# Intuitively, why are the curves of exponential, log, and parabolic functions all smooth, even though the gradient is being changed at every point?

Why are the curves of exponential, log, and parabolic functions all smooth, even though the gradient is being changed at every point? Shouldn't it be much more choppy?

By the way, if possible, can this be explained intuitively (not too rigorously), and without Calculus? Because I want to understand this, but I haven't learnt Calculus yet.

• Smooth means that the function and its derivative(s) are continuous. If the gradient did not change, that would be to say the derivative is constant. Ther eis a difference between constant and continuous, you know? – Hagen von Eitzen May 22 '18 at 4:53
• "Algebraic curves" sounds like a bit of an overkill for a tag. – user541686 May 22 '18 at 20:19
• Why do you think they should be choppy? Can you work through the logic? My intuition says there's no reason to assume that the curves can't be smooth, but that's because my intuition has been honed by having taken calculus. It will be easier for us to help if we understand where your intuition is taking you, and then we can work with that. – Cort Ammon May 22 '18 at 21:55
• the answer is literally in "at every point". "at every point" is literally the definition of "smooth". if it changes only every, say, 0.234 units, it would not be smooth. the literal definition of "smooth" is that it changes at every point. – Fattie May 23 '18 at 13:15

# "At every point"

The gradient might change "at every point", but you need to remember that those points can be arbitrarily close to each other (see "real numbers"):

When you reduce the distance between the sampling points for $e^x, x^2, \ln(x), \sin(x)$, the gradient changes also become smaller. After a few iterations, the screen resolution isn't high enough to show any change anymore and the curves look smooth.

On the other hand, the $|x|$ curve (absolute value, the green curve on the graph) doesn't change anymore as soon as $x = 0$ is plotted: there's an abrupt change of gradient around $x = 0$, even at a very high resolution. At $x = 0$, it's not possible to define a gradient for this curve.

If you zoom infinitely on the smooth curves, they will look just like straight lines. If you zoom on $|x|$ at $x=0$, you'll always see the sharp corner:

However small $\varepsilon$ is, going from $x=-\varepsilon$ to $x = \varepsilon$ will change the gradient of $|x|$ from $-1$ to $1$.

• @EthanChan It means given any real number there is no "next" real number for which the gradient changes and produce a corner in the graph. – Henricus V. May 22 '18 at 11:26
• @EthanChan: If you zoom infinitely on those smooth curves, they will look just like straight lines. If you zoom on $|x|$ at $x=0$, you'll always see the sharp corner. – Eric Duminil May 22 '18 at 11:58
• @EthanChan: Any material realization of the curve will always be choppy since nature appears to be discrete. What Eric is talking about is the abstract representation of the curve. You are saying the gradient changes between points A and B. The "point" is, that it does change between A and B, but not suddenly. You neglected all the points between A and B, and so on. – Joris Weimar May 23 '18 at 2:21
• Additionally to what has been previously said, consider two points of a smooth curve, let's say $x<y$, with $f'(x)<f'(y)$. On these points we have different gredients, but this is not the concept of smoothness. What it really is is the following: Can we find a point $z$ very-very close to $x$ such that the curve has a gradient $f;(z)$ very-very close to $f'(x)$? In all these smooth functions, this is possible and... Wait! Why can you find something very-very close? Because it's the real line! It, by its construction, has no gaps, which should not be confused with the property of density... – Βασίλης Μάρκος May 23 '18 at 10:57
• Great answer. +1 for the "zoom" animation especially. I think it provides a suggestive visual for what we mean by smoothness. However, I see that you do not emphasise that $|x|$ cannot consistently be said to have a gradient at $x=0$. It is not the case that the curve is not differentiable because there's an abrupt change of gradient -- I don't even see how that helps beyond heuristic reasoning. There are smooth curves that change gradient suddenly too. They are still smooth. The point about non-differentiability is borne out clearly in your second animation, but not in your words. – Allawonder May 23 '18 at 13:12

Welcome to the subtleties of the real line.

I believe your perplexity here comes from sources similar to those which generated the ancient struggle to understand the continuous vs. the discrete. It is true that many things we consider as smooth in real life are only smooth to our crude senses; when observed under sufficiently powerful microscopes one sees varying degrees of roughness. Eventually, it boils down to whether there are actual infinities in the real world. But let us come back to curves in the plane.

You can't actually visualise these things, or imagine them precisely, because they involve infinity. For example, you cannot just comprehend the idea that there is no real number next to any fixed real number, say $0$. What real number is "next to" it? -- None. The question is meaningless in this context. This is what will keep you from having many headaches struggling to visualise inherently non-visual things. However, one can only understand these things logically.

So, to get to your question. I believe, first of all, that you've not yet grasped the fact that for plane curves, there is no next point. Thus, the gradient doesn't really change -- we only usually think of these things in terms of motion, but it should be understood as a rough, heuristic image, which under thorough analysis doesn't stand. Instead think of the gradient of, say, $\exp x$ as being different at every point on it. Now if you try to imagine this, you can't help thinking discretely; that's what we've ever experienced -- one by one aggregates. But in the real line there is no next point. Well, one way I've personally tried to come close to visualising this (it's actually impossible, but one can try to have a rough visual) is to imagine it as an infinitely elastic string, so that between any two points you always have other points so that there are no gaps (this is not true in real life, even for things that look continuous -- eventually you reach the atomic structure, where gaps abound, etc.).

So, the word "smooth" here is much stronger than how we use it in real life. The scientist who approximates water as a continuum, for example, knows it is not really so. But in mathematics it is actually the case that the real line contains no gaps, no matter how deep you delve into it (this is known as completeness). So thinking of the slope as changing from point to point is only an approximate heuristic for imagining these things. Any mathematician knows that there's nothing like point to point like we usually mean that phrase in real life. It's simply impossible to move from point to point, passing through every intermediate point, in $\mathbb R$ -- you always have to jump over some (this was the key assumption in Zeno's famous arguments against motion -- he assumed, taking things at face value, that spacetime is continuous in the strong mathematical sense). So functions on $\mathbb R$ are actually way subtler than they seem -- this actually occupied mathematicians who developed analysis in the 19th century.

I don't know that these things can ever be understood intuitively, for no one has never experienced the infinite in real life. We can only grasp it abstractly and be content with that -- or trouble oneself to no end.

In summary, smooth functions on $\mathbb R$ can be approximated at every point (you can't visualise this -- don't even try) by straight line functions because as you zoom in on an arbitrary point on their graph, the graph looks more and more like a straight line -- this is what we mean by saying they are differentiable. If you want to think of it in terms of motion, then it is really impossible, for the second derivative measures the limiting rate of change of slope at each point. So, if you think of the slope as changing from point to point, you can't help imagining a broken curve since we cannot just imagine the continuous. IOW, the graphs of those functions are "smooth" because they are differentiable at all points where they are defined. It makes no matter that the gradients may be different at each point. The triangle wave, for example, is "choppy" because there are some points at which it cannot be approximated by a linear function (i.e., no matter how much you zoom into it about a peak or trough, it always looks like a bent line, not a straight one).

Think of y as your car's position on a street. Think of x as time. Your speed will then be the derivative of position, so the steepness of the curve. Any function with 'breaks' will be like a ride with jolts, or even star trek beaming going on. But it is totally possible to have a smooth ride with your foot on the gas, changing speed all the time (accelerating).

Because the derivative also changes smoothly.

It's hard to put it intuitively, but imagine throwing a ball down a steep hill, it accelerates all the time (second derivative), so the speed is changing all the time (first derivative), but the position (function) is still changing smoothly (the ball doesn't suddenly teleport), unless it hits a rock (not smooth).

In calculus you would later learn that second and higher order curves ( whose representing function can be also a series of order two or more) are smooth as they are amenable to continuous differentiations.

Smooth is often defined as, “a function that has continuous derivatives up to some desired order over some domain,” or similar. Often, the desired order is two, when we want a curve to look “smooth” to the human eye. A discontinuous function will have gaps, a function with a discontinuous first derivative will have sharp jagged edges, and a function with a discontinuous second derivative has obvious inflection points where it looks glued-together from different pieces.

The exponential function is a good example: if $f(x) = e^x$, $f = f' = f'' = ...$ and the derivative of any order is positive and continuous at any real number. The result is a smooth curve. Or, if $f(x) = \sin x$ and $g(x) = \cos x$, $f' = -g$ and $f'' = -g' = f$. The first derivative, second derivative, third and so on are all continuous, so you get a smooth curve.

In contrast, when you look at $f(x) = |x|$, which has a first derivative over $x<0$ and a different one over $x>0$, but none at $x = 0$, the graph has a sharp spike at 0. A little more complicated is when you try to graph its antiderivative $F$ (choosing the constant of integration $C = 0$):

$$F(x) = \begin{cases} \frac{-x^2}{2} & \mid & x < 0 \\ \frac{x^2}{2} & \mid & x \geq 0 \end{cases}$$

So this might be “smooth enough” for some purposes (those where the “some desired order” in the definition I gave is one; that is, where a continuous first derivative is all you care about). But you can see that there’s an inflection point at zero where the second derivative doesn’t exist.

One domain where this has practical applications is when we fit a curve to points on a computer. The standard method of doing this is with cubic-spline interpolation. (That is, divide the path into intervals between the points whose values we know, and solve for piecewise parametric equations $x(t)$ and $y(t)$ between those endpoints. Cubic-spline interpolation means that they are all cubic polynomials. Frequently, we do something else that is mathematically equivalent to this but faster, such as adding a weighted sum of control points.) In order for the result to look smooth, we want the first and second derivatives to be continuous, and this nearly always is enough to look good: software engineers very rarely use higher-degree polynomials to approximate a curve. This is probably how your computer is displaying the strokes of the letters in the font you are reading right now.

• The function $|x|$ doesn't have a derivative at $x=0$. – Allawonder May 22 '18 at 16:05
• @Allawonder Good point. I had better rephrase. – Davislor May 22 '18 at 17:01