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
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 where it crosses the x-axis at we simply retain the term and substitute the root into the other terms, getting
as the best-fitting parabola at
Now consider another way to think about this:
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 (the U-shaped piece, since the sideways U-shaped piece involves writing as a function of ) is as shown below.
To see this, we need to find
After a little trial and error, I found it was simplest to use the substitution and so rewrite the equation for the Folium of Descartes by using the substitution which results in
Now it is easy to see that as we have 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 looking at the curve
shown below when
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 with yet to be determined. This results in
Now observe that when we have
so that Thus, in our case with we see that 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
whose graph is shown below.
Here, the curve passes through the x-axis at with what appears to be a linear pass-through. This suggests, given our previous work, the substitution which results in
We don’t have much luck with here. But if we move the to the other side and factor, we get
Nice! Just divide through by to obtain
Now a simple calculation reveals that And sure enough, the line does the trick:
Then I decided to change the exponent again by considering
Here is the graph of the curve when
It seems we have two roots this time, with linear pass-throughs. Let’s try the same idea again, making the substitution moving the over, factoring, and dividing through by This results in
It is not difficult to calculate that
Now things become a bit more interesting when is even, since there is always a root at in this case. Here, we make the substitution move the over, and divide by resulting in
But since is even, then is a factor of so we have
Substituting in this equation gives
which immediately gives as well! This is a curious coincidence, for which I have no nice geometrical explanation. The case when 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
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.