My Everlasting Hunger For Obtaining Knowledge Will Never Subside. My Mind Will Forever Be A Bundled Mess

Littlewood polynomials are polynomials all of whose coefficients are either +1 or −1 (so even 0 is not allowed). If you take all Littlewood polynomials up to a certain degree, calculate all their (complex) roots, and plot those roots in the complex plane, then you get a beautiful fractal-like structure above.

The image is slightly misleading, because the “holes” on the unit circle tend to completely fill in if the degree goes up. Intuitively, the holes mean that complex numbers on the unit circle that are close to low-degree roots of unity are hard to approximate by low-degree Littlewood polynomials (unless they already are roots of unity).

In particular the structure at the edge of the ring is deeply interesting. Notice the familiarity with the dragon curve?

Moving blog! @chaoticaldynamics

Hi everyone! I’m moving my blog to @chaoticaldynamics

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AERIAL SHOTS OF SOUTH AFRICA, ZACK SECKLER

This stunning photo-essay by New York City-based photographer, Zack Seckler, took him seven consecutive days to photograph over 2,000 miles of South African terrain. The end result is a breathtaking look at Africa’s majestic landscapes, colors, and creatures that inhabit it. 

Grow with us @ Instagram.com/wetheurban

The topologist’s sine curve.

Limts: f(x)=sin(1/x) is a rare example of a function with a non-existent one-sided limit. More technically, f(x)=sin(1/x) is defined for all numbers greater than zero, yet the limit as x approaches zero from the right of f(x)=sin(1/x) does not exist. This can be reasoned by considering the value of f at x-values near zero. Informally, f(near zero) could be 1 f(just a bit closer to zero) could be -1 so f(numbers near zero) does not seem to settle on a single y-value.

Continuity: Note that f is continuous for all numbers greater than zero but not continuous at x=0 since f is undefined there. Even if we were to “fill in the bad point” and let f(0)=0, the function would still not be continuous at zero! (note this is the natural choice as sin(0)=0). We can see that the adjusted f is still not continuous at zero since the sequence x_n=1/(pi/2+npi) converges but f(1/x_n) is the sequence (-1)^n which does not converge. This is similar to the argument above. In other words, closing in on x=0, we can keep finding x values such that f(x)=-1 and f(x)=1.

Topology: In topology, the topologist’s sine curve is a classic example of a space that is connected but not path connected. This space is formed in R^2 by taking the graph of f(x)=sin(1/x) together with its limit points (the line segment on the y-axis [-1,1], the red line on the second image). The graph of f is connected to this line segment as f and the segment cannot be sepearted by an open disc (no matter how small). This can be informally reasoned by the zooming illustration in the second image. But the space is not path connected by the sequence argument above (there is no path to the point (0,0)).

Image credits: http://mathworld.wolfram.com/TopologistsSineCurve.html and https://simomaths.wordpress.com/2013/03/10/topology-locally-connected-and-locally-path-connected-spaces/

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“To dare in life is to make yourself vulnerable to the possibility of failure. Most of us don’t welcome failure. So instead we avoid taking risks. We compromise, taking cold comfort in the assumption that we’ve removed the possibility of failure as we buckle up in the passenger seat and let life take the wheel. The truth is, there’s no avoiding failure. While failure may never feel good, failure in a life of compromise can be twice as devastating.”

— Ryder Carroll, The Bullet Journal Method (via kxowledge)