I have been trying to analyse the meaning of higher order derivatives and their geometrical significance.

Given a function $f(x)$ what are the unique geometric interpretation of its higher orders?

  • $f'(x)$ - slope (rate of change of x w.r.t y)
  • $f''(x)$ - convexity / concavity of $f(x)$ based on $sign(f''(x))$
  • $f'''(x)$ - increasing / decreasing slope of $f'(x)$
  • $f''''(x)$ - ??
  • ...
  • $f^n(x)$ - ??

I apologize if this question is rudimentary.


There are many ways to answer this. I think my favorite is this:

The Taylor expansion of a a function $f$ at $a$ is:

$$ f(x) = f(a) + f^\prime (a)(x-a) + \frac{f^{\prime\prime}(a)}{2}(x-a)^2 + ... +\frac{f^{(n)}(a)}{n!}(x-a)^n+...$$

What you can do with this is take a function $f$ around some point $a$, then eliminate the information you get from the first $n$ derivatives by subtracting the first $n$ terms of the Taylor expansion, and then the lowest nonzero polynomial approximation you can get is, in a sense, the information that the $(n+1)$th derivative holds.

Let's try to match what you already know to this approach:

  • $f(a)$ is the most important information when trying to figure our how $f$ behaves around $a$ - it tells you where it's all located, and it's in fact the approximation of $f$ with a polynomial of degree $0$.
  • $f^\prime(a)$ is the result of eliminating all information given by $f(a)$ (by subtracting it), and then trying to polynomially approximate the function. So in fact $f^\prime(x)$ is answering the question "What straight line does $f$ most look like around $a$?". If this line is ascending you can say that $f$ is ascending, you can say that the slope of the line can be very naturally thought of as the slope of $f$ at $a$, and so on.
  • $f^{\prime\prime}(a)$. Once you get bored comparing $f$ to a straight line, you can eliminate this part too by subtracting $f^\prime(a) (x-a)$ and then the next lowest degree nontrivial polynomial approximation you can get is of degree $2$ - a parabola, and it's coefficient is (proportional to) $f^{\prime\prime}(a)$. And again, if $f^{\prime\prime}(x)>0$, the parabola is "smiling" and the function can be said to be concave, and if $f^{\prime\prime}(x)<0$, the parabola is "frowning" and $f$ can be said to be convex, the higher the coefficient the stronger the concavity / convexity, etc.
  • $f^{\prime\prime\prime}(a)$. This is where it gets less intuitive, because you can subtract the previous approximation, but now you have two problems: First, you already have a lot of information on the function, so how this changes what you know can be harder to visualize. Second, It's not immediately intuitive what it means to be similar to a certain monomial of degree $3$. It can, however, be a little clearer when $f^{\prime\prime}(a)=0$. In this case, $f$ is neither concave nor convex at $a$, but if the third derivative is positive, then it should be similar to $x^3$, meaning it's more convex on the left and more concave on the right, and if $f^{\prime\prime\prime}(a)<0$, then it should be similar to $-x^3$, that is, concave on the left and convex on the right. This is generally true, as the third derivative can be looked at as a measure of "how much more concave / convex the function gets passing around $a$", but this can be a little hard to visualize.
  • $f^{(n)}(a)$. In the same manner as before, this can be thought of as a measure of what $n$-degree monomial is $f$ most similar to, discarding all information from lower degree polynomial approximations. I think this is hard for us to visualize as a quality for the same reason that it's hard to tell what is the true "qualitative" difference between $x^2$ and $x^4$, maybe because our intuition is naturally built to make do with good approximations, and the value, slope and concavity / convexity of a function at a point already describe a very good approximation for most "everyday" cases.

It can also sometimes be helpful to look at the physical interpretation of the derivatives of motion: $f(x)$ is where you are, $f^\prime(x)$ is your velocity, $f^{\prime\prime}(x)$ is your acceleration, $f^{\prime\prime\prime}(x)$ is the rate in which your acceleration changes (for example, if you're in a car and you're gradually stepping on the gas), etc.

I think this is a really good question, because understanding concepts like this can give very powerful intuition, and I can't say I have any geometric intuition of any higher derivatives, but the coefficients of the Taylor series prevent this from keeping me up at night.

  • $\begingroup$ this is a brilliant answer. I really appreciate the effort you took to reply to the post. Indeed the Taylor series does aid in developing a strong intuition of higher degrees. I believe training our mind to interpret higher degrees is a fruitful endeavour in understanding the fabric of nature. $\endgroup$ – Jugesh Sundram Jan 19 '13 at 3:28

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