Why is the plot of $f(t)=\frac{\partial}{\partial t}\left\{\sin(\sin(\pi t))\right\}$ so similar to a triangle wave?

Question

I was playing around the other day and I found that the function

$$t\to f(t), f(t)=\frac{\partial}{\partial t}\left\{\sin(\sin(\pi t))\right\}$$ seemed to be very close to the triangle wave. Is there some intuitive explanation for this?

Answer 1

The reference to Fourier series is useful to understand the "curious" behaviour. So, before any attempt to explanation, a brief reminder is necessary.

The Fourier series of the triangle wave such as drawn on Mathreadler's graph is : $$y(x)=\sum_{n=1}^\infty a_n\cos(\pi nx)$$ $$a_n=2\int_0^1 \pi(1-2t)\cos(\pi nt)dt=\frac{8}{\pi n^2}\sin\left(\frac{\pi}{2n}\right)\left(\sin\left(\frac{\pi}{2n}\right)-\pi n\cos\left(\frac{\pi}{2n}\right)\right)$$ $$a_0=a_2=a_4=…=0$$ $$a_1\simeq 2.546\quad;\quad a_3\simeq 0.283\quad;\quad a_5\simeq 0.102\quad;\quad a_7\simeq 0.052\quad;\quad \text{etc.}$$ If we take the first term only, the curve is a pure sinusoid $\quad 2.546\cos(\pi x)$ : The black curve on the next figure.

If we take more and more terms, the curve tends to the triangle wave function.

Two terms $\quad 2.546\cos(\pi x)+0.283\cos(3\pi x)$ : The green curve.

Three terms $\quad 2.546\cos(\pi x)+0.283\cos(3\pi x)+0.102\cos(5\pi x)+$ : The red curve.

Four terms $\quad 2.546\cos(\pi x)+0.283\cos(3\pi x)+0.102\cos(5\pi x)+0.052\cos(7\pi x)$ : The blue curve.

This make understand that one can obtain a curve looking like the triangle wave with only few terms, if we are not demanding much accuracy.

For example, how to have an "almost triangle wave" with only two terms and such that $y(-1)=-\pi$ and $y(0)=\pi$ and $y(1)=-\pi$ , etc. ? $$y(x)=A\cos(\pi x)+(\pi-A)\cos(3\pi x)$$

The mean square error is obtained for the minimum value of $$\int_0^1 \left(A\cos(\pi x)+(\pi-A)\cos(3\pi x)-\pi(1-2x) \right)^2dx$$ The derivative with respect to $A$ is null. After a few calculus the result is $$A\simeq 2.703$$ $$y(x)=2.703\cos(\pi x)+0.439\cos(3\pi x)$$

This allows to understand that is is not difficult to find many functions which are "almost triangle wave".

One of the simplest is $y(x)=2.703\cos(\pi x)+0.439\cos(3\pi x)$. But many others, based on more complicated formulas can do as well or even better. The Fourier series will have the first terms close to the above terms.

For example the function $$y(x)=\pi \cos(\sin(\pi x))\cos(\pi x)$$ which is the derivative of $\sin(\sin(\pi x))$

has the Fourier series with coefficients : $$a_0=a_2=a_4=…=0$$ $$a_1\simeq 2.765\quad;\quad a_3\simeq 0.369\quad;\quad a_5\simeq 0.008\quad;\quad \text{etc.}$$

We see that the coefficients are in the correct range for an "almost triangle wave", but slightly less accurate than the above example.

Of course it is a nice achievement to have found the function $y(x)=\pi \cos(\sin(\pi x))\cos(\pi x)$. But we must be aware that a lot of functions are likely to have a similar behaviour, as "almost triangle wave".