\, \left(x + (2\, \pi)/(\omega)\right)\right)\end{aligned}`.

Note how replacing
`x` with
`\displaystyle \left(x + (2\, \pi)/(\omega)\right)` does not change the value of
`\sin(\omega\, x)`. Therefore, the period of
`\sin(\omega\, x)\!` would be
`\displaystyle (2\, \pi)/(\omega)`.

The question requires that replacing the
`x` with
`(x + 1)` should not change the value of
`\lambda(x)`. In other words, the period of
`\lambda` should be
`1`. If
`\lambda(x)\!` is indeed in the form
`\sin(\omega\, x)`, the value of
`\omega` should ensure that
`\displaystyle (2\, \pi)/(\omega) = 1`. Therefore,
`\omega = 2\,\pi`.

It would take a few more transformations to ensure that for all integers
`n`,
`\displaystyle \lambda(n) = 0` and
`\displaystyle \lambda\left(n + (1)/(2)\right) = 1`.

One possible approach is to make each
`n` a trough of this function, and each
`\displaystyle\left(n + (1)/(2)\right)` a crest of this function. The corresponding sine wave would have a range of only
`1`. However, without any transformation,
`y = \sin(2\, \pi\, x)` would have a range of two.

Therefore, it would be necessary to compress
`y = \sin(2\, \pi\, x)` vertically by a factor of
`2`, so that its range meets the needs.

`y = \displaystyle (1)/(2)\, \sin(2\, \pi\, x)` is the output of one such compression and has the correct period and range. However, its troughs would be at
`y = -\displaystyle (1)/(2)` (rather than
`y = 0`) whereas its crests are at
`\displaystyle y = (1)/(2)` (rather than
`y = 1`.) It would be necessary to shift this function upwards by
`\displaystyle (1)/(2)` unit for the troughs and crests to meet match the ones in the requirements.

`\begin{aligned} y &= \displaystyle (1)/(2)\, \sin(2\, \pi\, x) + (1)/(2) \\ &= (1)/(2)\, \sin(2\, \pi\, x) - (1)/(2) + 1 = (1)/(2)\left(\sin(2\, \pi\, x) - 1\right) + 1\end{aligned}`.