Question

You have given a function λ : R → R with the following properties (x ∈ R, n ∈ N):

Answer 1

Answer:

One possible combination:

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

$\displaystyle q(x) = \frac{1}{2}$.

$\displaystyle \lambda(x) = \frac{1}{2}\, \sin(2\, \pi\, x) - \frac{1}{2} + 1 = \frac{1}{2}\, \sin(2\, \pi\, x) + \frac{1}{2}$.

Step-by-step 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 $\sin(\omega\, x)$.

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

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

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 $\displaystyle \frac{2\, \pi}{\omega} = 1$. Therefore, $\omega = 2\,\pi$.

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

One possible approach is to make each $n$ a trough of this function, and each $\displaystyle\left(n + \frac{1}{2}\right)$ 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.

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

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