# Good Question 8 – or not?

Today’s question is not a good question. It’s a bad question.

But sometimes a bad question can become a good one.

This one leads first to a discussion of units, then to all sorts of calculus.

Here’s the question a teacher sent me this week taken from his textbook:

The normal monthly rainfall at the Seattle-Tacoma airport can be approximated by the model $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$, where R is measured in inches and t is the time in months, t = 1 being January. Use integration to approximate the normal annual rainfall.  Hint: Integrate over the interval [0,12].

Of course, with the hint it’s not difficult to know what to do and that makes it less than a good question right there. The answer is $\displaystyle \int_{0}^{12}{R(t)dt=37.4736}$ inches. You could quit here and go on to the next question, but …

Then a student asked. “If R is in inches shouldn’t be in units of the integral be inch-months, since the unit of an integral is the unit of the integrand times the units of the independent variable?”  Well, yes, they should. So, what’s up with that?

Also, the teacher figured that the integral of a rate is an amount and our answer is an amount, so why isn’t the integrand a rate?

The only answer I could come up with is that the statement “R is measured in inches” is incorrect; R should be measured in inches /month. The opening phrase “normal monthly rainfall” also seems to point to the correct units for R being inches/month.

Problem solved; or maybe does this lead to a different concern?

The teacher pointed out that R(6) = 0.7658 inches is a reasonable answer for the amount of rain in June whereas $\displaystyle \int_{0}^{6}{R(t)dt=}20.4786$ is not.

If R is a rate, then the amount of rain that falls in June (t = 6) is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$.

From here on we will assume that R is a rate with units of inches/month. Here are the individual monthly rates calculated with a CAS.

The total amount of rainfall (second line above) appears be R(1) + R(2) + R(3) + … +R(12) = 37.4742. This is very close to the amount calculated by integration.

The slight difference of 0.0006 is not a round off error.

Remember, behind every definite integral there is a Riemann sum!

Again, the units are the problem. Why does the sum of the monthly rates seem to give the total amount?  The reason is that the terms of the sequence above are actually the values of a right-side Riemann sum of the rate, R(t), over the interval [0,12] with 12 equal subdivisions of width 1 (month) each with the 1’s left out as 1’s often are. Therefore, their sum should come close to the total yearly rainfall, but it is really just an approximation of it.

The actual total for any month, n, is given by $\displaystyle \int_{n-1}^{n}{r(t)}dt$. For example the amount of rain that falls in June is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$ inches.

Here is the sequence of the actual monthly rainfall values in inches, and their sum.

This agrees with the integral. Why? Because one of  the properties of integrals tell us that $\displaystyle \sum\limits_{n=1}^{12}{\int_{n-1}^{n}{r(t)dt}}=\int_{0}^{12}{r(t)dt}$.

Another instructive thing with this integral is this: The function $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$ is periodic with a period of  $\frac{2\pi }{0.524}\approx 11.9908\approx 12$. So the sine function takes on (almost) all its values in a year, as you would expect. Since the sine values all but cancel each other out

$\displaystyle \int_{0}^{12}{3.121+2.399\sin \left( 0.524t+1.377 \right)dt}\approx \int_{0}^{12}{3.121dt=3.121\left( 12-0 \right)=37.452}$. Close!

The total rainfall divided by 12 is $\frac{37.452}{12}=3.121$ this must be close to the average rainfall each month. The average rainfall is $\displaystyle \frac{1}{12}\int_{0}^{12}{R\left( t \right)dt}=3.1228$ inches. Close, again!

So, there you have it. Is this a good question or not? We considered all these concepts while working not just with an equation but with numbers from a poorly stated problem:

• Unit analysis
• Integration by technology
• Realizing that a pretty good approximation is not correct, due again to units.
• A Riemann sum approximation in a real situation that comes very close to the value by integration
• Using a property of a periodic function to greatly simplify an integral
• Finding average value two ways

So, it turned out to be a sunny day in Seattle.

.

# Calculators

First some history and then an opinion

I remember buying my first electronic calculator in the late 1960s. It did addition, subtraction, multiplication, and division, and could remember one number. It displayed 8 digits and had a special button that displayed the next eight digits. I remember using those next eight digits never. To buy it I had to drive 40 minutes and spend \$70 – expensive even today.

The square root of 743 computed using the algorithm discussed in the post. The third iteration (fourth answer) is correct to 10 digits.

With it I learned an iterative algorithm for finding square roots: guess the root, divide the guess into the number, average the quotient and the guess, repeat using the average as the new guess.  You could do it all without writing anything down. (See the illustration on a modern calculator – accurate to 8 decimals in only 3 iterations (fourth answer), but then I could find the next 8 with the special button.)

Since then, I’ve had lots of calculators of all sorts.

Graphing calculators hit the general market around 1989 or 1990. This was the same time as the “reform calculus” movement. The College Board announced that the AP calculus exams would require graphing calculators in 1995 – five years to get the country ready.

The College Board held intensive training immediately following the reading. These were the TICAP conferences (Technology Intensive Calculus for Advanced Placement). Half the readers were invited for the first year and the other half for the second, then more for the third year.

Casio, Hewlett-Packard, Texas Instruments all gave participants calculators to use take home. Sharpe lent them calculators (and we haven’t heard of Sharpe since). Sample lessons were taught using Hewlett-Packard CAS calculators and then the same lesson was taught using TI-81s. The HP computer algebra system calculators, with far more features but using the far more complicated reverse Polish notation entry system, lost in the completion to the simpler to use, but less sophisticated TI-81s.

The teachers were not all happy. A friend of mine, due to retire in 2-3 years gave up his AP calculus classes early so he would not need to learn the calculators. Others embraced technology. The AP program forced the graphing calculator into high schools where they were used to improve learning and instruction. Yet even today not all high schools have embraced technology.

The calculator makers, especially Texas Instruments, provided print materials, software, workshops and conferences that helped teachers learn how to use graphing calculators in their classes at all levels.

Technology, as a way to teach, learn, and most importantly, do mathematics, caught on big time. And that was and is a good thing.

I think graphing calculators are very quickly becoming obsolete and should be phased out.

Technology has bypassed graphing calculators. Tablet computers, PCs, Macs, iPads, and the like, even smart phones, can do everything graphing calculators can do. They are more versatile. The larger screens are easier to see and can show more information without crowding.

The initial investment may be more than for a graphing calculator, but once purchased the apps are relatively cheap. There are many free apps that not only do computations and graphing, but CAS operations as well. Interactive geometry and statistics apps are also available.

These, along with online textbooks and internet access, put everything students need to learn math literally at their fingertips. Graphs and other results can be easily copied and printed, or pasted into note-taking apps.

One disadvantage is the initial cost for the hardware (but of course many students already have the hardware). The other disadvantage is the ability to communicate and find help both in the room and around the world during tests. Photographing the questions for later use by others is another concern.  I think (hope) it is just a matter of time before this problem can be overcome perhaps with an app that allows access only to the apps the teachers allow for tests.

Technology, like time, marches on.

# More Gold

Last week I discussed Lin McMullin’s Theorem (that would be me). The theorem finds where the Golden Ratio shows up in any fourth-degree polynomial.  Shortly after that began making the rounds, I got an e-mail from David Tschappat who was then a senior in college. He is now a doctoral student working for an online learning solutions provider.

He related the discovery of a similar appearance of the Golden Ratio in cubic polynomials! Here is his theorem and two other related facts about cubic polynomials (lemmas really) as developed by me.

All cubic polynomials are symmetric to their point of inflection. Therefore, we can simplify matters by translating any cubic so that its point of inflection is at the origin. The general equation will then be $f\left( x \right)=k{{x}^{3}}-3{{h}^{2}}kx$.

I choose this rather unusual form so that the x-coordinates of the extreme points will be “nice” variables, namely h and –h the zeros of ${f}'\left( x \right)=3k{{x}^{2}}-3{{h}^{2}}k=3k\left( x-h)(x+h \right)$. The leading coefficient k will eventually divide out and not be a concern.

Now consider two other values x = –2h and x = 2h. So, assuming h > 0, there are five evenly spaced values $-2h<-h<0 that we will consider.

The first two results I will leave as an exercise, as they say:

1. $f\left( -2h \right)=f\left( h \right)$ and $f\left( -h \right)=f\left( 2h \right)$. This means that the extreme values are the same as a point on the graph 3h units away on the other side of the y-axis. This is mildly interesting in itself. The reason it is mentioned id that we will use these points soon.
2. ${f}'\left( -2h \right)={f}'\left( 2h \right)$ which is really true for any points at equal distances on opposite sides of the origin. This is obvious from the symmetry of the cubic.

Now comes the interesting and strange part. An equation of a line through one extreme point with a slope equal to that at the point with the same y-coordinate on the other side of the origin (or for that matter the point h units farther away on the same side of the y-axis) is

$y=f\left( -h \right)+{f}'\left( 2h \right)\left( x-\left( -h \right) \right)$

$y=\left( k{{\left( -h \right)}^{3}}-3{{h}^{2}}k\left( -h \right) \right)+\left( 3k{{\left( 2h \right)}^{2}}-3{{h}^{2}}k \right)\left( x+h \right)$

$y=k\left( 11{{h}^{3}}+9{{h}^{2}}x \right)$

Now find where this line intersect the original cubic by solving $y=f\left( x \right)$. The equation can be solved by hand. We know that x = h is one solution. Synthetic division will give a quadratic and the quadratic formula will do the rest.

$k\left( 11{{h}^{3}}+9{{h}^{2}}x \right)=k{{x}^{3}}-3{{h}^{2}}kx$

$0={{x}^{3}}-12{{h}^{2}}x-11{{h}^{3}}$

$0=\left( x-h \right)\left( {{x}^{2}}+hx-11{{h}^{2}} \right)$

$\displaystyle x=h,\frac{1+3\sqrt{5}}{2}h,\frac{1-3\sqrt{5}}{2}h$

What? You were expecting the Golden Ratio? Not to worry; it’s there!

$\displaystyle \Phi \left( -h \right)+\phi \left( 2h \right)=\frac{1+\sqrt{5}}{2}\left( -h \right)+\frac{1-\sqrt{5}}{2}\left( 2h \right)=\frac{1-3\sqrt{5}}{2}h$

$\displaystyle \Phi \left( 2h \right)+\phi \left( -h \right)=\frac{1+\sqrt{5}}{2}\left( 2h \right)+\frac{1-\sqrt{5}}{2}\left( -h \right)=\frac{1+3\sqrt{5}}{2}h$

Where $\displaystyle \Phi =\frac{1+\sqrt{5}}{2}$ is the Golden Ratio and $\displaystyle \phi ={{\Phi }^{-1}}=\frac{1-\sqrt{5}}{2}$ is the reciprocal of the Golden Ratio.

A similar computation using the other extreme point gives similar results.

Yes, the Golden Ratio is there; I don’t know why it should be there, but it is.

Thank you, David.

# Lin McMullin’s Theorem

Mathematics more often tends to delight when it exhibits an unanticipated result rather than conforming to … expectations. In addition, the pleasure derived from mathematics is related in many cases to the surprise felt upon the perception of totally unexpected relationships and unities.

– Mario Livio, The Golden Ratio

I have a theorem named after me. I did not name it, but I did prove it – well more like I tripped over it. It is a calculus related idea. Here is how it came about. I say “came about” because as you will see I did not set out to prove it. I just sort of fell in my lap as I was working on something else.

I was trying to do an animation of an idea that I had heard about: If you have a fourth degree, or quartic, polynomial with a “W” shape it has two points of inflection. If you draw a line through the points of inflection three regions enclosed by the line and the polynomial’s graph are formed. The areas of these regions are in the ratio of 1:2:1. In order to make the animation work I needed the general coordinates of the four points where the line intersects the quartic.

The straightforward way to proceed would be to write a general fourth degree polynomial,

$f\left( x \right)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}$

differentiate it twice to find the second derivative. Then find the zeros of the second derivative (by the quadratic formula), write the equation of the line through them, and then find where else the line intersects the quartic. Without even starting I realized that even with a CAS the algebra and equation solving was going to be really fun (Not!). So, I decided on an alternative approach.

I decided to let the zeros of the second derivative be x = a and x = b, then at least they would be easy to work with. Then the second derivative is $\displaystyle {{f}'}'\left( x \right)=12{{c}_{4}}\left( x-a \right)\left( x-b \right)$ where the ${{c}_{4}}$  is the leading coefficient of the quartic and the 12 comes from differentiating twice.

I integrated to get the first derivative and added ${{c}_{1}}$, the coefficient of the linear term, as the constant of integration. I integrated again and added ${{c}_{0}}$, the constant term. as the constant of integration. This resulted in the original quartic function:

$\displaystyle f\left( x \right)={{c}_{4}}{{x}^{4}}-2\left( a+b \right){{c}_{4}}{{x}^{3}}+6ab{{c}_{4}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}$

Then I wrote the equation of y(xthe line through the points of inflection. It is too long to copy, but you may see it in the screen capture at the end of the post. (That is what is nice about a CAS: you really do not have to worry about how complicated things are.)

Then I solved the equation $f\left( x \right)=y\left( x \right)$. Two of the solutions are x = a and x = b as I expected. (This means that you could do synthetic division by hand since you know two of the roots.) The other two I did not expect. They are:

$\displaystyle {{x}_{1}}=\frac{1+\sqrt{5}}{2}a+\frac{1-\sqrt{5}}{2}b$ and $\displaystyle {{x}_{2}}=\frac{1+\sqrt{5}}{2}b+\frac{1-\sqrt{5}}{2}a$

And that’s when I stopped astonished! Those numbers are the Golden Ratio $\Phi =\frac{1+\sqrt{5}}{2}$,  and its reciprocal $\phi ={{\Phi }^{-1}}=\frac{1-\sqrt{5}}{2}$. So the roots are $\Phi a+\phi b$ and $\Phi b+\phi a$. How did they get there?

To this day I have no idea why the Golden Ratio should be so involved with quartic polynomials, but there they are in every quartic!

There were no assumptions made about a and b – they could be Complex numbers. In that case there are no points of inflection, but the “line” and the quartic still will have the same value at the four points.

October 16, 2022 Update: If the solutions of $\displaystyle {y}''=0$ are the Complex conjugates $\displaystyle a=\alpha +\beta i$ and $\displaystyle b=\alpha -\beta i$ then $\displaystyle {{x}_{1}}=\alpha +\sqrt{5}\beta i$ and $\displaystyle {{x}_{2}}=\alpha -\sqrt{5}\beta i$. When graphed on an Argand diagram the four points are collinear on the vertical line at $\displaystyle x=\alpha$

This was all in 2013 and until just this year I never checked the ratio of the areas. (They check.)

Here is a CAS printout of the entire computation.

An interactive Desmos demo of this can be found here

October 16, 2022, Update: If the solutions of $\displaystyle {y}''=0$ are the Complex conjugates $\displaystyle a=\alpha +\beta i$ and $\displaystyle b=\alpha -\beta i$ then $\displaystyle {{x}_{1}}=\alpha +\sqrt{5}\beta i$ and $\displaystyle {{x}_{2}}=\alpha -\sqrt{5}\beta i$. When graphed on an Argand diagram the four points are collinear on the vertical line at $\displaystyle x=\alpha$

November 10, 2020, Update: I received an email this week form Dominique Laurain, a computer science and applied math engineer from France, who describes himself as a mathematics hobbyist. He discovered another interesting relationship between the coordinates of the x-coordinates of the four points on the line described above. The four points on the line through the points of inflection of a fourth degree polynomial with Real roots in the drawing above from left to right are p, q, r, and s. The coordinates of the points of inflection are a and as in the post above.

One, of several, cross-ratios of four points with x-coordinates of p, q r, and s is defined as

$\displaystyle (p,s;r,q)=\frac{{\left( {r-p} \right)\left( {q-s} \right)}}{{\left( {r-s} \right)\left( {q-p} \right)}}$.

Dominique Laurin discovered that $\displaystyle (p,s;r,q)=\frac{{\left( {r-p} \right)\left( {q-s} \right)}}{{\left( {r-s} \right)\left( {q-p} \right)}}={{\Phi }^{4}}$

The computation is shown in the figure below.

There are 24 other cross-ratios depending on the order of the points. In groups of 4, the 24 possible orders are equal to 6 related values. See the cross-ratio link above above. For example, the cross-ratio in the order (s, p; r, q) is ${{\varphi }^{4}}$

Also, $(q,r;s,p)={{\Phi }^{4}}$

Other interesting information:

The Golden Ratio also appears in cubic equations. See the Tashappat – McMullin theorem here.

Quartic Polynomials and the Golden Ratio” by Harald Totland of the Royal Norwegian Naval Academy. (June 2009)

Speaking of the Golden Ratio, the Calculus Humor website has a nice feature on the Golden Ratio in logos. To view it click here.

Revised and updated July 20 and 23, 2017, November 10, 2020

# Experimenting with CAS – Chain Rule

Discovering things in mathematics can be facilitated by using a computer algebra system (CAS) available on many handheld calculators and computer apps. A CAS can provide good data with which to draw conclusions. You can do “experiments” by producing the results with a CAS and looking for patterns. As an example, let’s look at how you and your students might discover the chain rule for derivatives.

One of the ways you could introduce the chain rule is to ask your class to differentiate something like (3x + 7)2. Not knowing about the chain rule, just about the only way to proceed is to expand the expression to 9x2 + 42x + 49 and differentiate that: 18x + 42 and then factor 6(3x + 7). Then you show how this relates to the power rule and where the “extra” factor of 3 comes from differentiating the (3x + 7).  You really cannot a much more complicated example, say a third or fourth power, because the algebra gets complicated very fast.

Or does it?

Suggest your students use a CAS to do the example above this time using the third power. The output might look like this:

But even better: what we want is just the answer. Who cares about all the algebra in between? Try a few powers until the pattern become obvious.

Now we have some good data to work with. Can you guess the pattern?

Nor sure where the “extra” factor of 3 comes from? Try changing the 3 in the original and keep the exponent the same.

Now can you guess the chain rule? See if what you thought is right by changing only the inside exponent.

Then you can try some others:

You can count on the CAS giving you the correct data (answers). Do enough experiments until the chain rule pattern becomes clear.

But I think the big thing is not the chain rule, but that the students are learning how to experiment in mathematics situations. In these we started by changing only the outside power. Then we kept the power the same power and changed the coefficient of the linear factor. Then we changed the power inside power, each time seeing if our tentative rule for differentiating composite function was correct and adjusting it if it was not. Finally we tried a variety of different expressions. You change things. Not big things but little things. You don’t jump from one trial to something very different, only something a little different.

You can do the same thing for the product rule, the quotient rule, maybe some integration rules and so on. You have accomplished your goal when the students can produce the data they need without your suggestions.

But be aware: sometimes this can lead to unexpected results. Does the pattern hold here?

Or here?

Hint: $\frac{7}{3.2}=2.1875$  and $3{{\left( 3.2 \right)}^{3}}=98.304$

Revised 8-25-17