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Basically, I'd like to model sin x, but make it's derivative tend towards 0, so as x increases, it becomes a constant y = 0. The function begins like a typical sin x function, but slowly the fluctuation decreases until it isn't there anymore. If this works as I'm trying to have it work, I think some constant between 0 and 1 multiplying that derivative could also re-establish the normal sin x function.

I've been messing around with desmos graphing calculator, somehow trying to make sin x a result of some equation containing it's derivative, but I haven't been able to make much progress. I have taken classes in differential and integral calculus and linear algebra.

Edit: Sorry for the lack of rigour, I'm struggling to formulate the question properly

Edit2: User @ElliotG has provided me with the exact equation I'm looking to obtain, or atleast one that perfectly describes the idea of what the equation I'm looking for looks like: $\frac{sin x}{1 + x^2}$. The way I'd describe this function is that it is like sin x, but having its derivative constantly decreasing until it reaches 0. What I'd be interested in is: could there have been a way of obtaining a similar equation to $\frac{sin x}{1 + x^2}$ having in mind that what we want to do is have sin x as if its derivative was tending to 0? So all we start by knowing is how we want the function to behave without knowing what it looks like.

shintuku
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    I don't understand the question, but the function $\frac{\sin x}{1+x^2}$ might be of interest. – pancini Mar 19 '20 at 03:58
  • I honestly can't tell what's being asked here - can you clarify? It sounds like you're looking for a function which "models" $\sin$ somehow, but what exactly does that mean? – Noah Schweber Mar 19 '20 at 03:58
  • @ElliotG this is exactly what I'm looking for! But do you think there would have been some way of writing this equation in terms of the derivative of $sin x$? Because the way I would describe $\frac{sin x}{1+x^2}$ is that it is the same equation as $sin x$, but it's derivative is decreasing constantly – shintuku Mar 19 '20 at 04:03
  • Sorry to all for the lack of rigour, I'm struggling to formulate my question properly – shintuku Mar 19 '20 at 04:03
  • $\int \frac{d(sinx)}{dx}$ – NoLand'sMan Mar 19 '20 at 04:06
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    $e^{-\alpha x} \sin x$, where $\alpha <<1$. – mjw Mar 19 '20 at 04:08
  • The derivative of $\sin x$ is $\cos x$. In any case, the derivative of your function will probably look like the original function. – pancini Mar 19 '20 at 04:09
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    'The function begins like a typical sin x function, but slowly the fluctuation decreases until it isn't there anymore.' Are you asking for a damped $\sin$ function? If so, then $e^{-x}(A\cos({kx+\phi}))$ works, and it contains the derivative of $\sin x$. – Lt. Commander. Data Mar 19 '20 at 04:09
  • @mjw this is also a very nice equation that models what I mean to obtain, but what is the reasoning behind it? does it have some form of relationship to the derivative of $sin x$? – shintuku Mar 19 '20 at 04:20
  • @NamanKumar this too is very similar! Would you happen to know if there is a reasoning why exactly this function sort of gets to the idea of a sin x with a derivative approaching 0? – shintuku Mar 19 '20 at 04:22
  • @shintuku Yes, this equation is the solution to the differential equation of a damped harmonic oscillator. It's an elementary concept in physics. Reading about oscillators will tell you how this equation was derived and the rationale behind it. – Lt. Commander. Data Mar 19 '20 at 04:25
  • @NamanKumar I've been reading up on differential equations and it does look to be somewhat closer to what I've been looking for... thanks a lot for pointing that out to me – shintuku Mar 19 '20 at 04:33

2 Answers2

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Let $$f(x) = e^{-\alpha x} \sin x \quad \alpha <<1$$

Here is a graph of this function, and also another nice choice that was suggested $g(x)=\frac{\sin x}{1+\alpha^2 x^2}$, both with $\alpha = \frac{1}{10}$.

enter image description here

The derivatives are:

$$f^\prime(x)=e^{-\alpha x} [\cos x - \alpha \sin x] \textrm { and } g^\prime(x)=\frac{\cos x}{1+\alpha^2 x^2} -\frac{2x \sin x}{\alpha(1+\alpha^2 x^2)^2}$$

Here is a picture of the derivatives $f^\prime(x)$ and $g^\prime(x)$

$f$ and $f^\prime$ are in blue and $g$ and $g^\prime$ are in yellow.

enter image description here

mjw
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  • This is of great help! But tell me, how exactly are you coming up with the idea of writing a function that both contains an integral and a derivative? Is there some reasoning behind it? – shintuku Mar 19 '20 at 04:24
  • Well, the first thing I thought of was an exponentially damped sinusoid. That is $f(x)=e^{-\alpha x}\sin x$. After that, just computed the derivative and graphed the function and its derivative. In the comments, there was another interesting suggestion, so I did the same for that. Seems that you are looking for a function that goes to zero as $x\rightarrow \infty$ and its derivative also goes to zero multiplied by $\sin x$. There are lots of choices. These are just two that come up often enough. – mjw Mar 19 '20 at 04:28
  • Thank you for the answer and the graphs! – shintuku Mar 19 '20 at 04:31
  • You are welcome! – mjw Mar 19 '20 at 04:32
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Any function of the form $$y(t)=Ae^{-\gamma t}\sin(\omega t+\phi)$$ works. For example, substituting $\gamma = 0.2$, $A=1$, $\omega = 1$ and $\phi = 0$, we get the following curve:

Function

which satisfies your constraints.

Your question also pertained to why such a function works, and the reason is that this function was developed in physics as a model for the damped Harmonic Oscillator. This was initially intended as a solution to the following form of Newton's second law: $$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=0$$ and subsequently introducing an additional damping force. The equation is a method to model oscillators like pendulums in realistic scenarios (where constraints like air resistance prevent the oscillator from continuing indefinitely).

Lt. Commander. Data
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    This answer is also worth being accepted, thank you very much! I will be looking into the harmonic oscillator, especially that equation it is meant to solve, which seems to somehow lead to the harmonic oscillator. – shintuku Mar 19 '20 at 04:59