Is “locally linear” an appropriate description of a differentiable function?

Question

In this answer on meta, Pete L. Clark said:

I think the question concerns the idea that a differentiable curve becomes more and more like a straight line segment the closer one zooms in on its graph. (And I must say that I regard part of this confusion as an artifact of badly written recent calculus books who describe this phenomenon as "local linearity". Ugh!)

So, what's wrong with calling it "local linearity"? (Examples of the specific language from some relatively recent books follow.)

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From Finney, Demana, Waits, and Kennedy's Calculus: Graphical, Numerical, Algebraic, 1st ed, p107:

A good way to think of differentiable functions is that they are locally linear; that is, a function that is differentiable at a closely resembles its own tangent line very close to a.

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From Hughes-Hallett, Gleason, et al's Calculus: Single Variable, 2nd ed, pp138-9:

When we zoom in on the graph of a differentiable function, it looks like a straight line. In fact, the graph is not exactly a straight line when we zoom in; however, its deviation from straightness is so small that it can't be detected by the naked eye.

Following that, there is discussion of the tangent line approximation, then a theorem titled "Differentiability and Local Linearity" (the first time "local linearity"/"locally linear" appears) stating that if a function f is differentiable at a, then the limit as x goes to a of the quotient of the error in the tangent line approximation and the difference between x and a goes to 0.

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Ostebee and Zorn's Calculus from Graphical, Numerical, and Symbolic Points of View, 1st ed, p110:

Remarkably, the just-illustrated strategy of zooming in to estimate slope almost always works. Zooming in on the graph of almost any calculus function $f$, at almost any point $(a,f(a))$, eventually produces what looks like a straight line with slope $f'(a)$. A function with this property is sometimes called locally linear (or locally straight) at $x=a$. [Margin note: These aren't formal definitions, just descriptive phrases.] Local linearity says, in effect, that $f$ "looks like a line" near $x=a$ and therefore has a well-defined slope at $x=a$.

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(I did not find the term "local linearity" or "locally linear" at a quick glance in Stewart's Calculus: Concepts and Contexts, 2nd ed, or Leithold's The Calculus 7; the rest of the calculus books I have on hand predate the inclusion of graphing calculators/software in textbooks, so are not suitable for comparison.)

Answer 1

None of the books you quoted actually called differentiability "local linearity"; they just used as a good analogy. It is a good analogy, but it is not a good definition. A lot of mathematical terminology, especially from topology, uses the word "local," but it is almost always used with the same purpose. A locally compact set is one where points have arbitrarily small compact neighborhoods. A locally connected set is one where points have arbitrarily small connected neighborhoods. On the other hand, a non-linear differentiable function is not linear on any neighborhood of a point about which it is differentiable. It just looks linear.

Of course, such functions do have tangent lines, which is an equivalent definition of differentiability. I don't think that a tangent counts as local linearity, though.

If you're asking whether people learning calculus should be taught the words "local linearity" instead of "differentiability," I mean, that isn't really much more helpful than using the analogy but keeping the terminology. If you're asking whether we should use the analogy at all, I don't see why not, as long as it's clear that linearity is just an approximation.

EDIT: "Locally linear" only describes differentiable functions $\mathbb{R}\rightarrow\mathbb{R}$. "Differentiable" can be extended to functions between arbitrary differential manifolds. I think it's better to keep the extensible definition.

Answer 2

The "microscope" or "zooming in" explanation of differentiation is basically nonsense. It contradicts the notion of zooming in actually used in calculus (asymptotic expansion of a function near a point, revealing finer and finer information). Magnification matches the mathematics only at order 0 and 1, that is, it captures linear information but also destroys the higher order picture.

Suppose that $f(0)=0$. The $N$-fold magnification of the graph near 0 is the plot of $Nf(x/N)$. This is the straight line $xf'(0)$ to within an error that shrinks to an invisible size as the picture is "zoomed" by increasing $N$. This will, literally, linearize any differentiable function. The higher order approximations -- whose effect is beautifully visible in a fixed scale picture of the first several Taylor polynomials for $f(x)$ near 0 all superimposed on the same coordinate system -- are made indistinguishable from a straight line by flattening everything with the magnification. This is a step backward considering that the difference between a tangent line and an osculating circle is easily illustrated with any curve, and many people see this picture in some form before learning calculus.

What one actually wants to display are the effects away from the point, with the higher approximations being visibly closer to graph over a longer range. The "zooming" picture may be defensible if presented together with these more informative images, but by itself as an explanation of the derivative as a local linear approximation, it is using diagrams that tell a story opposite to the mathematical concepts involved.

Answer 3

@Paul and @Issac's comments on a wrong answer I had initially posted made me see why "linear on zooming in" is the right picture of differentiability. Suppose $f$ is some function and

$$g(x)=f(a)+m(x-a)$$

is a proposed linear approximation to it at a point $a$.

Suppose I zoom in on the graph of $f$ and $g$ by a multiple $\lambda>0$ and look at a point $h$ distance apart from $a$ in my zoomed coordinates. Then the vertical error in the approximation in my zoomed coordinates is

$$e(\lambda)=\lambda[f(a+h/\lambda)-g(a+h/\lambda)] =\lambda[f(a+h/\lambda)-f(a)]-mh$$

If I keep zooming the approximation will look better and better only if $\lim_{\lambda \to \infty} e(\lambda)=0$. This will be the case only if $f$ is differentiable at $a$ and $f'(a)=m$.