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Email Address: [ ] Follow Search for: [ ] [Search] * RSS - Posts * RSS - Comments Recent Posts * Polyhedra Graphics, I * The Puzzle Cabaret * Largest Number Found! * Recursively Generated Spirals * My Polyhedra Textbook, II Archives * April 2021 * December 2020 * June 2020 * November 2019 * June 2019 * May 2019 * December 2018 * November 2018 * October 2018 * September 2018 * August 2018 * July 2018 * June 2018 * May 2018 * April 2018 * March 2018 * February 2018 * January 2018 * December 2017 * November 2017 * October 2017 * September 2017 * August 2017 * July 2017 * June 2017 * May 2017 * April 2017 * March 2017 * February 2017 * January 2017 * December 2016 * November 2016 * October 2016 * September 2016 * August 2016 * July 2016 * June 2016 * May 2016 * April 2016 * March 2016 * February 2016 * January 2016 * December 2015 * November 2015 * October 2015 * September 2015 The Geometry of Polynomials I recently needed to make a short demo lecture, and I thought I'd share it with you. I'm sure I'm not the first one to notice this, but I hadn't seen it before and I thought it was an interesting way to look at the behavior of polynomials where they cross the x-axis. The idea is to give a geometrical meaning to an algebraic procedure: factoring polynomials. What is the geometry of the different factors of a polynomial? Let's look at an example in some detail: f(x)=2(x-4)(x-1)^2.poly0b Now let's start looking at the behavior near the roots of this polynomial. poly0c Near x=1, the graph of the cubic looks like a parabola -- and that may not be so surprising given that the factor (x-1) occurs quadratically. poly0d And near x=4, the graph passes through the x-axis like a line -- and we see a linear factor of (x-4) in our polynomial. But which parabola, and which line? It's actually pretty easy to figure out. Here is an annotated slide which illustrates the idea. Day137poly1 All you need to do is set aside the quadratic factor of (x-1)^2, and substitute the root, x=1, in the remaining terms of the polynomial, then simplify. In this example, we see that the cubic behaves like the parabola y=-6(x-1)^2 near the root x=1. Note the scales on the axes; if they were the same, the parabola would have appeared much narrower. We perform a similar calculation at the root x=4. Day137poly2 Just isolate the linear factor (x-4), substitute x=4 in the remaining terms of the polynomial, and then simplify. Thus, the line y=18(x-4) best describes the behavior of the graph of the polynomial as it passes through the x-axis. Again, note the scale on the axes. We can actually use this idea to help us sketch graphs of polynomials when they're in factored form. Consider the polynomial f(x)=x(x+1)^2 (x-2)^3. Begin by sketching the three approximations near the roots of the polynomial. This slide also shows the calculation for the cubic approximation. Day137poly3.png Now you can begin sketching the graph, starting from the left, being careful to closely follow the parabola as you bounce off the x-axis at x=-1. poly1d Continue, following the red line as you pass through the origin, and then the cubic as you pass through x=2. Of course you'd need to plot a few points to know just where to start and end; this just shows how you would use the approximations near the roots to help you sketch a graph of a polynomial. poly1f Why does this work? It is not difficult to see, but here we need a little calculus. Let's look, in general, at the behavior of f(x)=p (x)(x-a)^n near the root x=a. Given what we've just been observing, we'd guess that the best approximation near x=a would just be y=p(a) (x-a)^n. Just what does "best approximation" mean? One way to think about approximating, calculuswise, is matching derivatives -- just think of Maclaurin or Taylor series. My claim is that the first n derivatives of f(x)=p(x)(x-a)^n and y=p(a)(x-a)^n match at x=a. First, observe that the first n-1 derivatives of both of these functions at x=a must be 0. This is because (x-a) will always be a factor -- since at most n-1 derivatives are taken, there is no way for the (x-a)^n term to completely "disappear." But what happens when the nth derivative is taken? Clearly, the nth derivative of p(a)(x-a)^n at x=a is just n!p(a). What about the nth derivative of f(x)=p(x)(x-a)^n? Thinking about the product rule in general, we see that the form of the nth derivative must be f^{(n)}(x)=n!p(x)+ (x-a)(\text{terms involving derivatives of } p(x)). When a derivative of p(x) is taken, that means one factor of (x-a) survives. So when we take f^{(n)}(a), we also get n!p(a). This makes the nth derivatives match as well. And since the first n derivatives of p(x) (x-a)^n and p(a)(x-a)^n match, we see that p(a)(x-a)^n is the best n th degree approximation near the root x=a. I might call this observation the geometry of polynomials. Well, perhaps not the entire geometry of polynomials.... But I find that any time algebra can be illustrated graphically, students' understanding gets just a little deeper. Those who have been reading my blog for a while will be unsurprised at my geometrical approach to algebra (or my geometrical approach to anything, for that matter). Of course a lot of algebra was invented just to describe geometry -- take the Cartesian coordinate plane, for instance. So it's time for algebra to reclaim its geometrical heritage. I shall continue to be part of this important endeavor, for however long it takes.... Share this: * Twitter * Facebook * Like this: Like Loading... Related Published by [27c6b7] Vince Matsko Mathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math View all posts by Vince Matsko Posted on March 17, 2018Author Vince MatskoCategories Calculus, Geometry: 2D, PedagogyTags Day137 6 thoughts on "The Geometry of Polynomials" 1. Pingback: Calculus: Linear Approximations, I - Creativity in Mathematics 2. [cdb944] William Meisel says: April 4, 2018 at 4:51 pm Vince - this is really interesting and I am surprised that I have never come across it anywhere before. Does a similar thing work for functions of two variables? I'm thinking of things that are particularly tough to sketch by hand, like Descarte's Folium (x^3 + y^3 = 3xy). I have read articles that suggest using something called Newton's Polygon to sketch these, but I will admit I have not been able to follow the steps that clearly. (My intuition is that this method will not generalize to higher dimensions, because I guess you would have to use partial derivatives, but I am often wrong.) LikeLike Reply 3. [27c6b7] Vince Matsko says: April 5, 2018 at 10:50 pm William - yes, I got it to work with the Folium of Descartes! And in several other cases that I tried as well. Too much to go into here, but I plan to write a more detailed blog post on what I found out in a few weeks (this weekend I'll be writing about the Bay Area Mathematical Artists). Sorry to keep you in suspense....but thanks for the inspiration! LikeLiked by 1 person Reply 4. Pingback: Calculus: The Geometry of Polynomials, II - Creativity in Mathematics 5. Pingback: 100 Posts! - Creativity in Mathematics 6. [7e8814] Rachel Campos says: April 12, 2020 at 3:46 am Well written post. LikeLike Reply Leave a Reply Cancel reply Enter your comment here... 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