1
$\begingroup$

I noticed that the shape of a cubic graph of the equation $y=ax^3+0x^2+cx+0$ is similar to the shape of the graph of 1 period of $-\sin(x)$. Is there a cubic equation that exactly matches the graph of $-\sin(x)$ from $-\pi\le x \le\pi$?

I came fairly close simply through trial and error. The equation $y=.155x^3-x$ gets me a close match from $-\pi/2 < x < \pi/2$.

The $-\sin(x)$ has these 4 points. ${(-\pi,0),(-\pi/2, 1),(0,0),(\pi/2,-1),(\pi,0)}$

I calculated the values for a cubic equation containing all 4 of those points. The equation is $y=\frac{8}{3\pi ^3}x^3-\frac{8}{3\pi }x$. However, the equation is not as close as the one I manually found. While it does have all 4 points, the points of $(-\pi/2,1)$ and $(\pi/2,-1)$ are not relative maximum and minimums of the cubic function.

So is it even possible to match the two graphs exactly?

My graphs can be seen at https://www.desmos.com/calculator/aqux4jrvix

Thanks.

$\endgroup$
  • $\begingroup$ Any cubic which has $0,\pm \pi$ as roots has the form $\lambda x(x^2 -\pi^2)$ for some $\lambda \in \mathbb{R}$. But such a cubic has its local extrema at $\pm \pi / \sqrt{3}$ not at $\pm \pi / 2$. $\endgroup$ – dxiv Dec 14 '16 at 20:51
3
$\begingroup$

Suppose there existed some cubic function $f(x)$ such that $f(x) = \sin x$ for $x \in (-\pi, \pi)$. Then we could conclude that over the interval $(-\pi,\pi)$, $f'(x) = \sin'(x) = \cos x$. However, $f'(x)$ must be a polynomial of degree $2$ by the power rule, and $\cos x$ can't be modeled by a parabola on $(-\pi,\pi)$.

In case you don't believe me, just differentiate 3 more times to get $f''''(x) = \sin(x)$ for $x \in (-\pi, \pi)$ again; but $f''''(x)$ is obviously everywhere 0; which is definitely not $\sin x$

$\endgroup$
1
$\begingroup$

You can never find a polynomial that would match $\sin x$ on an open interval. The naïve reasoning would be to say that in the contrary case noone would talk about $\sin$, we would be talking about that polynomial.

THe correct way to prove that is to consider the derivatives. After a ceratin order $N$ all derivatives of a polynomial are identically zero. Yet the $N$-th derivative of $\sin x$ can not be identically zero for obvious reasons. Therefore, there is no polynomials that match $\sin x$ exactly.

Yet there is still hope. The first point is that $\sin x$ is an analytical function, and hence it can be approximated by its Taylor polynomial, wikipedia would be a good start. $$\sin x = x - \frac{x^3}{6}+\dots.$$ The second point would be to fix an interval, say, $[-\pi, pi]$, fix a degree $N$ and then try to find a polynomial $P$ of degree $N$ such that sup-norm of difference is minimised: $$\sup_{|x|\le \pi} |\sin x - P(x)|$$ This is what you essentially were trying to do.

The third one would be to fix $N$ of points $x_i$ where you want $P(x_i)=\sin x_i$. Such a polynomial exists, you can even prove that its degree can be chosen to be at most $N-1$.

There are many other approaches to finding a polynomial that "approximates" a trigonometric function, the choice of definition of the word "approximates" is completely up to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.