## Calculus: Hyperbolic Trigonometry, IV

Of course, there is always more to say about hyperbolic trigonometry….  Next, we’ll look at what is usually called the logistic curve, which is the solution to the differential equation

$\dfrac{dP}{dt}=kP(C-P),\quad P(0)\ \text{given}.$

The logistic curve comes up in the usual chapter on differential equations, and is an example of population growth.  Without going into too many details (since the emphasis is on hyperbolic trigonometry), $k$ is a constant which influences how fast the population grows, and $C$ is called the carrying capacity of the environment.

Note that when $P$ is very small, $C-P\approx C,$ and so the population growth is almost exponential.  But when $P(t)$ gets very close to $C,$ then $dP/dT\approx0,$ and so population growth slows down.  And of course when $P(t)=C,$ growth stops — hence calling $C$ the carrying capacity of the environment.  It represents the largest population the environment can sustain.

Here is an example of such a curve where $C=500,$ $k=0.02,$ and $P(0)=50.$

Notice the S shape, obtained from a curve rapidly growing when the population is small. It happens that the population grows fastest at half the carrying capacity, and then growth slows to zero as the carrying capacity is reached.

Skipping the details (simple separation of variables), the solution to this differential equation is given by

$P(t)=\dfrac{C}{1+Ae^{-kCt}},\qquad A=\dfrac{C-P(0)}{P(0)}.$

I will digress for a moment, however, to mention partial fractions (as I step on my calculus soapbox).  I have mentioned elsewhere that incomprehensible chapter in calculus textbooks:  Techniques of Integration.  Pedagogically a disaster for so many reasons.

The first time I address partial fractions is when summing telescoping series, such as

$\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+1)}.$

It really is necessary.  But I only go so far as to be able to sum such series.  (Note:  I do series as the middle third of Calculus II, rather than the end.  A colleague suggested that students are more tired near the end of the course, which is better for a more technique-oriented discussion of the solution to differential equations, which typically comes before series.)

You also need partial fractions to solve the differential equation for the logistic curve, which is when I revisit the topic.  After finding the logistic curve, we talk about partial fractions in more detail.  The point is that students see some motivation for the method of partial fractions — which they decidedly don’t in a chapter on techniques of integration.

OK, time to step off the soapbox and talk about hyperbolic trigonometry….  The punch line is that the logistic curve is actually a scaled and shifted hyperbolic tangent curve!  Of course it looks like a hyperbolic tangent, but let’s take a moment to see why.

We first use the definitions of $\sinh u$ and $\cosh u$ to write

$\tanh u=\dfrac{\sinh h}{\cosh u}=1-\dfrac2{1+e^{2u}}.$

This results in

$\dfrac2{1+e^{2u}}=1-\tanh u.$

You can see the form of the equation of the logistic curve starting to take shape.  Since the hyperbolic tangent has horizontal tangents at $y=-1$ and $y=1,$ we need to scale by a factor of $C/2$ so that the asymptotes of the logistic curve are $C$ units apart:

$\dfrac C{1+e^{2u}}=\dfrac{C}2\left(1-\tanh u\right).$

Note that this puts the horizontal asymptotes of the function at $y=0$ and $y=C.$

To take into account the initial population, we need a horizontal shift, since otherwise the initial population would be $C/2.$ We can accomplish this be replacing $\tanh u$ with $\tanh(u+\varphi):$

$\dfrac C{1+e^{2\varphi} e^{2u}}=\dfrac C2(1-\tanh(u+\varphi)).$

We’re almost done at this point:  we simply need

$e^{2\varphi}=A,\qquad 2u=-kCt.$

Solving and substituting back results in

$P(t)=\dfrac C2\left(1-\tanh\left(\dfrac{-kCt+\ln A}2\right)\right),$

which, since $\tanh$ is an odd function, becomes

$P(t)=\dfrac C2\left(1+\tanh\left(\dfrac{kCt-\ln A}2\right)\right).$

And there it is!  The logistic curve as a scaled, shifted hyperbolic tangent.

Now what does showing this accomplish?  I can’t give you a definite answer from the point of view of the students.  But for me, it is a way to tie two seemingly unrelated concepts — hyperbolic trigonometry and solution of differential equations by separation of variables — together in a way that is not entirely contrived (as so many calculus textbook problems are).

I would love to perform the following experiment:  work out the solution to the differential equation together as a guided discussion, and then prompt students to suggest functions this curve “looks like.”  Of course the $\arctan$ might be suggested, but how would we relate this to the exponential function?

Eventually we’d tease out the hyperbolic tangent, since this function actually does involve the exponential function.  Then I’d move into an inquiry-based lesson:  give the students the equation of a logistic curve, and have them work out the conversion to the hyperbolic tangent.

And as is typical in such an approach, I would put students into groups, and go around the classroom and nudge them along.  See what happens.

I say that yes, calculus students should be able to do this.  I recently sent an email about pedagogy in calculus which, among other things, addressed the question:  What do calculus students really need to know?

There is no room to adequately address that important question here, but in today’s context, I would say this:  I think it is more important for a student to be able to rewrite $P(t)$ as a hyperbolic tangent than it is for them to know how to sketch the graph of $P(t).$

Why?  Because it is trivial to graph functions, now.  Type the formula into Desmos.  But how to interpret the graph?  Rewrite it?  Analyze it?  Draw conclusions from it?  We need to focus on what is no longer necessary, and what is now indispensable.  To my knowledge, no one has successfully done this.

I think it is about time for that to change….

## Calculus: Hyperbolic Trigonometry, III

We continue where we left off on the last post about hyperbolic trigonometry.  Recall that we ended by finding an antiderivative for $\sec(x)$ using the hyperbolic trigonometric substitution $\sec(\theta)=\cosh(u).$  Today, we’ll look at this substitution in more depth.

The functional relationship between $\theta$ and $u$ is described by the gudermannian function, defined by

$\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.$

This is not at all obvious, so we’ll look at the derivation of this rather surprising-looking formula.  It’s the only formula I’m aware of which involves both the arctangent and the exponential function.  We remark (as we did in the last post) that we restrict $\theta$ to the interval $(-\pi/2,\pi/2)$ so that this relationship is in fact invertible.

We use a technique similar to that used to derive a formula for the inverse hyperbolic cosine.  First, write

$\sec\theta=\cosh u=\dfrac{e^u+e^{-u}}2,$

and then multiply through by $e^u$ to obtain the quadratic

$(e^u)^2-2\sec(\theta)e^u+1=0.$

$e^u=\sec\theta\pm\tan\theta.$

Which sign should we choose?  We note that $\theta$ and $u$ increase together, so that because $e^u$ is an increasing function of $u,$ then $\sec\theta\pm\tan\theta$ must be an increasing function of $\theta.$ It is not difficult to see that we must choose “plus,” so that $e^u=\sec\theta+\tan\theta,$ and consequently

$u=\ln(\sec\theta+\tan\theta).$

We remark that no absolute values are required here; this point was discussed in the previous post.

Now to solve for $\theta.$  The trick is to use a lesser-known trigonometric identity:

$\sec\theta+\tan\theta=\tan\left(\dfrac\pi4+\dfrac\theta2\right).$

There is such a nice geometrical proof of this identity, I can’t help but include it.  Start with the usual right triangle, and extend the segment of length $\tan\theta$ by $\sec\theta$ in order to form an isosceles triangle.  Thus,

$\tan(\theta+\alpha)=\sec\theta+\tan\theta.$

To find $\alpha,$ observe that $\beta$ is supplementary to both $2\alpha$ and $\pi/2-\theta,$ so that

$2\alpha=\dfrac\pi2-\theta,$

which easily implies

$\alpha=\dfrac\pi4-\dfrac\theta2.$

Therefore

$\theta+\alpha=\dfrac\pi4+\dfrac\theta2,$

which is precisely what we need to prove the identity.

Now we substitute back into the previous expression for $u,$ which results in

$u=\ln\tan\left(\dfrac\pi4+\dfrac\theta2\right).$

This may be solved for $\theta,$ giving

$\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.$

So let’s see how to use this to relate circular and hyperbolic trigonometric functions.  We have

$\sec(\text{gd}\,u)=\dfrac1{\cos(2\arctan(e^u)-\pi/2)},$

which after using the usual circular trigonometric identities, becomes

$\sec(\text{gd}\,u)=\dfrac{e^u+e^{-u}}2=\cosh u.$

It is also an easy exercise to see that

$\dfrac{d}{du}\,\text{gd}\,u=\text{sech}\, u.$

So revisiting the integral

$\displaystyle\int\sec\theta\,d\theta,$

we may alternatively make the substitution $\theta=\text{gd}\,u,$ giving

$\displaystyle\int\sec\theta\,d\theta=\int\cosh u\,(\text{sech}\, u\,du)=\int du,$

which is the same simple integral we saw in the previous post.

What about the other trigonometric functions?  Certainly we know that $\cos(\text{gd}\,u)=\text{sech}\,u.$  Again using the usual circular trigonometric identities, we can show that

$\sin(\text{gd}\,u)=\tanh u.$

Knowing these three relationships, the rest are easy to find: $\tan(\text{gd}\,u)=\sinh u,$ $\cot(\text{gd}\,u)=\text{csch}\,u,$ and $\csc(\text{gd}\,u)=\text{coth}\,u.$

I think that the gudermannian function should be more widely known.  On the face of it, circular and hyperbolic trigonometric functions are very different beasts — but they relate to each other in very interesting ways, in my opinion.

I will admit that I don’t teach students about the gudermannian function as part of a typical calculus course.  Again, there is the issue of time:  as you are well aware, students finishing one course in the calculus sequence must be adequately prepared for the next course in the sequence.

So what I do is this:  I put the exercises on the gudermannian function as extra challenge problems.  Then, if a student is already familiar with hyperbolic trigonometry, they can push a little further to learn about the gudermannian.

Not many students take on the challenge — but there are always one or two who will visit my office hours with questions.  Such a treat for a mathematics professor!  But I feel it is always necessary to give something to the very best students to chew on, so they’re not bored.  The gudermannian does the trick as far as hyperbolic trigonometry is concerned….

As a parting note, I’d like to leave you with a few more exercises which I include in my “challenge” question on the gudermannian.  I hope you enjoy working them out!

1.  Show that $\tanh\left(\dfrac x2\right)=\tan\left(\dfrac 12\text{gd}\,x\right).$
2. Show that $e^x=\dfrac{1+\tan(\frac12\text{gd}\,x)}{1-\tan(\frac12\text{gd}\,x)}.$
3. Show that if $h$ is the inverse of the gudermannian function, then $h'(x)=\sec x.$

## Bay Area Mathematical Artists, VIII

Yesterday was our final meeting of the Bay Area Mathematical Artists for the academic year 2017–2018.  We were back at Santa Clara University to visit their virtual reality lab, the Imaginarium!  It was an amazing visit, coordinated by Frank Farris, Tom Banchoff (who is visiting Santa Clara University this semester), and Imaginarium director Max Sims.

While we were waiting for everyone to arrive, Tom generously provided donuts for us to snack on.  But not until we were given a short presentation on a very special property of donuts — the “two-piece” property, which Tom discussed in his doctoral thesis.

A perfectly smooth donut — this is important, since bumps or ridges are problematic — has the property that if you take a knife and make a planar cut, you always get exactly two pieces.

Of course any convex shape automatically has this property.  Donuts, however, are not convex, so they are special in this regard.  Tom remarked that discussing this makes good breakfast conversation, since smooth bagels also have the two-piece property.  But a sufficiently curved banana does not (think of the letter U and make a horizontal cut to get three pieces).

Once everyone arrived, Tom began with a presentation of his forays into computer graphics and animations.  The first video he showed was made in 1968 with computer scientist Charles Strauss, also at Brown, which showed a torus turning inside out.

What was fascinating was how the video was made.  Keep in mind this was long before platforms like Processing were available….  Even though each frame was black-and-white and consisted of line drawings of a torus in various different configurations, it still took about a minute to create each frame.

Then, each frame had to be photographed — using film, if you remember what that is.  Then another minute, another photograph.  Next, the film was sent on to Boston, where it was processed into a movie.  The movie was sent back — and hopefully, it was just what you wanted….

Fast-forward to 1978, where the subject of the video was rotating cubes and hypercubes. I won’t try to describe it in words, but you can actually see the video online here.

It looks like the graphics in this movie were color graphics!  But no, that wasn’t available yet.  You see color because each frame used four photographs — each taken with a different filter on.  These were overlaid to create a color graphic.  Very creative, and again, certainly much more work than it would take today.  Keep in mind that 1978 was forty years ago….

There were still three movies to go!  Next, Tom showed us his 1985 movie about the hypersphere.  We then got to see Tom’s animation of donut slicing, which was the virtual version of our initial demonstration.  Finally, the last video was from 1999, which was a rotation of the flat torus.

The movies and talk took us halfway through the afternoon.  Now it was time to try on the headsets and have our very own virtual reality experience!

We learned from Max Sims a little about the technical challenges of creating a good VR experience.  First, 90 frames per second is ideal — contrast that with 30 frames per second when creating movies with a platform like Processing.  And second, the response time of the headsets needs to be no longer than 11 milliseconds — that is, when you turn your head to look at something, the image has to change that fast — or else you’ll get motion sickness.

The way the lab was set up, everyone got their own computer (although it took me three tries to find one which had everything working properly).  You can see people with headsets on and controls in their hands.

It is very hard to describe what it feels like with the headset on without experiencing it for yourself.  But when you put on the goggles, you’re in a 360-degree visual environment.  That is, when you turn your head to the right, you are seeing what’s to the right of your visual range in the VR simulation.  You can even look down and see what’s underneath you!  It was a little dizzying to look down and see a stream running underneath you, since that meant you were suspended in midair….

The hand controls were different for every experience.  Some allowed you to move things, or select different types of building blocks to make 3D images, or select a different simulation.  It took a bit of getting used to, and I didn’t fully get the hang of it in the hour we had to play around.  But Max said we were welcome back to try out the lab again, and I do intend to take him up on his offer.

After our incredible experience, we went out for Thai — sixteen of us, this time!  We spent over two hours at dinner, which is typical.  These gatherings bring together a diverse group of people interested in all aspects of mathematics and art, and it seems we never run out of things to talk about.

Then the drive home.  I was driving Nick, and we were amazed by the cloud formations we saw all along the way.  Despite my rather dirty windshield, Nick snapped this gorgeous pic.

The clouds were like this for most of the drive home, and were a very fitting end to our last gathering of the semester.

Plans are underway to continue meeting throughout the summer, so stay tuned!

## Calculus: Hyperbolic Trigonometry, II

Now on to some calculus involving hyperbolic trigonometry!  Today, we’ll look at trigonometric substitutions involving hyperbolic functions.

$\displaystyle\int\sqrt{1+x^2}\,dx.$

The usual technique involving circular trigonometric functions is to put $x=\tan(\theta),$ so that $dx=\sec^2(\theta)\,d\theta,$ and the integral transforms to

$\displaystyle\int\sec^3(\theta)\,d\theta.$

In general, we note that when taking square roots, a negative sign is sometimes needed if the limits of the integral demand it.

This integral requires integration by parts, and ultimately evaluating the integral

$\displaystyle\int\sec(\theta)\,d\theta.$

And how is this done?  I shudder when calculus textbooks write

$\displaystyle\int \sec(\theta)\cdot\dfrac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}\,d\theta=\ldots$

How does one motivate that “trick” to aspiring calculus students?  Of course the textbooks never do.

Now let’s see how to approach the original integral using a hyperbolic substitution.  We substitute $x=\sinh(u),$ so that $dx=\cosh(u)\,du$ and $\sqrt{1+x^2}=\cosh(u).$  Note well that taking the positive square root is always correct, since $\cosh(u)$ is always positive!

This results in the integral

$\displaystyle\int\cosh^2(u)\,du=\displaystyle\int\dfrac{1+\cosh(2u)}2\,du,$

which is quite simple to evaluate:

$\dfrac12u+\dfrac14\sinh(2u)+C.$

Now $u=\hbox{arcsinh}(x),$ and

$\sinh(2u)=2\sinh(u)\cosh(u)=2x\sqrt{1+x^2}.$

Recall from last week that we derived an explicit formula for $\hbox{arcsinh}(x),$ and so our integral finally becomes

$\dfrac12\left(\ln(x+\sqrt{1+x^2})+x\sqrt{1+x^2}\right)+C.$

You likely noticed that using a hyperbolic substitution is no more complicated than using the circular substitution $x=\sin(\theta).$  What this means is — no need to ever integrate

$\displaystyle\int\tan^m(\theta)\sec^n(\theta)\,d\theta$

again!  Frankly, I no longer teach integrals involving $\tan(\theta)$ and $\sec(\theta)$ which involve integration by parts.  Simply put, it is not a good use of time.  I think it is far better to introduce students to hyperbolic trigonometric substitution.

Now let’s take a look at the integral

$\displaystyle\int\sqrt{x^2-1}\,dx.$

The usual technique?  Substitute $x=\sec(\theta),$ and transform the integral into

$\displaystyle\int\tan^2(\theta)\sec(\theta)\,d\theta.$

Sigh.  Those irksome tangents and secants.  A messy integration by parts again.

But not so using $x=\cosh(u).$  We get $dx=\sinh(u)\,du$ and $\sqrt{x^2-1}=\sinh(u)$ (here, a negative square root may be necessary).

We rewrite as

$\displaystyle\int\sinh^2(u)\,du=\displaystyle\int\dfrac{\cosh(2u)-1}2\,du.$

This results in

$\dfrac14\sinh(2u)-\dfrac u2+C=\dfrac12(\sinh(u)\cosh(u)-u)+C.$

All we need now is a formula for $\hbox{arccosh}(x),$ which may be found using the same technique we used last week for $\hbox{arcsinh}(x):$

$\hbox{arccosh}(x)=\ln(x+\sqrt{x^2-1}).$

Thus, our integral evaluates to

$\dfrac12(x\sqrt{x^2-1}-\ln(x+\sqrt{x^2-1}))+C.$

We remark that the integral

$\displaystyle\int\sqrt{1-x^2}\,dx$

is easily evaluated using the substitution $x=\sin(\theta).$  Thus, integrals of the forms $\sqrt{1+x^2},$ $\sqrt{x^2-1},$ and $\sqrt{1-x^2}$ may be computed by using the substitutions $x=\sinh(u),$ $x=\cosh(u),$ and $x=\sin(\theta),$ respectively.  It bears repeating:  no more integrals involving powers of tangents and secants!

One of the neatest applications of hyperbolic trigonometric substitution is using it to find

$\displaystyle\int\sec(\theta)\,d\theta$

without resorting to a completely unmotivated trick.  Yes, I saved the best for last….

So how do we proceed?  Let’s think by analogy.  Why did the substitution $x=\sinh(u)$ work above?  For the same reason $x=\tan(\theta)$ works: we can simplify $\sqrt{1+x^2}$ using one of the following two identities:

$1+\tan^2(\theta)=\sec^2(\theta)\ \hbox{ or }\ 1+\sinh^2(u)=\cosh^2(u).$

So $\sinh(u)$ is playing the role of $\tan(\theta),$ and $\cosh(u)$ is playing the role of $\sec(\theta).$  What does that suggest?  Try using the substitution $\sec(\theta)=\cosh(u)$!

No, it’s not the first think you’d think of, but it makes sense.  Comparing the use of circular and hyperbolic trigonometric substitutions, the analogy is fairly straightforward, in my opinion.  There’s much more motivation here than in calculus textbooks.

So with $\sec(\theta)=\cosh(u),$ we have

$\sec(\theta)\tan(\theta)\,d\theta=\sinh(u)\,du.$

But notice that $\tan(\theta)=\sinh(u)$ — just look at the above identities and compare. We remark that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2),$ then as a result of the asymptotic behavior, the substitution $\sec(\theta)=\cosh(u)$ gives a bijection between the graphs of $\sec(\theta)$ and $\cosh(u),$ and between the graphs of $\tan(\theta)$ and $\sinh(u).$ In this case, the signs are always correct — $\tan(\theta)$ and $\sinh(u)$ always have the same sign.

So this means that

$\sec(\theta)\,d\theta=du.$

What could be simpler?

Thus, our integral becomes

$\displaystyle\int\,du=u+C.$

But

$u=\hbox{arccosh}(\sec(\theta))=\ln(\sec(\theta)+\tan(\theta)).$

Thus,

$\displaystyle\int \sec(\theta)\,d\theta=\ln(\sec(\theta)+\tan(\theta))+C.$

Voila!

We note that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2)$ as discussed above,  then we always have $\sec(\theta)+\tan(\theta)>0,$ so there is no need to put the argument of the logarithm in absolute values.

Well, I’ve done my best to convince you of the wonder of hyperbolic trigonometric substitutions!  If integrating $\sec(\theta)$ didn’t do it, well, that’s the best I’ve got.

The next installment of hyperbolic trigonometry?  The Gudermannian function!  What’s that, you ask?  You’ll have to wait until next time — or I suppose you can just google it….

## Calculus: Hyperbolic Trigonometry, I

love hyperbolic trigonometry.  I always include it when I teach calculus, as I think it is important for students to see.  Why?

1.  Many applications in the sciences use hyperbolic trigonometry; for example, the use of Laplace transforms in solving differential equations, various applications in physics, modeling population growth (the logistic model is a hyperbolic tangent curve);
2. Hyperbolic trigonometric substitutions are, in many instances, easier than circular trigonometric substitutions, especially when a substitution involving $\tan(x)$ or $\sec(x)$ is involved;
3. Students get to see another form of trigonometry, and compare the new form with the old;
4. Hyperbolic trigonometry is fun.

OK, maybe that last reason is a bit of hyperbole (though not for me).

Not everyone thinks this way.  I once had a colleague who told me she did not teach hyperbolic trigonometry because it wasn’t on the AP exam.  What do you say to someone who says that?  I dunno….

In any case, I want to introduce the subject here for you, and show you some interesting aspects of hyperbolic trigonometry.  I’m going to stray from my habit of not discussing things you can find anywhere online, since in order to get to the better stuff, you need to know the basics.  I’ll move fairly quickly through the introductory concepts, though.

The hyperbolic cosine and sine are defined by

$\cosh(x)=\dfrac{e^x+e^{-x}}2,\quad\sinh(x)=\dfrac{e^x-e^{-x}}2,\quad x\in\mathbb{R}.$

I will admit that when I introduce this definition, I don’t have an accessible, simple motivation for doing so.  I usually say we’ll learn a lot more as we work with these definitions, so if anyone has a good idea in this regard, I’d be interested to hear it.

The graphs of these curves are shown below.

The graph of $\cosh(x)$ is shown in blue, and the graph of $\sinh(x)$ is shown in red.  The dashed orange graph is $y=e^{x}/2,$ which is easily seen to be asymptotic to both graphs.

Parallels to the circular trigonometric functions are already apparent:  $y=\cosh(x)$ is an even function, just like $y=\cos(x).$  Similarly, $\sinh(x)$ is odd, just like $\sin(x).$

Another parallel which is only slight less apparent is the fundamental relationship

$\cosh^2(x)-\sinh^2(x)=1.$

Thus, $(\cosh(x),\sinh(x))$ lies on a unit hyperbola, much like $(\cos(x),\sin(x))$ lies on a unit circle.

While there isn’t a simple parallel with circular trigonometry, there is an interesting way to characterize $\cosh(x)$ and $\sinh(x).$  Recall that given any function $f(x),$ we may define

$E(x)=\dfrac{f(x)+f(-x)}2,\quad O(x)=\dfrac{f(x)-f(-x)}2$

to be the even and odd parts of $f(x),$ respectively.  So we might simply say that $\cosh(x)$ and $\sinh(x)$ are the even and odd parts of $e^x,$ respectively.

There are also many properties of the hyperbolic trigonometric functions which are reminiscent of their circular counterparts.  For example, we have

$\sinh(2x)=2\sinh(x)\cosh(x)$

and

$\sinh(x+y)=\sinh(x)\cosh(y)+\sinh(y)\cosh(x).$

None of these are especially difficult to prove using the definitions.  It turns out that while there are many similarities, there are subtle differences.  For example,

$\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y).$

That is, while some circular trigonometric formulas become hyperbolic just by changing $\cos(x)$ to $\cosh(x)$ and $\sin(x)$ to $\sinh(x),$ sometimes changes of sign are necessary.

These changes of sign from circular formulas are typical when working with hyperbolic trigonometry.  One particularly interesting place the change of sign arises is when considering differential equations, although given that I’m bringing hyperbolic trigonometry into a calculus class, I don’t emphasize this relationship.  But recall that $\cos(x)$ is the unique solution to the differential equation

$y''+y=0,\quad y(0)=1,\quad y'(0)=0.$

Similarly, we see that $\cosh(x)$ is the unique solution to the differential equation

$y''-y=0,\quad y(0)=1,\quad y'(0)=0.$

Again, the parallel is striking, and the difference subtle.

Of course it is straightforward to see from the definitions that $(\cosh(x))'=\sinh(x)$ and $(\sinh(x))'=\cosh(x).$  Gone are the days of remembering signs when differentiating and integrating trigonometric functions!  This is one feature of hyperbolic trigonometric functions which students always appreciate….

Another nice feature is how well-behaved the hyperbolic tangent is (as opposed to needing to consider vertical asymptotes in the case of $\tan(x)$).  Below is the graph of $y=\tanh(x)=\sinh(x)/\cosh(x).$

The horizontal asymptotes are easily calculated from the definitions.  This looks suspiciously like the curves obtained when modeling logistic growth in populations; that is, finding solutions to

$\dfrac{dP}{dt}=kP(C-P).$

In fact, these logistic curves are hyperbolic tangents, which we will address in more detail in a later post.

One of the most interesting things about hyperbolic trigonometric functions is that their inverses have closed formulas — in striking contrast to their circular counterparts.  I usually have students work this out, either in class or as homework; the derivation is quite nice, so I’ll outline it here.

So let’s consider solving the equation $x=\sinh(y)$ for $y.$  Begin with the definition:

$x=\dfrac{e^y-e^{-y}}2.$

The critical observation is that this is actually a quadratic in $e^y:$

$(e^y)^2-2xe^y-1=0.$

All that is necessary is to solve this quadratic equation to yield

$e^y=x\pm\sqrt{1+x^2},$

and note that $x-\sqrt{1+x^2}$ is always negative, so that we must choose the positive sign.  Thus,

$y=\hbox{arcsinh}(x)=\ln(x+\sqrt{1+x^2}).$

And this is just the beginning!  At this stage, I also offer more thought-provoking questions like, “Which is larger, $\cosh(\ln(42))$ or $\ln(\cosh(42))?$  These get students working with the definitions and thinking about asymptotic behavior.

Next week, I’ll go into more depth about the calculus of hyperbolic trigonometric functions.  Stay tuned!

## Calculus: The Geometry of Polynomials, II

The original post on The Geometry of Polynomials generated rather more interest that usual.  One reader, William Meisel, commented that he wondered if something similar worked for curves like the Folium of Descartes, given by the equation

$x^3+y^3=3xy,$

and whose graph looks like:

I replied that yes, I had success, and what I found out would make a nice follow-up post rather than just a reply to his comment.  So let’s go!

Just a brief refresher:  if, for example, we wanted to describe the behavior of $y=2(x-4)(x-1)^2$ where it crosses the x-axis at $x=1,$ we simply retain the $(x-1)^2$ term and substitute the root $x=1$ into the other terms, getting

$y=2(1-4)(x-1)^2=-6(x-1)^2$

as the best-fitting parabola at $x=1.$

$\displaystyle\lim_{x\to1}\dfrac y{(x-1)^2}=-6.$

For examples like the polynomial above, this limit is always trivial, and is essentially a simple substitution.

What happens when we try to evaluate a similar limit with the Folium of Descartes?  It seems that a good approximation to this curve at $x=0$ (the U-shaped piece, since the sideways U-shaped piece involves writing $x$ as a function of $y$) is $y=x^2/3,$ as shown below.

To see this, we need to find

$\displaystyle\lim_{x\to0}\dfrac y{x^2}.$

After a little trial and error, I found it was simplest to use the substitution $z=y/x^2,$ and so rewrite the equation for the Folium of Descartes by using the substitution $y=x^2z,$ which results in

$1+x^3z^3=3z.$

Now it is easy to see that as $x\to0,$ we have $z\to1/3,$ giving us a good quadratic approximation at the origin.

Success!  So I thought I’d try some more examples, and see how they worked out.  I first just changed the exponent of $x,$ looking at the curve

$x^n+y^3=3xy,$

shown below when $n=6.$

What would be a best approximation near the origin?  You can almost eyeball a fifth-degree approximation here, but let’s assume we don’t know the appropriate power and make the substitution $y=x^kz,$ with $k$ yet to be determined. This results in

$x^{3k-n}z^3+1=3zx^{k+1-n}.$

Now observe that when $k=n-1,$ we have

$x^{2n-3}z^3+1=3z,$

so that $\displaystyle\lim_{x\to0}z=1/3.$ Thus, in our case with $n=6,$ we see that $y=x^5/3$ is a good approximation to the curve near the origin.  The graph below shows just how good an approximation it is.

OK, I thought to myself, maybe I just got lucky.  Maybe introduce a change which will really alter the nature of the curve, such as

$x^3+y^3=3xy+1,$

whose graph is shown below.

Here, the curve passes through the x-axis at $x=1,$ with what appears to be a linear pass-through.  This suggests, given our previous work, the substitution $y=(x-1)z,$ which results in

$x^3+(x-1)^3z^3=3x(x-1)z+1.$

We don’t have much luck with $\displaystyle\lim_{x\to1}z$ here.  But if we move the $1$ to the other side and factor, we get

$(x-1)(x^2+x+1)+(x-1)^3z^3=3x(x-1)z.$

Nice!  Just divide through by $x-1$ to obtain

$x^2+x+1+(x-1)^2z=3xz.$

Now a simple calculation reveals that $\displaystyle\lim_{x\to1}z=1.$ And sure enough, the line $y=x-1$ does the trick:

Then I decided to change the exponent again by considering

$x^n+y^3=3xy+1.$

Here is the graph of the curve when $n=6:$

It seems we have two roots this time, with linear pass-throughs.  Let’s try the same idea again, making the substitution $y=(x-1)z,$ moving the $1$ over, factoring, and dividing through by $x-1.$  This results in

$x^{n-1}+x^{n-2}+\cdots+1+(x-1)^2z^3=3xz.$

It is not difficult to calculate that $\displaystyle\lim_{x\to1}z=n/3.$

Now things become a bit more interesting when $n$ is even, since there is always a root at $x=-1$ in this case.  Here, we make the substitution $y=(x+1)z,$ move the $1$ over, and divide by $x+1,$ resulting in

$\dfrac{x^n-1}{x+1}+(x+1)^2z^3=3xz.$

But since $n$ is even, then $x^2-1$ is a factor of $x^n-1,$ so we have

$(x-1)(x^{n-2}+x^{n-4}+\cdots+x^2+1)+(x+1)^2z^3=3xz.$

Substituting $x=-1$ in this equation gives

$-2\left(\dfrac n2\right)=3(-1)z,$

which immediately gives  $\displaystyle\lim_{x\to1}z=n/3$ as well!  This is a curious coincidence, for which I have no nice geometrical explanation.  The case when $n=6$ is illustrated below.

This is where I stopped — but I was truly surprised that everything I tried actually worked.  I did a cursory online search for Taylor series of implicitly defined functions, but this seems to be much less popular than series for $y=f(x).$

Anyone more familiar with this topic care to chime in?  I really enjoyed this brief exploration, and I’m grateful that William Meisel asked his question about the Folium of Descartes.  These are certainly instances of a larger phenomenon, but I feel the statement and proof of any theorem will be somewhat more complicated than the analogous results for explicitly defined functions.

And if you find some neat examples, post a comment!  I’d enjoy writing another follow-up post if there is continued interested in this topic.

## Bay Area Mathematical Artists, VII

We had yet another amazing meeting of the Bay Area Mathematical Artists yesterday!  Just two speakers — but even so, we went a half-hour over our usual 5:00 ending time.

Our first presenter was Stan Isaacs.  There was no real title to his presentation, but he brought another set of puzzles from his vast collection to share.  He was highlighting puzzles created by Wayne Daniel.

Below you’ll see one of the puzzles disassembled.  The craftsmanship is simply remarkable.

If you look carefully, you’ll see what’s going on.  The outer pieces make an icosahedron, and when you take those off, a dodecahedron, then a cube…a wooden puzzle of nested Platonic solids!  The pieces all fit together so perfectly.  Stan is looking forward to an exhibition of Wayne’s work at the International Puzzle Party in San Diego later on this year.  For more information, contact Stan at stan@isaacs.com.

Our second speaker was Scott Kim (www.scottkim.com), who’s presentation was entitled Motley Dissections.  What is a motley dissection?  The most famous example is the problem of the squared square — that is, dissecting a square with an integer side length into smaller squares with integer side lengths, but with all the squares different sizes.

One property of such a dissection is that no two edges of squares meet exactly corner to corner.  In other words, edges always overlap in some way.

But there are of course many other motley dissections.  For example, below you see a motley dissection of one rectangle into five, one pentagon into eleven, and finally, one hexagon into a triangle, square, pentagon and hexagon.

Look carefully, and you’ll see that no single edge in any of these dissections exactly matches any other.  For these decompositions, Scott has proved they are minimal — so, for example, there is no motley dissection of one pentagon to ten or fewer.  The proofs are not exactly elegant, but they serve their purpose.  He also mentioned that he credits Donald Knuth with the term motley dissection, who used the term in a phone conversation not all that long ago.

Can you cube the cube?  That is, can you take a cube and subdivide it into cubes which are all different?  Scott showed us a simple proof that you can’t.  But, it turns out, you can box the box.  In other words, if the length, width, and height of the larger box and all the smaller boxes may be different, then it is possible to box the box.

Next week, Scott is off to the Gathering 4 Gardner in Atlanta, and will be giving his talk on Motley Dissections there.  He planned an activity where participants actually build a boxed box — and we were his test audience!

He created some very elaborate transparencies with detailed instructions for cutting out and assembling.  There were a very few suggestions for improvement, and Scott was happy to know about them — after all, it is rare that something works out perfectly the first time.  So now, his success at G4G in Atlanta is assured….

We were so into creating these boxed boxes, that we happily stayed until 5:30 until we had two boxes completed.

I should mention that Scott also discussed something he terms pseudo-duals in two, three, and even four dimensions!  There isn’t room to go into those now, but you can contact him through his website for more information.

As usual, we went out to dinner afterwards — and we gravitated towards our favorite Thai place again.  The dinner conversation was truly exceptional this evening, revolving around an animated conversation between Scott Kim and magician Mark Mitton (www.markmitton.com).

The conversation was concerned with the way we perceive mathematics here in the U.S., and how that influences the educational system.  Simply put, there is a lot to be desired.

One example Scott and Mark mentioned was the National Mathematics Festival (http://www.nationalmathfestival.org).  Tens of thousands of kids and parents have fun doing mathematics.  Then the next week, they go back to their schools and keep learning math the same — usually, unfortunately, boring — way it’s always been learned.

So why does the National Mathematics Festival have to be a one-off event?  It doesn’t!  Scott is actively engaged in a program he’s created where he goes into an elementary school at lunchtime one day a week and let’s kids play with math games and puzzles.

Why this model?  Teachers need no extra prep time, kids don’t need to stay after school, and so everyone can participate with very little needed as far as additional resources are concerned.  He’s hoping to create a package that he can export to any school anywhere where with minimal effort, so that children can be exposed to the joy of mathematics on a regular basis.

Mark was interested in Scott’s model:  consider your Needs (improving the perception of mathematics), be aware of the Forces at play (unenlightened administrators, for example, and many other subtle forces at work, as Mark explained), and then decide upon Actions to take to move the Work (applied, pure, and recreational mathematics) forward.

The bottom line:  we all know about this problem of attitudes toward mathematics and mathematics education, but no one really knows what to do about it.  For Scott, it’s just another puzzle to solve.  There are solutions.  And he is going to find one.

We talked for over two hours about these ideas, and everyone chimed in at one time or another.  Yes, my summary is very brief, I know, but I hope you get the idea of the type of conversation we had.

Stay tuned, since we are planning an upcoming meeting where we focus on Scott’s model and work towards a solution.  Another theme throughout the conversation was that mathematics is not an activity done in isolation — it is a communal activity.  So the Needs will not be addressed by a single individual, rather a group, and likely involving many members of many diverse communities.

A solution is out there.  It will take a lot of grit to find it.  But mathematicians have got grit in spades.