Complex Numbers for Rotating, Translating, and Scaling the Plane June 7th, 2009
Patrick Stein

A good friend of mine recently discovered some of the fun things you can do with complex numbers if you’re using them to represent points in the plane. Yesterday, I re-read a passage by Tony Smith about why one should be interested in Clifford algebras. Tony Smith’s passage included all of the fun one can have with the complex plane and extends it to three, four, five, and more dimensions. I thought, I should segue from the complex numbers in the plane to Clifford algebras to quaternions in 3-space to Clifford algebras again in a series of posts here.

What are Complex Numbers

Say you’re playing around with polynomials. You start playing with the equation z^2 - 1 = 0. WIth a little fiddling, you find this is equivalent to z^2 = 1. Then, you take the square root of both sides to find that z = \pm \sqrt{1} = \pm 1. We started with a polynomial equation in one variable in which the highest exponent was two and we found two answers.

Pounding your chest and sounding your barbaric yawp, you move on to z^2 + 1 = 0. This should be easy, right? With the same fiddling, we find z^2 = -1 and then z = \pm \sqrt{-1}.

Uh-oh. What do we do now? We can’t think of any number that when multiplied by itself gives us a negative number. If we start with zero, we end with zero. If we multiply a positive number by itself, we get a positive number. If we multiply a negative number by itself, we get a positive number. Again!

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Trying to Short-Stop Iterated Functions May 15th, 2009
Patrick Stein

If a function is pretty fun, why not do it again?

My post yesterday about the Mandelbrot set got me thinking again about iterated functions. With the Mandelbrot set, you start with some complex number c and with z_0 = 0. Then, you generate z_{i+1} by doing z_{i+1} = z_i^2 + c.

Here’s another way to write this iteration. Let f_c(z) = z^2 + c. Let z_0 = 0 and z_1 = f_c(z_0) and z_2 = f_c(z_1) = f_c(f_c(z_0)) = f_c^2(z_0). In general, we write f_c^n(z) to mean f_c(f_c(\ldots f_c(z))) when there are n copies of f_c. It is easy to verify then that f_c^{a}(f_c^{b}(z)) = f_c^{a+b}(x).

The Jam

fibers For that post, I generated the image linked at the right here. Each level of that image represents one iteration of f_c^{i+1}(0) = f_c(f_c^i(0)) with each strand made up of four different f_c functions with neighboring c’s.

One of the annoying things about that image is that I went directly from one iteration to the next. I had pondered using conventional splines or other polynomials to interpolate between iterations in an effort to smooth out the transitions. I didn’t go to the trouble for it though because any straightforward interpolation would be every bit as fake as the linear version. There is no provision in the iteration for f_c^{\alpha} unless \alpha is a non-negative integer.

Now the question is, can we fake one? Can we make some other function g_c(z) so that g_c^2(z) = f_c(z)?

Faking easy functions

Let’s start with an easier function. In fact, let’s start with the easiest function: f(x) = 0. It is obvious that f^n(x) = f(x) for all positive integers n. As such, if we let g(x) = 0, it is trivially true that g^2(x) = f(x). But, that was too easy.

Maybe f(x) = x will be harder? No. Again, f^2(x) = f(f(x)) = f(x).

Faking translations

Let’s move up to something a little trickier. Let’s say that f(x) = x + c for some constant c. Now, we see that f^2(x) = f(f(x)) = f(x + c) = (x + c) + c = x + 2c. What are we going to try for g(x) then? Let’s try the obvious: g(x) = x + \frac{c}{2}. Hooray! g^2(x) = f(x). So, it makes sense then to think of f^{1/2}(x) as x + \frac{c}{2}. Similarly, we can think of f^{1/n}(x) as x + \frac{c}{n} for all positive integers n. In this way, we can make values f^q(x) for any positive rational number q.

Faking translations with scaling

Now, we’re cooking. Let’s up the ante. Let’s try f(x) = ax + b. What are we going to try for g(x) then? Let’s guess we can still use roughly the same form: g(x) = \alpha x + \beta. Then, g^2(x) = \alpha^2 x + \alpha \beta + \beta. If we want g^2(x) = f(x), then we need \alpha^2 = a and (\alpha + 1)\beta = b. This means that \alpha = \sqrt{a} and \beta = \frac{b}{1 + \sqrt{a}}. In particular, notice that if a and b are both real numbers with a less than zero, then \alpha is pure imaginary and \beta is complex.

TTFN

I have much more to say about iterated functions, but I will save some for the next iteration. Next time, I will start with that last case and calculate f^q(x) for rational q.

Finding Better Polynomials May 12th, 2009
Patrick Stein

Some time ago, I wrote a small domain-specific language for finding polynomials given their value or the value of their derivatives at particular points.

It occurred to me shortly after writing that code that I could easily extend it to include the value of its integral over a certain range. I didn’t get to tackling that right away, and that was a Good Thing. In the intervening time, it occurred to me that I could extend it to moments as well.

So, are you looking for a polynomial that is zero at both zero and one, has a derivative of zero at one, has a second derivative of negative one at one, whose integral from zero to one is one, and whose mean (first moment centered at zero) on the interval zero to one is one fourth?

(polynomial-to-string
  (calculate-polynomial-subject-to
    (value :at 0 :equals 0)
    (value :at 1 :equals 0)
    (derivative :at 1 :equals 0)
    (nth-derivative 2 :at 1 :equals -1)
    (integral :from 0 :to 1 :equals 1)
    (mean :from 0 :to 1 :equals 1/4)))

Well, that would be f(x) = \frac{149}{4}x - 163x^2 + 265x^3 - 190x^4 + \frac{203}{4}x^5. (For some reason though, gnuplot thinks f(x) = -1, so no graph for you…)

Here is the source file.

graph
Edit: I realized that gnuplot was off by one because it was truncating the fractions. So, I just changed the 4’s in the denominators to 4.0’s and Bob’s your uncle.

 

Phony Physics (a.k.a. Fun with Interpolation) April 3rd, 2009
Patrick Stein

In a previous post, I mentioned looking for a polynomial for an application. I am working on an application that involves clicking or dragging tiles around. Once you release the tile, I want it to snap to where it’s supposed to be.

The easiest way to do this is to just warp it to where it’s supposed to go. This isn’t very visually pleasing. The next best thing is to set up a time-interval on which it moves into place. Then, linearly interpolate from where it is to where it’s going (here with t normalized to range between zero and one):

(1-t)x_0 + tx_1

That’s much better than just warping, but it doesn’t have any sort of fade-in or fade-out. It instantaneously has its maximum velocity and then instantaneously stops at the end.

Typically, then one uses a cubic spline to scale the t value before interpolating to get a speed up in the beginning and then slow down in the end.
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Find the Polynomial You’ve Been Looking For April 2nd, 2009
Patrick Stein

Often, I want to have a polynomial that meets certain criteria. For most applications in the past, I figured it all out by hand.

Today, I was having trouble getting the conditions specified enough to get the sort of polynomial that I was expecting. After finding the coefficients for several sixth degree polynomials in a row, I figured I should instead be able to do something like this with the proper lisp:

(calculate-polynomial-subject-to
    (value :at 0 :equals 0)
    (derivative :at 0 :equals 0)
    (nth-derivative 2 :at 0 :equals 0)
    (value :at 1 :equals 1)
    (derivative :at 1 :equals 0)
    (nth-derivative 2 :at 1 :equals 0)
    (value :at 1/2 :equals 3/4))

That’s all done now (polynomials.lisp). So, for the record, the above is: 26x^3 - 63x^4 + 54x^5 - 16x^6. That is still not quite the polynomial I want for this application, but it’s close. A few minutes of Lisp saved me hours of whiteboard work.

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