11/15/2007, 10:20 AM
I'm still working on graphs, but it's not easy to strike a balance between using enough grid points to make a smooth graph and using few enough to keep memory usage within 1.5 GB and compute time to a fraction of an hour.
By the way, these are half-sized versions of the ones I originally made. I was going to attach them (without embedding them) for those who wanted to download them, but they are over 1 MB each, and I can't attach them at that size. I suppose I can email them if anyone wants to see the larger versions, but the small versions preserve most of the interesting details.
Okay, to establish (hopefully) clearly how I'm calculating the regular slog, here's a view at z=2, zoomed in pretty far. Note that each curve is almost a circle. Indeed, the further we drill down, the more they will look like circles.
An explanation is due here. Each circle is a contour of constant real part. No, they are not evenly spaced, nor even exponentially spaced. I will explain the spacing later.
Essentially, I iteratively exponentiated zero, for reference 57 times. I picked a number that puts a blue ring going through the origin, based on my palette, if you're wondering why 57. There is still a fair amount of inaccuracy here, but not enough to show up on the graph.
Once I had the 57th iterate of 0, I found the 58th (i.e., the 57th iterate of 1), and then subtracted each from 2. The distances from 2 are decreasing exponentially, so I take a natural logarithm of each, calling them l0 and l1 respectively (for log of "zero" and log of "one"). I then subtract, giving -0.36653913, which is very near the predicted value of ln(ln(2)). I'll call this value dl (for difference of logarithmcs).
I then create a grid of values (not evenly spaced), with real parts on the interval \( (l0,l0+dl) \), and imaginary part on the interval \( \left(0, \text{abs}(\pi / dl)) \). When I exponentiate (base e) the values in this grid, I get half a hollow disk, which I then subtract (point by point) from 2. Its inner radius passes through the 57th exponentiation of 1, the outer radius passes through the 57th exponentiation of 0. It's only a half of a hollow disk. I don't actually calculate the points above the real line; rather, I simply graph the lower half plane and mirror it to get the other half.
Then, I can simply take the logarithm of this grid, point by point, to get each successive layer moving out from the fixed point. Here, I've zoomed out by a factor of 10.
Don't be fooled by the small shift in the position of the colors: we've gone more than a full cycle through the palette. Note the scale on the axes. The red circle that was clipped and only visible in the corners is now the red circle that fits comfortably between the text labels closest to the origin.
By the way, these are half-sized versions of the ones I originally made. I was going to attach them (without embedding them) for those who wanted to download them, but they are over 1 MB each, and I can't attach them at that size. I suppose I can email them if anyone wants to see the larger versions, but the small versions preserve most of the interesting details.
Okay, to establish (hopefully) clearly how I'm calculating the regular slog, here's a view at z=2, zoomed in pretty far. Note that each curve is almost a circle. Indeed, the further we drill down, the more they will look like circles.
An explanation is due here. Each circle is a contour of constant real part. No, they are not evenly spaced, nor even exponentially spaced. I will explain the spacing later.
Essentially, I iteratively exponentiated zero, for reference 57 times. I picked a number that puts a blue ring going through the origin, based on my palette, if you're wondering why 57. There is still a fair amount of inaccuracy here, but not enough to show up on the graph.
Once I had the 57th iterate of 0, I found the 58th (i.e., the 57th iterate of 1), and then subtracted each from 2. The distances from 2 are decreasing exponentially, so I take a natural logarithm of each, calling them l0 and l1 respectively (for log of "zero" and log of "one"). I then subtract, giving -0.36653913, which is very near the predicted value of ln(ln(2)). I'll call this value dl (for difference of logarithmcs).
I then create a grid of values (not evenly spaced), with real parts on the interval \( (l0,l0+dl) \), and imaginary part on the interval \( \left(0, \text{abs}(\pi / dl)) \). When I exponentiate (base e) the values in this grid, I get half a hollow disk, which I then subtract (point by point) from 2. Its inner radius passes through the 57th exponentiation of 1, the outer radius passes through the 57th exponentiation of 0. It's only a half of a hollow disk. I don't actually calculate the points above the real line; rather, I simply graph the lower half plane and mirror it to get the other half.
Then, I can simply take the logarithm of this grid, point by point, to get each successive layer moving out from the fixed point. Here, I've zoomed out by a factor of 10.
Don't be fooled by the small shift in the position of the colors: we've gone more than a full cycle through the palette. Note the scale on the axes. The red circle that was clipped and only visible in the corners is now the red circle that fits comfortably between the text labels closest to the origin.
~ Jay Daniel Fox
