# College Math Teaching

## January 15, 2019

### Calculus series: derivatives

Filed under: calculus, derivatives — collegemathteaching @ 3:36 am

Reminder: this series is NOT for the student who is attempting to learn calculus for the first time.

Derivatives This is dealing with differentiable functions $f: R^1 \rightarrow R^1$ and no, I will NOT be talking about maps between tangent bundles. Yes, my differential geometry and differential topology courses were on the order of 30 years ago or so. ðŸ™‚

In calculus 1, we typically use the following definitions for the derivative of a function at a point: $lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} = lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} = f'(a)$. This is opposed to the derivative function which can be thought of as the one dimensional gradient of $f$.

The first definition is easier to use for some calculations, say, calculating the derivative of $f(x) = x ^{\frac{p}{q}}$ at a point. (hint, if you need one: use $u = x^{\frac{1}{q}}$ then it is easier to factor). It can be used for proving a special case of the chain rule as well (the case there we are evaluating $f$ at $x = a$ and $f(x) = f(t)$ for at most a finite number of points near $a$.)

When introducing this concept, the binomial expansion theorem is very handy to use for many of the calculations.

Now there is another definition for the derivative that is helpful when proving the chain rule (sans restrictions).

Note that as $h \rightarrow 0$ we have $|\frac{f(a+h)-f(a)}{h} - f'(a)| < \epsilon$. We can now view $\epsilon$ as a function of $h$ which goes to zero as $h$ does.

That is, $f(a+h) = hf'(a) + f(a) + \frac{\epsilon}{h}$ where $\frac{\epsilon}{h} \rightarrow 0$ and $f'(a)$ is the best linear approximation for $f$ at $x = a$.

We’ll talk about the chain rule a bit later.

But what about the derivative and examples?

It is common to develop intuition for the derivative as applied to nice, smooth..ok, analytic functions. And this might be a fine thing to do for beginning calculus students. But future math majors might benefit from being exposed to just a bit more so I’ll give some examples.

Now, of course, being differentiable at a point means being continuous there (the limit of the numerator of the difference quotient must go to zero for the derivative to exist). And we all know examples of a function being continuous at a point but not being differentiable there. Examples: $|x|, x^{\frac{1}{3}}, x^{\frac{2}{3}}$ are all continuous at zero but none are differentiable there; these give examples of a corner, vertical tangent and a cusp respectively.

But for many of the piecewise defined examples, say, $f(x) = x$ for $x < 0$ and $x^2 + x$ for $x \geq 0$ the derivative fails to exist because the respective derivative functions fail to be continuous at $x =0$; the same is true of the other stated examples.

And of course, we can show that $x^{\frac{3k +2}{3}}$ has $k$ continuous derivatives at the origin but not $k+1$ derivatives.

But what about a function with a discontinuous derivative? Try $f(x) = x^2 sin(\frac{1}{x})$ for $x \neq 0$ and zero at $x =0$. It is easy to see that the derivative exists for all $x$ but the first derivative fails to be continuous at the origin.

The derivative is $0$ at $x = 0$ and $2x sin(\frac{1}{x}) -cos(\frac{1}{x})$ for $x \neq 0$ which is not continuous at the origin.

Ok, what about a function that is differentiable at a single point only? There are different constructions, but if $f(x) = x^2$ for $x$ rational, $x^3$ for $x$ irrational is both continuous and, yes, differentiable at $x = 0$ (nice application of the Squeeze Theorem on the difference quotient).

Yes, there are everywhere continuous, nowhere differentiable functions.

## January 14, 2019

### New series in calculus: nuances and deeper explanations/examples

Filed under: calculus, cantor set — Tags: — collegemathteaching @ 3:07 am

Though I’ve been busy both learning and creating new mathematics (that is, teaching “new to me” courses and writing papers to submit for publication) I have not written much here. I’ve decided to write up some notes on, yes, calculus. These notes are NOT for the average student who is learning for the first time but rather for the busy TA or new instructor; it is just to get the juices flowing. Someday I might decide to write these notes up more formally and create something like “an instructor’s guide to calculus.”

I’ll pick topics that we often talk about and expand on them, giving suggested examples and proofs.

First example: Continuity. Of course, we say $f$ is continuous at $x = a$ if $lim_{x \rightarrow a} f(x) = f(a)$ which means that the limit exists and is equal to the function evaluated at the point. In analysis notation: for all $\epsilon > 0$ there exists $\delta > 0$ such that $|f(a)-f(x)| < \epsilon$ whenever $|a-x| < \delta$.

Of course, I see this as “for every open $U$ containing $f(a)$, $f^{-1}(U)$ is an open set. But never mind that for now.

So, what are some decent examples other than the usual “jump discontinuities” and “asymptotes” examples?

A function that is continuous at exactly one point: try $f(x) = x$ for $x$ rational and $f(x) = x^2$ for $x$ irrational.

A function that oscillates infinitely often near a point but is continuous: $f(x) = xsin(\frac{1}{x})$ for $x \neq 0$ and zero at $x = 0$.

A bounded unction with a non-jump discontinuity but is continuous for all $x \neq 0$: $f(x) = sin(\frac{1}{x})$ for $x \neq 0$ and zero at $x = 0$.

An unbounded function without an asymptote but is continuous for all $x \neq 0$ $f(x) = \frac{1}{x} sin(\frac{1}{x})$ for $x \neq 0$ and zero at $x = 0$.

A nowhere continuous function: $f(x) = 1$ for $x$ rational, and $0$ for $x$ irrational.

If you want an advanced example which blows the “a function is continuous if its graph can be drawn without lifting the pencil off of the paper, try the Cantor function. (this function is continuous on $[0,1]$, has derivative equal to zero almost everywhere, and yet increases from 0 to 1.