1
$\begingroup$

We always speak of the derivative as being the “best linear approximation”. And we also speak of linearizing. However, what does this really mean? For a given function $F$, what conditions on it make the claim “the derivative is the best linear approximation to $F$” true?

Are there functions that can’t be “locally linear” or locally approximated? If so, are these just mostly pathological, and we don’t have any interest in them (e.g. they don’t really show up in math)?

Are there important functions or mathematical objects that don’t really subject themselves well to the tools of analysis and approximation? (I understand this is a very broad and vague question.) I mean, there may be mathematical objects that we don’t know whether they are amenable to such efforts, but are there (important) objects where we are sure they definitely aren’t? Sort of like how abstract algebra/Galois theory showed the limitations of using radicals, giving rise to the notion of unsolvability?

$\endgroup$
  • 4
    $\begingroup$ Are you asking about non-differentiable functions? $\endgroup$ – Theoretical Economist May 1 at 1:20
  • 4
    $\begingroup$ the concept of a differentiable function is the formalisation of the intuitive idea of "locally well approximated by a linear function" and really the whole of differential calculus is devoted to approximating by a linear function/space etc. Take a look at math.stackexchange.com/a/3650987/568204 this recent answer of mine. So unless you have a specifically different meaning of "local linear approximation" in mind, the relevant concept is that of a non-differentiable function. And there are plenty of those (for example the absolute value function on the real line). $\endgroup$ – peek-a-boo May 1 at 1:30
  • $\begingroup$ @TheoreticalEconomist I suppose that's the obvious answer lol, sorry I know my question was vague. I was just thinking about the general concept of approximating nonlinear functions by linear functions, and I was wondering why this is valid if many nonlinear functions may not be differentiable? $\endgroup$ – twosigma May 1 at 3:38
  • $\begingroup$ @peek-a-boo Thanks! I can see how that is true for differentiable functions $\endgroup$ – twosigma May 1 at 3:41
5
$\begingroup$

However, what does this really mean? For a given function F, what conditions on it make the claim “the derivative is the best linear approximation to F” true?

It is unconditionally true. It is well-explained in the link of the comment.

Are there functions that can’t be “locally linear” or locally approximated?

Yes, a lot of them.

If so, are these just mostly pathological, and we don’t have any interest in them (e.g. they don’t really show up in math)?

No, they show up a lot and lot and lots of times in math, physics, everywhere.

Are there important functions or mathematical objects that don’t really subject themselves well to the tools of analysis and approximation?

Just consider a euclidean norm function : $f(x)=\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$. This is not differentiable at the origin. (i.e. cannot be linearly approximated) Without this, you will not even able to talk about what is distance between two given points. Some other nontrivial examples will be heaviside step function (this is not even continuous), and dirac-delta function (actually this is not even a function! If you are interested, look for the distribution theory).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Adding on to this great answer, there are also continuous functions which are nowhere differentiable (eg: the Wierstrauss Function, wikiwand.com/en/Weierstrass_function). However, these do tend to be more pathological $\endgroup$ – memerson May 1 at 2:30
  • $\begingroup$ Probably the simplest example of a non-differentiable but otherwise nice function is $x\mapsto|x|$, the $n=1$ case of the Euclidean norm in this answer. It has no linear approximation near $0$. $\endgroup$ – Andreas Blass May 1 at 2:35

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.