Tangent Sequences of a Cubic

Old mathematicians never die;
they just go off on a tangent.


Recently re-introduced to a pretty property of the tangents to cubics from Ross Honsberger's book, More Mathematical Morsels (Dolciani Mathematical Expositions).  I will illustrate a simple part of the property, then provide a source for a really nice extension that seems to be pretty new.

As the magician says, take a cubic, take any cubic... OK, so graph any y=f(x) where f(x) is a third degree polynomial. (you really don't have to graph it, but it might help you see a second property).  

1)  Pick any point other than the inflection point and record the x-value as A

2) Write the equation of the tangent line at that point
3)  Find where the tangent line intersects f(x) , call this B
4)  Repeat the process starting at Point B to get a point C,
5)  Do it again starting at C to get point D...
Look at the sequence A, B, C, D... What do you observe?


Ok, so let's practice a little integration. Find the area between the tangents and the curve.  Now think of the ratio of the two areas big/little...
Focus deeply, I'm reading your mind...
AHA, the ratio was 16.

"How does he do it?"

There is an additional nice extension about this idea explained by Alexander Bogomolny at his Cut-the-Knot web site.  Enjoy.

The Sharp Edge Between Infinity and "Finity"

After my problem on the steel rail got me in trouble, I have decided to stick to a purely theoretical problem that has no messy reality features to get in the way...


I recently came across two limit problems that got me to thinking about how narrow the line between infinity and the finite can be....

Try the following two limits.....

Both can be simplified by the old "multiply by one" trick ... letting the "one" in this case be the conjugate of the binomial ...For the first, it works like this...

Ok, sorry, I got those last two images reversed... the middle one is the simplest........
Which leads to the conclusion that the limit is 1/2... Ok, not terribly difficult, so what...???

 But if we do the same sort of thing with the second, we get :

Yikes, now the Limit has shot off to infinity... The difference between a one and a two for that leading coefficient has made a huge difference. At first we might think that we can find some number a such that :

has a finite limite between 1/2 and infinity, But when you try it.... if a is greater than 1, the limit is infinity, and if a is less than one, the limit is negative infinity, and in between, balanced on a razor's edge is a=1, which has a limit of 1/2...

Just goes to show that 1/2 is the average of Plus and minus infinity,

Archimedes and Calculus....

I showed my calculus kids that Archimedes knew the area between the curve y=x² and the x-axis from x=-2 to x=2, which is one of those really early things you evaluate as you learn the wonders of the definite integral. Actually, what he knew was the area between the curve y=4-x² and the x-axis; and he knew how to subtract one area from another. I pointed out to my students that it was an easy extension of the triangle area formula A=1/2 bh. To find the area inside a parabola which was cut by a chord perpendicular to the axis of symmetry. You just have to change the constant... and the area is 2/3 b h where the base is the length of the chord and the height is the perpendicular distance from the chord to the vertex of the parabola.


Now Archimedes actually knew more (lots more) than that... for example he knew that the same rule applied to any chord through a parabola if you measured the height as the greatest perpendicular distance from the chord to the arc of the parabola.

I’ve been researching a little about his writing, and can add that to figure out the above, he became the first person to evaluate an infinite geometric series(or at least the first to do it in writing and leave it where we could find it).

What Archimedes actually learned was that if you drew a chord to a parabola, the area of the section between the chord and the parabola is 4/3 the area of the largest inscribed triangle with the chord as a base.( and the triangle is 1/2 b h so 1/2 of 4/3 gives the 2/3... focus children).. In the picture I have shown the graph of y = 4-x² cut by the chord y=x+2 as an example.


The triangle has vertices at (-2,0), (1,3), and (-.5, 3.75){side bar problem.... can you show that the greatest perpendicular distance between the chord and the parabola will always be at a point where the tangent to the parabola was the same as the slope of the chord?}. He did this by showing that if you inscribed two more triangles in the sections of the parabola outside two sides of the triangle given, that they would add up to ¼ of the given triangle. And then if you draw four more outside these two, they will add up to 1/16 of the first. He then set out to show that 1 + ¼ + 1/16 + 1/64…. = 4/3. 1800 years before the invention of calculus, he used the idea of a limit to show

1 + 1/3 (1) = 4/3

1 + ¼ + 1/3 (1/4) = 4/3

1 + ¼ + 1/16 + 1/3 (1/16) = 4/3 and extended this to show

1 + ¼ + 1/16 + … 1/ (4n) + 1/3 (1/(4n) = 4/3 and of course, as n goes to infinity,1/(4n) also goes to zero, leaving the sum 4/3.


Simply incredible.... Archimedes....wow, .. Just know children.....He was NOT like us.

I will Derive

OK, one for all my weary calculus students, trying to make it to the end of the year after the AP exam... OK, you think the singing is bad, wait 'till you see the dancing.