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Here's a plot of a function:

enter image description here

Nothing special about this function - it's just $y = \sin{x}$. It's immediately clear, even without doing any calculations, which parts of the curve has positive first derivative, and which parts has positive second derivative. Parts of the curve that are increasing have positive first derivative, so e.g. between $-1$ and $+1$ the first derivative must be positive. Similarly if the function is concave up or down, it must have negative and positive second derivative respectively. So we can immediately say that between $0$ and $3$ the second derivative is negative.

How can I quickly tell where the 3rd and 4th derivatives are positive or negative? In this case it's easy to just do the math, but I'm looking to interpret what a result for "the fourth derivative is positive" or "the fourth derivative is more positive at $x = 3$ than at $x = 4$" says about the shape of the underlying function. Strictly, I am only interested in the fourth derivative, but I'm guessing that insight with the third derivative will help illustrate what happens with the fourth derivative.

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  • $\begingroup$ At the very least, you know if something switches from concave up to down, the third derivative was negative at that point, and vice versa for the opposite trend. You could mentally keep track of the switching of even these switches to find the sign of the fourth derivative at specific points and try to use intermediate value theorem from there. But geometric properties of higher order derivatives is encoded in algebraic geometric quantities (like curvature of curvature) which are difficult to see. The order of the term is inversely proportional to the length scale it is dominant. $\endgroup$
    – Ninad Munshi
    Commented Jun 3, 2021 at 8:30
  • $\begingroup$ I don't think I can, at a glance, see whether an arbitrary section of the graph of a polynomial is a quadratic or a cubic or a quartic (apart from counting local extrema and inflection points and hoping I can see them all). And I am pretty certain that I can't tell much about the third degree coefficient on a quartic graph. $\endgroup$
    – Arthur
    Commented Jun 3, 2021 at 8:35

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No - it is not visually possible. The following chart is made up of parabolae with zero third and fourth derivatives almost everywhere, and looks almost identical to yours. One of them is $y=\frac{4x}{\pi}-\frac{4x^2}{\pi^2}$ and the others effectively repeat that one

enter image description here

We could go further and take double my approximation and then subtract your sine curve. This would have third and fourth derivatives opposite in sign to yours almost everywhere. On $[0,\pi]$ it would be $y=\frac{8x}{\pi}-\frac{8x^2}{\pi^2} -\sin(x)$

As a visual challenge, can tell which of the following is your sine curve and which has the opposite signs of third and fourth derivatives? (My initial suggestion is the black curve between them)

enter image description here

Well done if you spotted red as the sine curve and blue as the alternative

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  • $\begingroup$ In the figures, is it a single function? $y = 4x/\pi -4x^2/\pi^2$ is a quadratic function, so it shouldn't oscillate, which makes it seem possible that if there is a way to tell, one would have to look further than the domain $[0, \pi]$. $\endgroup$
    – Allure
    Commented Jun 3, 2021 at 12:06
  • $\begingroup$ @Allure My first parabolae curve is a function, and continuous and differentiable everywhere, and infinitely differentiable everywhere except where it crosses the $x$-axis. It is not a polynomial function, but a succession of piecewise polynomial functions. Its purpose is to demonstrate the difficulty of visualising the sign of third and fourth derivatives. I would suggest that even the jump in its second derivative at the $x$-axis is difficult to spot. $\endgroup$
    – Henry
    Commented Jun 3, 2021 at 13:30

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