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:

One possible combination:

[tex]p(x) = \sin(2\, \pi \, x) - 1[/tex].

[tex]\displaystyle q(x) = \frac{1}{2}[/tex].

[tex]\displaystyle \lambda(x) = \frac{1}{2}\, \sin(2\, \pi\, x) - \frac{1}{2} + 1 = \frac{1}{2}\, \sin(2\, \pi\, x) + \frac{1}{2}[/tex].

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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