Good question. "Infinity" is not a real number, so the usual understanding of the multiplication operator (which takes two real numbers and produces a real number) does not apply. You can try to extend the real numbers by adding two objects called positive and negative infinity, defined in terms of the limit of an unbounded sequence. So positive infinity is something greater than any finite number (and negative infinity), and vice-versa. Then things like dividing something by infinity, or adding infinity to something, follow naturally from limit operations, and have the values that you'd expect.
Unfortunately there's still no help for "0 * infinity" and other so-called indeterminate forms. This kind of infinity is defined as the limit of any unbounded sequence, but unlike the defined forms like "1 / infinity = 0", the value of "0 * infinity" will depend on the details of the particular sequence. Consider: "x / infinity = 0" is true for any constant, finite x; substituting it into "0 * infinity" gives "(x / infinity) * infinity". The infinities cancel (since we can let them represent the same unbounded sequence), leaving "0 * infinity = x", for any x that you like.
So my preferred answer to the question "What is 0 * infinity" is: mu. It can take on any value, but only because you've thrown away the information about what unbounded sequence, exactly, is represented by this infinity. It means you need to go back and analyze the original problem more carefully. In the context of this particular problem, the question is if a finite bandwidth, divided into an infinite number number of infinitely small channels, gives you infinite bandwidth. Stated precisely this way it's almost obvious. If N is the number of channels and B is the total bandwidth, then the bandwidth of each channel is "B/N". What happens to the total of the channels' bandwidths as N approaches infinity? Well, N goes to infinity and B/N goes to zero, which is where the question "0 * infinity" comes from. But if we keep the information about the original sequences, "N * (B/N)" is just B, and nothing else.
Wrong. The ISS is in constant free fall, far from zero G.
I don't believe there is a difference.
So they fed an LCS with some sample data? OK, par-for-the-course. I'm far more interested in how they generated those '1 million' pre-labelled test images in the first place.
I read the paper; it was clever. They used a standard motion capture setup with their actor(s) going through several hundred different movements. Since their algorithm is stateless, they could analyze the motion and produce many distinct poses from each movement. Each pose was then "retargeted" (a well known technique in animation; example) onto many different 3D models of people of varying height, body type, etc., before finally being rendered into a perfectly labeled depth map.
They went through several iterations of this process:
Even if you never re-watch 99% of it, it's still worth keeping the whole 100%, because there is no way to know in advance which 1% you will want to re-watch.
I don't think anyone spends a lot of time going through their old media collection, casting about for things to revisit (unless they are very bored). My most common use case is when something suddenly reminds me of a specific song or movie that I last saw 5 years ago, or it comes up in conversation, and makes me want to go back and watch it again. If I then discover that I destroyed it in the Great Purge of 2003, it makes me a very sad panda.
A million watts? Why can't this "spent" fuel be used as an energy source?
Given that the Fukushima nuclear power plant produced somewhere around 4500MW total, and that you would be able to recover at best maybe 50% of that 1MW heat energy as electricity, and still have to deal with the radiation (meaniing dedicate an expensive reactor and containment system to house it), it just isn't worth it.
(while we're at it - where are ships with hulls overlooking Archimedes' principle? It's 2k+ years old, surely we should be able to ignore it by now...)
Here you go
To understand a program you must become both the machine and the program.