# College Math Teaching

## October 4, 2018

### When is it ok to lie to students? part I

Filed under: calculus, derivatives, pedagogy — collegemathteaching @ 9:32 pm

We’ve arrived at logarithms in our calculus class, and, of course, I explained that $ln(ab) = ln(a) + ln(b)$ only holds for $a, b > 0$. That is all well and good.
And yes, I explained that expressions like $f(x)^{g(x)}$ only makes sense when $f(x) > 0$

But then I went ahead and did a problem of the following type: given $f(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1}$ by using logarithmic differentiation, $f'(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1} (\frac{3}{x} + 2x -tan(x) -\frac{4x^3}{x^4+ 1})$

And you KNOW exactly what I did. Right?

Note that $f$ is differentiable for all $x$ and, well, the derivative *should* be continuous for all $x$ but..is it? Well, up to inessential singularities, it is. You see: the second factor is not defined for $x = 0, x = \frac{\pi}{2} \pm k \pi$, etc.

Well, let’s multiply it out and obtain: $f'(x) = \frac{3x^2 e^{x^2} cos(x)}{x^4 + 1} + \frac{2x^4 e^{x^2} cos(x)}{x^4 + 1} - \frac{x^3 e^{x^2} sin(x)}{x^4 + 1}-\frac{4x^6 e^{x^2} cos(x)}{(x^4 + 1)^2}$

So, there is that. We might induce inessential singularities.

And there is the following: in the process of finding the derivative to begin with we did: $ln(\frac{x^3 e^{x^2} cos(x)}{x^4 + 1}) = ln(x^3) + ln(e^{x^2}) + ln(cos(x)) - ln(x^4 + 1)$ and that expansion is valid only for $x \in (0, \frac{\pi}{2}) \cup (\frac{5\pi}{2}, \frac{7\pi}{2}) \cup ....$ because we need $x^3 > 0$ and $cos(x) > 0$.

But the derivative formula works anyway. So what is the formula?

It is: if $f = \prod_{j=1}^k f_j$ where $f_j$ is differentiable, then $f' = \sum_{i=1}^k f'_i \prod_{j =1, j \neq i}^k f_j$ and verifying this is an easy exercise in induction.

But the logarithmic differentiation is really just a motivating idea that works for positive functions.

To make this complete: we’ll now tackle $y = f(x)^{g(x)}$ where it is essential that $f(x) > 0$.

Rewrite $y = e^{ln(f(x)^{g(x)})} = e^{g(x)ln(f(x))}$

Then $y' = e^{g(x)ln(f(x))} (g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)}) = f(x)^{g(x)}(g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)})$

This formula is a bit of a universal one. Let’s examine two special cases.

Suppose $g(x) = k$ some constant. Then $g'(x) =0$ and the formula becomes $y = f(x)^k(k \frac{f'(x)}{f(x)}) = kf(x)^{k-1}f'(x)$ which is just the usual constant power rule with the chain rule.

Now suppose $f(x) = a$ for some positive constant. Then $f'(x) = 0$ and the formula becomes $y = a^{g(x)}(ln(a)g'(x))$ which is the usual exponential function differentiation formula combined with the chain rule.

## September 8, 2018

### Proving a differentiation formula for f(x) = x ^(p/q) with algebra

Filed under: calculus, derivatives, elementary mathematics, pedagogy — collegemathteaching @ 1:55 am

Yes, I know that the proper way to do this is to prove the derivative formula for $f(x) = x^n$ and then use, say, the implicit function theorem or perhaps the chain rule.

But an early question asked students to use the difference quotient method to find the derivative function (ok, the “gradient”) for $f(x) = x^{\frac{3}{2}}$ And yes, one way to do this is to simplify the difference quotient $\frac{t^{\frac{3}{2}} -x^{\frac{3}{2}} }{t-x}$ by factoring $t^{\frac{1}{2}} -x^{\frac{1}{2}}$ from both the numerator and the denominator of the difference quotient. But this is rather ad-hoc, I think.

So what would one do with, say, $f(x) = x^{\frac{p}{q}}$ where $p, q$ are positive integers?

One way: look at the difference quotient: $\frac{t^{\frac{p}{q}}-x^{\frac{p}{q}}}{t-x}$ and do the following (before attempting a limit, of course): let $u= t^{\frac{1}{q}}, v =x^{\frac{1}{q}}$ at which our difference quotient becomes: $\frac{u^p-v^p}{u^q -v^q}$

Now it is clear that $u-v$ is a common factor..but HOW it factors is essential.

So let’s look at a little bit of elementary algebra: one can show: $x^{n+1} - y^{n+1} = (x-y) (x^n + x^{n-1}y + x^{n-2}y^2 + ...+ xy^{n-1} + y^n)$ $= (x-y)\sum^{n}_{i=0} x^{n-i}y^i$ (hint: very much like the geometric sum proof).

Using this: $\frac{u^p-v^p}{u^q -v^q} = \frac{(u-v)\sum^{p-1}_{i=0} u^{p-1-i}v^i}{(u-v)\sum^{q-1}_{i=0} u^{q-1-i}v^i}=\frac{\sum^{p-1}_{i=0} u^{p-1-i}v^i}{\sum^{q-1}_{i=0} u^{q-1-i}v^i}$ Now as $t \rightarrow x$ we have $u \rightarrow v$ (for the purposes of substitution) so we end up with: $\frac{\sum^{p-1}_{i=0} v^{p-1-i}v^i}{\sum^{q-1}_{i=0} v^{q-1-i}v^i} = \frac{pv^{p-1}}{qv^{q-1}} = \frac{p}{q}v^{p-q}$ (the number of terms is easy to count).

Now back substitute to obtain $\frac{p}{q} x^{\frac{(p-q)}{q}} = \frac{p}{q} x^{\frac{p}{q}-1}$ which, of course, is the familiar formula.

Note that this algebraic identity could have been used for the old $f(x) = x^n$ case to begin with.

## August 28, 2018

### Commentary: what does it mean to “graduate from college”?

Filed under: editorial — collegemathteaching @ 1:21 am

Recently, an Oregon university touted graduating someone with Down’s syndrome:

Walking across the stage at graduation was more than just a personal accomplishment for Cody Sullivan as he became Oregon’s first student with Down syndrome to complete four years of college.

Sullivan, 22, received his certificate of achievement at the Concordia University graduation ceremony last month, declaring that while assignments and curriculum were modified for his learning abilities, Sullivan completed all the relevant coursework to make him an official college graduate.

It is every interestingly worded: “certificate of achievement” and “assignments and curriculum were modified for his learning abilities”.

This represents a different point of view than I have.

When a teach a course, getting a certain grade in a course requires that the person getting grade to master certain concepts and skills at a certain level. Those requirements are NOT modified for someone’s learning ability. And getting a degree in a certain subject means (or should mean) that one has established a certain competency in that said subject.

But, well, I wonder if we are moving toward a “meeting a certain competency level isn’t relevant” anymore and just giving “you were here and did stuff” certificates.

There was a time when I thought “aptitude matters” but, well?

### Conditional Probability in the news..

Filed under: probability — Tags: , — collegemathteaching @ 1:11 am

I am going to stay in my lane here and not weigh in on a social science issue. But I will comment on this article, which I was alerted to here. This is from the Atlantic article:

When the ACLU report came out in 2017, Dyer told the Fresno Bee the findings of racial disparities were “without merit” but also said that the disproportionate use of force corresponds with high crime populations. At the end of our conversation, Dyer pointed to a printout he brought with him, a list of the department’s “most wanted” people. “We can’t plug in a bunch of white guys,” he said. “You know who’s shooting black people? Black people. It’s black-on-black crime.”

But so-called “black-on-black crime” as an explanation for heightened policing of black communities has been widely debunked. A recent study by the U.S. Department of Justice found that, overwhelmingly, violent crimes are committed by people who are the same race as their victims. “Black-on-black” crime rates, the study found, are comparable to “white-on-white” crime rates.

So, just what did that “recent study” find? I put a link to it, but basically, it said that most white crime victims were the victim of a white criminal and that most black victims were the victim of a black criminal. THAT is their “debunking”. That is a conditional probability: GIVEN that you were a crime victim to begin with, then the perpetrator was probably of the same race.

That says nothing about how likely a white or a black person was to be a crime victim to being with. From the blog post critiquing the Atlantic article:

What the rest of us mean by “black-on-black crime rate” is the overall rate at which blacks victimize others or the rate at which they are victimized themselves––which, for homicide, has ranged from 6 to 8 times higher than for whites in recent decades. Homicide is the leading cause of death for black boys/men aged 15-19, 20-24, and 25-34, according to the CDC. That fact cannot be said about any other ethnicity/age combination. Blacks only make up 14% of the population. But about half of the murdered bodies that turn up in this country are black bodies (to use a phrase in vogue on the identitarian Left), year in and year out.

In short, blacks are far more often to be the crime victim too. Even the study that the Atlantic article linked to shows this.

Anyhow, that is a nice example of conditional probability.

## August 27, 2018

### On teaching limits poorly

Filed under: calculus, pedagogy — Tags: — collegemathteaching @ 4:52 pm

I will be talking about teaching limits in a first year calculus class.

The textbook our department is using does the typical: It APPEARS to be making the claim that the limit of the given function is 4 as $x$ approaches 2 because, well, 4 is between $f(2.001)$ and $f(1.999)$. But, there are an uncountable number of numbers between those two values; one really needs that the function in question “preserves integers” in order to give a good reason to “guess” that the limit is indeed 4.

I think that the important thing here is that the range is being squeezed as the domain gets squeezed, and, in my honest opinion, THAT is the point of limits: the limit exists when one can tighten the range tolerance by sufficiently tightening the domain tolerance.

But, in general, it is impossible to guess the limit without extra information about the function (e. g. maps integers to integers, etc.)

## August 20, 2018

### Algebra for Calculus I: equations and inequalities

Filed under: basic algebra, calculus, pedagogy — collegemathteaching @ 9:24 pm

It seems simple enough: solve $3x+ 4 = 7$ or $\frac{2}{x-5} \leq 3$.

So what do we tell our students to do? We might say things like “with an equation we must do the same thing to both sides of the equation (other than multiply both sides by zero)” and with an inequality, “we have to remember to reverse the inequality if we, say, multiply both sides by a negative number or if we take the reciprocal”.

And, of course, we need to check afterwards to see if we haven’t improperly expanded the solution set.

But what is really going on? A moment’s thought will reveal that what we are doing is applying the appropriate function to both sides of the equation/inequality.

And, depending on what we are doing, we want to ensure that the function that we are applying is one-to-one and taking note if the function is increasing or decreasing in the event we are solving an inequality.

Example: $x + \sqrt{x+2} = 4$ Now the standard way is to subtract $x$ from both sides (which is a one to one function..subtract constant number) which yields $\sqrt{x+2} = 4-x$. Now we might say “square both sides” to obtain $x+2 = 16-8x+x^2 \rightarrow x^2-9x+ 14 = 0 \rightarrow (x-7)(x-2) = 0$ but only $x = 2$ works. But the function that does that, the “squaring function”, is NOT one to one. Think of it this way: if we have $x = y$ and we then square both sides we now have $x^2 = y^2$ which has the original solution $x = y$ and $x = -y$. So in our example, the extraneous solution occurs because $(\sqrt{7+2})^2 = (4-7)^2$ but $\sqrt{7+2} \neq -3$.

If you want to have more fun, try a function that isn’t even close to being one to one; e. g. solve $x + \frac{1}{4} =\frac{1}{2}$ by taking the sine of both sides. 🙂

(yes, I know, NO ONE would want to do that).

As far as inequalities: the idea is to remember that if one applies a one-to-one function on both sides, one should note if the function is increasing or decreasing.

Example: $2 \geq e^{-x} \rightarrow ln(2) \geq -x \rightarrow ln(\frac{1}{2}) \leq x$. We did the switch when the function that we applied ( $f(x) = -x$ was decreasing.)

Example: solving $|x+9| \geq 8$ requires that we use the conditional definition for absolute value and reconcile our two answers: $x+ 9 \geq 8$ and $-x-9 \geq 8$ which leads to the union of $x \geq -1$ or $x \leq -17$

The fun starts when the function that we apply is neither decreasing nor increasing. Example: $sin(x) \geq \frac{1}{2}$ Needless to say, the $arcsin(x)$ function, by itself, is inadequate without adjusting for periodicity.

## August 3, 2018

Filed under: academia, editorial, research — collegemathteaching @ 12:52 am

Ok, it is nearing the end of the summer and I feel as if I am nearing the end of a paper that I have been working on for some time. Yes, I am confident that it will get accepted somewhere, though I will submit it to my “first choice” journal when it is ready to go. I have 6 diagrams to draw up, put in, and then to do yet another grammar/spelling/consistent usage check.

Part of this “comes with the territory” of trying to stay active when teaching at a non-research intensive school; one tends to tackle such projects in “modules” and then try to put them together in a seamless fashion.

But that isn’t my rant.

My rant (which might seem strange to younger faculty):

A long time ago, one would work on a paper and write it longhand and ..if you were a professor, have the technical secretary type it up. Or one would just use a word processor of some sort and make up your Greek characters by hand. You’d submit it, and if it were accepted, the publisher would have it typeset.

Now: YOU are expected to do the typesetting and that can be very time consuming. YOU are expected to make camera ready diagrams.

And guess what: you aren’t paid for your article. The editor isn’t paid. The referee(s) isn’t (aren’t) paid. But the journal still charges subscription fees, sometimes outrageously high fees. And these are the standard journals, not the “fly by night” predatory journals.

This is another case where the professor’s workload went up, someone else’s expense went down, and the professor received no extra benefit.

Yes, I know, “cry me a river”, blah, blah, blah. But in this respect, academia HAS changed and not for the better.

## June 20, 2018

### Editorial: one major disconnect between us and much of the public ..

Filed under: editorial — Tags: , , , , — collegemathteaching @ 1:52 am

The University of Chicago decided to stop requiring the ACT/SAT of its applicants. Now never in a million years would I give a suggestion to the University of Chicago (or any other elite school) as to what their admissions/applicant policies should be.

But there is a broader “scrap the college entrance exams” movement out there and much of the justification you hear is just complete nonsense. Example: “we have data that says the high school gpa is a better predictor of X”. (X meaning “first year success”, or “graduation”) Now that may be true, but why stop with just one bit of information if the second bit, taken together, increases predictive power?

And there is a second claim from those who admit that not all high schools are created equal, and an A in, say, high school calculus in one school might mean less than an A from another school: admitting that the quality of high schools vary means that you are just punishing the students from the academically weaker high schools a second time when you use a college entrance exam.

That claim misses the point entirely. Many schools (like ours) uses the score, at least in part, for placement purposes (we aren’t that tough to get into). And we have have decades of data that shows that, yes, the math ACT score matters, in terms of success in first year calculus. This isn’t our school (it is the University of Michigan), but we have very similar results.

And this brings us to the disconnect in attitudes.

1. We use scores to determine if the student has a reasonable probability of success in, say, a freshman calculus course. Now of course, sometimes someone under the cut-off has success. But if you give too much benefit of the doubt to prospective students, your DFW rate (D’s, F’s, Withdraws) will climb and administrators such be made aware of the trade-off.

2. We also understand that aptitude matters. There are many (more than you think) that aptitude has no role, or a very minor role (“you can do anything you want to do if you put your mind to it”, etc.) and some who embrace “blank slate” thinking (to them, aptitude is a fiction).

I suppose that people who REALLY believe “2” believe that, say, recruiting plays no role in the success of college sports team..a good coach can just draw from the student body and win games.

3. Part of the role of, say, the calculus sequence is to identify those who have a good probability of success in certain majors. Let’s face it; if you really can’t calculate $\frac{d}{dx}sin(2x)$ you have no business being an engineer. Yes, on rare occasion, I’ve had students flunk my class in science/engineering calculus class because they really could not do that.

## June 18, 2018

### And my “clever proof” is dashed

Filed under: complex variables, editorial, knot theory, numerical methods, topology — Tags: , — collegemathteaching @ 6:03 pm

It has been a while since I posted here, though I have been regularly posting in my complex variables class blog last semester.

And for those who like complex variables and numerical analysis, this is an exciting, interesting development.

But as to the title of my post: I was working to finish up a proof that one kind of wild knot is not “equivalent” to a different kind of wild knot and I had developed a proof (so I think) that the complement of one knot contains an infinite collection of inequivalent tori (whose solid tori contain the knot non-trivially) whereas the other kind of knot can only have a finite number of such tori. I still like the proof.

But it turns out that there is already an invariant that does the trick nicely..hence I can shorten and simplify the paper.

But dang it..I liked my (now irrelevant to my intended result) result!

## April 24, 2018

### And I trolled my complex variables class

Filed under: advanced mathematics, analysis, class room experiment, complex variables — collegemathteaching @ 6:34 pm

One question on my last exam: find the Laurent series for $\frac{1}{z + 2i}$ centered at $z = -2i$ which converges on the punctured disk $|z+2i| > 0$. And yes, about half the class missed it.

I am truly evil.