-1
$\begingroup$

The following are examples of enveloping functions.

$$f(x) = x\cdot\sin(x)$$

enter image description here

$$f(x) = x^2 \cdot \sin(x)$$

enter image description here

$$f(x) = \frac{1}{x} \cdot \sin(x)$$

enter image description here

It seems that given a function $f(x) = g(x)\cdot \sin(x)$, we have a sinusoidal curve enveloped by $g(x)$ and $-g(x)$.

  • Is this true?
  • If it is, what is the intuition behind it?
  • Is there a more general case for these enveloping functions?

Edit: On further thinking, I assume it has something to do with the amplitude of the function, as we know that $n\sin(x)$ vertically scales the function by $n$. Similarly, perhaps $g(x)$ vertically scales $\sin(x)$. Maybe this is still a rather loose explanation, so I'd love a more detailed mathematical one.

$\endgroup$
4
  • $\begingroup$ How do you define "envelop"? $\endgroup$ Commented May 18, 2022 at 9:11
  • $\begingroup$ @5xum I don't really have a well-defined mathematical definition but I'd love to be informed of one. $\endgroup$ Commented May 18, 2022 at 9:16
  • $\begingroup$ If $-1\le h(x) \le 1$ and $g(x)\gt 0$ it will be the case that $-g(x) \le g(x) h(x) \le g(x)$ multiplying the first inequality by a positive number $\endgroup$ Commented May 18, 2022 at 9:22
  • $\begingroup$ @Henry Thank you! It seems that this simple fact can provide really good intuition for what the graph of the product of $2$ functions will look like. $\endgroup$ Commented May 18, 2022 at 9:33

2 Answers 2

Reset to default
0
$\begingroup$

If it is, what is the intuition behind it?

For real arguments we have $|\sin x| \leqslant 1$ and therefore, the graph of $$g(x) = f(x)\sin x$$ runs somewhere between $f(x)$ and $-f(x)$, what you called "enveloping". Likewise, the graph of $g$ will never be above $|f(x)|$ or below $-|f(x)|$.

Is there a more general case for these enveloping functions?

Well, every time you draw a graph of some function $g(x) = f(x)\cdot h(x)$, then it depends both on $f$ and on $h$. In particular, if one function is bounded by a constant like $|h(x)|\leqslant H$, then the graph of $g(x)$ runs between $Hf(x)$ and $-Hf(x)$.

The example where one function is, say, sine is visually appealing because sine is hitting its maxima and minima frequently, which leads to some visual "guide". You can increase the effect by increasing the frequency of sine, e.g. by plotting $f(x)\sin(100x)$.

$\endgroup$
0
$\begingroup$

Your question heavily depends on what exactly you mean by the word "envelop" means.

From your images, it seems as if, when saying that the function $f$ is enveloped by the functions $h$ and $g$, you mean that the graph of the function $f$ is in the region "between" the graphs of $h$ and $g$.

To that end, we can define the concept of envelopment as:

If $f,g,h$ are all functions from $\mathbb R$ to $\mathbb R$, then $f$ is enveloped by $g$ and $h$ if, for all $x\in\mathbb R$, we have either $h(x)\leq f(x)\leq g(x)$ or $g(x)\leq f(x)\leq g(x)$.

Now, if the above definition indeed describes your intuitive concept of "envelopment", then the question of "Do the functions $g(x)$ and $-g(x)$ always envelop the function $f(x) = g(x)\cdot \sin(x)$" is really simple to answer.

The answer is yes, and the proof is really simple. Since we know that $\sin x \in [-1,1]$, it should be clear that $\sin x \cdot g(x)\in [-|g(x)|, |g(x)|]$ no matter what $g(x)$ is. If $g(x)\geq 0$, then this means $-g(x)\leq \sin x g(x) \leq g(x)$, otherwise, $g(x)\leq \sin x g(x)\leq g(x)$ will be true.

$\endgroup$
1
  • $\begingroup$ Thank you for the definition. $\endgroup$ Commented May 18, 2022 at 9:53

You must log in to answer this question.