# We Don't Need No Stinking Derivatives

September 13, 2007 5:53 AM Subscribe

Obscure Mathfilter: In the arcane past of mathematics, there is something known as a "Newton Polygon", which theoretically can be used to help sketch functions of two variables, say, x^3 + y^3 =3xy, for instance. But --

exactly HOW you use the polygon information to help sketch the graph has never been clearly explained to me.

Does anyone know how to do this?

Related -- there was a post last May or June in (i think) one of the Math Blogs that may have related to this. It was a shortcut approach to sketching polynomials without using derivatives, that just looked at the exponents of each term of the polynomial, and then connected them together using some rules. At the time, I thought "I'll come back and look at this in detail when I get a chance", but, alas, I cannot find it again. (My google-fu is not strong.) Does anyone know what I am talking about?

Alternatively, how would you sketch f(x, y) by hand, especially if it cannot be explicilty solved for x or y? (I realize all sorts of programs will do this for you; I am interested in by hand techniques.)

exactly HOW you use the polygon information to help sketch the graph has never been clearly explained to me.

Does anyone know how to do this?

Related -- there was a post last May or June in (i think) one of the Math Blogs that may have related to this. It was a shortcut approach to sketching polynomials without using derivatives, that just looked at the exponents of each term of the polynomial, and then connected them together using some rules. At the time, I thought "I'll come back and look at this in detail when I get a chance", but, alas, I cannot find it again. (My google-fu is not strong.) Does anyone know what I am talking about?

Alternatively, how would you sketch f(x, y) by hand, especially if it cannot be explicilty solved for x or y? (I realize all sorts of programs will do this for you; I am interested in by hand techniques.)

Response by poster: Bluefly -- that is indeed the post I was talking about. I

posted by wittgenstein at 7:07 AM on September 13, 2007

**also**think it doesn't seem to always work -- I was hoping by now there might be more details in the comments to the original post.posted by wittgenstein at 7:07 AM on September 13, 2007

In my multivariable calculus class, we would let z (which is f(x,y) in this case) equal something, so as to reduce the problem to two variables. For instance:

What does x^3 + y^3 - 3xy = 0 look like?

How about x^3 + y^3 - 3xy = 1?

And so on: x^3 + y^3 - 3xy = -1?

Once you've sketched a few lines like that, then you have a contour graph in two dimensions that will help you visualize what the third dimension will look like. I don't think that's a newton polygon, but that's one way to sketch graph.

I guess you'd still have to solve that nasty equation. I don't remember if the professor ever had an example with that high of powers. If it was just squares, then it would be a circle, minus something else.

posted by philomathoholic at 7:36 AM on September 13, 2007

What does x^3 + y^3 - 3xy = 0 look like?

How about x^3 + y^3 - 3xy = 1?

And so on: x^3 + y^3 - 3xy = -1?

Once you've sketched a few lines like that, then you have a contour graph in two dimensions that will help you visualize what the third dimension will look like. I don't think that's a newton polygon, but that's one way to sketch graph.

I guess you'd still have to solve that nasty equation. I don't remember if the professor ever had an example with that high of powers. If it was just squares, then it would be a circle, minus something else.

posted by philomathoholic at 7:36 AM on September 13, 2007

Best answer: To graph it, I'd first make some general observations. The curve has no x- or y-intercepts except the origin. There are no points with x and y both greater than 3; there are no points in the third quadrant at all. The second and fourth quadrants are symmetric.

Then I'd use implicit differentiation to find dy/dx = (y-x^2)/(x-y^2). This says the slope is zero where y=x^2, which we plug back in to find that it happens when x=0 or x is one of the three real roots of x^3-3x+1. The first gives y=0, each of the others gives a cubic defining three values of y.

So already I've got 10 points I more or less understand. (If I really wanted to get a handle on them I'd probably use Newton's method to approximate them.)

Repeat the process for the points with undefined slope (or just use the obvious x-y symmetry). Note that there are (at least) two branches at (0,0) -- one with horizontal tangent, one with vertical.

Now you've got plenty of information for a decent sketch.

That's not how this stuff works.

posted by gleuschk at 9:03 AM on September 13, 2007 [1 favorite]

Then I'd use implicit differentiation to find dy/dx = (y-x^2)/(x-y^2). This says the slope is zero where y=x^2, which we plug back in to find that it happens when x=0 or x is one of the three real roots of x^3-3x+1. The first gives y=0, each of the others gives a cubic defining three values of y.

So already I've got 10 points I more or less understand. (If I really wanted to get a handle on them I'd probably use Newton's method to approximate them.)

Repeat the process for the points with undefined slope (or just use the obvious x-y symmetry). Note that there are (at least) two branches at (0,0) -- one with horizontal tangent, one with vertical.

Now you've got plenty of information for a decent sketch.

*If it was just squares, then it would be a circle, minus something else.*That's not how this stuff works.

posted by gleuschk at 9:03 AM on September 13, 2007 [1 favorite]

This thread is closed to new comments.

posted by bluefly at 6:37 AM on September 13, 2007