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:

Day141Folium1

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.

Now consider another way to think about this:

\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.

Day141Folium2

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.

Day141Folium3.png

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.

Day141Folium4.png

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.

Day141Folium5

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:

Day141Folium6

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:

Day141Folium7

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.

Day141Folium8

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.

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Vince Matsko

Mathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math

2 thoughts on “Calculus: The Geometry of Polynomials, II”

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