Infinite Newton

Apropos Newton’s method in the complex plane, what happens when the degree of the polynomial goes to infinity?

Towards infinity

Obviously there will be more zeros, so there will be more attractors and we should expect the boundaries of the basins of attraction to become messier. But it is not entirely clear where the action will be, so it would be useful to compress the entire complex plane into a convenient square.

How do you depict the entire complex plane? While I have always liked the Riemann sphere here I tried mapping x+yi to [\tanh(x),\tanh(y)]. The origin is unchanged, and infinity becomes the edges of the square [-1,1]\times [-1,1]. This is not a conformal map, so things will get squished near the edges.

For color, I used (1/2)+(1/2)[\tanh(|z-1|-1), \tanh(|z+1|-1), \tanh(|z-i|-1)] to map complex coordinates to RGB. This makes the color depend on the distance to 1, -1 and i, making infinity white and zero some drab color (the -1 terms at the end determines the overall color range).

Here is the animated result:

What is going on? As I scale up the size of the leading term from zero, the root created by adding that term moves in from infinity towards the center, making the new basin of attraction grow. This behavior has been described in this post on dancing zeros. The zeros also tend to cluster towards the unit circle, crowding together and distributing themselves evenly. That distribution is the reason for the the colorful “flowers” that surround white spots (poles of the Newton formula, corresponding to zeros of the derivative of the polynomial). The central blob is just the attractor of the most “solid” zero, corresponding to the linear and constant terms of the polynomial.

The jostling is amusing: it looks like the roots do repel each other. This is presumably because close roots require a sharp turn of the function, but the “turning radius” is set by the coefficients that tend to be of order unity. Getting degenerate roots requires coefficients to be in a set of measure zero, so it is rare. Near-degenerate roots exist in a positive measure set surrounding that set, but it is still “small” compared to the general case.

At infinity

So what happens if we let the degree go to infinity? As I previously mentioned, the generic behaviour of \sum_{n=1}^\infty a_n z^n where a_n is independent random numbers is a lacunary function. So we should not expect anything outside the unit circle. Inside the circle there will be poles, so there will be copies of the undefined outside region (because of Great Picard’s Theorem (meromorphic version)). Since the function is analytic these copies will be conformal mappings of the exterior and hence roughly circular. There will also be zeros, and these will have their own basins of attraction. A few of the central ones dominate, but there is an infinite number of attractors as we approach the circular border which is crammed with poles and zeros.

Since we now know we will only deal with the unit disk, we can avoid transforming the entire plane and enjoy the results:

Attractors for random 10,000-degree polynomial.
Attractors for random 10,000-degree polynomial.
Attractors for random 10,000-degree polynomial.
Attractors for random 10,000-degree polynomial.

What happens here is that the white regions represents places where points get mapped onto the undefined outside, while the colored fractal regions are the attraction basins for the zeros. And between them there is a truly wild boundary. In the vanilla z^3+1 fractal every point on the boundary is a meeting point of the three basins, a tri-point. Here there is an infinite number of attractors: the boundary consists of points where an infinite number of different attractors meet.

Checking my predictions for 2016

Last year I made a number of predictions for 2016 to see how well calibrated I am. Here is the results:

Prediction Correct?
No nuclear war: 99% 1
No terrorist attack in the USA will kill > 100 people: 95% 1 (Orlando: 50)
I will be involved in at least one published/accepted-to-publish research paper by the end of 2015: 95% 1
Vesuvius will not have a major eruption: 95% 1
I will remain at my same job through the end of 2015: 90% 1
MAX IV in Lund delivers X-rays: 90% 1
Andart II will remain active: 90% 1
Israel will not get in a large-scale war (ie >100 Israeli deaths) with any Arab state: 90% 1
US will not get involved in any new major war with death toll of > 100 US soldiers: 90% 1
New Zeeland has not decided to change current flag at end of year: 85% 1
No multi-country Ebola outbreak: 80% 1
Assad will remain President of Syria: 80% 1
ISIS will control less territory than it does right now: 80% 1
North Korea’s government will survive the year without large civil war/revolt: 80% 1
The US NSABB will allow gain of function funding: 80% 1 [Their report suggests review before funding, currently it is up to the White House to respond. ]


US presidential election: democratic win: 75% 0
A general election will be held in Spain: 75% 1
Syria’s civil war will not end this year: 75% 1
There will be no NEO with Torino Scale >0 on 31 Dec 2016: 75% 0 (2016 XP23 showed up on the scale according to JPL, but NEODyS Risk List gives it a zero.)
The Atlantic basin ACE will be below 96.2: 70% 0 (ACE estimate on Jan 1 is 132)
Sweden does not get a seat on the UN Security Council: 70% 0
Bitcoin will end the year higher than $200: 70% 1
Another major eurozone crisis: 70% 0
Brent crude oil will end the year lower than $60 a barrel: 70% 1
I will actually apply for a UK citizenship: 65% 0
UK referendum votes to stay in EU: 65% 0
China will have a GDP growth above 5%: 65% 1
Evidence for supersymmetry: 60% 0
UK larger GDP than France: 60% 1 (although it is a close call; estimates put France at 2421.68 and UK at 2848.76 – quite possibly this might change)
France GDP growth rate less than 2%: 60% 1
I will have made significant progress (4+ chapters) on my book: 55% 0
Iran nuclear deal holding: 50% 1
Apple buys Tesla: 50% 0
The Nikkei index ends up above 20,000: 50% 0 (nearly; the Dec 20 max was 19,494)

Overall, my Brier score is 0.1521. Which doesn’t feel too bad.

Plotting the results (where I bin together things in [0.5,0.55], [0.5,0.65], [0.7 0.75], [0.8,0.85], [0.9,0.99] bins) give this calibration plot:

Plot of average correctness of my predictions for 2016 as a function of confidence.
Plot of average correctness of my predictions for 2016 as a function of confidence (blue). Red line is perfect calibration.

Overall, I did great on my “sure bets” and fairly weakly on my less certain bets. I did not have enough questions to make this very statistically solid (coming up with good prediction questions is hard!), but the overall shape suggests that I am a bit overconfident, which is not surprising.

Time to come up with good 2017 prediction questions.