Mathematics, 14.11.2020 14:00, carlinryan

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).

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answered: quinn2999

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 .

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 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 with (x + 1) should not change the value of $\lambda(x)$ . In other words, the period of $\lambda$ should be . 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 , $\displaystyle \lambda(n) = 0$ and $\displaystyle \lambda\left(n + \frac{1}{2}\right) = 1$ .

One possible approach is to make each 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 . 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 , 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.

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