College Math Teaching

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.


June 15, 2016

Elementary Math in the news: elections

Filed under: calculus, elementary mathematics, news — Tags: — collegemathteaching @ 9:11 pm

Ok, mostly I am trying to avoid writing up the painful details of a proposed mathematics paper.
But I do follow elections relatively closely. In the California Democratic primary, CNN called the election for Hillary Clinton late on June 7; at the time she lead Bernie Sanders 1,940,588-1,502,043, which is a margin of 438,537 votes. Percentage wise, the lead was 55.8-43.2, or 12.6 percentage points.

But due to mail in balloting and provisional ballot counting, there were still many votes to count. As of this morning, the totals were:

2,360,266-1,887,178 for a numerical lead of 473,088 votes. Percentage wise, the lead was 55.1-44.0, or 11.1 percentage points.

So, the lead grew numerically, but shrunk percentage wise.

“Big deal”, you say? Well, from reading social media, it is not obvious (to some) how a lead can grow numerically but shrink as a percentage.

Conceptually, it is pretty easy to explain: suppose one has an election involving 1100 voters who MUST choose between candidates. Say the first 100 votes that are counted happened to come from a strongly pro-Hillary group, and the tally after 100 was 90 Hillary, 10 Bernie. Then suppose the next 1000 was closer, say 550 for Hillary and 450 for Bernie. Then the lead grew by 100 votes (80 to 180) but the percentage lead shrunk from 80 percentage points to a 16.36 percentage point lead (58.18 to 41.82 percent). And it is easy to see that if the rest of the vote was really 55 percent Hillary, her percent of the vote would asymptotically shrink to close to 55 percent as the number of votes counted went up.

So, how might one have students model it? Let H(t), B(t) be increasing functions of t which represent the number of votes for Hillary and Bernie as a function of time. Assume no mistakes, hence H(t), B(t) can be assumed to be increasing functions. So we want a case there D(t) = H(t)-B(t) is an increasing function but P(t) = \frac{H(t)}{H(t)+ B(t)} decreases with time.

Without calculus: rewrite P(t) = \frac{1}{1+\frac{B(t)}{H(t)}} and note that P(t) decreases as \frac{B(t)}{H(t)} increases; that is, as B(t) outgrows H(t) . But H(t) must continue to outgrow B(t) . That is, the new ballots must still include more Hillary Bernie ballots, but the ratio of Bernie ballots to Hillary ballots must be going down.

If we use some calculus, we see that H'(t) must exceed B'(t) but to make P(t) decrease, use the quotient rule plus a tiny bit of algebra to conclude that H'(t)B(t)-B'(t)H(t) must be negative, or that \frac{B'(t)}{B(t)} > \frac{H'(t)}{H(t)} . That is, the Bernie ballots must be growing at a higher percentage rate than the Hillary ballots are.

None of this is surprising, but it might let the students get a feel of what derivatives are and what proportional change means.

December 22, 2015

Multi leaf polar graphs and total area…

Filed under: calculus, elementary mathematics, integrals — Tags: , — collegemathteaching @ 4:07 am

I saw polar coordinate calculus for the first time in 1977. I’ve taught calculus as a TA and as a professor since 1987. And yet, I’ve never thought of this simple little fact.

Consider r(\theta) = sin(n \theta), 0 \theta \ 2 \pi . Now it is well know that the area formula (area enclosed by a polar graph, assuming no “doubling”, self intersections, etc.) is A = \frac{1}{2} \int^b_a (r(\theta))^2 d \theta

Now the leaved roses have the following types of graphs: n leaves if n is odd, and 2n leaves if n is even (in the odd case, the graph doubles itself).




So here is the question: how much total area is covered by the graph (all the leaves put together, do NOT count “overlapping”)?

Well, for n an integer, the answer is: \frac{\pi}{4} if n is odd, and \frac{\pi}{2} if n is even! That’s it! Want to know why?

Do the integral: if n is odd, our total area is \frac{n}{2}\int^{\frac{\pi}{n}}_0 (sin(n \theta)^2 d\theta = \frac{n}{2}\int^{\frac{\pi}{n}}_0 \frac{1}{2} + cos(2n\theta) d\theta =\frac{\pi}{4} . If n is even, we have the same integral but the outside coefficient is \frac{2n}{2} = n which is the only difference. Aside from parity, the number of leaves does not matter as to the total area!

Now the fun starts when one considers a fractional multiple of \theta and I might ponder that some.

August 1, 2015

Interest Theory: discounting

Filed under: applied mathematics, elementary mathematics — Tags: , — collegemathteaching @ 10:29 pm

Some time ago, I served in the U. S. Navy. The world “Navy” was said to be an acronym for Never Again Volunteer Yourself. But I forgot that and volunteered to teach a class on Mathematical interest theory. That means, of course, I have to learn some of this, and so I am going over a classic text and doing the homework.

The math itself is pretty simple, but some of the concepts seem strange to me at this time. So, I’ll be using this as “self study” prior to the start of the semester, and perhaps I’ll put more notes up as I go along.

By the way, if you are interested in the notes for my undergraduate topology class, you can find them here.

Discounting: concepts, etc. (from this text) (Kellison)

Initial concept:

Suppose you borrow 100 dollars for one year at 8 percent interest. So at time 0 you have 100 dollars and at time 1, you pay back 100 + (100)(.08) = 108.
Now let’s do something similar via “discounting”. The contract is for 100 dollars and the rate is an 8 percent discount. The bank takes their 8 percent AT THE START and you end up with 92 dollars at time zero and pay back 100 at time 1.

So the difference is: in interest, the interest is paid upon pay back, and so the amount function is: A(t) = (1+it)A(0) . In the discount situation we have A(1)(1-d(1)) = A(0) where d is the discount rate. So the amount function is A(t) = \frac{A(0)}{1-dt} where t \in [0, \frac{1}{d})

If we used compound interest, we’d have A(t) = (1+i)^tA(0) and in compound discount we’d have A(t) = \frac{A(0)}{(1-d)^t}

This leads to some interesting concepts.

First of all, there is the “equivalence concept”. Think about the above example: if getting 92 dollars now lead to 100 dollars after one period, what interest rate would that be? Of course it would be \frac{8}{92} = .087. So what we’d have is this: i = \frac{d}{1-d} or d = \frac{i}{1+i} .

Effective rates: this is only of interest in the “simple interest” or “simple discount” situation.

Let’s start with simple interest. The amount function is of the form A(t) = (1 +it)A(0) . The idea is that if you invest, say, 100 dollars earning, say, 5 percent simple interest (NO compounding), then in one year you get 5 dollars of interest, 2 years, 10 dollars of interest, 3 years 15 dollars of interest, etc. You can see the problem here; say at the end of year one your account was worth 105 dollars and at the end of year 2, it was worth 110 dollars. So, in effect, your 105 dollars earned 5 dollars interest in the second year. Effectively, you earned a lower rate in year 2. It got worse in year 3 (110 earned only 5 dollars).

So the EFFECTIVE INTEREST in period n is \frac{A(n) - A(n-1)}{A(n-1)} = \frac{1 + ni)-(1+(n-1)i)}{1+(n-1)i}=\frac{i}{1+(n-1)i} which you can see goes to zero as n goes to infinity.

Effective discount works in a similar manner, though we divide by the amount at the end of the period, rather than the beginning of it:

\frac{A(n)-A(n-1)}{A(n)} = \frac{\frac{1}{1-nd} - \frac{1}{1-(n-1)d}}{\frac{1}{1-nd}} = \frac{d}{1-(n-1)d}

May 31, 2015

And a Fields Medalist makes me feel better

Filed under: calculus, editorial, elementary mathematics, popular mathematics, topology — Tags: — collegemathteaching @ 10:30 pm

I have subscribed to Terence Tao’s blog.

His latest post is about a clever observation about…calculus: in particular is is about calculating:

\frac{d^{k+1}}{dx^{k+1}}(1+x^2)^{\frac{k}{2}} for k \in \{1, 2, 3, ... \} . Try this yourself and surf to his post to see the “slick, 3 line proof”.

But that really isn’t the point of this post.

This is the point: I often delight in finding something “fun” and “new to me” about an established area. I thought “well, that is because I am too dumb to do the really hard stuff.” (Yes, I’ve published, but my results are not Annals of Mathematics caliber stuff. ๐Ÿ™‚ )

But I see that even the smartest, most accomplished among us can delight in the fun, simple things.

That makes me feel better.

Side note: I haven’t published much on this blog lately, mostly because I’ve been busy updating this one. It is a blog giving notes for my undergraduate topology class. That class was time consuming, but I had the teaching time of my life. I hope that my students enjoyed it too.

April 8, 2015

How my running (and walking and swimming and weight lifting and sports experience) helps with my teaching

Filed under: algebra, editorial, elementary mathematics — Tags: — collegemathteaching @ 4:52 pm

If you teach at an institution that has a competitive sports team, you’ll probably notice that the coaches spend time on recruiting. It is easy to see why: though athletes train hard to improve their performances, their inherent athletic ability provides an upper bound of how well they will do.

I played sports in high school, but wasn’t within an AU of being able to compete at the college level, any division. I remember summer wrestling; those recruited to wrestle for our team basically had their way with me on the mat.

In my current sports, I always do poorly in competition. For example, in my best running marathon, the winner beat me by 74 minutes! (winning time was 2:19).

It wasn’t that I didn’t try or didn’t train: it was that because I am a poor natural athlete, training only “moves the needle” just a bit, and not nearly enough for me to be competitive.

A coach could give me this workout or that workout…and get angry with me. But I have athletic limitations.

The same principle applies in mathematics.

Right now I am teaching the second semester of calculus for non-technical majors.

One question was: find the maximum and minimum of T(t) = 55 -21 cos(\frac{2 \pi (t-32)}{365}) . They were told that this function modeled daily temperature where t was in days, t=0 on January 1.

Now I asked the class some questions. And, well, let’s just say that they didn’t just recognize what the various terms and factors meant.

Now we took the derivative to find the local maximum and local minimum values and most of them got \frac{42 \pi}{365}sin(\frac{2 \pi (t-32)}{365}) . Now we set this equal to zero and all of them that we got zero when the argument was 0 or an integral multiple of \pi .

But now, when we had \frac{2 \pi (t-32)}{365} = 0 I said “of course, this gives us the solution t = 32 . And you guessed it…one of the students asked “why”. It took about a minute of explanation for her to see it. I kid you not.

So, I reminded myself of what it must have been like for my sports coaches in high school…..what it was like for them to work with me.

January 23, 2015

Making a math professor happy…

Filed under: calculus, class room experiment, elementary mathematics — Tags: , — collegemathteaching @ 10:28 pm

Calculus III: we are talking about polar curves. I give the usual lesson about how to graph r = sin(2 \theta) and r = sin(3 \theta) and give the usual “if n is even, the graph of r = sin (n \theta) has 2n petals and if n is odd, it has n petals.

Question: “does that mean it is impossible to have a graph with 6 petals then”? ๐Ÿ™‚

Yes, one can have intersecting petals and one try: r = |sin(3 \theta) | . But you aren’t going to get it without a trick of some sort.


October 3, 2014

Gaps in my mathematics education

Filed under: calculus, editorial, elementary mathematics — Tags: , , — collegemathteaching @ 1:19 pm

I’ve spoken about the many gaps in my mathematics education; I’ve written about a few. But in these cases, I was writing about the gaps at, say, the senior undergraduate to beginning graduate level.

I admit that I’ve enjoyed filling in some of these.

But, I also…have…elementary level gaps that I frequently overlook.

In my case: I never learned trigonometry all that well; I had forgotten about the laws of cosines and sines. And I had forgotten how to derive the following types of formulae: sin(A+B) = sin(A)cos(B) + cos(A)sin(b), cos(A+B) = cos(A)cos(B) - sin(A)sin(B) .

So, I spent a few minutes going over these old facts.


They aren’t hard but I am a bit surprised that I let my basic ignorance continue on this long.

October 2, 2014

ARGH!!! I got stuck at the board…

Filed under: calculus, elementary mathematics, pedagogy — Tags: , , — collegemathteaching @ 5:51 pm

Related rate problem that required the “law of cosines”, which…is a trig rule that I never bothered to learn and couldn’t derive on the spot.

ARRRRGGGHHHH!!!!!!!!! (even after 20+ years, even AFTER preparing, things like this happen from time to time).

Now, of course, I won’t rest until I’ve learned those stupid rules. ๐Ÿ™‚

I nailed the rest of them though.

Note: a student pulled out the manual and, given the diagram, finished it while I worked on another problem. He showed me the answer and I gave him a fist bump.

August 27, 2014

Nice collection of Math GIFs

Filed under: basic algebra, calculus, elementary mathematics, pedagogy — Tags: — collegemathteaching @ 12:25 am

No, these GIF won’t explain better than an instructor, but I found many of them to be fun.


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